Question

At ${50}^{o}C,$ the self-ionisation constant (ionic product) of $N{H}_3$ is given as,
$K_{N{H}_3}=\left[N{H}_4^{+}\right]\left[N{H}_2^{-}\right]={10}^{-30}{M}^2$.
How many $N{H}_2^{-}$ ions are present per $m{m}^3$ of pure liquid ammonia?

• A. $600ions/m{m}^3$
• B. $6\times {10}^{5}ions/m{m}^3$
• C. $6\times {10}^{4}ions/m{m}^3$
• D. $6\times {10}^{3}ions/m{m}^3$

Answer 1

The correct option is A $600ions/m{m}^3$

$2N{H}_3\left(l\right)\rightleftharpoons N{H}_4^{+}(aq.)+N{H}_2^{-}(aq.)\left(\text{self ionization}\right)$

$K_{N{H}_3}=\left[N{H}_4^{+}\right]\left[N{H}_2^{-}\right]={10}^{-30}\therefore \left[N{H}_4^{+}\right]=\left[N{H}_2^{-}\right]$
Thus, $[N{H}_2^{-}{]}^2={10}^{-30}$
$\left[N{H}_2^{-}\right]={10}^{-15}mol/lit\left[N{H}_2^{-}\right]=\frac{{10}^{-15}}{{10}^6}mol/m{m}^3(\because 1lit={10}^{6}m{m}^3)$
we know, $1mol\approx 6.0\times {10}^{23}ions$
$\left[N{H}_2^{-}\right]={10}^{-21}\times 6\times {10}^{23}ions/m{m}^3=600ions/m{m}^3$