JEE Main 2022 — Chemistry Physical Chemistry
50mL50\mathrm{mL}50mL of 0.1MCH3COOH0.1M{\mathrm{CH}}_{3}\mathrm{COOH}0.1MCH3COOH is being titrated against 0.1MNaOH0.1M\mathrm{NaOH}0.1MNaOH. When 25mL25\mathrm{mL}25mL of NaOH\mathrm{NaOH}NaOH has been added, the pH\mathrm{pH}pH of the solution will be____×10−2\times {10}^{-2}×10−2. (Nearest integer)
(Given : pKa(CH3COOH)=4.76{\mathrm{pK}}_{a}({\mathrm{CH}}_{3}\mathrm{COOH})=4.76pKa(CH3COOH)=4.76)
log2=0.30\mathrm{log}2=0.30log2=0.30
log3=0.48\mathrm{log}3=0.48log3=0.48
log5=0.69\mathrm{log}5=0.69log5=0.69
log7=0.84\mathrm{log}7=0.84log7=0.84
log11=1.04\mathrm{log}11=1.04log11=1.04
Held on 26 Jun 2022 · Verified 6 Jul 2026.
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CH3COOH(aq)+NaOH(aq.)→CH3COONa(aq)+H2O(l){\mathrm{CH}}_{3}\mathrm{COOH}(\mathrm{aq})+\mathrm{NaOH}(\mathrm{aq}.)\rightarrow {\mathrm{CH}}_{3}\mathrm{COONa}(\mathrm{aq})+{H}_{2}O(l)CH3COOH(aq)+NaOH(aq.)→CH3COONa(aq)+H2O(l)
5mmole2.5mmolesinitially2.5mmoles02.5mmolesafterreaction5\mathrm{mmole}2.5\mathrm{mmoles}\mathrm{initially} 2.5m\mathrm{moles}02.5\mathrm{mmoles}\mathrm{after}\mathrm{reaction}5mmole2.5mmolesinitially2.5mmoles02.5mmolesafterreaction
Resultant solution is acidic buffer solution with same concentration of acid and salts.
pH=pKa+log[salt][acid]\mathrm{pH}=\mathrm{pKa}+\mathrm{log}\frac{[\mathrm{salt}]}{[\mathrm{acid}]}pH=pKa+log[acid][salt]
So, pH\mathrm{pH}pH of solution pH=pKa=4.76=476×10−2\mathrm{pH}=\mathrm{pKa}=4.76=476\times {10}^{-2}pH=pKa=4.76=476×10−2
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