40 mL of a mixture of and HCl (aqueous solution) is titrated against 0.1 M NaOH solution conductometrically. Which of the following statement is correct? [2025]

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Solution

Q.
40 mL of a mixture of ${CH}_{3} COOH$ and HCl (aqueous solution) is titrated against 0.1 M NaOH solution conductometrically. Which of the following statement is correct
[2025]

1 The concentration of ${CH}_{3} COOH$ in the original mixture is 0.005 M.

2 The concentration of HCl in the original mixture is 0.005 M.

3 ${CH}_{3} COOH$ is neutralised first followed by neutralisation of HCl.

4 Point ‘C’ indicates the complete neutralisation of HCl.

Ans.
(2)

First NaOH reacts with HCl. Point B indicates complete neutralization of HCl. 2 mL of NaOH is used to neutralize HCl.

When NaOH completely neutralizes HCl:

$mmol of NaOH = mmol of HCl$

$M_{NaOH} \times V_{NaOH} (that reacts with HCl) = M_{HCl} \times V_{HCl}$

$0.1 \times 2 = M_{HCl} \times 40$

$M_{HCl} = 0.005 M$

When HCl is completely neutralized, then NaOH reacts with ${CH}_{3} COOH$ . Thus, (5−2) mL = 3 mL of NaOH reacts with ${CH}_{3} COOH$ . Point C indicates complete neutralization of ${CH}_{3} COOH$ .

When NaOH completely neutralizes ${CH}_{3} COOH$ :

$mmol of NaOH = {mmol of CH}_{3} COOH$

$M_{N a O H} \times V_{N a O H}$ (that reacts with ${CH}_{3} COOH$ )

$= M_{C H_{3} C O O H} \times V_{C H_{3} C O O H}$

$0.1 \times 3$ $= M_{C H_{3} C O O H} \times 40$

$M_{C H_{3} C O O H} = 0.0075 M$

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