The solubility of AgCN in a buffer solution of pH = 3.0 is () — Ionic Equilibrium Chemistry Question
Question
The solubility of AgCN in a buffer solution of pH = 3.0 is ($K_{sp} \text{ of AgCN} = 1.2 \times 10^{-16}; K_a \text{ of HCN} = 4.8 \times 10^{-10}$)
💡 Solution & Explanation
In an acidic buffer, dissolution is $\text{AgCN}(s) + \text{H}^+ \rightleftharpoons \text{Ag}^+ + \text{HCN}$. The equilibrium constant $K_{eq} = \frac{K_{sp}}{K_a} = \frac{1.2 \times 10^{-16}}{4.8 \times 10^{-10}} = 2.5 \times 10^{-7}$. Since pH = 3.0, $[\text{H}^+] = 10^{-3} \text{ M} \gg K_a$, meaning nearly all dissolved cyanide forms HCN. Let solubility be $S$, so $[\text{Ag}^+] = S$ and $[\text{HCN}] \approx S$. $K_{eq} = \frac{S^2}{[\text{H}^+]} \implies 2.5 \times 10^{-7} = \frac{S^2}{10^{-3}} \implies S^2 = 2.5 \times 10^{-10}$. $S = \sqrt{2.5 \times 10^{-10}} = 1.58 \times 10^{-5} \text{ M}$.