Calculate pH of solution whose 100 ml contains 0.2 g NaOH dissolved in it.

100 ml solution of NaOH solution

= 0.2 g NaOH

$\therefore$  1000 ml solution of NaOH contains = 2 g NaOH

Normality of solution = $\frac{2}{40} = 0.05 \mathrm{N}=\left[{\mathrm{OH}}^{-}\right]$

$\therefore \left[{\mathrm{H}}^{+}\right] = \frac{{10}^{-14}}{0.05}=\frac{1}{5}\times {10}^{-12}=2\times {10}^{-13}$

$$\begin{gathered} {\mathrm{P}}^{\mathrm{H}}=-\log (2\times {10}^{-13}) \\ =13-\log 2 \\ =12.699 \end{gathered}$$