Chemistry

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• UNIT 1: PROPERTIES AND USES OF TRANSITION METALS

  UNIT 1: PROPERTIES AND USES OF TRANSITION
  METALS
  Key unit competence:
  The learner should be able to explain the properties and uses of transition metals.
  Learning objectives
  At the end of this unit , students will be able to:
  • Discuss qualitatively the propertie of transition elements;
  • Explain the principle of ligan exchange;
  • State the rules of naming complex ions and stereoisomers;
  • Describe reactions of transition metals;
  • State the use of transition metals;
  • Relate the electronic configurations to special properties of transition metals;
  • Predict the shape of the complex compounds of transition metal cations;
  

  • Perform the confirmatory tests for transition metal ions.

  Introductory Activity
  The following photos show how some elements play a big role in our daily lives.
  

  Observe these objects carefully.

  

  Most of the metals in the periodic table belong in the d-block of transition metals.
  They are hard and strong, and many of them are very familiar to us. For instance, zinc
  is in brass instruments like trumpets and tubas. Have you ever heard of the element
  

  “scandium” before? But you’ve interacted with it if you have ever ridden a bicycle.

  1.1. Definition and electronic configuration of transition metals.
  Activity 1.1
  1. Write the electronic configuration of the following atoms and ions:
  a. Ca(Z=20)          b. Ca ²⁺           c. Na(Z=11)          d. Na⁺
  2. Referring to the portion of periodic table in this book,
  a. Write the electronic configuration of the elements from Sc to Zn.
  b. Point out any difference between the electronic configuration of the
  above elements and that of other elements in s and p blocks
  

  3. Define the term transition metal.

  According to IUPAC system, a transition metal is “an element whose atom has a
  partially filled d sub-shell, or which can give rise to cations with an incomplete
  d-orbitrals”.
  Transition metals are located between groups 1& 2 (s-block) and group 13 (p-block)
  on the periodic table. The elements are also called d-block elements because their
  

  valence electrons are in d-orbitals.

  

  The properties of transition elements are between the highly reactive metallic
  elements of the s-block which generally form ionic compounds and the less reactive
  elements of the p-block which form covalent compounds. Transition metals form
  ionic compounds as well as covalent compounds.
  
  The first 3 rows, i.e. period 4, period 5 and period 6, are called first transition series,
  second transition series and third transition series respectively. The metals of the
  first series are all hard and dense, good conductors of heat and electricity.
  This block is known as the transition metals because some of their properties show a
  gradual change between the active metals in s-block and p-block where non-metals
  are found.
  Electronic configuration is the arrangement of electrons in orbitals around the
  nucleus. The electronic structure of the first transition series is shown in the table
  

  below:

  

  When building electronic structure of transition metals, 4s orbital is filled before 3d
  orbitals.
  The transition elements are stable when their d-orbitals are filled (d¹⁰) or when their
  d-orbitals are half filled (d⁵).This explains the electronic structure of copper, [Ar]
  4s¹3d¹⁰ instead of
  [Ar] 4s²3d⁹. The same applies for Cr: [Ar] 4s¹3d⁵ and not [Ar] 4s²3d⁴.
  In order to attain that stability an electron can jump from 4s orbital to 3d orbital
  because those two orbitals are close in energy.
  This also explains why Fe²⁺ with 3d⁶ is easily oxidized to Fe³⁺ with 3d⁵ and Mn²⁺ with
  3d⁵ is resistant to oxidation to Mn³⁺ with 3d^4.
  Transition metals form ions by losing electrons first from the 4s sub-shell rather than
  

  the 3d sub-shell. Hence electronic configuration of Fe, Fe²+ and Fe³⁺ are the following:

             

  The 4s electrons are removed before 3d electrons. This is because the 3d electrons
  are inner while the 4s electrons are outer therefore the outer electrons (4s) have to
  be removed before the inner electrons.
  
  Checking up 1.1
  1. Explain the difference between the electronic configuration of transition
  elements and that of main group elements.
  2. 2.Why d-block metals are so called transition metals?
  1.2. Properties of the transition metals
  1.2.1. Melting and boiling points
  Activity 1.2 (a)
  Experiment: Investigation of the melting point of transition metals compared
  with s-block elements
  Materials: Potassium or Rubidium metal and copper or iron metal, pair of tongs,
  spatulas, bunsen burner and match box.
  Procedure:
  1. Take a half filled spatula of
  a. Potassium (K) or Na, Rb, or Cs
  b. Iron turnings or very small piece of copper sheet (which can fit on a
  spatula)
  2. Heat both spatulas on the Bunsen burner flame
  3. Write down the observations
  4. What can you conclude about your findings?
  The melting points and the molar enthalpies of fusion of the transition metals are
  both high in comparison to main group elements. Most of the transition metals have
  melting points above 1000^oC; mercury is liquid at room temperature.
  This is due to the high number of valence electrons that increases the electrostatic
  attraction force between those electrons and the metallic cations, hence increasing
  

  the strength of the metallic bond and the melting point.

  Table 1.1: Melting and boiling points of the 1^st series of Transition Metals

  
  

  Checking up 1.2 (a)
  Compare and comment on the melting points of transition metals and those of
  

  s-block metals.

  1.2.2. Densities and atomic/metallic radii
  Activity 1.2 (b)
  Procedure for practical:
  1. a. Take a magnesium ribbon and a copper foil of the same size (if
  possible you may use their turnings)
  b. You weigh those two samples using an electronic balance. And record
  their masses
  2. a. Take aluminum foil and copper foil of the same size (if possible you
  may use their turnings)
  b. You weigh those two samples using an electronic balance. Record their
  masses
  3. Comment on your observations by explaining why their masses are
  different and yet they have the same size.
  4. Use the internet or any book or even this one to interpret the data
  given about metallic radii of the first series transition metals. From your
  research, compare metallic radii of transition metals and those of main
  

  group elements.

  The transition elements are much denser than the s-block elements and show in
  general a gradual increase in density from left to right in a period as you can see
  below from scandium to copper. This trend in density can be explained by a decrease
  

  in metallic radii coupled with the relative increase in atomic mass

  Table 1.2: Density/g cm ^-3 of the first transition series

  
  

  Table1.3: Metallic radii of the first transition series
  

  
  

  Checking up 1.2 (b)
  The metallic radius of vanadium is smaller than that of titanium. Explain this
  

  statement.

  1.2.3. Ionization energies
  Activity 1.2 (c)
  Use this book or any other source (textbook or search engine) to interpret/ analyze
  the summary about ionization energies of the transition metals (first series). From
  your findings, compare
  
  a. Ionization energies of those transition metals.
  b. Ionization energies of transition metals and those of main group elements
  
  The ionization energy of transition metals is related to the energies of its d orbitals,
  its ease of oxidation, and its basicity.In simplest terms, the greater the ionization
  energy of a metal, the harder it is to pull an electron from it.
  
  As the number of protons increases across a period (or row) from left to right of
  the periodic table, the first ionization energies of the transition-metal elements are
  relatively the same, while that for the main-group elements increases.
  
  In moving across the series of metals from scandium to zinc, a small change in the
  values of the first and second ionization energies is observed. This is due to the buildup
  of electrons in the immediately underlying d-sub-shells that efficiently shields
  the 4s electrons from the nucleus and minimizing the increase in effective nuclear
  

  charge from element to element.

  Table 1.4: First, second and third ionization energies of 1st Series Transition
  

  metals /kJ mol^-1

  
  

  The figure 1.3 below shows the first iazonisation energies for transition metals of 1st,
  

  2nd and 3rd rows (series).

  

  In general, ionization energy increases as we move from left to right across the
  period. Notable dips occur at row 1, group 10 (Ni) and row 3, group 7 (Re).
  
  Checking up 1.2 (c)
  Briefly explain the following observations:
  a. The first ionization energy of cobalt is only slightly larger than the first
  ionization energy of iron.
  b. The third ionization energy of iron is much lower than the 3rd ionization
  

  energy of Mn.

  1.2.4. Transition elements have variable oxidation states
  Activity 1.2 (d)
  Use this book or any other source (textbook or search engine) to
  a. Explain the term oxidation number
  b. Compare the oxidation numbers of transition metals (first series) and those
  of main group elements.
  c. Analyze the stability of ions formed by transition metals (first series).
  
  Oxidation state is a number assigned to an element in chemical combination which
  represents the number of electrons lost or gained.The transition elements from
  Titanium to Copper all form ions with two or more oxidation states. In most cases,
  this is the result of losing the two electrons of 4s orbital and electeons in 3d orbitals.
  The 4s electrons are lost first because they are in the highest energy level. However,
  because the 3d and 4s energy level are so close in energy, the 3d electrons can also
  be lost when an atom forms a stable ion. The common oxidation states shown by the
  

  first transition series are:

  Table 1.5: The oxidation states shown by the transition metals (series)
  

  
  

  • The common stable oxidation states for those transition metals with variable
  oxidation states are bolded and underlined.
  • The oxidation state corresponding to a full or half-filled d-orbital is energetically
  stable. For example, Fe³⁺ is more stable than Fe²⁺and Mn²⁺ is more stable than
  Mn^3+.
  • However, in most compounds and solutions, copper exist as Cu²⁺ ion rather
  than Cu⁺ ion. Meaning that the former is more stable than the latter. The
  

  explanation of this is beyond this level.

  Checking up 1.2 (d)
  

  Which gaseous ion is more stable, Mn²⁺ or Mn³⁺? Explain why.

  1.2.5. Most transition metals and their compounds have high ability of being
  catalyst
  Activity 1.2 (e)
  Practicals:
  1. Preparation of oxygen using hydrogen peroxide, H₂O₂, without a catalyst
  a. Put 10 mL of H₂O₂ in a conical flask (Pyrex preferably)
  b. Heat the conical flask for about 5 min
  c. Write down the observation in A.
  2. Preparation of oxygen using hydrogen peroxide, H₂O₂, with MnO₂ as a
  catalyst
  a. Put10 mL of H₂O₂ in a conical flask(Pyrex preferably)
  b. Put a very small amount of MnO₂ in the conical flask
  c. Heat the conical flask for about 5 min
  d. Write down the observation in (B)
  

  Question: What is the role of MnO₂ in the above experiment?

  A catalyst is a substance that can speed up (positive catalyst) or that can slow down
  (negative catalyst) the rate of reaction and is found unchanged at the end of the
  reaction. But generally the term catalyst is used for the substance that helps in
  accelerating the rate of the reaction. A catalyst that speeds up the reaction provides
  another pathway with lower activation energy.
  In some catalytic process, transition metal ions undergo changes in their oxidation
  states but are regenerated at the end of the reaction.
  The reasons for transition metals to work as catalysts:
  • Presence of empty d orbitals which enable transition metal ions (or atoms) to
  form temporary bonds with reactant molecules at the surface of a catalyst and
  weakens the bond in the reactant molecules
  • Variable oxidation states which allow them to work as catalysts in the reactions
  

  involving the transfer of electrons.

  Table 1.6: Reactions catalysed by transition metals

  
  

  Checking up 1.2 (e)
  

  Explain why s-block metals and their compounds are not used as catalysts

  1.2.6. Most transition metal ions are paramagnetic
  Activity 1.2(f)
  

  Given the following materials:

  

  1. Organize yourself in to group to find the objects shown in the photo
  above.
  2. Using a magnet, classify the above materials into two groups as shown in
  the table below.
  Objects attracted by a magnet
  Objects not attracted by a magnet
  3. Research, using any relevant source (textbook or internet), to identify in
  which metal the objects A to E are made
  4. Research to know why some objects are attracted by a magnet while
  others are not
  Paramagnetism is a property of substances to be attracted in a magnetic field.
  Substances which are not attracted (i.e slightly repelled) in a magnetic field are
  said to be diamagnetic.Transition metal ions show paramagnetism because of the
  presence of unpaired electrons in their 3d arbitrals.
  The greater the number of unpaired electrons, the stronger the paramagnetism;
  that is the reason why:
  • Fe³⁺ is more paramagnetic than Fe²⁺ because Fe³⁺ has five unpaired electrons
  while Fe²⁺ has four unpaired.
  • Sc³⁺ and Zn²⁺ are not paramagnetic, they are diamagnetic because they do not
  have unpaired electrons.
  Other examples of paramagnetic substances are: Cr, Mn, CuSO_4, Fe, Co, Ni, Pt.
  Examples of diamagnetic substances are: Zn, Cu⁺, Au⁺, TiO_2.
  

  Checking up 1.2 (f)
  Predict whether the following substances are paramagnetic or not. Explain
  a. CuSO₄
  b. Co
  c. Ca
  

  d. Cr

  1.2.7. Formation of alloys
  Activity 1.2 (g)
  Observe the trophies/or other objects made in the materials below and compare
  their appearances with the elements from which they are derived.
  a. Bronze with copper
  b. Stainless steel with iron
  c. Pewter and copper
  You can use the internet, books (including this one) or any other relevant source
  to find the figures of the above objects, the elements they are made from and
  

  their uses.

  An alloy is a homogenious solid mixture (solid solution) made by combining two or
  more elements where at least one is a metal.
  
  Importance of alloying:
  • Increase of the strength of a metal,
  • Resistance to corrosion,
  • Gives to the metal a good appearance
  Generally, alloys are needed and used to improve the quality of the required
  material. For example, brass (alloy of zinc and copper) is much stronger than either
  

  pure copper or pure zinc. Pure gold is too soft to be used in some applications.

  Table 1.7: The properties and uses of some common alloys formed by transition
  

  metals (first series)

  

  

  

  Checking up 1.2 (g)
  1. Explain why alloys are said to be solid solutions.
  

  2. Give the importance of alloying

  1.2.8. Formation of complex ions
  Activity 1.2
  Use this book or any other source (library textbook or internet) to analyze and
  discuss on the following. You have to take note on what to be presented to share
  with your colleagues and teacher.
  a. What is a ligand?
  b. State the types of ligands
  c. The geometry of complexes
  
  A complex or coordination compound is a chemical species made of a central metal
  (cation or neutral) bonded to other chemical species called ligands by coordination
  or dative bonds. A complex may be neutral, positively or negatively charged.
  Transition metal form complexes because of:
  • Their small and highly charged ions,
  • The presence of vacant (empty) d-orbitals which can accommodate lone pair
  

  of electrons donated by other groups (ligands)

  The general formula of a complex is: [ML_n]^y
  Where:
  • M-metal ion or atom
  • L-Ligand
  • n-the number of ligands surrounding the metal
  • y-the charge of the complex; [ML_n] indicates a neutral complex.
  -Coordination number of a complex: is the number of coordinate bonds on the
  

  central metal in a complex.

  -Ligand: It is a species (anion or a molecule) that is bonded to a central metal ion
  or atom in a complex. A ligand should have at least one lone pair of electrons
  to form a coordinate bond.
  Ligands are classified depending on the number of sites at which one molecule
  of a ligand is coordinated to the central metallic atom; the ligands are classified
  as monodentate (or unidentate), ambidentate and polydentate (or multidentate)
  lingards.
  
  a. Monodentate ligands
  The ligands which have only one donor atom or are coordinated through one
  electron pair are called monodentate ligands because they have only one tooth with
  which to attach themselves to the central cation or atom. Such ligands are coordinated
  to the central metal at one site or by one metal-ligand bond only. These ligands may
  

  be neutral molecules or in anionic form.

  The table below provides examples of some monodentate ligands.
  

  Table 1.8: some monodentate ligands

  
  

  Ligands that can use different sites to coordinate to the central metal are called
  “ambidentate”: e.g. CN- and NC-(see table above).
  
  Notice that a ligand with a donor atom that possesses 2 lone pairs of electrons, such
  as H₂Ö:, is not bidentate, since it cannot use both lone pairs simultaneously to bind
  to the metal because of the steric effect.
  
  b. Polydentate ligands
  These may be bidentate, tridentate, tetradentate, pentadentate, and hexadentate
  ligands if the number of donor atoms present in one molecule of the ligand attached
  with the central metallic atom is 2, 3, 4, 5, and 6 respectively. Thus one molecule of
  

  these ligands is coordinated to the central metallic atom at 2, 3, 4, 5, and 6 sites

  respectively. In other words, we can say that one molecule of these ligands makes 2,
  

  3, 4, 5, and 6 metal-ligand coordinate bonds respectively.

  

  • Tetradentate

                              

  • Hexadentate
  The structure shows that it has two neutral N- atoms and four negatively charged Oatoms
  as its donor atoms which can form coordinate bonds with a transition metal
  

  ion.

                                        

  The complex ions which form between polydentate ligands and cations are known
  as chelates or chelated complexes.
  
  In general, polydentate ligands form more stable complexes than monodentate
  ligands. The stability of complex is much enhanced by chelation. A polydentate
  

  ligand can hold the central cation more strongly.

  Examples of complexes:
  • Copper (II) ions have a coordination number of four in most of its complexes:
  [Cu(H₂O)₄]^2+, [Cu(NH₃)₄]²⁺, [CuCl₄]²⁺, [Cu(NH₂-(CH₂)₂-NH₂)₂]²⁺, …

  

  • Most of ions have coordination number of 6.
  [Cr(H₂O)₆]³⁺ , [Cr(NH₃)₆]³⁺ , [Cr(H₂O)₄Cl₂]- , …
  • Very few ions have a coordination number of 2: [Ag(NH₃)₂]+, [Ag(CN)₂]-, [CuCl₂]-,
  …
  Geometry of complexes
  Complexes have a variety of geometries or shapes, but the most common geometries
  are the following:
  • Complexes with coordination number 2 adopt a linear shape. Example:
  [Ag(NH₃]₂+: [H₃N-Ag-NH₃]+
  
  The complexes having coordination number of 2 are linear since minimises ligand
  

  repulsion.

  

  • Complexes with coordination number 4 generall adopt a tetrahedral shape.
  

  But few of them can form a square planar shape.

  Examples:
  [Zn(NH₃)₄]²⁺, [NiCl₄]^2- and some few others adopt a square planar shapes, examples:
  [Cu(NH₃)4]²⁺ , [Ni(CN)₄]^2-,[CuCl₄]^2-,[CoCl₄]^2-,…
  
  The square plannar geometry is characteristic of transition metal ions with eight d
  

  electrons in the valence shell, such as platinum(II)and gold(III).

                                                  
  Copper (II) and cobalt (II) ions have four chloride ions bonded to them rather than
  six, because the chloride ions are too big to fit any more around the central metal
  ion.
  

   
  • Complexes with coordination number 6 adopt an octahedral shape.
  Example: [Cr(NH₃)₆]^3+.
  These ions have four of the ligands in one plane, with the fifth one above the plane,
  and the sixth one below the plane.
  

  
  
  Checking up 1.2
  1. What do you understand by :
  a. Coordination number.
  b. Ligand.
  2. Give the main types of ligands and give an example for each
  3. Say if the following statement is correct or wrong and justify: The
  coordination number equals the number of ligands bonded to the central
  metal.
  

  1.2.9. Many transition metal ions and their compounds are coloured
  Activity 1.2 (i)
  Experiment 1: Observation of the colors of transition elements
  Apparatus: Test tubes, droppers, spatula, test tube holders.
  Chemicals: NaCl, CaCl₂, FeSO₄, Fe₂ (SO₄)₃, KMnO₄, K₂Cr₂O₇ ,distilled water, Cr₂(SO₄)₃.
  1. What are the colours of the compounds above?
  2. Determine the oxidation states of each metal in the above compounds?
  3. a. Take an endful spatula of each product given above and put each in a test
  tube.
  b. Put 10 mL of distilled water in each test tube.
  c. Write down the colours of solutions formed and conclude.
  
  Experiment 2: Investigation of ligand exchange reactions involving copper (II) ions, Cu2+
  Apparatus: Test tubes, droppers, spatula, test tube holders.
  Chemicals: Copper (II) sulphate, concentrated hydrochloric acid, concentrated ammonia
  solution and distilled water.
  Procedure:
  1. Use a spatula to place a small amount of anhydrous copper (II) sulphate in a test
  tube.
  2. Add 10 drops of distilled water to the anhydrous copper (II) sulphate and shake
  3. To the test tube in step 2, add concentrated ammonia solution drop by
  drop while shaking the test tube until there is no further change. Record all
  observations.
  4. Repeat steps 1 and 2
  5. To the test tube from step 4, add concentrated hydrochloric acid drop by drop
  while shaking until there is no further change. Record all observations.
  
  Points for discussion:
  1. What happens when anhydrous copper (II) sulphate is dissolved in water?
  2. Describe what is observed when concentrated ammonia is added dropwise to
  an aqueous solution of copper (II) sulphate.Write balanced equations for each
  observation if possible
  3. Describe what happens when concentrated hydrochloric acid is added to an
  aqueous solution of copper (II) sulphate. Write balanced equation(s) for the
  observation(s) made.
  4. State any other possible observation(s) for this experiment.
  

  The formation of colored ions by transition elements is associated with the presence
  

  of incompletely filled 3d orbitals.

  

  

  This property has its origin in the excitation of d electrons from lower energy
  d-orbitals to higher energy. In fact, when the central metal is surrounded by ligands,
  these cause d orbitals to be split into groups of higher and lower energy orbitals.
  When electrons fill d-orbitals, they fill first of all the lower energy orbitals; if there is
  free space in higher energy d-orbitals, an electron can be excited from lower energy
  d-orbitals to higher energy d-orbitals by absorbing a portion of light corresponding
  to a given colour, the remaining color light is the white light minus the absorbed
  colour.
  
  
  When a coloured object is hit by white light, the object absorbs some colour and
  the colour transmitted or reflected by the object is the colour which has not been
  absorbed. The observed colour is called complementary colour.
  When a metal cation has full d-orbitals, such as Cu⁺or Zn²⁺or no electron in d orbital,
  such as Sc³⁺.
  
  Table1.9: Complementarities of colors observed and absorbed when light is
  

  emitted

  

  

  The colour of a particular transition metal ion depends upon two factors:
  

  • The nature of the ligand

  
  

  The principle of ligand exchange
  Complexing reactions involve competitions between different ligands for central
  metal. A more powerful ligand displaces a less powerful ligand from a complex.
  During the process there is a change in colour.
  

  Here below is a list of some ligands in increasing order of strength.

  
  The above series are called the spectrochemical series and shows that cynide ion
  and carbon monoxide are very strong ligands
  The stability of a complex ion is measured by its stability constant. The higher the
  

  stability constant of a complex, the more stable is the complex.

  Checking up 1.2 (i)
  Predict whether each of the following ion forms coloured compounds and explain
  your reasoning: Fe²⁺, Mn⁷⁺, K⁺
  
  1.3. The anomalous properties of Zinc and Scandiu
  Activity 1.3
  From the information you have learnt about the properties of transition metals,
  Suggest the difference between the properties of Zn and Sc and other transition
  metals. You can consult different sources (books or internet) to provide enough
  information.
  
  On the basis of the properties of transition metals, zinc and scandium are not
  

  considered as typical transition metals even though they are members of the d-block.
  

  Zinc:
  

  • It has a complete d-orbital.
  

  • Zinc forms only the colourless Zn²⁺ ion, isoelectronic with the Ga³⁺ion, with 10
  

  electrons in the 3d subshell.
  • Zinc and its compounds are not paramagnetic
  Scandium:
  • Has one oxidation state,+3
  • Sc forms only the colourless Sc³⁺ion, isoelectronic with the Ca²⁺ ion, with no
  electrons in the 3d subshell.
  • Its compounds are diamagnetic
  • It forms compounds containing ions with a completely empty 3d subshell.
  
  Checking up 1.3
  Give any one property by which Zn differs from Sc
  

  
  1.4. Naming of complex ions and isomerism in of transition
  metal complexes
  1.4.1. Naming of complex ions
  Activity 1.4 (a):
  1. Name the following molecules and explain the basis /principle used to
  name them.
  a. CaBr₂
  b. CCl₄
  c. SF₆
  2. Analyze the IUPAC rules for naming complex ions in the summary in this
  book or using any other source (textbook or search engine) and apply
  them by naming the following:
  a. [CuCl₄]^2-
  b. [Cu(H₂O)₆]²⁺
  c. [Cr(NH₃)₃(H₂O)₃]Cl₃
  d. [Pt(NH₃)₂Cl₂]
  e. (NH₄)₂[Ni(C₂O₄)₂(H₂O)₂]
  Naming molecules requires the knowledge of certain rules, such as how to name
  cations, anions, where to start from when both a cation and an anion are combined
  in an ionic molecule or when two non metals are combined in a covalent molecule.
  Like other compounds, complex compounds/ions are named by following a set of
  rules. You are familiar with some of them and the new ones can be understood and
  applied easily.
  1. In simple metal compounds, the metal is named first then the anion.
  Example: CaCl₂: calcium chloride
  2. In naming the complex:
  a. Name the ligands first, in alphabetical order, then the metal atom or cation,
  followed by its oxidation state written between brackets as Roman number,
  though the metal atom or cation is written before the ligands in the chemical
  formula.
  Example: [CuBr₄]^2-: Tetrabromocuprate (II) ion
  

  The names of some common ligands are listed in the table below:

  Table 1.10: Names of common ligands

  

  b. Greek prefixes are used to indicate the number of each type of ligand in the
  complex:
  

  The numerical greek prefixes are listed in the following table:

  Table 1.11: Greek numerical prefixes

  
  

  c. After naming the ligands, name the central metal.
  • If the complex bears a positive charge (cationic complex), the metal is named
  by its usual name.
  Example: Cu: Copper                               Pt: Platinum
  
  If the complex bears a negative charge (anionic complex), the name of the metal
  ends with the suffix –ate
  Example: Co in a complex anion is called cobaltate and Pt is called platinate.
  For some metals, the Latin names are used in the complex anions e.g. Fe is called
  

  ferrate (not ironate). See table below:

  Table 1.12: Latin names of some transition metals in anionic complexes

  
  

  1. For historic reasons, some coordination compounds are called by their common
  names.
  Example: Fe(CN)₆³- and Fe(CN)₆^4- are named ferricyanide and ferrocyanide
  respectively, and Fe(CO)₅ is called iron carbonyl.
  2. To name a neutral complex molecule, follow the rules of naming a complex
  cation. Example: [Cr(NH₃)₃Cl₃]: triamminetrichlorochromium (III)
  You can have a compound where both the cation and the anion are complex ions.
  Notice how the name of the metal differs even though they are the same metal ions.
  Remember: Name the cation before the anion.
  Example: [Ag(NH₃)₂][Ag(CN)₂] is diamminesilver(I)dicyanoargentate(I)
  Note that:
  • The names are written as a one word: Tetraamminecopper (II), not Tetraammine
  copper (II).
  • Complex ions formula is written between square brackets and the charge of the
  ion as superscript outside the brackets: [Cu(NH₃)₄]²⁺. When oppositely charged
  ions approach the complex ion, a neutral molecule can be obtained:
  [Cu(NH₃)₄]²⁺2Cl^- or simply, [Cu(NH₃)₄]Cl₂: tetraamminecopper(II)chloride.
  The ions outside the square brackets are known as “counter ions”.
  
  Checking Up 1.4 (a):
  1. Complete the table below using the names of the given metals when they
  are in anionic complexes
  Element
  

  Name in an anionic complex

  

  2. Give the systematic names for the following complex ions/compounds:
  a. [Cr(NH₃)₃(H₂O)₃]³⁺
  b. [Co(H₂NCH₂CH₂NH₂)₃]₂(SO₄)₃
  c. K₄[Fe(CN)₆]
  

  d. Fe(CO)₅

  1.4.2. Isomerism in complexes
  Activity 1.4 (b):
  1. Discuss on the following questions:
  a. What do you understand by the term “isomerism”?
  b. Is there any relationship between isomers and isomerism?
  c. Give examples of molecules that can exist as isomers and explain their
  isomerism
  2. Read and discuss the summary below to understand how complex ions/
  compounds exhibit isomers
  3. Present your findings to your colleagues and teacher to share your
  understanding.
  
  Isomers are chemical species that have the same molecular formulal, but different
  molecular structures or different arrangements of atoms or groups of atoms in
  space. Isomerism among transition metal complexes arises as a result of different
  arrangements of their constituent ligands around the metal.
  The diagram below shows the different categories of isomerism in transition metal
  

  complexes.

  

  In this unit, we are specifically concerned with ‘stereoisomerism’ which gives rise to
  isomers known as “stereoisomers”. Stereoisomers have the same structural formulal
  but different arrangements of ligands in space.
  They are classified in two categories: geometrical isomers and optical isomers.
  
  

  1. Geometrical isomers
  Coordination complexes, with two different ligands in the cis and trans positions
  from a ligand of interest, form isomers.
  

  For example, the square planar, diammine dichloroplatinum (II) Pt(NH₃)₂Cl₂),can be

                       presented as follows:

  

  The octahedral [Co(NH₃)₄Cl₂]⁺ ion can also have geometrical isomers.

  

  Different geometrical isomers are different chemical compounds. They exhibit
  different properties, even though they have the same formula. For example, the two
  isomers of [Co(NH₃)₄Cl₂]NO₃ differ in color; the cis form is violet, and the trans form is
  green. Furthermore, these isomers have different dipole moments, solubilities and
  reactivities.
  
  2. Optical isomers (enantiomers)
  Optical isomers are non-superimposable mirror images of each other. A classic
  example of this is your two hands (left and right); hold them face-to-face: one is the
  mirror image of the other. Now try to superimpose them one over another: they
  are non-superimposable (only the middle fingers superimpose one over the other.
  Chemical compounds that behave like the hands are called “chiral”, in reference to
  the Greek word for hands.
  
  Optical isomers are very important in organic and biochemistry because living
  systems often incorporate one specific optical isomer and not the other.
  
  Unlike geometric isomers, optical isomers have identical physical properties (boiling
  point, polarity, solubility, etc.). Optical isomers differ only in the way they affect
  polarized light and how they react with other optical isomers.
  
  1. For coordination complexes, many coordination compounds such as
  [M(en)₃]ⁿ⁺ [in which Mⁿ⁺ is a central metal ion such as iron(III) or cobalt(II)]
  form enantiomers, as shown in figure below.These two isomers will react
  differently with other optical isomers. For example, DNA helices are optical
  isomers, and the form that occurs in nature (right-handed DNA) will bind
  

  to only one isomer of [M(en)₃]ⁿ⁺ and not the other.

  

  Checking up 1.4 (b):
  1. The geometric isomer of [Pt(NH₃)₂Cl₂] is shown in the figure below. Draw
  

  the other geometric isomer and give its full name.

                          

  2. Draw the ion trans-diaqua-trans-dibromo-trans-dichlorocobaltate (II).
  3. Sketch the arrangement of bonds in the complexes
  a. Hexaaquacobalt(III) ion
  b. Hexacyanoferrate (III) ion
  c. Diamminesilver (I) ion
  d. The complex compound tetracarbonylnickel (0).
  4. The compound [NiCl₂(NH₃)₂] has cis-trans isomers. These have a complex
  non-ionic structure.
  a. Does [NiCl₂(NH₃)₂] have a tetrahedral or a square-planar structure?
  Explain your answer.
  b. Draw the cis and trans isomers for [NiCl₂(NH₃)₂].
  5. Early in the 20^th century, the German scientist Werner succeeded in
  clarifying the situation concerning the five compounds of PtCl₄^- and
  ammonia. The properties of these compounds are listed in the table
  

  below.

  

  a. What is the oxidation state of Pt in each of the compounds A-E?
  b. The co-ordination number of Pt in each compound is six. Write a right formula for each
  of the five compounds. Show the complex ion and the other ions and/or molecules
  present.
  c. Each of the compounds forms an octahedral complex ion. Draw the structures for the
  complex ions in A, B, C, and D.
  

  d. Which of the complex ions in (c) have isomers?

  1.5. The Chemistry of individual transition metals
  Activity 1.5
  Using the library and internet or other textbooks, make your own research and make
  presentation of the results of your research:
  1. On how each of the transition metals (first series) reacts with each of the
  following substances
  a. Oxygen
  a. Water
  c. Hydrochloric acid
  d. Sodium hydroxide
  e. Chlorine
  2. On the uses and their corresponding properties for each of the above
  

  transition metals.

  1.5.1. Scandium
  Scandium is a silvery-white solid. It melts at 1539^oC and boils at 2748^oC. Its density
  is about 3.0.
  1. Chemical reactions
  a. Reaction of scandium with air
  Scandium tarnishes in air, and burns readily, forming scandium (III) oxide, Sc₂O₃.
  4 Sc(s) + 3 O₂(g)  ——→     2 Sc₂O₃(s)
  
  b. Reaction of scandium with water
  When finely divided, or heated, scandium dissolves in water, forming Sc (III)
  hydroxide and hydrogen gas, H₂.
  2 Sc(s) + 6 H₂O(l) ——→2 Sc(OH)₃(aq) + 3 H₂(g)
  
  c. Reaction of scandium with acids
  Scandium dissolves readily in dilute hydrochloric acid, forming Sc(III) ions and
  hydrogen gas, H_2.
  2 Sc(s) + 6 HCl(aq) ——→2 Sc³⁺(aq) + 6 Cl−(aq) + 3 H₂(g)
  
  d. Reaction of scandium with halogens
  Scandium reacts with the halogens, forming the corresponding Sc(III) halides
  2 Sc(s) + 3 F₂(g)——→ 2 ScF₃(s)
  2 Sc(s) + 3 Cl₂(g) ——→2 ScCl₃(s)
  2 Sc(s) + 3 Br₂(g)——→ 2 ScBr₃(s)
  2 Sc(s) + 3 I₂(g)——→ 2 ScI₃(s)
  
  2. Uses
  • Scandium has as low density (2.99 g/cm³) asaluminium (2.7 g/cm³) but a much
  higher melting point.
  • An aluminium-scandium alloy has been used in fighter planes, high-end
  bicycle frames and baseball bats.
  • Scandium iodide is added to mercury vapour lamps to produce a highly
  efficient light source resembling sunlight. These lamps help TV cameras to
  reproduce colour well when filming indoors or at night-time.
  
  1.5.2. Titanium
  Titanium is a gray, solid with a density of about 4.50. It melts at 1667^oC and boils at
  

  3285^oC.

  1. Chemical reactions
  a. Reaction of titanium with air
  Titanium does not react with air under normal conditions. If brought to burn,
  

  titanium will react with both oxygen, O₂, and nitrogen, N₂.

                   

  b. Reaction of titanium with water
  Titanium does not react with water, under normal conditions. If the water is heated
  to steam, it will react with titanium, forming titanium(IV) oxide, TiO₂, and hydrogen,
  

  H²

  

  c. Reaction of titanium with acids
  Titanium does not react with most acids, under normal conditions. It will react with
  hot hydrochloric acid, and it reacts with HF, forming Ti(III) complexes and hydrogen
  gas, H₂.
  
  d. Reaction of titanium with bases
  Titanium does not appear to react with alkalis, under normal conditions, even when
  heated.
  e. Reaction of titanium with halogens
  Titanium reacts with halogens, when heated, forming the corresponding titanium(IV)
  halides
                    
  2. Uses
  • Titanium is as strong as steel but much less dense. It is therefore important as
  an alloying agent with many metals including aluminium, molybdenum and
  iron. These alloys are mainly used in aircraft, spacecraft and missiles because
  of their low density and ability to withstand extremes of temperature. They are
  also used in golf clubs, laptops, bicycles and crutches.
  • Power plant condensers use titanium pipes because of their resistance to
  corrosion. Because titanium has excellent resistance to corrosion in seawater,
  it is used in desalination plants and to protect the hulls of ships, submarines
  and other structures exposed to seawater.
  • Titanium metal connects well with bone, so it has found surgical applications
  such as in joint replacements (especially hip joints) and tooth implants.
  • The largest use of titanium is in the form of titanium (IV) oxide. It is extensively
  used as a pigment in house paint, artists’ paint, plastics, enamels and paper.
  It is a bright white pigment with excellent covering power. It is also a good
  reflector of infrared radiation and so is used in solar observatories where heat
  causes poor visibility.
  
  1.5.3. Vanadium
  Vanadium is a grey, solid with a density of about 6.11. It melts at 1915^oC and boils at
  3350^oC. It is insoluble in water at room temperature.
  
  1. Chemical reactions
  a. Reaction of vanadium with air
  Vanadium metal reacts with excess oxygen, O₂, upon heating to form vanadium (V)
  oxide, V₂O₅. When prepared in this way, V₂O₅ is sometimes contaminated by other
  vanadium oxides.
  
  b. Reaction of vanadium with water
  Vanadium does not react with water, under normal conditions.
  
  c. Reaction of vanadium with bases
  Vanadium metal is resistant to attack by molten alkali.
  In strong alkaline solutions (pH > 13), Vanadium (V) exists as colourless
  orthovanadate ions, VO₄³⁻.
  
  d. Reaction of vanadium with halogens
  Vanadium reacts with fluorine, F₂, when heated, forming vanadium (V) fluoride
  
  2. Uses
  • About 80% of the vanadium produced is used as a steel additive. Vanadium-
  steel alloys are very tough and are used for spanners, armour plate, axles,
  
  piston rods and crankshafts. Less than 1% of vanadium, and as little chromium,
  makes steel shock resistant and vibration resistant. Vanadium alloys are used
  in nuclear reactors because of vanadium’s low neutron-absorbing properties.
  • Vanadium (V) oxide is used as a pigment for ceramics and glass, as a catalyst
  and in producing superconducting magnets.
  
  1.5.4. Chromium
  Chromium is a silver gray metal with density of about 7.14. It melts at 1900^oC and
  boils at 2690^oC. Chromium is insoluble in water at room temperature.
  
  1. Chemical reactions
  a. Reaction of chromium with air
  Chromium metal does not react with air at room temperature. Heated clean
  chromium is oxidized superficially in air to green solid, chromium (II) oxide.
  
  b. Reaction of chromium with water
  Normally, Chromium metal does not react with water at room temperature. When
  red hot, it reacts with steam to form chromium (II) oxide.
  
  c. Reaction of chromium with acids
  Metallic chromium dissolves in dilute hydrochloric acid forming Cr(II) and hydrogen
  gas, H₂. In aqueous solution, Cr(II) is present as the complex ion [Cr(OH₂)₆]²⁺.
  
  Similar results are seen for sulphuric acid but pure samples of chromium may be
  resistant to attack.
  Chromium metal is not dissolved by nitric acid, HNO₃ but is passivated instead.
  
  d. Reaction of chromium with hydroxide ions
  Chromium dissolves rapidly in hot concentrated aqueous alkali forming a blue

  solution containing chromium (II) ion and hydrogen gas is evolved.

  

  Similar results are seen for sulphuric acid but pure samples of chromium may be
  resistant to attack.
  Chromium metal is not dissolved by nitric acid, HNO₃ but is passivated instead.
  
  d. Reaction of chromium with hydroxide ions
  Chromium dissolves rapidly in hot concentrated aqueous alkali forming a blue
  solution containing chromium (II) ion and hydrogen gas is evolved.
  
  e. Reaction of chromium with halogens
  Chromium reacts directly with fluorine, F₂, at 400°C and 200-300 atmospheres to
  form chromium (VI) fluoride, CrF₆.
  
  Under milder conditions, chromium (V) fluoride, CrF₅, is formed.
  
  Under milder conditions, chromium metal reacts with the halogens to form
  

  chromium tri halides or chromium (III) halides:

  

  2. Uses
  • Chromium is used to harden steel, to manufacture stainless steel (resists to
  corrosion) and to produce several alloys.
  • Chromium plating can be used to give a polished mirror finish to steel.
  Chromium-plated car and lorry parts, such as bumpers, were once very
  common. It is also possible to chromium plate plastics, which are often used
  in bathroom fittings.
  • About 90% of all leather is tanned using chromium. However, the waste
  effluent is toxic so alternatives are being investigated.
  • Chromium compounds are used as industrial catalysts and pigments (in
  bright green, yellow, red and orange colours). Rubies get their red colour from
  chromium, and glass treated with chromium has an emerald green colour.
  • Chromium (IV) oxide is used in magnetic tapes for sound/video recording.
  • Chromium is used in the control of cholesterol and help insulin sugar control
  in blood.
  

  1.5.5. Manganese
  Manganese is a grey-white solid with a slightly red colour. Its density is about
  7.44^oC. Manganese melts at 1244^oC and boils at 2060^oC. It is insoluble in water but
  soluble in diluted acids, at room temperature.
  
  1. Chemical reactions
  a. Reaction of manganese with air
  Manganese is not very reactive with air. The surface of manganese lumps oxidizes
  a little. Finely divided manganese metal burns in air. In oxygen the oxide Mn₃O₄ is
  formed and in nitrogen the nitride Mn₃N₂ is formed.
  
  b. Reaction of manganese with water
  

  Manganese reacts slowly with water to form manganese (IV) oxide:

  

  c. Reaction of manganese with acids
  Manganese dissolves readily in dilute sulphuric acid, forming a colorless solution of
  Mn(II) ions and hydrogen gas, H₂.
  
  d. Reaction of manganese with halogens
  Manganese reacts with the halogens, forming the corresponding manganese (II)
  

  halides. For fluoride, manganese (III) fluoride is also formed.

  

  2. Uses
  • Manganese is too brittle to be of much use as a pure metal. It is mainly used
  in alloys, such as steel. Steel contains about 1% manganese, to increase the
  strength and also improve workability and resistance to wear. Manganese
  steel contains about 13% manganese. This is extremely strong and is used for
  railway tracks, safes, rifle barrels and prison bars.
  • Drinks cans are made of an alloy of aluminium with 1.5% manganese, to improve
  resistance to corrosion. With aluminium, antimony and copper it forms
  highly magnetic alloys.
  • Manganese (IV) oxide is used as a catalyst, a rubber additive and to decolourise
  glass that is coloured green by iron impurities. Manganese (IV) oxide is a powerful
  oxidising agent and is used in quantitative analysis. It is also used to make
  Fertilizers and ceramics.
  • Manganese sulphate is used to make a fungicide.
  

  1.5.6. Iron
  Iron is a grey to black, odourless metal with density 7.874. It melts at 1535 ^oC and
  

  boils at 2750 ^oC.

  1. Chemical reactions
  a. Reaction of iron with air
  Iron reacts with oxygen, O_2, forming Fe (II) and Fe(III) oxides. The oxide layer does not
  passivate the surface. Finely divided iron, e.g. powder or iron wool, can burn:
  
  b. Reaction of iron with water
  Air-free water has little effect upon iron metal. However, iron metal reacts in moist
  air by oxidation to give a hydrated iron oxide. This does not protect the iron surface
  to further reaction since it flakes off, exposing more iron metal to oxidation. This
  process is called rusting.
  
  c. Reaction of iron with acids
  Iron metal dissolves readily in dilute sulphuric acid in the absence of oxygen forming
  Fe(II) ions and H₂. In aqueous solution Fe(II) is present as the complex [Fe(H₂O)₆]²⁺.
  
  Concentrated nitric acid, HNO₃, reacts on the surface of iron and passivates the
  

  surface (makes it unreactive).

  d. Reaction of iron with halogens
  

  Iron reacts with excess of the halogens, F₂, Cl₂, and Br₂, to form Fe(III) halides.

  

  2. Uses
  • Iron is an enigma – it rusts easily, yet it is the most important of all metals. 90%
  of all metal that is refined today is iron. Most is used to manufacture steel, used
  in civil engineering (reinforced concrete, girders etc) and in manufacturing.
  • Alloy steels are carbon steels with other additives such as nickel, chromium,
  vanadium, tungsten and manganese. These are stronger and tougher than
  carbon steels and have a huge variety of applications including bridges, electricity
  pylons, bicycle chains, cutting tools and rifle barrels.
  

  • Stainless steel is very resistant to corrosion. It contains at least 10.5% chromi
  um. Other metals such as nickel, molybdenum, titanium and copper are added
  to enhance its strength and workability. It is used in architecture, bearings,
  cutlery, surgical instruments and jewellery.
  • Cast iron contains 3–5% carbon. It is used for pipes, valves and pumps. It is not
  as tough as steel but it is cheaper.
  • Magnets can be made of iron and its alloys and compounds.
  • Iron catalysts are used in the Haber process for producing ammonia, and in the
  Fischer–Tropsch process for converting syngas (hydrogen and carbon monoxide)
  into liquid fuels.
  • Iron plays an important role in the transfer of oxygen by hemoglobin. Each
  hemoglobin binds four iron atoms. Iron in hemoglobin binds with oxygen as
  it passes through the blood vessels in the lungs and releases it in the tissues.
  
  1.5.7. Cobalt
  Cobalt is a dark grey metal with a density of 8.90. It is insoluble in water at room
  temperature.
  1. Chemical reactions
  a. Reaction of cobalt with air
  Cobalt does not react readily with air. Upon heating the oxide Co₃O₄ is formed, and if
  

  the reaction is carried out above 900°C, the result is cobalt (II) oxide, CoO.

  
  Cobalt does not react directly with nitrogen, N₂.
  
  b. Reaction of cobalt with water
  Cold water has little effect upon cobalt metal. The reaction between red hot cobalt
  metal and steam produces cobalt (II) oxide, CoO.
  
  c. Reaction of cobalt with acids
  Cobalt metal dissolves slowly in dilute sulphuric acid to form solutions containing
  the hydrated Co(II) ion together with hydrogen gas, H2. The actual occurrence of Co
  (II) in aqueous solution is as the complex ion [Co(OH₂)₆]²⁺.
  
  It also dissolves in dilute nitric acid to form cobalt (II) nitrate and oxides of nitrogen.
  
  (where NO_x stands for any oxide of nitrogen, i.e, NO, NO₂, …)
  
  Concentrated nitric acid renders it passive due to the formation of oxide layer Co₃O₄
  

  which is insoluble in the acid.

  d. Reaction of cobalt with halogens
  Metallic cobalt reacts with halogens, forming cobalt (II) halides.
  
  2. Uses
  • Cobalt, like iron, can be magnetized and so is used to make magnets. It is alloyed
  with aluminium and nickel to make particularly powerful magnets.
  • Other alloys of cobalt are used in jet turbines and gas turbine generators at
  high temperature.
  • Cobalt metal is sometimes used in electroplating because of its attractive appearance,
  hardness and resistance to corrosion.
  • Cobalt salts have been used for centuries to produce brilliant blue colours in
  paint, porcelain, glass, pottery and enamels.
  • Cobalt is an essential trace element and found at the centre of the vitamin
  B₁₂ (cobalmin, C₆₃H₈₈CoN₁₄O₁₄P). It contains a cobalt(III) ion and is necessary for
  the prevention of pernicious anaemia and the formation of red blood corpuscles,
  but it is involved many other functions too.
  

  1.5.8. Nickel
  Nickel is a grey solid metal with density of about 8.9. It melts at 1455^oC and boils at
  2920^oC.
  1. Chemical reactions
  a. Reaction of nickel with air
  Nickel does not react with oxygen, O₂, at room temperature, under normal conditions.
  Finely divided nickel can burn in oxygen, forming nickel (II) oxide, NiO.
  
  b. Reaction of nickel with water
  Nickel metal does not react with water under normal conditions. Nickel (II) ion
  complexes with water under acidic and neutral conditions forming a light green
  hexaqua nickel ion: [Ni(H₂O)₆]²⁺(_aq)
  
  In basic condition, nickel hydroxide precipitates:
  
  c. Reaction of nickel with acids
  Nickel metal dissolves slowly in dilute sulphuric acid to form the aquated Ni(II) ion
  and hydrogen, H₂.
  
  The strongly oxidizing concentrated nitric acid, HNO₃, reacts on the surface of
  nickel and passivates the surface.
  
  d. Reaction of nickel with hydroxide ions
  Metallic nickel does not react with aqueous sodium hydroxide.
  
  e. eaction of nickel with halogens
  Nickel reacts slowly with halogens, forming the corresponding dihalides.
  
  2. Uses
  • Nickel resists corrosion and is used to plate other metals to protect them. It is,
  however, mainly used in making alloys such as stainless steel. Nichrome is an
  alloy of nickel and chromium with small amounts of silicon, manganese and
  iron. It resists corrosion, even when red hot, so is used in toasters and electric
  ovens. A copper-nickel alloy is commonly used in desalination plants, which
  convert seawater into fresh water. Nickel steel is used for armour plating. Other
  alloys of nickel are used in boat propeller shafts and turbine blades.
  • Nickel is used in batteries, including rechargeable nickel-cadmium batteries
  and nickel-metal hydride batteries used in hybrid vehicles.
  • Nickel has a long history of being used in coins. The US five-cent piece (known
  as a ‘nickel’) is 25% nickel and 75% copper.
  • Finely divided nickel is used as a catalyst for hydrogenating vegetable oils.
  Adding nickel to glass gives it a green colour.
  
  1.5.9. Copper
  Copper is a light pink to red (shiny-reddish) metal of density 8.95 g/cm³. It melts at
  1083^oC and boils at 2570^oC.
  
  

  1. Chemical reactions
  a. Reaction of copper with air
  Copper metal is stable in air under normal conditions. When heated until red hot,
  

  copper metal and oxygen react to form Cu₂O.

