S5: Chemistry

    Key unit competency:

 Be able to prepare standard solutions and use them to determine concentration of 

other solutions by titration.

Learning objectives:
• Define the terms standard solution and primary standard solution.
• Explain the properties of a standard primary solution.
• Explain the titration process, emphasising the need for precise measurements.
• Prepare solutions with different concentrations.
• Properly use the burettes, pipettes during titration.
• Interpret the experimental data obtained by titration and report.
• Carry out acid-base, redox titrations and do calculations involved.
• Develop a team approach and a sense of responsibility in performing the 
   experiments of titration.
• Respect of procedure in practical experiment.
• Develop a culture of orderliness in performing practical experiments.

• Appreciate the use of appropriate measurements in daily life.

  Introductory Activity

             

Observe the above photo and attempt the following questions:

1. For the bottle of PRIMUS, and that of MUTZIG, we find on the bottle of its 
    labels 5% alcohol and 5.5% alcohol respectively.
a. Explain the meaning of 5% and 5.5% alcohol.
b. 5% alcohol corresponds to which volume of alcohol of the total 
    volume of 72cl of the bottle of PRIMUS.
c. Calculate the volume of alcohol corresponding to 5.5% alcohol of the 
     bottle of MUTZIG of the total volume of 65cl.
2 . On the label of the bottle of AGASHYA JUICE we find that dilution is 1:5. 

    Explain this ratio and explain also the purpose of diluting substances.

11.1. Definition of standard solution and primary standard solution.

Activity 11.1
1. You have a bag full of rice. How do you determine its weigth?
2. You have a jerrycan half-full of a liquid substance, how are you going to 
    proceed to know the exact quantity of the liquid?
3. You are given a basic solution, NaOH (aq), and you are requested to 

      determine its concentration. What do you need to do that?

In analytical chemistry, a standard solution is a solution containing a precisely known 
concentration of an element or a substance and used to determine the unknown 
concentration of other solutions. A known weight of solute is dissolved to make a 
specific volume. It is prepared using a standard substance, such as a primary standard.

A primary standard is defined as a substance or compound used to prepare standard 
solutions by actually weighing a known mass, dissolving it, and diluting to a definite 

volume.

Or a substance, which is chemically stable in aqueous solution and its concentration 
remains constant with change in time such that it can be used to standardize other 
solutions. 

Some important examples of primary standard are;
               
Standard solutions are normally used in titrations to determine unknown concentration 

of another substance.

   Checking up 11.1

Differentiate between standard solution and primary standard solution  

  11.2. Properties of a primary standard solution.     

  Activity 11.2
Do research and find out the role and characteristics of a good primary standard.

A good primary standard meets the following criteria:
• High level of purity
• High stability
• Be readily soluble in water
• High equivalent weight (to reduce error from mass measurements)
• Not hygroscopic (to reduce changes in mass in humid versus dry environments)
• Non-toxic
• Inexpensive and readily available 
• React instantaneously, stoichiometrically and irreversibly with other substances 
   i.e. should not have interfering products during titration.
• It should not get affected by carbon dioxide in air
  Molar concentrations are the most useful in chemical reaction calculations because 
  they directly relate the moles of solute to the volume of solution. The formula for 

  molarity is:   

               

Checking up 11.2

Discuss the properties of a good primary standard

11.3. Preparation of standard solutions

Activity 11.3

Describe how you would prepare 1L of a 1M solution of sodium chloride. The gram 

formula weight of sodium chloride is 58.5g/mol 

The preparation of the solution requires a reagent that is so to say the quantity to be 
weighed for mass if the reagent is in solid state ; The volume to be pipetted using 
pipette if the solute is a liquid and then after dissolve it in water, So the solution can be 

prepared by two methods such as dissolution method and dilution method.

In the preparation of solution, glasses, volumetric flask, pipette, glass rod, measuring 
cylinder, analytical balance, spatula, beakers, magnetic stirrer and other laboratory 

devices are used.

