JEE Main 2020ChemistryIonic EquilibriumHardNumerical

JEE Main 2020 — Ionic Equilibrium Question with Solution

JEE Main 2020 (07 Jan Shift 2)

Question

3 g of acetic acid is added to 250 mL of 0.1 M HCl and the solution made up to 500 mL. To 20 mL of this solution 12 mL of 5M NaOH is added. The pH of the solution is ________
[Given: pKa of acetic acid =4.75, molar mass of acetic acid 60 g/mol,log⁡3=0.4771 , Neglect any changes in volume]

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Show full solutionCorrect answer: 5.22
Correct answer
5.22

Step-by-step explanation

millimole of acetic acid in 20mL=2
millimole of HCl in 20mL=1
millimole of  NaOH=2.5


pH=PKa+log⁡3/22
=4.74+log⁡3
=4.74+0.48=5.22

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About this question

This is a previous-year question from JEE Main 2020, 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.