  
  

  b. Reaction of copper with water
  Copper does not react with water in all conditions.
  
  c. Reaction of copper with acids
  Copper is not dissolved by non-oxidizing dilute acids such as dilute
  H₂SO₄ and HCl to produce hydrogen gas. This is why it is called a
  ‘noble metal’. Other noble metals include gold, silver and platinum.
  But copper metal dissolves in dilute and concentrated nitric acid, HNO₃ to form
  copper (II) nitrate and oxides of nitrogen. Here nitric acid acts as an oxidising agent.
  
  It also reacts with hot concentrated sulphuric acid to form copper (II) sulfate, sulphur
  

  dioxide gas and water. But normally, sulphuric acid is not an oxidising acid!

  

  d. Reaction of copper with halogens
  

  Metallic copper metal reacts with the halogens forming corresponding dihalides.

  

  2. Uses
  • Historically, copper was the first metal to be worked by people. The discovery
  that it could be hardened with a little tin to form the alloy bronze gave the
  name to the Bronze Age.
  • Traditionally it has been one of the metals used to make coins, along with silver
  and gold. However, it is the most common of the three and therefore the
  least valued. All US coins are now copper alloys, and gun metals also contain
  copper.
  • Most copper is used in electrical equipment such as wiring and motors. This is
  because it conducts both heat and electricity very well, and can be drawn into
  wires. It also has uses in construction (for example roofing and plumbing), and
  industrial machinery (such as heat exchangers).
  • Copper sulphate is used widely as an agricultural poison and as an algaecide
  in water purification.
  • Copper compounds, such as Fehling’s solution, are used in chemical tests for
  sugar detection.
  • Copper helps in storing iron, is involved in production of pigments for colouring
  hair, skin and eyes.
  
  1.5.10. Zinc
  Zinc is a grey solid with a density of 7.14 g/cm³. It melts at 419.5 ^oC and boils at 907
  ^oC.
  
  1. Chemical reactions
  a. Reaction of zinc with air
  Zinc reacts with oxygen in moist air. The metal burns in air with a blue-green flame to
  

  form zinc (II) oxide, a material that goes from white to yellow on prolonged heating.

  

  b. Reaction of zinc with water
  Zinc is unaffected with cold water. However, elemental zinc will reduce steam at
  

  high temperatures:

  

  c. Reaction of zinc with acids
  Zinc metal dissolves slowly in dilute sulphuric acid to form Zn(II) ions and hydrogen,
  H₂. In aqueous solution the Zn (II) ion is present as the complex ion [Zn(H₂O)₆]²⁺.
  
  When zinc reacts with oxidizing acids like HNO₃, no hydrogen gas is evolved.
  
  
  2. Uses
  • Mostly, zinc is used to galvanise other metals, such as ironsheets (amabati), to
  prevent corrosion. Galvanised steel is used for car bodies, street lamp posts,
  safety barriers and suspension bridges.Many houses in Rwanda are covered by
  galvanized iron sheets (amabati).
  • Large quantities of zinc are used to produce die-castings, which are important
  in the automobile, electrical and hardware industries.
  • Zinc is also used in alloys such as brass, nickel silver and aluminium solder.
  • Zinc oxide is widely used in the manufacture of many products such as paints,
  rubber, cosmetics, pharmaceuticals, plastics, inks, soaps, batteries, textiles
  and electrical equipment. Zinc sulphide is used in making luminous paints,
  fluorescent lights and x-ray screens.
  • It is a component of insulin.
  
  Checking up 1.5
  1. State what is observed and write an equation, for the reaction that would
  take place when
  a. Copper is added to hot concentrated sulphuric acid.
  b. Chromium is dissolved rapidly in hot concentrated aqueous alkali
  c. Nickel (II) ions complexes react with water under acidic and neutral
  conditions.
  d. Powdered zinc is dissolved in hot aqueous alkali.
  2. State at least one property that makes that:
  a. An aluminum - scandium alloy be used in fighter planes, high-end bicycle
  frames and baseball bats.
  

  b. Many alloys of titanium with aluminium, molybdenum and iron be mainly
  

  used in aircraft, spacecraft and missiles.
  c. Vanadium-steel alloys be used for armour plate, axles, piston rods and
  crankshafts.
  d. Alternatives of tanning leather using chromium be investigated.
  e. Manganese steel be used for railway tracks, safes, rifle barrels and prison
  bars.
  f. Iron be considered as an enigma.
  g. Cobalt be necessary for the prevention of pernicious anaemia and the
  formation of red blood corpuscles.
  h. Nichrome be used in toasters and electric ovens.
  i. Most copper be used in electrical equipment such as wiring and motors.
  j. Galvanised steel be used for car bodies, street lamp posts, safety barriers
  and suspension bridges.
  
  Assignment
  Question 3 is given to you as an assignment. You can use any source to carry
  out research in order to gain and provide relevant information to be presented
  comfortably.
  3. The following figures show objects made in different transition metals.
  Observe them and complete the table with the main transition metal
  which forms the objects, its two important properties and other two uses
  (apart from that shown by the figure).
  
  
  1.6. Identification of transition metal ions
  Activity 1.6
  Given a substance Y which contains one cation (from transition metal) and one
  anion,identify the cation and anion in Y. Carry out the following tests on Y , record
  

  your observations and deductions in the table below. Identify any gas evolves.

  

  • The cation in Y is …………
  • The anion in Y is ……………
  • Write the ionic equations for the reactions in test (i) and test (ii)
  ……………………
  Different transition metals have different colors. Also, different charges, or cations
  of one transition metal can give different colors. Another factor is the natural of
  the ligand. The same cations of a transition metal can produce a different color
  depending on the ligand it binds to.
  Many compounds containing transition metals have certain characteristic colours
  and thus, by observing a compound, we can not identify it.
  • Appearance or colour of different solid compounds containing transition
  

  metals

  Table 1.13: Colours of different solid compounds containing transition metals
  (first series)
  
  • Colours of aqueous solutions of some transition metal ions
  In aqueous solutions where water molecules are the ligands, the colours of some
  metal ions observed are listed in the table below:
  Table 1.14: Colours of different aqueous solutions containing some transition
  

  metals (first series)

  • Action of heat on solid compounds containing transition metal ions
  

  Table 1.15: Colours of different solid compounds containing transition metals
  

  (first series) due to action of heat.

  Note: On heating the following temporary colour changes may also occur:
  

  
  

  • Effect of aqueous sodium hydroxide and aqueous ammonia on solutions
  containing transition metal ions
  The hydroxides of transition metals are precipitated from solutions of the metal ions
  by the addition of hydroxide ions or ammonia. The colour of the precipitate can
  often be used to identify the metal present. The precipitates formed are gelatinous
  

  and often coloured and some form soluble complex ions with excess ammonia.

  a. To about 1cm³ of the solution containing the positive ion (cation), add
  

  2M aqueous sodium hydroxide dropwise until in excess

  

  

  b. To about 1cm³ of the solution containing the positive ion (cation), add
  

  2M aqueous ammonia dropwise until in excess

  
  

  
  

  
  

  Confirmatory tests for some transition metal ions
  Confirmatory tests are the tests required to confirm the analysis. Generally, a
  confirmatory test is used after other tests have been carried out to isolate/identify
  the ion. In order to confirm the ion without any dought
  
  a. Zinc ions
  i. Addition of little solid ammonium chloride followed by disodium
  hydrogen phosphate solution to a solution of zinc ions gives a white
  precipitate. The precipitate dissolves in excess ammonia or dilute
  mineral acids.
  ii. Addition of potassium ferrocyanide solution to a solution of zinc ions
  gives a white precipitate.
  
  b. Chromium ions
  To a solution of chromium (III) ions, add excess aqueous sodium hydroxide followed
  by little hydrogen peroxide and boil the resultant mixture. A yellow solution of a
  chromate is formed.
  
  
  Treatment of the yellow solution with:
  i. Lead (II) ethanoate or Lead(II)nitrate solution gives a yellow precipitate of
  Lead(II) chromate. Pb²⁺(aq) + CrO₄^2-(aq) →PbCrO4(s)
  ii. Barium nitrate (or chloride) solution gives a yellow precipitate of barium
  

  chromate.

  

  c. Manganese (II) ions
  To the solution of manganese (II) ions, add little concentrated nitric acid followed
  by little solid lead(IV) oxide or solid sodium bismuthate(V) and boil the mixture. A
  purple solution is formed due to MnO₄^- ion.
  
  d. Iron (II) ions
  i. Addition of potassium hexacyanoferrate (III) solution to a solution of iron
  (II) ions gives a dark blue precipitate .
  ii. Addition of few drops of concentrated nitric acid to a solution of iron (II)
  ions gives a yellowish solution due to iron (III) ions formed. The solution
  gives positive test for iron (III) ions.
  e. Iron(III) ions
  i. Addition of potassium hexacyanoferrate (II) solution to a solution of
  iron(III) ions gives a dark blue precipitate
  ii. Addition of potassium thiocyanate or ammonium thiocyanate
  solution to a solution of iron (III) ions gives a blood red coloration.
  f. Cobalt (II) ions
  Addition of potassium thiocyanate or ammonium thiocyanate solution to a solution
  of cobalt(II) ions gives a blue colored product of potassium cobalt(II) tetrathiocyanate.
  
  g. Nickel (II) ions
  i. Addition of potassium cyanide solution gives a yellow-green
  precipitate of Nickel(II) cyanide. The precipitate dissolves in excess
  reagent to form a dark yellow solution tetracyanonickel (II) ion.
  

  
  ii. Addition of aqueous ammonia followed by 2 to 3 drops of
  dimethylglyoxime solution to a solution of nickel (II) ions gives a red
  precipitate. The formation of this precipitate may sometimes require
  that the solution mixture would be warmed.
  h. Copper (II) ions
  In addition to use of aqueous ammonia, the copper(II) ions can be confirmed by
  addition of the following reagents to an aqueous solution of copper(II) ions:
  i. Potassium iodide solution: A white precipitate of copper (I) iodide
  stained brown with free iodine.
  
  Brown color fades on addition of sodium thiosulphate solution due to the reaction
  

  below:

  

  Checking up 1.6 (a)
  Given a substance K which contains one cation and one anion, carry out the
  following tests on K and record your observations and deductions in the table
  below. Identify any gas evolved.
  
  • The cation present in the compound K is ……………
  

  • The anion present in the compound K is ……………

  Checking up 1.6 (b)
  You are provided with substance D which contains one cation and one anion.
  You are required to identify the cation and anion in D. Carry out the following
  tests, record your observations and deductions in the table below. Identify any
  gas evolved.
  
  
  Checking up 1.6 (c)
  Aqueous sodium hydroxide is added separately to solutions of salts of the
  transition metals A, B and C. Identify A, B and C from the following observations.
  A: The white precipitate which appears is soluble in an excess of aqueous sodium
  hydroxide and also in aqueous ammonia.
  B: The blue precipitate which appears is insoluble in an excess of aqueous sodium
  hydroxide but dissolves in aqueous ammonia to form a deep blue solution.
  C: The green precipitate which appears is insoluble in an excess of aqueous
  

  sodium hydroxide and also in aqueous ammonia.

  END UNIT ASSESSMENT
  a. Multiple choice questions: Write the Roman number corresponding to the
  correct answer.
  1. Which of the following elements is not a transition metal?
  i. Copper
  ii. Nickel
  iii. Iron
  iv. Magnesium
  2. Which of the following complexes is linear?
  i. [Ag(NH₃)_2]+
  ii. [CoCl₄]^2-
  iii. [Pt(NH3)₂Cl²]
  iv. [CuCl₄]^2-
  3. Which of the following ions does not form coloured solutions?
  i. Cu⁺
  ii. Mn²⁺
  iii. Cr³⁺
  iv. Co²⁺
  

  4. Which of the following reactions of Cu²⁺ is an example of a chelation
  

  reaction?
  

  i. [Cu(H₂O)₆]²⁺ + 2OH- → [Cu(H₂O)₄(OH)₂] + 2H₂O
  
  2. What is the characteristic of electron configurations of transition metals?
  3. Which electrons, 3d or 4s, have the lowest ionization energies in a
  transition metal?
  4. a. Name any three transition metals that are essential to the biological
  system.
  b. Why do you think transition metals form coordination compounds that
  have covalent bonds?
  5. Name the following coordination compounds using systematic
  nomenclature.
  a. [Co(H₂O)₆]Cl₂
  b. [Cr(NH₃)₆](NO₃)₃
  c. K₄[Fe(CN)₆]
  d. Na[Au(CN)₄]
  e. [Co(H₂O)₂(en)₂]Cl₃
  6. a. (i) What is meant by the term co-ordinate bond?
  (ii) Explain why co-ordinate bonds can be formed between transition
  metal ions and water molecules.
  b. What name is given to any ligand that can form two co-ordinate bonds
  to one metal ion? Give an example of such a ligand.
  7. In order to determine the concentration of a solution of cobalt(II) chloride,
  a 25.0 cm³ sample was titrated with a 0.0168 M solution of EDTA^4-; 36.2
  cm3 were required to reach the end-point. The reaction occurring in the
  titration is:
  [Co(H₂O)₆]²⁺ + EDTA⁴–——→ [Co(EDTA)]²– + 6 H₂O
  a. What type of ligand is EDTA4–?
  b. Calculate the molar concentration of the cobalt (II) chloride solution.
  8. The ethanedioate (oxalate) ion,C₂O₄² , acts as a bidentate ligand. This ligand
  forms an octahedral complex with iron (III) ions.
  a. Deduce the formula of this complex and draw its structure showing all
  the coordinate bonds present.
  b. Give the name of a naturally-occurring in human body complex
  compound which contains iron.
  c. What is theimportant function of this complex compound?
  9. The compound [Co(NH₃)₄Cl₂]Cl contains both chloride ions and ammonia
  molecules as ligands.
  a. State why chloride ions and ammonia molecules can behave as
  ligands.
  b. What is the oxidation state and the co-ordination number of cobalt in
  this complex compound?
  10. a. Suggest why the compound [Co(NH₃)₆]Cl₃ has a different colour from
  that of [Co(NH₃)₄Cl₂]Cl.
  b. Name and give the formula of an ammonia complex used to distinguish
  between aldehydes and ketones.
  11. Chloride ions form the tetrahedral complex ion [AlCl4]– but fluoride
  ions form the octahedral complex ion [AlF6]3-. Suggest a reason for this
  difference.
  

  URLs: 3Files: 4Labels: 2
• UNIT 2:EXTRACTION OF METALS

  UNIT 2: EXTRACTION OF METALS
  Key unit competence
  To be able to: Relate the properties of metals to their methods of extraction and uses
  and suggest preventive measures to dangers associated with their extraction.
  Learning objectives
  At the end of this unit , students will be able to:
  • Describe the extraction of copper, iron, sodium, tantalum, zinc, wolfram and
  aluminium;
  • Outline and explain the uses of copper, iron, tantalum, zinc, wolfram, and tin
  ore (cassiterite);
  • Explain the issues associated with the extraction of metals and preventive
  measures;
  • Relate the properties of metals to their methods of extraction.

  Introductory activity
  1. a. Do you know any metal and mineral extracted in Rwanda?
  b. If yes, name them?
  c. Give the applications of those metal and mineral .
  2. Most of the metals are found in nature, not as pure metals, but as
  compounds, i.e. combined with other chemical elements. Such metals are
  extracted from their compounds using chemical reactions. The following
  setup shows one example of a laboratory chemical reaction. Analyse it and
  follow the procedure to be able to interpret the results.

    

  Procedure
  1. Transfer one spatula measure of copper (II) oxide to a hard-glass test-tube.
  2. Carefully add one spatula of charcoal powder on top of the copper (II)
  oxide without any mixing.
  3. Strongly heat these two layers for 5 minutes in a Bunsen flame.
  4. Allow the tube to cool and then look closely at where the two powders
  meet in the test tube.
  Questions
  1. Describe the solid copper (II) oxide before heating.
  2. Name the gas formed in the reaction.
  3. Describe the solid remaining after heating.
  4. Name the solid formed in the reaction.
  5. Write the word and symbol equations for the reactions.
  6. What does this reaction tell you about the relative reactivities of carbon
  and copper?
  7. Explain why this reaction is a redox reaction.

  The chemical substances in the earth’s crust obtained by mining are called
  minerals. Minerals, which are source of metal economically needed, are called
  “Ore”. The unwanted impurities present in ore are called gangue. A native metal is
  any metal that is found in its metallic form, either pure or alloyed, in nature (example:
  Gold). The entire process of extraction of metal from its ore is called metallurgy.

  Many metals are found in the Earth’s crust as ores. An ore is usually a compound
  of the metal mixed with impurities. An ore is any naturally-occurring source of a
  metal that you can economically extract the metal from. Most metals exist in in
  nature as compounds, usually oxides or sulphides.

  Sulphide ores cannot be converted directly into the metal. Instead they must be
  converted to the oxide. This is achieved by roasting them in air. Roasting involves
  heating of ore in presence of air below melting point of metal in reverberatory
  furnace.

  Reverberatory furnace, in zinc, copper, tin and nickel production, is a furnace used
  for smelting or refining in which the fuel is not in direct contact with the ore but
  heats it by a flame blown over it from another chamber. In steel making, this process,
  now largely obsolete, is called the open-hearth process.

   

  In this process, volatile impurities escape leaving behind metal oxide and sulphur
  dioxide (metal sulphide is converted to metal oxide).
  
  This process causes problems because of the large quantity of sulphur dioxide
  produced. Sulphur dioxide is one of the principal causes of acid rain; hence SO2 is
  one of the air pollutants.
  However if the sulphur dioxide can be collected before being released into the
  atmosphere, it can be used to make sulphuric acid.

  Metals in their compounds are in oxidized form, i.e. have lost electron(s) and bear
  a positive charge. In order to get them as metals, they need to gain electrons: this
  process or reaction of gaining electrons is called “reduction”. Reduction of metals
  can be carried out using a chemical reducing agent such as carbon (coke, charcoal)
  or electricity.

  When the mineral is dug up, a method must be used to separate the metal from the
  rest of the ore. This is called extraction of the metal.

  2.1. Relating the properties of metals to their methods of extraction
  Activity 2.1
  Make a research (in books or internet) in order .to:
  1. Find different methods of extraction of metals referring to the reactivity.
  2. Explain why the ores have to be concentrated and different ways that can
  be used.
  2.1.1. Methods of metal extraction according to their properties
  A number of methods are used to extract metals from their ores. The best method
  to use depends on a number of factors. These factors are based on the properties of the metal.
  The main factor here is the reactivity of the metal (the position it takes in the reactivity
  series). Let us use the following question method to clarify these factors.
  • Will the method successfully extract the metal? This depends on the reactivity
  of the metal.
  • How much do the reactants cost? Raw materials vary widely in cost.
  • What purity is needed, and are the purification methods expensive? Some
  metals are not useful unless very pure, others are useful impure.
  • How much energy does the process use? High temperatures and electrolysis
  use a lot of energy.
  • How efficiently, and in what quantities, can the metal be made? Continuous
  processes are more efficient than batch processes.
  • Are there any environmental considerations? Some processes produce a lot
  of pollutants.

  Different methods of metal extraction will be considered in this unit. Some examples
  are given below:
  • Reduction of metal oxides with carbon
  Carbon (as coke or charcoal) is cheap. It not only acts as a reducing agent, but it
  also acts as the fuel to provide heat for the process. However, in some cases (for
  example with aluminium) the temperature needed for carbon reduction is too high
  to be economic - so a different method has to be used. Carbon may also be left in
  the metal as an impurity. Sometimes this can be removed afterwards (for example,
  in the extraction of iron); sometimes it cannot (for example in producing titanium),
  and a different method would have to be used in cases like this.
  • Electrolysis of the metal ore
  This method is used to extract metals which are difficult to reduce by chemical
  agents.These metals include aluminium or metals which are difficult to reduce such
  as Group 1 and 2 metals.
  Reduction of the metal oxide with hydrogen
  Hydrogen can be used as a reducing agent, it is also used in purification of cpper.
  This is the main method for the extraction of tungsten from its oxide.
  Sustainable use of natural resources: As these extraction processes are expensive
  and the supply of ore is not infinite, it is essential to recycle the metals as much
  as possible. Recycling of metals is one way of conservation and sustainable use of
  natural resources.
  Metallurgical operations are sources of pollution, water and air pollution. Measures

  must be taken to eliminate or at least to minimise that kind of pollution.

  

  2.1.2. Concentrating the ore
  Concentrating the ore is also called “Dressing” or “Benefaction of ore”. The ore from
  which the metal will be extracted has to be prepared before extraction. Concentrating
  the ore is getting rid of as much of the unwanted rocky material as possible before
  the ore is converted into the metal.This simply means “Removal of gangue from
  ore”. This may be done by chemical or physical means.
  • By physical processes
  Physical operations use physical techniques that do not change the chemical nature
  of the minerals; these techniques are based on physical properties such as: density,
  magnetic properties, etc… In many cases, it is possible to separate the metal
  compound from unwanted rocky material by physical means.
  - Mechanical sorting (Hand picking): this involves use of hands to pick the
  gangue and breaking away the gangue using a hammer.
  - Magnetic separation: the ore is crushed and a very strong magnet is used to
  sort out magnetic materials from non-magnetic materials. This method is for
  example used to separate wolframite from cassiterite where cassiterite being
  non-magnetic is not attracted by the magnet.
  - Washing: the ores are usually denser than the gangue, which may be washed

  away in a stream of water on an inclined table. Examples of ores separated in
  this way are galena and limestone; cassiterite from silicates.
  - Froth flotation method: this process is important in treating many sulphide
  ores like galena, zinc blende (zinc sulphide), copper pyrites, etc. Separation is
  based on different abilities of mineral and gangue particles to be moistened
  by water. During this process, the ore is first powdered and fed into a large
  concentration tank containing water and a suitable oil (frothing agents).

  
  

                     http://(Source: https://blog.byjus.com/extraction-of-zinc/)

  The mixture is agitated by blowing air through at a high pressure. The sulphide ores
  rise to the surface in the froth while the gangue sinks to the bottom. The froth is
  skimmed off the surface, and an acid is added to break up the froth. The concentrated
  ore is filtered and dried.
  
  

  • By chemical processes.
  For example, pure aluminium oxide is obtained from bauxite by a process involving
  a reaction with sodium hydroxide solution. Some copper ores can be converted
  into copper (II) sulphate solution by leaving the crushed ore in contact with dilute
  sulphuric acid for a long time and then copper can be extracted from the copper (II)
  sulphate solution.
  - Leaching with aqueous solvents: in this method the finely powdered ore is
  treated with a suitable reagent that dissolves the ore but not impurities, for
  example:
  1. Bauxite is crushed and digested with sodium hydroxide solution
  2. Zinc ores can be leached with dilute sulphuric acid and electrolysed

  - Roasting in air and Calcination: Here the ore is powdered and roasted in air
  to drive off the water (for the hydrated ores) and other volatile substances. For
  example,
  Carbonates decompose to release carbon dioxide.
  Roasting: Sulphides release sulphur dioxide gas:
               2ZnS(s) + 3O₂(g) → 2ZnO(s) + 2SO₂(g)
  - Calcination: It is a process heating the ore strongly either in limited supply of air
  ih the absence in air
                ZnCO₃(s) → ZnO(s) + CO₂(g)
  - Smelting is the process by which a metal is obtained, either as the element or
  as a simple compound, from its oxide ore by heating beyond the melting point,
  ordinarily in the presence of reducing agents, such as coke.
                      CuO + C → Cu +CO

  Checking up 2.1
  1. What is an ore?
  2. What is the difference between an ore and a mineral?
  3. Are ores a finite resource?
  4. Are ores renewable?
  5. When is carbon used for extraction?
  6. Name a metal that could be extracted from its ore using carbon.
  7. When is electrolysis used for extraction?
  8. What do you understand with dressing of ore, smelting, froth flotation and
  gangue?
  9. Name two metals that can only be extracted by electrolysis.
  10. Suggest a reason why iron is extracted using carbon rather than by
  electrolysis.
  11. State three factors which determine the choice of reduction method
  used for the extraction of metals from their ores.
  12. Complete with the following words: iron, extracted, crust, gold, native,
  elemental, reduced
  “Metals come from the Earth’s …….. Some metals like ……. are very unreactive
  and are found as ………. in their ………state. Metals such as zinc, lead and
  ………are found combined with oxygen in compounds. These metals can be
  ……… using chemical reactions. The metal oxides are ………as oxygen is
  removed from the compound”.
  13. Why most of metals are produced by reduction method ?

  2.2. Methods of extraction of Copper
  Activity 2.2
  Use the search engine and the library:
  1. To find out the name of the two main ores of Copper.
  2. To describe the full process in copper metal is extracted from its ores.
  3. To demonstrate how copper is useful in our daily life.
  We are going to deal with the extraction of copper from its ores, its purification by
  electrolysis, and some of its uses.

  2.2.1. Extracting copper from its ores
  There are two main copper ore types of interest, copper oxide ores and copper
  sulphide ores. Both ore types can be economically mined, however, the most
  common source of copper ore is the sulphide ore mineral chalcopyrite also known
  as copper pyrites, which accounts for about 50 percent of copper production.
  
  

  Table 2.1: Some main ores of copper

  
  

  The ores typically contain low percentages of copper and have to be concentrated
  by, for example, froth flotation before processing.
  The method used to extract copper from its ores depends on the nature of the ore.
  Sulphide ores such as chalcopyrite are converted to copper by a different method
  from silicate, carbonate or sulphate ores. Copper, Cu, is mainly extracted from the
  ore chalcopyrite, CuFeS₂, in a three stage process.

  • In the first stage, chalcopyrite is heated with silicon dioxide and oxygen
  2 CuFeS₂ + 2 SiO₂ + 4 O₂→ Cu₂S + 2 FeSiO₃ + 3 SO₂
  • In the second stage, the copper (I) sulphide is roasted with oxygen at a high
  temperature in a reverberatory furnace giving copper (II) oxide.
              Cu₂S + 2 O₂→2 CuO + SO₂
  • In the third stage, the copper (II) oxide is reduced by heating with carbon.
              CuO + C → Cu +CO
  The end product of this is called blister copper - a porous brittle form of copper,
  about 98 - 99.5% pure.
  
  

  2.2.2. Purification of copper
  When copper is made from sulphide ores by the first method above, it is impure.
  The blister copper is first treated to remove any remaining sulphur (trapped as
  bubbles of sulphur dioxide in the copper - hence “blister copper”) and then cast into
  anodes for refining using electrolysis(electrolytic refining).The purification uses an
  electrolyte of copper (II) sulphate solution, impure copper anodes, and strips of high purity copper for the cathodes.
  

  The diagram shows a very simplified view of a cell.
  

  

                 

  
  

  • Any metal in the impure anode which is below copper in the electrochemical

  series (reactivity series) does not go into solution as ions. It stays as a metal and

  falls to the bottom of the cell as “anode sludge” together with any unreactive

  material left over from the ore. The anode sludge may contain valuable metals

  such as silver and gold.

  • Metals above copper in the electrochemical series (like zinc) will form ions at

  the anode and go into solution. However, they will not get discharged at the

  cathode provided their concentration does not get too high.

  Extracting copper from other ores

  Copper can be extracted from non-sulphide ores by a different process involving

  three separate stages:

  Step 1: Reaction of the ore (over quite a long time and on a huge scale) with a

  dilute acid such as dilute sulphuric acid to produce a very dilute copper (II) sulphate

  solution.

  Step 2: Concentration of the copper (II) sulphate solution by solvent extraction.

  The very dilute solution is brought into contact with a relatively small amount of an

  organic solvent containing a substance which will bind with copper (II) ions so that

  they are removed from the dilute solution. The solvent must not mix with water.

  Copper (II) ions are removed again from the organic solvent by reaction with fresh

  sulphuric acid, producing a much more concentrated copper (II) sulphate solution

  than before.

  Step 3: Electrolysis of the new solution. Copper (II) ions are deposited as copper

  on the cathode. The anodes for this process were traditionally lead-based alloys, but

  newer methods use titanium or stainless steel. The cathode is either a strip of very

  pure copper or stainless steel.

  

  2.3. Methods of extraction of Iron

  
  

  Activity 2.3

  Using your own research, use the available resources (chemistry books, notebooks,

  internet...) to make a succinct summary of:

  1. The main ores of iron.

  2. The full process involved in the blast furnace while the iron is being

  extracted (Include the diagram, the raw materials introduced and the

  role of each and the equations of the main reactions occurring).

  3. The main different forms of iron, its properties and their respective uses.

  The common ores of iron are iron oxides, and these can be reduced to iron by

  heating them with carbon in the form of coke. Coke is produced by heating coal

  in the absence of air. Coke is cheap and provides both the reducing agent for the

  reaction and also the heat source - as you will see below.

  

  

  2.3.3. Extraction

  Three substances are needed to enable extraction of iron from its ore. The combined

  mixture is called the “charge”:

  • Iron ore

  • Limestone (calcium carbonate).

  • Coke - mainly carbon.

  The charge is placed into the blast furnace. Hot air is blasted through the bottom.

  Several reactions take place before the iron is finally produced. This is a continuous

  process that needs a high temperature. The high temperature is produced by

  burning the carbon in a blast of hot air.

  Oxygen in the air reacts with coke to give carbon dioxide.

  

  

  2.3.5. Different forms of iron

  • Pig iron (Cast iron after removing some impurities), an alloy of iron that contains

  2 to 4 percent carbon, along with varying amounts of silicon and manganese

  and traces of impurities such as sulfur and phosphorus.

  - It is made by reducing iron ore in a blast furnace.

  - It has a low tensile strength

  - It is used in making gates, pipes, lamp posts where high strength is not needed.

  • Wrought iron is a soft, ductile, fibrous variety that is produced from a semifused

  mass of relatively pure iron globules (haematite) partially surrounded by

  slag. It usually contains less than 0.1 percent carbon and 1 or 2 percent slag.

  - It is purer than cast iron.

  - It is fibrous and tough

  - It can be welded (joined by hammering when red hot)

  - It is malleable and ductile

  - It is used for making sheets, wire and nails.

  - Steel is an alloy of iron, carbon, manganese, nickel and vanadium.

  2.3.6. Alloys of iron and their uses

  
  

  Checking up 2.3

  1. Iron is extracted using the Blast Furnace

  a. What is introduced into the top of the blast furnace?

  b. What is the source of heat used in the blast furnace? Write the involved

  equation

  c. What is the main reducing agent and how is it produced? (give an equation)

  d. How is the iron oxide reduced by this reducing agent? (give an equation)

  e. What is the other reducing agent and how does it reduce the iron oxide?

  (give an equation)

  2. The blast furnace is a continuous process. What does this mean and why

  is it advantageous?

  2.4. Methods of extraction of tin
  

  Activity 2.4

  Make a research using the appropriate books and internet to:

  1. Find out the main tin ore and state 2 regions in Rwanda where it is mined.

  2. Describe the process taken to extract tin from its principal ore.

  3. State three properties and uses of tin.

  
  

  
  

  
  

  
  

  prevent corrosion. In modern times tin is used in many alloys. The first alloy,

  used in large scale since 3000 BC, was bronze, an alloy of tin and copper.

  Most notably tin/lead soft solders, typically containing 60% or more of tin.

  • Another large application for tin is corrosion-resistant tin plating of steel.

  • Because of its low toxicity, tin-plated metal is also used for food packaging,

  giving the name to tin cans, which are made mostly of steel.

  • Due to its resistance to atmospheric corrosion and low melting point, it can be

  used to make sheet glass.

  Checking up 2.4

  1. Name the main ore of tin.

  2. Explain the extraction of tin from tin oxide.

  3. Tin metal is obtained by removing oxygen from the metal oxide. What

  name do we give to this chemical reaction?

  4. Explain why tin is said to be very useful.

  2.5. Methods of extraction of Zinc

  Activity 2.5

  Use the library or other search engine to

  1. Find out the main ores of zinc.

  2. Describe all steps followed to obtain the purified zinc from its sulphide

  ore.

  3. Demonstrate how zinc is so useful.

   

  concentrated zinc sulphide ores settles on the surface leaving behind the impurities

  in water.

  b. Process of roasting

  The concentrated ore is then treated at 900oC in the presence of excess air, on the

  base of a reverberatory furnace.This process of heating is called roasting. During

  this process, zinc oxide is obtained from the zinc sulphide ore. The equation for this

  process is:

  

  

  

  Checking up 2.5

  1. Recall 2 uses of zinc (refer to unit 1)

  2. Describe the extraction of zinc from zinc oxide.

  3. Zinc is extracted by removing oxygen from zinc oxide.

  a. What name is given to a reaction in which oxygen is removed from a

  substance?

  b. Explain how oxygen can be removed from zinc oxide to make zinc.

  4. Describe different methods by which the impurities in zinc extracted can

  be removed.

  2.6. Methods of extraction of Sodium

  Activity 2.6

  You are requested to use all resources about sodium (books, internet, textbooks).

  1. Discuss on the importance of sodium and its compound in Chemistry

  and in everyday life.

  2. Sodium metal does not occur in Free State in nature. Explain why these

  statement?

  3. Use the available resources to describe the way this valuable metal can be

  obtained at large scale (include the mining process, separation methods

  from its compounds, reaction equations where necessary, the main

  drawbacks that can be encountered during this process and the way to

  overcome them).

  On industrial scale sodium metal is extracted by “Down’s Process”. Down’s Process is

  based on the electrolysis of fused salt (NaCl). This compound is found all around

  the world, dissolved in sea water. It is also mined from the ore called rock salt, which

  is made up mainly of sodium chloride (NaCl).

  2.6.1. Extraction using Down’s Cell

  The salt is found mixed with insoluble impurities of sand and bits of rock. Once the

  dissolved rock salt has been filtered, we are left with a solution of sodium chloride

  in water. You might think that this could be electrolysed to extract the sodium.

  However, because sodium is more reactive than the hydrogen in the water,

  hydrogen would be produced at the cathode instead of sodium. Instead, sodium

  chloride is crystallized out of the solution and melted. The molten sodium chloride

  is then electrolysed to extract the sodium metal. This is carried out in a Down’s cell as

  shown in figure below.

  

  

  

  2.6.4. Uses of sodium

  • Molten sodium is used as a coolant in some types of nuclear reactor. Its high

  thermal conductivity and low melting temperature and the fact that its boiling

  temperature is much higher than that of water make sodium suitable for this

  purpose.

  • Sodium wire is used in electrical circuits for special applications. It is very

  flexible and has a high electrical conductivity. The wire is coated with plastics

  to exclude moisture.

  

  • In countries that experience very cold winter, NaCl is spread on the roads to

  de-ice the roads.

  Checking up 2.6

  1. Why is sodium not extracted by carbon reduction process? What is the

  method used?

  2. What difficulties arise in the extraction of sodium from its ore?

  3. Choose the suitable answer. In sodium extraction, fused (molten) NaCl is

  used rather than the dissolved (aqueous) NaCl because:

  a. Obtaining molten NaCl is easy than obtaining aqueous NaCl.

  b. In molten NaCl, the ions are free to move and not in aqueous NaCl.

  c. Molten sodium NaCl is an electrolyte and not the aqueous NaCl.

  

  2.7. Methods of extraction of Aluminium

  Activity 2.7

  Research in library textbook or search engine about aluminium and its extraction

  and make a good summary containing the following:

  1. The main ores of aluminium.

  2. Why aluminium extraction is suitable to be done by electrolytic method

  rather than reduction with carbon.

  3. The way the main ore is purified to have the pure aluminium oxide

  4. The process by which aluminium is extracted from the purified ore

  by electrolytic method (include the diagram, reaction equations at

  electrodes and the way to solve some difficulties which may arise during

  this extraction process)

  5. The properties of aluminium which makes it to be very important in our

  daily uses and the uses of aluminium.

  

  

  

  

  

  min (Al, Cu, Mg and Mn). Aluminium alloys are used in aircraft and other transportation

  vehicles because of its low density.

  • It is used as jewellery because it is a shiny metal, it has a good appearance.

  • Resists corrosion because of the strong thin layer of aluminium oxide on its

  surface. This layer can be strengthened further by anodising the aluminium.

  • It is used for making window frames.

  • Aluminium foil is used for wrapping cigarettes, food, etc because it reflects

  light.

  • Drink cans because it is not toxic.

  • Aluminium utensils are extensively used for household purposes.

  • Aluminium is used to produce metals such as chromium and manganese from

  their ores (aluminothermic process).

  

  2.7.4. Recycling of aluminium

  Aluminum is one of the most recycled materials in the world. This is due to high

  cost of new aluminium that involves high cost of electricity needed to produce

  aluminium. Recycled aluminium cost less in production and constitutes another

  way of sustainable management of our natural resources.

  • The consumer throws aluminium cans and foil into a recycle bin.

  • The aluminium is then collected and taken to a treatment plant.

  • In the treatment plant the aluminium is sorted and cleaned ready for

  reprocessing.

  • It then goes through a re-melt process and turns into molten aluminium, this

  removes the coatings and inks that may be present on the aluminium.

  • The aluminium is then made into large blocks called ingots.

  • The ingots are sent to mills where they are rolled out, this gives the aluminium

  greater flexibility and strength.

  • This is then made into aluminium products such as cans, chocolate wrapping

  and ready meal packaging.

  • In as little as 6 weeks, the recycled aluminium products are then sent back to

  the shops ready to be used again.

  Checking up 2.7

  Consider the electrolytic extraction of aluminium.

  1. How is aluminium ore called?

  2. Explain why cryolithe is added to aluminium oxide

  3. Describe the process by which the ore of aluminium is purified.

  4. Write half-equations for the reactions at each electrode, and write an

  overall equation for the reaction.

  5. State what each electrode is made of.

  6. Explain why

  a. The anodes need to be regularly replaced.

  b. The electrolysis of aluminium oxide is expensive.

  c. Aluminium is recycled.

  7. Give three uses of aluminium and the properties responsible for each use.

  2.8. Methods of extraction of Wolfram (Tungsten)

  Activity 2.8

  1. Visit the nearby mining sites of wolfram and make a field report.

  2. Research using internet and some books and make a summary about the

  following:

  a. The ores of wolfram.

  b. Where in Rwanda do we find wolfram?

  c. How wolfram extraction differs from that of zinc in terms of reduction.

  d. Full description of the extraction process of tungsten from its ore.

  e. The main uses of tungsten.

  Tungsten ore is a rock from which the element tungsten can be economically

  

  2.8.2. The extraction process

  

  2.8.3. Advantages and disadvantages of the process

  a. Advantages:

  • It produces very pure tungsten

  • Hydrogen is a cheap reagent

  b. Disadvantages:

  • The energy cost are high

  • Using a flammable gas such as hydrogen at high temperatures is very dangerous

  2.8.4. Uses

  • Tungsten is mostly used in light bulb filaments which heat up to 2000ºC, when

  many other metals would vaporise, particularly at the pressures found inside

  light bulbs. This is because it has a very high melting point.

  • Tungsten has the highest melting point of all metals and is alloyed with

  other metals to strengthen them. Tungsten and its alloys are used in many

  high-temperature applications, such as arc-welding electrodes and heating

  elements in high-temperature furnaces.

  • Tungsten carbide (WC) is extremely hard and is very important to the metalworking,

  mining and petroleum industries. It is made by mixing tungsten

  powder and carbon powder and heating to 2200°C. It makes excellent cutting

  and drilling tools, including a new ‘painless’ dental drill which spins at ultrahigh

  speeds.

  • Calcium and magnesium tungstates are widely used in fluorescent lighting.

  Tungsten mill products are either tungsten metal products, such as lighting

  filaments, electrodes, electrical and electronic contacts, wires, sheets, rods etc

  or tungsten alloys.

  • Due to tungsten’s ability to keep its shape at high temperatures, tungsten

  filaments are now also used in a variety of household applications, including

  lamps, floodlights, heating elements in electrical furnaces, microwave ovens,

  x-ray tubes and cathode-ray tubes (CRTs) in computer monitors and television

  sets.

  • The metal tolerance to intense heat also makes it ideal for thermocouples and

  electrical contacts in electric arc furnaces and welding equipment.

  

  Checking up 2.8

  1. Nowadays, there is a special reduction method used to extract Tungsten

  from its ores.

  a. State 2 main ores of tungsten.

  b. Write the balanced equation of the reduction reaction of tungsten (II)

  oxide.

  2. Suggest the reason why ore concentrating plants are always located in

  close proximity to the mine.

  3. Give 2 widely known physical properties of tungsten and the uses

  associated to these properties.

  2.9. Methods of extraction of tantalum

  Activity 2.9

  1. Visit the nearby mining sites of tantalum and make a field report.

  2. The picture below shows the miners at work in Rwanda. The metal being

  extracted here is very important in modern area. According to Merchant

  and Consulting Ltd (2018), Rwanda is the world’s largest producer of this

  metal which is called “tantalum”.

  

  Research using any relevant source of information about tantalum:

  a. To find out the physical properties of tantalum metal and its main uses.

  b. To make a description of the extraction of tantalum from its main ore

  tantalite.

  Tantalum is a hard, heavy, shiny, grayish-blue metal that is very stable, almost

  impervious (impermeable) to air, water and all but a few acids. It has the third highest

  melting point of all elements (over 3000 oC), and its primary use is in capacitors for

  electronic applications, and for vacuum furnace parts.

  It is classified as a “refractory” metal, which means it can sustain high temperatures

  and resist corrosion. It is a good conductor of heat and electricity, which makes it

  useful in various electronics. Pure tantalum can be drawn into fine wire filament,

  which is used to evaporate other metals.

  2.9.1. Where tantalum is found

  Tantalum is found in hard rock deposits such as granites, carbonites and pegmatites

  (igneous rock that consists of coarse granite). The chief tantalum ores are tantalite

  [(Fe, Mn)

   

  Tantalum is an important component in many modern technologies, and is used in

  capacitors for everything from computers to mobile phones.

  Despite its importance in the world today, tantalum mining takes place in very few

  countries and mining it is difficult. Only four countries produced tantalum in 2016,

  and most was mined in the Democratic Republic of Congo (DRC). Rwanda, Brazil

  and China were the other top countries for tantalum mining in that year. Sites are

  being identified for future development, and existing sites are being evaluated for

  expansion.

  2.9.2. How tantalum is mined

  Tantalum comes from the processing and refining of tantalite. Tantalite is the

  common name for any mineral ore containing tantalum. Most tantalum mines are

  open pit; some are underground.

  The process of mining tantalum involves blasting, crushing and transporting the

  resulting ore to begin the process of freeing the tantalum. Before transportation, the

  ore is concentrated at or near the mine site, to increase the percentage (by weight)

  of tantalum oxide and niobium. The material is concentrated through wet gravity

  techniques, gravity, electrostatic and electromagnetic processes.

  2.9.3. How tantalum is processed

  The tantalum concentrate is transported to the processor for chemical processing.

  The concentrate is then treated with a mixture of hydrofluoric and sulphuric acids

  at high temperatures. This causes the tantalum and niobium to dissolve as fluorides.

  Numerous impurities are also dissolved. Other elements, such as silicon, iron,

  manganese, titanium, zirconium, uranium and thorium, are generally present and

  processed for other uses.

  The concentrate is broken down into a slurry (A slurry is a watery mixture of insoluble

  matter such as mud, lime, or plaster of Paris). The slurry is filtered and further

  processed by solvent extraction. Using methyl isobutyl ketone (MIBK), or liquid ion

  exchange using an amine extract in kerosenes, produces highly purified solutions of

  tantalum and niobium. In this way, the tantalum oxide is obtained which is finally

  reduced with molten sodium to produce tantalum metal in powder form.

  It can then be compacted (as it is for capacitors) to final shape, or may be melted

  (and refined) in an electron beam furnace.

  2.9.4. Uses for tantalum

  • Tantalum is used to make electrolytic conductors, aircraft engines, vacuum

  furnace parts, nuclear reactors and missile parts.

  • Tantalum is unaffected by body fluids, and is non-irritating, which makes it

  useful for surgical appliances.

  • It is common in the production of cell phones, personal computers, igniter

  chips in car air bags, cutting tools, drill bits, teeth for excavators, bullets and

  heat shields.

  • Because the metal is an electrical conductor, it is useful in many consumer

  electronics, such as microprocessors for plasma televisions.

  Checking up 2.9

  In pairs, answer the following questions:

  1. Where tantalum is found in Rwanda?

  2. How tantalum is mined.

  3. How tantalum is processed

  4. Give 3 uses of tantalum.

  5. What is the role of methyl isobutyl ketone (MIBK) in tantalum processing?

  2.10. Dangers associated with extraction of metals

  Activity 2.10

  If you have ever visited a mining site, remember all the processes involved and

  suggest all possible common dangers associated with the extraction of metals.

  The following are some of the dangers associated with metals extraction. Especially,

  these hazards are found in smelting and refining and in addition mining.

  1. Injuries

  Metal extraction industry has a higher rate of injuries than most other industries.

  Sources of these injuries include: splattering and spills of molten metaland slag

  resulting in burns; gas explosions and explosions from contact of molten metal

  with water; collisions with moving vehicles; falls of heavy objects; falls from a

  height,slipping and tripping injuries from obstruction of floors and passageways.

  Precautions

  Adequate training, appropriate personal protective equipment (PPE) [for example,

  hard hats, safety shoes, work gloves and protective clothing]; good storage,

  housekeeping and equipment maintenance; traffic rules for moving equipment

  (including defined routes and an effective signal and warning system); and a fall

  protection programme.

  2. Heat illnesses

  Heat stress illnesses such as heat stroke are a common hazard, primarily due to

  infrared radiation from furnaces and molten metal. This is especially a problem when

  strenuous work must be done in hot environments.

  Prevention of heat illnesses

  Water screens or air curtains in front of furnaces, spot cooling, enclosed airconditioned

  booths, heat-protective clothing and air-cooled suits, allowing

  sufficient time for acclimatization, work breaks in cool areas and an adequate supply

  of beverages for frequent drinking.

  3. Pollutions

  Mining operations are major contributors to the pollution of our environment such

  as: air pollution, water pollution, degradation of landscape, etc.

  Exposure to a wide variety of hazardous dusts, fumes, gases and other chemicals

  can occur during smelting and refining operations. Crushing and grinding ore in

  particular can result in high exposures to silica and toxic metal dusts (containing

  lead, arsenic and cadmium, for example). There can also be dust exposures during

  furnace maintenance operations. During smelting operations, metal fumes can

  be a major problem especially risk of developing respiratory system illness. In

  metallurgical operations, the carbon dioxide released has more dangers especially

  the global warming because it is a greenhouse gas.

  

  Control

  Dust and fume emissions can be controlled by enclosure, automation of processes,

  local and dilution exhaust ventilation, wetting down of materials, reduced handling

  of materials and other process changes. Where these are not adequate, respiratory

  protection would be needed.

  Many smelting operations involve the production of large amounts of sulphur

  dioxide from sulphide ores and carbon monoxide from combustion processes.

  Dilution and local exhaust ventilation (LEV) are essential.

  Sulphuric acid is produced as a by-product of smelting operations and is used in

  electrolytic refining and leaching of metals. Exposure can occur both to the liquid

  and to sulphuric acid mists. Skin and eye protection and LEV are needed.

  The extraction of some metals can have special dangers. Examples include fluorides

  in aluminium smelting, arsenic in copper, etc.

  These processes require their own special precautions

  Other dangers

  • Glare and infrared radiation from furnaces and molten metal can cause eye

  damage including cataracts. Proper goggles and face shields should be worn.

  High levels of infrared radiation may also cause skin burns unless protective

  clothing is worn.

  • High noise levels from crushing and grinding ore, gas discharge blowers and

  high-power electric furnaces can cause hearing loss. If the source of the noise

  cannot be enclosed or isolated, then hearing protectors should be worn. A

  hearing conservation program including audiometric testing and training

  should be instituted.

  • Electrical hazards can occur during electrolytic processes. Precautions include

  proper electrical maintenance with lockout/tagout procedures; insulated

  gloves, clothing and tools; and ground fault circuit interrupters where

  needed.

  • Manual lifting and handling of materials can cause back and upper extremity

  injuries. Mechanical lifting aids and proper training in lifting methods can

  reduce this problem.

  CASE OF RWANDA
  

  According to Rwanda Environment Management Authority (REMA), mining

  activities often impact significantly on the environment. For instance, sand

  collecting and quarrying are already shown some significant environmental

  impacts, including resource depletion, energy consumption, waste generation

  and emissions of air pollutants. The dangers to human life and health associated

  with mining include the displacement of people, land use changes, dust and

  noise pollution.

  In fact, the preparation of ores which uses a lot of water constitutes a major

  pollutant of stream water in Rwanda. For example, the waters draining the

  mining sectors of Rutongo and Gatumba pollute the rivers of Nyabarongo

  and Nyabugogo by sediments of clay and sand which they transport over long

  distances. It is this considerable mineral load which partly gives them the brown

  colour that is characteristic of the rivers in Rwanda. Mining and quarrying

  produce massive rejects which appear in nature in the form of enormous lots of

  earth and rocks. Erosion from rain water transports the mineral residue towards

  the valleys where streams are filled and covered by the residue which may be

  toxic to biodiversity.

  Checking up 2.10

  1. State 5 sources of injuries associated with the extractions of metals

  2. Give 3 ways you can prevent from heat illnesses.

  3. Describe 2 ways you can control the dangers associated with the

  chemicals in extraction of metals.

