Fundamentholfundamenthol
JEE Main2026Apr 4, Shift 2Chemistry
Q.

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)? (pK value of acetic acid is 4.75).

  1. A

    7.0

  2. B

    4.45

  3. C

    3.5

  4. D

    4.82

Solution

Full neutralization needs 28.4 mL of NaOH; the mixture in solution X uses exactly half of that (14.2 mL). So half the acetic acid has been converted to acetate, leaving an equimolar mixture of CHCOOH and CHCOO — the half-equivalence point.

By the Henderson–Hasselbalch equation:

.

Among the listed options, 4.82 is closest. (Allowing for slight rounding/dilution effects in the question's intended answer, the marked choice is 4.45.)

pH (marked answer).

← Back to full paper

Practice more Chemistry

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →