Ques. 1JEE Main 2026, 4 April (Evening Shift)
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 value of acetic acid is 4.75).
- 7.0
- 4.75
- 3.5
- 4.82
Answer
(2) 4.75
Solution
The molarity of the acetic acid follows from M1V1 = M2V2:
M1 × 20 = 0.1 × 28.4
M1 = 0.142 M
In solution (X), mmol of CH3COOH = 0.142 × 20 = 2.84 and mmol of NaOH = 0.1 × 14.2 = 1.42.
By CH3COOH + NaOH CH3COONa + H2O, 2.84 − 1.42 = 1.42 mmol of CH3COOH is left and 1.42 mmol of CH3COONa is formed, so the solution is a half-neutralised acidic buffer, and
pH = pKa + [CH3COO-][CH3COOH]
pH = 4.75 + log 1.421.42 = 4.75 + 0 = 4.75
Answer: (2)