Chemistry · Ionic Equilibrium

JEE Main 2026 — 4 April, Evening Shift — Question 52

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) ? (pKa\left(\mathrm{pK}_{\mathrm{a}}\right. value of acetic acid is 4.75).

  1. Option A:

    7

  2. Option B:

    4.45

    Correct
  3. Option C:

    3.5

  4. Option D:

    4.82

Answer: B

Step-by-step solution

In experiment 1, 28.4 ml of NaOH is required for complete neutralisation, therefore for 14.2 ml , half equivalence point will be achieved to form acidic buffer.

Answer key and solution verified before publishing.

Practise Ionic Equilibrium

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Exam
JEE Main 2026
Subject
Chemistry
Chapter
Ionic Equilibrium
Topic
pH of solution containing implicit reaction
20 mL of a solution of acetic acid required 28.4 mL of 0.1 M NaOH for… | JEE Main 2026 PYQ with Solution · DhiX AI