Calculate the percentage ionization of 0.01 M acetic acid in 0.1 M HCl, K_a of acetic acid is 1.8 × 10^–5

${\mathrm{K}}_{\mathrm{a}}=\frac{\mathrm{x}(0.1+\mathrm{x})}{(0.01-\mathrm{x})}=1.8\times {10}^{-5}$
$\therefore \mathrm{x} \mathrm{is} \mathrm{very} \mathrm{low},$
So, $\frac{\mathrm{x}(0.1+\mathrm{x})}{(0.01-\mathrm{x})}\approx \mathrm{x}\times \frac{0.1}{0.01}=1.8\times {10}^{-5}$
$\Rightarrow \mathrm{x}=1.8\times {10}^{-6}$
$\begin{array}{l}\text{ Now }\mathrm{\%}\text{ ionization }=\frac{\text{ Number of moles ionized }}{\text{ Number of moles taken }}\times 100\\ =\frac{\mathrm{x}}{0.01}\times 100=0.018\mathrm{\%}\end{array}$