What is the aqueous ammonia concentration of a solution prepared by dissolving 0.15 mole of in 1 L o — Ionic Equilibrium Chemistry Question
Question
What is the aqueous ammonia concentration of a solution prepared by dissolving 0.15 mole of $\text{NH}_4^+\text{CH}_3\text{COO}^-$ in 1 L of water? Given: $K_a (\text{CH}_3\text{COOH}) = 1.8 \times 10^{-5}$; $K_b (\text{NH}_4\text{OH}) = 1.8 \times 10^{-5}$.
💡 Solution & Explanation
The salt undergoes both cationic and anionic hydrolysis: $\text{NH}_4^+ + \text{CH}_3\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4\text{OH} + \text{CH}_3\text{COOH}$. Because $K_a = K_b$, $[\text{NH}_4\text{OH}] = [\text{CH}_3\text{COOH}] = x$. The hydrolysis constant is $K_h = \frac{K_w}{K_a K_b} = \frac{10^{-14}}{(1.8 \times 10^{-5})^2} \approx 3.086 \times 10^{-5}$. Using $K_h \approx \frac{x^2}{C^2}$ gives $x = C \sqrt{K_h} = 0.15 \times \sqrt{3.086 \times 10^{-5}} = 0.15 \times 5.55 \times 10^{-3} \approx 8.3 \times 10^{-4} \text{ M}$. Therefore, correct answer is A.