Chemistry - Ionic Equilibrium Question with Solution | TestHub

Let NaOH concentration is C M .

$\begin{array}{rl}{H}_{2}O\rightleftharpoons H^{+}+O{H}^{-}& \\ xM(C+x)M& \\ pH=-\log x=7.20& \\ \Rightarrow \log x=-7.20=\log \frac{10^{-7}}{1.6}& \\ \therefore x=\frac{10^{-7}}{1.6}& \end{array}$

And, $\mathrm{pOH}=-\log (\mathrm{C}+\mathrm{x})=14-7.20$

or, $\log (\mathrm{C}+\mathrm{x})=-6.80=-7+0.20=\log (1.6\times 10^{-7})$

or, $\mathrm{C}+\mathrm{x}=1.6\times 10^{-7}$.

$\therefore \mathrm{C}=1.6\times 10^{-7}-\frac{10^{-7}}{1.6}=9.75\times 10^{-8}\mathrm{M}$.