0.4 grams of sodium hydroxide is dissolved in 500 ml solution. The solution is now diluted by adding 500 ml of water. ${\mathrm{p}}^{\mathrm{H}}$ of the resultant solution is

$$\begin{gathered} {\mathrm{N}}_{\mathrm{b}}=\left[{\mathrm{OH}}^{-}\right]=\frac{0.4}{40}\times \frac{1000}{500}; \\ =\frac{0.8}{40}\mathrm{N} ; \\ {\mathrm{V}}_1{\mathrm{N}}_1={\mathrm{V}}_2{\mathrm{N}}_2; \\ 500\left(\frac{0.8}{40}\right)=1000\times {\mathrm{N}}_{\mathrm{b}}; \\ {\mathrm{N}}_{\mathrm{b}}=\left[{\mathrm{OH}}^{-}\right]={10}^{-2}; \\ {\mathrm{p}}^{\mathrm{OH}}=2; \\ {\mathrm{p}}^{\mathrm{H}}=12; \\ \mathrm{Alternate} \mathrm{method}; \\ \mathrm{Final} \mathrm{volume} \mathrm{solution}=1000 \mathrm{ml}; \\ \therefore \left[{\mathrm{OH}}^{-}\right]=\frac{0.4}{40}\times \frac{1000}{1000}={10}^{-2} \mathrm{M}; \\ {\mathrm{p}}^{\mathrm{OH}}=2 ; \\ {\mathrm{p}}^{\mathrm{H}}=12; \end{gathered}$$