Chemistry

LEARNING OBJECTIVES

After reading this unit, you will be able to:

•define solubility.

•describe factors that affect solubility.

•explain the concept of unsaturated, saturated and supersaturated solutions.

•explain the solubility curves of different salt solutions.

•describe different methods of preparing soluble and insoluble salts.

•name the sources and uses of salts in daily life.

KNOWLEDGE GAIN

Liquid crystals are matter that has properties between those of liquid and solid crystal. Liquid crystals can be found naturally.For example, many proteins and cell membranes are liquid crystals. Other well-known example of liquid crystals is solution of soap.

ACTIVITY 6.1: Demonstrating Unsaturated, Saturated and Supersaturated Solutions

Materials Required

Three hard glass beakers (500 ml), sugar, water, and spoon.

Procedure

•Fill all three beakers with equal amount of water.

•Label them A, B and C.

•Add one teaspoon sugar in beaker A, two teaspoons sugar in beaker B, and three teaspoons in beaker C.

•Stir with spoon.

•Observe the beakers.

•Heat the beaker C and observe the beaker.

•Let beaker C sit uncovered for some days to evaporate 1/2 of water.

•Take beaker B and hang a string inside it.

•Leave beaker B for 45–60 minutes undisturbed.

•Observe the beaker B, after 45–60 minutes.

•Cover the beaker B to avoid further evaporation and let it sit for some days.

•Observe beaker B again.Did you see crystals?

In Activity 6.1,

•Solution in Beaker A is unsaturated.

•Solution in Beaker B is unsaturated.

•Solution in beaker C is saturated.

After heating, sugar left in beaker C also dissolves completely, so the solution in beaker C is unsaturated.

After evaporation ½ of water, beaker C contains 3 teaspoons of dissolved sugar. Now the solution in beaker C is supersaturated.

Definition

•Solubility: When a substance dissolves in water or in other liquids (solvent), it is said to be soluble in it. The properties of a substance by which it tends to dissolve in water (or any other solvent) is called its solubility.

•Unsaturated solution: A solution that contains less solute that can dissolve at a given temperature is an unsaturated solution.

•Saturated solution: A solution that contains as much solute as can dissolve in the given solvent at a given temperature is a saturated solution.

•Supersaturated solution: A solution that contains the maximum amount of solute at an elevated temperature is a supersaturated.

  6.1 SATURATED AND UNSATURATED SOLUTIONS

ACTIVITY 6.2: Testing Whether a Given Solution of a Substance is Unsaturated, Saturated, or Supersaturated, at a Particular Temperature

Prepare an unsaturated, saturated and a supersaturated solution of common salt separately in three beakers. Now, add a small amount of the crystal of common salt turn by turn to each of the solution.

•If it dissolves and the concentration of the solution increases, it is unsaturated.

•If it does not dissolve even on vigorous stirring and the concentration of the solution remains the same, it is saturated.

•If it grows in size and the concentration of the solution falls, it is supersaturated.

If a well-powdered solute, such as copper sulphate is added little by little to a definite volume of water at room temperature with constant stirring, it dissolves continuously, till a point comes, when no more of the copper sulphate will dissolve. The excess of the copper sulphate settles down to the bottom. The solution at this stage is said to be a saturated solution at that temperature. Thus, a saturated solution at a particular temperature contains as much solute as it can dissolve at that temperature.

At that temperature, solvent and solute are in a state of equilibrium. This state of equilibrium or saturation can be changed either by adding more water (solvent) to it or by increasing the temperature of solution. For example, the saturated solution at a given temperature becomes unsaturated when heated because more of solute will be required to make the solution saturated at higher temperature. Therefore, different amounts of the same solute are required to prepare saturated solutions at different temperatures. In other words, a solution saturated at one temperature may not be saturated at another temperature.

A solution which is unable to dissolve any more of the solute at a particular temperature is called a saturated solution at that temperature. At every step, before obtaining a saturated solution, a solution is always unsaturated with respect to the solute. An unsaturated solution contains less solute than it can dissolve at that temperature. It becomes more unsaturated with the rise of temperature

.A solution which can dissolve more solute at a given temperature is called unsaturated.

EXERCISE 6.1

1. How can state of equilibrium be changed?

2. A solution which can dissolve more solute at a given temperature is called unsaturated.

                                                                                                              (True or False)

3. In your own words, define saturated solution.

4. An unsaturated solution contains _____________ solute than it can dissolve at that temperature.

5. Distinguish between saturated and unsaturated solutions.

 6.2 SUPERSATURATED SOLUTIONS

A saturated solution prepared at a higher temperature contains more of a given solute than does a saturated solution prepared at a lower temperature. So, when a saturated solution prepared at a higher temperature is cooled to room temperature, it will tend to throw out the excess of the solid from the solution in the form of crystals. It takes, however, sometime for the excess solute to come out. During this interval, solution holds in it more solute than is required to saturate it. Such a solution is called a supersaturated solution. Thus, a supersaturated solution is one which has more of the solute than a saturated solution requires.

