On decreasing the pH from 7 to 2 , the solubility of a sparingly soluble salt ( MX ) of a weak acid ( HX )…

At $\mathrm{pH}=7$, i.e., the solubility in water is $\mathrm{S}$,

The solubility product, ${\mathrm{K}}_{\mathrm{sp}}={\mathrm{S}}^2 ----->1$

Assume the solubility of sparingly soluble salt is $\mathrm{at} \mathrm{pH}=2$

$$\begin{gathered} \mathrm{MX}\rightleftharpoons {\mathrm{M}}^{\oplus }+{\mathrm{X}}^{\ominus } \\ {\mathrm{S}}_1 {\mathrm{S}}_1-\mathrm{x} \end{gathered}$$

The weak dissociation reaction is as follows,

$$\begin{gathered} {\mathrm{X}}^{\ominus }+{\mathrm{H}}^{\oplus }\rightleftharpoons \mathrm{HX} \\ {\mathrm{S}}_1-\mathrm{x} {10}^{-2} \mathrm{x} \end{gathered}$$

Assume the acid dissociation constant is ${\mathrm{K}}_{\mathrm{a}}$.

Now, $\frac{1}{{\mathrm{K}}_{\mathrm{a}}}=\frac{\left[\mathrm{HX}\right]}{\left[{\mathrm{H}}^{+}\right]{\displaystyle \left[{\mathrm{X}}^{-}\right]}}$

${\mathrm{K}}_{\mathrm{sp}}={\mathrm{S}}_1\left({\mathrm{S}}_1-\mathrm{x}\right)$

$\frac{1}{{\mathrm{K}}_{\mathrm{a}}}=\frac{{\mathrm{S}}_1}{\left[{\mathrm{H}}^{+}\right]{\displaystyle {\mathrm{S}}_1-\mathrm{x}}}$ assume ${\mathrm{S}}_1\simeq \mathrm{x}$ 

$$${\mathrm{S}}_1-\mathrm{x}=\frac{{\mathrm{K}}_{\mathrm{sp}}}{{\mathrm{S}}_1}$

$$\begin{gathered} \frac{1}{{\mathrm{K}}_{\mathrm{a}}}=\frac{{{\mathrm{S}}_1}^2}{\left[{\mathrm{H}}^{+}\right]{\displaystyle {\mathrm{K}}_{\mathrm{sp}}}} \\ \Rightarrow {{\mathrm{S}}_1}^2 =\frac{{10}^{-2}{\mathrm{K}}_{\mathrm{sp}}}{{\mathrm{K}}_{\mathrm{a}}}------->2 \end{gathered}$$

Now from 1 and 2,

$$\begin{gathered} \frac{{{\mathrm{S}}_1}^2}{{\mathrm{S}}^2}=\frac{{10}^{-2}}{{\mathrm{K}}_{\mathrm{a}}} \\ \Rightarrow {\mathrm{K}}_{\mathrm{a}}=\frac{{10}^{-2}\times {10}^{-8}}{{10}^{-6}} \\ \Rightarrow {\mathrm{K}}_{\mathrm{a}}={10}^{-4} \\ \Rightarrow {\mathrm{pK}}_{\mathrm{a}}=4 \end{gathered}$$