If the pH of 0.1 M acetic acid is 2.87, the percentage of ionization of acetic acid is 

If pH of 0.1 N acetic acid is 2.87, the percentage of ionization is 

$pH=-log\left[H^{+}\right]$

$\left[H^{+}\right]={10}^{-2.87}$=$1.35\times {10}^{-3}$,

$C{H}_{3}COOH\leftrightharpoons C{H}_{3}CO{O}^{-} + H^{+}$

[H⁺] = $\mathrm{\alpha C}$

%$\mathrm{\alpha }$ = $\frac{1.35\times {10}^{-3}}{0.1}\times 100$=1.35