  4. Suggest a way of protecting mining workers against the risk of lung

  disease due to dust at their working place?

  2.11. Preventive measures associated with metal extractions

  Activity 2.11

  We already know that there are many dangers associated with the extraction of

  metals.

  Recommend measures that must be taken to prevent from these dangers (risks).

  Effort should be made by mining componies or planned for the future, to eliminate

  or minimize the environmental problems associated with metal extractions include:

  1. The potential sources of air contaminants should be enclosed and isolated

  2. Brief, for any operation related to metal extraction, measures must be

  adopted to protect the workers in particular and the environment in

  general by:

  • Elimination or reduction of air polluting gases.

  • Avoiding water and soil pollution.

  • Protect the landscape.

  • Using cleaner production techniques i.e Minimize sources of pollution and use

  of energy.

  3. Adopt technology that minimizes wastes produced through process-reengineering/

  recycling

  Checking up 2.11

  1. 1.In extraction of metals, the best and least costly preventive measures

  are those taken at the planning stage of a new process of extraction.

  Explain the main aspects that should be taken into account.

  2. 2.what are the main sources of pollution in metallurgy?

  END UNIT ASSESSMENT

  1. Which metal is extracted from Bauxite?

  a. Tin

  b. Tantalum

  c. Copper

  d. Aluminum

  2. Brass is

  a. An Element

  b. A Compound

  c. A Mixture

  d. An Alloy

  3. Bronze is an alloy of

  a. Copper and Zinc

  b. Lead and Copper

  c. Copper and Tin

  d. Barium, Zinc and Iron

  4. Which of the following metals is often found in pure state?

  a. Copper

  b. Iron

  c. Gold

  d. Aluminum

  5. Which metal is extracted from Haematite?

  a. Tin

  b. Iron

  c. Manganese

  d. Cadmium

  6. Rocks rich in metals with economic value are known as

  a. Metalloids

  b. Ores

  c. Allotropes

  d. Slag

  7. An alloy is a

  a. Compound of three elements

  b. Homogeneus mixture of two or more metals

  c. Heterogeneous mixture

  d. Element in impure form

  8. If a metal ore is called "pyrites" then it most probably has

  a. Chlorine

  b. Oxygen

  c. Sulphur

  d. Nitrogen

  9. Often to prevent corrosion, metals are galvanized by covering them with a

  layer of

  a. Copper

  b. Sodium

  c. Zinc

  d. Tin

  10. What is not true about Tantalum?

  a. It is classified as a "refractory" metal

  b. Tantalum oxide is reduced with molten sodium to produce tantalum

  metal in powder form.

  c. Its ore minerals include scheelite

  d. It is found in hard rock deposits such as granites, carbonites and

  pegmatites

  11. Complete with the terms applied in the Extraction of iron from haematite

  in industry (Blast Furnace).

  a. Raw materials: ___________________________________

  A mixture of ___________, __________ and _____________ is added at

  the top of the furnace. ________ air is blown into the furnace from the

  bottom. A chain of chemical reactions occur:

  b. Carbon reacts with oxygen in air to form ______________.

  Equation: _______________________________________________

  c. The hot carbon dioxide rises in the furnace and is reduced by _________

  to form __________. Equation: ____________________________

  d. Carbon monoxide is a ____________ agent. It ___________ iron(III) oxide

  in haematite to form hot molten ___________.

  Equation: ______________________________________________

  The hot molten iron is then rum out from the bottom of the furnace.

  e. The formula of limestone: ________________

  Limestone breaks up into____________ and_______________ when

  heated.

  Equation: ________________________________________________

  f. Calcium oxide helps to remove ___________ (the impurities) to form a

  liquid ‘_______’. Equation: ________________________________

  12. a. By giving reagents and conditions, state three different methods

  of extracting metals starting from their oxides. In each case, write

  equation(s) to illustrate the extraction of an appropriate metal.

  b. i. Why are metals more usually extracted from their oxides rather than

  from any other compound?

  ii. State two environmental problems associated with the extraction of metals

  from their oxides or sulphides and give the chemical responsible for each

  problem.

  13. a. Give the ores of iron.

  b. Explain the extraction of iron from its ores.

  14. a. Give 2 or 3 uses of aluminium, copper, zinc, and iron.

  b. Give two reasons why the extraction of aluminium is expensive.

  15. Tungsten is prepared in a pure form by high temperature reduction of

  tungsten (VI) oxide with hydrogen.

  a. Construct an equation for this reaction.

  b. Suggest why carbon is not used as the reducing agent.

  c. Suggest one advantage (other than purity of the product) and one

  disadvantage of using hydrogen as the reducing agent on an industrial

  scale.

  16. Zinc and copper are extracted in the same way as iron (in blast furnace)

  but exist as their sulphide ores.

  a. How is the sulphide ore converted into an oxide and what is the problem

  with this process? (give an equation)

  b. Why can aluminium not be extracted in this way?

  c. Why can tungsten not be extracted in this way?

  17. Copper is a widely used metal. The main ore of copper contains copper

  sulphide. Copper can be extracted from copper sulfide in a three stage

  process.

  a. In the first stage of extraction the copper sulfide is heated in air.

  i. Balance the symbol equation for the reaction. Cu2S + O2 → CuO +

  SO2

  ii. Explain why there would be an environmental problem if the gas from

  this reaction were allowed to escape into the atmosphere.

  b. In the second stage copper oxide, CuO, is reduced using carbon.

  Describe and explain what happens during this reaction.

  c. During the third stage the copper can be purified as shown in the

  diagram.

  

  i. What is the name of the type of process used for this purification?

  ii. Give one use of purified copper.

  d. Copper-rich ores are running out. New ways of extracting copper from

  low grade ores are being researched. Recycling of copper may be better

  than extracting copper from its ores. Explain why.

• UNIT 3: NPK AS COMPONENTS OF FERTILIZERS

  Key unit competency:

   To be able to analyze the components of quality Fertilizers and their benefits, effects of misuse and dangers associated with the substandard fertilizers. 

  Learning objectives 

  At the end of this unit , students will be able to:

   • State the major constituents of fertilizers; 

  • Identify the characteristics of good fertilizer; 

  • Briefly describe the manufacture of fertilizers; 

  • State the advantages and disadvantages of  using fertilizers; 

  • Interpret the labels on the fertilizer containers; 

  • Classify the fertilizers in terms of composition.

  Introductory Activity

  A plot of land has been divided into two parts and in both irish potatoes have been cultivated by two cultivators. 
  

  One of them harvested 2000 kg of irish potatoes of big size and the other harvested 50kg of irish potatoes of small size. 

  Given that on both plots of  land, the following work has been done at the same time 
  

  • Cultivation,
  • planting
  • Hoeing (or weeding)
  • Harvesting 
  

  Suggest reason(s) which caused the difference in the harvest.

  The total population in Rwanda was estimated at 11.3 million people in 2016, according to the latest census figures. Looking back, in the year of 1960, Rwanda had a population of 2.9 million people. Rwanda’s population will shoot to 18.2 million people by 2050 at an average growth rate of 2.3 %, the United Nations Population Fund (UNFPA) has projected. These statistics show that the population of Rwanda is going on increasing but as we know the area of Rwanda is not increasing. That is why Fertilizers and other agricultural techniques are needed for the population of Rwanda to be capable of feeding itself and even feed some other population in the region. 

  3.1. Types of Fertilizers

  Activity 3.1 

  1.  a. What is the role of fertilizers

  b. Name any examples of Fertilizers you have ever heard 

  2. Using this book or any other book or internet, read and analyse the content about the types of fertilizers and make a summary to be presented to the class. 

  A fertilizer is any material, organic or inorganic, that is used to supply nutrients to the soil. 

  There exist types of Fertilizers: 

  1. Natural Fertilizers(or organic Fertilizers) 

  2. Artificial Fertilizers (or chemical Fertilizers) 

  3.1.1. Natural Fertilizers 
  

  The name organic fertilizer refers to materials used as fertilizer that occur regularly in nature, usually as a by-product or end product of a naturally occurring process. They are made from remains of dead plants, wastes from animals or they can be minerals. Examples include manures and minerals. Manure is an organic material that is used to fertilize land. 
  

  1. Farmyard manure: animal manure that consists of feces 

  2. Green manure: is a term used to describe specific plant or crop varieties that are grown and turned into the soil to improve its overall quality. 

  3. Compost manure: is organic matter that has been decomposed and recycled as a fertilizer and soil amendment. 

  4. Minerals: Mineral mined powdered limestone, rock phosphate and sodium nitrate, are inorganic compounds which are energetically intensive to harvest and are approved for usage in organic agriculture in minimal amount. 
  

  3.1.2. Artificial Fertilizers 

  They are fertilizers which are chemically synthesized which contain one or more of the major elements required by plants for good growth.

  Examples: Urea, N.P.K, ammonium dihydrogen phosphate, NH4 (H2PO4),… 

  

  

  Checking up 3.1 

  Give the two main types of Fertilizers and discuss the pros and cons of using one or another  type of fertilizer.

  3.2. Components of a fertilizer

  Activity 3.2 
  

  1. Name any nutrients you know that plants need in order to grow 

  2. Using this book, any other books or internet do a research and find out 

  a. The types of nutrients and give any three nutrients in each category that plants need for their growth and classify the nutrients depending on how plants need them

  b. Any two roles for each nutrients for the plant growth.

  First it is important to understand that all industrial Fertilizers, by convention, regardless of type and specific use, have something called a NPK ratio.The NPK ratio will be prominently labeled on the package and indicates the percentage of major (or primary) nutrients the fertilizer contains. Example: Urea is a fertilizer with an NPKratio of 46-00-00. 
  

  The nutrients of plants are classified into three types namely: 
  

  • Major nutrients

   • Secondary nutrients 

  • Micronutrients 
  

  3.2.1. The major nutrients 
  

  The major nutrients for soil are nitrogen (N), phosphorus (P), and potassium (K). These major nutrients usually are lacking or insuffiscient in the soil because plants consume these nutrients in large amounts for their growth and survival.

  The letter N represents the actual nitrogen content in the fertilizer by percentage mass while P and K represent the amount of oxide in the form of phosphorus (V) oxide (P2O5) and potassium oxide (K2O) respectively.

  

  

  3.2.2. Secondary nutrients 

  Now, in the category of secondary nutrients, are calcium (Ca), magnesium (Mg), and Sulphur (S). As, these nutrients are generally enough in the soil, so fertilization is not always needed. Also, large amounts of Calcium are added when lime is applied to acidic soils. In fact, Sulphur is usually found in sufficient amounts from the slow decomposition of soil. 

  3.2.3. Micronutrients 

  In fact, micronutrients are those elements essential for plant growth which are needed but in only very small (micro) quantities. These elements are even called minor elements or trace elements. The common micro nutrients are boron (B), copper (Cu), iron (Fe), chlorine (Cl), manganese (Mn), molybdenum (Mo) and zinc (Zn). In fact, recycling organic matter such as grass clippings and tree leaves is an excellent way of providing micro nutrients to growing plants.

  

  

  3.3. The manufacture of Fertilizers

  Activity 3.3 

  1. Write reactions for the formation of the following compounds

  a. Ammonium sulphate

  b. Potassium sulphate

  c. Ammonium nitrate

  d. Urea 

  e. Ammonium phosphate 

  2. Using this book or any other book or internet, read and analyse the content about the manufacture of the following Fertilizers and make a summary to be presented to the class; 

  Ammonium sulphate, potassium sulphate, ammonium nitrate, urea, and phosphates 

  3. Rwanda has a resources that can be used to produce an industrial fertilizer; name that resource.

  

  

  

  

  

  supply phosphorus to the plants. These minerals are, therefore, converted into soluble materials, by reacting them with sulphuric acid, or phosphoric acid or nitric acid. 

  

  

  

  

  

  Characteristics of a good fertilizer

  A good fertilizer should have the following characteristics:

  It should contain the required nutrients, in such a form that they can be assimilated by the plants. It should be cheap. 

  It should be soluble in water. 

  It should be stable, so that it may be available for a long time for the growing plant. 

  It should not be injurious to the plants. 

  It should be able to correct the acidity of the soil.

  Not pollutant

  

  3.4. Disadvantages of the use of organic and inorganic Fertilizers

  Activity 3.4 

  “The use of fertilizers is a harm to humanity”, yes or not. Explain

  3.4.1. Organic Fertilizers 

  The use of organic fertilizer may have many advantages but also it may have some disadvantages 

  a.  Advantages 

  1. The manures add organic matter (called humus) to the soil which restores the soil texture for better retention of water and for aeration of soil. For example, organic matter present in the manures increases the water holding capacity in sandy soils and drainage in clay soil.

   2. The organic matter of manures provides food for the soil organisms (decomposers such as bacteria, fungi, etc.) which help in making nutrients available to plants. 

  3. Nutrient release: slow and consistent at a natural rate that plants are able to use. No danger of over concentration of any element, since microbes must break down the material. 

  4. Trace minerals: typically present in a broad range, providing more balanced nutrition to the plant. 

  5. They will not burn: safe for all plants with no danger of burning due to salt concentration. 

  6. Long lasting: does not leach out since the organic matter binds to the soil particles where the roots have access to it. 

  7. Fewer applications required: once a healthy soil condition is reached, it is easier to maintain that level with less work

  8. Controlled growth: does not over-stimulate to exceptional growth which can cause problems and more work. 

  b. Disadvantages

   1. Many organic products produce inconsistent results. 

  2. The level of nutrients present in organic fertilizer is often low. 

  3. The time of their preparation is too long. 

  4. Eutrophication 

  3.4.2. Inorganic Fertilizers 

  The use of inorganic fertilizers may have many advantages but also it may have some disadvantages

  a. Advantages 

  1. Chemical Fertilizers are made with synthetic ingredients designed to stimulate plant growth. 

  2. Commercial chemical Fertilizers have the advantage of predictability and reliability 

  3. Formulations are blended with accuracy and you can buy different blends for different types of plants; commercial formulated Fertilizers allow you to know exactly which nutrients you’re giving your plants, rather than guessing at the composition of organic formulas 

  b. Disadvantages 

  1. They can burn plants 

  2. They require a specific timetable of application and watering because of fast release of nutrients 

  3. Groundwater 

  • Increased nitrate levels increase the risks of blue baby syndrome, a rare form of anaemia which affects babies below 6 months of age. The cause is the oxidation by nitrite ions of Fe2+ in haemoglobin to Fe3+. The oxidized hemoglobin cannot bind oxygen, and the baby turns blue from lack of oxygen. Conditions in the digestive tracks of young children are more favourable to the bacteria which reduce nitrates to nitrites than those in adults. 

  • Another hazard of chemical Fertilizers is that carcinogenic nitrosoamines (yellow oil substance) may be formed in the human digestive track by the conversion of nitrate into nitrite. The nitrite produced in the stomach it combines with HCl to produce nitrous acid. Nitrous acid can react with any secondary amine in foods to form nitrosoamines and the reaction of nitrite with amino acids. 

  4. Repeated use or excess use of the same fertilizer producing acidic ions (NH₄+). Example of such a fertilizer is (NH4)2SO₄.

  5. Repeated use or excess use of the same fertilizer producing basic ions. Example of such a fertilizer is CaCO_3. 

  _{6. Warm temperat}ures and high rain fall: Cations such as Ca⁺⁺, Mg⁺⁺, K⁺ which are essential to living organisms, are leached (dissolved) from the soil profile, leaving behind more stable materials rich in Fe and Al oxides. This natural weathering process makes soils acid.

  • Man-made processes also contribute significantly to soil acidity. For example, Sulphur dioxide (SO₂) and nitrogen oxides (NOx) released primarily by industrial activities react with water to form acid rain, which acidifies soils, particularly forest soils with. 

  • Organic acids from plants during decomposition; 

  • CO₂ from root respiration and microbial respiration. 

  Effects of acid soil

  • Major effects of extremes in pH levels include gaps in nutrient availability and the presence of high concentrations of minerals that are harmful to plants. In very alkaline soil, certain micronutrients such as zinc and copper become chemically unavailable to plants. In very acidic soil, macronutrients such as calcium, magnesium and phosphorous are not absorbed while others reach toxic levels, 

  • Acid soil, particularly in the subsurface, will also restrict root access to water and nutrients. 

  • In addition to affecting how nutrients are dispensed to growing plants, pH levels also influence microorganism activity that contributes to the decomposition of organic materials. A neutral pH is ideal for microbial action that produces chemical changes in soil, making nitrogen, sulfur and phosphorus more available. A pH that is either too high or too low may also interfere with the effectiveness of pesticides by changing their basic composition or weakening their ability to kill unwanted insects. 

  Plant growth and most soil processes, including nutrient availability and microbial activity, are favoured by a soil pH range of 5.5 – 8. Example: The optimal pH range for most plants is between 5.5 and 7.0. The optimal pH range for some plants is between is given in the table below.

  

  

  For soils the pH should be maintained at above 5.5 in the topsoil and 4.8 in the subsurface. 

  Eutrophication

  The undesirable overgrowth of vegetation caused by high concentration of plants nutrients (Nitrogen and Phosphorous) in bodies of water (lakes, rivers,...)
  

  As consequence, water plants (e.g: water hyacinth: amarebe) grow more vigorously and this prevents the sun light from reaching the water and stops photosynthesis of aquatic plants which provide oxygen in the water to animals needed then animals die, deposits of organic matter on the bottom of the lake build up.

  When lake water is enriched with nutrients (e.g.: nitrates and phosphates), algal flourish, and produce an algae bloom, a green scum with an unpleasant smell. When algal die they are decomposed by aerobic bacteria. When the oxygen content falls too low to support aerobic bacteria, anaerobic bacteria take over. They convert the dead matter into unpleasant-smelling decay products and debris which falls to the bottom. Gradually, a layer of dead plant material builds up on the bottom of the lake. The lowering of the oxygen concentration leads to the death of aquatic animals (fish, crabs,…….)

  Checking up 3.4 

  1. Ammonia itself can be used as a fertilizer but has some disadvantages. Explain the disadvantages of using ammonia as a fertilizer. 

  2. Give any two advantages of the use of 

  a. Natural Fertilizers 

  c. Artificial Fertilizers 

  3. Give any two causes of acid soils

  3.5. Dangers of the use of substandard Fertilizers

  Activity 3.5 

  Using books or internet find out the dangers of substandard fertilizers

  Sub-standard fertilizer means any fertilizer which does not conform to the required NPK ratio. 

  Example: A fertilizer may be labelled 16-00-00, while the real NPK ratio is for example 25-00-00, 10-00-05, etc 
  

  Using these Fertilizers can lead to:

  • Soil pollution (basic soil or acidic soil) due to accumulation of ions which are acidic or basic • Poor growth of plants 

  • Poor harvest 

  • Eutrophication 

  • Fertilizer burn: leaf scorch resulting from over-fertilization, usually referring to excess nitrogen salts. Fertilizer burn is the result of desiccation of plant tissues due to osmotic stress, creating a state of hypertonicity. 

  In order to reduce the effects of substandard fertilizers different measure can be taken; 

  • Standardization of the fertilizer before use

  • Production of fertilizers in Rwanda, as this will help us to choose good minerals (where necessary) in producing fertilizers

   • Use of chemical fertilizers with coated pellets so that nutrients are released slowly 

  • Regular watering

  

  You provided with the following

  1. A Solution prepared by mixing 5.0 g of a sample of ammonium sulphate fertilizer which were warmed with sodium hydroxide and the ammonia evolved was absorbed in 100 cm3 of 0.5moldm-3 sulphuric acid 

  2. 1M sodium hydroxide 

  Procedure

  a. Fill the burette with solution of sodium hydroxide

  b. Pipette 20 cm3 of solutions of the prepared solution in (1), in conical flask. Add 2-3 drops of methyl orange indicator.

  c. Titrate this solution with sodium hydroxide from the burette until the indicator changes colour (indicator changes from pink to yellow).

  d. Record the results in the table. 

  

  

  

  

  

  

  

• UNIT 4 :BENZENE

  UNIT 4: BENZENE
  Key unit competence:
  To be able to relate the chemistry and uses of benzene to its nature and structure
  Learning objectives
  At the end of this unit , students will be able to:
  • State the physical properties of benzene;
  • Describe the uses of benzene;
  • Outtline the preparations of benzene;
  • Describe the chemical properties of benzene;
  • State the conditions required for different reactions;
  • Relate the conditions for the reactions of benzene to its chemical stability;
  • Illutrate the mechanism of electrophilic substitutions on benzene.

  Introductory Activity
  From your prior studies in organic chemistry, it is known that carbon is tetravalent
  while hydrogen is monovalent and compounds constituted by the two elements
  are known as ‘hydrocarbons’. The structures and chemistry of the hydrocarbons
  reflects to their uses as fuels and starting materials for many substances important
  in life such as pharmaceutical drugs, solvents, packaging materials, clothes and
  so on. In this activity you need to follow instructions given to explain how the
  structure of a substance determines its chemical properties and uses.
  1. Write down the molecular formulae for these five hydrocarbons
  a. A molecule with 6 carbon atoms and 14 hydrogen atoms
  b. A molecule with 6 carbon atoms and 12 hydrogen atoms
  c. A molecule with 6 carbon atoms and 10 hydrogen atoms
  d. A molecule of 6 carbon atoms with 8 hydrogen atoms
  e. A molecule of 6 carbon atoms with 6 hydrogen atoms
  2. From the molecules in 1) above, choose molecule(s) that fit(s) in the
  description provided, and then draw its (their) structural formula (e).
  a. Unsaturated hydrocarbon (s) that decolorize (s) bromine water and
  alkaline potassium manganate (VII)

  b. Saturated hydrocarbon (s)

  c. Hydrocarbon (s) with empirical formula of CH
  d. Unsaturated hydrocarbon (s) which do (es) not decolorize bromine
  water and potassium manganate (VII).
  e. Unsaturated hydrocarbon (s) that form(s) a white precipitate when
  treated with ammoniacal silver nitrate and forms a reddish-brown
  precipitate when treated with ammoniacal copper (I) chloride.
  3. It is known that unsaturated hydrocarbons decolourise both bromine
  water and alkaline potassium manganate (VII). Explain any assumption
  you can suggest about the compound in question 2.d)

  Some or all people are unique in their living attitudes and values. But being unique
  does not mean to be isolated from others as people need each other in order to
  complement and build a strong nation.
  

  This is true for benzene. From the above activity question 3 you may have been stuck
  while discussing on why this unsaturated compound has properties that are different
  from other unsaturated hydrocarbons provided within the same activity. But this
  does not mean that it is quite different from them. It will share some properties with
  others but exhibit its identity or its unique properties from others.
  

  In this unit, you will discover what makes benzene resistant towards some reactants
  and its importance will be highlighted.

  4.1. Structure of benzene
  Activity 4.1
  • Research in books or search engine about the structure of benzene.
  • Read and make a summary on the historical development of benzene’s
  structure.
  Michael Faraday was the first to isolate benzene from coal. Benzene was found to
  have the molecular formula of C₆H_6. However, its structural formula posed a problem
  for many years.
  For example, you can work out the structures of compounds whose molecular
  formula is C₆H₆ and see how many you can find.

  The structure of benzene must be only one, in which all the six hydrogen atoms
  occupy equivalent positions. This was discovered by Friedrich August Kékulé Von
  Stradonitz while daydreaming of a snake seizing its own tail. From this, he proposed a
  ring structure of six carbon atoms with double bonds alternating with single bonds.
  

  Furthermore, X-ray diffraction studies, first carried out by Kathleen Lonsdale, showed
  that benzene is planar and all its C-C bonds are of the same length (0.139 nm which is
  intermediate of C-C single bond and C=C double bond in alkenes) and bond angles
  of the same size (120^o).

  

  By comparing benzene with alkenes, the following points are noticed:
  

  • Benzene fixes 3 moles of hydrogen, thus it has 3 double bonds,
  • Benzene does not decolourise bromine water or acidified potassium
  manganate (VII) and does not turn green the acidified potassium dichromate,
  • Benzene does not react with water and hydracids under normal conditions.
  

  From the above points it can be easily noticed that benzene is not quite an alkene,
  due to its double bonds which do not occupy fixed positions. This change of positions
  of the double bonds is referred to as ‘resonance’.

  

  The sp² hybridized orbitals of carbon is involved in sigma bond formation with other
  two carbon atoms and one hydrogen atom to make a hexagonal ring. The remaining
  unhybrid p-orbital is involved in side-ways overlapping with a neighbor carbon
  atom to form a pi-bond. Since there is an equal probability of making the pi-bond
  with either neighbor carbon atom, pi-electron remains delocalized over six carbon
  atoms of the ring.

  

  Checking up 4.1
  Discuss and provide appropriate answers for the following questions:
  1. a. Benzene has the molecular formula C₆H₆. Draw the Kekulé structure for
  this showing all the atoms.
  b. Draw the skeletal structure showing the way the Kekulé structure is normally
  drawn.
  2. How does the structure of benzene differ from the cyclohexane structure?
  
  

  34.2. Physical properties, uses and toxicity of benzene
  Activity 4.2
  • Using the same resources (books or internet) as in activity 4.1, make a
  research about the main points that should be talked about while discussing
  the physical properties of any substance.
  • Then, make a summary to be presented about properties, uses and toxicity
  of benzene.
  

  Benzene has the following physical properties:
  • Benzene is a colourless volatile liquid with an aromatic (pleasant/sweet) smell.
  • Benzene boils at 80.1 °C
  • Benzene melts at 5.5°C.
  • Like other aromatic hydrocarbons (arenes) benzene is insoluble in water.
  • It is less dense than water (specific gravity or relative density is 0.88).. Describe the Structure and Bonding of Benzene

  Benzene has many uses:
  It has been used by chemists since 1800 because it is a good solvent for other organic
  compounds. Benzene itself is an excellent solvent for certain elements, such as
  sulphur, phosphorus, and iodine. It is found in crude oil. It is used to make plastics,
  resins, synthetic fibers, rubber, lubricants, dyes, detergents, drugs and pesticides.
  
  

  Benzene is highly toxic and is said to be carcinogenic.
  A person exposed for long time to benzene (even at low levels), can develop anaemia
  and leukaemia.
  

  Benzene is formed in both natural and synthetic processes. Natural sources of
  benzene include volcanoes and forest fires. It is a component of crude oil, petrol and
  cigarette smoke.

  

  Checking up 4.2
  1. Benzene is flammable and carcinogenic. What do you understand by
  the term “carcinogenic”?
  2. What advice can you give to your friend who smokes?

  4.3. Preparation of benzene
  Activity 4.3
  • Some of the reaction of all alkanes and alkynes discussed in senior five lead
  to the formation of benzene. Use the following examples to describe how
  each of the following conversions can be carried out
  1. From CH CH to C₆H₆
  2. C6H6 from n-hexane
  3. Ethanol to C₆H₆
  • To add other methods used to prepare benzene and to be able to describe
  them, use the same sources (books/search engines) as in previous activities
  to discuss about all the methods that can be used to obtain benzene.
  • Take a note to share with others.
  • Some of the reactions of alkanes and alkynes discussed in senior five lead to
  the formation

  All the raw materials provided in the activity above are from the topics covered
  in senior five, so hopefully you performed them very well. The methods used for
  preparing benzene are based on reduction reaction and decomposition reaction and
  even addition reaction.
  
  

  1. Industrial preparation (on large scale)
  a. From petroleum oils: By catalytic reformation of petroleum products
  By fractional distillation followed by reforming. Fraction of naphtha is
  heated over Cr₂O₃ – Al₂O₃ at 500-550^oC and 15atm pressure (aromatisation).

  

  When platinum is used at 15 atm pressure at 500oC, the process is called
  ‘platforming’.
  
  

  b. By converting methylbenzene into benzene
  Methylbenzene is much less commercially valuable than benzene. The
  methyl group can be removed from the ring by a process known as
  “demethylation”.
  The methylbenzene is mixed with hydrogen at a temperature of between 550 and

  650°C, and a pressure between 30 and 50 atmospheres, with a mixture of silicon

  dioxide and aluminium oxide as catalyst.

  

  c. From ethyne
  When ethyne is heated in the presence of iron as catalyst or organo-Nickel, it undergoes
  cyclization.

  

  2. Laboratory preparation
  a. From benzoic acid
  In this method benzoic acid is heated with soda lime.

  

  b. From benzenediazonium salt
  In this method, the benzenediazonium salt formed by reacting phenylamine
  with sodium nitrite and a mineral acid is treated with hyposphorous acid
  (H₃PO₂) and water.

  

  d. From cyclohexane
  When cyclohexane is heated with Palladium or Platinum as catalyst and with
  sulfur, it undergoes dehydrogenation forming benzene. When cyclohexane
  is heated with sulphur, benzene is also produced.

  

  Checking up 4.3
  Discuss and describe how you can obtain benzene starting with inorganic
  reagents, showing necessary conditions at every step.

  4.4. Chemical stability of benzene
  Activity 4.4
  In chemical energetics (senior five), you learnt many forms of enthalpy changes
  that take place when various reactions take place. In this activity, you have to
  use some of the concepts of these enthalpy changes in order to understand the
  stability of benzene. By following instructions provided as questions and using
  the following data:
  Enthalpy change of atomization of carbon, C(s):       +715 kJ (mol of C atoms)^-1
  Enthalpy change of atomization of hydrogen,            H₂(g): +218 kJ mol^-1
  Bond energy of C=C (average):                                 610 kJ mol^-1
  Bond energy of C-C (average):                                 346 kJ mol^-1
  Bond energy of C-H (average):                                 413 kJ mol^-1
  
  Discuss and work out the enthalpy change of formation of benzene by the
  

  following stages.

  1. Calculate the energy needed to produce
  a. Six moles of gaseous carbon atoms from C(s)
  b. Six moles of gaseous hydrogen atoms from H₂(g)
  2. Calculate the energy released when
  a. Three moles of C-C bonds are formed from gaseous atoms
  b. Three moles of C=C bonds are formed from gaseous atoms
  c. Six moles of C-H bonds are formed from gaseous atoms.
  3. Use your answers to [1] and [2] to calculate the total energy change when
  a mole of gaseous benzene is formed from its elements.
  4. Compare your answer with experimental value of +82 kJ mol^-1.
  5. Now, use the available resources (books or internet) to find out what you
  

  can present about the stability of benzene.

  Benzene is an aromatic compound with molecular formula of C₆H₆. It is a planar
  

  hexagonal ring with three pi-bonds in an alternate manner.

  The delocalization of pi-electrons in benzene molecule provides extra stability which
  is known as ‘aromaticity’. Due to this aromaticity, benzene is more stable than
  expected as compared to aliphatic alkenes or the cyclic alkenes with three double
  bonds. Thus, it does not undergo addition reaction like alkenes do. In other words,
  benzene is less reactive than alkenes for addition reactions as this type of reactions
  can be responsible for loss of aromaticity (or resonance or stability). Benzene reacts
  preferably through substitution reactions in which one of its bonded H-atoms is
  

  replaced by an electrophile.

  Benzene is not the only aromatic molecule known (it is the smallest aromatic
  molecule, others include naphthalene, anthracite,…). Thus, for a molecule to be
  aromatic, it has to fulfill the following criteria:
  It must be cyclic and flat
  It must be conjugated (i.e, all atoms around the ring must be able to participate in
  pi-bonding through resonance)
  It must have pi-delocalised electrons (4n + 2), where n (number of benzene rings),
  n= 0,1,2,3,4,5,6. This is known as Huckel’s rule.
  
  The stability of benzene can be explained on the basis of resonance in the
  molecule. There are two possible resonance structures (or forms) of benzene
  molecule that are in equilibrium. Thus, an approaching reagent (such as bromine for
  instance) can not be attracted to a double bond before the structures changes. The
  resonance hybrid of benzene molecule is represented with a circle at the center of
  

  hexagonal ring of carbon atoms as shown below:

  

  Another measurement of stability of benzene is the tendency of benzene to undergo
  electrophilic substitution reactions rather than electrophilic addition reactions as
  alkenes. The regular-hexagonal planar ring of benzene is attributed to resonance
  stabilization of this conjugated cyclic alkene. Two resonance structure of benzene is
  responsible for the extra stability of molecule. The presence of the p electron cloud
  makes a negative zone that could be attacked by electrophilic reagents, by giving
  

  electrophilic substitution reactions.

  Thermochemical data show that benzene does not have true double bonds. The
  theoretical heat of formation of gaseous benzene, taking into consideration 3
  double bonds, is +252 kJ/mol while experimental value is +82 kJ/mol, therefore
  the true structure is more stable by 170 kJ/mol than cyclohexa-1,3,5-triene (Kekulé
  

  structure).

  The enthalpy of hydrogenation of cyclohexane is -120 kJ/mol.

      

  Therefore, since Kekulé (cyclohexa-1,3,5-triene) structure has 3 double bonds, the
  

  expected heat of hydrogenation is 3 times i.e. 3 x (-120) kJ/mol = -360 kJ/mol.

  

  However the experimental enthalpy of hydrogenation of benzene is only -208 kJ/
  mol! Therefore benzene is more stable by 152 kJ/mol than it would be if it was
  cyclohexa-1,3,5-triene. This stabilization energy is called delocalisation energy or
  

  resonance energy.

  

  Note: Because of this extra stability, benzene:

  •    Does not undergo reactions with halogens and halogen acids which are characteristic of alkene,

  •    Does not react with water in the presence of H+ and does not react with acidified KMnO4

  •    Cannot be represented by these structures because of its inertness

  

  •    Under drastic conditions, it however reacts with Cl2 or Br2 in the presence of ultraviolet light/light or halogen carrier,

  •    Reacts so fast with oxygen, by producing yellow luminous flame which is sooty.

  Checking up 4.4
  

  Refer to your results from the activity 4.4 to discuss and conclude on this:
  Do your results support that real benzene is more or less stable than the Kekule structure? Explain your answer.

  4.5. Reactions of Benzene

  Activity 4.5 From the previous topics discussed in this unit, you have found that benzene has some uniqueness from aliphatic unsaturated compounds.
  Use the same resources to find out
  •    How benzene reacts and

  •    Its reactions with different substances and their respective mechanisms

  As seen in the previous discussions, since benzene contains carbon-carbon double bonds, it might be expected to undergo electrophilic addition reactions readily as it is the case for alkenes. This is not the case, however, and benzene does not decolourise bromine water. Neither does it readily undergo any other addition reactions.

  The reason for this is that the delocalized system in benzene is stable, and addition reactions would break up this delocalization and lead to the formation of the products which are less stable than benzene itself. Benzene thus tends to undergo electrophilic substitution reactions rather than addition reactions.

  4.5.1. Electrophilic aromatic substitution reactions
  Aromatic compounds undergo substitution reactions with electrophiles in which one or more hydrogens of the benzene ring are substituted.

  Since the reagents and conditions employed in these reactions are electrophilic, these reactions are commonly referred to as “Electrophilic Aromatic Substitution”. The catalysts and co-reagents serve to generate the strong electrophilic species needed to perform the initial step of the substitution.

  Many substitution reactions of benzene have been observed and the five most useful are listed below.

  The specific electrophile in each type of reaction is listed in the right hand column.

  

  All electrophilic substitution reactions of benzene follow the same mechanism. After the formation of the electrophile, a two-step mechanism has been proposed for these electrophilic substitution reactions.

  In the first, slow or rate-determining step the electrophile forms a sigma-bond to the benzene ring, generating a positively charged benzenonium intermediate. In the second, fast step, a proton is removed from this intermediate, yielding a substituted benzene ring.i

  Briefly, electrophilic aromatic substitution reaction is realised in 3 steps:

  1. Electrophile formation

  2. Attack of the ring by electrophiles

  3. Deprotonation = loss of H+

  1. Halogenation Benzene reacts with chlorine or bromine in the presence of a catalyst, replacing one of the hydrogen atoms on the ring by a chlorine or bromine atom.

  •    The reactions happen at room temperature.

  •    The catalyst has to be a Lewis acid known as “halogen carrier”. The most commonly used catalysts are: aluminium (or iron) chloride, AlCl3/ FeCl3 or aluminium (or iron) bromide, AlBr3/ FeBr3 if you are reacting benzene with bromine.
     
  Example: The reaction with chlorine ( Chlorination)
  The reaction between benzene and chlorine in the presence of either aluminium chloride or iron gives chlorobenzene.

  2. Friedel-craft-acylation Acylation involves substituting an acyl group, RCO-, into a benzene ring.
  145Chemistry Senior Six Student Book
  The most reactive substance containing an acyl group is an acyl chloride (also is known as an acid chloride). These have the general formula of RCOCl.

  

  2. Friedel-craft-acylation Acylation involves substituting an acyl group, RCO-, into a benzene ring.
  145Chemistry Senior Six Student Book
  The most reactive substance containing an acyl group is an acyl chloride (also is known as an acid chloride). These have the general formula of RCOCl.

  

  3. Friedel-Crafts Alkylation

  This reaction involves substituting an alkyl group into a benzene ring. Hydrogen on the ring is replaced by a group like methyl or ethyl and so on.

  a. Using haloalkanes

  Benzene reacts with chloroalkanes in the presence of anhydrous AlCl3  or FeCl3  as a catalyst under reflux at 50oC to form alkylbenzenes

  

  b. Using alkenes

  Alkylbenzenes other than methylbenzene can be formed by reacting benzene with alkenes in the presence of HCl and AlCl3, under reflux at temperatures below 50oC. Mechanism:
  Step 1: The alkene reacts with the HCl in the same way as in electrophilic addition reactions:

  The carbocation behaves as the electrophile.
  Step 2 and Step 3 proceed in the same way as in the alkylation reaction described above.
  The overall reaction can be written as follows:

  

  The more stable cation gives the major product, methylethylbenzene (or isopropylbenzene).

  

  

  

  5. Nitration

  Nitration happens when one (or more) of the hydrogen atoms on the benzene ring is replaced by a nitro group, -NO2. Benzene is treated with a 50:50 mixture of concentrated nitric acid and concentrated sulphuric acid at a temperature not exceeding 50°C. The mixture is held at this temperature for about half an hour. Yellow oily nitrobenzene is formed.

  C6H6 + HNO3 → C6H5NO2 + H2O

  The concentrated sulphuric acid is acting as a catalyst and so is not written into the equations.
  Mechanism:
  Step 1: Nitric acid is a less strong acid than sulphuric acid, and acts as a base as the electrophile is formed.  

           H2SO4 + HNO3→  H2O + NO2+ + HSO4

  Step 2: The NO2+ is the electrophile and attacks the delocalised ring, breaking it temporarily:

  

  Step 3: The delocalised system then reforms itself by pulling in the electrons from the C-H bond. The H+ recombines with the HSO4- to form H2SO4.

  

  The overall reaction is C6H6 + HNO3→ C6H5NO2 + H2O

  The sulphuric acid behaves as a catalyst. The product is known as nitrobenzene.
  4.5.2. Some addition reactions and combustion reaction

  The benzene ring can undergo addition reaction under drastic conditions, breaking down its resonance

  

  3. Combustion reaction

  As other hydrocarbons, benzene burns in air forming carbon dioxide (or carbon monoxide in a limited supply of air) and water.

  Checking up 4.5

  Discuss and find out the answers for the following questions: Benzene can be nitrated to form nitrobenzene, C6H5NO2.
  a. Draw the structural formula for benzene and give its empirical formula
  b. State the reagents needed for the nitration of benzene
  c. An electrophile is formed during the nitration of benzene

  i. What is the formula of this electrophile?

  ii. Write an equation for the production of the electrophile iii. Use curly arrows to show the mechanism for the nitration of benzene

  C6H6(l) + 15/2 O2(g) → 6 CO2(g) + 3 H2O(l)  or  C6H6(l) + 9/2O2(g) → 6 CO(g) + 3 H2O(l)
  

  Checking up 4.5 Discuss and find out the answers for the following questions: Benzene can be nitrated to form nitrobenzene, C6H5NO2.
  a. Draw the structural formula for benzene and give its empirical formula
  b. State the reagents needed for the nitration of benzene
  c. An electrophile is formed during the nitration of benzene

  i. What is the formula of this electrophile?

  ii. Write an equation for the production of the electrophile

  iii. Use curly arrows to show the mechanism for the nitration of benzene

  4.6. Nomenclature and positional isomerism in derivatives of benzene
  Activity 4.6 1. Name  the following molecules:

  a. CH3CH2CH(CH3)CH3

  b. ClCH2CH2CHOHCH3

  c. CH3CH(C6H5)CH2CH2CH3

  d. C6H5NO2

  e. C6H4ClBr

  2. Discuss about rules for naming aromatic compounds in this book or any other source (textbook or internet). Then, make a summary to be presented.

  As you have seen from the previous lessons of this unit, benzene and its derivatives are referred to as aromatic compounds. The following diagram provides the structures of some aromatic compounds starting with benzene with one ring and then others with more than one ring and their respective names:

  

  

  Some benzene derivatives have their traditional or popular names such as the following:

  

  •    Di-substituted benzene derivatives with the prefixes “Ortho- or o-” for substituent groups on adjacent carbons (e.g, C1 and C2) in benzene ring. “Meta- or m-” for substituents separated by one carbon atom (e.g, C1 and C3). “Para- or p-” for substituent groups on carbons on opposite sides of ring (e.g, C1 and C4). The positions on the benzene ring are as follows:

  

  Benzene derivatives consisting of two substituents attached to the ring could be distinguished among three positional isomers (ortho- , meta- and para- isomers).
  These are named either by numbers or by using non numerical prefixes (ortho, meta and para).
  Notice that there are 2 identical ortho positions (2, 6), and 2 identical meta positions (3,5).

  

  

  

  Checking up 4.6

  Discuss and provide appropriate answers to the following questions:
  

  1. You are provided with C6H4Br2. Give three different structural formulae of isomers of C6H4Br2 and name them.
  

  2. Provide all the structures and names of compounds having the same molecular formula as C6H5NO3.

  
  

  
  

  
  

  
  

  
  

  
  

  
  

• UNIT 5: DERIVATIVES OF BENZENE

  Key unit competency

  The learner should be able to relate aromatic ketones, aldehydes, carboxylic acids and amines to their chemical activity.

  Learning objectives

  At the end of this unit , students will be able to:

  • Explain the effects of substituent groups on the benzene ring;

  • Give systematic names of aromatic compounds;

  • Describe the preparation and reactions of phenol, benzoic acid, benzaldehyde, phenyl ethanone and phenylamine;

  • State the uses of phenols;

  • Describe the reactions of phenol, aromatic carbonyl compounds and carbox-ylic acids;

  • Describe chemical properties of phenylamines;

  • Explain the azo-coupling reactions of phenylamine in manufacture of dyes and indicators;

  • Test and compare th acidity of phenol with alcohols and carboxylic acids;• Test for the presence of phenol in a given solution;

  • Compare and contrast the alkalinity of phenylamines with aliphatic amines and ammonia.;

  • Perform experiments on the reactions of phenol and phenylamine

  

  The simplest and most important member of aromatic hydrocarbons is benzene (C₆H₆). The benzene ring is particular because of its stability and certain properties.Many important chemical compounds are derived from benzene by replacing one or more of its hydrogen atoms with another functional group. It is a typical compound from which many of compounds of common properties derive.

  Some examples of derivatives of benzene are given below:

  

  

  5.1. Effect of substituent groups on the benzene ring

  

  

  The nature of a substituent already present in the benzene ring, not only determines the position of the next incoming group but also influences the rate of the second substitution reaction compared to the rate of substitution in benzene itself.A substituent might increase the rate of the second substitution, i.e. make the ring more reactive relative to benzene. Another group if present in benzene ring could decrease the rate of further substitution, i.e. make the ring less reactive compared to benzene.

  Groups already on the ring affect both the rate of the reaction and the site of attack. We say, therefore, that substituent groups affect both reactivity and orientation in electrophilic aromatic substitutions.

  5.1.1. Deactivating and activating substituents

  We can divide substituent groups into two classes according to their influence on the reactivity of the ring. The substituents which cause the compounds to undergo second substitution faster than benzene are called Activating Substituents (electron-releasing groups); they increase the electronic density on the benzene ring.On the other side, substituents which retard the rate of further substitution are referred to as Deactivating Substituents (electron-withdrawing groups); they decrease the electronic density on the benzene ring.

  

  5.1.2. Directing the incoming substituents

  During the formation of monosubstituted products in benzene, the electrophile can be attached at any position on the benzene ring. But, when the monosubstituted product is to be converted into disubstituted one, the existing substituent present in the ring directs the incoming group to a particular position. This is referred to as directive influence of the group. Depending on their directive influence, various groups (substituents) can be divided into two categories:

  • Ortho and Para Directors

  • Meta Directors

  a. Ortho and para directors

  These direct the new substituents to enter the ring primarily in Ortho and Para positions to themselves. These groups increase the electron density at the ring. Thus the reactivity of benzene ring towards electrophilic substitution reactions increases. For example if we carry out nitration of toluene, the mixture of ortho and paranitrotoluenes is formed

  

  b.Meta directors

  These direct the new substituents to enter the ring primarily in Meta position to themselves. For example, the nitration of benzoic acid produces m-nitrobenzene.

  

  These groups withdraw the electrons from benzene ring through resonance effect, reducing the electron density at the benzene ring. They decrease the reactivity of benzene ring towards electrophilic substitution reaction and make it less susceptible to the electrophilic attack.

  It has been found experimentally that in general ortho-para directing substituents activate the benzene ring and thus enhance the rate of reaction with electrophiles. On the contrary, the meta directing substituents deactivate the ring and retard the rate of reaction as compared to unsubstituted benzene.

  • Why activating substituents (Activators) have ortho and para directing properties?

  When the substituent present in the ring, has one or more lone pairs of electrons on the atom attached to the ring, it interacts with pi-electron system of the ring and it acts as electron donor (electron-donating substituent).

  
  

  The presence of an electron-donating group such as –OH or -NH2 causes further electrophilic substitution in ortho-para positions and also activates the ring to electrophilic attack.

  
  

  Let us take the example of phenol (C6H5-OH) and aniline (C6H5-NH2) which have available electron pairs on the atom directly attached to benzene ring. Thus phenol and aniline exhibit resonance and can be represented as hybrid of the following forms:
  

  
  

  

  In the above two examples, positions 2 and 4 are relatively richer in electrons than position 3 and this makes them susceptible to electrophilic attack. The electrophile would attack the ring preferentially at ortho and para positions where the electron density relatively is greater as compared to the meta positions. The second electrophile will be directed where sites are negatively charged, i.e ortho and para positions.

  From the above considerations, we conclude that all groups which are electron-donating are ortho-para directing and facilitate electrophilic substitution in the benzene ring.

  • Why Deactivating Substituents (Deactivators) have meta-directing prop-erties?

  When the substituent has at least one strongly electronegative atom and a multiple bond in conjugation with benzene ring, the substituent acts as electron- withdrawing substituent.

  Consider the nitrobenzene which contains –NO2 is able to exist as the following resonance forms:

  

  In the above example, it may be noted that resonance causes the decrease of electron density in the ring of nitrobenzene, and specifically at the ortho and para positions.

  In general, the electron withdrawing substituents decrease the electron density of benzene ring and thereby act as deactivators and meta-directors.

  • Anomalous Behaviour of Halogen Substituents

  The resonance effect enables the halogen substituents to act as ortho and para director. It is also expected to activate the ring to electrophilic attack, but on the contrary it is a ring deactivator. This is attributed to the very high electronegativity of the halogens due to which they withdraw electrons so strongly that they deactivate the benzene ring.

  While the resonance effect accounts for the ability of halogen to donate electrons to ortho and para positions, the combination of the two effects makes the halogenated benzene deactivated.

  

  5.2. Phenol

  

  The phenols are organic compounds with one or more  -OH-OH-  groups directly attached to a carbon atom in a benzene ring. The following are examples of phenols:

  

  Phenols occupy an important position in the modern synthetic organic chemistry for the preparation of dyes, antioxidants, phenolic resins and certain pharmaceutical products.

  The most important member in this family is phenol (hydroxybenzene):

  

  Phenol (hydroxybenzene) is a colorless crystalline solid which melts at 43oC and boils at 182oC. On exposure to air or light, it becomes coloured due to oxidation.

  Phenol is soluble in organic solvents and slightly soluble in water at room temperature, but infinitely soluble above 66 °C.

  Phenol exhibits intermolecular hydrogen bonding and this makes its melting point higher than that of hydrocarbons of comparable molecular mass.

  5.2.1. Sources and preparations of phenol

  Phenols are common in nature; examples include tyrosine, one of the standard amino acids found in most proteins. Many of the more complex phenols used as flavourings and aromas are obtained from essential oils of plants. Other phenols obtained from plants include thymol, isolated from thyme, and eugenol, isolated from cloves.

  Phenol, the cresols (methylphenols), and other simple alkylated phenols can be obtained from the distillation of coal tar or crude petroleum

  Phenol can be prepared:

  a. From benzenesulfonic acid

  In this method, benzenesulphonic acid obtained from sulphonation of benzene reacts with sodium hydroxide to produce phenol.

