JEE Main 2022 — Chemistry Physical Chemistry

2022integermedium

$50\mathrm{mL}$ of $0.1M{\mathrm{CH}}_{3}\mathrm{COOH}$ is being titrated against $0.1M\mathrm{NaOH}$ . When $25\mathrm{mL}$ of $\mathrm{NaOH}$ has been added, the $\mathrm{pH}$ of the solution will be____ $\times {10}^{-2}$ . (Nearest integer)

(Given : ${\mathrm{pK}}_{a}({\mathrm{CH}}_{3}\mathrm{COOH})=4.76$ )

$\mathrm{log}2=0.30$

$\mathrm{log}3=0.48$

$\mathrm{log}5=0.69$

$\mathrm{log}7=0.84$

$\mathrm{log}11=1.04$

Official previous-year question

Held on 26 Jun 2022 · Verified 6 Jul 2026.

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Solution

${\mathrm{CH}}_{3}\mathrm{COOH}(\mathrm{aq})+\mathrm{NaOH}(\mathrm{aq}.)\rightarrow {\mathrm{CH}}_{3}\mathrm{COONa}(\mathrm{aq})+{H}_{2}O(l)$

$5\mathrm{mmole}2.5\mathrm{mmoles}\mathrm{initially} 2.5m\mathrm{moles}02.5\mathrm{mmoles}\mathrm{after}\mathrm{reaction}$

Resultant solution is acidic buffer solution with same concentration of acid and salts.

$\mathrm{pH}=\mathrm{pKa}+\mathrm{log}\frac{[\mathrm{salt}]}{[\mathrm{acid}]}$

So, $\mathrm{pH}$ of solution $\mathrm{pH}=\mathrm{pKa}=4.76=476\times {10}^{-2}$

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Solution

${\mathrm{CH}}_{3}\mathrm{COOH}(\mathrm{aq})+\mathrm{NaOH}(\mathrm{aq}.)\rightarrow {\mathrm{CH}}_{3}\mathrm{COONa}(\mathrm{aq})+{H}_{2}O(l)$

$5\mathrm{mmole}2.5\mathrm{mmoles}\mathrm{initially} 2.5m\mathrm{moles}02.5\mathrm{mmoles}\mathrm{after}\mathrm{reaction}$

Resultant solution is acidic buffer solution with same concentration of acid and salts.

$\mathrm{pH}=\mathrm{pKa}+\mathrm{log}\frac{[\mathrm{salt}]}{[\mathrm{acid}]}$

So, $\mathrm{pH}$ of solution $\mathrm{pH}=\mathrm{pKa}=4.76=476\times {10}^{-2}$

JEE Main 2022 — Chemistry Physical Chemistry

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JEE Main 2022 — Chemistry Physical Chemistry

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Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry