${\mathrm{p}}^{\mathrm{H}}$ of the mixture containing 200ml of 0.1M ${\mathrm{CH}}_3\mathrm{COOH}$ and 100ml of 0.15M $\mathrm{KOH}$is $\left({\mathrm{p}}^{{\mathrm{k}}_{\mathrm{a}}} \mathrm{of} {\mathrm{CH}}_3\mathrm{COOH}=4.8\right)$

Number of milli equivalents of ${\mathrm{CH}}_3\mathrm{COOH}$ = 200(0.1M)1 = 20

Number of milli equivalents of KOH = 100(0.15M)1 = 15

Excess WA + limited SB acts like an acidic buffer

$\underset{\left.\begin{array}{l}15 \mathrm{meq}\\ 20\mathrm{meq}\end{array}\right\}\begin{array}{c}5\mathrm{meq}\\ \mathrm{excess}\end{array}}{\mathrm{WA}} + \underset{15 \mathrm{meq}}{\mathrm{SB}} \to \underset{15 \mathrm{meq}}{\mathrm{salt}} + \mathrm{water}$

The solution contains 5 meq of WA and  15 meq of Salt. 

$\begin{array}{l}\therefore {\mathrm{p}}^{\mathrm{H}}={\mathrm{p}}^{\mathrm{ka}}+\log \frac{{\mathrm{V}}_{\mathrm{s}}{\mathrm{N}}_{\mathrm{s}}}{{\mathrm{V}}_{\mathrm{a}}{\mathrm{N}}_{\mathrm{a}}}\\ {\mathrm{p}}^{\mathrm{H}}=4.8+\log \frac{15}{5}\\ {\mathrm{p}}^{\mathrm{H}}=4.8+\log 3\end{array}$