Question

Auto-ionisation of liquid $N{H}_3$ is

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

with $K_{N{H}_3}=\left[N{H}_4^{\oplus }\right]\left[N{H}_2^{⊝}\right]={10}^{-30}at-{50}^{o}C$

Number of atomic ions $\left(N{H}_2^{⊝}\right)$, present per $m{m}^3$ of pure liquied $N{H}_3$ is:

• A. $602$
• B. $301$
• C. $200$
• D. $100$

Answer 1

The correct option is A $602$

$K_{N{H}_3}=\left[N{H}_4^{\oplus }\right]\left[N{H}_2^{⊝}\right]={10}^{-30}$

$\left[N{H}_4^{\oplus }\right]=\left[N{H}_2^{⊝}\right]$

So $\left[N{H}_2^{⊝}\right]={10}^{-15}$

So no of ions in $m{m}^3$ = ${10}^{-15}\times N_o\times {10}^{-6}=6.02\times {10}^2=602$