Conductometric titration of a mixture of strong and weak acids with strong base

The mixture contains:

• $\mathrm{HCl}$ = strong acid
• $\mathrm{CH_3COOH}$ = weak acid

It is titrated with $0.1\,\mathrm{M}$ NaOH.

Order of neutralisation

In conductometric titration, $\mathrm{HCl}$ is neutralised first because it is fully ionised and contributes highly mobile $\mathrm{H^+}$ ions.

Reaction: $\mathrm{HCl + NaOH \rightarrow NaCl + H_2O}$

After all HCl is consumed, acetic acid starts getting neutralised: $\mathrm{CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O}$

Hence, option C is false.

Interpretation of conductometric graph

For such a mixture, the conductance:

• decreases initially as highly mobile $\mathrm{H^+}$ from HCl are replaced by less mobile $\mathrm{Na^+}$
• then changes differently when $\mathrm{CH_3COOH}$ is neutralised
• then rises sharply after complete neutralisation due to excess $\mathrm{OH^-}$

Therefore, the first break point corresponds to complete neutralisation of HCl.

So if point $C$ is the second equivalence point (usual labeling in such graphs), it does not indicate only HCl neutralisation. Thus D is false.

Using the graph information

From the standard conductometric titration graph for this question, the first equivalence point corresponds to volume of NaOH used for HCl only.

If this first equivalence volume is $2\,\mathrm{mL}$, then moles of HCl in $40\,\mathrm{mL}$ mixture are: $n_{\mathrm{HCl}} = 0.1 \times 2\times 10^{-3} = 2\times 10^{-4}\,\mathrm{mol}$

Therefore concentration of HCl is: $[\mathrm{HCl}] = \frac{2\times 10^{-4}}{40\times 10^{-3}} = 5\times 10^{-3}\,\mathrm{M} = 0.005\,\mathrm{M}$

So option B is correct.

Check option A

Option A says acetic acid concentration is $0.005\,\mathrm{M}$, but from the graph acetic acid corresponds to the additional NaOH consumed after the first equivalence point, not $2\,\mathrm{mL}$. Hence A is false.

Final conclusion

The correct statement is: $\boxed{\text{B: The concentration of } \mathrm{HCl} \text{ in the original mixture is } 0.005\,M}$