The solubility of Pb(OH) in water is M. The solubility of Pb(OH) in a buffer solution of pH = 8 is — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

First, find $K_{sp}$ from the solubility in pure water: $K_{sp} = 4S^3 = 4(6.0 \times 10^{-6})^3 = 8.64 \times 10^{-16}$. In a buffer of pH = 8, $\text{pOH} = 6$, so $[\text{OH}^-] = 10^{-6} \text{ M}$. The new solubility $S'$ is simply $[\text{Pb}^{2+}]$, given by $S' = \frac{K_{sp}}{[\text{OH}^-]^2} = \frac{8.64 \times 10^{-16}}{(10^{-6})^2} = \frac{8.64 \times 10^{-16}}{10^{-12}} = 8.64 \times 10^{-4} \text{ M}$.