pH of a ${10}^{-10}$ M NaOH is nearest to

 As the solution is very dilute $\left[{\mathrm{OH}}^{-}\right] \mathrm{of} \mathrm{water} \mathrm{also} \mathrm{to} \mathrm{be} \mathrm{taken} \mathrm{into} \mathrm{consideration}$

Total conc. of $\left[{\mathrm{OH}}^{-}\right]$ in solution

$= {\left[{\mathrm{OH}}^{-}\right]}_{(\mathrm{from} \mathrm{water})} + {\left[\mathrm{OH}\right]}_{(\mathrm{from} \mathrm{NaOH})}$

$= {{10}^{-7}}_{\left(\mathrm{approx}\right)} + {10}^{-10} = {10}^{-7}\left[1+0.001\right]$

$$\begin{gathered} =1.001\times {10}^{-7} \\ {\mathrm{p}}^{\mathrm{OH}}=-\log \left[1.001\times {10}^{-7}\right] \\ =7-\log 1.001 \\ =7-0.0004=6.999 \\ {\mathrm{p}}^{\mathrm{H}}=14-6.999=7.001,\mathrm{slightly} \mathrm{greater} \mathrm{than} 7 \end{gathered}$$