JEE Main 2025 · 3 April · evening · Single correct · +4 / −1
40 mL of a mixture of CH₃ COOH and HCl (aqueous solution) is titrated against 0.1 M NaOH solution conductometrically. Which of the following statement is correct?
- (a) The concentration of CH₃ COOH in the original mixture is 0.005 M
- (b) The concentration of HCl in the original mixture is 0.005 M
- (c) CH₃ COOH is neutralised first followed by neutralisation of HCl
- (d) Point ' C ' indicates the complete neutralisation of HCl
Answer and solution
Answer (b) The concentration of HCl in the original mixture is 0.005 M
Solution
Chemistry ma’am · 4 steps
Read the graph one segment at a time: HCl finishes at B, acetic acid at C, and the volumes give the moles.
- Order of neutralisation. NaOH reacts with the strong acid HCl first. The weak acid CH₃COOH, which is barely ionised, reacts only after all the HCl is used up.
Segment What happens Conductance A to B (0 to 2.0 mL) Fast H⁺ from HCl is replaced by slower Na⁺ falls steeply B to C (2.0 to 5.0 mL) Weakly ionised CH₃COOH turns into CH₃COO⁻ and Na⁺ ions rises slowly C to D (beyond 5.0 mL) Excess Na⁺ and fast OH⁻ build up rises steeply - HCl concentration. HCl uses 2.0 mL of 0.1 M NaOH, that is 0.1 × 2.0 = 0.2 mmol. In 40 mL this is 0.2/40 = 0.005 M.
- Acetic acid check. CH₃COOH uses 5.0 - 2.0 = 3.0 mL, that is 0.3 mmol, so its concentration is 0.3/40 = 0.0075 M, not 0.005 M.
- So. The concentration of HCl in the original mixture is 0.005 M, which is option (b).
The idea to keep. In a mixed-acid conductometric titration, the strong acid is neutralised first. Each break in the graph marks one acid finishing, and the volume between two breaks belongs to that acid alone.
Answer from the NTA final answer key. Solution by PrepWiser.