What is the [${\mathrm{H}}^{+}$] of a 0.075 M solution of the acid HA?

Equilibrium	Constant, ${\mathrm{K}}_{\mathrm{a}}$
HA	$4.8\times {10}^{-8}$

$\underset{\underset{0.075(1-\mathrm{\alpha })}{0.075}}{\mathrm{HA}}\rightleftharpoons \underset{\underset{0.075\mathrm{\alpha }}{ }}{{\mathrm{H}}^{+}}+{\underset{\underset{0.075\mathrm{\alpha }}{ }}{\mathrm{A}}}^{-}$

${\mathrm{k}}_{\mathrm{a}}=4.8\times {10}^{-8}=\frac{0.075\mathrm{\alpha }\times 0.075\mathrm{\alpha }}{0.075(1-\mathrm{\alpha })}$

$\begin{array}{r}4.8\times {10}^{-8}=\frac{0.075{\mathrm{\alpha }}^2}{1}\\ \mathrm{\alpha }=\sqrt{\frac{4.8\times {10}^{-8}}{0.075}}\end{array}$

$\sqrt{{\mathrm{K}}_{\mathrm{a}}\cdot \mathrm{C}}$

$\begin{array}{l}\left[{\mathrm{H}}^{+}\right]=0.075\alpha =\sqrt{4.8\times {10}^{-8}\times 0.075}\\ \left[{\mathrm{H}}^{+}\right]=6\times {1\mathrm{O}}^{-5}\end{array}$