Conductometric Titration Curves Matching for Ionic Conductivities

To determine the correct matching between the titrations in List-I and the conductance curves in List-II, we analyze the change in total ionic conductance before and after the equivalence point for each case using the given limiting ionic conductivities ($\Lambda_0$).

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1. Pair (P): Titrate $\text{KCl}$, Titrant $\text{AgNO}_3$

• Before Equivalence Point:
  The precipitation reaction occurring in solution is: $\text{Cl}^- + \text{Ag}^+ \rightarrow \text{AgCl}(s)$ As $\text{AgNO}_3$ is added, $\text{Cl}^-$ ions are replaced by $\text{NO}_3^-$ ions while $\text{K}^+$ ions remain unchanged.
  The net change in conductivity per mole of added titrant is: $\Delta \Lambda_1 = \Lambda_0(\text{NO}_3^-) - \Lambda_0(\text{Cl}^-) = 7.2 - 7.6 = -0.4 \text{ mS m}^2 \text{ mol}^{-1}$ Since $\Delta \Lambda_1 < 0$, the conductance decreases slightly before the equivalence point.

• After Equivalence Point:
  Adding excess $\text{AgNO}_3$ introduces unreacted $\text{Ag}^+$ and $\text{NO}_3^-$ ions into the solution.
  The change in conductivity per mole of added titrant is: $\Delta \Lambda_2 = \Lambda_0(\text{Ag}^+) + \Lambda_0(\text{NO}_3^-) = 6.2 + 7.2 = +13.4 \text{ mS m}^2 \text{ mol}^{-1}$ Thus, the conductance increases after the equivalence point.

• Matching Graph: This behavior (slight decrease followed by an increase) corresponds to Graph (3).
  $\text{P} \rightarrow 3$

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2. Pair (Q): Titrate $\text{AgNO}_3$, Titrant $\text{KCl}$

• Before Equivalence Point:
  The precipitation reaction occurring in solution is: $\text{Ag}^+ + \text{Cl}^- \rightarrow \text{AgCl}(s)$ As $\text{KCl}$ is added, $\text{Ag}^+$ ions are replaced by $\text{K}^+$ ions while $\text{NO}_3^-$ ions remain unchanged.
  The net change in conductivity per mole of added titrant is: $\Delta \Lambda_1 = \Lambda_0(\text{K}^+) - \Lambda_0(\text{Ag}^+) = 7.4 - 6.2 = +1.2 \text{ mS m}^2 \text{ mol}^{-1}$ Since $\Delta \Lambda_1 > 0$, the conductance increases slightly before the equivalence point.

• After Equivalence Point:
  Adding excess $\text{KCl}$ introduces unreacted $\text{K}^+$ and $\text{Cl}^-$ ions.
  The change in conductivity per mole of added titrant is: $\Delta \Lambda_2 = \Lambda_0(\text{K}^+) + \Lambda_0(\text{Cl}^-) = 7.4 + 7.6 = +15.0 \text{ mS m}^2 \text{ mol}^{-1}$ Thus, the conductance increases at a steeper slope after the equivalence point.

• Matching Graph: This behavior (slight increase followed by a steeper increase) corresponds to Graph (4).
  $\text{Q} \rightarrow 4$

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3. Pair (R): Titrate $\text{NaOH}$, Titrant $\text{HCl}$

• Before Equivalence Point:
  The neutralization reaction is: $\text{OH}^- + \text{H}^+ \rightarrow \text{H}_2\text{O}(l)$ $\text{OH}^-$ ions are replaced by $\text{Cl}^-$ ions.
  The net change in conductivity per mole of added titrant is: $\Delta \Lambda_1 = \Lambda_0(\text{Cl}^-) - \Lambda_0(\text{OH}^-) = 7.6 - 19.9 = -12.3 \text{ mS m}^2 \text{ mol}^{-1}$ This results in a steep decrease in conductance before the equivalence point.

• After Equivalence Point:
  Adding excess strong acid $\text{HCl}$ introduces highly conducting $\text{H}^+$ ions along with $\text{Cl}^-$ ions: $\Delta \Lambda_2 = \Lambda_0(\text{H}^+) + \Lambda_0(\text{Cl}^-) = 35.0 + 7.6 = +42.6 \text{ mS m}^2 \text{ mol}^{-1}$ This leads to a steep increase in conductance.

• Matching Graph: This forms a classic V-shaped curve, corresponding to Graph (2).
  $\text{R} \rightarrow 2$

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4. Pair (S): Titrate $\text{NaOH}$, Titrant $\text{CH}_3\text{COOH}$

• Before Equivalence Point:
  The neutralization reaction is: $\text{OH}^- + \text{CH}_3\text{COOH} \rightarrow \text{CH}_3\text{COO}^- + \text{H}_2\text{O}(l)$ $\text{OH}^-$ ions are replaced by $\text{CH}_3\text{COO}^-$ ions.
  The net change in conductivity per mole of added titrant is: $\Delta \Lambda_1 = \Lambda_0(\text{CH}_3\text{COO}^-) - \Lambda_0(\text{OH}^-) = 4.1 - 19.9 = -15.8 \text{ mS m}^2 \text{ mol}^{-1}$ This causes a steep decrease in conductance before the equivalence point.

• After Equivalence Point:
  Excess weak acid $\text{CH}_3\text{COOH}$ is added. Due to the presence of acetate ions ($\text{CH}_3\text{COO}^-$) from the salt formed, further dissociation of the weak acid is suppressed by the common-ion effect. Thus, the concentration of free ions remains practically constant, causing the curve to become nearly horizontal/flat.

• Matching Graph: This behavior corresponds to Graph (5).
  $\text{S} \rightarrow 5$

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Summary of Matching:

$\text{P-3, Q-4, R-2, S-5}$

This corresponds to Option (C).