In an attempted determination of the solubility product constant of , the solubility of this compoun — Ionic Equilibrium Chemistry Question
Question
In an attempted determination of the solubility product constant of $\text{Tl}_2\text{S}$, the solubility of this compound in pure $\text{CO}_2$ free water was determined as $2.0 \times 10^{-6}\text{ M}$. Assume that the dissolved sulphide hydrolyses almost completely to $\text{HS}^-$ and that the further hydrolysis to $\text{H}_2\text{S}$ can be neglected, what is the computed $K_{sp}$? For $\text{H}_2\text{S}$, $K_{a1} = 1.4 \times 10^{-7}$, $K_{a2} = 1.0 \times 10^{-14}$
💡 Solution & Explanation
The dissolution is $\text{Tl}_2\text{S(s)} \rightleftharpoons 2\text{Tl}^+ + \text{S}^{2-}$. With solubility $S = 2.0 \times 10^{-6}\text{ M}$, $[\text{Tl}^+] = 2S = 4.0 \times 10^{-6}\text{ M}$. The $\text{S}^{2-}$ ion undergoes complete hydrolysis: $\text{S}^{2-} + \text{H}_2\text{O} \rightleftharpoons \text{HS}^- + \text{OH}^-$, so $[\text{HS}^-] \approx S = 2.0 \times 10^{-6}\text{ M}$ and $[\text{OH}^-] \approx 2.0 \times 10^{-6}\text{ M}$. To find the unhydrolyzed $[\text{S}^{2-}]$, we use the hydrolysis constant $K_h = \frac{K_w}{K_{a2}} = \frac{10^{-14}}{1.0 \times 10^{-14}} = 1.0$. Also, $K_h = \frac{[\text{HS}^-][\text{OH}^-]}{[\text{S}^{2-}]} \implies 1.0 = \frac{(2.0 \times 10^{-6})(2.0 \times 10^{-6})}{[\text{S}^{2-}]} \implies [\text{S}^{2-}] = 4.0 \times 10^{-12}\text{ M}$. Finally, $K_{sp} = [\text{Tl}^+]^2[\text{S}^{2-}] = (4.0 \times 10^{-6})^2 (4.0 \times 10^{-12}) = 64.0 \times 10^{-24} = 6.4 \times 10^{-23}$. Therefore, correct answer is A.