The supersaturated solution in contact with the solid solute starts depositing the solute so as to reach the concentration of a saturated solution. Thus, in a supersaturated solution the concentration of the solute falls when it comes in contact with the solute.

A solution in which more solute is dissolved by increasing the temperature of a saturated solution is called supersaturated solution.

6.2.1 Dilute and Concentrated Solutions

ACTIVITY 6.3: Identifying Dilute and Concentrated Solution

Take equal amount of water in two beakers. Put a spoonful of sugar in one beaker and three teaspoons in the other. Stir with the glass rod.

Can you tell which solution is more sweet?

The solution containing three teaspoonful of sugar is more sweeter. This solution is concentrated whereas another one is dilute.The quantity of solute dissolved in a certain quantity of solvent, by weight or volume denotes the concentration of the solution. Accordingly two types of solutions are known – dilute and concentrated.

A solution containing relatively small amount of solute in a fixed amount of solvent or compared to that of the solvent is a dilute solution.

The solution made by mixing a teaspoon of sugar in a cup of water is a dilute solution.

Solution containing relatively more quantity or large amount of solute in the fixed amount of solvent is a concentrated solution. The solution made by mixing three teaspoons of sugar in a cup of water is a concentrated solution.

EXERCISE 6.2

1. What do you mean by supersaturated solution?

2. How can you distinguish between dilute and concentrated solutions?

3. Explain the formation of crystals

6.3 FACTORS INFLUENCING SOLUBILITY OF DIFFERENT SALTS

The solubility of solid (solutes) in liquid (solvent) depends on

i) Nature of solute. Dissolution of solid solutes in liquids can be summed up in a phrase “like dissolves like”. Some solutes such as NaCl, KCl, KNO₃, etc., have larger solubilities in water. On the other hand, solids such as I₂ and S₈ are not soluble in water they dissolve in CCl₄, CS₂.

(ii) Temperature. Temperature has a direct effect on solubility. The solubility of most of the ionic compounds increases with increase in temperature. On the other hand, some compounds dissolve better by decreasing the temperature.

Note:The solubility of solids and liquids increases with increase in temperature.The solubility of gases always decreases with increase in temperature

Temperature can also increase the amount of solute that can be dissolved in a solvent. Generally speaking, as the temperature increases, more solute particles will dissolve.For example, when you add potassium chloride (KCl) to water, a solution is easily made. When you heat this solution and keep adding KCl, you will find that a large amount of KCl can be dissolved as the temperature keeps rising. This occurs because as the temperature increases the inter molecular forces can be easily broken and allow more solute to dissolve.

6.3.1 How does Solubility Change with Temperature?

Molecules in liquid are less closely packed and the inter molecular forces of attraction between the molecules are weaker than those in solids. The molecules of the liquids are constantly moving in different directions with different speeds. Thus, on heating, the kinetic energy of the molecules will increase more and move faster.

Molecules of the solid solute are held together by strong inter molecular force of attraction. So the molecules in solids are closely packed. When the temperature is increased, the kinetic energy of the molecules will increase which makes the molecules to move fast.Similarly, stirring the solution with a spoon or a glass rod helps to increase the kinetic energy of the molecules of solvent. Thus, on heating or stirring, the kinetic energy of the molecules of the solute increases. The molecules of the solute strike each other and get separated from each other. These separated solute molecules get mixed with the solvent molecules to form a solution.

When the molecules vibrate more at high temperature, the inter molecular force of attraction between the molecules weakens and the space between them increases. That’s why warm water dissolves more solute than cold water. Thus, the solubility of the substances increases with the rise of temperature.

Thus, the solubility of any solid in a solvent depends on temperature. The solubilities of most solutes like potassium nitrate, copper sulphate, ammonium chloride etc. increase with increase in temperature. The solubilities of sodium chloride and potassium chloride etc. increase slightly with an increase in temperature. Solubilities of some substances in water such as calcium sulphate, calcium hydroxide and sodium sulphate decrease with the rise of the temperature.

EXERCISE 6.3

1. Complete the phrase:Like dissolves ___________.

2. I₂ and S₈ are not soluble in __________ but they are soluble in __________.

3. The solubility of gases always increases with increase in temperature.

                                                                         (True or False) 

4. What are the factors that affect solubility?

5. The solubility of calcium sulphate in water

(a) decreases with rise of temperature     (b) increases with rise of temperature

(c) cannot be determined                         (d) first increases and then decreases with rise of temperature.