  

  

  

  5.2.2. Acidity of phenol

  

  The O-H bond is weaker in phenol than in alcohol. This is because the lone pair of electrons on the oxygen atom becomes associated with delocalized electrons of the ring. Because of this partial double bond develops between carbon and oxygen with the result that C-O bond is strengthened and the O-H weakened as the electronic density is displaced towards the ring. This gives to phenol a more acidic nature than alcohols.

  Therefore, phenol is more acidic than phenylmethanol and ethanol or cyclohexanol and this explains why phenol and not cyclohexanol nor phenylmethanol is soluble in sodium hydroxide solution. Because of this acidic property, phenol unlike alcohols reacts with alkali metal and sodium hydroxide or potassium hydroxide solution. Alcohols are not acidic enough to react with sodium hydroxide or potassium hydroxide solution but react only with alkali metal.

  

  Note: Phenol is a weaker acid it does not react with sodium carbonate or sodium hydrogen carbonate unlike carboxylic acids. The reaction with sodium carbonate or sodium hydrogen carbonate is also used to distinguish a carboxylic acid like benzoic acid from a phenol. Carboxylic acids give effervescence (liberate CO2) with sodium carbonate or sodium hydrogen carbonate while phenols do not.

  5.2.3. Reactions of phenols

  The reactions of phenol are of two types:

  • Reactions in which the O-H is broken;

  • Those involving the aromatic ring (Electrophilic substitution reactions).

  a. Esterification

  Phenols are weaker nucleophiles compared to alcohols because their lone pairs of electrons are partially delocalized over the benzene ring; that is why they do not form esters by direct reaction with carboxylic acids. However, the phenoxide ion (C6H5O-) is a better nucleophile and it reacts with acid derivatives such as acid chloride or acid anhydride, which are themselves more reactive than the parent acid:

  

  

  c. Electrophilic substitution reactions of phenol

  These reactions involve the replacement of hydrogen atoms at Para and Ortho positions of the ring, since the –OH is a para and ortho directing. The hydroxyl group increases the availability of electrons in the aromatic ring especially at para and ortho positions.

  Because of this ring activation, phenols react more readily with electrophiles than benzene itself.

  

  

  

  

  

  

  

  

  

  An aromatic hydrocarbon or arene (or sometimes aryl hydrocarbon)is a hydrocarbon with sigma bonds and delocalized pi electrons between carbon atoms forming a circle. In contrast, aliphatic hydrocarbons lack this delocalization

   

  5.3.1. Structure and nomenclature of aromatic hydrocarbons

  The trivial name of the parent monocyclic arene is benzene. The other members of this class are to a large extent assigned the systematic IUPAC names. However IUPAC have adopted the trivial names of lower arenes particularly, which have become popular by long usage. This has been done for brevity and convenience. Thus methylbenzene is invariably named as Toluene.

  a. Structure and nomenclature of aromatic alkanesIn the IUPAC system, arenes of this class are named in straight forward manner as substituted-benzenes.

  For example,

  
  

  When there are two substituents on the benzene ring, then positions are indicated by numbers, or by the prefixes ortho (o-), meta (m-) and para (p-). Thus the isomeric dimethylbenzenes are named as:

  

  if there are three or more substituent groups present on the ring, the arenes are preferably designated by IUPAC names. One of the groups is written at the to position of the hexagon, which becomes number 1. The six carbon atoms of the benzene are then numbered from 1 to 6 around the ring so that the substituents groups get the lower numbers. The substituent groups are preferably named in the alphabetical order. Thus,

  
  

  Note: The hydrocarbon group left after the removal of a hydrogen atom of the benzene itself is called phenyl group (C6H5-). The group left after the removal of a hydrogen atom of the CH3- group of toluene is called benzyl (C6H5-CH2-).

  b.Structure and nomenclature of some aromatic alkenes

  
  

   

  5.3.2. Reactions of alkylbenzenes

  Alkylbenzenes involve two parts of their structures in chemical reactions:

   

  • The side chain which can be oxidized by strong oxidizing agents like KMnO4 and Na2Cr2O7

  • The benzene ring which can take place in electrophilic substitution. In electro-philic substitution alkylbenzenes react more strongly than benzene itself since the alkyl groups donate electrons. Alkyl groups are ortho and para directing.

  a.   Oxidation of the side chain

  When alkylbenzenes are oxidized by powerful oxidizing agents (such as hot acidified or alkaline KMnO4 and Na2Cr2O7), the entire side chain, regardless of length, are oxidized to benzoic acid.

   For example,

  

  
  

  With weak oxidising agents such as acidic manganese dioxide (MnO2) or chromylchloride (CrO2Cl2), the side chain is oxidised to aldehyde (-CHO) group.

  
  

  b.Radical substitution

   Radical substitution takes place in three steps as for aliphatic alkanes: initiation, propagation and termination. The side chain substitution of hydrogen atom (s) occurs when chlorine or bromine is bubbled through boiling alkylbenzene in the presence of ultraviolet light or strong sunlight.

  

  Note that the above reaction may continue until all hydrogen atoms are replaced by halogen atoms.

  
  

  Note: Bromine gives similar products under similar conditions.

  

  5.4. Aromatic carbonyl compounds

  The compounds with carbonyl group attached to the benzene ring are known as aromatic aldehydes and aromatic ketones.

  

   5.4.1. Structure and nomenclature of aromatic carbonyl compounds

  a. Aromatic Aldehydes

  These carbonyl compounds contain a phenyl group in their structures. Aromatic aldehydes are of two types: (a) those in which the aldehyde group (CHO) is directly attached to a carbon of the aromatic ring; and (b) those in which the aldehyde group (CHO) is directly attached to a carbon of the side chain. Aldehydes of type (a) are called aromatic aldehydes, while those of type (b) are best regarded as aryl-substituted aliphatic aldehydes.
  

  

  Benzaldehyde is a typical aromatic aldehyde, while phenylacetaldehyde and cinnamic aldehyde are to be designated as aryl-substituted aliphatic aldehydes.

  
  

  The IUPAC name of an aromatic aldehyde is derived by dropping the ending –ene of the name of the parent hydrocarbon and appending the suffix –al. Thus

  

  Aromatic aldehydes are generally called by their common names which are derived from the names of the corresponding carboxylic acid by replacing the –ic or –oic acid by –aldehyde. Thus the name benzaldehyde is derived from benzoic acid; ortho-tolualdehyde is derived from ortho-toluic acid, and salicylaldehyde from salicylic acd.

  

  b.Aromatic ketones

  Ketones containing the carbonyl group attached to a benzene ring are named phenones. The individual name of such a ketone is derived by adding phenone to the stem formed by removing –ic from the name of the corresponding acid.

  

  The common names of aromatic ketones are obtained as usual by naming the alkyl or aryl groups attached to the carbonyl group, followed by the word ketone. These are given in the brackets above

  5.4.2. Preparation of aromatic carbonyl compounds

  a. Preparation of aromatic aldehydes

  Benzaldehyde is the simplest member in this family and it may be prepared by the following methods which are applicable to aromatic aldehydes in general. Benzaldehyde can be prepared using different methods. However, the main meth-od is oxidation of methylbenzene. The methods of preparation include:

  

  b.Preparation of aromatic ketones

  Although aromatic ketones may be prepared by any of the methods used for aliphatic ketones, they are generally prepared by Friedel-Crafts reaction between an aromatic hydrocarbon and acylchloride or acid anhydride. The benzene is treated with acylchloride or acid anhydride in the presence of a halogen carrier.

  • From acylchloride

  

  

  5.4.3. Reactions of aromatic carbonyl compounds

  The most typical reactions of the carbonyl groups are nucleophilic addition. In the carbon-oxygen double bond of the carbonyl group, oxygen is more electronegative than carbon, hence it has a strong tendency to pull electrons towards itself. This makes the carbon-oxygen double bond highly polar.

  The partially positive carbon atom can be attacked by nucleophiles. During the reaction, the carbon-oxygen bond gets broken and the net effect is that the carbonyl group undergoes addition reaction.

   

  The carbonyl group of aromatic aldehydes and ketones withdraws electrons from the benzene ring by inductive effect and resonance effect. Hence this group deactivates the benzene ring towards electrophilic substitution. The presence of the carbonyl group directs the substitution in meta-position. The electrophilic substitution of aromatic aldehydes and ketones is more difficult than electrophilic substitution of non substituted benzene.

  a.   Reactions of aromatic aldehydes

  Aromatic aldehydes present the same properties as aliphatic aldehydes. They give a positive test with Brady’s reagent (2, 4- Dinitrophenyl Hydrazine; observation: yellow or orange precipitate), with Fehling solution (observation: red solution brown precipitate), with Tollen’s reagent (observation: silver mirror) and with Schiff reagent (pink colour will be observed).

  Benzaldehyde is the typical and the simplest of aromatic aldehyde and will be used for illustrating the properties of this class of aromatic aldehydes. Benzaldehyde undergoes chemical reactions involving the side chain and benzene ring.

   

  i.   Benzaldehyde is not oxidized as readily as aliphatic aldehydes of oxidizing agents. While it reduces ammoniacal silver nitrate forming a silver mirror, it does not reduce Fehling’s solution. Nevertheless, benzaldehyde undergoe oxidation by atmospheric oxygen at ordinary temperature to form benzoic acid. This process known as autoxidation is catalyzed by light.

  

  
  

  vi. Benzaldehyde on treatment with concentrated sodium hydroxide solution gives benzyl alcohol and sodium benzoate this reaction is called Cannizzaro reaction:

  Note: Benzaldehyde condenses with hydroxylamine (NH2OH) and phenylhydrazine (C6H5NHNH2) to form benzaldoxime and phenylhydrazone respectively. Similarly, benzaldehyde reacts with hydrazine and semicarbazide to give hydrazine and semicarbazone respectively.

  Benzaldehyde is used for flavouring foods, scenting soaps, in the manufacturing of perfumes, in the preparation of dyes and in the synthesis of antibiotics.

  b. Reactions of aromatic ketones

  The reactivity of the carbonyl group in aromatic ketones is not greatly affected by the aryl groups attached to it. In consequence, they undergo the same general reactions as aliphatic ketones. However, they do not form the bisulphite compound, and in addition give the usual substitution reactions in the aromatic ring.

  
  

  Aromatic acids contain one or more carboxyl groups (–COOH) attached directly to the benzene ring. The acids in which the –COOH group is attached to the side-chain group may be regarded as aryl-substituted aliphatic acids. However, there are no characteristics differences in the behavior of the ring and side-chain acids. The term “aromatic acids” includes both classes of compounds.

  5.5.1. Structure and nomenclature of aromatic carboxylic acids

  Aromatic carboxylic acids are called by their common names or after the name of the parent hydrocarbon. Thus:

  

  

  5.5.2. Preparation of aromatic carboxylic acids

  Aromatic acids can be prepared by the same general methods which are available for aliphatic acids. In addition they may be obtained by oxidation of aromatic hydrocarbons having a side-chain.

  

  

  

  5.5.3. Reactions of aromatic carboxylic acids

  Aromatic acids are white crystalline solids, having higher melting points than aliphatic acids. They are slightly soluble in cold water but dissolve readily in hot water to form a colorless solution and on cooling, the acids recrystallise.

  Aromatic carboxylic acids with unsubstituted benzene ring are slightly stronger acids than the aliphatic acids. Thus benzoic acid is stronger acid than acetic acid.

  For the most part, the reactions of aromatic carboxylic acids are identical with those of aliphatic acids, the more important differences being in their rates.

  Benzoic acid, the simplest aromatic carboxylic acid is more acidic than phenol since it can react with Na2CO3 or NaHCO3 to liberate carbon dioxide gas. Its reactivity is attributed to its two parts: the carboxylic acid groupand the benzene ring.

  

  

  

  

  5.5.4 Uses benzoic acid and its derivatives

  • Benzoic acid is used in medicine, in the dye industry for making aniline blue.

  • Sodium benzoate being less toxic is used for preserving food products such as tomato ketchup and fruit juices.

  

  5.6. Aromatic amines

  

  The amino derivatives of the aromatic hydrocarbons are of two types:

  a. Aromatic amines or aryl amines in which the –NH2 group (or substituted –NH2 group) is attached directly to a carbon of the benzene ring.

  b. Aryl-alkyl or aralkyl amines in which the –NH2 group is attached to a carbon of the side-chain.

  Like the aliphatic amines, the aromatic amines may also be divided into primary, secondary and tertiary amines.

  5.6.1. Structure and nomenclature of aromatic amines

  in systematic naming, the numbering of carbon depends on the whole structure of the aromatic compound and the nature of the group attached to the benzene ring. In general, when more than one group are attached to the benzene ring, some groups are given priority as functional group.

  Here is a simple list for guidance on precedence (priority) of groups: COOH > COO- > COOR >COCl> CONH2 > CN > CHO > COR > OH >NH2 which means that from the list given, the amino is the least important, whereas the carboxylic acid group is the most important.

  

  
  

  5.6.2. Preparation of Phenylamine and uses of its derivatives

  a. Preparation of phenylamine

  i. Phenylamine, like other arylamines, can be prepared by the reduction of nitro compounds. The nitro compounds are treated with granulated tin, zinc or iron and HCl.

  

  

  b.Uses of aniline and its derivatives

  Derivatives of phenylamine or aniline are known as “Anilines”. These are employed in various fields of science and everyday life as given below:

  i. Anilines are used in the rubber industry for the processing of rubber material such as car tyres, balloons, gloves, etc.

  ii. It is used as a dyeing agent in the manufacture of clothes such as jeans, etc

  iii. It is employed in the production of drugs such as Paracetamol, Tylenol, Acetaminophen.

  iv. It is used as a pesticide and fungicides in the agricultural industry.

  v. It is utilized in the manufacture of polyurethane, which is in turn used in the making of plastics.

  5.6.3. Alkalinity of phenylamine

  Amines, both aliphatic and aromatic are basic. They are soluble in water. Phenylamine is weaker base than ammonia and aliphatic amines because the non-bonded electrons on the nitrogen atom are delocalized into the pi-system of the benzene ring. This makes the lone pair of electrons less available for reaction with a proton.

  The delocalization of the lone pair of electrons on the nitrogen atom in phenylamine can be represented by the following resonance forms:

  

  Thus aniline structure is greatly resonance stabilized.

  5.6.4. Reactions of phenylamine

  Aniline, also called phenylamine or aminobenze (C6H5NH2), is a colorless oily liquid, slightly soluble in water and soluble in organic solvents. It turns brown on exposure to sunlight.

  The aromatic amines in general give all the reactions of the amine group of aliphatic amines. However, the reactivity of the amine group is modified by delocalization of the non-bonded electrons on nitrogen atom into the pi-system of the aromatic ring.

  

  

  

  

  5.6.5. Reactions of diazonium salts

  Aromatic amines when treated with nitrous acid in cold mineral acid solution, give a very important class of compounds known as aryldiazonium salts. For example, aniline reacts with nitrous acid in hydrochloric acid solution at 0-50C to form a solution of benzenediazonium chloride. The reaction producing these salts is referred to as Diazotization.

  Diazonium salts give two types of reactions: (a) those in which the –N2X (where X is any anion, such as Cl-, Br-, NO3-, etc) is replaced by another univalent atom or group, with the liberation of nitrogen gas; (b) those in which the two nitrogen atoms are retained.

  

  c. Coupling reactions

  Diazonium salts act as electrophiles and react with highly activated benzene rings (like phenols and aromatic amines) to form brightly colored compounds called azo-compounds with the general formula: Ar-N=N-Ar, -N=N- is a linkage called azo linkage and responsible for the colour of the dye. These reactions are referred to as coupling reactions and they proceed via electrophilic substitution mechanism.

  

  A large number of benzene derivatives can be synthesized via diazonium salts. This will be illustrated by taking example of benzenediazonium chloride.

  Reactions of benzenediazonium chloride are summarized in the following scheme

  

  

• UNIT 6: POLYMERS AND POLYMERIZATION

  Key unit competence

  To be able to relate the types of polymers to their structural properties and uses

  Learning objectives
  At the end of this unit , students will be able to:

  •    Define the terms monomer, polymer and polymerization;

  •    Describe the formation of polymers;

  •    Describe addition and condensation polymerization;

  •    Explain the terms thermosetting and thermosoftening of the plastics;

  •    Discuss the advantages and disadventages of both natural and synthetic polymers;

  •    Explain the biodegradability property of polymers based on their chemical structure;

  •    Use equations to distinguish between condensation and addition polymerization;

  •    Prepare phenol-methanol polymer (Bakelite);

  •    Relate the structure and properties of polymers to their uses in the plastic and textile industries;

  •    Reduce polymer wastes by reusing, recycling and appropriate disposal;

  Introductory activity

  The images below show materials commonly used in our daily life.

  Observe carefully the following images in pairs, and answer the questions below:

  

  a. Give the name for each material and one of its uses
  b. Search in library (textbooks) and on internet, the chemical nature of each of the each material mentioned above
  c. Discuss the characteristics of the above materials and identify what do they have in common

  Nowadays the materials made of plastics such as fibers, plastic and rubber materials, are all around us and are commonly called polymers by chemists. Polymers are commonly used in household utensils, automobiles, clothes, furniture, spaceaircraft, biomedical and surgical components. Polymeric materials are light weight but can possess excellent mechanical properties and can be easily processed by different methods. In this unit you will learn more about polymers, their types and some important-synthetic and natural polymers.

  6.1. Definition of monomer, polymer and polymerization
  

  Activity 6.1

  1. Explain the following terms:

  a. Polymer
  b. Polymerization

  2. Identify the products made of polymers that are used at your home and mention at least 3 materials.

  6.1.1. Monomer
  The term monomer comes from mono “one” and meros “part”,which expresses a single unit or a small molecular subunit that can be chemically bind to another identical or different molecule to form larger molecule (polymer).The monomer is repeated in the polymer chain and it is the basic unit which makes up the polymer. For instance in the large compound formed by nA → -A n-  where A is a monomer and the polymer is given by the repeated monomers in the chain; i.e. –A-A-A-A-A-A-. The larger molecules such as carbohydrates, lipids, nucleic acids and proteins are found in living systems, like our own bodies.

  

  6.1.2. Polymer
  A polymer is a large molecule (macromolecule or giant molecule) composed of smaller molecules (monomers) linked together by intermolecular covalent bonds. Polymers have a high molecular weight in the range of 103 to 107. A polymer can be represented as (–An–) or (– A – A – A – A – A-……) which is a polymer of the monomer A.

  

  Table 6.1: Some examples of polymers and monomers

  
  

  6.1.3. Polymerization

  Polymerization is the process in which monomer units are linked by chemical reaction to form long chains (polymers).

  For example, a gaseous compound, Butadiene, with a molecular weight of 54 g/ mole combines nearly 4000 times by polymerization and gives a polymer, known as polybutadiene.

  Butadiene + butadiene + butadiene + … + butadiene → Polybutadiene

  Note: 1.The degree of polymerization is defined as the number of monomeric units in a macromolecule or polymer or oligomer (a polymer consisting of few number of monomers units) molecule.
             2. A polymer formed by identical monomers is called homopolymrer whilea polymer formed by different monomers is a copolymer.

  Preparation protocol of plastic from milk

  Experimental activity for the preparation of plastic from Milk Materials

  •    500ml 2% milk

  •    60ml vinegar

  •    One beaker of 100 mL

  •    2 beakers of 1 L each

  •    1 spoon

  •    1 strainer

  •    Aluminum foil

  •    Thermometer

  •    Hot plate or Bunsen burner and stand

  •    Matches if using a Bunsen burner

  Procedure
  1. Assemble all materials and chemicals

  2. Turn on the Bunsen burner or hotplate.

  3. Place beaker of milk over heat.  Stir constantly.

  4. Place thermometer into the milk.  Heat milk until it reaches 37ºC. 

  5. Remove from heat and immediately add vinegar, stirring constantly.  Here, the teacher should ask the students what they think will happen when the vinegar is added.

  6. The solution will rapidly separate into curds suspended in a clear yellow liquid.

  7. Strain the solution through the strainer into the empty 1L beaker.  Hold up the beaker to see the clear liquid.

  8. Scoop the curd onto a large piece of aluminum foil and press out into a thin layer.  Pass the aluminum foil around the class. Then dry especially overnight

  Milk is a colloid, which is defined as a suspension of large molecules of proteins in water.  Essentially, milk is a suspension of protein globules in water. 

  These proteins can undergo polymerization to create a natural plastic, as the casein molecules are associated together in long chains.  Proteins are generally unstable and are prone to unfolding, which changes the natural state of the protein.  This process is called denaturing.  The addition of acid, in this case vinegar, causes the casein protein to unfold and rearrange into the long chains of a polymer.  The process then causes the casein to precipitate out of the milk, leaving a clear watery substance behind.  The casein can then be formed into various shapes before drying.  In our experiment, thermal energy in the form of heat was applied to speed the process and cause a more complete separation.

  

  

  There two types of polymerization reaction: Addition polymerization and condensation polymerization.

  6.2.1. Addition polymerization

  Addition polymerization is a process where monomers are linked together to form a polymer, without the loss of atoms from the molecules. When the monomer molecules add up to form the polymer, the process is called “Addition polymerization”.
  

  This involves the combination together of monomer units to give new product (polymer ) having the same empirical formula to the monomer but having relative molecular weight.The monomer units are usually unsaturated compounds.
  

  Some examples of addition polymers are polyethylene, polypropylene, polystyrene, polyvinyl acetate, polyvinyl chloride (PVC), rubber, polytetrafluoroethylene (Teflon), etc.
  

  
  

  Example 2: Formation of PVC (polyvinylchloride)

  PVC (polyvinyl chloride) is found in plastic wrap, simulated leather, water pipes, and garden hoses, it is formed from vinyl chloride (H₂C=CHCl).

  Vinyl is a common name for ethylene, PVC is formed from the following reaction:

  

  Example 3: Formation of rubber: Rubber can be natural or synthetic

  Rubber is a natural polymer of isoprene (2-methyl-1,3-butadiene). It is formed by a linear 1,4- addition polymer of isoprene.

  Natural rubber has elastic properties because it has the ability to return to its original shape after being stretched or deformed. Therefore it is known as Elastomer. Natural rubber is prepared from latex which is a colloidal solution of rubber in water.

   Addition polymerization takes place in three steps like initiation, propagation and termination reading to the formation of polymer.

  i. Chain initiation

  A peroxide molecule breaks up into two reactive free radicals. Light or heat can provide the energy needed for this process. The chain is initiated by free radicals produced by reaction between some of the ethene and the oxygen initiator from peroxide.

  The process of initiation involves two parts: generation of initiator free radical and initiation of polymerization reaction.

  

  The second part of initiation occurs when the free radical initiator attacks and attaches to a monomer molecule. This forms a new free radical, which is called the activated monomer as indicated in the chemical equation below:

  

  ii. Chain propagation

  During this step, there is successive addition of large number of monomer molecules to form polymer free radical chain.In the propagation phase, the newly-formed activated monomer attacks and attaches to the double bond of another monomer molecule. This addition occurs again and again to make the long polymer chain.

  

  iii. Chain termination

  Termination step involves the coupling of two free radicals in order to produce a final molecule. The process is a termination step because no new free radicals are formed.

  

  6.2.2. Condensation polymerization

  Condensation polymerization is a process where two or more monomers chemically combine to form a polymer with elimination of simple molecules like water, ammonia, hydrogen chloride, alcohol, etc.

  In this type of reaction, each monomer generally contains two functional groups for the condensation reaction to take place.

  There are two main types:  polyamides which are formed between a diol with a dicarboxylic acid and polyesters which are formed when a dicarboxylic acid and a diamine react to form nylons.The condensation polymers include Nylons, Terylene, Kevlar polymer, proteins, cellulose, Bakelite, Dacron, etc.

  Example 1: FORMATION OF POLYAMIDES

  This type of polymers is formed by the result of generation of amide bonds in the polymerization reaction.

  i. Formation of Nylon

  Nylon 6 and nylon- 6, 6 are important examples for this type of polymers nylon 6 is synthesized from e-caprolactam, which on heating decomposes into 6-aminohexanoic acid that polymerizes into nylon 6. Here the number 6 represents the number of carbon atoms present in the monomer unit.

  

  Nylon -6,6 is produced by the condensation reaction between two monomer units adipic acid and 1,6-hexanediamine in the presence of heat. This is formed from a sixcarbon diacid and a six-carbon diamine as shown below.

  

  ii. Formation of Kevlar

  Kevlar is formed from the polymerization of benzene -1,4-diamine and benzene-1,4dioic acid as follows:

  

  In that polymerization –NH₂ group of hexamethylenediamine reacts with –COOH group of adipic acid to form –NH-CO- amide linkage with elimination of H₂O.

  Polyamides such as nylon-6,6 and Kevlar are widely used in clothing. Kevlar has some remarkable properties,including fire resistance and higher strength than steel. It is used to make protective clothing-for example for fighters,bulletproof vests and helmets.

  Example 2: FORMATION OF POLYESTERS
  

  i. Formation of Terylene

  

  ii. Formation of Dacron

  It is made from dimethyl -1,4-benzene dicarboxylate and 1,2-ethane diol: Dimethyl -1,4-benzene dicarboxylate + 1,2-ethane diol → Dacron + methanol

  

  Other polymer formation:
  

  Formation of bakelite:

  Bakelite    (polyoxybenzylmethyleneglycolanhydride)    is    the    oldest    synthetic    polymer.    It    is    formed    from    phenol    and    formaldehyde    in    the    presence    of    either    an    acid    or    a    base    catalyst.    The    initial    product    could    be    Novalac,    then    novalac    on    heating    with    formaldehyde    undergoes    cross    linking    to    form    bakelite.

  

  

  Experiment on the preparation of phenol formaldehyde resin (bakelite) Chemicals

  •    Glacial acetic acid,

  •    40% formaldehyde solution,

  •    Phenol, conc. H₂SO₄

  Apparatus Required:
  •    Glass rod,

  •    beakers,

  •    funnel,

  •    measuring cylinder,

  •    dropper

  •    Filter paper

  Procedure:
  1. Place 5ml of glacial acetic acid and 2.5ml of 40% formaldehyde solution in a 500ml beaker and add 2 grams of phenol.

  2. Add few ml of conc. Sulphuric acid into the mixture carefully. Within 5 min, a large mass of plastic is formed

  3. The residue obtained is washed several times with distilled water, and filtered product is dried and yield is calculated.

  Conclusion:

  A mixture of phenol and formaldehyde are allowed to react in the presence of a catalyst. The process involves formation of methylene bridges in ortho, para or both ortho and para positions. This results first in the formation of linear polymer (Called NOVALAC) and then in to cross-linked polymer called phenol-formaldehyde resin or bakelite.

  Protein formation:

  Polymerization of amino acids:

  •    Two amino acids can be linked by a condensation reaction (by removing of water molecule)

  •    Peptide bond is formed between the carbon atom in the acid group and the nitrogen atom in the amine group.

  •    The result molecule is called a dipeptide

  •    A chain of amino acids can be built up in this way and is called polypeptide

  •    A protein may just contain one polypeptide or may have two or more chains that interact

  

  The polyester Dacron and the polyamide Nylon-6,6, are two examples of synthetic condensation polymers. Some differences between addition polymerization and condensation polymerization are summarized in Table 6.2.

  

  

  

  6.3.1 Classes of polymers

  In general, polymers are classified into two classes such as natural polymers and synthetic polymers.

  6.3.1.1 Natural polymers

  Natural polymers are those that are obtained from natural sources. They are made naturally and are found in plants and animals or other living organisms. For examples: cotton, silk, wool, natural rubber, cellulose and proteins.

  a. Silk

  Silk is a fine continuous protein fiber produced by various insects’ larvae usually for cocoons.It is mainly a lustrous elastic fiber produced by silkworms and used for textiles.

  For example, orb-weaving spiders produce a variety of different silks with diverse properties, each tailored to achieve a certain task. Most arthropod species produce silks used for building structures to capture prey and protect their offspring against environmental hazards.

  b. Cotton

  Cotton is soft, fluffy staple fiber that grows in a boll, or protective case, around the seeds of the cotton plants. Cotton is natural cellulosic fiber.

  c. Proteins

  Proteins are highly complex substance made up of hundreds of thousands of smaller units called amino acids which are attached to one another to make along chain. There are around 20 different amino acids that occur naturally in proteins.  Many proteins act as enzymes that catalyze biochemical reactions and are essential to metabolic functions.

  d. Natural rubber

  Natural rubber is an elastic material obtained from the latex sap of trees that can be vulcanized and finished into a variety of products. It is a hydrocarbon polymer of isoprene (2-methylbuta-1,3-diene) obtained from latex. Latex is an emulsion of rubber particles in water that coagulate by addition of ethanoic acid.

  

  e. Cellulose

  Cellulose is an insoluble substance which is the main constituent of plant cell walls and the vegetable fibres such as cotton. It is a polysaccharide consisting of chains of glucose monomers.

  6.3.1.2 Synthetic polymers

  A part from the natural polymers, there are also synthetic polymers which are synthesized in the laboratory. They are manufactured from lower mass molecular compounds. Synthetic polymers (polymer made by two different monomers) are copolymers formed when many molecules of buta-1,3-diene or its derivatives chemical are joined with unsaturated compound. They can also be either thermosetting or thermoplastic polymers.

  Examples: Polyethene is a polymer formed by linking together a large number of ethene molecules. PVC, Nylons, Terylene and Polystyrene.etc

  6.3.2 Types of polymers
  The polymers can be classified into three main different types such as

  (i) plastics,

  (ii) rubber, and

  (iii) fibers.

  a. Plastics
  The plastics are types of polymers that are the most commonly used. Plastics are polymerized organic substances, solid of high molar mass, which at some time in its manufacture can be shaped by flow. They are electrical and thermal insulators. Their advantage is recycling and this allows them to be used many times. Plastics are materials that can be softened (melted) by heat and re-formed (molded) into another shape. The disadvantage of plastics is in their temperature resistance as they get quite fast soft and loose mechanical properties.

  Examples: fibers, Polyethylene, Teflon, Plexiglas, PVC, etc.

  b. Rubbers

  Rubber is a tough elastic polymeric substance made from the latex of a tropical plant or synthetically made.  There are two types of rubbers; natural rubber and synthetic rubber.Rubbers are soft and springy and return to their original shape after being deformed.

  •    Natural rubber

  Natural rubber is an elastic material obtained from the latex sap of trees that can be vulcanized and finished into a variety of products. Natural rubber is extracted from rubber producing plants, most notably the tree “Hevea brasiliensis”, which originates from South America.

  Natural rubber is a polymer of the monomer 2-methylbuta-1,3-diene (isoprene). Poly(2-methylbuta-1,3-diene) can exist in cis- and trans- isomeric forms. Natural rubber is the cis-form.

  •    Synthetic rubber

  A synthetic rubber is any artificial elastomer (man-made polymer having elastic properties)

  There are several synthetic rubbers in production. These are produced in a similar way to plastics, by a chemical process known as polymerization. They include neoprene, Buna rubbers, and butyl rubber. Synthetic rubbers have usually been developed with specific properties for specialist applications. Synthetic rubber can be made from polymerizing buta -1,3-diene, CH₂=CH-CH=CH=CH₂.

  The synthetic Rubber is an important addition of polymers that are obtained by polymerizing a mixture of two or more monomers. An example is styrene-butadiene rubber (SBR), a synthetic rubber formed by a mixture of 1,3-butadiene and styrene in a 3 to 1 ratio, respectively.

  

  The combination of monomer units gives a new polymer product having the same empirical formula to the monomer but having a higher molecular weight. The monomer units are usually unsaturated compounds (i.e. alkenes and their derivatives).

  Alkenes can be made to join together in the presence of high pressure and by adding a suitable catalyst. The π-bond breaks and the molecules are held together.

  c. Fibers

  Fiber (from Latin Fibra) is a natural or synthetic substance that is significantly longer. Fibers are often used in the manufacture of other materials. In manufacture of strongest engineering materials often fibers are incorporated, for example carbon fiber and ultra-high molecular-weight polyethylene.

  Fibers are strong polymers that do not change shape easily. They are made into thin, strong threads which can be woven together, Nylon is an example.

  •    Natural fiber

  Natural fibers are substances produced by plants and animals that can be turned into filament, thread or rope and further be woven, woolen, matted or bound.

  •    Synthetic fiber

  A man made textile fibers including usually those made from natural materials such as rayon and acetate as well as fully synthetic fibers (such as nylon or acrylic fibers).

  Checking up 6.3

  1. Polymers found in natural materials can be formed by the reaction between amino acids

  a. Deduce the formula of the product formed when two molecules of alanine, CH3CH(NH2)COOH react together and deduce the name of linkage present in the product.
  b. Give the name of the type of naturally occurring polymer containing this linkage.

  2. Write the structural formula of;

  a. Polypropylene (PP)
  b. Polyvinyl chloride (PVC)
  c. Polystyrene (PS)
  d.  Nylon 6, nylon-6,6
  e. Teflon (polytetrafluoroethylene

  

  Polymers or in general plastics have different properties depending on their nature. Among the properties, they can be thermosetting or thermosoftening whereas on the other side they can be biodegradable or non-biodegradable.

  6.4.1. Thermosoftening and thermosetting properties

  Thermosoftening (thermoplastics) and thermosetting (thermosets) are properties of polymers on how they soften on heating and harden on cooling.

  a. Properties of Thermosoftening

  polymers Thermosoftening polymers have weak intermolecular forces and they can be remolded into new shapes. They can be softened between 65 oC and 200oC and can be returned to their original state by heating.

  At higher temperatures thermoplastic becomes liquid and suitable for injection molding. After cooling, melt harden and keep a given shape. Disadvantage of thermoplastic is in their temperature resistance, meaning that they get quite fast soft and loose mechanical properties. Thermoplastic have linear and complex molecules.

  Thermosoftening is a property by which some polymers can be softened on heating. This allow them to cool and harden, and then can be resoftened many times.The

  

  For the thermosoftening polymers, the Van der Waal’s forces between the chains are often very strong and the polymers have relatively high melting and boiling points.  Due to their variable chain length, most polymers have different Van der Waal’s forces and these polymers tend to melt gradually over a range of temperatures rather than sharply at a fixed temperature. As the chains are not rigidly held in place by each other, polymers tend to be reasonably soft.

  The density and strength of addition polymers varies widely and they depend to a certain extent on the length of the hydrocarbon chain, but depend much more strongly on the nature and extent of the branching on the chain.

  Polymers which have very few branches are very compact and the chains can thus pack together very efficiently.examples P.V.C, Dacron, Polypropene. Etc

  b. Properties of Thermosetting polymers

  Thermosetting Polymers are some polymers which cannot be reshaped once heated as they are completely decomposed. Thermosetting polymers include phenol-formaldehyde, urea-aldehyde, silicones and allyls. Thermosetting Polymers have cross-links (covalent bonds between chains) that do not break on heating and they comprise three dimensional network structure. The greater the degree of crosslinking makes the polymer more rigid. These polymers are generally insoluble in solvents and have good heat resistance quality.

  Thermosetting polymers are generally stronger than thermoplastic polymers due to strong covalent linkage between polymer chains.
  When thermosetting plastics or thermosets are molded, covalent bonds are formed between the chains. They are more brittle in nature and their shape is permanent. Once they are softened, they cannot be returned to their original state by heating. Thermosetting polymers include phenol-formaldehyde, urea-aldehyde, silicones and allyls.The following is the structure of thermosetting polymers;

  

  Advantages and disadvantages of the above mentioned properties:

  . Advantages

  - Thermoplastics are convenient for manufacturers to use and they are not expensive. They are even recyclable.
  - Thermosets retain their strength and shape even when heated, they have high heat resistance and structural integrity. .   

  Disadvantages

  - Thermoplasts melt and some degrade in direct sunlight or under high U.V light levels. Many suffer from creep, a relaxation of the material under long term loading. They tend to fracture rather than deform under high stress.
  - Thermosets absorb moisture and toxicity easily.
  - They are not recyclable.

  6.4.2. Biodegradable and non-biodegradable properties

  These are the properties of polymers depending on how they react overtime as a result of biological activity especially to be broken by microorganisms.
  If they do not respond on the degradation or decomposition, they are said to be non-biodegradable polymers

  a. Biodegradable polymers

  Biodegradable polymers are the polymers that are fully decomposed into carbon dioxide, methane, water, biomass and inorganic compounds under aerobic or anaerobic conditions and the action of living organisms. Therefore the biodegradable properties are all characteristics of some polymers to be decomposed under aerobic or anaerobic conditions and action of living organisms.

  Biodegradation or biotic degradation is a specific property of certain plastic materials. Microorganisms (bacteria, fungi, algae) recognize polymers as a source of organic compounds (e.g. simple monosaccharides, amino acids, etc.) and energy that sustain them.

  Biodegradable plastics are plastics that will fully decompose to carbon dioxide, methane, water, biomass and inorganic compounds under aerobic or anaerobic conditions and the action of living organisms. Plastics are typically composed of artificial synthetic polymers.

  For biodegradation to happen there are two reactions that can allow it to proceed: biodegradation based on oxidation and the other based on hydrolysis.

  Those reactions can occur either simultaneously or successively. The decomposition of condensation polymers (example: polyesters and polyamides) take place through hydrolysis, while polymers with carbon atoms only in main chain (example: polyvinyl alcohol, lignin) decompose by oxidation which may be followed by hydrolysis of products of oxidation.

  The advantage of biodegradable plastics is that they decompose into natural substances and do not require separate collection, sorting, recycling or any other final waste solution (disposal at landfills or burning) as is the case with nonbiodegradable plastics.

  

  The biodegradability of condensation polymers may compromise their effectiveness, since physical and chemical durability is one of the reasons for their widespread use. A balance must be struck between practical durability and long-term biodegradability. The degradable polymers are applied in many areas (Table 6.3).

  

  b. Properties of non-biodegradable polymers

  Most of the major synthesized polymers are non-biodegradable. All kind of plastics and synthetic fibres are non-biodegradable.They are polymers which are resistant to environmental degradation thus accumulate in form of waste. These polymers cannot be changed to a harmless natural state by the action of bacteria, and may therefore damage the environment.

  Since the chains of non-biodegradable polymers are non-polar, addition polymers are insoluble in water. Their intermolecular forces are strong and the chains are often tangled, they are generally insoluble in non-polar solvents as well.
  In fact the long saturated hydrocarbon chains result in polyalkenes being very unreactive generally, as they cannot react with electrophiles, nucleophiles or undergo addition reactions.

  This results in their widespread use as inert materials; they are commonly used as insulators, packaging and in making containers.
  However their low reactivity means that they are not easily decomposed in nature and as a result have a very long lifetime. Such substances are said to be nonbiodegradable, and constitute an environmental hazard as they are very persistent in nature and thus difficult to dispose of.

  Checking up 6.4

  1. Polyethene is a non-biodegradable plastic.

  a. Explain the term bio-degradable
  b. Give one environmental benefit of using biodegradable plastics.
  c. Developing biodegradable plastics involves compromise. Suggest one factor that requires careful consideration and explain your choice. 

  2. Multiple choice questions (choose the letter corresponding to the right answer);

  A.The word ‘polymer’ meant for material made from ______________.

  a. Single entity

  b. Two entities

  c. Multiple entities

  d. Any entity

  B. One of characteristic properties of polymer material __________ .

  a. High temperature stability

  b. High mechanical strength

  c. High elongation

  d. Low hardness

  C. Polymers are ___________ in nature.

  e. Organic

  f. Inorganic

  g. Both (a) and (b) h. None

  D. These polymers cannot be recycled:

  a. Thermoplasts

  b. Thermosets

  c. Elastomers

  d. All polymers

  E. In general, strongest polymer group is __________ .

  a. Thermoplasts

  b. Thermosets

  c. Elastomers

  d. All polymers

  F. These polymers consist of coil-like polymer chains:

  a. Thermoplasts

  b. Thermosets

  c. Elastomers

  d. All polymers.

  G. Strong covalent bonds exists between polymer chains in __________ .

  a. Thermoplasts

  b. Thermosets

  c. Elastomers

  d. All polymers

  H. Following is the unique to polymeric materials:

  a. Elasticity

  b. Viscoelasticity

  c. Plasticity

  d. None.

  I. Elastic deformation in polymers is due to _____________ .

  a. Slight adjust of molecular chains

  b. Slippage of molecular chains

  c. Straightening of molecular chains d.

  d. Severe of Covalent bonds

  J. Kevlar is commercial name for ___________ .

  a. Glass fibers

  b. Carbon fibers

  c. Aramid fibers d.

  d) Cermets 

  

  Vulcanization is process used to convert natural rubber or related polymers to improve its resilience, elasticity and durability by heating them with Sulphur or other equivalent curatives or accelerators.  During this process, the rubber undergoes a multiple series of chemical change

  

  The vulcanization process was discovered in 1839 by Charles Goodyear in USA and Thomas Hancock in England. Both discovered the use of Sulfur and White Lead as a vulcanization system for Natural Rubber.

  Vulcanization of rubbers by sulfur alone is an extremely slow and inefficient process. The chemical reaction between sulfur and the Rubber Hydrocarbon occurs mainly at the C = C (double bonds) and each cross-link requires 40 to 55 sulphur atoms (in the absence of accelerator.

  

  Importance of vulcanization is that, it converts the raw rubber into more durable rubber with high tensile strength. It can withstand high temperature between 40 and 1000 centigrade.

  By heating rubber with sulphur, sulphur atoms are introduced between the chains and improve its elasticity .The properties of rubber improved by vulcanization include tensile strength; elasticity; hardness; tear strength; abrasion resistance; and resistance to solvents.

  Another important of vulcanization of rubber is the cross links that may be derived from carbon to carbon or through an oxygen atom or through sulphur atom, or through all the three, therefore, percentage elongation at break decreases. The vulcanization decreases the tendency of water absorption of rubber.

  Checking up 6.5

  1. a. The vulcanization is a process used to convert the natural rubber into more durable materials. Explain five important uses of vulcanization.   

  b. Name two substances added to natural rubber during vulcanization process as: i. Accelerators ii. Fillers iii. Anti-oxidants

  2. a.  What is vulcanized rubber?       

  c. Give two useful items of vulcanized rubber

  3. Describe briefly the process of vulcanization of rubber.

  

  Polymers are widely used materials in our daily life. To date, the importance of polymers is highlighted in their applications in different areas of sciences, technologies and industry from basic uses to biopolymers and therapeutic polymers.

  6.6.1. Uses of polymers

  Polymers found many uses in our daily life:

  Plastics are inexpensive, lightweight, strong, durable, corrosion-resistant materials, with high thermal and electrical insulation properties.
  In general due to the properties of polymers, they are used to make a vast array of products that bring medical and technological advances, energy savings and numerous other societal benefits.
  The following Table 6.4 shows the uses of the following commonly known polymers:

  Table 6.4: Uses of common used polymers

  

  

  

  6.6.2. Effects of polymers on environment

  Polymers or polymeric materials have eased the life of people but there are dangerous to the environment.

  However, concerns about usage and disposal are diverse and include accumulation of waste in landfills and in natural habitats, physical problems for wildlife resulting from ingestion or entanglement in plastic, the leaching of chemicals from plastic products and the potential for plastics to transfer chemicals to wildlife and humans. In the polymerization of compounds, some additive chemicals can be potentially toxic (for example lead and tributyl tin in polyvinyl chloride, PVC). Further substantial quantities of plastic have accumulated in the natural environment and in landfills.

  The rural areas are more prone to this type of contamination and the related effects, as a majority of the people from these areas as there is over use of plastics on a large scale. Discarded plastic contaminates a wide range of natural terrestrial, freshwater and marine habitats.

  When dumped in landfills, plastic materials interact with water and form hazardous chemicals which may be toxic to humans and other aquatic organisms. If these compounds seep down towards groundwater aquifers, they degrade the water quality, leading to groundwater pollution.

  Many of plastics waste lead to the formation of persistent organic pollutants (POPs), compounds which are very dangerous to the whole environment. These compounds persist in the environment and due to their property of bio-accumulation; they have high levels of toxicity in the food chain. Blockage due to plastic accumulation may form breeding grounds for mosquitoes and other harmful vector insects, which might cause numerous diseases in humans.

  Burning plastic leads to the contamination of the atmosphere, due to the release of other poisonous chemicals, leading to air pollution. Recycling them requires carefull attention as they lead to the development of skin and respiratory problems due to inhalation of toxic chemicals.

  When plastic is burned they release toxic chemicals that are deposited in soil and surface water and on plants.
  Non-biodegradable polymers or long term biodegradable materials, especially plastic bags when they in the soil, they do not allow rain water for penetration which causes soil erosion.

  NB: Rwanda has taken a tremendous decision of stopping the use of plastic bags in the country, which has got promising result for the environment and the users in general.

  Checking up 6.6

  1. Give one large scale use for polyester polymers and state the property of polyesters on which the use depends.

  2. Enumerate three environmental problems caused by the widespread use of plastics as polymers.

  

  Plastics have transformed everyday life; its usage is increasing gradually all over the world. It is evident that plastics bring many societal benefits and offer future technological and medical advances. However, concerns about usage and disposal are diverse and include accumulation of waste in landfills and in natural habitats, physical problems for wildlife resulting from ingestion or entanglement in plastic, the leaching of chemicals from plastic products and the potential for plastics to transfer chemicals to wildlife and humans.

  The management of waste materials including polymers involves improvement of effects caused by these materials.  A major part of plastic produced is used to make some disposable items of packaging or other short-lived products that are discarded within a period of manufacture. Due to the durability of the polymers, we have seen that substantial quantities of discarded end-of-life plastics are accumulating as debris in landfills and in natural habitats worldwide. A number of waste prevention techniques such as reuse, recycling and disposal of plastics are discussed below.

  a. Reuse

  This is a process of using the polymers more than once. This encompasses the entire spectrum of used goods. Spanning from collectables, antiques and memorabilia to general used goods retail and wholesale. Dealing in secondhand items typically involves the salvage of used items and may dismantle into components. Beyond salvage and to enhance reuse the industry includes repair and refurbish, remanufacturing.

  There is considerable scope for re-use of polymer materials for the transport of goods, and for potential re-use or re-manufacture from some plastic components in high-value consumer goods such as vehicles and electronic equipment.

  b. Recycle

  Recycling is one of the most important activities that can be applied to reduce the impacts of the plastics used in industry. Recycling is a waste-management strategy; it is a technique that can be used to reduce the environmental impact and resource depletion.

  Recycling provides opportunities to reduce oil usage, carbon dioxide emissions and the quantities of waste requiring disposal. Today’s recycling industry has evolved largely into a service industry involved in the collection, sorting, processing and transportation of waste streams and by-products.

  Different polymer materials need to be collected, separated and cleaned. Further, they are melted down before being changed into a new material. Some plastics cannot be melted – they burn or harden instead of melting. It is even more difficult to recycle these plastics as they can only be used in the same shape as they were originally cast.

  The biodegradable materials can be recycled, broken down into their original components and reused, but they still need to be collected, separated and cleaned.

  Plastic materials can be recycled in a variety of ways and the ease of recycling varies among polymer type, package design and product type.

  c.  Disposal

  Old polymers disposal is the action of getting rid of old polymers. The disposal of non-biodegradable polymers is a significant problem.

  There are three options:

  i. Burying in landfill sites

  This is widespread in all developed countries but is a completely unsustainable practice, as each landfill site will eventually fill up. Landfill sites are also unsightly and unhygienic.

  iv. Burning

  This is a technique to dispose some materials. However, burning polymers releases greenhouses gases such as carbon dioxide and can also release toxic gases, depending on exactly type of polymer being burned.

  Checking up 6.7

  List the advantages and disadvantages for each of the method to deal with the old polymers

  END UNIT ASSESSMENT

  1. Explain the terms crosslinking and thermosetting with reference to condensation polymers.  

  For what purposes are thermosetting polymers suitable?  

  2.  a. How the chemical inertness of poly(ethene) arise?      

       b. How does it increase the usefulness of the material?
        c.  How does it affect the disposal of waste polyethene?

  3. a. What type of functional group joins the repeating units in nylon?

           b.  In what way does the structure of nylon resemble that of a polypeptide?
            c.  What type of interaction takes place between polymer molecules which contain the functional group present in polypeptide?

  4. Identify 3 examples of synthetic polymers

  5. What is the benefit of cross-linking polymer chains?

  6. a) Give one example for each of the following type of polymer. Write the structural formula of the polymer and monomer unit. Give at least one use of the polymer named

  i. Natural addition polymer

  ii. Synthetic addition polymer

  iii. Natural condensation polymer

  iv. Synthetic condensation polymer        

  b. State the role of each of the following in the manufacture of plastics

  i.  Fillers

  ii.  Plasticizers

  c. i.  what is meant by vulcanization?

   ii. Discuss the effect of vulcanization on rubber molecules and state how it affects the physical properties of rubber.
  iii. Name the monomer units in natural rubber
  iv. Name one commercial synthetic rubber. Write equation to show how it is formed and give one use of it

  
  

• UNIT 7: SOLVENT EXTRACTION AND COLLIGATIVE PROPERTIES

  Key unit competency
  

  
  

  The learner should be able to apply partition and Raoult’s law to separate mixtures, determine the molecular and formula masses of compounds using colligative properties.

