$CH_3COOH\rightleftharpoons CH_3COO^-+ H^+$
$[H^+]_{\text{cân bằng}}= x$
$\Rightarrow K_a=\frac{[CH_3COO^-][H^+]}{[CH_3COOH]}=\frac{x^2}{0,01-x}=10^{-4,67}$
$\Leftrightarrow x=4,52.10^{-4}$
$\Rightarrow \alpha=\frac{4,52.10^{-4}.100}{0,01}= 4,5\%$
=> chọn B