Chemical And Ionic Equilibrium Result Question 3
3. The pH of a $0.02$ $M$ $ NH_4 Cl$ solution will be [Given $K_b\left(\mathrm{NH}_4 \mathrm{OH}\right)=10^{-5}$ and $\log 2=0.301$ ]
(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 pH of the solution is
$\mathrm{pH}=7-\frac{1}{2}\left(\mathrm{p} K_b+\log C\right)$
Given, $K_b\left(\mathrm{NH}_4 \mathrm{OH}\right)=10^{-5}$
$\begin{aligned} & \therefore \quad \mathrm{p} K_b=-\log K_b=-\log \left(10^{-5}\right)=5 \\ & C=\text { concentration of salt solution }=0.02 \mathrm{M} \\ & =2 \times 10^{-2} \mathrm{M} \\ & \end{aligned}$
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
$\begin{aligned} & =7-\frac{1}{2}\left[5+\log \left(2 \times 10^{-2}\right)\right] \\ & =7-\frac{1}{2}[5+\log 2-2] \\ & =7-\frac{1}{2}[5+0.301-2]=7-1.65=5.35 \end{aligned}$
BETA