6.4 SOLUBILITY CURVE

The variation in the solubility of any given substance with change of temperature is shown by solubility curve. The graph showing relationship between temperature and solubility of the substance at different temperatures is called a solubility curve

To draw solubility curves, temperature is represented along the X-axis and solubility along the Y-axis. Various solubility points plotted are connected by a smooth curve which is a solubility curve.

The solubility of copper sulphate at different temperatures is given in Table 6.2.

The solubility curve drawn for table 6.2 data is shown below.

The solubility curves of some substances is shown in Figure 6.3.

The solubility of sodium chloride rises a little with rise of temperature. The solubility of lead nitrate, potassium nitrate and sodium nitrate increases rapidly with increase of temperature.

The solubility of calcium sulphate and calcium hydroxide decreases with rise of temperature.The general shape of the curve indicates the rate of change in the solubility with a rise in temperature. A steep curve, e.g., that of potassium nitrate shows that solubility increases rapidly with rise of temperature. A flat curve, e.g., that of sodium chloride indicates that solubility increases slowly with the rise of temperature.

After studying the solubility curve, the following information can be obtained.

1. The solubility of a substance at a particular temperature can be determined.

2. The solubility of a given substance at any temperature can be determined.

3. The solubility curve helps us predict which substance will crystallize out first from a solution containing two or more solutes.

4. The solubility curve helps us compare the solubilities of different substances at the same temperature.

5. It brings the change in the composition of a solute substance.

6. It gives a clear idea that solubility of substance changes with the temperature.

Example 1

Answer the following questions using the solubility graph below

1. How much sodium nitrate will dissolve at 30°C?

2. Which solid is most soluble at 60°C?

3. Which solid is least soluble at 40°C?

4. At what temperature will 60 g of sodium sulphate dissolve in 100 g of water?

Solution

1. Looking at the following solubility graph, draw a line up (vertically) from 30°C until it touches the NaNO₃ line. Following this, locate the point on Y-axis to find the amount of NaNO₃ that dissolves. Therefore, approximately 95 g of NaNO₃ will dissolve in 100 g of water at 30°C.

2. The highest line at 60°C is the green line (NaNO₃), therefore it is the most soluble at 60°C.

3. The lowest line at 40°C is the purple line (NaCl), therefore NaCl is the least soluble at 40°C.

4. Looking at the solubility graph on page 137, draw a line over (horizontally) from60 g until it touches the Na₂SO₄ line. Following this, locate the point on X-axis to find the temperature at which 60 g of Na₂SO₄ will dissolve. Therefore, 60 g of Na₂SO₄will dissolve in 100 g of water at 50°C

6.5 CALCULATION OF SOLUBILITY

ACTIVITY 6.4: Determining the Solubilities of Potassium Sulphate, Sodium Chloride and Sodium Nitrate at Room Temperature

•Take 100 grams of water in each of three separate beakers at room temperature.

•Take the weights of all the three beakers with water separately.

•Add potassium sulphate to the first, common salt to the second and potassium nitrate to the third beaker.

•Add and stir well to dissolve the required solute to make each solution saturated.

•Take the weight of each beaker containing saturated solution.

•Calculate the amount of solute added in each beaker containing 100 grams of water to make it a saturated solution at room temperature

Calculation

Weight of the solvent in each beaker = x grams

Weight of the saturated solution in each beaker = y grams\ The weight of the solute = (y – x) grams.

In other words, the difference in the weight of the two is the amount of a solute required to form a saturated solution at room temperature. If the above activity is done properly, following result will be obtained.

The result of this activity shows that for different substances, the amount of solute needed to make a saturated solution in the same solvent is different at a particular temperature.

The amount of solute (in grams) required to form a saturated solution in 100 grams of solvent (water) at a particular temperature is called the solubility of the substance at that temperature.

The above activity shows that solubility of sodium chloride is 35 at 20°C, that of potassium nitrate is 60 at 20°C and that of potassium sulphate is 12 at 20°C.

Here,

Different substances have different solubilities. Solubilities of some common substances in water at 20°C and 30°C are given in table below.

             Solubilities of Some Common Substances

Above table shows clearly that the solubility of the solute increases with the rise in temperature, but the solubility of sodium chloride increases slightly with the rise in temperature.

6.5.1 Solved Numerical

Example 1

The solubility of a solute at 30°C is 40. What amount of water is required to make saturated solution of 80 grams of a solute?

Solution

Weight of a solute = 80 grams

Solubility at 30°C = 40

Weight of solvent (water) = ?

