Answer
(a) 1.97
(b) 0.67
Work Step by Step
(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:
$K_a = \frac{[H_3O^+][NH_3]}{[N{H_4}^+]}$
$5.556 \times 10^{-10} = \frac{7.943 \times 10^{-10}*[NH_3]}{[N{H_4}^+]}$
$\frac{5.556 \times 10^{-10}}{7.943 \times 10^{-10}} = \frac{[NH_3]}{[N{H_4}^+]}$
$0.67 = \frac{[NH_3]}{[N{H_4}^+]}$
