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

Given Weight = 4 gms of NaOH

GMW = 23 + 16 + 1 = 40 units

No. of moles of NaOH $=\frac{\mathrm{Weight} \mathrm{of} \mathrm{NaOH}}{\mathrm{GMW}}=\frac{4}{40}=\frac{1}{10}=0.1$

$\mathrm{M}=\frac{\mathrm{n}}{\mathrm{V}}=\frac{0.1\mathrm{M}}{1}=0.1 \mathrm{M}$

$\mathrm{NaOH}\rightleftharpoons {\mathrm{Na}}^{+}+{\mathrm{OH}}^{-}$

0.1                    0           0 

0                      0.1         0.1

$\left[{\mathrm{OH}}^{-}\right]=0.1={10}^{-1}$

$\mathrm{POH}=-{\log }_{10}^{\left[{\mathrm{OH}}^{-}\right]}=1$

$\mathrm{POH}+\mathrm{PH}=14\Rightarrow \mathrm{PH}=14-1=13$

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