According to the formula,Solubility

Example 2

7 grams of saturated solution of salt saturated at 60°C is evaporated to dryness; 2 grams of white residue is left behind. What is the solubility of salt at that temperature?

Solution

Weight of saturated solution = 7 grams

Weight of solute (Salt) = 2 grams

Solubility of salt at 60°C = ?

To find the weight of solvent (water),

Weight of Solution

= Weight of Solute + Weight of Solvent

Weight of Solvent

= Weight of Solution – Weight of Solute

Weight of Solvent

= 7 grams – 2 grams = 5 grams

To find solubility, according to the formula,Solubility

The solubility of salt at 60°C is 40.

Example 3

At 30°C, 7 grams of sugar dissolves in 5 grams of water to form a saturated solution. Find the solubility of sugar?

Solution

Given, Mass of solute (sugar) = 7 grams

Mass of solvent (water) = 5 grams

Solubility at 30°C = ?

According to the formula,

Solubility

The solubility of sugar at 30°C is 140.

Example 4

How much copper sulphate will be dissolved in 30 grams of its saturated solution. Solubility of the salt at 30°C is 35. Also find the weight of solvent.

Solution

Weight of saturated solution of copper sulphate = 30 grams

Solubility of copper sulphate at 30°C = 35

Let the amount of solute (copper sulphate) be x grams

Amount of solvent = (30 – x) grams

Solubility of copper sulphate = 35

According to the formula,

Therefore, weight of solvent in 30 grams of saturated solution at 30°C

= Weight of saturated solution – Weight of solute (x)

= 30 grams –7.78 grams

= 22.22 grams

6.6  DIFFERENT WAYS OF PREPARING NORMAL SALTS

6.6.1 Preparing Salts from Reaction of Acid with Active Metal

ACTIVITY 6.5: Preparation of Zinc sulphate ZnSO4.7H2O (White Vitriol)

•Take about 40 cm³ of 4H2SO4 in a 250 cm³ beaker and add about 10 g of zinc granules to the acid. Zinc reacts with dilute H₂SO₄ and hydrogen gas is evolved.

•When the evolution of gas ceases, it indicates that the reaction is over. Filter the solution to remove un reacted zinc.

•The solution thus obtained is heated in a china dish and concentrated to crystallization point.

•Cool the solution. On cooling, needle shaped crystals of white vitriol(ZnSO₄.7H₂O) are obtained. The crystals are separated and dried between the folds of a paper.

Zinc sulphate is a water soluble salt. It can be prepared in the laboratory by the action of dilute H₂SO₄ on zinc granules.

6.6.2 Preparing Salts from Reaction of Acid with Carbonates and Hydrogen Carbonate

ACTIVITY 6.6: Preparation of Sodium Chloride

•Take two test tubes, label them A and B.

•Take about 0.5 g sodium carbonate (Na₂CO₃) in test tube A and about 0.5 g of sodium hydrogen carbonate (NaHCO₃) in test tube B.

•Add about 2 ml of dilute HCl to both the test tubes.

•What do you observe?

The reactions occurring in Activity 6.6 are written as:

Test tube A:

Test tube B:

EXPERIMENT 1

Aim

To prepare calcium chloride by the action of an acid on an insoluble carbonate

Materials Required

•measuring cylinder

• hydrochloric acid (50 cm3)

•250 cm3 beaker (2)

• calcium carbonate powder (6 g)

•filter funnel

• filter paper

•conical flask

• evaporating dish

•Bunsen burner

• wire gauze

•glass spatula

• plastic vial

•tripod stand

• electronic top pan balance

Procedure

•Take 50 cm³ dilute hydrochloric acid in a beaker.

•Using an electronic top pan balance, weigh 6.0 g of calcium carbonate in a small plastic bottle.

•Add a spatula of calcium carbonate from the plastic bottle into the acid in the beaker. Stir the mixture until all the calcium carbonate has dissolved before adding more calcium carbonate.

•Continue to add more calcium carbonate to the acid until no more can dissolve and some of it is left at the bottom of the beaker.

•Filter off the excess calcium carbonate with a piece of filter paper and a funnel.

•Collect the filtrate and place it into an evaporating dish. Obtain crystals of calcium chloride from the solution by evaporation followed by crystallization.

Conclusion

Calcium chloride is a salt of calcium and chlorine. Anhydrous calcium chloride is solid white powder at room temperature.

6.6.3 Preparation of Salts from Reaction of Acids with Metal Oxide

ACTIVITY 6.7: Preparation of Copper Sulphate CuSO₄ . 5H2O(Blue Vitriol)

•Take about 50 cm3 of dilute H2SO4 in a 250 cm3 beaker and warm it gently.

