MCQ
$0.1\, M$ formic acid solution is titrated against $0.1 \,M \,NaOH$ solution. What would be the difference in $pH$ between $1/5$ and $4/5$ stages of neutralization of acid ?
- A$2 \log \,3/4$
- B$2 \log\, 1/5$
- C$\log\, 1/3$
- ✓$2 \log \,4$
✓
Answer
Correct option: D.
$2 \log \,4$
d
When acetic acid is titrated with NaOH it will behaave as Buffer solution.
When acetic acid is titrated with NaOH it will behaave as Buffer solution.
We know in case of acidic buffer $\mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{[\mathrm{Salt}]}{[\mathrm{Acid}]}$
$\mathrm{pK}_{\mathrm{a}}$ of acetic acid $=4.77$
In case of $\frac{1}{5}$ th Titration $-$
$[$ Salt $]=\frac{x}{5}$
[Acid $]=\frac{4 x}{5}$
$\mathrm{pH}_{1}=4.77+\log \frac{1}{4}$
In case of $\frac{4}{5}$ th Titration -
$\mathrm{pH}_{2}=4.77+\log \frac{4}{1}$
Difference in $\mathrm{pH}=2 \log 4$
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