$100$ml of $0.2$M $HCl$ are mixed with $100$ ml of $0.2MC{H}_{3}COONa$, the $pH$ of the resulting solution would be nearly:

The correct option is B $0.7$
Total reaction is $0.2$ moles of $H^{+}$ reacts with $0.2$ moles of $C{H}_{3}COONa$ to give $0.2$ moles of $C{H}_{3}COOH$ & water.

$\underset{0.2}{HCl}+\underset{0.2}{C{H}_{3}COONa}⟶\underset{0.2}{C{H}_{3}COOH}+NaCl$

The concentration of $C{H}_{3}COOH$ is $0.2M$.

Hence $H=-\log (2\times {10}^{-01})$

$pH=-\log 2+\log 10$

$=0.698$

$\simeq 0.7$ .