Four grams of $\mathrm{NaOH}$ solid are dissolved in just enough water to make 1 litre of solution. What is the $\left[{\mathrm{H}}^{+}\right]$of the solution?

 Moles of $\mathrm{NaOH}=\frac{4}{40}=0.1$

$\mathrm{NaOH}$dissolves into water, thus dissociating into its respective ions as

$\mathrm{NaOH}\left(\text{ aq. }\right)\rightleftharpoons {\mathrm{Na}}^{+}\text{(aq.) }+{\mathrm{Cl}}^{-}\text{(aq.) }$

Hence, $\left[{\mathrm{OH}}^{-}\right]=\frac{0.1}{1}=0.1\mathrm{moles}/$ litre

Since, $\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]={10}^{-14}\mathrm{moles}/$ litre

Therefore, $\left[{\mathrm{H}}^{+}\right]={10}^{-13}$ moles/litre

Hence the correct answer (D) ${10}^{-13}\mathrm{moles}/\mathrm{litre}$.