Given: Ag(NH) Ag + 2NH, and at 298 K. If ammonia is added to a water solution containing excess of A — Ionic Equilibrium Chemistry Question
Question
Given: Ag(NH$_3$)$_2^+ \rightleftharpoons$ Ag$^+$ + 2NH$_3$, $K_c = 7.2 \times 10^{-8}$ and $K_{sp} \text{ of AgCl} = 1.8 \times 10^{-10}$ at 298 K. If ammonia is added to a water solution containing excess of AgCl(s) only, calculate the concentration of the complex in 1.0 M aqueous ammonia.
💡 Solution & Explanation
The reaction is $\text{AgCl}(s) + 2\text{NH}_3 \rightleftharpoons \text{Ag(NH}_3)_2^+ + \text{Cl}^-$. $K_{eq} = \frac{K_{sp}}{K_c} = \frac{1.8 \times 10^{-10}}{7.2 \times 10^{-8}} = 2.5 \times 10^{-3}$. Let the solubility be $S$. Then $[\text{Ag(NH}_3)_2^+] = S$, $[\text{Cl}^-] = S$, and $[\text{NH}_3] = 1.0 - 2S$. $K_{eq} = \frac{S^2}{(1.0 - 2S)^2} = 2.5 \times 10^{-3}$. Taking the square root, $\frac{S}{1.0 - 2S} = 0.05$. Solving for $S$: $S = 0.05 - 0.1S \implies 1.1S = 0.05 \implies S = \frac{0.05}{1.1} \approx 0.04545 \text{ M}$. Thus, the concentration of the complex is $0.0455 \text{ M}$.