The solubility product of Mg(OH) is . The pH of an aqueous saturated solution of Mg(OH) is () — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

$K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2$. Since $[\text{Mg}^{2+}] = [\text{OH}^-]/2$, we have $([\text{OH}^-]/2)[\text{OH}^-]^2 = 9.0 \times 10^{-12}$, which simplifies to $[\text{OH}^-]^3 = 18 \times 10^{-12}$. Taking the log of both sides: $3 \log[\text{OH}^-] = \log 18 - 12 = \log(1.8 \times 10) - 12 = 0.26 + 1 - 12 = -10.74$. Thus, $\log[\text{OH}^-] = -3.58$, meaning $\text{pOH} = 3.58$. The $\text{pH} = 14 - 3.58 = 10.42$.