## Chemistry 10th Edition

(a) 1. Since $N{H_4}^+$ is the conjugate acid of $NH_3$ , we can calculate its kb by using this equation: $K_b * K_a = K_w = 10^{-14}$ $1.8\times 10^{- 5} * K_a = 10^{-14}$ $K_a = \frac{10^{-14}}{ 1.8\times 10^{- 5}}$ $K_a = 5.556\times 10^{- 10}$ 2. Calculate $[H_3O^+]$: $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 9.55}$ $[H_3O^+] = 2.818 \times 10^{- 10}$ 3. Write the $K_a$ equation, and find the ratio: $K_a = \frac{[H_3O^+][NH_3]}{[N{H_4}^+]}$ $5.556 \times 10^{-10} = \frac{2.818 \times 10^{-10}*[NH_3]}{[N{H_4}^+]}$ $\frac{5.556 \times 10^{-10}}{2.818 \times 10^{-10}} = \frac{[NH_3]}{[N{H_4}^+]}$ $1.97 = \frac{[NH_3]}{[N{H_4}^+]}$ (b) 1. We've already calculated the $K_a$ for $N{H_4}^+$ 2. Calculate $[H_3O^+]$: $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 9.1}$ $[H_3O^+] = 7.943 \times 10^{- 10}M$ 3. Write the $K_a$ equation, and find the ratio: $K_a = \frac{[H_3O^+][NH_3]}{[N{H_4}^+]}$ $5.556 \times 10^{-10} = \frac{7.943 \times 10^{-10}*[NH_3]}{[N{H_4}^+]}$ $\frac{5.556 \times 10^{-10}}{7.943 \times 10^{-10}} = \frac{[NH_3]}{[N{H_4}^+]}$ $0.67 = \frac{[NH_3]}{[N{H_4}^+]}$