JEE MAIN - Chemistry (2025 - 3rd April Evening Shift - No. 19)
40 mL of a mixture of $\mathrm{CH}_3 \mathrm{COOH}$ and HCl (aqueous solution) is titrated against 0.1 M NaOH solution conductometrically. Which of the following statement is correct?
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The concentration of $\mathrm{CH}_3 \mathrm{COOH}$ in the original mixture is 0.005 M
The concentration of HCl in the original mixture is 0.005 M
$\mathrm{CH}_3 \mathrm{COOH}$ is neutralised first followed by neutralisation of HCl
Point ' C ' indicates the complete neutralisation of HCl
Explanation
For HCl:
$ \text{Moles of HCl} = \text{Moles of NaOH used} $
$ M \times 40 = 0.1 \times 2 $
$ M = 0.005 $
For $\mathrm{CH}_3 \mathrm{COOH}$:
$ \text{Moles of } \mathrm{CH}_3 \mathrm{COOH} = \text{Moles of NaOH used} $
$ M \times 40 = 0.1 \times 3 $
$ M = 0.0075 $
Since HCl is a strong acid, it is neutralized first.
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