What is the equilibrium constant of the reaction: Fe(OH)(s) + 3HO Fe + 6HO? of Fe(OH) — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

The overall reaction can be constructed from two equilibria. First, dissolution: $\text{Fe(OH)}_3\text{(s)} \rightleftharpoons \text{Fe}^{3+}\text{(aq)} + 3\text{OH}^-\text{(aq)}$ with $K_{sp} = 4 \times 10^{-38}$. Second, neutralization of hydroxide: $3\text{H}_3\text{O}^+ + 3\text{OH}^- \rightleftharpoons 6\text{H}_2\text{O(l)}$, whose equilibrium constant is $(1/K_w)^3 = (10^{14})^3 = 10^{42}$. The overall equilibrium constant is the product: $K = K_{sp} \times (1/K_w)^3 = 4 \times 10^{-38} \times 10^{42} = 4 \times 10^4$.