11.3.1. Preparation of standard solution by dissolution of solids

Activity 11.3.1 

 Preparation of 250 mL of 1M NaOH solution

 Materials

• Beaker                            - balance         - sodium hydroxide solid
• Spatula                           - funnel           - glass rod

• Volumetric flask          - stopper        - distilled water

Procedure:
i. Calculate the mass of sodium hydroxide needed 
ii. Weigh a clean beaker, and record its mass  with   a clean spatula, add 
sodium hydroxide until the combined mass of weighing beaker and 
sodium hydroxide is 
iii. Add about 50 cm3of distilled water, stir with a glass rod until all solids 
     have dissolved.
iv. Pour all the solution carefully through a funnel into a volumetric flask; 
     wash all the solution out of the beaker and off the glass rod.
v. Add distilled water until the level is about  below the graduation 
    mark on the graduated flask. Add the rest of distilled water from a washing 
    bottle until the bottom of the meniscus is at the level of mark when viewed 
    at eye level.
vi. Insert the stopper of the flask and invert the flask several times to mix the 
     solution.
vii. Label the solution.

Scope: This method is applied for solute in solid state and you should be able to 
determine the mass required from calculation to be weighed and provide distilled 

water to dissolve the solute.

Examples:

1. Describe in details how you can prepare the following solution: 50 mL of NaOH, 10%.

Answer:
10% means that in 100 mL of solution only 10 g are pure in NaOH, So 50 mL of NaOH 
will be prepared by taking the mass of NaOH, dissolving it in water and madding up to 

50 ml.

             

Procedure:
• Weigh 5g of NaOH accurately using glass watch, spatula and analytical balance.
• Dissolve it in a volumetric flask of 50 mL containing already little water and mix 
   using a baguette and shake till you get homogeneous mixture (you should take 
   care since it is an exothermic reaction).
• Top up using distilled water and shake again and cover your solution.

• Label your solution: 10% NaOH; 50 mL and the date of preparation.

2. Describe in details how you can prepare    solution.

Answer:

Step 1: Calculations

Calculate the amount of anhydrous sodium carbonate required to be dissolved in 

 of solution. i.e.:

             

Step 2: Weighing
• Weigh a clean empty beaker and record its mass, 
• Using a clean spatula add the mass of pure anhydrous sodium carbonate equal 
     to the calculated mass. Let it be  
• Actual mass of the carbonate transferred into the beaker = g

Note: Not the entire sample gets transferred into the beaker since part of it sticks on 

            the walls of the weighing bottle.

Step 3: Procedure
• Using a wash bottle, carefully add about of distilled water. Stir using a 
  glass rod until the entire solid has dissolved.

• Carefully pour all the solution through a filter funnel into a volumetric flask. Wash 
   off all the solution out of the beaker and off the glass rod; ensure that all the 
   washings run into the volumetric flask.

• Add distilled water until the level of the volumetric is about 
 below the mark. Add the rest of the distilled water drop by drop using
 a dropping  pipette until the bottom of the meniscus is level with the 250cm3
 mark when viewed at eye level. 

• Insert the stopper of the flask and invert the flask several times to mix the solution.

Note: Since the concentration of solution has been determined, thus the 

solution prepared is a standard solution.

   Checking up 11.3.1

Describe in details, how you can prepare the following solutions:
a. 
b. 250 mL acidified potassium dichromate 1N

c. 100 mL oxalic acid 5g/L. 

11.3.2. Preparation of standard solution by dilution         

Activity 11.3.2

Materials
• Measuring cylinder 
• Volumetric flask
• Stopper
• Concentrated 

• Distilled water

Notice: Take care when mixing water and acid. When you mix acid with water, it is 
extremely important to add the acid to the water rather than the other way around. 
It is because acid and water react in a vigorous exothermic reaction, releasing heat, 
sometimes boiling the liquid. If you add acid to water, the water is unlikely to splash 
up, but even if it did, it is less likely to hurt you than if you add water to acid. When 

water is added to acid, the water boils and the acid may splatter and splash.

Apart from, dissolution method, we can prepare a solution by dilution from the stock 
solution. This consists of reducing the concentration of a concentrated stock solution 
to less concentrated solution by adding water.

The rationale is how we can prepare the desired solution of concentration C, with the 
volume V, from the stock solution whose concentration is .The question consists of 
determination of the volume  to be pipetted from stock solution and diluted using .

Xml of water to get V. So, from the conservation principle of matter, the quantity of the 

solute before and after dilution should be the same.

