Question

Consider the following equilibrium reaction:

$N{H}_3+N{H}_3\rightleftharpoons N{H}_4^{+}+{N{H}_2}^{-}$

Find the ionic product of the above equilibrium reaction?

Answer 1

$N{H}_3$ is self ionised. So, the ionic product is the equilibrium constant of the reaction.

$2N{H}_3\rightleftharpoons N{H}_4^{+}+{N{H}_2}^{-}$

$K_{N{H}_3}=\frac{\left[N{H}_4^{+}\right]\left[N{H}_2^{-}\right]}{{\left[N{H}_3\right]}^2}$

Since a little of the $N{H}_3$ is ionised, therefore its concentration remains virtually unchanged.

So, the Ionic product of equillibrium is $K_{N{H}_3}=\left[N{H}_4{]}^{+}\right[N{H}_2{]}^{-}$