Chemistry - Ionic Equilibrium Question with Solution | TestHub

Concentration of $\mathrm{NaOH}$ given is $0.001\mathrm{M}$

$\left[{\mathrm{OH}}^{-}\right]=0.001=1\times {10}^{-3}$ 

${\mathrm{p}}^{\mathrm{OH}}=-\log \left[{\mathrm{OH}}^{-}\right]$

$=-\log \left(1\times {10}^{-3}\right)=$

$$\begin{gathered} =3 \\ {\mathrm{p}}^{\mathrm{H}}=14-{\mathrm{p}}^{\mathrm{OH}} \end{gathered}$$

$=14-3=11$

Hence, the ${\mathrm{p}}^{\mathrm{H}}$ of the given solution of $\mathrm{NaOH}$ is 11.