8.1 Calculating Solubility of a salt

3.1 Degree of Ionization (Dissociation)

5.1 Strong acid
5.2 Weak acid
5.3 Strong Acid + Weak Acid
5.4 Two weak acids
5.5 Dibasic and polyprotic weak acids

6.1 Salts of weak acid + strong base
6.2 Salt of weak base + strong acid
6.3 Salt of weak acid and weak base

7.1 Acidic buffer
7.2 Basic buffer
7.3 Salt buffer

8.1 Calculating Solubility of a salt

$$AgC{l_{(s)}} \rightleftharpoons A{g^ + } + C{l^ - }$$

Let the solubility of a salt be $x$ moles per litre, then at equilibrium, the concentration of ${A{g^ + }}$ equals $x$ moles per litre and that of ${C{l^ - }}$ is also $x$ moles per litre

$$\therefore {x^2} = {K_{sp}}$$

b) Solubility of salt (${AgCl}$) of strong acid and a strong base in a solution which already contains a common ion. $$\left[ {A{g^ + }} \right] = a{\text{ }}mol{L^{ - 1}}$$

$${AgCl} \rightleftharpoons {A{g^ + }} + {C{l^ - }}$$

$x' + a$ ${x'}$

Let the solubility of ${AgCl}$ in presence of common ion be ${x'}$. Therefore,

$$(x' + a) \times x' = {K_{sp}}$$

The ${x'}$ calculated here will be less than the value of $x$ calculated earlier.

$${C{H_3}COOAg} \rightleftharpoons {C{H_3}CO{O^ - }} + A{g^ + }$$

$x-y$ $x$

Since it is a salt of weak acid and strong base, it will also get hydrolysed.

$$C{H_3}CO{O^ - }{\text{ + }}{H_2}O\rightleftharpoons C{H_3}COOH{\text{ + }}O{H^ - }$$

$x-y$ $y$ $y$

Let the solubility of the salt be $x{\text{ }}mol{L^{ - 1}}$ and the amount of ${C{H_3}CO{O^ - }}$ getting hydrolyzed be $y{\text{ }}mol{L^{ - 1}}$. Therefore,

$$(x - y) \times x = {K_{sp}}$$

$$\frac{{{y^2}}}{{x - y}} = {K_h} = \frac{{{K_w}}}{{{K_a}}}$$

Solving these two equations, we can find the value of $x$

Example 8. Calculate the solubility for $A{B_2}$ type salt.

Solution. $$A{B_{{2_{(s)}}}}\rightleftharpoons {A^{ + 2}} + 2{B^ - }$$

1 0 0

1 $s$ $2s$

$${K_{sp}} = \left[ {{A^{ + 2}}} \right]{\left[ {{B^ - }} \right]^2} = s \times {(2s)^2} = s \times 4{s^2} = 4{s^2}$$

$$\therefore s = \sqrt {\frac{{{K_{sp}}}}{4}} $$

Example 9. Compound $AB$ gets ionized $80\% $. What should be the solubility of $AB$, calculate if its solubility product is $6.4 \times {10^{ - 9}}$ ?

Solution. $$AB \rightleftharpoons {A^{ + }} + 2{B^ - }$$

$s$ $0.8s$ $0.8s$

$${K_{sp}} = \left[ {{A^{ + }}} \right]\left[ {{B^ - }} \right]$$

or, $$6.4 \times {10^{ - 9}} = 0.8s \times 0.8s$$$$0.64{s^2} = 6.4 \times {10^{ - 9}}$$$$\therefore s = {10^4}$$