What is the hydrolysis constant, , for ? — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

The solution pH is 3.80, so $[\text{H}^+] = 10^{-3.8} = 1.6 \times 10^{-4}\text{ M}$. The hydrolysis reaction is $\text{PuO}_2^{2+} + \text{H}_2\text{O} \rightleftharpoons \text{PuO}_2\text{OH}^+ + \text{H}^+$. For this equilibrium, $K_h = \frac{[\text{H}^+][\text{PuO}_2\text{OH}^+]}{[\text{PuO}_2^{2+}]}$. Since $[\text{H}^+] = [\text{PuO}_2\text{OH}^+]$, $K_h = \frac{(1.6 \times 10^{-4})^2}{0.010} = \frac{2.56 \times 10^{-8}}{0.010} = 2.56 \times 10^{-6} \approx 2.6 \times 10^{-6}$. Therefore, correct answer is A.