•Add black copper (II) oxide to the above acid in small portions. Initially, Copper (II) oxide reacts with the acid and dissolves. When no more of the oxide dissolves, the solution is filtered to remove un reacted copper (II) oxide.

•The bluish filtrate is concentrated by heating in a china dish. When the hot saturated solution is obtained, the solution is allowed to cool undisturbed.

•On cooling, blue crystals of blue vitriol (CuSO₄.5H₂O) are formed. The crystals are separated and dried between the folds of a filter paper.Copper sulphate can be prepared by the reaction of dilute H₂SO₄ with copper (II) oxide

In this activity, you will observe that black copper oxide (CuO) dissolves in dilute sulphuric acid (H2SO4) to form a blue copper sulphate solution.On heating the copper sulphate solution, a saturated solution is obtained. On cooling the saturated solution, we get the crystals of copper sulphate.

Note: All metal oxides react with acids but there is not always a colour change. Magnesium Oxide is a white powder. The salt produced when it reacts with acid is also white.

6.6.4 Preparing Salts from Reaction of Acids with Bases

ACTIVITY 6.8: Preparation of Sodium Sulphate

•Take about 2 ml of dilute NaOH solution in a test tube and add two drops of phenolphthalein solution.

•What is the color of the solution?•Add dilute sulphuric acid solution to the above solution drop by drop.

•Is there any color change for the reaction mixture?

•Why did the color of phenolphthalein change after the addition of an acid?

•Now add a few drops of NaOH to the above mixture.

•Does the pink color of phenolphthalein reappear?

•Why do you think this has happened?

In Activity 6.8, you will observe that the effect of a base is nullified by an acid and The reaction between an acid and a base to give a salt and water is known as a neutralization reaction. In general, a neutralization reaction can be written as:

Base + Acid → Salt + Water

ACTIVITY 6.9: Preparation of Lead Carbonate

Materials Required

Sodium carbonate, lead nitrate test tubes, glass rod

Procedure

•Make a solution of sodium carbonate with water.

•Make a solution of lead nitrate with water.

•Take 20 ml solution of sodium carbonate solution in a test tube.

•Add 20 ml lead nitrate solution to the test tube.

•Observe the mixture (reaction).

Reaction

Observation In this activity, you will observe that a solid white precipitate is produced. This is lead carbonate. The lead nitrate and sodium carbonate have exchanged their ions. This is an example of double replacement reaction.

                     We have already studied the neutralization reaction in Unit 5.

6.6.5 Preparation of Salt from Metal Oxides and Non-metal Oxide

Salts are also prepared from the reaction of metal oxides with non-metal oxides. Some examples are:

(i) Sodium oxide reacts with carbon dioxide to give sodium carbonate. The reaction takes place at a temperature of 500–550°C.

Na₂O(s) + CO₂(g) → Na₂CO₃(s)

(ii) Potassium oxide reacts with carbon dioxide to form potassium carbonate. The reaction takes place at a temperature of 400–450°C.

K₂O(s) + CO₂(g) → K₂CO₃(s)

(iii) Calcium Oxide reacts with sulphur trioxide to form calcium sulphate.

CaO(s) + SO₃(l) → CaSO₄(s)

(iv) Magnesium oxide reacts with carbon dioxide to form magnesium carbonate.

MgO(s) + CO₂(g) → MgCO₃(s)

Note:Metal oxide reacts with carbon dioxide at very high temperature.6.6.6 Preparing Salt with a Base and Non-metal Oxide

6.6.6 Preparing Salt with a Base and Non-metal Oxide

ACTIVITY 6.10: Preparation of Calcium Carbonate

•Take calcium hydroxide solution in a test tube.

•Pass carbon dioxide gas through lime water (calcium hydroxide solution) and record your observations.

Reaction

Note: The salts obtained by reaction of hydrochloric acid are called chlorides.

The salts obtained by reaction of sulphuric acid are called sulphates.

The salts obtained by reaction of nitric acidare called nitrates.

The salts obtained by reaction of carbonic acid are called carbonates.

The salts obtained by reaction of acetic acidare called acetates.

EXERCISE 6.4

1. Copper oxide dissolves in dilute sulphuric acid to form __________.

2. Write the molecular formula of

(a) White vitriol (b) Blue vitriol

3. Metal oxide reacts with carbon dioxide at very high temperature.

                                                   (True or False)

4. Complete the following reactions:

(a) CaO(s) + SO₃(g) →?

(b) MgO(s) + CO₂(g) →?

5. The salts obtained by reaction of sulphuric acid are

(a) chlorides (b) carbonates(c) acetates(d) none of these.

