The solubility of metal sulphide in saturated solution of (concentration ) can be represented as:<br — Ionic Equilibrium Chemistry Question
Question
The solubility of metal sulphide in saturated solution of $\text{H}_2\text{S}$ (concentration $= 0.1\text{ M}$) can be represented as:<br>$\text{MS(s)} + 2\text{H}^+\text{(aq)} \rightleftharpoons \text{M}^{2+}\text{(aq)} + \text{H}_2\text{S(aq)}$;<br>$K_{eq} = \frac{[\text{M}^{2+}][\text{H}_2\text{S}]}{[\text{H}^+]^2}$<br>The values of $K_{eq}$ for the metal sulphides, MnS, ZnS, CoS and PbS are $3 \times 10^{10}$, $3 \times 10^{-2}$, 3 and $3 \times 10^{-7}$, respectively. If the concentration of each metal ion in a saturated solution of $\text{H}_2\text{S}$ is 0.01 M, which metal sulphide(s) will precipitate at $[\text{H}^+] = 1.0\text{ M}$?
💡 Solution & Explanation
Precipitation of MS corresponds to the reverse of the given dissolution reaction, which is favored when the reaction quotient $Q < K_{eq}$ is false. Wait, $Q = \frac{[\text{M}^{2+}][\text{H}_2\text{S}]}{[\text{H}^+]^2}$. If $Q > K_{eq}$, the reaction shifts backward to form solid MS (precipitation). Calculating $Q$: $Q = \frac{(0.01)(0.1)}{(1.0)^2} = 10^{-3}$. We need to find which sulphide has $10^{-3} > K_{eq}$. Among the choices: $3 \times 10^{10}$ (MnS), $3 \times 10^{-2}$ (ZnS), 3 (CoS) are all greater than $10^{-3}$. Only $3 \times 10^{-7}$ (PbS) is smaller than $10^{-3}$. Therefore, only PbS will precipitate.