       

       

       

Procedure:
• Pipette only 54.35 ml of sulphuric acid accurately using a pipette from a stock 
  solution.
• Pour them gently in the flatted balloon of 1L containing already little water and 
   shake. Note that you should take much care since the reaction is exothermic 
    and sulphuric acid is harmful to skin remember that we pour acid to water 
    not water to acid that is A-W not W-A.
• Top up to 1L using distilled water and shake again in order to homogenize the 
   solution.
• Cover the solution, then label it:    and date of preparation.

2. Calculate the volume of  that would be required to prepare 

                   

Checking up 11.3.2

A 0.2N solution was diluted by addition of 200 mL of water.
 Calculate the dilution factor if the solution is diluted from 0.2N to 0.05N.

 Determine the final volume of the solution after dilution.

11.4. Simple acid-base titrations

Activity 11.4

You are provided with 

Materials

- Burette - Indicator (phenolphthalein)
- Measuring cylinder - Washing bottle
- Conical flask - Beakers

- Retort stand - Funnel

Procedure

a. Using a pipette, transfer 10 mL of     into a conical flask. 
b. Add three drops of phenolphthalein indicator and titrate it with 
   from the burette.
c. Repeat the titration until you obtain consistent values.

d. Record your readings in the table below:

     

a. Calculate the average volume of  used.
b. Calculate the number of moles of HCl that react with 
c. Calculate the molarity of   in 10 mL.

Titration is the controlled addition and measurement of the amount of a solution 
of known concentration required to react completely with a measured amount of a 

solution of unknown concentration.

Acid-base titration

It is the determination of the concentration of an acid or base by exactly neutralizing 

the acid or base of known concentration

Alkalimetry and acidimetry

• Alkalimetry is the specialized analytic use of acid-base titration to determine 
                            the concentration of a basic substance. 
• Acidimetry is the same concept of specialized analytic use of acid-base titration 

                          to determine the concentration of an acidic substance.

Equivalence point

The point at which the two solutions used in a titration are present in chemically 
equivalent amount is the equivalence point. At this point the moles of two solutions 

will be equal.

             

Indicators and pH-meters can be used to determine the equivalence point. The 
point in a titration at which an indicator changes color is called the end-point of the 

titration.

Equipment's and set up (Figure 11.1) of materials for Titration

The common equipment used in a titration are:

• Burette
• Pipette
• pH-indicator/acid-base indicator
• White tile: used to see a color change in the solution(a white paper can also be 
  used)
• Conical flask (Erlenmeyer flask)
• Titrant: a standard solution of known concentration 

• Analyte: a solution of unknown concentration

                   

How to perform titrations

Knowing the use of pipette and burettes and how to handle them, the following points 

are useful in order for a correct titration to be done:

1. The apparatus should be arranged as shown in the above Figure.
2. The burette tap is opened with the left hand and the right hand is used to
     shake the conical flask. 
3. The equivalence-point is reached when the indicator just changes permanently
      the colour.
4. At the end point, the level of the titrant is read on the burette
5. The titration is now repeated, three more times are recommended. Towards 
    the end-point, the titrant is added dropwise to avoid overshooting. 

Notice: Before titration, check if the tip of the burette is filled with the titrant, and doesn’t 
contain bulb of air. If there is a bulb of air, a quick opening and closing of the tap will expel

the air out of the burette.

Choice of indicators in acid-base titrations

When the technique of acid-base titration is extended to a wide variety of acidic and 
alkaline solution, care needs to be taken about the choice of indicator for any given 

reaction.

The choice of an inappropriate indicator would lead to incorrect results, and it is 
therefore extremely important that the indicator is chosen carefully.

The principle on which a choice of indicator is made concerns the strength of the acid 
or base involved in the reaction. Note that the strength of an acid or base is not to be 
confused with the concentration of its solution. Example of strong and weak acids and 

bases and choice of indicator are given in the Table below.

                            Table 11.1. Examples of strong/weak acids and bases

          

 Indicators which are suitable for particular types of acid-base titrations are given in 

 the Table 11.2.

      Table 11.2. Examples of indicators

     

          The results are summarized in a table as shown below.

             Table 11.3. Sample results

     

        Notice:
                • Burette readings should be written to two decimal places (for burette having 

                  precision up to hundredth)

Average title should be obtained using values which differ not by more 

. Consistent values :24.00 and 24.10

Average volume  

Examples: 
1) Suppose 20.00 mL of     NaOH is required to reach the end-point in the 
titration of 10.0 mL of HCl of unknown concentration. How can these titration data 

be used to determine the molarity of the acidic solution.

