Calculate in a solution originally having and . For , and . — Ionic Equilibrium Chemistry Question
Question
Calculate $[\text{S}^{2-}]$ in a solution originally having $0.1 \text{ M} – \text{HCl}$ and $0.2 \text{ M} – \text{H}_2\text{S}$. For $\text{H}_2\text{S}$, $K_{a1} = 1.4 \times 10^{-7}$ and $K_{a2} = 1.0 \times 10^{-14}$.
💡 Solution & Explanation
The strong acid HCl provides $[H^+] = 0.1 \text{ M}$ which heavily suppresses the dissociation of $\text{H}_2\text{S}$. The overall dissociation constant for $\text{H}_2\text{S} \rightleftharpoons 2\text{H}^+ + \text{S}^{2-}$ is $K_{eq} = K_{a1} K_{a2} = (1.4 \times 10^{-7})(1.0 \times 10^{-14}) = 1.4 \times 10^{-21}$. Substituting into the equilibrium expression: $K_{eq} = \frac{[\text{H}^+]^2[\text{S}^{2-}]}{[\text{H}_2\text{S}]} \implies 1.4 \times 10^{-21} = \frac{(0.1)^2 [\text{S}^{2-}]}{0.2}$. Solving for $[\text{S}^{2-}]$ yields $\frac{1.4 \times 10^{-21} \times 0.2}{0.01} = 2.8 \times 10^{-20} \text{ M}$. Therefore, correct answer is B.