The ionization constant of in water is at . The rate constant for the reaction of and to form and at — Ionic Equilibrium Chemistry Question
Question
The ionization constant of $\text{NH}_4^+$ in water is $5.6 \times 10^{-10} \text{ mol L}^{-1} \text{s}^{-1}$ at $25^\circ \text{C}$. The rate constant for the reaction of $\text{NH}_4^+$ and $\text{OH}^-$ to form $\text{NH}_3$ and $\text{H}_2\text{O}$ at $25^\circ \text{C}$ is $3.4 \times 10^{10} \text{ L mol}^{-1} \text{s}^{-1}$. The rate constant for proton transfer from water to $\text{NH}_3$ at $25^\circ \text{C}$ is
💡 Solution & Explanation
The given ionization constant of $\text{NH}_4^+$ is structurally its $K_a = 5.6 \times 10^{-10} \text{ M}$. The base dissociation constant of $\text{NH}_3$ is $K_b = K_w / K_a = 10^{-14} / (5.6 \times 10^{-10}) = 1.7857 \times 10^{-5} \text{ M}$. For the reaction $\text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^-$, the equilibrium constant $K_b$ is the ratio of forward ($k_f$) to backward ($k_b$) rate constants. Given $k_b = 3.4 \times 10^{10} \text{ L mol}^{-1} \text{s}^{-1}$, we have $k_f = K_b \times k_b = (1.7857 \times 10^{-5}) \times (3.4 \times 10^{10}) = 6.07 \times 10^5 \text{ s}^{-1}$. Therefore, correct answer is A.