Answer:
Begin with the balanced neutralization reaction equation. From the equation, 

determine the chemically equivalent amount of HCl and NaOH. 

       

Calculating the number of moles of NaOH used in the titration;

      

 

  

      

            Table 11.4. Burette readings
 

 Each value or entry in the table must be recorded or written to two decimal places.

Different initial readings should be used. Initial reading in each experiment should be 

correctly subtracted from the final reading. 

Questions:

  

Answer:

  

       

         

         

         11.5. Titration involving redox reactions

          Activity 11.5

1. What does differentiate redox reactions from other reactions?

2. Balance the following redox equations using half-equation method:

            

   11.5.1. Titrations with potassium manganate (VII); 

    Potassium manganate (VII),is a useful oxidizing agent in a sufficiently acidic 
medium normally used in the reactions such as:
• Oxidation of iron (II) to iron III
• Oxidation of ethanedioate (oxalates) or oxalic acid to carbon dioxide.
• Oxidation of nitrites to nitrates
• Oxidation of hydrogen peroxide to oxygen, etc.
During the reactions, the manganate (VII) ions, which is purple, is reduced to 
manganese(II),  which is pale pink but practically colorless.
Thus, titrations involving potassium manganate (VII), do not require an indicator. 
  acts as its own indicator.
The end-point of titration is detected when the solution shows a permanent faint pink 
color, which is as a result of slightly excess of potassium manganate (VII). 
For reduction of the potassium manganate (VII) to be complete, sufficient acid should 
be used and the solution added slowly, not rapidly. Use of insufficient acid results in 
precipitation of manganese (IV) oxide, which appears as a brown precipitate. 
The suitable acid for this process is sulphuric acid. Hydrochloric acid is unsuitable 
because the  ion is a stronger oxidizing agent than it oxidizes the chloride 

ion in the acid to molecular chlorine.

     Nitric acid is also not used to acidify the solution of potassium manganate (VII) since it 
     is also a powerful oxidizing agent. In potassium manganate (VII) titrations, an indicator 
    is used only when the reducing agent forms a colorless solution which makes it difficult 
    to observe the pink color.
    11.5.1. Titration of  by Potassium manganate (VII), 
      Ativity 11.5.1

    Balance the following redox equations using half-equations method:

    

Most redox titration labs utilize an iron (II) salt, frequently iron (II) sulphate 
heptahydrate. This compound is used for two reasons. First, it has a large molar mass, 
making it a good (but not the best) primary standard. Second, its solution is near 
colorless, so the pink/purple endpoint is easy to detect. Color proves to be an issue 
when dealing with many other compounds. Iron (II) salts are an excellent choice for a 

titration utilizing potassium permanganate.  

    

    

    Answer:

    Write balanced equation:

    
                   

         Answer
        Iron dissolves in dilute sulphuric acid by the reaction equation:
        
               

         

         11.5.2. Titration of oxalic acid or oxalates by potassium manganate (VII)

                        Answer

          

          The reducing agent is the oxalate ion, Oxalic acid is a white crystalline 

           solid of formula 

The reaction between the oxalate ion and manganate (VII) ion is kinetically slow at 
temperature below Titrations at room temperature take too long time for the 
purple color of the manganate (VII) ion to disappear. In the oxalate ion, carbon has the 
oxidation state of +3; when reacted with potassium permanganate the oxidation state 
changes into +4. Titrations involving the oxalates are a little more involved, but the 
results were very good. The reaction below describes the overall redox reaction:

       

          

          Answer:

         Overall equation is:

          

           Procedure

            i. A fixed mass 2.4 g of calcite is weighed and dissolved in dilute hydrochloric acid
            ii. To this solution is then added ammonia solution until the solution is alkaline 
                 and this is followed by addition of ammonium ethanedioate (oxalate) to 

                 precipitate all calcium ions as calcium oxalate

         iii. The precipitate is then filtered off, washed and then dissolved in a minimum 
               dilute sulphuric acid solution and then solution made up to in a 
               volumetric flask with distilled water. 
       iv. of the resultant solution in (iii) is pipetted and to it is added 20 cm3
             of diluted sulphuric acid and the mixture is heated to about 70 °C.

