JEE Advanced 2023 Paper 1, Chemistry Section 2 Q3: Solubility Equilibria Of Sparingly Soluble Salts

Solubility at pH 7. In neutral water ${\text{X}}^{-}$ is not consumed by protonation, so $s_1=\sqrt{K_{sp}}=10^{-4}$ mol/L, giving $K_{sp}=10^{-8}$.

Solubility at pH 2. Write the two coupled equilibria with $s$ = total solubility of MX:

$\text{MX(s)}\rightleftharpoons {\text{M}}^{+}+{\text{X}}^{-}$ with $K_{sp}=[{\text{M}}^{+}][{\text{X}}^{-}]$.

${\text{X}}^{-}+{\text{H}}^{+}\rightleftharpoons \text{HX}$ with $1/K_a=[\text{HX}]/([{\text{X}}^{-}][{\text{H}}^{+}])$.

Mass balance: $[{\text{M}}^{+}]=s$ and $[{\text{X}}^{-}]+[\text{HX}]=s$. At pH = 2 the protonation is extensive, so most X${}^{-}$ converts to HX; let $[{\text{X}}^{-}]=y$ and $[\text{HX}]\approx s-y$. From the protonation equilibrium $\frac{s-y}{y\cdot 10^{-2}}=\frac{1}{K_a}$, giving $y=\frac{(s-y)K_a}{10^{-2}}$ and the $K_{sp}$ condition $s\cdot y=K_{sp}$.

Combine. When the acid is weak enough that $s-y\approx s$, we get $y=\frac{s{K}_a}{10^{-2}}$ and substituting into $K_{sp}=sy$ yields $$s=\frac{\sqrt{K_{sp}\cdot 10^{-2}}}{\sqrt{K_a}}\cdot \frac{1}{\sqrt{1}}=\frac{\sqrt{K_{sp}}}{10\sqrt{K_a}}.$$

Use the given solubility ratio. $\frac{s}{s_1}=\frac{10^{-3}}{10^{-4}}=10$. With $s_1=\sqrt{K_{sp}}$, this gives $\frac{1}{10\sqrt{K_a}}=10\Rightarrow \sqrt{K_a}=10^{-2}\Rightarrow K_a=10^{-4}$.

Therefore $p{K}_a=4$. Answer: (B).