6.7 USES AND SOURCES OF SALT

Salt is all around us—underground and on the earth’s surface in the dried up residues of ancient seas. Some salts have even arrived from outer space in meteors. But our biggest sources of salts are seas and oceans.There are many different types and grades of salt and a number of different methods of production. White salt is produced by

evaporating ‘solution-mined’ brine in pressure vessels. The rock salt we use for constructing roads comes from mining ancient deposits. In some countries the natural energy of the sun is used to evaporate brine produced from sea water.

6.7.1 Sodium Chloride (NaCl)

Sodium chloride (NaCl) is known as common salt. It is present in sea water along with many other salts. Sodium chloride is separated from these salts. It also exists as solid salt, in the form of rocks, in several parts of the world. The large crystals of sodium chloride obtained from rocks are generally brown in colour due to impurities present in them. This is known as Rock salt. Rock salt is mined just like any other mineral. Common salt is an important component of our food. It is also used for the extraction of sodium metal.

6.7.2 Washing Soda (Na₂CO₃.10H₂O)

Washing soda is prepared from sodium chloride by the Ammonia-soda process or Solvay process. In this process, CO₂ gas is bubbled through a brine solution saturated with ammonia. It results in the formation of sodium hydrogen carbonate.

NaHCO₃ so formed precipitates out in the presence of excess of sodium chloride. It is filtered off and then ignited to get sodium carbonate (Na₂CO₃).

Anhydrous sodium carbonate thus formed is called soda ash. When soda ash is dissolved in water and subjected to crystallization, the crystals that separate out are of sodium carbonate decahydrate (Na₂CO₃.10H₂O) which is also known as washing soda.

Uses of Sodium Carbonate

•Large quantities of sodium carbonate are used in the manufacture of glass, borax, soap and caustic soda.

•It is used in the paper, paints and textile industries.

•It is used for softening hard water. It removes temporary as well as permanent hardness.

•It is used for washing purposes in the laundry.

•As an important laboratory reagent both in qualitative and quantitative analysis.

6.7.3 Baking Soda, Sodium Hydrogen-carbonate (NaHCO₃)

Sodium hydrogen carbonate is obtained as a primary product in the Solvay’s process for the manufacture of Na₂CO₃.

It can also be prepared by passing CO₂ through a solution of Na₂CO₃ whereby NaHCO₃ being less soluble crystallizes out.

Uses of Sodium Hydrogen carbonate

•It is used as a component of baking powder. In addition to sodium hydrogen carbonate, baking powder contains mild edible acid such as tartaric acid or some other similar acidic compound. When mixed with water and heated, sodium hydrogen carbonate reacts with the acidic component of the baking powder producing CO₂ which causes bread or cake to swell and become light. Tartaric acid present in baking powder neutralizes sodium carbonate which would otherwise impart bitter taste to the food item.

•It is used in soda-acid fire extinguishers.Soda-acid fire extinguishers contain a solution of NaHCO₃ and H₂SO₄. These two chemicals are brought into contact by pressing a knob or by inverting the extinguisher. CO₂ is produced which forces a stream of effervescing gas on the fire. CO₂ surrounds the combustible substance and cuts off the supply of oxygen in air. Thus, the fire gets extinguished.

•It is used in medicines. It acts as mild antiseptic for skin infections. It is also present as an ingredient in antacids. Being alkaline it neutralizes excess acid in the stomach.

EXERCISE 6.5

1. Write two uses of sodium carbonate.

2. What is the molecular formula of washing soda?

3. The chemical name of baking soda is __________.

4. Sodium chloride is present as an ingredient in antacids.

5. The crystals of sodium chloride obtained from rocks are generally white in color.

                                                                                                            (True or False)

6.8 IDENTIFICATION OF IONS

6.8.1 Identification of Cations

For identification of cations (positive ions), dilute sodium hydroxide (NaOH) solution and dilute ammonia solution are used.

Most cations combine with hydroxide ions to form insoluble metal hydroxide.

Mⁿ⁺(aq) + nOH^–(aq) → M(OH)_n(s)

Note: For the identification of both cations and anions, precipitation reactions are used. The chemical reagent used for the precipitation of insoluble salt of the concerned ion is called precipitating reagent.

NH⁺₄ ions do not form a precipitate with both aqueous sodium hydroxide and aqueous ammonia. When a mixture of a strong alkali (NaOH) and ammonium ions are heated, ammonia gas is given out                     

Ammonia gas can be detected by its characteristics smell. It changes moist red litmus paper into blue.

Ca²⁺ ions form a white precipitate with aqueous sodium hydroxide but not with aqueous ammonia. The precipitate does not dissolve in excess sodium hydroxide.