       v. The hot solution is then titrated with 0.02M potassium permanganate solution

       Results: of the oxalate solution required  of 0.02M   
                 solution  for complete reaction which is detected by the intense purple colour.
                    
                 a. Write down the overall equation for the above sequence of analysis

                 b. Calculate the percentage of purity of calcite 

         Answer:

         

            

Checking up 11.5.2

              

              11.5.3. Titration of hydrogen peroxide by potassium manganate (VII), 

               Activity 11.5.3

                 

Hydrogen peroxide is a powerful reducing agent in the presence of a strong oxidizing 
agent such as potassium permanganate. For this reason dilute concentrations of both 
compounds must be used. A common laboratory experimentally determines the 
concentration of commercially sold hydrogen peroxide (about 3%) via titration with 
potassium permanganate (usually prepared to be approximately 0.02 M). Concentrated 

solutions of these reactants will react explosively and must be avoided.

               

               Example:

          

          

             Answer

            Write the balanced equation

             

               

Note: The advantages of potassium permanganate as oxidizing agent are:
i. It acts as its own indicator (i.e. purple color)
ii. The crystals are obtained at high state of purity
iii. The crystals are anhydrous and not deliquescent 
iv. It has a fairly high relative molecular mass
v. A wide range of substances can be oxidized by it.

 However, potassium permanganate has disadvantages:
i. The crystals are not very soluble in water
ii. The compound is decomposed by the light
iii. The compound is reduced by water and organic matter from atmosphere

iv. The meniscus of the solution may be difficult to see 

Checking up 11.5.3

    

    11.5.4. Titration with potassium dichromate 

    Activity 11.5.4

Discuss the advantages and disadvantages of using  as an oxidizing agent in titration.

Potassium dichromate (VI) unlike potassium manganate (VII) is a primary standard. In 
the titration of potassium dichromate (VI), the color change is from orange to green
and this makes it not possible to detect the sharp end-point.

 Thus, an indicator is necessary for this reaction. Therefore redox indicator such as 
barium diphenylamine sulphonate must be used, which changes its colour to blue at 

the end point.       

Potassium dichromate (VI) is a weaker oxidizing agent than potassium manganate (VII). 
As such it cannot oxidize the chloride ion to chlorine, and therefore hydrochloric acid 
in addition to sulphuric acid can be used to acidify solution of potassium dichromate 

(VI).    

The reactions of potassium dichromate (VI) are similar to those of potassium manganate 
(VII), however at the end-point the color is green due to the presence of chromium 

(III) ions that is;

          

    a. Determination of the concentration of iodide ions.

   Since iodide ions can be oxidized to molecular iodine by dichromate (VI) ion, 

   potassium dichromate (VI) can be used as a primary standard in this reaction.

       

      The amount of iodine liberated in the reaction is determined by titrating it against a 

      standard solution of sodium thiosulphate using starch indicator.

 b. Analysis of iron (II) salts 

     

      Overall equation is:

        

 c. Analysis of nitrite salts

       

Advantages of  over 

 Unlike is used as a primary standard i.e. it can always be obtained pure; 
it is stable under ordinary conditions, its aqueous solution can be stored for long time 

if sufficiently protected from evaporation.

Examples 2 :

          

          Answer:

                

    Examples 2:

           

            Answer:

           Write balanced equation

             

       Checking up 11.5.4

       

        11.5.5. Titrations involving sodium thiosulphate and iodine

         Activity 11.5.5

          

          Sodium thiosulphate is a white crystalline solid with the formula
          It is not a primary standard as the water content of the crystals is variable. 
          The solid acts as a reducing agent in a redox reaction and the reducing agent being the 

           thiosulphate ion,

           

 However, the thiosulphate ion is very sensitive to acid (weak or strong). In presence 
of an acid (hydrogen ions), it decomposes to sulphur dioxide and elemental sulphur 

(which settles as a yellow precipitate). 

        

Therefore, solutions of sodium thiosulphate should never be acidified. 

The thiosulphate ions reduce iodine according to the reaction below:

         

Overall equation is:

           

a. Detection of the end point:

In titrations involving iodine and sodium thiosulphate, starch is normally used as an 

indicator.

