$\mathrm{N}_{\text {Mix }} \mathrm{V}_{\text {Mix }}=\underset{\text { (Base) }}{\mathrm{N}_1 \mathrm{~V}_1}-\underset{\text { (Acid) }}{\mathrm{N}_2 \mathrm{~V}_2}\ldots(i)$
Given,
$\begin{aligned}
\mathrm{M}_1 & =0.2 \mathrm{M}, \mathrm{N}_1=0.2 \mathrm{~N} \\
\mathrm{M}_2 & =0.1 \mathrm{M}, \mathrm{N}_2=0.1 \mathrm{~N} \\
\mathrm{~V}_1 & =50 \mathrm{ml}, \\
\mathrm{V}_2 & =50 \mathrm{ml}, \mathrm{V}_{\text {mix }}=100 \mathrm{ml}
\end{aligned}$
On substituting the given values is equation. (i), We get
$\begin{aligned}
\mathrm{N}_{\text {mix }} \times 100 & =50 \times 0.2-50 \times 0.1 \\
\mathrm{~N}_{\text {mix }} & =\frac{5}{100}=5 \times 10^{-2} \\
\therefore \quad\left[\mathrm{OH}^{-}\right] & =5 \times 10^{-2}
\end{aligned}$
Now,
$\begin{aligned}
{ }_{\mathrm{P}} \mathrm{OH} & =-\log \left[\mathrm{OH}^{-}\right]=-\log \left[5 \times 10^{-2}\right] \\
& =2-\log 5=1.3
\end{aligned}$
Also, $\mathrm{pH}+\mathrm{pOH}=14$
$\mathrm{pH}=14-1.3=12.7$