Small amount of freshly precipitated magnesium hydroxides are stirred vigorously in a buffer solutio — Ionic Equilibrium Chemistry Question
Question
Small amount of freshly precipitated magnesium hydroxides are stirred vigorously in a buffer solution containing 0.25 M of $\text{NH}_4\text{Cl}$ and 0.05 M of $\text{NH}_4\text{OH}$. $[\text{Mg}^{2+}]$ in the resulting solution is ($K_b \text{ for NH}_4\text{OH} = 2.0 \times 10^{-5} \text{ and } K_{sp} \text{ of Mg(OH)}_2 = 8.0 \times 10^{-12}$)
💡 Solution & Explanation
The concentration of hydroxide in the buffer is $[\text{OH}^-] = K_b \times \frac{[\text{Base}]}{[\text{Salt}]} = 2.0 \times 10^{-5} \times \left(\frac{0.05}{0.25}\right) = 4.0 \times 10^{-6}\text{ M}$. At equilibrium with the freshly precipitated $\text{Mg(OH)}_2$, $[\text{Mg}^{2+}][\text{OH}^-]^2 = K_{sp}$. So, $[\text{Mg}^{2+}](4.0 \times 10^{-6})^2 = 8.0 \times 10^{-12} \implies [\text{Mg}^{2+}](16 \times 10^{-12}) = 8.0 \times 10^{-12}$. Solving for $[\text{Mg}^{2+}]$ yields $0.5\text{ M}$.