The solubility of AgCN in solution, assuming no complex formation, is — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

The dissolution reaction is $\text{AgCN(s)} \rightleftharpoons \text{Ag}^+ + \text{CN}^-$. Given $K_{sp} = 1.0 \times 10^{-16}$. In a $0.02\text{ M KCN}$ solution, the $\text{CN}^-$ concentration is dominated by the KCN, so $[\text{CN}^-] \approx 0.02\text{ M}$. The solubility is represented by $[\text{Ag}^+] = \frac{K_{sp}}{[\text{CN}^-]} = \frac{1.0 \times 10^{-16}}{0.02} = 5.0 \times 10^{-15}\text{ M}$. Therefore, correct answer is B.