What is the solubility of MnS in pure water, assuming hydrolysis of ions? of , and for . () — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

For $\text{MnS} \rightleftharpoons \text{Mn}^{2+} + \text{S}^{2-}$, $K_{sp} = 2.5 \times 10^{-10}$. The $\text{S}^{2-}$ ion highly hydrolyzes: $\text{S}^{2-} + \text{H}_2\text{O} \rightleftharpoons \text{HS}^- + \text{OH}^-$, with $K_h = \frac{K_w}{K_{a2}} = \frac{10^{-14}}{10^{-14}} = 1$. Let the solubility be $S$. Then $[\text{Mn}^{2+}] = S$. Because hydrolysis is very extensive, almost all $\text{S}^{2-}$ converts to $\text{HS}^-$, and $[\text{HS}^-] \approx [\text{OH}^-] \approx S$. From $K_h = 1 = \frac{[\text{HS}^-][\text{OH}^-]}{[\text{S}^{2-}]}$, we get $[\text{S}^{2-}] = [\text{HS}^-]^2 \approx S^2$. The $K_{sp}$ expression simplifies to $K_{sp} = [\text{Mn}^{2+}][\text{S}^{2-}] \approx S(S^2) = S^3$. Solving $S^3 = 2.5 \times 10^{-10} = 250 \times 10^{-12}$. Given the hint $0.63^3 = 0.25$, we scale it to $6.3^3 \approx 250$. Thus, $S = 6.3 \times 10^{-4}\text{ M}$. Therefore, correct answer is A.