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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$Ag\left( s \right)\left| {AgBr\left( s \right)\,\left| {B{r^ - }\left( {0.01\,M} \right)} \right|\,\left| {{I^ - }\left( {0.02\,M} \right)} \right|\,AgI\left( s \right)} \right|Ag\left( s \right)$
the correct information is
[Given : $K_{sp}\,\left( {AgBr} \right) = 4 \times {10^{ - 13}}$ ,
$K_{sp}\,\left( {AgI} \right)$ $ = 8 \times {10^{ - 17}},\frac{{2.303\,RT}}{F} = 0.06\,V,\,\log \,2 = 0.3]$
$\mathrm{SO}_3, \mathrm{H}_2 \mathrm{SO}_3, \mathrm{SOCl}_2, \mathrm{SF}_4, \mathrm{BaSO}_4, \mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_7$
$\xrightarrow{{Mg/THF}}A\mathop {\xrightarrow{{(i)\,C{H_3}CHO}}}\limits_{(ii)\,aq.N{H_4}Cl} B$
($en =$ ethylenediamine)
[Given : $\sqrt{3}=1.73$ ]