The pH of0.1 M CuSO₄ solution is $\left(\text{ Given : }{K}_{\mathrm{a}}\left({\mathrm{Cu}}^{2+}\right)=1.0\times {10}^{-8}\mathrm{M}\right)$

${\mathrm{Cu}}^{2+}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons \mathrm{Cu}(\mathrm{OH}{)}^{+}+{\mathrm{H}}^{+}$

$K_{\mathrm{h}}=\frac{\left[\mathrm{Cu}(\mathrm{OH}{)}^{+}\right]\left[{\mathrm{H}}^{+}\right]}{\left[{\mathrm{Cu}}^{2+}\right]}=\frac{{\left[{\mathrm{H}}^{+}\right]}^2}{{\left[{\mathrm{Cu}}^{2+}\right]}_0-\left[{\mathrm{H}}^{+}\right]}=\frac{{\left[{\mathrm{H}}^{+}\right]}^2}{{\left[{\mathrm{Cu}}^{2+}\right]}_0}$

$\left[{\mathrm{H}}^{+}\right]={\left(K_{\mathrm{h}}{\left[{\mathrm{Cu}}^{2+}\right]}_0\right)}^{1/2}={\left[\left(1.0\times {10}^{-8}\mathrm{M}\right)\left(0.1\mathrm{M}\right)\right]}^{1/2}=3.16\times {10}^{-5}\mathrm{M}$

$\mathrm{pH}=-\log \left(3.15\times {10}^{-5}\right)=4.5$