JEE Chemistry 2025 Question with Solution

Given: A mixture of CH$_3$COOH and HCl of volume $40 \, \text{mL}$ is titrated with $0.1 \, \text{M}$ NaOH conductometrically.

Find: The correct statement among the given options.

In a conductometric titration of a strong acid and a weak acid with a strong base, the strong acid HCl is neutralised first. This is because highly mobile $\text{H}^+$ ions are replaced by less mobile $\text{Na}^+$ ions, so conductance decreases first.

Thus, region A-B corresponds to neutralisation of HCl:

$$\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}$$

From the graph, complete neutralisation of HCl occurs at point B, where volume of NaOH used is $2.0 \, \text{mL}$.

Moles of NaOH used for HCl:

$$0.1 \times \frac{2.0}{1000} = 0.0002 \; \text{mol}$$

Therefore, moles of HCl in the original mixture are also:

$$0.0002 \; \text{mol}$$

Concentration of HCl in $40 \, \text{mL} = 0.04 \, \text{L}$:

$$\frac{0.0002}{0.04} = 0.005 \; \text{M}$$

Now, from B-C, CH$_3$COOH is neutralised:

$$\text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O}$$

The NaOH used for CH$_3$COOH is:

$$(5.0 - 2.0) \, \text{mL} = 3.0 \, \text{mL}$$

So its concentration is:

$$\frac{0.1 \times 3.0/1000}{0.04} = 0.0075 \; \text{M}$$

Hence:

• Statement A is incorrect.
• Statement B is correct.
• Statement C is incorrect because HCl is neutralised first.
• Statement D is incorrect because point B, not C, indicates complete neutralisation of HCl.

Therefore, the correct option is B.

Given: The curve has slope changes at $2.0 \, \text{mL}$ and $5.0 \, \text{mL}$ of NaOH added.

Find: Which statement matches the conductometric behaviour.

The first break point marks exhaustion of the strong acid. Therefore:

$$\text{At } 2.0 \, \text{mL}, \; \text{HCl is completely neutralised.}$$

The second break point marks exhaustion of the weak acid. Therefore:

$$\text{At } 5.0 \, \text{mL}, \; \text{CH}_3\text{COOH is completely neutralised.}$$

For HCl:

$$\text{moles} = 0.1 \times 2.0 \times 10^{-3} = 2.0 \times 10^{-4} \; \text{mol}$$ $$\text{Concentration} = \frac{2.0 \times 10^{-4}}{4.0 \times 10^{-2}} = 5.0 \times 10^{-3} \; \text{M}$$

For CH$_3$COOH:

$$\text{moles} = 0.1 \times 3.0 \times 10^{-3} = 3.0 \times 10^{-4} \; \text{mol}$$ $$\text{Concentration} = \frac{3.0 \times 10^{-4}}{4.0 \times 10^{-2}} = 7.5 \times 10^{-3} \; \text{M}$$

So the only correct statement is that the concentration of HCl in the original mixture is $0.005 \, \text{M}$. The correct option is B.