  
  

  Learning objectives
  

  At the end of this unit , students will be able to:

  • Define partition coefficient and solvent extraction;

  • State the Raoult’s law;

  • State and explain the advantages of carrying out distillation processes under reduced pressures;#

  • Discuss the chemical principles upon which simple distillation and steam dis-tillation are based;

  • State examples of the applications of the distillation methods used in various industries;

  • Describe the effect of the solute on vapor pressure, boiling and freezing points of the solvent;

  • Explain colligative processes;

  • Carry out separation experiments based on the solute partitioning between two immiscible solvents;

  • Calculate the amount of the solute extracted from the solvent;

  • Interpret boiling point and vapor pressure composition curves of both ideal and non-ideal mixture;

  • Calculate the molecular mass of substances using steam distillation;

  • Calculate molecular mass of polymers using colligative properties and steam distillation;

  • Interpret the boiling point composition curves of azeotropic mixtures;

  • Apply Raoult’s Lawto calculate vapor pressure of given solutions and mole fractions;

  • Carry out experiments to explain colligative properties;

  • Apply colligative properties to determine the molecular mass of the solute in a solution.
  

  Introductory activity

  1. Describe the phase diagram for water.

  2. Have you ever heard the term “solvent extraction”? If yes can you tell what it is about and give an example of a solvent  extraction?

  3. Describe the process of producing banana juice? Is there any application of solvent extraction?

  4. Take a tea bag and put it in water (cold or hot); what do you observe? Explain your observation

  5. You are provided with the following materials: separating funnel, 1,1,1-trichloroethane, iodine, water, retort stand and its accessories:

  Procedure:

  Put water into the separating funnel and add 1,1,1-trichloroethane in the water and shake well.

  Add iodine in the mixture and shake well, after standing, observe well.

  What happens to the iodine? How do you compare the coloration of the two phases? What explanation can you give?

  When traditional healers crush leaves of a healing plant, mix with water, filter and give you the filtrated mixture as medicine, they have applied solvent extraction technique, using water as solvent, to extract the medicinal ingredients.

  When you prepare your cup of tea, putting a tea bag in the hot water, the color of water change to brown black; you have solvent extracted a certain number of substances present in the tea leaves by hot water, and left other in the tea residue.

  Solvent extraction is the separation of a particular substance from a mixture by dissolving that substance in a solvent that will dissolve it, but which will not dissolve any other substance in the mixture.

  Let us start with two solvent systems of water and an organic liquid, two immiscible solvents.

  When water and an organic liquid are shaken together, and then allowed to stand, the liquids separate into a two phase system with the more dense liquid on the bottom.

  If a solute is added to a system of two liquid layers, made up of two immiscible components, then the solute will distribute itself between the two layers so that the ratio of the concentration in one solvent to the concentration in the second solvent remains constant at constant temperature. A quantitative measure of how a component will distribute between the two phases is called the distribution or partition coefficient.

  7.1. Definition of partition coefficient and solvent extraction

  The ability of a solute (inorganic or organic) to distribute itself between an aqueous solution and an immiscible organic solvent has long been applied to separation and purification of solutes either by extraction into the organic phase, leaving undesirable substances in the aqueous phase; or by extraction of the undesirable substances into the organic phase, leaving the desirable solute in the aqueous phase. Species that prefer the organic phase (e.g., most organic compounds) are said to be lipophilic (“liking fat”) or hydrophobic (“disliking water”), while the species that prefer water are said to be hydrophilic (“ liking water”) or lipophobic (“disliking fat”).

  7.1.1. Partition coefficient

  Partition means”divided into two parts with a boundary or an interface”.When a solute is added to two immiscible solvents in contact with each other at constant temperature, the solute gets distributed or partitioned between the two solvents with different equilibrium concentrations.

  For example, when iodine is added to water and carbon tetrachloride, it distributes in such a way that the equilibrium ratio of the concentrations of iodine in the two solvents is constant at any given temperature. If and are the concentrations of iodine in water and carbon tetrachloride respectively, then

                                

  The constant is called the distribution or partition coefficient of the solute between the two solvents at a given temperature. The value of depends on the nature of the solute and the solvent pair. The equation above is the mathematical form of the Nernst Distribution Law (or Nernst’s Partition law) or simply Distribution law or Partition law.

  This law states that at constant temperature, when different quantities of a solute are allowed to distribute between two immiscible solvents in contact with each other then at equilibrium the ratio of the concentration of the solute in two
  

  layers is constant. The solutions where this law applies are called “ideal solutions”. However, if the amount of the solute added is sufficiently small, then the distribution coefficient is relatively independent of the concentration.

  7.1.2. Solvent extraction

  The term solvent extraction refers to the distribution of a solute between two immiscible liquid phases in contact with each other, i.e., a two-phase distribution of solute.

  Solvent extraction has become a very powerful method of separation for various reasons. One amongst them is, it is very simple, rapid, selective and sensitive. This method does no need any kind of sophisticated instrument apart from a separating funnel.

  It is the partial removal of a solute from one liquid (usually water) into another immiscible liquid (e.g. ether). This is a method of separating one component from a mixture. It is based on finding a solvent which dissolves the desired component much better than it does for any of the others. Solvent extraction is also called liquid-liquid extraction and partitioning.

  For the solvent extraction to succeed, the following considerations are recommended:

  • the solvent to be used must be a good solvent for the solute and chemically unreactive with the solute

  • the solvent must be immiscible with water or any other solvent involved

  • the solvent must be volatile, to facilitate the separation with the solute at the end of the operation; example, the most used solvent is ether that fulfills those requirements,

  Activity 7.1:

  In an experiment, 100 of 0.10mold aqueous propanoic acid was shaken with 50 of organic solvent which is immiscible with water, until equilibrium was reached. 10.0 of aqueous layer required 12.0 of 0.02mold NaOH for reaction in a titration.

  Questions:

     a. Calculate the concentration of propanoic acid in aqueous layer and in organic layer.

     b. Calculate the equilibrium ratio between the concentration of propanoic acid in organic layer and in aqueous layer;

    c. What is the name of that ratio?

  Multiple extractions

  The process of extraction when carried out with the total amount of the given solvent in a single operation is referred to as simple extraction. But to recover the maximum amount of the substance from a solution, the extraction is made in two or more successive operations using small portions of the solvent provided. This is called multiple extractions. In such a process the aqueous solution is first extracted with a portion of the solvent in a separatory funnel. The aqueous layer from which some substance has been removed is then transferred to another funnel. This is shaken with a second portion of the solvent. Similarly, the aqueous layer from the second extraction is treated with a third portion of solvent, and so on.

                       

  Example:

  The partition coefficient of a substance X between methylbenzene and water is 8.0, X being more soluble in methylbenzene than water.

  The mass of X extracted from an aqueous solution containing 6.0 g of X may be calculated as follows:

  If shaken with one portion of 100cm3 of methylbenzene:

  Let x be the mass of X extracted.

  Mass of X remaining in aqueous solution= (6.0-x)g

   

  If shaken with two successive portions of 50 cm3 of methylbenzene:

  

  Mass of X extracted in (i) is less than that in (ii). This shows that it is more efficient to use a solvent in several smaller portions than all at once when extracting a solute from another solvent.

  Limitation of Nernst Distribution Law

  The law is valid only when the molecular state of the solute is the same in both the solvents. If the solute undergoes dissociation or association in any one of the solvents, then in such cases the distribution law no longer holds.

  

  7.2. Raoult’s law and colligative properties

  7.2.1 Raoult’s law and ideal solutions

  
  

  Activity 7.2
  

  (a) Observe well the figure below and answer the following questions:

  1. What is vapour pressure?

  2. What effect the solutes have on the vapor pressure?

    
  

  Let’s consider the figures above; the figure at the left side shows that some molecules exist as gaseous molecules above the liquid phase. Those molecules exert a pressure on the surface of the liquid. That pressure is called “vapour pressure” of the solvent or liquid. Equilibrium will be reached when there is balance between the number of molecules leaving the liquid phase to enter the vapour phase and the number of molecules leaving the vapour phase to enter the liquid phase. Vapour pressure increases with increasing temperature. When the vapour pressure is equal to atmospheric pressure, the liquid starts to boil.

  
  

  On the right side of the figure, we see lesser gaseous molecules in the container with nonvolatile solute, and still lesser in the container with a volatile solute. In both cases we see that introduction of a solute into a solvent causes a decrease in its vapour pressure.

  
  

  Another observation is that in the container with two volatile liquids, the total vapour pressure is the sum of the vapour pressures of the two solvents; in this case the vapour pressure of each solvent is called “partial vapour pressure”.

  
  

  Ideal solution

  
  

  
  

  The French chemist, François -Marie Raoult, in 1886 gave the quantitative relationship between vapour pressures and mole fractions of components. The relationship is known as the Raoult’s law which states that” the partial vapour pressure of a volatile component of ideal solution is product of its mole fraction and vapour pressure of the pure component at constant temperature.
  

  
  

  
  

  
  

  
  

  Raoult’s law is applicable only if the liquids are miscible

  
  

  
  

  
  

  A solution which obeys Raoult’s law over the entire range of concentration at all temperature is known as an ideal solution.

  
  

  Raoult’s law describes the behavior of ideal solutions of completely miscible and volatile liquids. Ideal solution is defined as the one in which each component obeys Raoult’s law over the entire concentration range. Besides obeying Raoult’s law, an ideal

  solution shows two more properties:

         a. When such solution is prepared, no heat is evolved or absorbed, i.e.

         

            b. The volume of such a solution is the sum of the volumes of its components,

           

  In fact, liquid-pairs rarely form ideal solutions. Benzene-toluene, ethene chloride-ethene bromide and carbon tetrachloride-silicon tetrachloride are few examples of liquid pairs which form very nearly ideal solutions.

  A binary solution of components A and B will behave ideally when intermolecular force between a molecule of A and a molecule of B (A-B interactions) is same as intermolecular force between two molecules of A (A-A interactions) or the force acting between two molecules of B (B-B interactions). This means that one component likes itself as much as it likes the other component. This is due to similarity in their structure that their solutions behave ideally.

  At any fixed temperature, the vapour phase is always richer in the more volatile component compared to the solution phase. In other words, mole fraction of the more volatile component is always greater in the vapour phase than in the solution phase. The composition of vapour phase in equilibrium with the solution is determined by the partial pressure of components.

  

  i. Calculate the vapour pressure of the solution prepared by mixing 25.5 g of chloroform and 40g of dichloromethane at 25°C.

  ii. Calculate the mole fraction of each component in vapour phase.

  Answer:

    i. The number of moles in the solution:

  

  Total number of moles in the solution

   

  Mole fractions in the solution:

             

  The vapour pressure of the solution is 347.9 mm Hg.

                     ii. The vapour pressure of each component in vapour phase:

                  

  Consider and the mole fractions of components A and B respectively in the vapour phase

  

  The mole fraction in vapour phase is given by the following equation:

             

  Since dichloromethane is more volatile than chloroform and the vapour phase is also richer in dichloromethane, it may be concluded that at equilibrium, vapour pressure will be always rich in the component which is more volatile.

  Another type of solutions is solids in liquid solution, in which we take the solid as the solute and the liquid as the solvent. Generally, the solute is nonvolatile in nature and the vapour pressure is less than the pure vapour pressure of the solution.

  Examples:

                 • Solution of sugar, salt or glucose and water

                 • Solution of iodine and sulfur in carbon disulphide

  Vapour pressure produced by the solution of nonvolatile solute and solvent results in the vapour pressure of the solution solely from the solvent. This vapour pressure is lower than the vapour pressure of the pure solvent at the same temperature.

  The decrease in vapour pressure is due to:

  The surface area of the solution is occupied by both nonvolatile solute and the pure solvent particles which results in the reduction of the surface for the solvent particles. As evaporation is a surface phenomenon, the more the surface area is, the greater the evaporation and hence more the vapour pressure. In a pure solvent, there is more surface area available for the particles to vaporize, therefore have more vapour pressure. On the other hand, when we add a nonvolatile solute, the solvent particles get less surface area to escape and hence exert low vapour pressure.

  The number of particles escaping the surface is much greater in pure solvent than that of solution containing nonvolatile solute.

  Calculation of the vapour pressure of Solutions of Solids in Liquids

  Let’s A be the solvent and B a solute; according to Raoult’s law, the partial pressure of individual component is directly proportional to its mole fraction.

  

  Example:

  An aqueous solution of glucose is made by dissolving 10 g of glucose, (C6H12O6), in 90 g of water at 30oC. If the vapour pressure of pure water at 30oC is 32.8 mmHg, what would be the vapour pressure of the solution?

  

  Data:

  Mass of glucose: 10g

  Molar mass of glucose= 180g/mol

  Mass of water: 90g

  Molar mass of water: 18g/mol

  Vapour pressure of pure water: 32.8 mmHg

  Calculation of moles of water and glucose211 Mass of water:

     

  Note: The vapour pressure of nonvolatile component is negligeable.

  
  

  Non-ideal Solutions
  

  Activity 7.2 (b):
  

  You are given the following solutions:

  Chloroform-acetone, water-propanol, n-hexane-n-heptane, ethanol-chloroform, ethanol-cyclohexane, ethanol-acetone, phenol-aniline, benzene-toluene.

  Choose the solutions which obey Raoult’s law and explain why others do not obey it.

  Most of liquid-liquid solutions do not obey Raoult’s law over the entire range of concentrations. Such solutions are called non-ideal solutions. Besides Raoult’s law, non- ideal solutions do not obey the other two conditions: their formation is accompanied by changes of heat and volume. When non-ideal solutions are prepared, some energy is either released or absorbed and the volume of the solution is different from the sum of volumes of the components. The non-ideal behavior of most of liquid-liquid solutions is due to the intermolecular forces between molecules. Vapour pressure of non-ideal solutions is either higher or lower than what predicted by Raoult’s law. Based upon this, there are two types of non-ideal solutions namely, the non-ideal solutions with positive deviations and those with negative deviations.

  Most of the real mixtures are non- ideal.

  Non-ideal solutions with negative deviations

  In some liquid solutions, the observed vapour pressure of the solution is lower than the ideal value as expected from Raoult’s law. Such deviation is called negative deviation.

  
  

  
  

  Examples:

  i. Acetic acid-pyridine

  ii. Phenol-aniline

  iii. chloroform-acetone

  iv. water-nitric acid

  v. water- sulphuric acid

  vi. water-hydrochloric acid

  In a mixture of phenol and aniline, the intermolecular hydrogen bond between phenolic proton and lone pair on nitrogen atom of aniline is stronger than the respective intermolecular hydrogen bond between similar molecules; the molecules in the mixture are more strongly bonded than those in the pure liquids. Similarly, propanone-trichloromethane mixture forms a solution with negative deviation from Raoult’s law. When the liquids are mixed, hydrogen bonds are formed between the propanone and trichloromethane molecules (figure 7.7). There is no hydrogen bonding in pure liquid propanone or in pure liquid trichloromethane.

                                                 

  Non-ideal solutions with positive deviations

  The solutions whose vapour pressure is higher than that expected from Raoult’s law are said to show positive deviations.

  

  Examples of non-ideal solutions showing positive deviation from Raoult’s law:

  i. Benzene-cyclohexane

  ii. n-butane-n-heptane

  iii. carbon disulphide-acetone

  iv. carbon tetrachloride-benzene

  v. ethanol-acetone

  vi. ethanol-hexane

  vii. ethanol-chloroform

  viii. water-ethanol

  ix. water-propanol

  x. ethanol-cyclohexane

  Example:

  When hexane and ethanol are mixed, the ethanol molecules are separated. The hydrogen bonds are broken and much weaker Van der Waals forces hold the ethanol and hexane molecules together in the mixture. The ease of escape into the vapour is caused by the hexane molecules, which cannot form hydrogen bonds, coming between the ethanol molecules and disrupting the hydrogen bonding, making it easier for ethanol molecules to escape into the vapour.

                                 

  Azeotropic mixtures

  All non-ideal solutions with large deviations show either maxima or minima in their vapour pressure compositions curves. The solutions with positive deviations show maxima and those with negative deviations show minima.

  The solution corresponding to maximum or minimum has a unique property that the vapour formed on their evaporation has the same composition as the the liquid phase of the solution. Such solution boils at constant temperature and can not be separated by fractional distillation. Such a solution is called Azeotropic mixtures or azeotropes. The azeotrope is defined as the mixture of liquids which boils at constant temperature like a pure liquid and possesses the same composition of components in the solution as well as in vapour phase.

  The term azeotrope stems from a Greek word meaning boiling without changing. Each azeotrope has a characteristic boiling point.

  Azeotropes formed by the solutions showing positive deviations from Raoult’s law are called minimum boiling azeotropes or positive azeotropes; such an azeotrope has the maximum vapour pressure and, therefore the minimum boiling point among all the mixtures formed by these two liquids. These azeotropes always have boiling point lower than either of the components.

  Examples: Ethanol-water mixture

  The fractional distillation is able to concentrate the alcohol to the best up to 95% by mass of alcohol. Once this composition is reached, the solution and vapour have the same composition, and no additional fractionation occurs. Other methods of separation have to be used for preparation of 100% ethanol.

                                      

  Azeotropes formed by the solutions showing negative deviations from Raoult’s law are called the maximum boiling azeotropes or negative azeotropes. These azeotropes have boiling points higher than either of the components.

  Azeotropes consisting of two constituents are called binary azeotropes. Those consisting of three constituents are called ternary azeotropes. Azeotropes of more than three constituents are also known.

    

  Examples of azeotropes with minimum boiling points and maximum boiling points:

  

  Checking up 7.2 (b)

  1. On the basis of information given below answer the following questions: Information:

  a. In bromoethane and chloroethane mixture intermolecular interactions of A-A and B-B type are nearly the same as A-B type interactions.

  b. In ethanol and acetone mixture A-A or B-B type intermolecular interactions are stronger than A-B type interactions.

  c. In chloroform and acetone mixture A-A or B-B type intermolecular interactions are weaker than A-B type interactions.

  Questions:

  1. a. Which solution will obey the Raoult’s law?

      b. Which solution will deviate from Raoult’s Law and which kind of deviation?

  2. On the basis of information given below, mark the correct option. Information: on adding acetone to methanol some of the hydrogen bonds between methanol molecules break.

  a. At specific composition methanol-acetone mixture forms minimum boiling azeotrope and will show positive deviation from Raoult’s law.

  b. At specific composition methanol-acetone mixture will form minimum boiling azeotrope and will show negative deviation from Raoult’s law.

  c. At specific composition methanol-acetone mixture will form maximum boiling azeotrope and will show negative deviation from Raoult’s law.

  Application of Raoult’s Law

  Activity 7.2(c):

  Use search internet or other resources and then discuss about Raoult’s law and its applications then write a report about your findings.

  Questions:

  1. Have you carried out a distillation? If yes, what the principle that underlies that separation technique?

  2. Have you heard about the production of a local and illegal alcohol called Kanyanga? If yes, do you have an idea of how it is made?

  3. Distinguish simple distillation from fractional distillation.

  Fractional distillation

  Distillation is a separation technique that separates liquid components of a solution according their boiling points. A simple distillation uses a simple condenser; it works well for most routine separation and purification procedures for organic liquids with large differences in their boiling points. When the boiling point differences of the components to be separated are not large, fractional distillation must be used. When an ideal solution of two liquids, such as benzene (boiling point 80°C) and toluene (boiling point 110°C), is distilled by simple distillation, the first vapor produced will be enriched in the lower-boiling component (benzene).

  However, when that initial vapour is condensed and analyzed, the distillate will not be pure benzene. The boiling point difference of benzene and toluene (30°C) is too small to achieve a complete separation by simple distillation. So, fractional distillation is an application of Raoult’s law illustrated in figure 7.10.

                     

  

       

  Industrial application of fractional distillation: crude oil refinery.

  Crude oil is a complex mixture of hydrocarbons of different chain lengths, from the short one such as , methane, to the very long ones with more than 20 carbon atoms (C-20). Due to their boiling points which are not very different, there are separated, through fractional distillation, into different fractions of hydrocarbons called: kerosene, naphtha, etc... (Fig.7.12) 

        

  In short: Condensation takes place in discs which increases surface area found in fractionating column. Further heating of the mixture produces more vapour, which in turns will reheat the condensed liquid in discs. During the distillation, the temperatures decrease progressively from the bottom to top of column.

  Steam distillation

  Activity 7.2 (d)

  When people suffer from flue, cold or any other respiratory infections, they collect Cyprus, eucalyptus, and other leaves. They boil them in water and expose themselves to the vapour released. Explain how they are cured.

  Distillation using water vapor is called steam distillation. Steam distillation can be used when the material to be distilled has a high boiling point and presents the risk that decomposition might occur if direct distillation is employed. The liquid is added to the still, and steam is passed through it. The solubility of the steam in the liquid must be very low. Most essential oils have been obtained by steam distillation, or, in the more general sense, by hydrodistillation. Essential, volatile, or ethereal oils are mixtures composed of a variety of volatile, liquid, or solid compounds that vary widely in concentration and boiling points. They are present in the interstices of vegetable tissues and can be extracted by hydrodistillation.

  Considering the manner in which the contact between water and the original matrix is promoted, a terminology that distinguishes three types of hydrodistillation has been developed:

       • Water distillation,

       • Steam distillation, and

       • Direct steam distillation.

  When the first method is employed, the material to be distilled comes in direct contact with boiling water.

  In the second method, the material is supported on a perforated grid or screen held some distance above the bottom of the still. In this case, low-pressure, saturated, wet steam rises through the material. The typical features of this method are that the steam is always fully saturated, wet, and never superheated, and the material is only in contact with steam, and not with boiling water. The last type of hydrodistillation, direct steam distillation, resembles the preceding type, except that no water is kept in the bottom of the still, but live steam, saturated or superheated, is passed through the sample, and the process is frequently maintained at higher than atmospheric pressures.

  Steam distillation is the most commonly used method for collecting essential oils. Use of this method is prevalent not only because it yields exceptionally pure and clean products, but because it allows for collection of temperature-sensitive aromatic compounds. Unlike simple distillation, steam distillation involves a pressurized system. When the system is pressurized, essential oils can be distilled at temperatures well below their normal boiling point; thus protecting the integrity of their delicate and complex chemical nature.

  When using steam distillation, it is vital to pay careful attention to the heat source. The temperature of the system must remain within a strict range; too low and the essential oils will not be distilled, too high and there is risk of damaging the essential oils or collecting unwanted, non-aromatic compounds. Temperatures required for optimal steam distillation typically fall between 60° C and 100° C. One benefit of steam distillation is that the temperature can be continuously adjusted and precisely controlled to ensure that the system always remains within the optimal temperature range. Similarly, pressure must also be rigidly controlled. To do so, steam distillation requires the use of a “closed system.” This means that the system is pressurized to a level above that of atmospheric pressure.

  Steam distillation is conducted in a distillation still that uses steam water to remove the essential oil from the plant material. Heat is applied to the water, which produces steam.

  The steam rises and moves through a chamber holding the plant material. As the steam forces its way through the plant material, it ruptures small glands that hold the essential oil. Since essential oils contain only aromatic compounds that are readily volatile, the essential oil is easily carried by the steam into the condensing tube. Once there, the liquid is condensed and accumulates in the collecting still. After removal from the still, a mixture of essential oil and water will be formed. Because essential oils are not water soluble, the mixture will naturally separate into two layers (figure 7.14). The watery layer is called hydrosol and is often sold as floral water.

   

  
  

  Applications

  Steam distillation is widely used in the manufacturing of essential oils, for instance perfumes. This method uses a plant material that contains of essential oils. Mainly orange oil is extracted on a large scale in industries using this method. Application of steam distillation can be found in the production of consumer food products and petroleum industries. They are used in separation of fatty acids from mixtures.

  Checking up 7.2 (d)

  Explain how the essential oils can be obtained from plants?

  7.2.2. Colligative properties

  Activity 7.3:1. The photos you see here have been taken in Europe or North America during winter.

        

  On the left, someone is pouring a liquid, an antifreeze in the radiator of his/her car.

  On the right, workers are spreading salt from a truck on a road covered by ice.

  Question: Do you have an idea why those persons are doing that?

  2. Take a container and divide it into two compartments with a thin membrane containing microscopic pores large enough to allow water molecules but not solute particles to pass through.Then add a concentrated salt solution to one compartment and a more dilute salt solution to the other. Initially, the two solutions levels start out the same.

  a. What is the name of the membrane?

  b. Observe well and after a while, write your observations.

  c. Name the process observed and explain it and explain how it can be stopped.

  d. Draw with labels the phenomenon.

  3. Have you ever heard these words “colligative properties”? If yes explain what they mean and give example to illustrate.

  The state of water depends on both pressure and temperature. At sea level, pure liquid water freezes to form solid ice at 0°C and boils to form vapor at 100°C.

  However, the addition of particles to water can affect its boiling and freezing points. Particles interfere with the ability of the water molecules to vaporize or freeze. The boiling temperature of water varies depending on the solutes it contains.

  Liquid solutions have physical properties significantly different from those of the pure solvent, a fact that has great practical importance. The properties of dilute solutions containing nonvolatile solute, which depend upon relative number of solute and solvent particles but no depending upon their nature, are called colligative properties.

  Definition of colligative properties of solutions

  Colligative properties are properties that depend only on the number of particles present in solution and not in any way on the nature of the solute particles. These properties are bound together by a common origin; they all depend on the number of solute particles present, regardless of whether they are atoms, ions, or molecules.

  The colligative properties are Vapour pressure lowering, Freezing point depression, Boiling point elevation and Osmotic pressure.

  Lowering of vapour pressure

  Molecules can escape from the surface of a liquid into the gas phase by evaporation. In closed container, the pressure exerted by the vapour in the space above the liquid, if the dynamic equilibrium between liquid and vapour is reached, is called the vapour pressure of the substance. The vapour pressure of a liquid is the pressure exerted by its vapour when the liquid and vapour phase are in dynamic equilibrium. A substance with no quantifiable vapour pressure is called nonvolatile and the one that exerts vapour pressure is called volatile.

  According to Raoult’s law, the vapour pressure exerted by a component of a solution at a given temperature is equal to the mole fraction of that component in the solution multiplied by the vapour pressure of that component in pure state.

  In a solution of nonvolatile solute in a volatile solvent the vapour pressure will be exerted by the solvent molecules only because the solute is non volatile i.e., does not go into vapour state. Therefore, the vapour pressure of its solution is always less than that of the pure solvent.

  In the pure solvent, the molecules occupy large surface and they escape easily; in the solution of nonvolatile solute and solvent, the solute molecules reduce the surface of the solvent molecules which lead to the reduction of molecules of solvent escaping. According to Raoult’s law:

        

  Equation (1) can therefore be rewritten as:

       

  The decrease in vapour pressure,P, is directly proportional to the solute concentration (measured in mole fraction).

  In the case of solution of two or more volatile components, the total vapour pressure over the solution is the sum of the partial vapour pressures of each volatile component. Consider the ideal solution containing two volatile liquid, 1 and 2.

  By Raoult’s law, the partial pressures of component 1 and 2 vapours above the solution are:

      

  Example:

  Calculate the vapour pressure of a solution made by dissolving 218 g of glucose (molar mass=180.2g/mol) in 460 mL of water at 30°C. What is the vapour pressure lowering? The vapour pressure of pure water at 30°C is 31.82 mm Hg. Assume the density of the solution is 1.00 g/mL.

  Answer:

  The vapour pressure of the solution:

  

       

  Boiling point elevation

  Boiling point of a liquid is the temperature at which its vapour pressure becomes equal to the atmospheric pressure. As the vapour pressure of a solution containing a nonvolatile solute is lower than that of the pure solvent, its boiling point is higher than that of the pure solvent, because to reach atmospheric pressure at which the solution should boil we need to raise the temperature. The solution has lower vapour pressure than the solvent at the same temperature, figure 7.15. The increase in boiling point is known as elevation in boiling point,

      

     

  Thus, the molal freezing point depression constant is equal to the depression in the freezing point produced when one mole of solute is dissolved in 1 kg of the solvent.

                                          
  

  Example:

  Ethylene glycol is common automobile antifreeze. It is water soluble and fairly nonvolatile (b.p 197°C). Calculate the freezing point and boiling point elevation of a solution containing 651 g of this substance in 2505 g of water. The molal mass of ethylene glycol (EG) is

  Answer:

  Calculate the molality of ethylene glycol (m):

   

  Osmotic pressure

  Osmosisis the phenomenon of spontaneous flow of the solvent molecules through a semipermeable membrane from pure solvent to solution or from a dilute solution to concentrated solution. It was first observed by Abbe Nollet. Some natural semipermeable membranes are animal bladder, cell membrane.

  You must have observed that if resins are soaked in water for some time, they swell. This is due to the flow of water into the resins through its skin which acts as a semipermeable membrane (permeable only to the solvent molecules). This phenomenon is also observed when two solutions of different concentrations in the same solvent are separated by a semipermeable membrane. In this case the solvent flows from a solution of lower concentration to a solution of higher concentration. The process continues till the concentrations of the solutions on both sides of the membrane become equal. The spontaneous flow of the solvent from a solution of lower concentration (or pure solvent) to a solution of higher concentration when the two are separated by a semipermeable membrane is known as osmosis. The pressure required to prevent osmosis by pure solvent is called osmotic pressure of solution, π.

  Thus, osmotic pressure may be defined as the excess pressure that must be applied to the solution side to just prevent the passage of pure solvent into it when the two are separated by a perfect semipermeable membrane.

  This is illustrated in figure 7.16. For dilute solutions, it has been found that osmotic pressure is proportional to the molarity, C of the solution at a given temperature, T.

  
  

  It two solutions of the same osmotic pressure are separated by a semipermeable membrane, there is no osmosis. The two solutions are called isotonic solution. Hypotonic solution is the less concentrated solution and hypertonic solution the more concentrated one.

  The osmotic pressure is a colligative property. It depends on the number of particles of solute present in the solution and not on their nature.

  

  Applications of colligative properties

  As seen in the introductory activity (7.4), one of the colligative properties, the freezing point depression is applied to de-ice roads in countries which experience very cold winters.

  Colligative properties of non electrolyte solutions provide a means of determination of molar masses and molecular formulae of dissolved solutes which are nonvolatile. Osmotic pressure and changes in freezing point, boiling point, and vapor pressure are directly proportional to the concentration of solute present. Consequently, we can use a measurement of one of these properties to determine the molar mass of the solute from the measurements. In practice, only freezing point depression and osmotic pressure are used because they show the most pronounced changes.

         • Determination of molar mass from relative lowering of vapour pressure

  Relative lowering of vapour pressure is the ratio of lowering in vapour pressure to vapour pressure of pure solvent. The relative lowering in vapour pressure of solution containing a nonvolatile solute is equal to the mole fraction of solute in the solution.

        

  

  The above expression is used to determine the molecular mass of the solute B, provided the relative lowering of vapour pressure of a solution of known concentration and molecular mass of solvent are known. Where, mass of Solute and solvent respectively. molecular weight of solute and solvent respectively. However, the determination of molecular mass by this method is often difficult because the accurate determination of lowering of vapour pressure is difficult. Ostwald and Walker method is used to determine the relative lowering of vapour pressure.

  Examples

  1. The relative lowering of vapour pressure produced by dissolving 7.2 g of a substance in 100g of water is 0.00715. What is the molecular mass of the substance?

  Substituting the values we get

  

  Molecular mass of the substance is 181.26 amu

  2. The vapour pressure of 5% aqueous solution of a nonvolatile organic substance at 373 K is 745 mm Hg. Calculate the molar mass of the substance. Solution

  

                         • Determination of molar mass from elevation of boiling point

  The molar mass of a non electrolyte can be calculated from the elevation of boiling point:

  where, is the mass of solute ( in grams), and is the molar mass of the solute (in g/mol) and, W1 is the mass of the solvent in Kg units.

  This gives

  

  The constant is called the molal elevation constant or ebullioscopic constant for the solvent. may be defined as the elevation in boiling point when one mole of a solute is dissolved in one kilogram of the solvent. BKis expressed in degree per molality.

  Examples

  1. A solution containing 4.2 grams of an organic compound in 50 grams of acetone shows an elevation of boiling point by 1.8 K. Determine the molar mass of the organic compound. of acetone=1.71 K kg /mol.Solution

  

  

  Therefore, molar mass of the solute is 58g/mol.

                 • Determination of molar mass from depression in freezing point

  The molar mass of a non-electrolyte solute can be obtained from the freezing point depression.

                                          

  where, is the mass of solute ( in grams), and is the molar mass of the solute (in g/mol) and, W1 is the mass of the solvent in Kg units.

  
  

  This gives

             

  
  

  The method of obtaining the molar mass of a solute from the freezing point depression is used for the substances which are affected by heat such as, proteins.

  
  

  The constant (fK), for a solution, is called molal depression constant or molal cryoscopic constant for the solvent. fKmay be defined as the depression in freezing point of a solution when one mole of a solute is dissolved in 1 kilogram of the solvent.
  

  
  

  Examples:

  100g of non-electrolyte solute dissolved in 50g of benzene lowered the freezing point of benzene by 0.40K. The freezing point depression constant of benzene is 5.12 K Kg/mol. Find the molar mass of the solute.

  
  

  Solution:

  
  

  Thus, molar mass of the solute is 256g/mol.

  
  

  This property is used in the countries that experience cold winter where salt is spread on the roads to de-ice the roads; salt mixed with ice decreases the freezing temperature of water and the ice liquefies or melts at temperatures below zero.

  
  

  • Determination of molar mass from osmotic pressure

  Because small pressures can be measured easily and accurately, osmotic pressure measurements provide a useful way of molar masses determination of large molecules, like protein, polymers and other macromolecules. The osmotic pressure method has the advantage over other methods as pressure measurement is around the room temperature and the molarity of the solution is used instead of molality. As compared to other colligative properties, its magnitude is large even for very dilute solutions. The technique of osmotic pressure for determination of molar mass of solutes is particularly useful for biomolecules as they are generally not stable at higher temperature and polymers have poor solubility.

  
  

  

  
  

  Where, d = density, R = gas constant

  
  

  T = temperature, Mm = molar mass of solute
  

  
  

  Note: Osmotic pressure method is preferred for determination of very high molecular mass because osmotic pressure effects for small mole fractions are much higher than the effects on freezing and boiling points i.e.osmotic pressure changes for very small mole fraction can be more easily determined than freezing point changes.

  
  

  Examples

  1. The osmotic pressure of aqueous solution of a certain protein was measured in order to determine the molar mass of the protein. The solution contained 3.50 mg of protein dissolved in sufficient water to form 5.00 ml of solution. The osmotic pressure of the solution at 25°C was found to be 1.54 torr. Calculate the molar mass of the protein.

  

  
  

  2. A 50.00 mL sample of an aqueous solution contains 1.08 g of human serum albumin, a blood plasma protein. The solution has an osmotic pressure of 5.85 torr at 298 K. what is the molar mass of the albumin?

  
  

  Answer:

  
  

  1atm=760mmHg=760torr= 101325Pa

  
  

  
  

  Reverse Osmosis and Water Purification

  If a pressure higher than the osmotic pressure is applied to the more concentrated solution side, the direction of flow of the solvent can be reversed. As a result, the pure solvent flows out of the solution through the semipermeable membrane. This process is called reverse osmosis. It is of great practical application as it is used for desalination of sea water to obtain pure water. When pressure more than osmotic pressure is applied, pure water is squeezed out of the sea water through the membrane.

  
  

  These days many countries, such as desert Arab countries, use desalination plants based on this technology to meet their potable water requirements.

   

  
  

  Checking up 7.3

  1. Why is the osmotic pressure measurement preferred for determining the molecular mass of proteins?

  2. The vapour pressure of pure benzene at certain temperature is 0.850 bar. A nonvolatile, non electrolyte solid weighing

       0.5 g is added to 39.0 g of benzene (molar mass = 78 g/mol). The vapour pressure of the solution is 0.845 bar. What
  

        is the molecular of the solid substance?

  3. Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of
  

      glucose is dissolved in 450 g of water.

  4. An organic liquid C has a freezing point of 198.5 °C and molal depression constant of 45.5°C/m. When 0.049 g of

       a substance A was dissolved in 0.521 g of C, the resultant mixture was found to freeze at 186°C. Calculate the molar

        mass of A.

  5. Water boils at 100°C at a pressure of 760 mm Hg. When the pressure is reduced to 660 mm Hg, water boils at 96°C.
  

       explain this observation.

  
  

  END UNIT ASSESSMENT

  1. Colligative properties are observed when:

      a. a non-volatile solid is dissolved in a volatile liquid

      b. a non-volatile liquid is dissolved in an another volatile liquid

      c. a gas is dissolved in a non-volatile liquidd. a volatile liquid is dissolved in an another volatile liquid
  

          Mark the correct option(s).

  
  

  2. Which of the following binary mixtures will have same composition in liquid and vapour phase?

     a. Benzene-Toluene

     b. Water-Nitric acid

     c. Water-Ethanol

     d. n-Hexane-n-Heptane

  
  

  3. Considering the following couples of solvents, predict which mixture will show a positive deviation from Raoult’s law.

     a. Methanol and acetone

     b. Chloroform and acetone

     c. Nitric acid and water

     d. Phenol and aniline

  
  

  4. Relative lowering of vapour pressure is a colligative property because

      a. It depends on the concentration of a non-electrolyte solute in solution and does not depend on the nature of the
  

           solute molecules.

      b. It depends on number of particles of electrolyte solute in solution and does no t depend on the nature of the

           solute molecules.

      c. It depends on the concentration of anon-electrolyte solute in solution as well as on the nature of the solute molecules.

      d. It depends on the concentration of an electrolyte or a non-electrolyte solute in solution as well as on the nature of
  

           solute molecules.Mark the correct option(s).

  
  

  5. If two liquids A and B form minimum boiling azeotrope at some specific composition:

      a. A-B interactions are stronger than those between A-A or B-B

      b. Vapour pressure of solution increases because more number of molecules of liquids A and B can escape from the solution

      c. Vapour pressure of solution decreases because less number of molecules of only one of the escape from the solution

      d. A-B interactions are weaker than those between A-A or B-B.Mark the correct option(s).

  
  

  6. Colligative properties depend on

     a. The nature of the solute particles dissolved in solution

     b. The number of solute particles in solution

     c. The physical properties of the solute particles dissolved in solution

     d. The nature of the solvent particles.Mark the correct option(s).

  
  

  7. If 0.500 grams of caffeine is dissolved in 100 mL of water, what percentage of caffeine can be separated from the water
  

       using a single 40 mL sample of methylene chloride? The distribution coefficient = 4.6

  
  

  8. Benzoic acid can be separated from water using octanol as the organic solvent. The distribution coefficient for
  

      this   water/octanol system is P = 1.87. Assuming that 1 gram of benzoic acid has been dissolved in 100 mL of

      water, how many 20 mL extractions must be done to extract 60+ percent of the benzoic acid from the water?

  
  

  9. A 0.100 gram sample of phthalic acid was dissolved in 100 mL of water. When 25 mL of diethyl ether was used to

      extract the phthalic acid, 0.042 grams of phthalic acid were recovered. What is the distribution coefficient for this extraction?

  
  

  10. What mass of ethylene glycol (C2H6O2, molar mass=62.1 g/mol), the main component of antifreeze, must be added
  

        to 10.0 L of water to produce a solution for use in a car’s radiator that freezes at -23.3°C? Assume the density of water

        is exactly 1g/ml. Kf=1.86°C.kg/mol

  
  

  11. A chemist is trying to identify a human hormone that controls metabolism by determining its molar mass. A sample

        weighing 0.546g was dissolved in 15.0 g benzene, and the freezing point depression was determined to be

        0.240°C. Calculate the molar mass of the hormone. Kf for benzene is 5.12°C.Kg/mol.

  
  

  12. To determine the molar mass of a certain protein, 1.00x 10-3 g of it dissolved in enough water to make 1.oo mL of
  

         solution. The osmotic pressure of this solution was found to be 1.12 torr at 25.0°C. Calculate the molar mass of protein.
  

  
  

  13. The molecular masses of polymers are determined by osmotic pressure method and not by measured other

         colligative properties. Give two reasons.
  

  
  

  14. Non-ideal solutions exhibit either positive or negative deviations from Raoult’s law. What are these deviations and why
  

        are they caused? Explain with one example for each type.
  
• UNIT 8: QUANTITATIVE CHEMICAL EQUILIBRIUM

  Key unit competence

  To write expressions and calculate the values of equilibrium constant, interpret the

  values of Kc in relation to the yield of the products in reversible reactions.

  Learning objectives

  At the end of this unit , students will be able to:

  • Explain how the temperature affects the magnitude of equilibrium constant

  Kc;

  • Derive the relationship between Kc and Kp;

  • Write expression for Kc and Kp;

  • Interpret the Kc values in relation to the yield of the reversible reactions;

  • Compare the Kc value with Qc value and predict if a reaction is at equilibrium

  or not.

  • Compare and interpret the values of Kc and Kp of different reactions;

  • Perform calculations involving equilibrium constants in terms of concentration(Kc), and partial prsessure (Kp).

  

  In most of the chemical reactions, the reactants are not completely converted into products. The reaction proceeds to a certain extent and reaches a state at which the concentrations of both reactants and products remain constant with time.

  Chemical equilibrium is the state in which both reactants and products of a chemical reaction are present in concentrations that do not change with time. Chemical equilibrium deals with the reversible reactions, which reach equilibrium state; where the forward reaction proceeds at the same rate as the reverse reaction. The scope of chemical equilibrium includes the study of characteristics and factors affecting the chemical equilibrium.

  8.1. Definition and characteristics of equilibrium constant Kc

  
  

  8.1.1. Definition

   The equilibrium constant expresses ratio between the concentrations of the products and the concentrations of the reactants under the given conditions. The equilibrium constant is symbolized by Kc , and the ratio Q expresses the value of the reaction quotient when the forward and reverse reactions occur at the same rate. When the reaction reaches the equilibrium, no change is observed between the chemical composition of the mixture with time and the enthalpy of the products is equal to that of the reagents while the Gibbs free energy change for the reaction is equal to zero. By definition: The equilibrium constant, Kc, is the ratio of the equilibrium concentrations of products over the equilibrium concentrations of reactants each raised to the power of their stoichiometric coefficients at given temperature.
  

  8.1.2. Characteristics of equilibrium

  Consider the following reversible reaction involving homogeneous system;

  
  

  Note: All the concentrations used to calculate Kc are measured at homogeneous

  equilibrium.

  Where, A, B, C, and D are chemical species, and a, b, c, and d are their respective

  stoichiometric coefficients.

  • If Kc value >> 1, the direct reaction is favored and the reaction mixture contains

  mostly products.

  • If Kc value <<1, the reverse reaction is favored and the reaction mixture

  contains mostly reactants.

  • If the Kc value is close to 1 (0.10 < Kc < 10), the mixture contains appreciable

  amounts of both reactants and products

  Examples: Write the equilibrium constant expression, Kc, for the following reactions:

  

  Note:

  i. The equilibrium can be approached from either direction (forward or

  backward direction).

  ii. The equilibrium constant Kc does not depend on the initial concentrations

  of reactants and products but it depends on temperature.

  The equilibrium is described as homogeneous if all species are in the same phase

  and heterogeneous if they are in different phases.

  Worked examples:

  Write the equilibrium constant expression, Kc, for the reactions

  

  Evaluating K_c  for a reaction:

  In order to determine the value of an equilibrium constant, when amounts of

  reactants are known at equilibrium. The amount of one of the substances in the

  equilibrium mixture is found by experiment. The amounts of the others can then

  be worked out from the stoichiometric equation as shown in the following worked example.

  Examples

  1. 2.0 mol of ethanoic acid and 2.0 mol of ethanol are mixed and allowed to come

  to equilibrium with the ethyl ethanoate and water they have produced. At

  equilibrium, the amount of ethanoic acid present is 0.67 mol. Calculate K_c.

  Solution

  
  

  
  

  therefore the concentration units cancel out, and K is a unitless number.

  Characteristics of the equilibrium constant Kc

  The equilibrium constant, K_c is governed by temperature, which is the only factor

  that can change the internal potential energy of the reactants or products.

  i. If the forward reaction is exothermic, K_c decreases with the increase in

  temperature.

  ii. If the forward reaction is endothermic, K_c increases with the increase in

  temperature.

  iii. The application of a catalyst to a reaction has no effect on a K_c value.

  
  

  8.2. Deriving equilibrium constant Kc
  

  
  

  8.2.1. Deriving equilibrium constant Kc from thermodynamic approach

  The equilibrium constant gives the rate of transformation and expresses the ratio

  of the concentrations between the products and the reagents under the given

  conditions.

  A relation can be derived between the change of standard free enthalpy and the

  equilibrium constant. Enthalpies and entropies are important thermodynamic

  functions used for a wide variety of applications. Enthalpy and entropy express the

  energy and the randomness of a reaction respectively. In chemistry, the temperature

  is an important factor that often can influence the rate of the reaction.

  The application of thermodynamic data is expanded due to the introduction of the

  Gibb’s energy, because of its link to the equilibrium constant, Kc.

   A relationship between the equilibrium constant, Kc, and Gibb’s energy, G, entropy

  S at temperature T have been introduced.

  

  

  

  8.3. Mass action law and equilibrium constant expression

  Activity 8.3

  1. Explain briefly the mass action law.

  2. By giving an example, deduce the unit of concentration.

  8.3.1. Mass action law

  In chemistry, all the species involved in the chemical reaction must be equal in both

  sides of reaction (reactants and products). This is the basic explanation of the mass

  action law which can be defined as follows: Law of Mass Action or Equilibrium law

  states that “The ratio of an equilibrium concentrations of products and reactants

  each raised to the power of their coefficients in the balanced stoichemetric

  equations is a constant, Kc, at a constant temperature".

  The molar concentration i.e. number of moles per litre is also called Active Mass. It is

  expressed by enclosing the symbols of formulae of the substance in brackets.

  For example, molar concentration of A is expressed as [A].

  
  

  According to law of mass action: the total mass of A and B involved in the collision

  must be equal to the total mass of new formed compounds

  
  

  
  

  8.4. Definition of equilibrium constant in terms of partial pressures “Kp”

  Activity 8.4

  1. Write down the ideal gas equation and explain all terms involved in.

  2. Explain the term “partial pressure’’

  For gaseous equilibria, the concentration of reactants and products can be expressed

  in terms of pressures of the different components. If the reaction mixture behaves as

  an ideal gas, the ideal gas equation is PV = nRT. The molar concentration [X], which is

  n/V, is then equal to P/RT. This shows that, at a given temperature when T is constant,

  the pressure of a component is proportional to its molar concentration.

  The equilibrium constant with the concentrations of reactants and products

  expressed in terms of molarity is expressed by Kc. 

  

     Determination of the partial pressure

  

  

  

  8.5. Relationship between Kc and Kp

  Activity 8.5

  Explain the terms Kc and Kp

  The pressure and molarity are related by the Ideal Gas Law, PV = nRT where

  P = nRT/V

  Consider the general reaction:

  

  

  Δn = moles of gaseous products - moles of gaseous reactants

  K_p= equilibrium constant in gaseous phase expressed in partial pressures

  K_c = equilibrium constant in molar concentrations

  Note: Kc = Kp when the number of gas molecules are the same in both sides

   (Δn = 0).

  Characteristics of equilibrium constant

  • The value of the equilibrium constant is independent of the original

  concentration of reactants.

  • The equilibrium constant has a definite value for every reaction at a particular

  temperature. However, it varies with change in temperature.

  • For a reversible reaction, the equilibrium constant for the forward reaction is

  inverse of the equilibrium constant for the backward reaction.

  • In general, K_forward reaction = 1/K’_backward reaction and vice-versa

  • The value of an equilibrium constant tells the extent to which a reaction proceeds in the forward or reverse direction.

  • The equilibrium constant is independent of the presence of catalyst.

  

  8.6. Calculations on Kc and Kp

  
  

  

  

  

  

  

  

  

  

  

  

  

  

  

  

  

  8.7. Comparison between reaction quotient Qc and equilibrium constant Kc

  Activity 8.7

  1. What do you understand by the term reaction quotient

  2. By giving an example differentiate the reaction quotient and reaction

  constant Kc.

  The reaction quotient, Q, is the resulting value when we substitute reactant and the

  product concentrations into the equilibrium expression. A reaction quotient (Qr or

  Q) is a function of the activities or concentrations of the chemical species involved

  in a chemical reaction.

  The difference is that Q can be calculated at any conditions (not only on equilibrium

  conditions) while the equilibrium constant is determined when there is a reversible

  reaction.

  Example:

  Consider for the following general equation:

  

  • If Q = Kc, the reaction is at equilibrium and the concentrations will not change

  since the rates of forward and backward reactions are equal.

  • If Q < Kc, the ratio of products over the reactants is too small and the reaction

  is not at equilibrium; the reaction will move toward the equilibrium by forming

  more products.

  • If Q > Kc, the ratio of products over the reactants is too large and the reaction

  is not at equilibrium; the reaction will move toward the equilibrium by forming

  more reactants. 

  

  

  

  

  

• UNIT 9: pH OF ACIDIC AND ALKALINE SOLUTIONS

  Key unit competency

  To be able to:

  • Prepare solutions, measure their pH, and calculate the pH of acidic and alkaline solutions.

