Question 14

Added on: Sep 19, 2026

MATRIX MATCHHARD

In a conductometric titration, small volume of titrant of higher concentration is added stepwise to a larger volume of titrate of much lower concentration, and the conductance is measured after each addition.

The limiting ionic conductivity ( $\Lambda_0$ ) values (in $\text{mS} \cdot \text{m}^2 \cdot \text{mol}^{-1}$ ) for different ions in aqueous solutions are given below:

\begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline \textbf{Ions} & \text{Ag}^+ & \text{K}^+ & \text{Na}^+ & \text{H}^+ & \text{NO}_3^- & \text{Cl}^- & \text{SO}_4^{2-} & \text{OH}^- & \text{CH}_3\text{COO}^- \\ \hline \mathbf{\Lambda_0} & 6.2 & 7.4 & 5.0 & 35.0 & 7.2 & 7.6 & 16.0 & 19.9 & 4.1 \\ \hline \end{array}

For different combinations of titrates and titrants given in List-I, the graphs of 'conductance' versus 'volume of titrant' are given in List-II.

Match each entry in List-I with the appropriate entry in List-II and choose the correct option.

List - I

Titrate: $KCl$ ;

Titrant: $AgNO_3$

Titrate: $AgNO_3$ ;

Titrant: $KCl$

Titrate: $NaOH$ ;

Titrant: $HCl$

Titrate: $NaOH$ ;

Titrant: $CH_3COOH$

List-II

(A)

P-4, Q-3, R-2, S-5

(B)

P-2, Q-4, R-3, S-1

(C)

P-3, Q-4, R-2, S-5

(D)

P-4, Q-3, R-2, S-1

Detailed Solution

The conductometric titration curves are determined by the relative limiting ionic conductivities ( $\Lambda_0$ ) of the ions being consumed and produced.

Case (P): $KCl$ + $AgNO_3 \rightarrow AgCl(s) + KNO_3$
- Before Equivalence: $Cl^-$ ( $\Lambda_0 = 7.6$ ) is replaced by $NO_3^-$ ( $\Lambda_0 = 7.2$ ). The net change in conductance is proportional to $7.2 - 7.6 = -0.4$ , which is a very slight decrease (almost horizontal).
- After Equivalence: Excess $AgNO_3$ is added. Conductance increases with a slope proportional to $\Lambda_0(Ag^+) + \Lambda_0(NO_3^-) = 6.2 + 7.2 = 13.4$ .
- This matches Graph (3).
Case (Q): $AgNO_3$ + $KCl \rightarrow AgCl(s) + KNO_3$
- Before Equivalence: $Ag^+$ ( $\Lambda_0 = 6.2$ ) is replaced by $K^+$ ( $\Lambda_0 = 7.4$ ). The net change is proportional to $7.4 - 6.2 = +1.2$ , causing a slight increase.
- After Equivalence: Excess $KCl$ is added. Conductance increases more steeply with a slope proportional to $\Lambda_0(K^+) + \Lambda_0(Cl^-) = 7.4 + 7.6 = 15.0$ .
- This matches Graph (4).
Case (R): $NaOH$ + $HCl \rightarrow NaCl + H_2O$
- Before Equivalence: $OH^-$ ( $\Lambda_0 = 19.9$ ) is replaced by $Cl^-$ ( $\Lambda_0 = 7.6$ ). Net change is $7.6 - 19.9 = -12.3$ . Conductance decreases sharply.
- After Equivalence: Excess $HCl$ is added. Conductance increases very steeply due to $H^+$ and $Cl^-$ . Slope $\propto 35.0 + 7.6 = 42.6$ .
- This matches Graph (2).
Case (S): $NaOH$ + $CH_3COOH \rightarrow CH_3COONa + H_2O$
- Before Equivalence: $OH^-$ ( $19.9$ ) is replaced by $CH_3COO^-$ ( $4.1$ ). Net change is $4.1 - 19.9 = -15.8$ . Conductance decreases sharply.
- After Equivalence: $CH_3COOH$ is a weak acid. Its dissociation is further suppressed by the common ion effect of $CH_3COO^-$ from the salt. The added titrant contributes negligible conductance, resulting in a near-horizontal line.
- This matches Graph (5).

Matching Result: P-3, Q-4, R-2, S-5. Correct Option: (C)

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