Chemical and Ionic Equilibrium 2 Question 68
3. The pH of a $0.02 \mathrm{M} \mathrm{NH}{4} \mathrm{Cl} K{b}\left(\mathrm{NH}_{4} \mathrm{OH}\right)=10^{-5} \left.\log 2=0.301\right]$
(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 $\mathrm{NH}{4} \mathrm{Cl}
\left(\mathrm{NH}{4} \mathrm{OH}\right) (\mathrm{HCl}) \mathrm{NH}_{4} \mathrm{Cl} (\mathrm{pH}<7) \mathrm{pH}$ of the solution is
Given,
C
Now,
On substituting the given values in above equation, we get