[અહીં $K_b\,(NH_4OH) = 10^{-5}$ અને $log\,2 = 0.301$ ]
\({H^ + } = \sqrt {\frac{{{K_w} \times C}}{{{K_b}}}} \)
\([{H^ + }] = \sqrt {\frac{{{{10}^{ - 14}} \times 2 \times {{10}^{ - 2}}}}{{{{10}^{ - 5}}}}} \)
\( - \log [{H^ + }] = 6 - \frac{1}{2}\log \,20\)
\(\therefore \,pH = 5.35\)