On decreasing the pH from 7 to 2 , the solubility of a sparingly...

At $pH = 7$, i.e., the solubility in water is $S$,

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

Assume the solubility of sparingly soluble salt is $atpH = 2$

$MX ⇌ {M}^{⊕} + {X}^{⊖}{S}_{1}{S}_{1} - x$

The weak dissociation reaction is as follows,

${X}^{⊖} + {H}^{⊕} ⇌ HX{S}_{1} - x{10}^{- 2}x$

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

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

${K}_{sp} = {S}_{1}\left({S}_{1} - x\right)$

$\dfrac{1}{{K}_{a}} = \dfrac{{S}_{1}}{\left[{H}^{+}\right]{S}_{1} - x}$ assume ${S}_{1} ≃ x$

${S}_{1} - x = \dfrac{{K}_{sp}}{{S}_{1}}$

$\dfrac{1}{{K}_{a}} = \dfrac{{{S}_{1}}^{2}}{\left[{H}^{+}\right]{K}_{sp}} ⇒ {{S}_{1}}^{2} = \dfrac{{10}^{- 2}{K}_{sp}}{{K}_{a}} - - - - - - - > 2$

Now from 1 and 2,

$\dfrac{{{S}_{1}}^{2}}{{S}^{2}} = \dfrac{{10}^{- 2}}{{K}_{a}} ⇒ {K}_{a} = \dfrac{{10}^{- 2} \times {10}^{- 8}}{{10}^{- 6}} ⇒ {K}_{a} = {10}^{- 4} ⇒ {pK}_{a} = 4$