Question

Liquid $N{H}_3$ ionises to a slight extent. At a certain temperature, its ionization constant $K_{SIC\left(N{H}_3\right)}={10}^{-30}$. The number of $N{H}_4^{+}$ ions present per $100c{m}^3$ of pure liquid is:

• A. ${10}^{-15}$
• B. $6.022\times {10}^8$
• C. $6.022\times {10}^7$
• D. $6.022\times {10}^6$

Answer 1

The correct option is C $6.022\times {10}^7$

Self ionisation of $N{H}_3$ takes place as shown below:

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

Since ionisation constant $={10}^{-30}$, molarity of $N{H}_4^{+}={10}^{-15}$.

So, the number of ions of $N{H}_4^{+}=1^{-15}\times 6.022\times {10}^{23}\times \frac{100}{1000}=6.022\times {10}^7$.