Question

Calculate the degree of ionisation and hydroxyl ion concentration in $0.2M$ ${NH}_3$ solution.

($K_b=1.85\times {10}^{-5}$ for $298K$)

Answer 1

$N{H}_3+H_{2}O\rightleftharpoons N{H}_4^{+}+O{H}^{-}$

$0.2$

$0.2(1-\alpha )$ $0.2\alpha$ $0.2\alpha$

$K_b=\frac{\left[N{H}_4^{+}\right]\left[{OH}^{-}\right]}{\left[N{H}_3\right]}=\frac{\left(C\alpha \right)\left(C\alpha \right)}{C(1-\alpha )}$

Assume $\alpha <0.01$ so $1-\alpha =1$

$1.85\times {10}^{-5}=0.2(\alpha {)}^2$

$\alpha =0.962\times {10}^{-2},\left[{OH}^{-}\right]=0.2\alpha =1.924\times {10}^{-3}M$