Question

At 50C, liquid ${NH}_3$ has ionic product is ${10}^{-30}$. How many amide $\left(N{H}_{2^{-}}\right)$ ions are present per ${mm}^3$ in pure liquid ${NH}_3$ ? (take $N_A=6\times {10}^{23}$)

Answer 1

Ionic product of $N{H}_3,K_{N{H}_3}=\left[N{H}_4^{+}\right]\left[N{H}_2^{-}\right]={10}^{-30}$

$\left[N{H}_2^{-}\right]=\sqrt{{10}^{-30}}={10}^{-15}M$

Since,

1 litre contains ${10}^{-15}moles$

then ${10}^{6}m{m}^3$ contains ${10}^{-15}moles$

Therefore $1m{m}^3$ contains $=\frac{{10}^{-15}}{{10}^6}={10}^{-21}moles$

Number of $N{H}^{2-}ions=6\times {10}^{23}\times {10}^{-21}=600ions$