  • Explain the concept of buffer solution, hydrolysis of salts and discuss its applications in manufacturing industry and biological processes.

  Learning objectives

  At the end of this unit , students will be able to:

  • Define the degree of ionization (α);

  • Define the terms Ka, pH, pKa, Kb, Pkb and Kw;

  • Write equations for salt hydrolysis reactions and the expression for the hydrolysis constant;

  • Explain how buffer solution control pH;

  • Explain the buffer capacity in relation to buffer range;

  • Describe the applications of buffer solution in domains such as biological processes, agriculture, natural system (e.g. lakes) and industrial manufacture of cosmetic and drugs.

  • Perform calculations involving pH, Ka, pKa, Kb, pKb and Kw;

  • Interpert th values of Ka and Kb in relation to the strength of acids and bases;

  • Prepare different solutions and appropriately use pH-meter to measure their pH;

  • Compare the strength of acids and bases on the same concentration using the values of Ka and Kb;

  • Relate the values of pH and pOH;

  • Calculate the pH and the hydrolysis constant of aqueous solutions of salts;

  • Perform experiments to show that hydrolysis of some salts results in neutral, acidic and alkaline solutions;

  • Prepare buffer solutions of different pH values;

  • Derive Henderson-Hasselbalch relation and use it to calculate the pH of buffer solution.

  Introductory activity

  
  

  Observe the above representations a, b, c, d, e and f which represent some useful products made of acids, bases and salts and answer to the following questions.

  1. Identify one chemical substance found in each product above.

  2. State whether the chemical substance identified above is an acid, a base or a salt.

  3. What do you comment about the pH of the chemical substance in (1) 

  Many fruit juices and other soft drinks contain acids and this can be identified by their sour or sharp taste. They contain many dissolved compounds of which acids are among. Some acids are chemicals that can be harmful or dangerous to human and should never be tested by drinking them (i.e. sulphuric acid used in car batteries). 

  Bases are products commonly used in our daily life such as soap, toothpaste, magnesium syrup, baking soda. Solutions of bases are slippery on touch because they attack oils on the skin and convert them into soaps. This makes them to be good cleaning agents. Bases dissolve in water to form solutions called alkaline solutions. 

  Acids can react with bases to form compounds known as salts which are of great importance in nature. For example ammonium chloride is used as an electrolyte in dry cells, calcium carbonate is used to manufacture cement, potassium nitrate is used as a fertilizer, magnesium sulphate and iron (II) sulphate used to manufacture of drugs.

  9.1. Degree of ionization in relation to strength of acids and bases

  Activity 9.1

  (a) Explain the following terms;

   (i) an acid

   (ii) a base

  (b) Give an example of an acid and a base which you encounter in your daily life.

  (c) Explain the difference between a strong acid and a strong base.

  (d) How is a weak acid different from a weak base?

  (e) Give an example of weak acids and weak bases.

  (f) Write down chemical equations to show ionization of a weak acid and weak base water.

  9.1.1. Acids and bases

  In the previous years of your study, an acid has been defined as a substance that donates hydrogen ions while a base was defined as a substance that accepts hydrogen ions. Acidity and alkalinity are measured with a logarithmic scale called pH. 

  • Strong and weak acids 

  When an acid is dissolved in water, there are more hydrogen ions than hydroxide ions in the solution meaning that the solution is acidic. We can distinguish strong and weak acids. 

  A strong acid is ionized completely when dissolved in water. For example hydrochloric acid ionizes completely in water to form hydrogen or hydroxonium ions and chloride ions as illustrated below. The direct arrow indicates that the compound is completely ionized.

  

  When a weak acid is dissolved in water, only a small proportion of it ionizes to form ions. For example ethanoic acid ionizes partially in water to form ethanoate ions and hydrogen or hydroxonium ions. The following illustration shows that the ethanoic acid as a weak acid dissolves partially in water, the direct and indirect arrows symbolize the partial ionization or an equilbrium dissocition reaction

  

  Other weak acids include sulphurous acid (H2 SO3 ), carbonic acid (H2 CO3 ), phosphoric acid (H3 PO4 ), methanoic acid (HCOOH) etc. In general the organic acids are known to be weak acids.

  The strength of an acid can be influenced by four factors namely: 

  (a)Bond strength: When the bond holding a hydrogen atom is strong, the compound formed is a weak acid since the bond cannot be easily broken to release hydrogen ions in solution. As an example, consider the halogen acids HF, HCl, HBr and HI. Their bond strength decreases with a decrease in electronegativity of the halogen atoms down the group 17. HI is a stronger acid than HCl or HBr because of a weak H – I bond due to its lower electronegativity of iodine atom than Cl or Br. This bond is easily broken down to release hydrogen ions in solution. HF is a weak acid due to its strong H – F bond caused by a highly electronegativity of fluorine atom. This bond is not easily broken down to release hydrogen ions in solution. 

  (b)Nature of the solvent: The more basic the solvent in which the acid is dissolved, the stronger the acid. This is because the proton released by the acid is easily accepted by the basic solvent. This enables the acid to continue ionizing in the solvent to produce hydrogen ions e.g. Benzoic acid is a stronger acid in aqueous ammonia than in water. 

  (c) Presence of a halogen atom in organic acids: The halogen atom increases acidity because, being more electronegative than carbon it tends to withdraw electrons towards itself (negative inductive effect). This reduces the electron density in the O – H bond thus weakening it. In solution, the O – H bond is easily broken down to release hydrogen ions.This explains why chloroethanoic acid is a stronger acid than ethanoic acid. 

  (d) Number of carbon atoms in organic acids: As the number of carbon atoms increases, the acidity of the organic acid decreases. This is because a long chain alkyl group pushes electrons towards the carboxyl group (positive inductive effect) increasing the electron density on the O – H bond. This makes the O – H bond stronger and not be easily broken down to release hydrogen ions in solution.This explains why methanoic acid is a stronger acid than ethanoic acid.

  • Strong and weak bases 

  When a base is dissolved in water, there are more hydroxide ions in the solution than hydrogen ions. This kind of solution is called alkaline. Strong and weak bases can be distinguished based on their ionization character. 

  When a strong base is dissolved in water, it ionizes completely. For example sodium hydroxide (NaOH) ionizes in water to form sodium ions and hydroxyl ions. Similarly, potassium hydroxide (KOH) is a strong base because it dissociates completely when dissolved in water as illustrated below:

  

  For weak base, they dissolve partially in water. For example ammonia solution ionizes partially in water to form ammonium ions and hydroxyl ions..

  

  Phenylamine (C6 H5 NH2 ) and Hydroxylamine (NH2 OH) are other examples of weak bases 

  9.1.2. Degree of ionization (α) 

  An acid or a base when dissolved in water, it ionizes completely or partially depending on its characteristic as strong or weak. 

  The degree of ionization can be defined as the ratio of the number of ionized molecules to the total number of molecules dissolved in water.It is symbolized by αa in the case of acid and αb in the case of base.

  
  

  When a weak acid or weak base is dissolved in water, partial ionization occurs such that equilibrium is set up between the undissolved molecules and the ions formed in water. This equilibrium is known as ionic equilibrium of an acid or base. The degree of ionization is a fraction ranging from 0 to 1. a is equal to zero for the insoluble substances or non-electrolytes (i.e. substances which do not ionize in water). a is equal to 1 for strong electrolytes such as strong acids or bases because they ionize completely in water. a is less than 1 in the case of weak electrolytes such as weak acids or bases because they ionize partially in water. The degree of ionization can also be expressed as percentage as shown by the relation below.
  

  
  

  The greater the degree of ionization, the stronger the acid or base. For example, the concentration of an acid that undergoes ionization is equal to the concentration of hydrogen ions formed as shown below.
  

  
  

  Example 

  What is the degree of dissociation of a weak acid in a 0.25M solution, given the concentration of H+ as 0.001 mol dm-3? 

  Solution
  

  
  

  Checking up 9.1 

  1. A 0.1moldm-3 methanoic acid solution contains 0.0042 moldm-3 of hydrogen ions. Calculate the percentage of the acid that is ionized. 

  2. A solution of 0.035M nitrous acid contains 0.0037M of hydrogen ions. Calculate the percentage ionization of the acid.
  

  9.2. Explanation of acid and base dissociation constants (Kand K)
  

  Activity 9.2

  1. Explain your understanding of the following expressions

   (a) Acid dissociation constant.

   (b) Base dissociation constant.

  2. (a) Draw an equation to show ionization of ethanoic acid in water.

   (b) Derive an expression for acid dissociation constant, Ka of ethanoic acid in water.

  3. (a) Express an equation for the ionization of methylamine in water.

   (b) Write the expression for the base dissociation constant, Kb of methylamine.

  Weak electrolytes such as weak acids or weak bases are partially ionized in water thus their dissociation is reversible and can reach equilibrium. Therefore the equilibrium constants can be used to explain the strength of acids and bases based on their ionization reactions. 

  9.2.1. Acid dissociation constant, Ka 

  The acid dissociation constant also known as acidity constant or acid ionization constant is the equilibrium constant for the ionization reaction of an acid. It is denoted by Ka . 

  This constant is a quantitative measure of the strength of the acid in solution in units of moldm-3. Acid dissociation constants are mostly associated with weak acids because strong acids are completely ionized in the aqueous solution and their Ka values are extremely large or infinity. 

  Consider a weak acid HA, its ionization is represented as follows:
  

  
  

  The equilibrium constant for the dissociation of HA is as follows:
  

  
  

  9.2.2. Base dissociation constant, Kb 

  The base dissociation constant also known as base ionization constant is the equilibrium constant for the ionization reaction of a weak base denoted by Kb . The equilibrium constant Kb measures the strength of the base in solution in units of moldm-3. Base dissociation constants are mostly associated with weak bases because strong bases are completely ionized in the aqueous solution and their Kb values are extremely large. 

  Consider a weak base, BOH undergoing ionization which is represented as follows:
  

  
  

  The base dissociation constant Kb can be expressed by applying the equilibrium law:
  

  
  

  Note: Kb provides a measure of the strength of the base. When Kb is large, the base is highly dissociated, and the base is strong. And when Kb is small, very little of the base is dissociated and the base is weak. 

  9.2.3 Relationship between equilibrium (dissociation) constant and the degree of ionization α; Ostwald’s dilution law

   The relationship between the equilibrium constant and the degree of dissociation of weak electrolytes has been introduced by Ostwald in 1888. 

  The Ostwald’s dilution law states that “the degree of ionization of a weak electrolyte is inversely proportional to the square root of the molar concentration of the electrolyte”.
  

  
  

  Where C is the molar concentration and Ka is the equilibrium constant.

   Consider the dissociation equilibrium of formic acid HCOOH which is a weak electrolyte in water.
  

  
  

  Where α is the degree of ionization and represents the fraction of the total concentration of HCOOH that exists in the completely ionized state. Further (1 – a) is the fraction of the total concentration of HCOOH in the unionized state. 

  Relationship between Ka and a. 

  If the total concentration of the acid is C at the initial step and its degree of ionization is aC; at the equilibrium Ca, Ca and C (1 – a) represent the concentration of H+, and HCOOH- respectively as shown below. 

  If a is too small, Ka = a2 C

  

      
  

  
  

  The higher the Ka value, the greater the degree of ionization and the stronger the acid.
  

  
  

  The higher the Kb value, the greater the degree of ionization and the stronger the base. 

  Worked examples 

  1. (a) A solution of a weak acid, HA contains 0.25M, given the concentration of H+ as 0.001 moldm-3.Calculate the dissociation constant of HA. (b) Determine the degree of ionization of this acid in 1M solution? 

  Solution:
  

  
  

  (b) Degree of dissociation in a 1 M solution:
  

  
  

  2. Calculate the degree of ionization of 0.1M acetic acid, CH3 COOH if its ionization constant Ka = 1.8 x10-5 moldm-3.
  

  
  

  Checking up 9.2

  1. Calculate the degree of ionization of a 0.04M ethanoic acid solution at 25 0C given that its Ka is 1.3 x 10-5 mol dm-3.

  2. The acid dissociation constant of a monobasic acid is 4.39 x 10-5 mol dm-3 at 25 0C.

   Calculate the degree of ionization of a 0.01M solution of the acid.

  3. (a) Write an equation for the ionization of methylamine in water.

   (b) Express the dissociation constant Kb for methylamine.

   (c) The hydroxyl ion concentration of a 1M methylamine solution is 0.04 moldm-3.

   Calculate the Kb for methylamine.

  The relationship between Ka and Kb is derived from the dissociation of conjugate acid and conjugate base. For example considering NH4 + and NH3 ; NH4 + is a conjugate acid and NH3 is a conjugate base. 

  The ionization of a conjugate acid in water is shown below:
  

  
  

  Ka and Kb are related each other through the ionization constant.
  

  When we multiply by–log10 to both sides, we get:
  

  
  

  Checking up 9.3

  1. (a) Ethyl ammonium ion,CH3CH2-NH3 + is a conjugate acid.Write an equation for the ionization of this ion in water.

   (b) Write the expression of acid dissociation constant, Ka ,of ethyl ammonium ions water.

  2. (a) Ethylamine, CH3CH2-NH2

   is a conjugate base. Write the chemical equation for ionization of ethylamine in water.

   (b) Write an expression for the base dissociation constant, Kb of ethylamine in water.

  3. From the reactions above, derive an expression to show the relationship between Ka and Kb

  9.4. Use of Ka or pKa and Kb or pKb to explain the strength of acids and bases 

  Activity 9.4 1. (a) Ethanoic acid has pKa of 4.77 at 25 0 C. What is meant by pKa of the acid? 

  (b) Given the following acids and their corresponding Ka values in the table below,

  

  State the strongest and the weakest acid and justify your answer. 

  2. The ionization of ethanoic acid in water is shown by the following equation :

  

  Use the above equation to write to relate the acid dissociation constant, Ka and the degree of ionization, a

  The acidic and basic dissociation constants are respectively represented by Ka and Kb . Based on the negative logarithm, pKa and pKb are derived from–log10Ka and –log10Kb respectively. Therefore, Ka , pKa , Kb and pKb are more helpful in predicting whether a species donates or accepts a proton.

  

  

  

  

  

  Checking up 9.4

  
  

  9.5. Explanation of ionic product of water
  Activity 9.5

  Water is a neutral liquid at pH = 7

  a. Explain why at 25 °C water has a pH equals to 7.

  b. (Demonstrate how the ionization of water molecules is made.

  c. What do you understand by “Ionic product of water ?

  Water is a weak electrolyte and neutral in nature. Pure water is commonly known as universal solvent and water takes part in many equilibrium reactions. Water ionizes as shown by the reaction equation as follows:
  

  
  

  
  

  
  

  The ionic product of water increases with temperature (Table 9.3). However at all temperatures, the concentration of H+ ions remains equal to the concentration of OH- ions in pure water.
  

  
  

  
  

  9.6. Definition and calculations of pH and pOH of acidic and alkaline solutions 
  Activity 9.6

  
  

  The pH is a scale commonly used to measure the degree of acidity or alkalinity of a solution.It is measured on a scale of 0 to 14. The term pH is derived from “p,” which is a mathematical symbol expressing the negative logarithm, and “H,” the chemical symbol for Hydrogen. In general pH can be defined as the negative logarithm of Hydrogen ion activity or pH = -log [H+] 

  Both pH and pOH are two methods that are used to describe the strength of acids, alkali, or ionizable salts (i.e. salt of a weak acid and strong base, salt of weak base and a strong acid or salt of weak acid and weak base).The pH is a measure of the hydrogen ion concentration of a solution.
  

  Within this section, the following terms will be commonly used such as [H+] = hydrogen ions concentration; [OH- ] = hydroxide ions concentration. 

  The pH can be expressed as the negative logarithm of the concentration of [H+] in a solution. The Solutions having a high concentration of hydrogen ions have a low pH and solutions with a low concentrations of H+ ions have a high pH. On the other hand, “pOH” is the negative of the logarithm of the concentration of hydroxide ions in a solution.It describes how alkaline a solution is; the more alkaline a solution is, the higher the concentration of hydroxide ions in the solution.
  

  Importance of pH and pOH 

  Knowing the pH and pOH is important because it helps us to find the concentration of hydrogen or hydroxide ions in the solution; it allows determining the acidity and alkalinity of the solution. While knowing the pH of a solution, the strength of the acid and base in the solution can be determined and the separation of strong and weak acids or strong and weak bases can be done based on the value of pH or pOH. 

  Mathematically pH or pOH can be expressed as follows:
  

  
  

  The Figure 9.1 shows the location of acidic, neutral and alkaline compounds based on pH scale while the Figure 9.2 shows the pH or pOH scale (from 0 to 14). Figure 9.3 shows examples of compounds corresponding to different values of the pH scale.
  

  
  

  
  

  The pH of strong acids is ranged from 1-3 and weak acids from 4-6. Strong bases have values of pH ranging from 11 to 14 and weak bases have values of pH ranging from8 to 10.
  

  
  

  Note: When a drop of hydrochloric acid is added to water, the concentration of hydrogen ions increases and pH decreases. The pH of an acidic solution is less than 7. On the other hand when sodium hydroxide is added to water, the added hydroxyl ions reduce the concentration of hydrogen ions and pH increases. pH of an alkaline solution is greater than 7. The Table 9.4 shows the variation of pH in function of the hydrogen ion concentrations.
  

  
  

  9.6.1. pH of strong acids 

  Strong acids like hydrochloric acid has a pH around 0 to 1.The lower the pH, the higher the concentration of hydrogen ions in the solution. A strong acid is one which completely dissociates into its ions in water. This makes calculating the hydrogen ion concentration, which is the basis of pH,easier than for weak acids. The following are examples on how to determine pH of a strong acid.
  

  1. Determine the pH of a 0.025 M solution of hydrobromic acid (HBr) solution 

  Hydrobromic acid or HBr, is a strong acidand will ionize completely in water by giving H+ and Br- ions. For every mole of HBr, there will be 1 mole of H+ and the concentration of H+ will be the same as the concentration of HBr. Therefore, [H+] = 0.025 M.
  

  
  

  
  

  
  

  
  

  
  

  
  

  
  

  
  

  
  

  
  

  
  

  
  

  
  

  
  

  
  

  9.7. Salt hydrolysis
  Activity 9.7.

  1. What do you understand by the term “a salt”?

  2. Explain two types of salts. Give examples in each case.

  3. Discuss how salts are formed. Explain any four methods of preparing salts.

  4. What do you understand by “salt hydrolysis”?

  5. Give four examples of salts which undergo hydrolysis.

  6. a. (i) Write an equation for the hydrolysis of sodium benzoate in water.

  (ii) Write an expression for the hydrolysis constant, Kh of sodium benzoate.

   b. A solution contains 0.2 moles of sodium benzoate per litre at 25°C.  

     Calculate the pH of the solution given that the hydrolysis constant, Kh of sodium benzoate is 1.6x10-10 mol dm-3 at 25°C.

                             

  In the previous sections, we have seen that the pH value can be used to determine the strength of acid or a base. When the concentrations of [H+] > [OH- ], the water becomes acidic and when [H+] < [OH- ], the water acquires basic nature. A salt hydrolysis is a phenomenon that is observed when there is change of the concentrations of hydrogen or hydroxyl ions in a solution. 

  A salt is a chemical substance formed when either part or all the ionizable hydrogen of an acid have been replaced by a metallic ion or ammonium radical.Salts are strong electrolytes because when dissolved in water they dissociate almost completely into two different ions such as cations and anions.
  

  Salt hydrolysis is defined as a reaction in which the cation or anion or both of a salt react with water to form an acidic or alkaline solution. 

  When the cations from the salt are more reactive than anions, they interact with water molecules detaching hydroxyl ions. This leaves hydrogen ions in solution making it acidic.
  

  
  

  When the anions from the salt are more reactive than cations, they interact with water removing hydrogen ions and releasing the hydroxyl ions in solution. This results in the alkaline solution.
  

  
  

  In general, the process of salt hydrolysis is the reverse of neutralization: Salt + Water ↔ Acid + Base 

  If acid is stronger than base, the solution is acidic and in case base is stronger than acid, the solution is alkaline. When both the acid and the base are either strong or weak respectively, the solution is generally neutral in nature. As the nature of the cation or the anion of the salt determines whether its solution will be acidic or basic, it is proper to divide the salts into four categories.
  

  (i) Salt of a strong base and a weak acid.

  (ii) Salt of weak base and strong acid.

  (iii) Salt of weak base and weak acid.

  (iv) Salt of a strong acid and a strong base.

  9.7.1. Salt of weak acid and strong base

  The solution produced by a strong base and a weak acid is basic. The anion reacts with water to form a weak acid and OH- ions. 

  
  

  Ethanoate ions disturb the ionic equilibrium of water where they accept hydrogen ions from water. This makes the resultant solution alkaline due to excess hydroxyl ions present. Such a solution has pH value greater than 7. The ions produced CH3 COO¯ in turn react with water to form a weak acid, CH3 COOH and OH¯ ions as follows:
  

  
  

  pH of the resultant solution on this hydrolysis is then calculated from, pH = 14 – pOH

  9.7.2. Salt of weak base and strong acid

  These are salts which undergo cation hydrolysis to form acidic solutions. Examples of such salts include;

                          
  

  
  

  Ammonium ions disturb the ionic equilibrium of water by accepting hydroxyl ions from water. This makes the resultant solution acidic due to excess hydrogen ions from water thus a pH value less than 7.
  

  
  

  9.7.3. Salt of weak base and weak acid 

  These are salts which undergo both cation and anion hydrolysis when dissolved in water. The nature of the resultant solution depends on the relative strength of the weak base and the weak acid. It may finally be acidic, alkaline or neutral. 

  E.g. if Kb of the anion is greater than Ka for the cation, the solution formed is alkaline because the anion is greatly hydrolyzed to produce more hydroxyl ions in solution. 

  When the Kb of the anion is less than Ka the cation, the resultant solution is acidic because the cation will be hydrolyzed to a greater extent producing excess hydrogen ions in solution. And if Kb is approximately equal to Ka , the resultant solution is neutral. 

  Examples of such salts include:
  

  
  

  
  

  9.7.4. Salt of strong acid and strong base

  Salts of strong acids and strong bases dissolve in water to give neutral solutions.

  For example; 

  

  Aqueous solutions of the salts consist of ions with very little affinity for hydrogen ions or hydroxyl ions in water. The ions of such salts are only hydrated (i.e. surrounded by water molecules). For example; Sodium chloride dissociates in water to give the anion Cl¯ .

  

  Since hydrochloric acid is a strong acid, Cl¯  is very weak base. Cl¯ is unable to accept a proton (H+) from an acid, particularly water. That is why Cl¯  does not hydrolyze. The pH of sodium chloride solution remains unaffected. 

  Hydrolysis constant, Kh and the degree of hydrolysis 

  Consider a salt, CH3 COONa whose concentration is Cs . The salt undergoes anion hydrolysis as follows:

  

                    

  Dissociation of the acid:

  

                       

  

  

  

  9.8. Buffer solution
  Activity 9.8

  1. Explain the term buffer solution.

  2. Analyze and differentiate the types of buffer solutions.

  3. Discuss how pH and pOH of a buffer solution can be calculated.

  9.8.1. Definition of buffer solution

  Buffer solution is a solution that resists a pH change when a small amount of a base or acid is added to it.

  A buffer solution consists of a solution of a weak acid and its salt with a strong base. This is called an acidic buffer. Examples of acidic buffers include: Ethanoic acid and  sodium ethanoate solution, carbonic acid and sodium hydrogen carbonate. 

  A buffer solution may also consist of a solution of a weak base and its salt of strong acid. This is called a basic or alkaline buffer. Examples of basic buffers include: Aqueous ammonia and ammonium chloride solution, aqueous ammonia and ammonium sulphate or ammonium nitrate.

   In general a buffer solution is an aqueous solution which involves the mixture of a weak acid and its conjugate base, or vice versa. 

  9.8.2. pH of buffer solution 

  The pH of a buffer solution or the concentration of the acid and base can be calculated using the Henderson-Hasselbalch equation. Henderson-Hasselbalch equation was given by Lawrence Joseph Henderson  (1878-1942) and Karl Albert Hasselbalch (1874-1962). 

  
  

  Taking negative logarithms to base 10 on both sides;
  

  
  

  This equation is known as the Henderson-Hasselbalch equation. Therefore, the pH of an acidic buffer depends on the relative concentration of the salt and the acid in the mixture.
  

  
  

  
  

  
  

  
  

  
  

  
  

  
  

  
  

  
  

  9.9. Preparation of buffer solutions of different pH
  Activity 9.9

  1. Write an expression to show Henderson-Hasselbalch equation.

  2. Explain how Henderson-Hasselbalch equation is used for the determination of the mass of salt to be dissolved when preparing an acidic buffer solution.

  3. State two methods of preparing buffer solutions of different pH.

  A buffer solution is a solution that resists a change in pH, because it contains species in solution able to react with any added acid or base.

  In general, preparing a buffer solution requires either a weak acid and a salt of the acid’s conjugate base or a weak base and a salt of the base’s conjugate acid.

  9.9.1. By mixing weak acid and its corresponding salt or weak base and its corresponding salt

  It is very important to prepare buffer solutions of known pH in the laboratory. Using the Henderson-Hasselbalch equation, there are two terms which determine the final pH of the solution that is pKa , whose value is responsible for the ‘coarse selection’ of pH and the ratio [conjugate base]/ [acid] that provides ‘fine tuning’ to the final pH.
  

  So, to prepare an acidic buffer solution of known pH, select an acid whose pKa is within the range of one unit of the desired pH. The ratio of salt to acid concentrations is then adjusted to achieve the desired pH.
  

  Example 

  Suppose you want to prepare an acidic buffer with a pH of 4.0. A suitable weak acid would be ethanoic acid CH3 COOH because its pKa is 4.8. The conjugate base is ethanoate ion CH3 COO- , which is provided by the sodium ethanoate salt, CH3 COONa. Ethanoic acid is available as a laboratory bench reagent with concentration of 1.0 mol dm-3. 

  The question is, what mass of sodium ethanoate should I add to the ethanoic acid to make this buffer solution? 

  Using the Henderson-Hasselbalch equation;

  
  

  Therefore, an acidic buffer solution of pH 4.0 will be prepared by dissolving 13 g of sodium ethanoate in 1.0 dm3 of 1.0 mol dm-3 ethanoic acid in a volumetric flask. Insert a calibrated pH meter to monitor the pH of the prepared acidic buffer solution. 

  Note that it is the ratio of acid to conjugate base that is important in determining the pH of the buffer solution not the concentrations. However, when more concentrated solutions are used, the buffer solution can efficiently react with added acid and base before becoming saturated.
  

  9.9.2. By partial neutralization 

  Buffers can also be prepared by the partial neutralization of a weak acid by a strong base through titration process.

  This is done by running excess weak acidic solution from the burette into a strong basic solution in the conical flask until half neutralization takes place. A weak acid is partially neutralized by the strong base to form salt of weak acid. This salt is formed together with excess of the weak acidic solution which makes the resultant solution a buffer solution. Do this while measuring the pH of the resultant buffer solutionwith a pH meter.

  For example reacting 40 ml of 1.0M propanoic acid (C2H5COOH) solution (pKa is 4.87) with 60.0 ml of 0.10 M sodium hydroxide solution (NaOH). This will produce a buffer solution consisting of propanoic acid and sodium propanoate of pH 4.13.

  Checking up 9.9

  The following materials and chemicals are provided by the technician laboratory.

  • 100ml Beaker,

  • 10ml measuring cylinder,

  • 50ml measuring cylinder,

  • Electronic weighing scale,

  • Calibrated pH meter,

  • Glass rod,

  • 100ml volumetric flask

  • Ethanoic acid,

  • sodium ethanoate and distilled water 

  Procedure

  1. Measure exactly 50 ml of distilled water to a 100 ml beaker.

  2. Using a measuring cylinder, add 5 ml of 0.3M ethanoic acid to the beaker.

  3. Then weigh exactly 0.3g sodium ethanoate (CH3COONa)

  4. Add a little of the sodium ethanoate at a time, stirring the mixture with a glass rod to dissolve.

  5. Insert the calibrated pH meter into the resultant solution in the beaker and measure its pH.

  6. Quantitatively transfer the buffer solution to a 100 ml volumetric flask.Add distilled water up to the mark. Cap and invert the flask twice to mix.

  
  

  9.10. Explanation of the working of buffer solutions
  Activity 9.10

  1. a. Explain what is meant by a buffer solution?

   b. How does buffer acts?

  2. Explain each of the following terms as used in the study of buffer solutions.

  a. Buffer capacity

  b. Buffer range

  3. Explain two applications of buffer solutions that are used in:

  a. Biological processes

  b. Manufacturing industry

  In a buffer solution, the weak acid and the conjugate base or weak base and its conjugate acid are responsible for controlling pH.

  9.10.1 Working process of an acidic buffer

  Consider a buffer solution made of ethanoic acid and sodium ethanoate solutions. Ethanoic acid is partially ionized because it is a weak acid whereas sodium ethanoate is fully ionized because it’s a strong electrolyte.

  

  When a small amount of a base is added, the added hydroxyl ions react with ethanoic acid to form water. This prevents an increase in the concentration of hydroxyl ions hence pH is kept constant.

  

  9.10.2. Working process of a basic buffer 

  Consider a buffer solution of aqueous ammonia and ammonium chloride. Aqueous ammonia is partially ionized while ammonium chloride is completely ionized because it is a strong electrolyte.

  

  Such a solution contains a few hydroxyl ions and a large proportion of ammonium ions from the salt. 

  When a small amount of a base is added, the added hydroxyl ions react with ammonium ions to form un ionized aqueous ammonia. This prevents any increase in the concentration of hydroxyl ions in the solution hence no change on pH.

  

  Similarly when a small amount of an acid is added, the added hydrogen ions react with un ionized aqueous ammonia to form water. This prevents an increase in the concentration of hydrogen ions thus keeping pH constant.

  

  9.10.3. Definition of buffer capacity and buffer range 

  Buffer capacity 

  This is a measure of the ability of a buffer solution to resist changes in pH when a base or acid is added. On addition of a base or acid to a buffer system, the effect on pH change can be large or small depending on the initial pH of the solution and it’s ability to resist that pH change. 

  Buffer capacity is therefore defined as the number of moles of acid or base which when added to one litre of a buffer solution changes its pH by 1. 

  Buffer capacity (β) has no units since it is a ratio of number of moles of acid or base added, to change in pH of buffer solution. 

  Mathematically;

  

  Buffer capacity is the efficiency of the buffer solution to control pH. The acid added to one litre of buffer solution changes its pH by only 1 unit.The buffer capacity is maximum at pH=pKa where [acid]=[conjugate base]. 

  Buffer range 

  Buffer range is the pH range within which the buffer solution is effective. 

  Buffer range corresponds to the change in pH of about ±1. For example, acetic acid/ sodium acetate buffer works at optimum pH of 4.8. This means that it would work from pH 3.8 on addition of an acid to pH 5.8 when a base is added. Beyond this pH range, the buffer has no capacity to buffer the solution. 

  The effective buffer range of buffer solutions is different and depends on the acid or base dissociation constant.

  

  

  9.10.4. Applications of buffer solutions 

  pH of various solutions has to be controlled either in industries or in cells of living organisms because any slight change may greatly affect the functioning of the whole system.

   In biological processes

  • Buffer solutions are used to maintain the pH of human blood constant at 7.4. Intravenous injections must be correctly buffered so that they do not change the pH of blood in humans. 

  • Proteins in living organisms act as buffers by controlling pH in body cells. They are composed of amino acids linked together in long chains. Each amino acid has two functional groups; the amino group and acidic carboxyl group. The carboxyl group donates hydrogen ion when pH in body cells is high (alkaline). Amino group accepts hydrogen ion from body cells when pH is low (Acidic). 

  • Buffers keep the correct pH for enzymes to work in many organisms. Enzymes work best in a specific range of pH. If this pH is not controlled, the rate of enzyme activity slows down or enzymes stop working due to denaturation. 

  • In bacteriological research, culture media are generally buffered to maintain the pH as bacteria required a constant pH all the time to grow. 

  In agricultural processes 

  • Plants grow well in soils with a narrow pH range. Some soils become acidic due to acidic rains which is a serious problem to plant or crop growth. Organic matter and mineral salts in a fertile soil act as buffers. It is important to maintain

   • pH of the soils for proper growth of vegetation and soil microorganisms.

   In natural systems

   • Water bodies such as lakes, rivers and streams are important habitats for aquatic organisms e.g. fish and young amphibians. They should have a stable pH to ensure survival of these organisms. Otherwise extreme pH may cause physical damage to the gills and fins of fish.

  In industries 

  • Buffer solutions are used in fermentation to control pH changes such that anaerobic fermentation bacteria are not killed. This prevents the solutions from becoming too acidic and spoiling the product. 

  • Buffer solutions are used in manufacture of cosmetics whose pH must be controlled to prevent adverse effects on body cells. 

  • They are used in the production of pharmaceutical drugs to prevent deterioration when the drug is administered or stored. Buffers ensure stability and clinical effectiveness of the drugs. 

  • Dyes used in textile industries are buffered in order to maintain colour strength in different fabrics after production. 

  • Buffers are used in leather industries to control pH during tanning and dyeing. This gives a product of fine texture and colour.

  

  END UNIT ASSESSMENT
  

  SECTION A: Multiple choice questions

  

  

  

  

  

  

  

• UNIT 10: INDICATORS AND TITRATION CURVES

  Key unit competence

  To be able to relate curves to the type of acid orbase titrated, properly choose and use indicators in acid and base titration

  Learning objectives

  At the end of this unit , students will be able to:

  • Explain how the indicator works;

  • Explain what is meant by the pH range of indicator;

  • State the criteria for the selection of acid-base indicator for the use in titrations;

  • Describe the changes in pH durind acid/base titrations;

  • Perform experiments to show that the effectivenees of different indicators is related to the pH changes which occur during titration;

  • Draw and interpret titration curves for various acid/base titrations;

  • Match the titration curve to the type of acid and base titrated;

  • Interpret pH curves of different titration reactions.

  Introductory Activity

  1. Explain the difference between acid and base.

  2. Discuss the relationship between end point and equivalence point during titration

  10.1. Definition of acid-base indicator

  Activity 10.1Different types of indicators used in the Laboratory are given below.

  a. Phenolphthalein indicator.

  b. Methyl orange.

  c. Litmus papers.

  d. Methyl red.

  e. Reagents of acid and base e.g. NaOH (aq) and HCl (aq)

  f. Carry out different tests on acid and base above using different indicators and write down your observations.

  An indicator is a substance which changes colour according to the nature of solution. An acid have indicator changes the colour according to wheather the medium is acidic or basic. Therefore it can be used to know if a solution is acidic or basic. It can also be used to determine the end point in a titration.

  The substances that change colour when the acidity of the solution changes are known as acid-base indicators. A very common indicator used is litmus, which is obtained from lichen. Litmus paper is prepared by soaking absorbent paper with litmus solution and then drying it.

  Acid-base indicators are used to find out the equivalence point of a titration.They change their colour within a certain pH range.

  The titration is a process commonly used to determine the concentration of an acid or a base. During the titration, there are two main key points:

  a. The amount of acid or a base mixed together must be controlled and measured

  b. There must be some way of determining when enough acid or a base has been added to neutralize the solution.

  The equivalence point informs us when enough acid or base has been added to neutralize the solution. The pH changes considerably near the equivalence point. A titration curve can be plotted to show the changes in pH.

  In the previous units, it has been seen that acids are substances which donate a proton and become electron acceptors. Bases are proton acceptors and electron donors. During the titration process the concentration of unknown (acid or a base) is determined. Indicators are weak acids whose conjugate base is of different color.Indicators are weak organic acids or bases which changes color whenever there is a change in pH.

  Indicators are used in titration solutions to indicate the completion of the acid-base reaction. Universal indicator is a mixture of indicators which give a gradual change in color over a wide pH range. The pH of a solution can be approximately identified when a few drops of universal indicator are mixed with the solution.

  Checking up 10.1

  Explain why each of the following procedures is used during acid –base titration in the laboratory.

  1. After cleaning the burette some of the solution to be put inside it is added to the burette and rolled around inside them and then discarded.

  2. You constantly swirl the reaction vessel as you perform the titration.

  3. You should always keep your eye in line with the meniscus of the burette when reading the burette volume.

  10.2. The pH range of indicators
  

  Activity 10.2

  1.Discuss your understanding of ionization of both weak acids and weak bases.

  2. Discuss ionization of both strong acids and strong bases.

  3. Explain the term conjugate base.

  Weak acids or weak bases are classifies as weak electrolytes which ionize partially in solution while strong acids or strong bases as strong electrolytes ionize completely in solution.

  

  

  

  At this point the equilibrium shifts to the right.

  Under low pH (acidic) the concentration of H+ ions is high and the equilibrium

  shifts to the left as predicted by Lechatelier’s principle.

  

  At a low value of pH, a weak acid indicator is almost completely in the HIn form, the color of which predominates. As the pH increases, the intensity of the color of HIn decreases and the equilibrium is pushed to the right. Therefore, the intensity of the color of In- increases. An indicator is most effective if the color change is distinct and over a small pH range.

  

  

  

  A good indicator for a specific acid-base titration has an endpoint with a pH at or near the pH of the equivalence point.

  10.2.3. Choice of indicators

  It is important to choose the correct indicator to be used. An indicator is appropriate for a given titration if the rapid change of pH at equivalence overlaps the pH range of the indicator. The table 10.2 summarizes some types of titrations and the choice of appropriate indicators for them

  

  The pH at the equivalence-point must be known for the proper selection of an indicator

  Note: Choose an indicator whose range lies on vertical section of the pH curve.

  Checking up 10.2

  1. Write the equation to show ionization of the weak acid HClO₂ and the expression for its equilibrium constant.

  2. How do you know that the Ka for any weak acid is a small number?

  10.3. Acid-base titration curves

  Activity 10.3

  Explain how Acid—Base indicators are used to determine the equivalence point of a titration.

  A titration curve is normally a plot of pH versus volume of titrant. It shows how the pH of an acid or base change as it is neutralized. For example as a base is added to acid, a gradual increase in pH will occur until the solution gets close to the equivalence point. Near the equivalence point, a rapid change in pH occurs. At the equivalence point, equivalent amounts of acid and base have been added and the pH will reflect which species are present. Beyond the equivalence point, when excess base is added an abrupt increase of PH is observed as a steep increase of the titration curve.

  Titrations can result in acidic, basic, or neutral solutions at the equivalence point.

  We can distinguish the following titration curves:

  a. Strong acids – strong base titration curve where strong acids and bases are completely converted to H3O+ and OH- in water solution. This titration results in a neutral solution at the equivalence point.

  b. Strong acids titrated with weak bases result in acidic solutions at the equivalence point.

  c. Weak acids titrated with strong bases result in a basic solution at the equivalence point.

  d. Weak base and weak acid

  Note: The following points are considered to plot pH titration curves:

  • use graph paper

  • orientation of axis

  • naming of axis

  • graduation of axis

   joining/matching pointsa. Titration curve for a strong acid neutralised by a strong base

  

  

  

  

  

  

  

• UNIT 11: SOLUBILITY AND SOLUBILITY PRODUCT OF SPARINGLY SOLUBLE SALTS

  UNIT 11: SOLUBILITY AND SOLUBILITY PRODUCT OF SPARINGLY SOLUBLE SALTS

  Key unit competence

  To be able to calculate the solubility product constant of sparingly soluble salts and deduce the applications of common ion effect in the industry.

  Learning objectives

  At the end of this unit , students will be able to:

  • Define the term solubility product Ksp;State and explain the factors that affect the solubility of sparingly soluble salts;

  • State and explain the applications of solubility product;

  • Explain common ion effect on the solubility of sparingly soluble salt;

  • Explain the relationship between kidney stone formation and solubility and the solubility product;

  • Explain the applications of the solubility product and the common ion effect;

  • Perform a simple experiment to determine the solubility product of sparingly soluble salt;

  • Write the equations of dissociation and Ksp expression for sparingly soluble salts;

  • Calculate the molar concentration of ions and Ksp values for sparingly soluble salts;

  • Relate the solubility product principle to the selective precipitation of sub-stances;

  • Use the values of Ksp and Qc to predict if a mixture of solutions will form a precipitate or not;

  • Relate the common ion effect to the solubility to the sparingly soluble salt.

  Introductory activity 11:1.

  1. Identify substances which are soluble in water from the following list CuSO₄, Pb (NO₃)₂, CaCl₂ Agl.

  2. Explain what happens when ionic salts dissolve in water.

  This unit introduces us to solubility equilibria which are found in saturated solutions of slightly soluble ionic salts. We will discuss the difference between solubility equilibria and dissociation equilibria and we will clarify some possible points of confusion in solving problem related to equilibria products.

  11.1. Definition of solubility and molar solubility

  Activity 11.1

  1. Differentiate between solubility and Molar solubility and state the units.

  2. What is the difference between solubility equilibria and other equilibria systems
  

  Solubility

  In the previous units, it has been mentioned that a solution is made by a solute which is dissolved in a solvent. Similarly, the solubility is a chemical property referring to the ability for a given substance, the solute, to dissolve in a solvent. In this section, we will deal with solubility which is a term that refers to the maximum amount of solid (either in moles or grams) that actually does dissolve in a solvent at equilibrium, producing ions; this amount can be calculated for a particular solid.Solubility may be considered to be an equilibrium between solid and ions in solution.

  In general, the solubility of a solid in water is given by the maximum number of grams or number of moles of the solid that will dissolve in 100g of water at a given temperature. Generally the solubility of salts increases with increased temperature.

  Solubility is most fundamentally expressed in molar (mol l–1 of solution) or molal (mol kg–1 of water) units. A solution that contains the maximum possible amount of the solute in said to be “saturated”.

  Molar solubility

  Molar solubility can be expressed as the number of moles of a solute that can be dissolved per liter of solution at saturation.

  Most ionic salts are soluble in water while some are sparingly soluble (often called insoluble)

  • All Nitrates,

  • Generally Group I carbonates, except lithium carbonate are soluble but group II carbononates are sparingly soluble.

  • All Chlorides except lead chloride, silver chloride and mercury chlorid

  • Sodium, Potassium, and Ammonium Sulphate are also soluble, the rest of the sulphate are insoluble.

  Checking up 11.1

  1. Insoluble substances such as AgCl, Ca₃(PO₄)₂ partly dissolve and they form a heterogenous mixture. Solubility of solutes often increases with increased temperature. Observe the graph below and answer the following questions.

  

  2. Explain the nature of the graph.

  3. Identify the letter corresponding to:

  a. Salt whose solubility is not affected by an increase in temperature.

  b. Salt which dissolves highly with increased temperature.

  c. Salt which moderately dissolve with increased temperature.

  11.2. Unsaturated, saturated and super saturated solutions

  Activity 11.2

  1. Dissolve two spatula full end of Copper II sulphate in 100cm3 of distilled water. State what is observed 2. Without changing the volume of water add extra spatula full of copper II sulphate crystals stepwise, state what is observed. Keep on stirring until there is no further change.

  All the beginning of the activity, you noticed that all the amount of the salt added to water dissolved completly. If more solid as added to an unsaturated solution, it dissolves until the solution is saturated (at which point there is solid in equilibrium with the solvated ions).

  A saturated solution is one which contains the maximum amount of dissolved solid at a particular temperature in the presence of undissolved solute. A solution will be saturated when the solid is in equilbrium with its ions dissolved.

  Checking up 11.2

  The term “solubility equilibria” refers to the kind of equilibria that exists in saturated solution of slightly soluble ionic solids.

  Explain what will be the concentration of the ions in a saturated solution.

  A super saturated solution contains more solute than a saturated solution. This kind of solution is instable and any disturbance, such as shaking, will cause the excess solute to precipitate, Adding a crystal of solute to a super saturated solution, the precipitation will be observed and will cause the crystal to grow bigger in size until the solubility equilibrium is achieved.

  11.3. Equations of the dissociation of sparingly soluble salts in water

  
  

  Activity 11.3

  Write equations to show the dissociation of the following substances in water.

  1. Barium sulphate.

  2. Silver Iodide.

  3. Silver carbonate.

  4. Calcium phosphate.

  A sparingly soluble solute is one which slightly dissolves in a given solvent. It partly dissociates into component ions.

  The main examples of sparingly soluble salts include; Barium sulphate, BaSO₄; and Silver chloride ,AgCl and lead(II) chrolide, Pbcl₂
  

  As stated previously, when salts dissolve in water, they dissociate into ions. H2_O

  
  

  Checking up 11.3

  1. Write balanced equations for the dissociation of some sparingly soluble ionic solts.

  2. Write down the ionic equations for each of the reactions in 1 above
  

  11.4. Definition of solubility product ksp

  Activity 11.4.3.
  

  Differentiate solubility and solubility product:4. State the units of Ksp.
  

  In the previous sections, we have seen that solubility of a solid is expressed as the concentration of the “dissolved solid” in a saturated solution. A solution is made of solute and solvent.

  The solubility product is the equilibrium constant expressed in terms of concentrations of the ions produced from a sparingly soluble solid in contact with a saturated solution.

  The equilibrium constant for the system is given the symbol (Ksp) where the sp added to k tell us that this equilibrium constant is a solubility product. Equilibrium is set up between the undissolved solid and the hydrated ions in solution. A precipitate will appear if the solubility product is exceeded for the system containing silver chloride.

  Examples of expression of Ksp for some compounds

  

  There is no denominator in the expression for the solubility product because the denominator is a pure solid and pure solids and liquids are never included in equilibrium constant expression.

  A solubility product is generally a special example of a heterogeneous equilibrium constant. It involves more than one phase; that is solids in contact with liquids.

  ii. Solubility product for lead II sulphate (PbSO₄)

  Suppose you made a saturated solution of lead (II) sulphate PbSO4 by shaking the solid with water until no more would dissolve. Lead (II) Sulphate is almost insoluble in water and a white solid would be observed suspended in the water; after sometime the solid will settle to the bottom.Lead (II) Sulphate is an ionic compound and some Lead (II )ions (Pb²⁺) and Sulphate ions(SO₄^2-) will break away from the lattice and go into solution; others which had broken off previously will return to attach themselves to the solid; this results into an equilibrium as follows.

  

  iii. Solubility product for lead II iodide (PbI₂)

  

  Note: In brief, the formula for the solubility products is the product of molar concentration of the respective ions, each concentration raised to the power of the stochiometric coefficient in the dissociation equation.Solubility products are only constant at a particular temperature which is usually 298K.

  Checking up 11.4

  Express the solubility product Ksp for the following sparingly soluble salts.

  a. AgI(s)

  b. CaSO₄(s)

  c. Ag₂CO₃ (s)

  d. Li₃PO₄(s).

  11.5. Relationship between solubility(s) and solubility product (Ksp)

  
  

  Activity 11.5

  
  

  1. Explain the meaning of the term solubility equilibrium.

  2. What is meant by the term solubility product?
  

  Solubility equilibrium refers to kind of equilibrium that exists in saturated solutions of sparingly soluble ionic salts.Solubility is normally expressed in moldm^-3 or (gdm^-3).The term solubility product refers to the numerical value of the equilibrium constant for the equation that represents the substance dissolved in water.

  Determination of the solubility product from solubility

  The solubility can be denoted by using ‘s’ and solubility product is expressed by Ksp

  The the term solubility product constant suggests that Ksp is related to the solubility of ionic solute . However ,this does not mean that Ksp and molar solubility ,molarity of solute in a saturated solution are equivalent .It means that one can determine Ksp,from molar solubility from Ksp.

  Consider the following reaction:

  

  Checking up 11.5

  A saturated solution of a sparingly soluble salt has the following equilibrium:

  

  a. Write the solubility product expression Ksp for this system.

  b. Using ‘s’ as solubility of the salt in moles per liter calculate Ksp for this system.

  11.6. Calculations involving solubility product

  In solubility equilibria calculations, it is usually expressed as grams of solute per liter of solution. Since Ksp is expressed in molar concentrations, solubility expressed in grams per litre must be converted into molar solubility.

  Activity 11.6

  Write solubility product expressions,Ksp for each of the following solubility equilibria:
  

  

  
  

  Molar solubility, solubility and solubility product all refer to a saturated solution.

  Worked examples:

  1. The solubility of calcium carbonate CaCO₃, at 298K is 6.9×10-3 moldm^-3. Calculate the solubility product at this temperature.

  Solution

  
  

  Key point

  Every mole of calcium carbonate which dissolves gives 1 mole of calcium ions and 1 mole of carbonate ions; in solution; so 6.9×10 ^-3 moles of calcium carbonate dissolves in 1dm3of solution, then there will be 6.9×10^-3 moles of calcium carbonate ions and 6.9×10^-3 moles of CO₃^2- ions in dm³ of the solution.
  

