What is the $\mathrm{pH}$ of$0.1M$$\mathrm{NaOH}$solution?

Step 1: Calculation of $\mathrm{pOH}$ of$0.1\mathrm{M}$$\mathrm{NaOH}$solution

$\mathrm{NaOH}$ is an effective electrolyte. In an aqueous solution, it totally dissociates.
$\mathrm{NaOH}\to {\mathrm{Na}}^{+}+{\mathrm{OH}}^{-}$
Hence, $\left[{\mathrm{OH}}^{-}\right]=\left[\mathrm{NaOH}\right]=0.1\mathrm{M}$
$\begin{array}{rcl}\mathrm{pOH}& =& -\log \left[{\mathrm{OH}}^{-}\right]\\ & \Rightarrow & \mathrm{pOH}=-\log 0.1\\ & \Rightarrow & \mathrm{pOH}=1\end{array}$

Step 2: Calculation of $\mathrm{pH}$ solution$0.1\mathrm{M}$$\mathrm{NaOH}$solution

Now,
$\begin{array}{rcl}\mathrm{pH}& =& 14-\mathrm{pOH}\\ & \Rightarrow & \mathrm{pH}=14-1\\ & \Rightarrow & \mathrm{pH}=13\end{array}$
Therefore, the $\mathrm{pH}$ of$0.1M$$\mathrm{NaOH}$solution is $13$.