40 mL of a mixture of $$CH_3COOH$$ and HCl (aqueous solution) is titrated against 0.1 M NaOH conductometrically. Which of the following statement is correct?

The conductometric titration curve represents the titration of a mixture of a strong acid (HCl) and a weak acid ($CH_3COOH$) with a strong base (NaOH).

The conductance of the solution depends on the number of ions present and their ionic mobility. Since the mobility of $$H^+$$ ions is exceptionally high, the conductance changes characteristically during the titration.

Initially, in segment AB, HCl is neutralized first because it is completely dissociated in solution. The highly mobile $$H^+$$ ions are replaced by the comparatively less mobile $$Na^+$$ ions from NaOH, causing a sharp decrease in conductance.

Point B, corresponding to $$2.0\ \text{mL}$$ of NaOH added, represents the equivalence point of HCl.

Beyond point B, in segment BC, NaOH starts neutralizing acetic acid:

$$CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O.$$

Since acetic acid is only weakly ionized whereas sodium acetate is completely dissociated, the number of ions in solution increases, producing a gradual increase in conductance.

Point C, corresponding to $$5.0\ \text{mL}$$ of NaOH added, represents the complete neutralization of acetic acid.

After point C, any additional NaOH contributes free $$OH^-$$ ions to the solution. Because $$OH^-$$ ions possess high mobility, the conductance rises sharply in segment CD.

Hence:

• Option (C) is incorrect because HCl is neutralized before $$CH_3COOH$$.
• Option (D) is incorrect because point C corresponds to the equivalence point of $$CH_3COOH$$, while HCl is neutralized at point B.

Now calculate the concentrations.

Given,

$$V_{\text{mixture}}=40\ \text{mL},$$

$$M_{\text{NaOH}}=0.1\ \text{M}.$$

For HCl,

$$M_{\text{HCl}}\times40=0.1\times2.0,$$

so

$$M_{\text{HCl}}=\frac{0.2}{40}=0.005\ \text{M}.$$

For acetic acid, the volume of NaOH consumed is

$$5.0-2.0=3.0\ \text{mL}.$$

Therefore,

$$M_{CH_3COOH}\times40=0.1\times3.0,$$

giving

$$M_{CH_3COOH}=\frac{0.3}{40}=0.0075\ \text{M}.$$

Thus,

• $$[HCl]=0.005\ \text{M}$$
• $$[CH_3COOH]=0.0075\ \text{M}$$

Hence, the correct statement is that the concentration of HCl is

$$0.005\ \text{M},$$

so the correct answer is option (B).