## Chemistry 9th Edition

a. $0.56 = \frac{[NH_3]}{[NH_4^+]}$ b. $0.35 = \frac{[NH_3]}{[NH_4^+]}$ c. $5.6 = \frac{[NH_3]}{[NH_4^+]}$ d. $2.2 = \frac{[NH_3]}{[NH_4^+]}$
1. Since $NH_4^+$ is the conjugate acid of $NH_3$ , we can calculate its $K_a$ by using this equation: $K_b * K_a = K_w = 10^{-14}$ $1.8\times 10^{- 5} * K_a = 10^{-14}$ $K_a = \frac{10^{-14}}{ 1.8\times 10^{- 5}}$ $K_a = 5.6\times 10^{- 10}$ 2. Calculate $[H_3O^+]$: $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 9.00}$ $[H_3O^+] = 1.0 \times 10^{- 9}$ ** The reaction for $NH_4^+$ acting as an acid is: $NH_4^+(aq) + H_2O(l) -- \gt NH_3(aq) + H_3O^+(aq)$ 3. Write the $K_a$ equation, and find the ratio: $K_a = \frac{[H_3O^+][NH_3]}{[NH_4^+]}$ $5.6 \times 10^{-10} = \frac{1.0 \times 10^{-9}*[NH_3]}{[NH_4^+]}$ $\frac{5.6 \times 10^{-10}}{1.0 \times 10^{-9}} = \frac{[NH_3]}{[NH_4^+]}$ $0.56 = \frac{[NH_3]}{[NH_4^+]}$ -------------- b. 1. As we have calculated: $K_a (NH_4^+) = 5.6\times 10^{- 10}$ 2. Calculate $[H_3O^+]$: $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 8.80}$ $[H_3O^+] = 1.6 \times 10^{- 9}$ 3. Write the $K_a$ equation, and find the ratio: $K_a = \frac{[H_3O^+][NH_3]}{[NH_4^+]}$ $5.6 \times 10^{-10} = \frac{1.6 \times 10^{-9}*[NH_3]}{[NH_4^+]}$ $\frac{5.6 \times 10^{-10}}{1.6 \times 10^{-9}} = \frac{[NH_3]}{[NH_4^+]}$ $0.35 = \frac{[NH_3]}{[NH_4^+]}$ ------------- c. 1. As we have calculated: $K_a (NH_4^+) = 5.6\times 10^{- 10}$ 2. Calculate $[H_3O^+]$: $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 10.00}$ $[H_3O^+] = 1.0 \times 10^{- 10}$ 3. Write the $K_a$ equation, and find the ratio: $K_a = \frac{[H_3O^+][NH_3]}{[NH_4^+]}$ $5.6 \times 10^{-10} = \frac{1.0 \times 10^{-10}*[NH_3]}{[NH_4^+]}$ $\frac{5.6 \times 10^{-10}}{1.0 \times 10^{-10}} = \frac{[NH_3]}{[NH_4^+]}$ $5.6 = \frac{[NH_3]}{[NH_4^+]}$ --------------- d. 1. As we have calculated: $K_a (NH_4^+) = 5.6\times 10^{- 10}$ 2. Calculate $[H_3O^+]$: $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 9.6}$ $[H_3O^+] = 2.5 \times 10^{- 10}$ 3. Write the $K_a$ equation, and find the ratio: $K_a = \frac{[H_3O^+][NH_3]}{[NH_4^+]}$ $5.6 \times 10^{-10} = \frac{2.5 \times 10^{-10}*[NH_3]}{[NH_4^+]}$ $\frac{5.6 \times 10^{-10}}{2.5 \times 10^{-10}} = \frac{[NH_3]}{[NH_4^+]}$ $2.2 = \frac{[NH_3]}{[NH_4^+]}$