Question

At $T(K)$, if the ionisation constant of ammonia in solution is $2.5\times 10^{-5}$, the $pH$ of $0.01M$ 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)

$K_a\times K_b=K_w=10^{-14}($ at $25^{\circ }C)$

where, $K_a=$ ionisation constant for acid $\left(N{H}_4^{+}\right)$

and $K_b=$ ionisation constant for base $\left(N{H}_3\right)$

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

i.e. $K_a\left(N{H}_4^{+}\right).K_b\left(N{H}_3\right)=10^{-14}$

$\therefore K_a\left(N{H}_4^{+}\right)=\frac{10^{-14}}{K_a\left(N{H}_4^{+}\right)}$

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

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

For $pH$ of $0.01M$ ammonia $\left(N{H}_3\right)$

$\therefore N{H}_{4}OH$ dissociate as :

$\underset{\text{At equilibrium}}{}\underset{0.01(1-\alpha )}{N{H}_{4}OH}\rightleftharpoons \underset{0.01\alpha }{N{H}_4^{+}}+\underset{0.01\alpha }{O{H}^{-}}$

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

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

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

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

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

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