Chemistry

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.

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.

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).

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:

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.

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

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

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)

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

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:

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.