JEE Main 2023ChemistryIonic EquilibriumMediumNumerical

JEE Main 2023Ionic Equilibrium Question with Solution

JEE Main 2023 (13 Apr Shift 2)

Question

20 mL of 0.1MNaOH is added to 50 mL of 0.1M acetic acid solution. The pH of the resulting solution is ×10-2. (Nearest integer) Given : pKaCH3COOH=4.76

log 2 = 0.30

log 3 = 0.48

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Show full solutionCorrect answer: 458
Correct answer
458

Step-by-step explanation

When a strong base is added to a weak acid solution, it results in the formation of a salt. Here, acid is present in a limiting reagent and base is present in excess amounts. So, by using the pH formula:

pH=pKa+logsaltacid

CH3COOH+NaOHCH3COONa+H2O5                2                                         3                 0                 2

pH=pKa+log23

pH=4.76+0.30-0.48

= 4.76  .18

= 4.58

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About this question

This is a previous-year question from JEE Main 2023, covering the Ionic Equilibrium chapter of Chemistry. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.