Question

At $T( K )$, if the ionisation constant of ammonia in solution is $2.5 \times 10^{-5}$, the $pH$ of $0.01 \,M$ ammonia solution and the ionisation constant of its conjugate acid respectively at that temperature are $(\log 2=0.30)$

• A. $10.7,4.0 \times 10^{-8}$
• B. $10.7,4.0 \times 10^{-10}$
• C. $3.3,4.0 \times 10^{-8}$
• D. $3.3,4.0 \times 10^{-10}$

Answer 1

Answer: (B)

$K_{a} \times K_{b}=K_{w}=10^{-14}\left(\right.$ at $\left.25^{\circ} C \right)$

where, $K_{a}=$ ionisation constant for acid $\left( NH _{4}^{+}\right)$

and $K_{b}=$ ionisation constant for base $\left( NH _{3}\right)$

$=2.5 \times 10^{-5}$

i.e. $K_{a}\left( NH _{4}^{+}\right) . K_{b}\left( NH _{3}\right)=10^{-14}$

$\therefore K_{a}\left( NH _{4}^{+}\right)=\frac{10^{-14}}{K_{a}\left( NH _{4}^{+}\right)}$

$K_{a}=\frac{10^{-14}}{2.5 \times 10^{-5}}=4 \times 10^{-10}$

i.e. ionisation constant of conjugate acid $=4 \times 10^{-8}$

For $pH$ of $0.01 M$ ammonia $\left( NH _{3}\right)$

$\therefore NH _{4} OH$ dissociate as :

$\underset{\text{At equilibrium}}{} \underset{0.01(1-\alpha)}{NH_{4} OH} \ce{<=>} \underset{ 0.01 \alpha}{NH^{+}_{4}} + \underset{ 0.01 \alpha}{OH^{-}}$

$(\therefore $ Concentration of ammonia solution $=0.01 M )$

Also, $\alpha=\sqrt{\frac{K_{b}}{C}}=\sqrt{\frac{2.5 \times 10^{-5}}{0.01}}=0.05$

and $\left[ OH ^{-}\right]=C \alpha=(0.01) \alpha$ or $\left[ OH ^{-}\right]=0.0005$

$\therefore pOH =-\log \left[ OH ^{-}\right]$

or, $pH =14- pOH =14+\log [(0.0 l ) \alpha]=14-3.3$

Thus, $pH =10.7$ and also ionisation constant $\left(K_{a}\right)=4 \times 10^{-10}$