JEE Main 2026 · ChemistryMultiple choiceSingle correctEasyCalculation

JEE Main 4 April 2026, Shift 2, Chemistry Q55

Question 55 of 75 in this shift, Chemistry question 5 of 25, Section A.

20 mL of a solution of acetic acid required 28.4 mL of 0.1 M NaOH for its neutralization. A solution (X) was prepared by mixing 20 mL of the above acetic acid and 14.2 mL of 0.1 M NaOH solution. What is the pH of the solution (X)? (

\mathrm{pK_a}

value of acetic acid is 4.75).

(1)7.0
(2)4.75Official answer
(3)3.5
(4)4.82

Official answer

Option 2

NTA final key.

Chapter: Equilibrium
Topic: pH and Buffers
Idea tested: Acidic Buffer pH Calculation

Same idea in other shifts

Asked 4× in all

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18 Mar 2021, Shift 1 · Q55In order to prepare a buffer solution of pH 5.74, sodium acetate is added to acetic acid. If the concentration of acetic acid in the…EasyNumerical value

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.