20 mL of 0.1 ~M ~NaOH is added to 50 ~mL of 0.1 ~M acetic acid solution. The pH of the resulting solution is _ _ _ _ _ _ × 10⁻² (Nearest integer) Given: pKa(CH₃ COOH)=4.76 log 2=0.30 log 3=0.48

Equilibrium: JEE Main 2023 Chemistry question

JEE Main 2023 · 13 Apr, Evening shift · Q11

JEE Main 202313 Apr, Evening shiftChemistryMediumEquilibrium123× since 2002

20 mL of

0.1 ~\mathrm{M} ~\mathrm{NaOH}

is added to

50 \mathrm{~mL}

0.1 ~\mathrm{M}

acetic acid solution. The

\mathrm{pH}

of the resulting solution is ___________

\times 10^{-2}

(Nearest integer) Given :

\mathrm{pKa}\left(\mathrm{CH}_{3} \mathrm{COOH}\right)=4.76

\log 2=0.30

\log 3=0.48

Your answerType your answer. Integers and decimals are accepted.

JEE Main 2023 Equilibrium Question with Answer (13 Apr Evening shift, Q11) | JEE Arc