Solid Magnesium hydroxide, , interacts with ammonium ions via the following proposed mechanism: . Ca — Ionic Equilibrium Chemistry Question
Question
Solid Magnesium hydroxide, $Mg(OH)_2$, interacts with ammonium ions via the following proposed mechanism: $Mg(OH)_2 (s) + 2NH_4^+ (aq) \rightleftharpoons Mg^{2+} (aq) + 2NH_3 (aq) + 2H_2O (l)$. Calculate the exact equilibrium constant ($K_c$) for this specific reaction mechanism at $25^\circ C$. (Given $K_{sp}$ for $Mg(OH)_2 = 1.0 \times 10^{-11}$ and $K_b$ for $NH_3 = 1.8 \times 10^{-5}$). If the answer is $y \times 10^{-2}$, enter the value of $y$ rounded to two decimal places.
💡 Solution & Explanation
The given reaction can be constructed from two fundamental equilibria. Eq 1 (Dissolution): $Mg(OH)_2 \rightleftharpoons Mg^{2+} + 2OH^-$ with constant $K_{sp} = 1.0 \times 10^{-11}$. Eq 2 (Reverse Base Ionization): $NH_4^+ + OH^- \rightleftharpoons NH_3 + H_2O$. The constant for the forward ionization of $NH_3$ is $K_b$, so the constant for $NH_4^+ + OH^-$ is $1/K_b$. Since the main reaction involves 2 moles of $NH_4^+$, we multiply Eq 2 by 2, making its constant $(1/K_b)^2$. The overall reaction constant $K_c = K_{sp} \times \left(\frac{1}{K_b}\right)^2 = \frac{1.0 \times 10^{-11}}{(1.8 \times 10^{-5})^2} = \frac{10 \times 10^{-12}}{3.24 \times 10^{-10}} \approx 3.08 \times 10^{-2}$. Thus, $y = 3.08$.