$[OH^-] = C\alpha $ $ = 2 \times {10^{ - 2}} \times \frac{{10}}{{100}} = 2 \times {10^{ - 3}}$
$pOH = -log [2 \times 10^{-3}] = -log2 + (- log 10^{-3})$
$= -0.3010 + 3 = 2.699 = 2.7$
$pH + pOH = 14 $
$pH = 14 - 2.7 = 11.3$
(આપેલ : $pK _{ b }\left( NH _3\right)=4.74,NH _3$ નું મોલર દળ $=17\, g\, mol ^{-1},NH _4 Cl$નું મોલર દળ $= 53.5\, g\, mol ^{-1}$