\({pH{\text{ }} = {\text{ }} - log{\text{ }}\left( {{H^ + }} \right)\;{H^ + } = {\text{ }} - antilog{\text{ }}3.7}\)
\({H{\text{ }} = {\text{ }}2{{10}^{ - 4}}\,mol/Lit}\)
$NH_3$ + $H_2O$ $\rightleftharpoons$ $NH_4^ + + O{H^ - }$
[આપેલ : $\sqrt{2}=1.41$ ]