Salts like AgCl, CaSO_{4}, Ag_{2}CrO_{4}, PbSO_{4}, PbCl_{2}, etc. are considered as insoluble. They are not completely insoluble in water because no salt is absolutely insoluble in water, i.e. they slightly dissolve in water. These salts are called sparingly soluble salts.

If a sparingly soluble salt is added to water, a very small extent dissolves and forms a saturated solution at a given temperature. At saturated solution, the ions in the solution collide and reproduce the solid phase. Ultimately a dynamic equilibrium is established between solute and ions in a solution.

In a saturated solution of a sparingly soluble salt at a fixed temperature, a dynamic equilibrium exists between the salt in the solid phase and ions present in the solution.

Let us consider a sparingly soluble salt AB is under dissolution to make a saturated solution at a fixed temperature, then following equilibrium holds good:

`AB\ \rightleftharpoons\ A^{+}\ +\ B^{-}`

Applying law of mass action at equilibrium,

`K=\frac{[A^{+}][B^{-}]}{[AB]}`

For a saturated solution at a constant temperature,

`\begin{align*} [AB]_{solid}&=constant\\ K&=\frac{[A^{+}][B^{-}]}{constant}\\ K\ constant &= [A^{+}][B^{-}]\\ K_{sp}&=[A^{+}][B^{-}] \end{align*}`

where K_{sp} is another constant called solubility product that can be defined as, “The product of ionic concentration for a saturated solution at a constant temperature”.

##### Relation between solubility and solubility product

**1. For AB type of salts (AgCl, AgBr, AgI, PbSO _{4}, BaSO_{4}, etc.)**:

Let us take an example of AgCl, where ‘s’ be the solubility in mol/lit.

`\begin{align*} \underset{S}{AgCl}\ &\rightleftharpoons\ \underset{S}{Ag^{+}} +\ \underset{S}{Cl^{-}}\\ K_{sp}&=[Ag^{+}][Cl^{-}]=s.s=s^{2}\\ s&=\sqrt{K_{sp}} \end{align*}`

**2. For A _{2}B or AB_{2} type of salt (Ag_{2}S, Ag_{2}CrO_{4}, CaF_{2}, BaF_{2}, etc.)**:

`\begin{align*} \underset{S}{Ag_{2}S}\ &\rightleftharpoons\ \underset{2S}{2Ag^{+}} +\ \underset{S}{S^{--}}\\ K_{sp}&=[Ag^{+}]^{2}[S^{--}]=(2s)^{2}.s=4s^{3}\\ s&=\sqrt[3]{\frac{K_{sp}}{4}} \end{align*}`

##### Solubility product principle

At a given temperature,

i. When the ionic product is less than the solubility product, the solution is unsaturated.

ii. When an ionic product is equal to a solubility product, the solution is saturated.

iii. When an ionic product is greater than the solubility product, the solution is supersaturated and precipitation takes place. This principle is called the solubility product principle or theory of precipitation.

##### Common Ion Effect

In a solution of a weak electrolyte, there exists a dynamic equilibrium between unionized molecules and ions present in the solution.

Let us consider a weak electrolyte AB is under the dissolution, following equilibrium takes place:

`\underset{unionized\ molecule}{AB}\ \rightleftharpoons\ A^{+} +\ B^{-}`

When a strong electrolyte that gives either A^{+} or B^{–} is added to the solution of AB, the equilibrium is disturbed. To maintain the equilibrium, the process shifts towards the backward direction according to Le-Chatelier’s principle i.e. ionization of AB is suppressed.

Thus, the suppression of ionization of weak electrolytes due to the addition of strong electrolytes having one of the ions the same as the weak electrolyte is called the common ion effect.

##### Examples

- When acetic acid (weak electrolyte) is added to the solution of sodium acetate (strong electrolyte), equilibrium exists as:

CH_{3}COOH ⇌ CH_{3}COO^{–}+ H^{+}

CH_{3}COONa ⇌ CH_{3}COO^{–}+ Na^{+}

Due to common ion (CH_{3}COO^{–}), the ionization of CH_{3}COOH is supressed due to which H^{+}concentration in solution is decreased. That is the acidic strength of the solution is also decreased.

- The basic strength of NH
_{4}OH decreases due to addition of NH_{4}Cl

NH_{4}OH ⇌ NH_{4}^{+}+ OH^{–}

NH_{4}Cl ⇌ NH_{4}^{+}+ Cl^{–}

Due to common ion (NH_{4}^{+}), the ionization of NH_{4}OH is supressed due to which OH^{–}concentration in solution is decreased. That is the basic strength of solution is also decreased.

##### Application of common ion effect and solubility product principle

##### 1. Application in qualitative analysis

i. Precipitation of group II metal cation ( Hg^{++}, Pb^{++},Bi^{++}, Cu^{++}, As^{++}, Sb^{++}, Sn^{++}): In group II, H_{2}S is passed through the salt solution acidified with dil. HCl. The dissociation takes place as:

H_{2}S ⇌ 2H^{+} + S^{—}

HCl ⇌ H^{+} + Cl^{–}

Here, the H+ ion suppresses the ionization of H_{2}S due to the common ion effect and hence the concentration of sulphide ion is lowered. As a result of this, the solubility product of the metal sulphide group is less than its ionic product. As a result, radicals of group II are precipitated as their sulphides.

