What is the concentration of aqueous ammonia in the solution prepared by dissolving of ammonium acet — Ionic Equilibrium Chemistry Question
Question
What is the concentration of aqueous ammonia in the solution prepared by dissolving $1.8\text{ moles}$ of ammonium acetate in $10\text{ L}$ water? ($K_a \text{ of } \text{CH}_3\text{COOH} = K_b \text{ of } \text{NH}_4\text{OH} = 1.8 \times 10^{-5}$)
💡 Solution & Explanation
Ammonium acetate is a salt of a weak acid and a weak base. The concentration is $C = 1.8 / 10 = 0.18\text{ M}$. The hydrolysis constant $K_h = \frac{K_w}{K_a K_b} = \frac{10^{-14}}{(1.8 \times 10^{-5})^2} = \frac{10^{-14}}{3.24 \times 10^{-10}} \approx 3.086 \times 10^{-5}$. The degree of hydrolysis $h = \sqrt{K_h} = \sqrt{3.086 \times 10^{-5}} \approx 5.55 \times 10^{-3}$. The concentration of the hydrolyzed aqueous ammonia is $[\text{NH}_4\text{OH}] = C \times h = 0.18 \times 5.55 \times 10^{-3} \approx 1.0 \times 10^{-3}\text{ M}$. Therefore, correct answer is A.