Addition of the starch produces iodine-starch complex which is blue. Further drop 
by drop addition of the thiosulphate solution is continued until the solution turns 
colorless which marks the end point of titration.

Thiosulphate titration has important applications in the laboratory and water 
treatment plants.

Potassium iodate (V), potassium dichromate (VI), iodine, potassium manganate (VII) 
solutions can be used to standardize sodium thiosulphate solutions.
For example, potassium iodate (V) reacts with iodide ions in presence of an acid to 

liberate iodine. 

          

   The liberated iodine is then titrated against standard sodium thiosulphate using the 

   starch indicator.

             

 b. Preparation of starch solution:

0.5g of soluble starch powder is mixed with water into a thin smooth paste which is 
then poured into of boiling water. The mixture is then boiled after for about a 
minute. The solution is then cooled and used when still fresh.

But starch solution may be preserved by addition of few crystals of mercuric iodide. 

Example 1 :

          

 Answer:      

    

          

Example 2 :

A weighed sample of impure solid potassium iodate (V) of 0.80g is dissolved in 
water and made up to   of solution in a volumetric flask.   of this 
solution is reacted with excess potassium iodide solution, to liberate iodine. 
Find the percentage of the purity of potassium iodate (V). In the titration, 
  sodium thiosulphate were needed to react with the liberated 

iodine.

Answer:

Write balanced equations:

           

           

  c. Finding the percentage of copper in copper ore or in its salt 

  There are two possible methods depending on the copper ore (or salts).

a. In the first method which is applicable to any copper salt, a solution of excess 
    potassium iodide is added to a solution of the copper salt. A dark brown mixture 
    is produced (white precipitate stained brown). The brown colour is due to iodine 

    liberated.

      

  The iodine liberated in the reaction is now titrated with a standard solution of sodium 

thiosulphate in the second reaction according to the following equation.

         

 b. The second applicable method is to liberate iodine from iodate (V) ions, then titrate 

       it with thiosulphate ions as indicated by the following equations: 

          

Examples:

 An experiment was carried out in a laboratory to determine the percentage of copper 
in sample of impure copper metal. Nitric acid was added to a sample of impure 
copper metal. The resulting copper (II) nitrate was reacted with excess of potassium 
iodide. The iodine liberated was titrated with a solution of sodium thiosulphate of 
concentration 0.480M. The volume of sodium thiosulphate required was 

Use the following equations in your calculations.

         

Answer:

          

d. Determination of the percentage of available chlorine in a liquid

The method consists of measuring a fixed volume of the original liquid and then dilute 
it to with distilled water. Aliquot portions of this solution is pipetted and added 

to an excess of KI solution acidified with ethanoic acid.

The iodine liberated is then titrated, for example, with a standard solution of sodium 
thiosulphate. This is a displacement (redox) reaction in which, aqueous potassium 
iodide reacts with chlorine to form potassium chloride and iodine. The equation for 

the reaction is:

          

Examples:

1. Concentration of chlorine in treated water for domestic use can be monitored by 
testing water samples. In one such test, excess potassium iodide (KI) was added to 
  sample of water.
       

   Answer:
        

Reducing agent is I-(iodide ions)

c. Equations:

     

In the above reaction, chlorine underwent both oxidation and reduction, hence it is a 
disproportionation reaction.
e. Starch solution, which changes colour from white to blue black.

Determination of the percentage of available chlorine in bleaching powder.

Procedure:
Solution was made by weighing 2.5g of bleaching powder. This is stirred in a little 
water and the supernatant milk suspension poured into volumetric flask.
The solid residue is then mixed again with little water and the fine suspension decanted 
into the volumetric flask.
This process is continued until most of the sample has been transferred.

The solution is then made up to  mark. 

 The flask is then shaken vigorously and immediately 
 of the suspension in pipetted into a clean conical flask. 
 of 2M sulphulic acid or 2M ethanoic acid is added followed by 
 of 0.5M potassium iodide. The liberated iodine is then titrated with solution FA2

 which is 0.1M sodium thiosulphate solutions using starch indicator.

Specimen results:

     

    

   Calculations:

    b. Calculate the number of moles of available chlorine in 
   c. Calculate the mass of the available chlorine in the sample of bleaching 
       powder weighed.

  d. Hence calculate the percentage of available chlorine bleaching powder.

