An amount of of solid is rendered soluble in by adding just sufficient excess cyanide ion to form an — Ionic Equilibrium Chemistry Question
Question
An amount of $0.01\text{ moles}$ of solid $\text{AgCN}$ is rendered soluble in $1\text{ L}$ by adding just sufficient excess cyanide ion to form $\text{Ag(CN)}_2^-$ and the concentration of free cyanide ion is $2.5 \times 10^{-7}\text{ M}$. Determine $[\text{Ag}^+]$ in the solution neglecting hydrolysis of cyanide ion. $K_{\text{diss}}$ for $\text{Ag(CN)}_2^- = 1.0 \times 10^{-20}$.
💡 Solution & Explanation
When $0.01\text{ moles}$ of $\text{AgCN}$ is dissolved to form the complex, the concentration of the complex $[\text{Ag(CN)}_2^-]$ is essentially $0.01\text{ M}$. The free cyanide ion concentration $[\text{CN}^-]$ is given as $2.5 \times 10^{-7}\text{ M}$. The dissociation reaction is $\text{Ag(CN)}_2^- \rightleftharpoons \text{Ag}^+ + 2\text{CN}^-$. The dissociation constant is $K_{\text{diss}} = \frac{[\text{Ag}^+][\text{CN}^-]^2}{[\text{Ag(CN)}_2^-]} = 1.0 \times 10^{-20}$. Substituting values: $\frac{[\text{Ag}^+](2.5 \times 10^{-7})^2}{0.01} = 1.0 \times 10^{-20} \implies [\text{Ag}^+](6.25 \times 10^{-14}) = 1.0 \times 10^{-22}$. Solving for $[\text{Ag}^+]$ gives $\frac{1.0 \times 10^{-22}}{6.25 \times 10^{-14}} = 1.6 \times 10^{-9}\text{ M}$. Therefore, correct answer is B.