Chemical and Ionic Equilibrium 2 Question 68

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:

  1. 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

$=7-\frac{1}{2}\left[5+\log \left(2 \times 10^{-2}\right)\right]$

$=7-\frac{1}{2}[5+\log 2-2]$

$=7-\frac{1}{2}[5+0.301-2]=7-1.65=5.35$