Solubility Product

Compounds such as AgCl, AgBr, AgI, BaSO₄ etc., when placed in a polar solvent such as water, dissolve to a very limited extent to produce ions in the solution. These sparingly soluble ionic substances (salts) establish an equilibrium between the solid phase and the ions in the solution. For example, for an ionic substance AB. the equilibrium is,

This equilibrium is characterised by an equilibrium constant K given by

As the concentration of the solid substance AB is constant at a particular temperature, no matter how much solid is in contact with the solution the above equilibrium (between ions and solid substance) can be described by a new constant.

K_sp expressed as,

K x constant = K_sp = [A⁺_(aq)][B^-_(aq)]

K_sp is called the solubility product constant or simply as solubility product of the salt concerned.

For a sparingly soluble ionic salt of the type AB₂ the dissolution leads to the equilibrium

Such an equilibrium may be represented by the solubility product constant (K_sp) given by, K_sp = [A²⁺_(aq)][B^-_(aq)] ²

For a salt of the type A_mB_n (s), dissolution leads to the dissociation according to the reaction,

where mp⁺ = nq^-.

The solubility product constant for the A_mB_n is then expressed as,

K_sp = [A^p+_(aq) ]^m + [ B^q-_(aq)]ⁿ.

Thus, the solubility product constant (K_sp) of a sparingly soluble salt is defined as the product of the molar concentrations of the ions in a saturated solution of the sparingly soluble salt, each raised to the power equal to the stoichiometric coefficient of the species, in the balanced chemical equation.

The K_sp values are generally very small. Such numbers are expressed in terms of ten raised to certain negative powers. For convenience, the solubility product constants (K_sp) can be expressed as p K_sp described by, pK_sp = -log K_sp.

Relationship between solubility and solubility product

This relationship may be derived in the following manner:

(1) For a salt of the type AB

Let the solubility of sparingly soluble salt AB at a certain temperature be 'S' moles per litre. Then

Therefore, [A⁺_(aq)] = S mol L^-1 and [B^-_(aq)] = S mol L^-1. The

solubility product constant is then expressed as,

K_sp = [A⁺_(aq)][B^-_(aq)] = (S mol L^-1) (S mol L^-1) = S₂ (mol L^-1)²

This gives,

Thus, for salts of the type AB (i.e. 1 : 1 type), the solubility of the salt at any temperature is given by

(2) For a salt of the type AB₂

A salt of AB₂ type (i.e.,1:2 type) dissociates as follows:

If the solubility of AB₂ is S mol L^-1, then

[A²⁺_(aq)] = S mol L^-1 and 2[B^-_(aq)] = 2S mol L^-1

Then, K_sp = (S) x (2S)²

or 4S³ = K_sp

or S = (K_sp /4)^1/3

Thus, the solubility product of a substance can be calculated if its solubility is known and vice versa. Since the solubility of a substance changes with temperature, K_sp also changes with temperature.

Problems

14. The solubility product of AgCl in water is 1.5 x 10^-10. Calculate its solubility in 0.01 M NaCl aqueous solution.

Solution

NaCl dissociates completely as:

Concentration of Cl^- ion in 0.01 M NaCl solution is [Cl^-] = 0.01 M.

Let solubility of AgCl in 0.01 M NaCl be 'x' mol L^-1, then

[Ag⁺] = 'x' mol L^-1, Cl^-(aq) = 'x' mol L^-1

Now, K_sp = [Ag⁺][Cl^-]

= 'x' x 0.01 = 0.01x

0.01 x = 1.5 x 10^-10

15. At 298 K, the solubility of silver chloride in water is 0.00188 g L^-1. What is its K_sp?

Solution

The equilibrium that exists in a solution of silver chloride is,

If S mol/L is the solubility of AgCl, then

K_sp = [Ag⁺_(aq)][Cl^-_(aq)] (S mol / L) (S mol / L) = S² mol² / L².

At 298 K, the solubility of AgCl in water is 0.00188 g/L. The molar mass of AgCl is 143.5 g/mol. So, Solubility of AgCl,

Then, K_sp (AgCl) = [Ag⁺][Cl^-] = S x S = S²

      = (1.31 x 10^-5 mol / L) ²= 1.7 x lO^-10 mol² / L²

while writing the K_sp values, the units are dropped. So, K_sp (AgCl) at 298K = 1.7 x 10^-10_.