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$\log\, 1/3$
- ✓$2 \log \,4$
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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Acidified $K _2 Cr _2 O _7$, alkaline $KMnO _4, CuSO _4, H _2 O _2, Cl _2, O _3, FeCl _3, HNO _3$ and $Na _2 S _2 O _3$. The total number of reagents that can oxidise aqueous iodide to iodine is