Zn²⁺ ions form white precipitate with both aqueous sodium hydroxide and aqueous ammonia. The precipitate dissolves in excess of both alkalis.

Mg²⁺ ions form a white precipitate with both aqueous sodium hydroxide and aqueous ammonia but the precipitate does not dissolve in excess of both alkalis.

Cu²⁺ ions form a blue precipitate with both aqueous sodium hydroxide and aqueous ammonia. The precipitate dissolves in excess ammonia to form a dark blue solution.

Fe²⁺ ions form a green precipitate with both aqueous sodium hydroxide and aqueous ammonia. The precipitate does not dissolve in excess alkali.

Fe³⁺ ions form a brown precipitate with both aqueous sodium hydroxide and aqueous ammonia. The precipitate does not dissolve in excess alkali.

Al³⁺ ions form a white precipitate with both aqueous sodium hydroxide and aqueous ammonia. The precipitate dissolves in excess aqueous sodium hydroxide but not in excess aqueous ammonia.

Figures 6.4 and 6.5 summarize the reaction of cations with aqueous sodium hydroxide and aqueous ammonia respectively.

EXERCISE 6.6

1. NH⁴⁺ ions form precipitate with aqueous sodium hydroxide. (True or False)

2. Fe²⁺ ions form __________ precipitate with aqueous sodium hydroxide.

3. How will you identify Cu²⁺ ions in a sample of solution?

4. The precipitate formed by Al³⁺ ion does not dissolve in

(a) water (b) excess aqueous sodium hydroxide (c) excess aqueous ammonia (d) All of these.

5. Fe³⁺ ions form a brown precipitate with __________ and __________.

6.8.2 Identification of Anions

The common chemical reagents used in the identification of anions are dilute hydrochloric acid (HCl), aqueous barium nitrate (Ba(NO₃)₂), aqueous silver nitrate (AgNO₃), dilute nitric acid (HNO₃) and dilute ammonia (NH₃) solution. Besides, red litmus is also used.

Cl^– ions form white precipitate when aqueous silver nitrate (AgNO₃) followed by excess dilute nitric acid (HNO₃) is added to salt solution. The precipitate does not dissolve in excess dilute nitric acid, but dissolves in excess aqueous ammonia. The white precipitate darkens if kept in sunlight for 10–15 minutes.

I^– ions form yellow precipitate when acidified with dilute nitric acid followed by aqueous lead nitrate [Pb(NO₃)₂] or silver nitrate. The precipitate does not dissolve in excess ammonia, but dissolves in hot water forming a colorless solution.NO^–₃ ion does not form any precipitate with chemical reagent when a mixture of aqueous sodium hydroxide and nitrate ion along with aluminium powder is heated, ammonia gas is given out.

or

Hydrogen gas is generated from the reaction between aluminium and sodium hydroxide.

Nitrate ion decomposes into nitrogen dioxide and oxygen when heated.

Nitrogen dioxide gas is brown in color whereas oxygen is a colorless gas. Oxygen gas is tested by a glowing splint.

Nitrates can be reduced by hydrogen to ammonia.

CO^2–₃ ion turns moist red litmus into blue. Effervescence (bubbles) of colorless and odorless carbon dioxide gas forms when carbonates react with dilute hydrochloric acid.

Carbon dioxide gas can be tested with lime water. CO₂ gas turns lime water milky. Excess carbon dioxide clears the milkiness.

SO^2–₃ ion form white precipitate with both aqueous barium nitrate [Ba(NO₃)₂] and aqueous barium chloride (BaCl2). The precipitate dissolves in both dilute hydrochloric acid and dilute nitric acid.SO2–4 ions form white precipitate with both aqueous barium chloride and aqueous barium nitrate. The precipitate neither dissolves in dilute nitric acid nor in hydrochloric acid. Figure 6.6 summarizes the reaction of anions with chemical reagent in order to identify them.

EXERCISE 6.7

1. How will you identify SO₃^2– ions in a solution?

2. I^– ions form __________ precipitate when acidified with dilute nitric acid followed by aqueous lead nitrate.

3. Name three reagents used in the identification of anions.

4. Nitrate ion decomposes into nitrogen dioxide and oxygen when heated.(True or False)

5. CO₃^2– ion turns moist __________ litmus into __________.

6.8.3 Identification of Gases

The following table gives the tests for identifying gases.

6.9 SUMMARY

•A solution which is unable to dissolve any more of the solute at a particular temperature is called a saturated solution at that temperature.

•A solution which can dissolve more solute at a given temperature is called an unsaturated.

•A solution containing relatively small amount of solute in a fixed amount of solvent or compared to that of the solvent is a dilute solution.

•Solution containing relatively more quantity or large amount of solute in the fixed amount of solvent is a concentrated solution.

