The pH of a solution obtained by mixing 100 ml of 0.2 M ${CH}_{3}COOH$ with 100 ml of 0.1 M $NaOH$ will be: ($p{K}_a$ for ${CH}_{3}COOH$ = 4.74 )

The correct option is C 8.87

0.2 M $C{H}_{3}COOH$ 0.2 M NaOH

$\therefore$ Both react as NaOH is strong base

$\therefore \frac{0.2}{2}MC{H}_{3}CO{O}^{-}$ is left in sals

$=01MC{H}_{3}CO{O}^{-}$ is left in sals

$C{H}_{3}CO{O}^{-}+H_{2}O\rightleftharpoons C{H}_{3}COOH+O{H}^{-}K=\frac{K_w}{K_a}$

$\alpha n\sqrt{\frac{k}{c}}$

$=\sqrt{\frac{10-74}{K_a\times 0.1}}$

$\Rightarrow {10}^{-7}\times \sqrt{\frac{10}{K_a}}$

$K_a\Rightarrow {10}^{-4.74}$ $(P{K}_a=4.74)$

$\therefore \alpha ={10}^{-7}\sqrt{\frac{10}{10-4.74}}$

$={10}^{-7}\times {10}^{+2.67}$

$={10}^{-4.33}<0.1$

$\therefore \left[O{H}^{-}\right]=C\alpha$

$={10}^{-1}\times {10}^{-4.33}$

$={10}^{-5.33}$

$\therefore pOH=5.33$

$pH=14-p01+$

$=8.67$