40 mL of a mixture of CH3COOH and HCl (aqueous solution) is titrated against 0.1 M NaOH solution conductometrically.…

Strategy: Analyze the conductometric titration curve. The first segment represents the neutralization of the strong acid ($\ce{HCl}$), and the second represents the weak acid ($\ce{CH3COOH}$).

Point Analysis:

• 0 to 2 mL: Conductan\ce drops rapidly as $\ce{H^+}$ (highly mobile) is repla\ced by $\ce{Na^+}$ (less mobile). $\ce{HCl}$ is being neutralized.
• 2 to 5 mL: Conductan\ce rises slowly/remains stable as $\ce{CH3COOH}$ forms $\ce{CH3COONa}$.
• Beyond 5 mL: Ex\cess $\ce{NaOH}$ causes conductan\ce to rise rapidly ($\ce{OH^-}$ is highly mobile).

Calculation:

• Vol for $\ce{HCl} = 2 \text{ mL}$. $M_{\ce{HCl}} \times 40 = 0.1 \times 2 \implies M_{\ce{HCl}} = 0.005 \text{ M}$. ✓ Correct.
• Vol for $\ce{CH3COOH} = 5 - 2 = 3 \text{ mL}$. $M_{\ce{acid}} \times 40 = 0.1 \times 3 \implies M_{\ce{acid}} = 0.0075 \text{ M}$.

$\boxed{\text{Answer: (b) }}$