Chemical and Ionic Equilibrium 2 Question 3
3. The pH of a $0.02 \mathrm{M} \mathrm{NH}{4} \mathrm{Cl}$ solution will be [Given $K{b}\left(\mathrm{NH}_{4} \mathrm{OH}\right)=10^{-5}$ and $\left.\log 2=0.301\right]$
(2019 Main, 10 April II)
(a) 4.65
(b) 2.65
(c) 5.35
(d) 4.35
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Answer:
Correct Answer: 3. (c)
Solution:
- Key Idea $\mathrm{NH}{4} \mathrm{Cl}$ is a salt of weak base $\left(\mathrm{NH}{4} \mathrm{OH}\right)$ and strong acid $(\mathrm{HCl})$. On hydrolysis, $\mathrm{NH}_{4} \mathrm{Cl}$ will produce an acidic solution $(\mathrm{pH}<7)$ and the expression of $\mathrm{pH}$ of the solution is
$$ \mathrm{pH}=7-\frac{1}{2}\left(\mathrm{p} K_{b}+\log C\right) $$
Given, $K_{b}\left(\mathrm{NH}_{4} \mathrm{OH}\right)=10^{-5}$
$\therefore \quad \mathrm{p} K_{b}=-\log K_{b}=-\log \left(10^{-5}\right)=5$
C $=$ concentration of salt solution $=0.02 \mathrm{M}$ $=2 \times 10^{-2} \mathrm{M}$
Now, $\mathrm{pH}=7-\frac{1}{2}\left(\mathrm{p} K_{b}+\log C\right)$
On substituting the given values in above equation, we get
$=7-\frac{1}{2}\left[5+\log \left(2 \times 10^{-2}\right)\right]$
$=7-\frac{1}{2}[5+\log 2-2]$
$=7-\frac{1}{2}[5+0.301-2]=7-1.65=5.35$