  
  

  2. The solubility of calcium sulphate CaSO₄ at 298K is 0.67g/dm³. Calculate the solubility product at this temperature. (O=16, S=32, Ca=40)

  solution

  

  Key point: number of moles=mass(g)/molar mass(gmol^-1)

  The concentration is given in g/dm³; Convert it to mol/dm³

  1 mole of CaSO₄ weighs:40+32+(16×4)=136g

  0.67 corresponds to 0.67/136=4.93×10^-3 mol/dm³

  The solubility of calcium sulphate is 4.93×10^-3 mol/dm³

  Each mole of calcium sulphate that dissolves produce 1 mole of Ca²⁺ ions and 1 mole of SO₄^2- ions in solution.

  

  3. The solubility of lead II chloride, PbCl₂ is 0.016moldm³ at 298k. Calculate the solubility product at this temperature.

  Solution

  

  Key point

  The ratio of the ions in the compound is 1:2; each mole of lead( II) chloride produces 1 mole of lead II, and [Pb²⁺]=0.016 moldm^-3. However, each mole of lead (II) chloride produces 2 moles of chloride ions in solution. If 0.016 mol of lead (II )chloride dissolves, there will be twice this amount of chloride ions present.

  

  4. The solubility of calcium phosphate, Ca₃(PO₄)₂ is 7.7× 10-4g/dm³ at 250c. Calculate the solubility product at this temperature. (O =16, P =31,Ca=40)

  Solution

  

  Key point

  Convert the concentration in g/dm³ to mol/dm³

  Number of moles= mass(g)/ molar mass(g)

  1 mole of Ca₃(PO₄)₂ weighs 310g.

  Concentration in moldm^-3=7.7×10-4/310=2.48×10^-6mol/dm³

  Each mole of calcium phosphate that dissolves produces 3 moles of calcium ions in solution and 2 moles of phosphate ions.

  

  Calculating solubility from solubility product

  Examples

  1. Calculate the solubility in mol/dm³ of silver chloride, AgCl, at 298K if its solubility product is 1.8×10^-10 mol²dm^-6

  Solution

  

  Key points

  For every mole of silver chloride that dissolves, the solution will contain 1 mole of Ag⁺(aq) and 1 mole of Cl-(aq) so if “s” moles dissolved, the solution will contain “s” moles of each ion.

  [Ag⁺]= s moldm^-3

  [Cl^-]= s moldm³

  Ksp= [Ag⁺][Cl-]

  
  

  1.8×10^-10=s x s

  
  

  

  S =1.3×10^-3mol/dm³

  2. Calculate the molar solubility of PbSO₄ (lead II sulphate), given its solubility product equal to 1.6×10^-8.

  Solution

  

  Key point

  One mole of lead II sulphate dissolves to produce 1 mole of a Pb²⁺ and 1 mole of SO₄^2-ions in solution.[Pb²⁺]=smoldm-3[SO₄^2-]=smoldm-3Where “s”= moldm^-3 PbSO₄ that dissolved (the molar solubility)

  Ksp=[ Pb²⁺][ SO₄^2-]=1.6×10^-8(s)(s)= 1.6×10^-8

  

  Molar solubility= 1.26×10^-4M

  3. Calculate the solubility in mol/dm³ of silverI sulphide, Ag₂S at 298K if its solubility product is 6.3×10^-51 mol³dm^-9

  Solution

  

  Key point

  
  

  For every one mole of silver I sulphide that dissolves, the solution will contain 2 moles of Ag⁺ and 1 mole of S^2-ions. We will call the solubility of silver I sulphide “s” moldm^-3.

  

  

  4. Calculate the solubility in gdm^-3 of chromium III hydroxide, Cr(OH)₃ at 250C if its solubility product is 1.0×10-33mol⁴dm^-12(H=1, O=16,Cr=52)

  Solution

  

  Key points

  
  

  Work out the solubility in moldm^-3Number of moles=mass(g)/ molar mass(g)1mole of Cr(OH)₂ produces 1 mole of Cr³⁺(aq) and 3 moles of OH^- ions

  
  

  

  Checking Up 11.6

  1. Calculate the molar solubility of Sr₃(AsO₄)₂ given its solubility product equal 4.29 x 10^-192. Given that the molar solubility of Ag₂CO₃ is equal to 1.27x10-4M. Calculate its solubility product.3. The Ksp of Mg(OH)₂ is 1.8 x 10^-11

  a. Calculate the solubility of Mg(OH)₂ in pure water.

  b. Calculate the molar solubility of Mg(OH)₂ in a solution of PH of 11.22.

  11.7. Definition and calculation of ionic product (QC)

  Activity 11.7

  a. Explain the meaning of the term ionic product.

  b. State the difference between ion product and solubility product.11.7. Definition and calculation of ionic product (QC)

  The ionic product (Q) of salt is the product of the concentrations of the ions in solution raised to the same power as in solubility product expression. Whereas the solubility product Ksp describes the equilibrium concentrations, the ion product describes concentrations that are not necessarily equilibrium concentrations. Ksp is applicable to saturated solutions only, whereas ion product is applicable to all types of solutions of any concentration.

  
  

  The ion product tells chemist whether a precipitate will form when solutions of two soluble salts are mixed.

  
  

  i. Q>Ksp, the solution is supersaturated and the ionic solid will precipitate

  
  

  ii. Q<Ksp, the solution is unsaturated and more of the ionic solid, if available will dissolve. No precipitate occurs

  
  

  iii. Q=Ksp, the solution is saturated, at equilibrium. Precipitate will not form, the rate of dissolution is equal to the rate of precipitation, no net change in the amount of dissolved solid will occur.
  

  
  

  
  

  The process of calculating the value of the ion product and comparing it with the magnitude of solubility determining whether a solution is unsaturated saturated or supersaturated.

  
  

  Strategy

  
  

  i. Write the balanced equilibrium equation for precipitation reaction and the expression for Ksp.

  
  

  ii. Determine the concentrations of all ions in solution when the solutions are mixed and use them to calculate the ion product(Q)

  
  

  iii. Compare the values of Q and Ksp to decide whether a precipitate will form.
  

  Worked examples

  1. Will a precipitate of lead II chloride be formed if 10 cm3 of 0.10 mol dm^-3lead II nitrate solution, Pb(NO₃)₂ is mixed with 10 cm3 of 0.20 mol dm-3hydrochloric acid, HCl? Ksp (PbCl₂) =1.6×10^-5 mol³dm^-9 at 298K.

  Solution

  

  When the solutions are mixed, all the ion concentrations will decrease. In this case each solution is being diluted from10cm³ to a total volume of 20cm³; so, each is diluted by a factor of 2.

  

  This answer is bigger than that solubility product. Because Q>Ksp we predict that PbC_l2 will precipitate when the two solutions are mixed PbCl₂ will continue to precipitate until the system reaches equilibrium; which occurs when;

  

  2. Will a precipitate of calcium hydroxide; Ca(OH)₂ form if 5.0cm3 of 0.05moldm-3sodium hydroxide solution; NaOH, is added to 5.0cm³ of 0.05moldm^-3 calcium chloride solution. CaCl₂? Ksp(Ca(OH)₂) = 5.5 × 10^-6 mol³ dm-9 at 298K

  Solution

  

  Mixing; dilutes the solutions by a factor of 2 ion concentration before reaction takes place is

  

  Q>Ksp calcium hydroxide will precipitate when the two solutions are mixed until the ion concentrations are sufficiently reduced ;( i.e. until the system reaches equilibrium)

  3. Will a precipitate of calcium hydroxide form if 5.0 cm³ of ammonia solution con-taining OH^- ions with concentration of 2.0 × 10^-3 mol/dm³ is added to 5.0 cm³ 0f 0.05 mol dm³ is added to 5.0 cm³ of 0.05 mol dm^-3 calcium chloride solution. CaCl₂? Ksp (Ca(OH)₂) = 5.5×10^-6 mol³dm^-9 at 298K.

  Solution

  

  11.8. Predicting precipitation reactions using the ionic product and Ksp

  Activity 11.8.

  Distinguish between heterogeneous and homogeneous mixtures.
  

  Homogeneous mixtures appear in one phase, Heterogeneous mixtures are in more than one phase.Sparingly soluble salts form heterogeneous mixtures involving formation of precipitates.Homogeneous matures do not form precipitates.

  A solubility product is a special example of a heterogeneous equilibrium constant, it involves more than one phase, a solid in contact with a liquid.Reagents such as Sodium hydroxide NaOH, and Ammonium hydroxide NH4OH can precipitate out insoluble salts from their solutions Examples include:

  

  

  

  Checking up 11.8

  
  

  Ammonia is a weak base which reacts with water to produce hydroxide ions. A 1.0 moldm^-3 solution of ammonia has [OH^-] = 4.2×10^-3 mol dm^-3. Ammonia solution was added to 10 cm³ 0.10 mol dm^-3 solution of magnesium nitrate, calcium nitrate, strontium nitrate and barium nitrate. (All the nitrates have the formula X(NO₃₎₂. The experiments were then repeated with 10 cm3 samples of a 1.0 mol dm-3solution of sodium hydroxide, where [OH^-] = 1.0 mol dm^-3. In which cases will a precipitate be formed?

  Solubility products (in mol³dm^-9);

  
  

  
  

  11.9. Separation of ions by fractional precipitation

  
  

  Suppose you are given an aqueous solutions containing two ions: Ag⁺ and K⁺. How can you proced to separate the two cations. Ions can be separated from each other basing on the solubility of their salts in water. This is done by using a reagent that forms a precipitate with one or a few of the ions in solution (fractional precipitation).

  
  

  Examples

  1. A solution containing both Ag⁺ and Cu2⁺If hydrochloric acid solution is added to this solution, AgCl (Ksp=1.8×10^-10) precipitates, while Cu₂⁺ remains in solution because CuCl₂ is soluble. The reagent HCl forms a precipitate with Ag⁺.

  

  2. Fractional precipitation of a carbonate (CO₃^-2) and a Chloride (Cl^-).A solution suspected to contain chloride and a carbonate is mixed with a solu-tion of Silver nitrate A solution of Silver nitrate precipitates out both the Carbonate and a Chloride forming Silver carbonate and Silver Chloride respectively.We can distinguish both the carbonate and a chloride by using concentrated nitric acid.

  Observations:Carbonate reacts to give carbon dioxide gas. There is no observable change with a Chloride.A solution contains 1.0 x10^-2 M Ag+ and 2.0 x10^-2 M Pb²⁺ When CI- ion is added to the solution, both AgCl (Ksp =1.8x10^-10) and PbCl₂ (Ksp =1.7x10^-3) precipitate from the solution.

  What concentration of Cl^- ions is necessary to begin the precipitation of each salt, and which salt precipitates first?

  Solution

  The salt requiring the lower concentration of CI^- ions will precipitate out first. For AgCl we have: Ksp [Ag⁺][Cl^-] = 1.8 x 10 ^-10 Because [Ag⁺] =1.0x10 ^-10 M, the greatest concentration of Cl^- ions that can be present without causing precipitation of AgCl can be calculated from the Ksp expression

  

  Checking up 11.9

  1. State one importance of fractional precipitation.

  2. Write ionic equations to show precipitation of silver shloride and silver carbonate when a solution of silver nitrate is added to their saturated solution.

  Activity 11.10

  Give example of pairs of ionic salts with common ions.

  11.10. Common ion effect and solubility

  
  

  It is known that the solubility of a sparingly soluble ionic substance is obviously decreased in a solution of another ionic compound when the two substances have an ion in common. This can be explained by Lechatelier’s principle. Common ion effect is the decrease of solubility of a sparingly soluble salt caused by the presence of a common ion.

  
  

  For example, molar solubility of BaSO₄ in pure water is 1.1x105M; but if BaSO₄ is dissolved in an aquous solution arleady containing SO₄^2- or Ba²⁺ ion, its solubility will be lower than 1.1x10^-5M

  
  

  

  If sulphuric acid (H₂SO₄) is added to the solution of BaSO₄, the concentration of SO₄^2- ions increased and shifts equilibrium to the left side of the reaction. As a result, barium gets precipitated as BaSO₄(s). H₂SO₄ is a strong acid and it dissociates completely as indicated in the equation:

  The sulphate ion (SO₄^2-) is the common ion among these two reactions thus when its concentration increases due to dissociation of sulphuric acid it shifts equilibrium to the left side of the reaction resulting in the precipitation of barium as BaSO4(s).
  

  

  This phenomenon of “Common Ion Effect” is widely used for the complete precipitation of soluble or sparingly soluble salts.
  The solubility product expression tells us that the equilibrium concentration of the cation and the anion are inversely related. When the concentration of the anion increases; the maximum concentration of cation needed for the precipitation to occur decreases; and vice versa, because why Ksp is constant. Consequently, the solubility of an ionic compound depends on the concentration of the other salt that contain the same ions as the given salt.

  
  

  Calculations on the common ion effect

  
  

  
  

  

  
  

  

  Checking up 11.10
  

  
  

  The Ksp of Ca(OH)₂ is 5.5x10-6
  a. Calculate the solubility of Ca(OH)₂ in pure water.
  

  b. Calculate the solubility of Ca(OH)₂ in 0.77M solution of Ba(OH)₂

  
  

  11.11. The pH and solubility
  

  
  

  Activity11.11
  

  1. Explain the meaning of the term pH
  

  2. Explain how pH affects the solubility of a solute.
  

  
  

  The term pH refers to the concentration of [H+] ions in solution.
  

  
  

  The pH can be found if the [H+] ions in a solution is known.

  
  

  Changes in pH
  

  
  

  Some sparingly soluble salts produce ions such as, PO₄^3-, C₂O₄^2-, OH-, CO₃^2- which behave like bases; these can combine with hydrogen ions to form weak and poorly dissociated acid molecules.

  
  

  

  This reduces the concentration of H+ ions in the solution and for such salts their solubility will always increase  with a decrease of PH or addition of a dilute acid to the solution, the excess H+ ions will combine with the base-like ions thereby reducing their concentration and allowing the dissolution of more salt.

  
  

  Checking up 11.11
  

  
  

  1. The solubility product of silver oxalate Ag₂C₂O₄ is given by the expression, Ksp = [Ag+] ² [C₂O₄^2-]. Explain how the solubility of silver oxalate would be affected if a few drops of the following solutions are added.

  
  

  a.  Concentrated ammonia solution
  

  b. Sodium oxalate
  

  
  

  2. Strontium hydroxide is sparingly soluble in water.
  

  
  

  a.  Write the equation for the solubility of strontium hydroxide in water.
  b.  Write an expression for the solubility product (Ksp) of strontium hydroxide.

  c. Sodium hydroxide was added to a saturated solution of strontium hydroxide.
  

  
  

  i. State what happened to the solubility of strontium hydroxide in water.

  
  

  ii. Explain your answer in (i) above.

  
  

  11.12. Complex ion formation and solubility
  

  
  

  Activity 11.12
  

  
  

  3. Explain what you understand by the following terms:
  

  
  

  a. Complex ion
  

  b. Ligand
  

  c. Coordination number
  

  d. Hydrated cation
  

  e. Lewis base
  

  
  

  2. Arrange the following ligands in order of stronger Lewis base,
  

  
  

  NH₃, Cl-,     H₂O,     CN

  
  

  

  

  Checking up 11.12
  

  1. State the factors that favour the formation of complex ions.
  

  2. Can the shape of the complex ion change after ligand displacement?

  Explain your answer.
  

  
  

  11.13. Applications of solubility product
  

  
  

  Activity 11.13
  

  According to your experience or by doing research, give example of application of the solubility product.

  
  

  Some areas where solubility product is used:

  
  

  a. Volumetric analysis:
  The concentration of Chloride ions in a solution of soluble Chloride salt can be determined by titration with a standard solution of silver nitrate Potassium chromate is used as an indicator. A  white precipitate of silver Chloride  forms first due to lower solubility of silver Chloride , the end point is indicated by the formation of a red precipitate of silver chromate   This can be seen from  the solubility product
  

  
  

  

  
  

  Checking up11.13

  
  

  The solubility product of Bi₂S₃ is listed as an unimaginably small number, 1x10^-97.
  

  
  

  1. Calculate the solubility ofBi₂S₃ in pure water.
  

  2. Suppose you tried to dissolve some Bi₂S₃in a solution of 0.50M Na2S how many grams of Bi₂S₃ would theoretically dissolve in one litre of this solution?

  
  

  END UNIT ASSESSMENT
  

  
  

  1. Consider two compounds A and B where A has a solubility product of 3.3x10^-3 while B has a solubility product of 4.4x10^-7.Can we conclude that compound has ahigher Molar solubility than B. Explain your answer?
  

  
  

  2. Write the solubility product expressions for the following solubility equilibria.

  
  

  

  3. Consider a compound with the formula   L₃X₄   that dissolve to produce L⁴⁺and X^3-ions in solution if the solubility product of this compound is 8.2x10^-24. Calculate the molar solubility of the compound.
  

  
  

  4. Explain what is meant by the term common ion effect and how is it accounted for by applying Lechatelier’s principle.
  

  
  

  5. Calculate the molar solubility of Ba(IO₃)² given that its Ksp is equal to 1.5x10^-9.

  
  

  
  

  
  

  
  
• UNIT 12:ELECTROCHEMICAL CELL AND APPLICA TIONS
• UNIT 13 FACTORS THAT AFFECT THE RATE OF REACTIONS
• UNIT 14:RATE LAWS AND MEASUREMENTS

  Key unit competency:

  Measure the rates of reaction and formulate simple rate equations using the experimental results.

  Learning objectives:
  By the end of this unit the learners should be able to:

  • State and explain kinetic conditions for a chemical reaction to take place.
  • Explain the effect of the temperature and catalysts on the rate of the reaction using Boltzmann distribution of energies (and of collision frequency).
  • Differentiate between SN1 and SN2 mechanisms.
  • State and explain the rate determining steps for multi-step reactions.
  • Deduce the order of reaction from appropriate experimental data.
  • Calculate the initial rates and the rate constants of reactions from the experimental data.
  • Perform practical activities to show how different reactions have different rates.
  • Interpret the graphs which show the change in activation energy with the catalyst.
  • Calculate the half-life of chemical reaction.
  • Perform practical activities to measure the rates of reaction by observing the changes in physical quantities (e.g. volume, mass and colour change).
  • Use collision theory to predict if the reaction will go faster or slower.
  • Construct rate equations of the form (Rate = k [A]n[B]m) limited to simple cases involving zero,

    first and second order reactions.
  • Interpret graphs of concentration against time and those of concentration against rate for zero

    and first order reactions.
  • Develop a spirit of team work, analysis, and self-confidence while discussing exercises

    and performing the experiments.
  • Appreciate the contributions of Arrhenius and Boltzmann on the effect temperature and activation

     energy of different substances and number of molecules

  Introductory activity

  One of the mission of chemistry consists in making substances from other substances. Referring to the following six figures; some processes are observed during a chemical reaction from each Figure.

  Observe these figures carefully and answer the related questions.

  

   

  (1)                                                                                              (2)

   

                   (3)                                                                            (4)

   

                                        (6)

                   (5)

  i. Suggest what is happening in each image.

  ii. Classify the processes into slow or fast processes, and explain why?

  14.1 Theories of reaction rates

  Activity 14.1

  1. Draw the energy profile diagram for

  a. An exothermic reaction

  b. An endothermic reaction

  2. Using textbook or internet read and analyze the content about collision theory of reaction rates and make a summary to be presented to the class.

  In this work you should be able to explain contribution of Arrhenius and Boltzmann on the effect of temperature and activation energy on rates of reaction.

  In the previous unit, the factors that influence the rate of a reaction have been discussed.

  It is known that during a chemical reaction, the reactant species are converted into the new substances called products. This means that during a chemical reaction, the concentration of reactants decreases as they are consumed and that of products increases as they are in formation.

  In a chemical reaction, how quickly or slowly reactants turn into products is called the rate of reaction.

  Two theories known as “collision theory” and transition state theory” explain the rate of reaction.

  14.1.1. Collision theory

  According to the collision theory, a chemical reaction takes place due to the effective collisions of reacting molecules. For an effective collision to happen, the reactant species must be oriented in space correctly to facilitate the breaking of old bonds and forming of new bonds. Molecules which did not participate in effective collisions do not favor the reaction. During a chemical reaction species that collide must possess also a minimum amount of kinetic energy called “activation energy (Ea)”. The activation energy varies depending on the reaction. Therefore, the rate of reaction depends on the activation energy; a higher activation energy means that fewer molecules will have sufficient energy to undergo an effective collision.

  Consider the reaction: A-A + B-B → 2(A-B)

  For the reaction to take place, A-A and B-B must possess the activation energy, Ea that is required for breaking their bonds in order to form the new compound 2(A-B). The energy changes that occur during a chemical reaction can be shown in a Figure 14.1.

  

  Figure 14.1: The energy profile diagram

  The activation energy, (Ea), is the minimum energy required to initiate a reaction. Only colliding molecules with energy greater than or equal to Ea can lead to the reaction process. When two reactant molecules approach one another, their velocity will be slowed down due to the repulsion between their electrons clouds. However, if the initial kinetic energy of the colliding molecules is high, they can overcome the repulsion force and approach one another leading to the formation of products by merging the electron clouds.

  For a chemical reaction to occur, there are three main conditions that must be fulfilled.

  i. The reactants must collide: no reaction is observed if there are no collisions between reactant compounds.
  ii. The molecules must have sufficient energy (activation energy) to initiate the reaction.
  iii. The molecules must have proper orientation. Unless the reactant particles possess this orientation when they collide, the collision will not be an effective one.

  Application examples:
  1. Consider the reaction: A + B→ AB
  As the two reactant molecules approach one another, their velocity will be slowed down due to the repulsion between their electron clouds.

  

  Repulsion between electron clouds

  If the initial kinetic energy of the colliding molecules is low, the molecules will repel one another and no product will be formed.

  

  If the initial kinetic energy of the colliding molecules is high, they can overcome the repulsion force and approach one another so that electron clouds merge to form the products.

  

  2. Consider the reaction (CO(g) + O₂(g) → CO₂(g)) represented as follows:

  

  

  These examples show that when an effective collision requires a proper orientation of the reacting

    particles to form the products.

  3. The following examples (a) and (b) show also that the probability of a reaction to occur depends

      not only on the collision energy but also on the spatial orientation of the molecules when they collide.

  

  

  The collision theory is based on the kinetic theory and assumes a collision between reactants before a reaction can take place. It explains how temperature and activation energy affect the rate of reaction.

  a. The speed of reaction and theory of collision
  The activation energy depends on the nature of chemical bonds, which are broken during the reaction.

    The stronger the bonds, the greater is the activation energy.

  For an elementary reaction, the collision theory shows that the speed (rate, R) of a reaction

     is given by R = f x p x z

  Where: R = the rate or speed of the reaction
               f = the fraction of molecules having kinetic energy that can cause a reaction
               p = the probable fraction of collisions with proper (effective) orientation
               z = the frequency of collision

  b. Effect of increase in temperature on the rate of reaction
  An increase in temperature typically increases the rate of reaction.

  An increase in temperature will raise the average kinetic energy of the reactant molecules.

  Therefore, a greater proportion of molecules will have the minimum energy necessary for an effective collision. For example, an increase in temperature by 10 °C (or 10 K), doubles the speed of reaction.

  From the kinetic theory, it is known that the kinetic energy (K) of a gas is directly proportional to its temperature.

  

  Where: m = mass of the gas
  v = speed of the gas molecule

  N_A = Avogadro’s number = 6.6023 x10²³ particles for one mole of a gas at standard conditions

  

               = The Boltzmann constant

  T = Absolute temperature in Kelvin (K)

  R = Universal (molar) gas constant = 0.08206 L atm/mol K or 8.314 m³Pa/mol K
  The numerical values of the gas constant, R, can be expressed using different units as

         indicated in Table 14.1.
  Table14.1. Gas constant R converted in various units

  

  When the temperature is increased, there is an increase in the total number of particles with energy

   equal to or greater than the activation energy (Ea).

  The combined effects cause the rate of reaction to increase as illustrated by the Maxwell-Boltzmam distribution curve (Figure 14.2).

  

  Figure 14.2: The Maxwell-Boltzmann distribution curve

  Note:
  • The area under the curve is the same in both cases because the number of molecules remains the same.
  • All the fractions add up to unity
  • The fraction of molecules with energy E ≥ Ea, is greatest at higher temperature, T₂.
  That is when temperature increases,

  i. the average kinetic energy of the molecules increases resulting in more collisions per unit time,
  ii. The fraction of molecules with E ≥ Ea increases
  At temperature T₁, the number of particles having energy equal to or greater than the activation energy (Ea) is given by the area ABC (Figure 14.2).
  At T₂ the number of particles having energy equal or greater to Ea increases about two times by the area ADF (Figure 14.2).

  c. Temperature, rate of reaction and Arrhenius equation
  The relationship between the rate constant (k) of a reaction and temperature (T) of the system was first proposed by Arrhenius in 1889 and it is commonly known as the Arrhenius equation. It is written as:

                                    (1)
  Where: k = rate constant
  Ea = activation energy
  R is the gas constant = 8.314 (J mol ^-1 K ^-1)
  T = absolute temperature in Kelvin degrees (0°C = 273 K)

  A = Arrhenius constant or frequency factor
  A = p · Z, where Z is the collision frequency and p is called the steric factor (always less than 1)

  and reflects the fraction of collisions with effective orientations.
  e^-Ea/RT represents the fraction of collisions with sufficient energy to produce a reaction.
  The Arrhenius equation (1) is often written using the natural logarithm of both sides in order to determine

  the activation energy (Ea) for a chemical process.

  

  From the equation (2), it is possible to determine the activation energy using the ratio of two rate constants (k1, k2) and the two corresponding reaction temperatures (T₁, T₂).

  The following equations may be obtained:

  

  

  Subtracting the equation (4) from the equation (3), we have:

  

  Simplifying the equation (5) and rearranging it gives:

  

  

  

  The activation energy can be deduced from the equation (8) if k₁, k₂, T₁ and T₂ are known.

  Example 1:
  For a reaction with activation energy of 55 kJ/mol, by what factor will the rate constant go up with a rise

    in temperature from 300 K to 310 K?

  

  The above example showed that an increase of the temperature leads to the increase of the rate of the reaction. The rate of the reaction doubles for any 10K rise in temperature.

  Example 2:
  What is the activation energy of a reaction whose rate quadruples when the temperature is raised from 293 K to 313 K?

  Solution:

  

  

  The Arrhenius relation can be plotted to obtain a straight line.

  The graph of ln k versus 1/T (Figure 14.3) is a straight line with a negative slope.

  It shows the dependence of the rate of reaction with temperature.

  

  Figure 14.3: The Arrhenius plot of ln k against 1/2

  A part from the collision theory that explain the effect of temperature on the rate of reaction, the transition state theory also is based on the temperature effect on the rate of reaction.

  14.1.2 The transition state theory

  The transition state theory has been developed by Henry Erying in 1935. The theory supposes that a collision between reactants does not lead immediately to products. This theory supposes that the molecules collide and remain held for a certain interval of time, forming intermediate species or activated complexes (transition state) whose energy is higher than the average energy of the reactants. These activated complexes get then dissociated by generating either the reaction products or the starting reactants.

  The activated complex is an unstable grouping of atoms, formed during a fruitful collision that breaks apart to form reaction product(s).
  The energy needed to form an activated complex is equal to or greater than the activation energy Ea of the reaction.

  Example:

  

  (* ) The exponent is used to designate the activated complex.
  The transition state theory provides a way to calculate the rate constant for the reaction.

  According to the transition state theory:

  (1) During collision, the attraction between reacting molecules decreases and the kinetic energy of the molecules is converted into potential energy.

  (2) When the molecules approach, the interaction of the electronic configuration allows the rearrangement

      of valence electrons.

  (3) A temporary bond between atoms A and B is formed where as the bond for a bond B-C is weakered, which leads to the formation of activated complex. The activated complex is momentary as it decomposes

    to give the products (A-B + C)

  

  

  Figure 14.4: The diagram of energy profile for an exothermic reaction

  The difference in potential energy between the reactants and the activated complex corresponds to the activation energy of the reactants, Ea
  Eb is the energy required for the products to give reactants (for reverse reaction)
  The change in enthalpy of the reaction ΔH can be found as follows:

  
  
  Example: Consider the diagram profile for the following reversible reaction:

  

  

  ΔHr_xn = Ea (rev) – Ea (fwd) > 0, the reaction is exothermic

  If the potential energy of the products is less to the energy of reactants, the obtained energy (for the transition state to change to products will be greater than the activation energy.. Hence the reaction will be exothermic. If the potential energy of products is greater than that of reactants, the energy released for the activated complex to give products will be less than the activation energy. Hence the reaction will be endothermic.

  For any reaction, there exist a barrier of energy (activation energy) which must be overcome for a reaction to take place.

  The use of a catalyst can reduce the activation energy as it provides an alternative reaction pathway with the minimum activation energy as a result, more reactant molecules possess the energy required for a successful collision. As the speed of the reaction is proportional to the effective collisions, the presence of a catalyst increases the speed of both forward and reverse reaction but it is regenerated after reaction.

  It should be noted that even if a catalyst lowers the activation energy, the difference in energy, ΔH, between products and reactants remains constant. For an endothermic reaction, the activation energy of the reactants is greater than that of products and ΔH of the reaction is negative (Figure 14.5).

  

  Figure 14.5: Potential energy for an endothermic reaction.
  Example:

  Consider the following diagram profile of the reaction:

  

  

  ΔH_rxn = Ea (rev) – Ea (fwd) < 0, the reaction is endothermic

  Checking up 14.1

  1. Given the following equation for the decomposition of nitrous oxide:

  2N₂O(g) → 2N₂(g) + O₂(g) and given that the activation energy for the forward reaction is + 251 kJ and

    that the change in enthalpy for the reaction is +167 kJ.

  a. Sketch and label a potential energy diagram for the decomposition of nitrous oxide:

  b. What is the activation energy for a reverse reaction?

  2. For the reaction between A-A and B-B to give AB +AB               i.e A-A + B-B → AB +AB,

  Draw a situation showing effective collision and another situation which is not effective.

  14. 2. Measuring the rates of reaction by observing the mass changes, colour changes and

            volume changes

  Activity 14.2

  Experiment 1: Measuring the rate of reaction by observing the volume change of the gas produced. Apparatuses and chemicals: 3.0M HCl, calcium carbonate in form of marble chips, graduated measuring cylinder or flexible graduated syringe, balance, funnel, con-ical flask, stop clock, stopper for closing the conical flask, retort stand .

  Procedure:

  1. Weigh 40g of marble chips (calcium carbonate) and put them in the conical flask

  

  2. Measure 100 cm³ of 3.0M HCl and transfer all the acid into the conical flask using a funnel. Close the mouth of the conical flask with a stopper

  3. Connect the conical flask to a flexible graduated syringe. Clamp the flexible graduated syringe.

  Start the stop clock and record the total volume of carbon dioxide collected in the flexible graduated syringe. Record your results in the table below:

  

  Measuring the rate of reaction by observing the volume change

  Questions:

  a. Plot a graph of gas volume of carbon dioxide evolved on y-axis against time on the x-axis.

  b. Determine the rate of evolution of gas at 10 seconds, 20 seconds, and 100 seconds. What can you conclude about the reaction rate as time progresses?

  Experiment 2: Measuring the rate of reaction by obsering color change

  You are provided with 0.2M HCl(aq) and 0.05M Na2S2O3(aq).By changing the concentration of Na2S2O3 and keeping constant the concen-tration of HCl.

  a. On a piece of paper, draw a cross (X) using a marker pen and place this paper under a conical flask.

  b. Complete the following table of results:

  

  i. Write an equation between Na2S2O3(aq) and HCl(aq).

  ii. Plot a graph of time for the cross to disappear against volume of Na2S2O3(aq)

  iii. Write down the color changes in the reaction above.

  iv. Describe the chemical test of a gas produced in the reaction

  Measuring the rates of reaction uses different techniques depend on the reaction.

  The rate of reaction can be measured by using the change in mass, color change, volume and other properties. By analyzing the reaction mixture at suitable time intervals, one can determine the concentration of both reactants and products at particular time, hence determining the reaction rate.

  Sometimes it is easier to measure the change in the amount of a reactant that has been used up; sometimes it is easier to measure the change

  in the amount of a product that has been produced.

  The method used to analyze the reaction mixture depends on the reaction under consideration.

  The concentration is not measured directly in many cases, it is determined based on the signal related

   to the change in concentration. For example in the case of the reaction producing the colour, the intensity

   of the colour can be measured and related to the amount of products formed. If a reaction gives off gas,

   the change in volume or in mass can be measured. In this case, we will discuss about the change in volume or mass and change in color to measure the rate s of the reaction.

  14.2.1. Measuring the rates of reaction by observing the volume changes
  To measure the reaction rate by change in volume of gas produced is a convenient method if one of the products is a gas. A gas syringe can be used for this purpose (Figure 14.7).

  

  Figure 14.7: The gas syringe method.

  This method consists of a ground glass and a plunger, which moves outwards as the gas collects and is calibrated to record the volume directly. As more gas is produced, the plunger is pushed out and the volume of the gas in the syringe can be recorded. By measuring the volume at different time intervals, we can plot the data by following the change in volume against time and hence determine the rate of the reaction (Figure 14.8).

  

  Figure 14.8: Graph showing the volume of gas collected against time.

  Examples of reactions that produce gas

  i. Reaction that produce hydrogen gas
  When a metal reacts with an acid, hydrogen gas is produced. For example, magnesium reacts with

   sulphuric acid to produce magnesium suphate and hydrogen as follows.

  

  The hydrogen can be collected in a test tube. A glowing splint can be used to test the presence of hydrogen. The ‘pop’ sound shows that hydrogen is present.

  ii. Reaction that produce carbon dioxide
  When a carbonate reacts with an acid, carbon dioxide gas is produced. When carbon dioxide is passed through limewater, it turns the limewater milky. A burning splint will also stop burning (be extinguished)

   in the presence of CO₂ gas. These are simple tests for the presence of carbon dioxide.

  

  iii. Reaction that produce oxygen
  Hydrogen peroxide decomposes in the presence of a manganese (IV) oxide as catalyst to produce oxygen and water.

  

  14.2.2. Measuring the rate of reaction by observing change in mass

  Many reactions involve a change in mass and it may be measured directly.

  For a reaction that produces gas, the decrease in mass can be measured by standing the reaction mixture directly on a balance (Figure 14.9).The mass loss indicates the amount of gas that has been produced and escaped from the reaction vessel. The method allows continuous reading and a graph can be plotted directly of sample mass and mass lost against time (Figures 14.10 and 14.11), respectively. The release of carbon dioxide from the reaction between a carbonate and a diluted acid can be measured by this method and the rate is determined

  CaCO³(s) + 2 HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g)

  To start the reaction, the flask is gently lent to one side, causing the card to fall and the marble chips and acid to mix. A piece of cotton wool placed in the neck of the flask allows carbon dioxide gas to escape. As the gas escapes the mass of the flask reduces within an interval of time. The rate of the reaction can be deduced as the change in mass over the change in time.

  

  Figure 14.9. Measurement of the reaction rate following change in mass of reactants against time

  Since the reaction rate is the change in the amount of a product or a reactant per unit time, any property

  that is related to amount of product or reactant present can be used to measure the rate of reaction.

  

  Figure 14.10. A graph of the sample mass versus time Figure 14.11. A graph of the mass loss versus time.

  property that is related to amount of product or reactant present can be used to measure the rate of reaction

  14.2.3. Measuring the rate of reaction by observing color change

  Measuring the rate of reaction by observing color change can be applied if one of the reactants or products is colored and gives characteristic absorption in the visible region (about 320-380 nm of wavelength). Sometimes, an indicator can be added to generate a coloured substance that can be followed in the reaction. A colorimeter or spectrophotometer can be used to measure the intensity of the light transmitted by the reaction compounds. As the concentration of the coloured substances increases, it means that it absorbs more light and is less transmitted.

  Amount of light absorbed depends on the amount of absorbing species present. If a reactant or product absorbs light, then the color intensity will vary as the amount of this species changes. An example is the reaction between sodium thiosulphate and hydrochloric acid.

  Na₂S₂O₃ (aq) + HCl(g) →SO₂(g) + 2NaCl(aq) +H₂O(l) + S(s)

  The solution changes its color from colorless to yellow due to sulphur precipitate which is formed, and the concentration of yellow precipitate increases with time as more sulphur is being formed indicating that the reaction has occurred.

  14.3. Experimental determination of orders of reaction and rate laws

  Activity 14.3

  Using this textbook or internet or any other relevant book, read and analyze the content about experimental determination of orders of reactions and rate laws and make summary to be presented in the class. In this work you should also do the exercises of calculation.

  • Rate law or rate equation
  The rate of the reaction changes with time. The rate of reaction does not depend on the choice of substance, but it does depend on the way in which the chemical reaction is written. Therefore, a rate must be specified with a specific time unit (Mole/L. sec or M.s^-1.

  The rate law or rate equation of a chemical reaction is a mathematical equation that links the reaction rate with concentration or pressure of reactants and a constant rate.

  In a chemical reaction, the stoichiometric coefficients indicate the reacting mole ratio of the reactant and the number of mole of product that can be obtained.

  Let us consider the following equation:

  aA+ bB→ cC+ dD

  In the reaction, a mole of A will combine with b moles of B to form c moles of C and d moles of D.

  The rate law of this reaction can be expressed as:

  Where:
  k: rate constant

  x and y are integers: small whole numbers (Usually 0,1 or 2) called orders of reaction

  [A] and [B] are molar concentrations of A and B respectively.

  The proportionality constant, k, is known as the rate constant and is specific for the reaction shown at a particular temperature. The rate constant changes with temperature and its units depend on the sum of the concentration term exponents in the rate law. The exponents (x and y) must be experimentally determined and do not necessarily correspond to the coefficients in the balanced chemical equation.

  Note: When the concentration of A is changed by a factor of x, the rate will increase by a factor of x. A reaction’s rate law may be determined by the initial rates method.

  • Reaction order
  The sum of the concentration term exponents in a rate law equation is known as its reaction order.  We can also refer to the relationship for each reactant in terms of its exponent as an order. All the concentrations of all reactants taking part in a reaction cannot determine the order of reaction. The order of reaction is given by the number of atoms or molecules whose concentrations vary during the chemical change. In this part, we will be dealing with zero order, first order, and second order reaction.

     • For a zero order reaction any change in concentration of reactant does not affect the rate of reaction.
     • For a first order reaction, increasing the concentration of reactant two or three times also increase the

         rate two or three times (i.e. if the concentration of a reactant changes by a given factor,

         the rate must also change by the same factor) .

      • For a second order reaction, doubling or tripling the concentration of a reactant increases the rate four

        or nine times respectively (i.e. any change in concentration of a reactant by a factor x changes the

        rate by x² times.

  Experimental determination of order of reaction
  The order of the reaction can be determined using the initial rate method.
  Let us consider the following reaction:

  aA + bB → cC + dD

  The rate law = k [A]ⁿ [B]^m

  Where n and m are the order of reactants A and B respectively. The order n can be equal to a or not, and the order m can also be equal to b or not.

  As the order of the reaction is determined by experiment, the initial rate method can be used to determine the order of the reaction which is the sum (n + m), n is the partial order with respect to the reactant A and m is the partial order with respect to the reactant B

  In order to determine the order with respect to A, two experiments are carried out separately keeping the concentration of B constant in both cases while changing the concentration of A. The initial rate for both the experiments are then determined.

  Any change in the initial rate can only be due to the change in the concentration of A and not B.

  To determine the order with respect to B, the concentration of A is taken as constant while the concentration of B is changing.

  Application examples:

  Example 1: The following results were obtained for Q run between A and B

  

  a. What is the order of reaction with respect to A and with respect to B.
  b. What is the rate equation for the reaction?
  c. Calculate the rate constant.

  Solution
  A+B →products

  The rate of reaction =k[A]ⁿ[B]^m

  

  Let us compare run (a) and run (b) where [A]is constant

  

  The order of reaction with respect to B=2

  

  

  The order of reaction with respect to A=1
  Rate equation =k [A]ⁿ [B]^m Then
  Rate equation =k [A]¹[B^]2
  Rate =[A][B]²

  

  Example 2: P, Q, and R reacted together to form products.

  P +Q +R → Products

  The table shows the results of the experiments carried out to investigate the kinetics of the reaction.

  

  a. Deduce the order of reaction with respect to P, Q and R.
  b. Write the rate equation
  c. Calculate the rate constant k for the reaction and state the units.

  Solution:
  a. Let us write the rate as follows:

  

  For finding the order of reaction with respect to P:

  Comparing experiments (1) or (2) where [Q] and [R] are constant, we have,

  

  Simplifying the equation we get

  2 = 2x
  x = 1

  

  Simplifying the equation we get
  4 = 2^y
  2^y = 2²
  y= 2

  For finding the order of reaction with respect to R

  Comparing experiments (3) or (4) where [P] and [R] are constant,, we have

  

  Simplifying the equation we get

  

  b) The rate equation,

  

  c. The rate constant can be found as follows:

  From the rate law

  

  You can use experiment (1)

  

  Checking up 14.3The table below shows the experimental data for the following reaction :

  A+2B → C

  

  a. Determine the order of reaction with respect to A and B.

  b. Write the rate equation for the reaction

  c. Calculate the rate constant for the reaction and give its units.

  d. Calculate the rate of reaction when the concentrations of A and B are

         8.50 x10-³ and 3.83 x10-³ mol dm-³, respectively.

  14.4. Relationship between reactant concentrations and time for zero, first and second order reaction

  Activity 14.4

  1. For hypothetical reaction: n A → B

  Write the equation of the:

  a. Average rate of reaction in terms of A

  b. Rate law for the reaction

  c. Compare the equation found in (a) and the equation found in (b)

  d. Replace the exponent n by the following figure and in each case integrate from 0 to the figure.

  i. 0

  ii. 1

  iii. 2

  2. Using this textbook or internet or any other relevant book, read and analyze the content about relation between reactant concentrations and time for zero order reaction, first order reaction and second order reaction and make summary to be presented to the class. In this work you should do the exercises of calculation and sketch different graphs for order 0, 1, and 2.

  Rate laws express the rate as a function of the reactant concentration. Rate laws

  can also be converted into the equations that express the concentrations of the reactants or products at any given time during the course of a reaction.

  
  14.4.1. Zero order reaction
  In a zero –order reaction, the rate is independent of the concentration of the reactant.
  Consider the reaction: A→ Products

  The rate law is R = k [A]⁰

  Let us consider [A]₀ as the initial concentration of A , [A]t as the concentration of A at time t and [A] as the concentration of A that has reacted. Therefore [A] = [A]₀ - [A]_t

  The rate of the reaction is expressed as the change in concentration over time.

  

  Integrating from the time t = 0 to time t = t, we have:

  

  For a zero order reaction, the concentration-time graph is a straight line, showing a constant rate

  (Figure 14. 12). The gradient of the line = k. The rate-concentration graph is a horizontal line (Figure 14.13).

  

  Figure 14.12: Concentration-time graph for a zero order reaction

  

  Figure 14.13: Rate-concentration graph for a zero order reaction

  Examples of zero order reactions are:
  The reaction of the iodination of propanone

  CH₃COCH₃(aq) + I2 (aq) → CH₃COCH₂I(aq) +HI(aq)

  The reaction is said to be zero order with respect to iodine.

  The rate law equation can be given as follows: R = k [CH₃COCH₃]

  1. The adsorption of gaseous reactants in gaseous reactions:

  Sometimes, reactions between gases are zero order with respect to one of the reactants. This often indicates that this reactant has been adsorbed on the surface of the vessel. The rate of reaction then depends on the frequency with which molecules of the adsorbed gas collide with the inside of the vessel. This frequency is proportional to the concentration of the non-adsorbed reactant.

  2. Reaction of sodium thiosulphate and HCl is zero order with respect to HCl

  14.4.2. First order reaction
  A first order reaction is a reaction whose rate depends on the concentration of a single reactant raised

   to the first power.
  Consider the first order reaction A →products
  The rate equation can be written as: R = k [A]

  

  Integrating for time t=o to t=t, we have

  

  

  

  Therefore for all first order reactions, half-life (t_1/2) is independent of initial concentration [A]₀ of the reactant.

  For a first order reaction, the graph concentration-time (Figure 14.14) is a curve showing the decreasing

  rate with concentration while the graph rate-concentration (Figure 14.15) is a straight line passing

   through the origin.

                                  

  Figure 14.15: Concentration-time graph for            Figure 14.16: Rate-concentration graph for a first order reaction a first order reaction.

  14.4.3. Second order reaction
  A second order reaction is a reaction involving two reacting species.
  For the following reaction: A+B → Products,

  

  Integrating from time, t = 0 to time, t = t:

  

  

  For a second order reaction, the graph concentration-time (Figure 14.17) is curve while the graph rate-concentration is a parabola. Characteristic of the square function (Figure 14.18).

                                                    

  Figure 14.17: Concentration-time graph for           Figure 14.18: Rate-concentration for a second order graph  for a second order reaction.                        reaction

  14.4.4. Pseudo-first order reaction

  It is known that the order of reaction depends on the dependency of the rate of reaction on the concentration of reactants. If the rate is independent of the concentrations of reactants, the order of reaction is zero.

  A reaction which is not first-order reaction naturally but made first order by increasing or decreasing the concentration of one or the other reactant is known as Pseudo first order reaction. Pseudo-first order reaction is a reaction dependent upon the concentrations of both the reactants, but one of the components is present in large excess and thus its concentration hardly changes as the reaction proceeds.
  Consider the reaction: A + B → Products

  This reaction is dependent upon the concentrations of both A and B but if the component B is present in large excess and the concentration of B is very high compared to that of A, the reaction is considered to be pseudo-first order reaction with respect to A. similarly, if the component A is in large excess and the concentration of A is very high as compared to that of B, the reaction is considered to be pseudo first order with respect to B.

  I. If [B] is in excess, the rate equation can be written as follows:   Rate   =k[A] [B]The concentration of B practically remains constant during the reaction, and therefore the rate law can

   be written:

  Rate =k’[A]

  Where the rate constant, k’= k[B]

  The real order of that reaction is 2 but in practice it will be order 1.

  Example:
  Let us consider the elementary reaction where water is in excess:

  CH₃CH₂COOCH₃ (l) + H₂O (l) + H+ (aq) → CH₃CH₂COOH(aq) +CH₃OH(aq)

  The rate law for the reaction is: Rate = k’ [CH₃CH₂COOCH₃], where k’ = k [H₂O]

  At low concentration of H₂O, water is first order. But when water is in excess, water does not affect the

   rate of reaction.

  • Rate constant and its units

  The rate of the reaction, as seen above, is the change in concentration over the change in time.

  Therefore, the reaction rate is expressed in M^s-1 or mol dm^-3 s^-1. The unit of concentration is mol/dm³ and

  that of time is seconds (s).

  The units of the rate constant depend on the form of the rate law in which it appears.
  Units of k for a zero order reaction

  

  Note: Time can be expressed in seconds, minutes, days, hours, months, years, etc.

  

  Units of k for a first order reaction

  

  Units of k for second order reaction

  

  Checking-up 14.4

  1. The half–life of a first-order reaction is 2.5 min, calculate the time taken for the amount of reactant to 

      decrease to10% its original value.

  2. Consider the following reaction: A + B  products

  This is first order with respect to A and second order with respect to B.

  a. Write an expression of the reaction.

  b. Draw and label graphs that would allow the rate constant to be determined from a series

  of experiments in which

  i. [A] is kept constant, but [B] is varied.

  ii. [B] is kept constant, but [A] is varied.

  14.5. Difference between order of reaction and molecularity

  Activity 14.5

  Using this textbook, internet or any other relevant book, read and analyze the content about order of reaction and molecularity and make summary to be pre-sented to the class.

   In this work you should also do give examples of reactions for each.

  Molecularity is the number of reacting species (e.g: molecules, ions) that participate (take part) simultaneously in the formation of the transition state. That is, the number of species involved in the rate determining step for the reactions occurring in stages.

  It can be the number of species that participate as reactants simultaneously in an elementary reaction or elementary processes (reaction occurring in single event or step).

  If a single molecule is involved simultaneously in an elementary reaction, the reaction is unimolecular.

  For example: N₂O₅→NO₃ + NO₂, One reactant molecule is decomposed into two product molecules.

  If two molecules are involved simultaneously in an elementary reaction, the reaction is bimolecular.

  Examples:
  i. NO (g) + O₃ (g) → NO₂ (g) + O₂ (g)
  ii. Mechanisms for the reactions of primary alkylhalides (1⁰ R-X ) with nucleophiles, S_N2, involve two molecules in the rate determining step.
  If three molecules are involved simultaneously in an elementary reaction, the reaction is termolecular.

  Termolecular reactions are less probable than unimolecular or bimolecular reactions and are rarely encountered because the chance that three particles collide at the same time, with proper orientation

    and sufficient energy, are considered extremely small.

  An example of termolecular reaction is thought to occur during the formation of ozone from oxygen in

   the outer atmosphere:

                          2 O₂ + N₂ → O₃ + O. + N₂*

  The rate laws for most reactions have the general form:

  Rate =k[reactant A]^x[reactant B]^y…..
  The exponents x and y in a rate law equation are called reaction orders.
  For the reaction aA + bB+ cC → Product…….

