Match Conductometric Titration Curves with Reactant Pairs

To determine the correct matching between List I and List II, we analyze the conductivity variations during each conductometric titration. The solution of $X$ is added gradually to the solution of $Y$ initially present in the beaker.

The electrical conductivity ($\kappa$) of a solution depends on the total concentration of ions present and their molar ionic conductivities ($\lambda^\circ$). The $H^+$ and $OH^-$ ions possess exceptionally high ionic mobilities ($\lambda^\circ(H^+) \approx 349.8\text{ S cm}^2\text{mol}^{-1}$ and $\lambda^\circ(OH^-) \approx 198.3\text{ S cm}^2\text{mol}^{-1}$) compared to other typical ions ($\approx 40 - 75\text{ S cm}^2\text{mol}^{-1}$).

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Step-by-Step Analysis:

P. $(C_2H_5)_3N\ (X) + CH_3COOH\ (Y)$

• Initial state: Solution $Y$ contains $CH_3COOH$, a weak acid (weak electrolyte) with low dissociation, resulting in a low initial conductivity.
• Before equivalence point: Addition of triethylamine, $(C_2H_5)_3N$, converts the weak acid into its salt, $[(C_2H_5)_3NH]^+[CH_3COO]^-$, which is a strong electrolyte. The increase in highly dissociated ions causes the conductivity to increase.
• After equivalence point: Addition of excess $(C_2H_5)_3N$ (a weak base) does not dissociate appreciably due to the weak nature of the base. Thus, the conductivity does not change much.
• Match: P $\rightarrow$ 3 (Conductivity increases and then does not change much)

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Q. $KI\ (0.1\text{ M})\ (X) + AgNO_3\ (0.01\text{ M})\ (Y)$

• Initial state: Solution $Y$ contains $AgNO_3$ (strong electrolyte).
• Before equivalence point: Addition of $KI$ causes precipitation of $AgI$: $Ag^+ + NO_3^- + K^+ + I^- \rightarrow AgI(s)\downarrow + K^+ + NO_3^-$ The highly mobile $Ag^+$ ions ($\lambda^\circ \approx 61.9\text{ S cm}^2\text{mol}^{-1}$) are replaced by $K^+$ ions ($\lambda^\circ \approx 73.5\text{ S cm}^2\text{mol}^{-1}$) of comparable ionic mobility. Thus, the conductivity does not change much.
• After equivalence point: Once precipitation is complete, excess $KI$ (a strong electrolyte) adds free $K^+$ and $I^-$ ions into the solution, causing the conductivity to increase.
• Match: Q $\rightarrow$ 4 (Conductivity does not change much and then increases)

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R. $CH_3COOH\ (X) + KOH\ (Y)$

• Initial state: Solution $Y$ contains $KOH$, a strong base with high initial conductivity due to the presence of highly mobile $OH^-$ ions.
• Before equivalence point: The neutralization reaction is: $K^+ + OH^- + CH_3COOH \rightarrow CH_3COO^- + K^+ + H_2O$ Highly mobile $OH^-$ ions are replaced by much less mobile acetate ions, $CH_3COO^-$, causing the conductivity to decrease.
• After equivalence point: Excess $CH_3COOH$ added is a weak acid, and its dissociation is further suppressed by the common ion effect from $CH_3COO^-$. Thus, the conductivity does not change much.
• Match: R $\rightarrow$ 2 (Conductivity decreases and then does not change much)

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S. $NaOH\ (X) + HI\ (Y)$

• Initial state: Solution $Y$ contains $HI$, a strong acid with very high initial conductivity due to highly mobile $H^+$ ions.
• Before equivalence point: The neutralization reaction is: $H^+ + I^- + Na^+ + OH^- \rightarrow Na^+ + I^- + H_2O$ Highly mobile $H^+$ ions are replaced by less mobile $Na^+$ ions, causing the conductivity to decrease sharply.
• After equivalence point: Excess $NaOH$ added introduces free $Na^+$ and highly mobile $OH^-$ ions, causing the conductivity to increase sharply.
• Match: S $\rightarrow$ 1 (Conductivity decreases and then increases)

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Conclusion:

Matching the results:

• $\text{P} \rightarrow 3$
• $\text{Q} \rightarrow 4$
• $\text{R} \rightarrow 2$
• $\text{S} \rightarrow 1$

This corresponds to Option A.