40 mL of a mixture of $\mathrm{CH}_3 \mathrm{COOH}$ and HCl (aqueous solution) is titrated against 0.1 M…

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?

  1. The concentration of $\mathrm{CH}_3 \mathrm{COOH}$ in the original mixture is 0.005 M
  2. The concentration of HCl in the original mixture is 0.005 M
  3. $\mathrm{CH}_3 \mathrm{COOH}$ is neutralised first followed by neutralisation of HCl
  4. Point ' C ' indicates the complete neutralisation HCl

Solution

From the given graph 2 ml NaOH solution is used for neutralisation of HCl and 3 ml NaOH solution is used for neutralisation of $\mathrm{CH}_3 \mathrm{COOH}$.
$\therefore$ Mole of $\mathrm{HCl}=$ Mole of NaOH used
$\begin{aligned}
& \mathrm{M} \times 40=0.1 \times 2 \\
& \mathrm{M}=0.005
\end{aligned}$
$\therefore$ Mole of $\mathrm{CH}_3 \mathrm{COOH}=$ Mole of NaOaH used
$\begin{aligned}
& \mathrm{M} \times 40=0.1 \times 3 \\
& \mathrm{M}=0.0075
\end{aligned}$
HCl is strong acid and will be neutralised first.

Asked in: JEE Main 2025 (03 Apr Shift 2)

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