The molar solubility of Zn(OH) in 1 M ammonia solution at room temperature is () — Ionic Equilibrium Chemistry Question
Question
The molar solubility of Zn(OH)$_2$ in 1 M ammonia solution at room temperature is ($K_{sp} \text{ of Zn(OH)}_2 = 1.6 \times 10^{-17}; K_{stab} \text{ of Zn(NH}_3\text{)}_4^{2+} = 1.6 \times 10^{10}$)
💡 Solution & Explanation
The dissolution reaction is $\text{Zn(OH)}_2(s) + 4\text{NH}_3 \rightleftharpoons \text{Zn(NH}_3)_4^{2+} + 2\text{OH}^-$. The equilibrium constant $K_{eq} = K_{sp} \times K_{stab} = (1.6 \times 10^{-17})(1.6 \times 10^{10}) = 2.56 \times 10^{-7}$. If molar solubility is $S$, then $[\text{Zn(NH}_3)_4^{2+}] = S$, $[\text{OH}^-] = 2S$, and $[\text{NH}_3] \approx 1 \text{ M}$. $K_{eq} = \frac{S(2S)^2}{(1)^4} = 4S^3 = 2.56 \times 10^{-7} = 256 \times 10^{-9}$. Solving for $S$, $S^3 = 64 \times 10^{-9} \implies S = 4 \times 10^{-3} \text{ M}$.