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What is $[H^+]$ in mol/L of a solution that is 0.20 M in $CH_3COONa$ and 0.10 M in $CH_3COOH$? $K_a$ for $CH_3COOH$ = $1.8\times10^{-5}$.
A
$9.0\times10^{-6}$
B
$3.5\times10^{-4}$
C
$1.1\times10^{-5}$
D
$1.8\times10^{-5}$
Detailed Solution
The solution is an acidic buffer: $CH_3COOH \rightleftharpoons CH_3COO^- + H^+$
$CH_3COONa$ is fully dissociated, giving $[CH_3COO^-] \approx 0.20$ M; because of this common ion the ionisation of the acid is suppressed, so $[CH_3COOH] \approx 0.10$ M.
$K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]}$
$[H^+] = \frac{K_a[CH_3COOH]}{[CH_3COO^-]} = \frac{1.8\times10^{-5}\times0.10}{0.20}$
$[H^+] = 9.0\times10^{-6}$ mol/L
$CH_3COONa$ is fully dissociated, giving $[CH_3COO^-] \approx 0.20$ M; because of this common ion the ionisation of the acid is suppressed, so $[CH_3COOH] \approx 0.10$ M.
$K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]}$
$[H^+] = \frac{K_a[CH_3COOH]}{[CH_3COO^-]} = \frac{1.8\times10^{-5}\times0.10}{0.20}$
$[H^+] = 9.0\times10^{-6}$ mol/L