The solubility product of is . The pH of saturated solution of is about — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

Let the solubility of $\text{Co(OH)}_3$ be $S$. The $K_{sp}$ expression is $K_{sp} = [\text{Co}^{3+}][\text{OH}^-]^3 = (S)(3S)^3 = 27S^4$. Solving for $S$: $27S^4 = 2.7 \times 10^{-43} = 27 \times 10^{-44} \implies S^4 = 10^{-44} \implies S = 10^{-11}\text{ M}$. The hydroxide concentration provided by the dissociation is $[\text{OH}^-] = 3S = 3 \times 10^{-11}\text{ M}$. Because this is vastly smaller than the hydroxide concentration naturally provided by the autoionization of water ($10^{-7}\text{ M}$), the pH is completely dictated by water. Thus, the solution pH remains practically neutral at 7.0. Therefore, correct answer is A.