If HCl is not added, the precipitation of group II, as well as group IIIB metal cation, occurs at the same time as their sulphides and selective precipitation cannot be achieved.

ii. Precipitation of group IIIA metal cation as their hydroxide (Fe^{+++}, Al^{+++},Cr^{+++}):

In this group, H_{2}S is passed through the salt solution containing NH_{4}Cl and NH_{4}OH. The dissociation takes place as:

NH_{4}OH ⇌ NH_{4}^{+} + OH^{–}

NH_{4}Cl ⇌ NH_{4}^{+} + Cl^{–}

Here, the common ion NH_{4} reduces the concentration of OH^{–} in the solution. The concentration of OH^{–} gets reduced to such an extent that the solubility product becomes less than their ionic product and these hydroxides get precipitated out.

##### 2. Purification of common salt

The impure salt, NaCl can be made pure by passing HCl gas in its saturated solution.

NaCl ⇌ Na^{+} + Cl^{–}

HCl ⇌ H^{+} + Cl^{–}

Due to common ion (Cl^{–}), the new product of ionic concentration will be greater than the solubility product of NaCl. Hence, NaCl gets precipitated while all the impurities remain in the solution.

##### Solved Numerical Examples

1. Calculate the solubility product of CaF_{2} at a certain temperature if its solubility at that temperature is 1.3 x 10^{-5} mol/lit.

solution:

`\begin{align*}CaF_{2}\ &\rightleftharpoons\ Ca^{++} +\ 2F^{-}\\ \left [ Ca \right ]^{+}&= 1.3\times 10^{-5}\ mol/lit\\ \left [ F \right ]^{-}&= 1.3\times 10^{-5}\ mol/lit\\ K_{sp}&=\left [ Ca \right ]^{++}\left [ F^{-} \right ]^{2}\\ &= 1.3\times 10^{-5} \times (2\times1.3\times 10^{-5})^{2}\\ &=8.79\times10^{-15} \end{align*}`

2. The solubility of CaCO_{3} is 4.3 x 10^{-5} gm/100ml at a certain temperature. Calculate its solubility product at that temperature.

solution:

Solubility 4.3 x 10^{-5} gm/100ml means 100 ml of solution contains 4.3 x 10^{-5} gm of CaCO_{3}.

`\begin{align*}Molarity &= \frac{W}{V}\times \frac{1000}{mol.\ wt.}\\ &= \frac{4.3\times 10^{-5}\times 1000}{100\times 100}=4.3\times10^{-6}M\\ CaCO_{3} &\rightleftharpoons\ Ca^{++} +\ CO_{3}^{--}\\ \left [ Ca^{++} \right ]&= 4.3\times 10^{-6}M\\ \left [ CO_{3}^{--} \right ]&= 4.3\times 10^{-6}M\\ K_{sp}&= \left [ Ca^{++} \right ]\left [ CO_{3}^{--} \right ]\\ &= 4.3\times 10^{-6}\times 4.3\times 10^{-6}\\ &= 1.89\times 10^{-11} \end{align*}`

3. The solubility product of CaF_{2} is 3.9 x 10^{-10} at a certain temperature. Calculate its solubility in 0.01 NaF solution.

solution:

NaF ⇌ Na^{+} + F^{–} For 0.01M NaF, [F^{–}] =0.01M

Again, CaF_{2} ⇌ Ca^{++} + 2F^{–}

Let solubility of CaF_{2} be x mol/lit.

So, [Ca^{++}] = x & [F^{–}] = 2x

`\begin{align*}K_{sp}&=[Ca^{++}][F^{-}]^{2}\\ or,\ 3.7\times 10^{-10}&=[x]\ [2x + 0.01]^{2}\\ or,\ 3.7\times 10^{-10}&=[x]\ [0.01]^{2}\\ \because 2x\ is\ very\ &small\ compared\ to\ 0.01.\\ \therefore x &= 3.9\times 10^{-6}\ mol/lit. \end{align*}`

4. What is the minimum volume of water required to dissolve 1gm of CaSO_{4} at 298K? (K_{sp} for CuSO_{4} is 9.1 x 10^{-6})

solution:

CaSO_{4}(s) ⇌ Ca^{++}(s) + SO_{4}^{—}(s)

K_{sp} = [Ca^{++}][SO4^{—}] = s.s = s^{2}

so, s = 3 x 10^{-3} M

gm/lit = Molarity x Mol.wt. = 3 x 10^{-3} x 136 = 0.408 gm/lt

0.408 gm of CaSO_{4} dissolved in 1000 ml water

1 gm of CaSO_{4} dissolved in 2.45 x 10^{3} ml water.

5. If the volume of 25 cm^{3} of 0.05M Ba(NO_{3})_{2} are mixed with 25 cm^{3} of 0.02M NaF. Will any BaF_{2} precipitate? (K_{sp} of BaF_{2} at 298 K = 1.7 x 10^{-6})

solution:

[Ba^{++}] after mixing = 0.05 x (25/50) = 2.5 x 10^{-2}M

[F^{–}] after mixing = 0.02 x (25/50 ) = 1.0 x 10^{-2}M

BaF_{2} ⇌ Ba^{++} + 2F^{–}

Ionic product = [Ba^{++}][F^{–}]^{2} = [2.5 x 10^{-2}][2 x 1.0 x 10^{-2}]^{2} = 1.0 x 10^{-5}

Solubility product = 1.7 x 10^{-6}

Since the ionic product is more than the solubility product, precipitation will occur.