  The order of a reaction is the sum of powers to which the concentration terms raised in the rate equation

  Rate =k[A]^x [B]y[C]^z

  For the reaction rate; x, y, z are the powers of the considered concentration of A, B and C respectively with respect to each reactant.
  x is the order of the reaction with respect to A
  y is the order of reaction with respect to B
  z is the order of reaction with respect to C

  The sum of x, y and z gives the overall order of reaction.
  x, y, z are obtained through experiments only and not from the number of moles as written in the stoichiometric equation.
  The difference between the order and molecularity of reaction are shown in the Table 14.1.
  Table 14.1. Differences between order of reaction and molecularity

  

  Checking-up 14.5

  It has been proposed that the conversion of ozone into O₂ proceeds by a two –step mechanism:

  O₃(g) → O₂ (g)+ O. (g)

  O₃(g) + O.(g) → 2O2(g)

  a. Describe the molecularity of each elementary reaction in this mechanism.

  b. Write the equation for the overall reaction

  c. Identify the intermediates

  14.6. Reaction mechanisms and kinetics

  Activity 14.6

  1. a. Five people have to run 100 m. Two of them do not perform well in running (they are slow in running). You are asked to organize them in order to perform that exercise so that the people leave the same place and reach the same destination at the same time.

  b. Differentiate between SN₁ and SN₂ reactions. Show also the mechanism of reaction.

  2. Using this textbook or internet or any other relevant book, read and analyse the content about the mechanism of reaction and make summary to be presented to the class

  Reaction rates provide information regarding how a chemical process occurs as well as the mechanism by which a reaction happens at molecular level. Most chemical reactions occur via a series of steps called the reaction mechanism. The individual steps, called elementary steps, cannot be observed directly. A reaction mechanism consists of a set of proposed elementary steps involving molecular species – reactants as well as reaction intermediates. A reaction mechanism explains how a given reaction might take place and from which a rate law can be derived, which must agree with the one determined experimentally.

  If the mechanism consists of more than one elementary step, the sum of these steps must be equal to the overall balanced equation for the reaction.

  A reaction mechanism describes in great detail the order in which bonds are broken and formed and the changes in relative positions of the atoms in the course of the reaction.

  The Table 14.2 summarizes the types of elementary steps and the rate laws that they follow. A and B represent the reactants or reaction intermediates. Typically these steps are usually either unimolecular or bimolecular.

  Table 14.2. Elementary steps and rate laws

  

  Example: N₂ (g) +3H₂ (g) →2NH₃ (g)
  Molecularity of reaction is equal to 1+3 = 4 but the order can be derived from the rate determining step.

  The reaction mechanism is the step-by-step process by which reactants actually become products.

  In the reaction mechanisms, some steps are fast while others are slow.

  A reaction cannot proceed faster than the rate of the slowest elementary step, the slowest step in a mechanism establishes the rate of the overall reaction which is known as the rate determining step.

  When a proposed mechanism consists of more than one elementary step, the one with the slowest rate

   will determine the overall rate of reaction.

  Examples:
  1. Consider the reaction: NO_2(g) + CO_(g) → NO_(g) + CO_2(g)
      This reaction is believed to take place in two steps.

  Step-1: NO₂ + NO2 → NO₃ + NO; (slow rate-determining)
  Step-2: NO₃ + CO → CO₂ + NO; (fast)
  The overall chemical equation is obtained by adding the two steps and canceling any common species

   to both sides.
  The overall reaction is given by:

  If the reaction follows a one elementary step mechanism, the rate law would be:
  Rate = k[NO₂][CO]However, the experimentally determined rate law is:
  Rate = k[NO₂]²

  The rate law for the rate-determining step: Rate = k₁[NO₂]², which is identical in form to the rate law obtained experimentally. The second step, which occurs very fast, does not influence the overall rate.

  An intermediate substance is neither a reactant nor a product in the overall reaction. It is formed in one step and consumed in the next step. In the above example, the intermediate is NO₃ because it is produced in the first step and consumed in the second one.

  2. Consider the reaction: 2NO₂(g) + F₂(g) → 2NO₂F(g)
  The experimental rate law is: Rate = k [NO₂][F₂]

  The reaction is first order with respect to each reactant.

  This reaction occurs follow two steps (slow and fast):

  

  F is an intermediate species, it is produced in the first step and consumed in the next step.
  Since the overall rate of the reaction is determined by the slow step, it seems logical that the observed rate law is Rate = k₁ [NO₂][F₂].

  SN₁ and SN₂ mechanisms and kinetics

  Chlorine, bromine, and iodine (X) possess a higher electronegativity than carbon.

  As a result, the bonding electrons in C-X bond are unevenly distributed. The carbon atom is partially positively charged (δ+) while the halogen atom is partially negatively charged (δ-).

  The polarity of carbon-halogen bonds forms the basis of two frequently found reaction types of this compound family, namely nucleophilic substitutions (S_N) reactions and elimination reactions.

  In a nucleophilic substance, or electron rich species, nucleophile, selectively attack a positive or partially positive charge of an atom or group of an atom or group of atoms to replace the leaving group.

  Nucleophilic substitutions (S_N) reactions can be of:

  Second-order reactions: Occur with primary alkyl halides.
   
  First-order reactions: Occur with tertiary alkyl halides.

  Secondary alkyl halides can undergo second-order reactions or first-order reactions.

  In both kinetic cases of substitutions, a leaving group (halide ion) is substituted by a nucleophile.

  1. Second-order nucleophilic substitution (SN² reactions)
  A nucleophile attacks a positively polarized carbon atom. The attack of the nucleophile results in the heterolytic cleavage of the carbon-ligand bond, where the bonding electron pair is completely passed

  onto the ligand (X). The substrate, along with the nucleophile, participates in the rate-determining step. Thus, the reaction rate depends on both the substrate and the nucleophile’s concentration. Therefore,

   this reaction type is called bimolecular nucleophilic substitution reaction (SN²).

  

  Example:
  CH₃CH₂Br + HO- → CH₃CH₂OH +Br –

  

  The rate of hydrolysis of a primary alkyl halide is proportional to the concentration of both

    halogenoalkane and hydroxide ions.

  This reaction is second order and its rate expression is, Rate = k[RX] [OH-]

  2. First-order nucleophilic substitution (SN₁ reactions)

  The carbon-halogen bond is cleaved and involves the halogenoalkane alone, forming a halide ion

  and a carbocation. The attack of the nucleophile on the carbocation yields the substitution product.

  Thus, the reaction rate depends only on the halogenoalkane concentration.

  Therefore, this reaction type is called unimolecular nucleophilic substitution reaction (SN¹ reaction).

  

  Example:
  Let us consider a hydrolysis reaction of a tertiary halogenoalkane:

  CH₃)₃CBr + NaOH → CH₃)₃COH + NaBr. The experimental results show that the reaction is first order

    with respect to the halogenoalkane:

  Rate=k[CH₃)₃CBr]
  [1]

  

  

  The rate of hydrolysis of a tertiary alkyl halide is proportional to the concentration of halogenoalkane

   but does not depend on hydroxide ions. The reaction is 1^st order with respect to the halogenoalkane.

  Rate =k[CH₃)₃CBr]

  Checking-up 14.6

  1. Differentiate between SN¹ and SN² reactions.

  2. The decomposition of nitrous oxide, N₂O ,is believed to occur by two –step mechanism:

  N₂O (g) +N₂(g) + O. (g) (slow)N₂O(g) +O (g) → N₂(g)+O₂(g) (fast)

  a. Write the equation for the overall reaction.

  b. Write the rate law for the overall reaction.

  END UNIT ASSESSMENT

  1. Collision theory states that particles must collide with each other in correct orientation and

      sufficient energy to

         A. attract

         B. repel

         C. react

         D. respond

  2. According to Collision theory, particles must

  A. collide every where

  B. collide with correct orientation

  C. must be kept under immense pressures

  D. shall not be below their melting points

  3. Some particles collide but bounce back afterwards it is called

  A. successful collision

  B. unsuccessful collision

  C. successful reaction

  D. unsuccessful reaction

  4. The rate law for a reaction is k [A][B]2Which one of the following statements is false?

  A. The reaction is first order in A.

  B. The reaction is second order in B.

  C. The reaction is second order overall.

  D. k is the reaction rate constant

  5. A reaction was found to be zero order in A.

  Increasing the concentration of A by a factor of 3 will cause the reaction rate to

  A. remain constant

  B. increase by a factor of 27

  C. increase by a factor of 9

  D. triple

  6. The rate law of the overall reaction is k[A][B]0.

  Which of the following will not increase the rate of the reaction?

  A. increasing the concentration of reactant A

  B. increasing the concentration of reactant B

  C. increasing the temperature of the reaction

  D. adding a catalyst for the reaction

  7. When a lit match is touched to the wick of a candle, the candle begins to burn.

  When the match is removed, the candle continues to burn. The match

  A. behaves as a catalyst

  B. supplies the activation energy

  C. is part of the rate determining step

  D. lowers the activation energy barrier

  8. Which step of a reaction is the rate-determining step?

  A. the fastest step

  B. the last step of the reaction mechanism

  C. the first step

  D. the slowest step

  9. Consider the following mechanism:

  Step 1: Cl. + O₃ → ClO. + O₂

  Step 2: O. + ClO→ Cl + O₂

  The reaction intermediate is

  A. Cl

  B. O₂

  C. O₃

  D. ClO.

  10. The table below shows the rates of reaction between substance A and B at different concentrations

  

  a. Determine the order with respect to A and B.

  b. Write the rate expression or rate law.

  c. Calculate the rate constant indicating clearly its units

  11. a. A reaction between two reactants, P and Q, gave the rate-concentration graphs shown below:

  

  b. The rate equation for a reaction between R and S is Rate = K[R]2[S]

  i. Determine the order with respect to R and S.

  ii. Sketch the rate-concentration graphs for reactants R and S.

  12. Consider the following reaction: A + B P + R

  

  i. Determine the rate equation and the rate constant.

  ii. Find the value of Z.

  13. In the reaction: P + Q + R → Product (s), deduce the order of reaction with respect to P, Q and R if:

  a. The rate does not change when [R] is doubled.

  b. The rate increases by 9 when [Q] is tripled

  c. The rate doubles when [P] is doubled

  14. In the reaction: A + B → C, Rate = [A][B]2. State how the rate will change if:

  a. The concentration of A is halved.also halved

  b. The concentration of A is tripled and that of B is doubled.

  15. In the reaction: 2NO + Br₂ →2NOBr, when the concentration of NO is reduced from 4x10^-2M to 2x10^-2M the rate falls by a factor of 4. What is the order of reaction with respect to NO? Give a reason to your answer.

  16. In the reaction: 2NO₂→ 2N₂ + O₂ , when the concentration of NO₂ decreases from 5x10^-3M to 2x10^-3M, the rate of decomposition falls by a factor of 2.5. What is the order of reaction? Give a reason to your answer.

  17. The reaction between 2-bromo–2-methylpropane (CH3)3C–Br and aqueous hydroxide ions OH– takes place by the following outline mechanism:

  Step 1 (CH₃)₃C–Br (CH₃)₃C+ + Br– (slow)

  Step 2 (CH₃)₃C+ + OH– (CH₃)₃C–OH (fast)

  a. What is the order of reaction with respect to (CH₃)₃C–Br?

  b. What is the total order of reaction?

  c. Write the rate equation for this reaction.

  18. a. Paracetamol has a biological half-life of 380 seconds.

  How long will it take for the level of paracetamol in the body to fall to one-sixteenth of its origi-nal value?

  b. 50% of a first order reaction is complete in in 23 minutes.

  Calculate the time required to complete 90% of the reaction.

• UNIT 15 RADIOACTIVITY

  Key unit competence:

  To be able to explain the importance and dangers of radioisotopes in everyday life.

  Learning objectives:

  • Explain the process of radioactivity.

  • Explain the properties of alpha, Beta and Gamma rays.

  • Develop awareness of the dangers of radioactive substances and nuclear

  weapons.

  • Write and balance nuclear reaction equations.

  • Compare and contrast chemical and nuclear reactions.

  • Explain half-lives of radioactive isotopes/radioisotopes.

  • Perform calculations involving the half- life of radioactive substances.

  • Explain the applications of radioisotopes in medicine, agriculture and industries.

  • Apply the calculations of half-life to determine the age of fossils.

  Introductory activity:

  1. Have you ever heard about Radioactivity?

  2. If yes, can you explain what it is?

  3. Can you mention some applications of Radioactivity?

  4. What is the difference between an ordinary chemical reaction and a

  nuclear reaction?

  You are provided with illustrations which are linked with radioactivity. Observe

  them and try to analyse them.

  

  

  The above photos show different ways through which radiations reach into

  our body (cells). The most familiar to you is radiation from sun rays! However,

  everything present in the pictures above, and many others that are not included,

  emits radiation. Now, discuss on the following points. Note: If you get stuck by

  some points, don’t be frustrated! You are allowed to visit any important document

  (textbook, search engine …) that can help you to find out the

  required solution.

  Point 1: Can you see or feel the presence of radiation?

  Point 2: How are we exposed to natural ionizing radiation? (Radiation with the

  ability to rip out one or several electrons from an atom or molecule is referred

  to as ionizing radiation and Radiation which does not have sufficient energy to

  damage atoms or molecules is called non-ionizing radiation).

  Point 3: What do you understand by “radioactive materials”? Do you think

  all of them are natural? Do you think all of them (i.e, radioactive materials) as

  harmful?

  The discovery of the electron towards the end of the nineteenth century was the

  starting point of new avenues of research in science, which were to give physicists

  an insight into the structure and nature of the atoms of matter. They discovered

  that there is a nuclear phenomenon in some elements that pushes them to emit

  radiations.

  This phenomenon was called radioactivity as proposed by Marie Curie to describe

  those emissions of nuclear radiation by some of the heavy elements.

  Radioactivity is an integral part of our environment. All living beings have been

  exposed to a constant flux of natural radiation on the surface of our friendly planet,

  but this radiation has no negative effect.

  Particles and rays are emitted when a nucleus of a radioactive isotope of element

  breaks down. They break down to acquire stability as all atoms want to be stable.

  

  We are constantly being bombarded by particles of cosmic radiations: several

  hundred go through our bodies every second. Rocks like granite, which have become

  symbols of permanence and durability (hence used in building), contain light traces

  of radioactive uranium. Sitting on or walking near a block of granite exposes you to

  many sources of radioactivity. Even the food we eat or the air we breathe contains

  radioactive elements (such as radon) either formed by the intervention of cosmic

  rays, or as old as the solar system itself. There is absolutely no way to escape from

  it: we are even radioactive! Eight thousand atoms of potassium-40 or carbon-14

  disintegrate in our bodies every second.

  

  15.1 Definition of radioactivity, radioisotopes and comparison

  between chemical and nuclear reactions.

  Activity 15.1

  1. What do you think about the terms ‘radioisotopes’ and ‘radioactivity’? How

  can you define the two terms from your understanding? From chemical

  and nuclear equations, describe any differences between these two types

  of equations

  2. Now use a search engine or textbooks (including even this one) to find

  out appropriate explanations on radioisotopes, radioactivity and nuclear

  equation, then make a good report to share with others.

  15.1.1. Definition of radioactivity and radioisotopes

  Radioactivity is a nuclear phenomenon. It is the process by which an unstable atomic

  nucleus changes into another more stable atomic nucleus by emitting energy in form

  of radiation. Substances which have the property of emission of radiation are called

  radioactive substances. Radioactivity is also known as radioactive disintegration

  or radioactive decay.

  A radioactive decay results when an atom with one type of nucleus, called “the

  parent radioactive nuclide” transforms into another atom with a different nucleus.

  The new product or element is named “the daughter nuclide”;, and thus the decay

  process results in “transmutation”. Transmutation, in this case, means creation of

  an atom of a new element. In this way, the energy or radiation emitted may take the

  form of particles such as alpha (α) or beta (β) particles.

  During the transmutation process, daughter nuclides are often in metastable or

  excited state; they lose energy in form of gamma ( γ ) ray to become de-excited.

  Gamma rays, here, can be compared to the heat of reaction that accompanies an

  exothermic reaction.

  In nuclear chemistry, the term ‘nuclide’ is used to designate a nucleus of an element.

  ‘Nucleons’ is the term used for nuclear particles such as protons and neutrons. In

  radiochemistry, the nucleon number stands for mass number (sum of protons and

  neutrons present in the nucleus of a given atom).

  Different nuclides, which have the same proton number but different nucleon

  numbers are called isotopes or isotopic nuclides.

  Radioisotopes or radioactive isotopes are the atoms of an element whose atomic

  nuclei undergo decay by emitting radiation(s).

  15.1.2. Comparison between chemical reaction and nuclear reaction

  Nuclear reaction and chemical reaction differ as shown in the following table:

  Table 15.1: Differences between a chemical reaction and a nuclear reaction

  
  

  Checking up 15.1:

  1. Explain what is meant by each of the following terms:

  a. Radioactive decay.

  b. Daughter nuclide.

  c. Transmutation.

  2. Differentiate between isotopes and radioactive isotopes

  3. Describe the following as pertaining to chemical reaction or nuclear

  reaction:

  a. Isotopes have the same chemical properties as they have the same

  number of electrons.

  b. Hydrogen nuclei are the reactants.

  c. Large amounts of energy are released.

  d. The mass is strictly conserved.

  15.2. Emission of alpha particles, beta particles and gamma

  rays and their properties and effect of electric and magnetic

  field on them.

  Activity 15.2

  1. State the types of radiation as revealed in the previous discussions.

  2. Use your search engine or any textbooks available to find out the

  properties the three types of radiation and the effect of electric and

  magnetic fields on them.

  15.2.1. Aplha particules, betaparticles gamma rays and their properties

  Different forms of radiation are emitted from an unstable nucleus as it decays. The

  main types of emitted particles are alpha particles, beta particles and gamma rays.

  The detailed information on each particle is provided below.

  a. Alpha partices

  An alpha particle contains two protons and two neutrons (so, its mass number, A=4

  and atomic number, Z=2). Because it has 2 protons, an alpha particle has a charge

  of 2+ (+2). That makes it identical to helium nucleus. In equations, it is written as

  the Greek letter “alpha (α)” or as the symbol for helium . The charge of an alpha

  particle was found experimentally by passing it in an electric field between two

  plates where it was attracted towards the negative plate.

  The main properties of an alpha particle are the following:

  • Alpha particle bears a positive charge of +2

  • It has a mass of 4 amu

  • It is deflected toward the negative pole of electric and magnetic fields. Look at

  Figure 15.4 below.

  • It affects a photographic plate and causes fluorescence on striking a fluorescent

  material.

  • It ionizes the gas through which it passes.

  • Not very penetrating; a very thin sheet of aluminium foil or a sheet of paper

  stops it.

  • It can be shielded by paper or clothing.

  • It destroys living cells and causes biological damage.

  • It is strongly ionizing

  The process of α-decay occurs commonly in nuclei with atomic number greater than

  83. The nuclei of these elements are extremely unstable due to the large number of

  neutrons and protons present. More information about stability of nuclei of atoms is

  discussed in this unit, section 15.4.

  When a nuclide decays by alpha emission, it loses 2 atomic number units and 4 mass

  units; in other words the daughter nuclide is the element located at 2 places before

  the parent nuclide in the periodic table.

  

  

  

  This shows that the number of neutrons decreases by one and the number of protons increases by one. 

  

  It is noteworthy that a β-emission results in the production of isobars (nuclides

  having the same mass numbers but different atomic numbers). The beta emission

  produces a new element with 1 more atomic number unit, or the element that

  directly follows in the periodic table.

  

  A positron is similar to an electron except that a positron has a positive (+1) charge.

  A positron is produced by an unstable nucleus when a proton is transformed into a

  neutron and a positron.

  This shows that the number of neutrons increases by one and the number of protons

  decreases by one; the element just preceding the parent nuclide in the periodic

  table is formed.

  

  Examples:

  
  

  c. Gamma rays

  Gamma rays, γ, are high-energy radiation released as an unstable nucleus undergoes

  a rearrangement of its particles to give a more stable, lower energy nucleus.

  Because gamma ray is an electromagnetic radiation, it has no mass and no charge.

  The main properties of gamma radiation (gamma ray) are the following:

  • It is an electromagnetic radiation of short wavelength and higher frequency,

  hence high energy

  • It is not deflected by electric and magnetic fields. Refer to figure 15.4 for more

  understanding.

  • It affects photographic plates.

  • Its ionizing power is very low compared to alpha-particles and beta-particles.

  • It travels at the same speed as that of light.

  • It has the greatest penetrating ability, 5000-10000 times that of alpha particles.

  • It causes fluorescence when they strike a fluorescent material.

  • It is diffracted by crystals.

  • It can be stopped by several inches (5cm thick piece) of lead or a thick concrete.

  • It can easily pass through the human body and cause immense biological

  damage.

  Normally, there are very few pure gamma emitters. In radiology, one of most

  commonly used gamma emitter is technetium (Tc). The excited state called

  “metastable technetium” is written as technetium-99m, Tc-99m. By emitting energy

  in the form of gamma rays, the excited nucleus becomes more stable.

  

  Stable nuclides are usually in the state of least energy or ground state. But these

  nuclides can be excited by particles or photon bombardment. The excited nucleus

  returns into the ground state by emission of excess energy as ϒ-rays.

  

  

  the properties of the particles discussed above.

  Table 15.2: Distinction between the properties of α, β and γ radiations (summary)

  
  

  Checking up 15.2

  1. How are an alpha particle and a helium nucleus similar? Different?

  2. Naturally occurring potassium consists of three isotopes: potassium-39,

  potassium-40 and

  Potassium-41.

  a. Write the atomic symbols for each isotope.

  b. In what ways are the isotopes similar and in what ways do they differ?

  

  15.3 Nuclear equations and radioactive decay series

  Activity 15.3

  1. Use your search engine or any available resources to learn about nuclear

  equations, how they are balanced and radioactive decay series and

  make a summary to be presented. While carrying out your research try to

  answer to question number 2 below.

  
  

  More stable

  A nuclear equation is balanced when the sum of the mass numbers and the sum

  of the atomic numbers of the particles and atoms on one side of the equation are

  equal to their counterparts on the other side.

  The changes in mass and atomic numbers of an atom that emits a radioactive particle

  are shown in the table below.

  

  15.3.1. Nuclear equations

  A nuclear reaction is represented by a nuclear equation as follows:

  Example 1:

  Radium-226 emits an alpha particle to form a new isotope whose mass number,

  atomic number and identity we must determine.

  Step 1: Write the incomplete nuclear equation.

  

  

  Step 3: Determine the missing atomic number

  
  

  Step 6: The equation is balanced.

  Note: Nuclear reactions involving hitting a nuclide by a particle such as proton

  or neutron as in the example 2 are referred to as ‘bombardment reactions’.

  Bombardment reactions are not natural; they are induced (or artificial) nuclear

  reactions.

  For the case of example 1, when a radioactive particle is emitted, the type of nuclear

  reaction is emission reaction; those are natural radioactive isotopes.

  15.3.2. Radioactive decay series

  Radioactive decay Series is the series of steps by which a radioactive nucleus

  decays into a non-radioactive nucleus. The element goes from radioactive to nonradioactive.

  A radioactive element disintegrates by emission of an α- or β-particle from the

  nucleus to form a new “daughter element.” This again disintegrates to give another

  “daughter element’. This is why the whole series of elements starting with the parent

  radioactive element to the stable end-product is called radioactive disintegration

  series or radioactive decay series as seen above.

  Naturally radioactive nuclides disintegrate to acquire stability. In nature there are

  four radioactive decays, that is, a series starting with a radioactive element and then

  ending with a reasonable stable element.

  The three series are Uranium, Thorium and Actinium series. Uranium series is the

  most important.

  The three series are similar because they all involve losses of alpha and beta particles

  ending with isotopes of Lead. Uranium series gives Lead-206, the most stable isotope

  of Lead; Thorium gives Lead-208 and actinium series gives Lead-207.

  The fourth series is the Neptunium series which leads to Bismuth-209.

  1. The uranium series

  It starts with the parent element Uranium-238 and ends with the stable element

  Lead-206. It derives its name from Uranium-238 which is the starting nuclide of the

  series and has the longest half-life. In the process 8 alpha and 6 beta particles are

  emitted before Lead-206 is attained. The whole process is shown belo

  

  

  

  15.4. Stability and instability of nuclei of atoms

  Activity 15.4

  1. What do you think can influence the stability of the nucleus of a given

  atom?

  1. Use a search engine or any available resources to understand and share

  everything about stability of nuclei of atoms.

  Nuclides can either be stable or unstable toward radioactivity but a nucleus that is

  unstable can become stable by undergoing a nuclear reaction or change.

  Most isotopes of elements up to atomic number 19 have stable nuclei. Elements

  with higher atomic numbers (20 to 83) consist of a mixture of isotopes, some may

  have unstable nuclei. Elements with atomic numbers of 84 and higher consist only

  of radioactive isotopes. So many protons and neutrons are crowded together in their

  nuclei that the strong repulsions between the protons make these nuclei unstable.

  Nuclear Stability is a concept that helps to identify the stability of an isotope. The

  ratio of neutrons to protons (n/p) is a good indicator to know if an isotope is

  radioactive or not.

  To help you understand this concept, there is a chart of the nuclides known as a

  Segre chart (see figure 15.5)

  This chart shows a plot of the number of neutrons versus the number of protons

  for known stable nuclides and shows the existence of the stability zone or band

  of stability. It can be observed from the chart that there are more neutrons than

  protons in stable nuclides with atomic number (Z) greater than 20 (Calcium).

  These extra neutrons are necessary for stability of the heavier nuclei.

  The excess neutrons act somewhat like nuclear glue. Atomic nuclei consist of protons

  and neutrons, which attract each other through the nuclear force, while protons

  repel each other via the electric force due to their positive charge. These two forces

  compete, leading to stability of various nuclei.

  Neutrons stabilize the nucleus, because they attract each other and protons, which

  helps offset the electrical repulsion between protons. As a result, as the number

  of protons increases, an increasing ratio of neutrons to protons is needed to form

  a stable nucleus. If there are too many or too few neutrons for a given number of

  protons, the resulting nucleus is not stable and it undergoes radioactive decay.

  

  

  Checking up 15.4

  1. What can you take as the basis to predict the stability of nuclei of atoms?

  2. Use your periodic table to provide examples (2 in each case) of

  a. Elements with naturally occurring stable isotopes.

  b. Elements with naturally radioisotopes.

  c. Elements with mixtures of both stable and unstable isotopes.

  Please specify the atomic number of each element you have provided

  15.5. Rate of decay of radioactive substances and half-life of a

  radioisotope

  
  

  15.5.1. Rate of decay of radioactive substances

  As seen in units 13 and 14, the time for different chemical reactions to be completed

  varies. In this unit, we are concerned with the calculation of decay rate for a decay

  process which is a first order reaction.

  Consider a simple case for a first daughter stable. Suppose that at time, t, there are

  N radioactive nuclides and dN disintegrations in a time dt. The rate of disintegration,

  dN/dt, is given by:

  
  

  
  

  Worked examples
  

  
  

  
  

  
  

  According to the radioactive decay law, the above information can be summarized

  by plotting the graph iodine-131 percentages against time (half-lives)

  
  

  From the decay curve, we can deduce a generalized formula that expresses how a

  fraction of a nuclide decreases rapidly (exponentially) as time increases.

  

  The half-life for a given isotope is always the same; it does not depend on how many

  atoms you have or on how long they have been sitting around. Each radioactive

  isotope decays at its own rate and therefore has its own half-life reason why it ranges

  from fractions of seconds to millions (and even billions) of years. For example, look

  at table 15.5.

  

  • The shorter the half-life, the faster the decay of nuclide and the longer the

  half-life, the slower the decay of the nuclide. Nuclides with short half-lives are

  often used in medicine as they do not stay in the body for a long time, hence

  causing minimal damage.

  • The rate of radioactive decay is unaffected by any chemical or physical change.

  Detection of radioactivity: devices that detect radiation include:

  • Cloud chamber: detects alpha and beta particles radiation, leaves a trail of

  ions in the water or ethanol vapor (gas) in the chamber.

  • Bubble chamber: detects alpha and beta particles radiation, holds a superheated

  liquid in which particles leave a path of bubbles if they are present.

  • Electroscope: Has leaves that repel or attract each other depending on the

  charge in the air. If the electroscope is given a negative charge the metal leaves

  separate from each other. A negatively charged electroscope discharges when

  ions in the air remove electrons from it, and consequently, a positively charged

  electroscope discharges when it takes electrons from the air around it. The

  rate of discharge of the electroscope is a measure of ions in the air and can be

  used as a basis of measurement and detection.

  • Geiger counter: measures radioactivity by producing an electric current when

  radiation is present, detects alpha, beta and gamma radiation.

  Worked examples

  1. If we start with 1.000 grams of Sr-90, 0.953 grams will remain after 2.0

  years.

  a. What is the half-life of strontium-90?

  b. How much strontium-90 will remain after 5.00 years?

  

  

  

  

  

  15.6. Uses of some radioisotopes

  Activity 15.6:

  1. According to your own understanding, how do you think radioactivity is

  important in daily life?

  2. Consult different sources (textbooks and search engines) to carry out a

  deep research on the uses of radioisotopes. Please take care because you

  will have to present your findings!

  Radioactive isotopes have a variety of applications. Generally, they are useful because

  either we can detect their radioactivity or we can use the energy they release.

  Radioactive isotopes are effective tracers because their radioactivity is easy to detect.

  A tracer is a substance that can be used to follow the pathway of that substance

  through some structures. For instance, leaks in underground water pipes can be

  discovered by running some tritium-containing water through the pipes and then

  using a Geiger counter to locate any radioactive tritium subsequently present in the

  ground around the pipes.

  Tracers can also be used to follow the steps of a complex chemical reaction. After

  incorporating radioactive atoms into reactant molecules, scientists can track where

  the atoms go by following their radioactivity. One excellent example of this is the

  use of carbon-14 to determine the steps involved in photosynthesis in plants.

  1. Radioactive Dating

  Radioactive isotopes are useful for establishing the ages of various objects. The

  half-life of radioactive isotopes is unaffected by any environmental factors, as seen

  above, so the isotope acts like an internal clock.

  

  2. In Medicine

  Radioactive isotopes have many medical applications in diagnosing and treating

  illness and diseases.

  When a radiologist wants to determine the condition of an organ in the body, the

  patient is given a radioisotope that is known to concentrate in that organ. After

  a patient receives a radioisotope, a scanner produces an image of the organ. The

  scanner moves slowly across the region of the body where the organ containing the

  radioisotope is located. The gamma lays emitted from the radioisotope in the organ

  are used to expose a photographic plate with a scan of the organ.

  One example of a diagnostic application is using radioactive iodine-131 to test for

  thyroid activity. The thyroid gland in the neck is one of the few places in the body

  with a high concentration of iodine.

  

  the next day a scanner is used to measure the amount of radioactivity in the thyroid

  gland. The amount of radioactive iodine that collects there is directly related to the

  activity of the thyroid, allowing radiologists to diagnose both hyperthyroidism and

  hypothyroidism.

  

  Iodine-131 has a half-life of only 8 days. So, it has low cell damage due to the

  minimum exposure. Technetium-99 can also be used to test thyroid function. Bones,

  the heart, the brain, the liver, the lungs, and many other organs can be imaged in

  similar ways by using the appropriate radioactive isotope.

  Other medical applications of radioisotopes include:

  • Radiation from Co-60 (γ-rays) is used to irradiate the tumours (for instance,

  diagnosis and treat thyroid disorders).

  • Iodine-125 (I-125) is used in treatment of brain cancer and in osteoporosis (a

  disease which causes bones to become weaker and easily broken) detection.

  • Iodine-131 (I-131) is used to diagnose and treat thyroid disorders, in treatment

  of Graves’ disease, goiter and prostate cancer.

  • Phosphorus-32 (P-32) is used in the treatment of leukaemia, excess red blood

  cells (tumours) and pancreatic cancer.

  • Technetium-99m is used in imaging of skeleton and heart muscle, brain, liver,

  heart, lungs, bones, spleen, kidney and thyroid. This is the most widely used

  radioisotope in nuclear medicine.

  • Cerium-141 (Ce-141) is used in gastrointestinal tract diagnosis and in measuring

  blood flow to the heart.

  • Sodium-24 (Na-24) in the form NaCl is used as a tracer in blood.

  • Strontium-85 (Sr-85) is used in detection of bone lesions and brain scans

  • Radio Gold (Au-198) is used in Liver disease diagnosis.

  • Radio iron (Fe-59) is used in Anemia diagnosis.

  • In addition, radioisotopes are also used in sterilization of medical devices.

  3. In Agriculture

  Obviously, we obtain food to eat and some drinks as a result of agriculture. But

  contaminated food causes some diseases. Thus, there are some radioisotopes that

  kill dangerous microorganisms present on food by “irradiation”.

  The radiation emitted by some radioactive substances can be used to kill

  microorganisms on a variety of foodstuffs thereby increasing the shelf life of these

  produces. Produces such as tomatoes, mushrooms, sprouts, and berries are irradiated

  with the emissions from cobalt-60 or caesium-137. This exposure kills a lot of the

  bacteria that could cause spoilage and so the produce stays longer. Eggs and some

  meat, such as beef, pork, and poultry, can also be irradiated. Normally, irradiation of

  food does not make it radioactive.

  By using known vintages (qualities of wines), oenologists (wine scientists) can

  construct a detailed analysis of the cesium-137 of various wines through the years.

  The verification of a wine’s vintage requires the measurement of the activity of

  cesium-137 in the wine. By measuring the current activity of cesium-137 in a sample

  of wine (the gamma rays from the radioactive decay pass through glass wine bottles

  easily, so there’s no need to open the bottle), comparing it to the known amount of

  cesium-137 from the vintage, and taking into account the time passed, researchers

  can collect evidence for or against a claimed wine vintage.

  In addition in plant research, radiation is used to develop new plant types to speed

  up the process of developing large amount of agricultural products. This involves

  insect control, drastic reduction of pest populations and, in some cases, elimination

  of insects by exposing the male ones to sterilizing doses of radiation. Radiation

  pellets are used in grain elevators to kill insects and rodents. Irradiation prolongs the

  shelf-life of foods by destroying bacteria, viruses, and molds as seen above.Other

  agricultural uses of radioisotopes include the following:

  • Radioactive phosphorus (P-32) is used in the study of metabolism of plants.

  • Radioactive sulphur (S-35) helps to study advantages and disadvantages of

  fungicides.

  • Pests and insects on crops can be killed by gamma - radiations.

  • Gamma - rays are used for preservation of milk, potatoes etc.

  • Yield of crops like carrot, root, apples or grapes can be increased by irradiation

  with radioisotopes.

  4. In Industry

  The applications of radioisotopes in industry are so many. Many types of thickness

  gauges exploit the fact that gamma rays are attenuated when they pass through the

  material. By measuring the number of gamma rays, the thickness can be determined.

  This process is used in common industrial applications such as:

  a. The automobile industry: to test steel quality in the manufacture of cars

  and to obtain the proper thickness of tin and aluminum

  b. The aircraft industry: to check for flaws in jet engines

  c. Road construction: to gauge the density of road surfaces and sub surfaces

  d. Pipeline companies: to test the strength of welds and leakage

  e. Oil, gas, and mining companies: to map the contours of test wells and

  mine bores, and

  f. Cable manufacturers: to check ski lift cables for cracks.

  The isotope 241Am is used in smoke detectors , in thickness gauges designed to

  measure and control metal foil thickness during manufacturing processes, to

  measure levels of toxic lead in dried paint samples, and to help determine where oil

  wells should be drilled.

  The isotope 252Cf (a neutron emitter) is used for neutron activation analysis, to

  inspect airline luggage for hidden explosives, to gauge the moisture content of soil

  and other materials, in bore hole logging in geology, and in human cervix-cancer

  therapy.

  Checking up 15.6

  1. Define tracer and give an example of how tracers work.

  2. Explain how radioactive dating works.

  3. Name two isotopes that have been used in radioactive dating.

  4. The current disintegration rate for carbon-14 is 14.0 Bq. A sample of

  burnt wood discovered in an archeological excavation is found to have

  a carbon-14 disintegration rate of 3.5 Bq. If the half-life of carbon-14 is

  5,730 years, approximately how old is the wood sample?

  5. Bone and bony structures contain calcium and phosphorus.

  a. Explain why the radioisotopes of calcium-47 and phosphorus-32

  would be used in the diagnosis and treatment of bone diseases.

  b. During nuclear tests, scientists were concerned that strontium-85,

  a radioactive product, would be harmful to the growth of bone in

  children. Explain

  15.7. Nuclear fission and fusion and their applications

  Activity 15.7:

  1. Have you ever heard about nuclear fission and fusion? If yes, explain the

  two terms and state any use of one of/both these processes

  2. From your prior knowledge, explain what you think about the difference

  between fission and fusion.

  3. Use your search engine, or any available source to read about nuclear

  fission and fusion and then make a summary to be presented to your

  colleagues.

  While many elements undergo radioactive decay naturally, some nuclear reactions

  are not spontaneous but are brought about when stable isotopes are bombarded

  with high-energy particles (like neutrons, α-particles, protons ...). Nuclear fission

  and fusion are good examples artificial radioactivity as they do not take place

  spontaneously.

  15.7.1. Nuclear fission and fusion

  1. NUCLEAR FISSION

  Nuclear fission is a process in which a large atomic nucleus is split into two smaller

  nuclei. Large nuclei obviously have a large number of protons. The close proximity of

  so many protons makes these nuclei unstable due to the repulsion forces between

  protons. Thus, the nucleus of the unstable isotope splits to form smaller atoms by

  bombardment with a suitable sub-atomic particle. Those stable isotopes that are

  bombarded by a neutron to undergo fission reactions (to become fissionable) are

  known to be “fertile radioisotopes”.

  

  The neutrons emitted, in this fission reaction, bombard more uranium nuclei

  available to form a “reaction chain”. This chain reaction is the basis of nuclear power.

  As uranium atoms continue to split, a significant amount of energy is released, in

  form of heat, from the reaction. This heat released is used to produce electricity (in a

  nuclear plant) or used in atomic/nuclear bombs.

  Other examples: nitrogen-14 and oxygen-16 undergoing alpha and neutron

  bombardment respectively.

  

  2. NUCLEAR FUSION

  Nuclear fusion is a process that consists of joining two atomic nuclei of smaller

  masses to form a single nucleus of a larger mass.

  A good example is the fusion of two “heavy” isotopes of hydrogen (deuterium,

  Hydrogen-2, and tritium, Hydrogen-3) into the element helium.

  

  However, very high temperatures and pressures are required for the fusion to take

  place because of the repulsion between the positive nuclei. Thus, for the fusion to

  take place the nuclei must have enough kinetic energy to overcome these repulsion

  forces between like charges.

  Like fission, in the fusion process large quantity of energy is liberated in the form of

  heat. This energy is also used in atomic/nuclear bombs (Hydrogen bomb).

  Note that it is very difficult to carry out nuclear fusion between two large nuclei due

  to the highly strong repulsion forces that are between their positively charged nuclei.

  Table 15.6 provides the differences between nuclear fission and nuclear fusion.

  

  15.7.2. Applications of fission and fusion

  Both fission and fusion are nuclear reactions that produce energy as described

  above. Fission is used in nuclear power reactors since it can be controlled, while

  fusion is not utilized to produce power since the reaction has not yet controlled up

  to now. The two processes have an important role in the past, present and future in

  energy creation.

  Example of nuclear energy calculation:

  Consider the following nuclear reaction:

  
  

  
  

  • Nuclear fission is non-renewable because uranium or other fissile nuclides

  needed for this process are not renewable.

  • For example: In Atomic Bomb, there are fission reactions of uranium. The energy

  produced is uncontrolled and this principle is used to manufacture the

  bombs and missiles. When controlled, the nuclear fission is also useful in the

  production of electricity.

  • Nuclear fusion energy (if mastered) could be renewable because hydrogen

  needed for this process is available in nature in large amount.

  • For example: In Hydrogen Bomb, the nuclear reaction involves the fusion of

  deuterium and tritium nuclei to form helium.

  
  

  Checking up 15.7:

  1. Classify the following as pertaining to nuclear fission, nuclear fusion, or

  both:

  a. Small nuclei combine to form larger nuclei.

  b. Large amounts of energy are released.

  c. Very high temperatures are needed for the reaction.

  2. In a fission reaction, U-235 bombarded with a neutron produces Sr-94,

  another small nucleus and 3 neutrons. Write the complete equation for

  this fission reaction

  15.8. Health hazards of radioactive substances
  

  Activity 15.8:

  1. Have you ever heard about health risks associated with nuclear radiations?

  If yes, state and explain them.

  2. Use your search engine or any available resources to you to find out how

  we are all exposed to radiation and the hazards that are brought to us by

  radiation.

  Radioactive materials in the environment, whether natural or artificial, do expose

  people to risks.

  This can happen in two ways:

  • The radiation from the material can damage the cells of the person directly.

  This is damage by irradiation.

  • Some of the radioactive materials can be swallowed or breathed in. While

  inside the body, the radiation it emits can produce damage. This is damage

  by contamination.

  Some health Hazards are: radiation burns, hair loss (temporary or permanent),

  cancer, reproductive sterility, mutations in offspring, etc.

  We are all exposed to low levels of radiation every day. Naturally occurring

  radioisotopes are part of atoms of wood, brick and concrete in our homes and

  the buildings of schools, hospitals, supermarkets, etc. This radioactivity is called

  “background radiation” and is present in the soil, in the food we eat, in the water we

  drink, and the air we breathe. For instance, one of the naturally occurring isotopes of

  present in any potassium-containing food. Other naturally occurring radioisotopes

  in air and food are carbon-14, radon-222, strontium-90 and iodine-131.

  In addition to naturally occurring radiation from construction materials in our homes,

  we are constantly exposed to radiation (cosmic rays) produced in space by the sun.

  The larger the dose of radiation received at one time, the greater the effect on the

  body. Exposure to low amount of radiation cannot be detected, but at medium

  levels the whole-body exposure produces a temporary decrease in number of

  white blood cells. If the exposure is very high, the person suffers the symptoms of

  radiation sickness such as nausea, vomiting, fatigue, and a reduction of white-cell

  count which can even be lowered to zero. So the victim suffers from diarrhea, hair

  loss, hair loss and infection. Too much exposure is expected to cause death.

  You can ask yourself how radiation (which is in the air all around you) might lead to

  lung cancer. This is explained by the presence of Radon gas (mainly from granite rock)

  which is the main source of background radiation, and which is in turn responsible

  for almost all the radiation we get exposed to over our lifetime. When we breathe in;

  some of the radioactive atoms in the air undergo radioactive decay and emit alpha,

  beta or gamma radiation. These radiations can collide with and ionize atoms in our

  lung tissue, which could ultimately lead to lung cancer.

  

  Figure 15.10: Ways through which radiations reach the body cells

  Checking up 15.8:

  1. List any three sources of natural radiation.

  2. What are some symptoms of radiation sicknesses?

  END UNIT ASSESSMENT

  I. MULTIPLE CHOICE QUESTIONS. Choose the letter corresponding to the

  appropriate answer

  1. Elements which emit natural radioactivity are known as:

  a. radio elements

  b. active elements

  c. radioactive elements

  d. nuclear elements

  2. Spontaneous emission of radiation by unstable nuclei is called

  a. positive radioactivity

  b. artificial radioactivity

  c. natural radioactivity

  d. nuclear elements

  3. The half-life of technetium-99 is 6 hours. How much of a 100 milligram

  sample of technetium-99 will remain after 30 hours?

  a. ? 12.5 mg

  b. ? 3.125 mg

  c. ? 6.25 mg

  d. ? 1.56 mg

  4. What change occurs in the nucleus of an atom when it undergoes beta

  emission?

  a. The outer shell of electrons is filled.

  b. The number of neutrons decreases by one.

  c. A high speed electron is produced.

  d. A proton is produced.

  e. There is a release of energy.

  5. Which one of the following does not occur in nuclear reaction?

  a. Nuclear radiation is released.

  b. There is a change in mass.

  c. It involves a rearrangement of electrons.

  d. New elements are made

  6 When beta decay occurs in a radioactive isotope, the atomic number (Z)

  always

  a. increases by one

  b. stays the same

  c. Increases by two

  d. decreases by one

  7. Which of the following statements is not correct concerning alphaparticles?

  a. they are composed of helium nuclei

  b. They are emitted from unstable nuclei

  c. they have a positive charge

  d. they can penetrate thick sheets of lead

  8. Which one of the following is true when a nucleus undergoes radioactive

  decay?

  a. a new element is always formed

  b. alpha-particles are always emitted

  c. beta-particles are always emitted

  d. the unstable nucleus loses energy

  9. Why would the occupants of a house fitted with smoke detectors

  containing americium–241 not be at risk from alpha radiation emitted by

  these devices?

  a. ? It has very low penetrating power through the air.

  b. Alpha radiation has very low ionizing power.

  c. the occupants wear protective clothing at all times

  d. Alpha radiation is not harmful

  10. Which of these metals is used as a shield against radioactive emissions?

  a. Uranium

  b. Lead

  c. Radium

  d. Gold

  

  

  

  b. Half-life

  c. Helium

  d. Longer

  e. Nuclear fission

  f. Nuclear fusion

  g. Radioactive

  h. Radioactive decay

  i. Radioactivity

  j. Transmutation

  1. An element that gives off nuclear radiation is known to be ….

  2. When an element changes to another, more stable element, there is …

  3. The amount of time for half the atoms in a radioactive sample to decay is

  referred to as …

  4. The more stable a nucleus is, the ______ its half-life.

  5. The process in which the nucleus of an unstable atom releases radiation in

  order to become stable is known as …

  6. The changing of an atom into another, more stable atom during decay is

  …

  7. The splitting of an atom into 2 smaller nuclei (nuclear power plant) is

  known as …

  8. An alpha particle is actually a nucleus of ____________.

  9. The name given to the several steps required to get a radioactive element

  to a stable element is the …

  10. The joining of 2 atoms to form a single, larger nucleus is known as …

  III. Short and long answer open questions

  1. Define radioactivity.

  2. Describe an alpha particle. What nucleus is it equivalent to?

  3. Plutonium has an atomic number of 94. Write the nuclear equation for the

  alpha particle emission of plutonium-244. What is the daughter isotope?

  4. Francium has an atomic number of 87. Write the nuclear equation for the

  alpha particle emission of francium-212. What is the daughter isotope?

  5. Write balanced equations for the following nuclear reactions:

  a. Nuclide carbon-14 undergoes beta decay

  b. Uranium-238 decays by alpha particle emission

  c. Carbon-11 decays by position emission

  d. Cobalt-60 decays by gamma radiation

  e. Gold-195 decays by electron capture

  6. a) Give values for a, b, c and d in the following nuclear equations:

  

  Calculate the mass number and atomic number of element Y.

  7. If radium-226 undergoes a series of decays that produce five α and four β

  particles, what is the final product?

  8. Strontium-90 is a beta particle emitter and has a half-life of 28.1 yrs.

  a. Write the decay equation for strontium-90.

  b. Calculate the decay constant

  9. A sample of a particular radioactive isotope is separated and monitored

  over a period of 15 hours. If it is found that 12.0 grams of the isotope

  remain after 4.2 hours and that 10.8 grams remain after 11.3 hours, what is

  the half-life of the isotope?

  10. What is the age BP of a bone fragment that shows an average of 2.9 dpm/

  gC in 2005? The carbon in living organisms undergoes an average of 15.3

  dpm/gC, and the half-life of 14C is 5730 years. (BP = Before Present, with

  the year 1950 used as the reference; dpm/gC = disintegrations per min.

  per gram Carbon)

  11. How much energy is released (in kJ) in the following fusion reactions to

  yield 1 mol of 4He or 3He?

  a. 2H + 3H → 4He + 1n

  b. 2H + 2H → 3He + 1n

  (The atomic masses are: 1H = 1.00782 u; 2H = 2.01410 u; 3H = 3.01605 u;

  3He = 3.01603 u; 4He = 4.00260 u and 1n = 1.008665 u)

  12. It has been estimated that 3.9 x 1023 kJ/s is radiated into space by the sun.

  What is the rate of the sun’s mass loss in kg/s?

  13. How much energy (in kJ) is produced in the fission reaction of 1.0 mol of

  uranium-235 according to the following equation?

  235U + 1n → 142Ba + 91Kr + 31n

  (The atomic masses are: 235U = 235.0439 u; 142Ba = 141.9164 u, 91Kr =

  90.9234 u; 1n = 1.00867 u

  14. a) State four uses of radioactive isotopes

  b) The half-life of cobalt-60 is 5.2 years. What fraction of cobalt-60 would

  remain after 26 years?

  c) The half-life of carbon-14 is 5600 years. Analysis of a fossil from a historical

  site showed that 6.25% of carbon-14 was present compared to living

  material. Calculate the age in years of the fossil.

  15. The half-life of uranium-238 is 4.5 billion

  years. What will be the 238U/206Pb atomic ratio in a rock that is 5.0 billion

  years old? (Assume that isotope lead-206 was not present initially)

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• ASSESSMENT