If the concentration of [ N H _ 4 ^ + ] in a solution having 0.02MN...

N{H}_{4}OH ⇋ N{H}_{4}^{+} + O{H}^{-}$<br /><br />${k}_{b} = \dfrac{\left[N{H}_{4}^{+}\right]\left[O{H}^{-}\right]}${\left[N{H}_{4}OH\right]}$

$⇒ 1.8 \times {10}^{- 5} = \dfrac{\left[N{H}_{4}^{+}\right]\left[0.01\right]}{\left[0.02\right]}$

{As O{H}^{-} will mainly come from \text{ Ca }\left(O{H}_{2}\right) only} $<br /><br />$⇒ \left[N{H}_{4}^{+}\right] = 36 \times {10}^{- 6} = a \times {10}^{- 6}$<br /><br />$⇒ a = 36