The hydronium ion concentration in $5 \mathrm{x} {10}^{-4}$ M aqueous solution of NaOH at 25° C, is :

Given:

The [${\mathrm{H}}^{+}$] concentration of the hydronium ion $=2.0\times {10}^{-4}\mathrm{M}$

We know that,

$\begin{array}{l}2{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{H}}_3{\mathrm{O}}^{+}+{\mathrm{OH}}^{-}\\ \mathrm{K}=\frac{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]}{\left[{\mathrm{H}}_2\mathrm{O}\right]}\end{array}$

$\begin{array}{l}{\mathrm{K}}_{\mathrm{W}}=\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]=1.0\times {10}^{-14}{\mathrm{M}}^2\\ \left[{\mathrm{OH}}^{-}\right]=\frac{\left[{\mathrm{K}}_{\mathrm{w}}\right]}{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}\end{array}$
 

$\begin{array}{l}\left[{\mathrm{OH}}^{-}\right]=\frac{1.0\times {10}^{-14}{\mathrm{M}}^2}{2.0\times {10}^{-4}\mathrm{M}}\\ \left[{\mathrm{OH}}^{-}\right]=5\times {10}^{-11}\mathrm{M}\end{array}$