Answer
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. $2.2 = \frac{[NH_3]}{[NH_4^+]}$
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 $[H_3O^+]$:
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 9.00}$
$[H_3O^+] = 1.0 \times 10^{- 9}$
** 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:
$K_a = \frac{[H_3O^+][NH_3]}{[NH_4^+]}$
$5.6 \times 10^{-10} = \frac{1.0 \times 10^{-9}*[NH_3]}{[NH_4^+]}$
$\frac{5.6 \times 10^{-10}}{1.0 \times 10^{-9}} = \frac{[NH_3]}{[NH_4^+]}$
$0.56 = \frac{[NH_3]}{[NH_4^+]}$
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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:
$K_a = \frac{[H_3O^+][NH_3]}{[NH_4^+]}$
$5.6 \times 10^{-10} = \frac{1.6 \times 10^{-9}*[NH_3]}{[NH_4^+]}$
$\frac{5.6 \times 10^{-10}}{1.6 \times 10^{-9}} = \frac{[NH_3]}{[NH_4^+]}$
$0.35 = \frac{[NH_3]}{[NH_4^+]}$
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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:
$K_a = \frac{[H_3O^+][NH_3]}{[NH_4^+]}$
$5.6 \times 10^{-10} = \frac{1.0 \times 10^{-10}*[NH_3]}{[NH_4^+]}$
$\frac{5.6 \times 10^{-10}}{1.0 \times 10^{-10}} = \frac{[NH_3]}{[NH_4^+]}$
$5.6 = \frac{[NH_3]}{[NH_4^+]}$
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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:
$K_a = \frac{[H_3O^+][NH_3]}{[NH_4^+]}$
$5.6 \times 10^{-10} = \frac{2.5 \times 10^{-10}*[NH_3]}{[NH_4^+]}$
$\frac{5.6 \times 10^{-10}}{2.5 \times 10^{-10}} = \frac{[NH_3]}{[NH_4^+]}$
$2.2 = \frac{[NH_3]}{[NH_4^+]}$