 Theory:

  

    Number of moles of iodine is equal to the amount of chlorine set free.

   Answers:

    

Checking up 11.5.5

             

                  

                 

11.6. Back titration

        Activity 11.6

Do research and explain the term back titration and discuss the applications of back 

titration.

This is a technique of determining the concentration of an unknown substance by 
calculating backwards. In this case a known amount of a titratable reagent (substance) 
X in excess amount is reacted with an unknown amount of substance Y to give products 
and at the end of the reaction, the excess of X is titrated with a standard solution of 

substance Z and this allows to determine the amount of Y that has reacted.

Applications of back titration:

1. Analysis of calcium carbonate materials (Examples: limestone, marble……..)
2. Determination of percentage of ammonia in an ammonium salt.

3. Determination of concentration of chromates and iodides in redox titration.

Example 1 :

     

    Answer: 

    Step 1: Write balanced equations

        

Step 2: Calculate initial number of moles of HCl

           

Step 3: Calculate number of HCl in excess solution

        

Step 4: Moles of HCl which reacts = 0.024- 0.008= 0.016mol

              Initial moles of reactants = 0.024mol 

               Number of moles in excess= 0.008mol

Step 5: Number of moles of HCl which react with  

                  

Example 2:

 50cm3 of 0.45M hydrochloric acid solution were added to 
 of sodium hydroxide solution and the resultant solution on titration required           
 of 0.2M sodium hydroxide solution for complete reaction. Calculate the:

a. Number of moles of hydrochloric acid that reacted with original sodium 
     hydroxide solution.

b. Molarity of the original sodium hydroxide solution.

Answer:

       

Total number of moles of HCl:

       

   Example 3:  

             

Answer:

            a. Equation:

            

    b. of the resultant solution contain 0.00478mole of NaOH   

           

    c. NaOH reacts with  Cl as shown below:

        

      

   d. From the reaction in (c) above

         

Checking up 11.6

           

   11.7. Applications of titration

11.7.1. Determination of the number of moles of water of crystallization

Activity 11.7.1

Do research and explain the term water of crystallization
Water of crystallization is defined as that definite amount of water
which a substance associate with on crystallizing out of an aqueous solution.
Many crystals cannot form without the presence of water i.e. water of hydration.

Examples of some hydrated substances

                

a. Weighing method

This method is preferably used when the crystals are neutral, and they cannot react 

with acid or basic solution.

Procedure:

    

 • Weigh the mass of a clean dry crucible with lid.
• Two or three grams of hydrated are added and the whole set up is 
   weighed.
• The crucible, lid and the content are heated on a pipe-clay triangle on a tripod 
  stand, gently at first and later strongly, to drive off water of crystallization.
• The set up is allowed to cool in a dessicator (to exclude moisture) and 
   reweighed.
• Heat again the crucible, lid and content and then cool as before and weigh 
  again.
• Repeat the process of heating, cooling and reweighing until a constant mass is 
   reached which shows that all water has been expelled.
• Using calculations, it is possible to get x moles of water of crystallization found 

   in one mole of substance. See the following example:

Example:

4.99g of hydrated copper (II) sulphate,   was strongly heated until 
all water of crystallization was eliminated. The mass of anhydrous copper 
sulphate was found to be 3.19g. Calculate the number of moles of water of 
crystallization contained in one mole of (Cu = 63.5, S = 32, O 

=16, H = 1)

Answer:

         

b. Using titration

       

Answer

Checking up 11.7.1

         

 11.7.2. Determination of atomic masses 

Activity 11.7.2

4.99g of hydrated copper (II) sulphate,  was strongly heated until 
all water of crystallization was eliminated. The mass of anhydrous copper 
sulphate was found to be 3.19g. Determine the number of moles of water of 

crystallization(x).

It is possible to find the atomic mass of the metal atom in titrating a basic compound 

which reacts with acidic solution.

Example:

     

Answer:

           

    

Checking up 11.7.2

        

11.8. End Unit Assessment

         1. Explain the following terms:
         a. Standard solution

         b. Primary standard.

2. Describe the characteristics of a primary standard.

       

        

        a. Calculate the value of x in the formula
       b. Calculate the value of y in the formula
       c. Calculate the value of z in the formula
       d. Calculate the molecular mass of the oxalate

       e. Calculate the number of water of crystallization n.