Answer
$[NH_3] = 0.54M$
Work Step by Step
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 $[H_3O^+]$:
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 9.12}$
$[H_3O^+] = 7.6 \times 10^{- 10}$
3. Write the $K_a$ equation, and find the ratio:
$K_a = \frac{[H_3O^+][NH_3]}{[N{H_4}^+]}$
$5.6 \times 10^{-10} = \frac{7.6 \times 10^{-10}*[NH_3]**}{0.732}$
$\frac{5.6 \times 10^{-10}}{7.6 \times 10^{-10}} = \frac{[NH_3]}{0.732}$
$0.74 * 0.732 = [NH_3]$
$[NH_3] = 0.54M$
** Notice, since the $K_a$ has a very low value, and the ammonium concentration is large, we can ignore the $NH_3$ produced by the dissociation of ammonium.