•The solubility of any solute in a solvent depends on temperature.

•The graph showing relationship between temperature and solubility of the substance at different temperature is called a solubility curve.

• The general shape of the curve indicates the rates of change in the solubility with a rise in temperature.

• Solubility =  

•For the identification of cations and anions, precipitation reactions are used.

•For the identification of cations (positive ions), dilute sodium hydroxide (NaOH) solution and dilute ammonia solution are used.

•The common chemical reagents used in the identification of anions are dilute hydrochloric acid (HCl), aqueous barium nitrate (Ba(NO₃)₂), aqueous silver nitrate (AgNO₃), dilute nitric acid (HNO₃) and dilute ammonia (NH₃) solution. Besides, red litmus is also used.

6.10 GLOSSARY

•Anhydrous: substance containing no water.

•Bauxite: an amorphous clay rock that is the chief commercial ore of aluminium.

•Borax:a white compound which occurs as a mineral in some alkaline salt deposits and is used in making glass and ceramics.

•Crystallization: a separation technique that is used to separate a solid that has dissolved in a liquid.

•Crystallise: form or cause to form crystals.

•Crystals: a piece of a homogeneous solid substance having a natural geometrically regular form.

•Excess: an amount of something that is more than necessary.

•Granules: a small compact particle of a substance.

•Mercerizing: a chemical treatment applied to cotton fibers.

•Phenolphthalein: a colorless crystalline solid (pink in alkaline solution) used as an acid-base indicator.

•Stirring: mixing thoroughly.

6.11 UNIT ASSESSMENT

I. Multiple Choice Questions

1. A solution which can dissolve more solute at a given temperature is called ______ .

(a) Saturated(b) Unsaturated (c) Supersaturated (d) none of these

2. Choose the correct statement

(a) With increase in temperature, the saturated solution becomes unsaturated.(b) The quantity of solute dissolved in a certain quantity of solvent denotes the concentration of solution. (c) Both (a) and (b) (d) None of these

3. Which of these is a dilute solution?

(a) A teaspoon of sugar in 200 ml water(b) Three teaspoons of sugar in 200 ml water (c) Both (a) and (b) (d) neither (a) nor (b)

4. The temperature at which reversal of solubility occurs is called

(a) critical temperature (b) transition temperature (c) both (a) and (b) (d) None of these

5. Which of the following is correct?

6. Sodium oxide reacts with carbon dioxide to give

(a) Carbonic acid (b) Sodium carbonate (c) Sodium chloride (d) Sulphuric acid

7. Complete the reaction:

8. Metal oxide reacts with carbon dioxide at

(a) room temperature (b) low temperature (c) very high temperature (d) does not react

9. Fe³⁺ ions form a ______ precipitate with both Na(OH) (aq) and NH₃ (aq)

(a) white (b) blue (c) green (d) brown

10. CO2 gas turns lime water to

(a) milky white (b) red (c) blue (d) yellow

II. Open Ended Questions

1. Distinguish between saturated and unsaturated solutions.

2. Explain supersaturated solution.

3. How can you prepare a supersaturated solution?

4. How can a saturated solution be changed to unsaturated?

5. Prepare zinc sulphate crystals from zinc metal and dilute sulphuric acid.

6. What are the factors that influence the solubility of salts?

7. Give two uses of solubility curves.

8. Write two uses of

(a) Washing soda(b) Baking soda

9. How will you identify Ca²⁺ ions and Fe²⁺ ions in a sample of solution?

10. How will you identify CO₃^2– ions in a solution?

III. Numerical

1. The solubility of a solute at 30°C is 20. What amount of water is required to make saturated solution of 80 grams of a solute?

2. At 30°C, 14 grams of sugar dissolves in 10 grams of water to form a saturated solution. Find the solubility of sugar?

IV. Practical-based Questions

1. Which beaker contains unsaturated solution?

2. Look at the solubility graph below and answer the questions that follow:

(i) How much sodium nitrate will dissolve at 60°C?

(a) 100 grams (b) 110 grams (c) 120 grams (d) 130 grams

(ii) Which solute is the most soluble at 80°C?

(a) Sodium Chloride (b) Sodium Sulphate (c) Sodium Nitrate (d) Potassium Nitrate

(iii) Which solute is the least soluble at 10°C?

(a) Potassium Nitrate (b) Sodium Chloride (c) Sodium Sulphate (d) Sodium Nitrate

(iv) 140 grams of solute is dissolved at 80°C is true for

(a) Sodium Nitrate (b) Potassium Nitrate (c) Sodium Sulphate (d) Sodium Chloride

        PROJECT

Carry out an experiment to show the method/process of crystallization.