Explanation

(c)
50 mL of 0.1 M $HCl = \frac{0.1 \times 50}{1000} = 5 \times 10^{- 3}$
50 mL of 0.2 M $NaOH = \frac{0.2 \times 50}{1000} = 10 \times 10^{- 3}$
Hence, after neutralisation $NaOH$ is left
$= 10 \times 10^{- 3} - 5 \times 10^{- 3}$
$= 5 \times 10^{- 3}$
Total volume= 100 cc
The concentration of $NaOH$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{5 \times 10^{- 3} \times 1000}{100} = 0.05\ M$
$\left\lbrack OH^{-} \right\rbrack = 0.05\ M = 5 \times 10^{- 2}M$
$pOH = - \log{\ \lbrack OH^{-}\rbrack}$
$= - \log{\ \lbrack 5 \times 10^{- 2}\rbrack}$
$= 1.3010$
$pH + pOH = 14$
$pH = 14 - 1.3010 = 12.699$