At 298 K,  ${\mathrm{P}}^{\mathrm{H}}$ of  4% ionized 0.01M ${\mathrm{NH}}_4\mathrm{OH}$ solution will be

$\begin{array}{l}\left[{\mathrm{OH}}^{-}\right]=\mathrm{C\alpha }\\ =0.01\times \frac{4}{100}\\ =4\times {10}^{-4}\\ {\mathrm{p}}^{\mathrm{OH}}=4-\log 4\\ {\mathrm{p}}^{\mathrm{H}}=14-4+\log 4\\ {\mathrm{p}}^{\mathrm{H}} =10.602\end{array}$