The solubility product of base is . Its solubility in the final solution is — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

The final solution has a pH of 8.0, so the $\text{pOH} = 6.0$, giving an $[\text{OH}^-] = 10^{-6}\text{ M}$. For the generic base $\text{A(OH)}_2$, the solubility product expression is $K_{sp} = [\text{A}^{2+}][\text{OH}^-]^2$. Letting the molar solubility be $S$, $[\text{A}^{2+}] = S$. Substituting the values gives $4.0 \times 10^{-30} = S(10^{-6})^2 = S(10^{-12})$. Solving for $S$ yields $S = \frac{4.0 \times 10^{-30}}{10^{-12}} = 4.0 \times 10^{-18}\text{ M}$. Therefore, correct answer is C.