The question asks to match four different titration reactions (List-I) with their corresponding conductivity variation curves (List-II). The conductivity of a solution depends on the concentration and molar ionic conductivity (mobility) of the ions present. The key principle is to analyze how these factors change as the titrant (X) is added to the analyte (Y).

Key Ionic Mobilities: H⁺ and OH⁻ ions have exceptionally high molar ionic conductivities compared to other ions. The general trend is λ°(H⁺) > λ°(OH⁻) > λ°(other ions).

Let's analyze each case:

P. $\mathop {(C{}_2{H_5}){}_3N}\limits_X$ + $\mathop {C{H_3}COOH}\limits_Y$ This is the titration of a weak acid (acetic acid, Y) with a weak base (triethylamine, X).

1. Initial Solution (Y): The initial solution is aqueous CH₃COOH, a weak acid. It is poorly dissociated, so the concentration of ions (H⁺ and CH₃COO⁻) is very low. Thus, the initial conductivity is low.
2. During Titration (adding X): The neutralization reaction is: $(C₂H₅)₃N + CH₃COOH → (C₂H₅)₃NH⁺ + CH₃COO⁻$ A salt, triethylammonium acetate, is formed. This salt is a strong electrolyte and is fully ionized. We are essentially replacing non-ionic weak acid and weak base molecules with ions. As the concentration of ions increases, the conductivity of the solution increases.
3. After Equivalence Point: We are adding excess weak base, (C₂H₅)₃N. Since it's a weak base, it dissociates very little. Additionally, the presence of the common ion (C₂H₅)₃NH⁺ from the salt suppresses its dissociation further. Therefore, the concentration of ions does not increase significantly, and the conductivity does not change much. Conclusion: The conductivity increases, then does not change much. This matches List-II, description 3.

Q. $\mathop {KI(0.1M)}\limits_X$ + $\mathop {AgN{O_3}(0.01M)}\limits_Y$ This is a precipitation titration of silver nitrate (Y) with potassium iodide (X).

1. Initial Solution (Y): The initial solution is aqueous AgNO₃, a strong electrolyte, fully dissociated into Ag⁺ and NO₃⁻ ions. It has a certain initial conductivity.
2. During Titration (adding X): The reaction is a precipitation reaction: $Ag⁺(aq) + NO₃⁻(aq) + K⁺(aq) + I⁻(aq) → AgI(s)↓ + K⁺(aq) + NO₃⁻(aq)$ The net effect is that Ag⁺ ions are removed from the solution and are replaced by K⁺ ions. The molar ionic conductivities of Ag⁺ (61.9 S cm² mol⁻¹) and K⁺ (73.5 S cm² mol⁻¹) are similar in magnitude. Therefore, the conductivity of the solution does not change much until the equivalence point.
3. After Equivalence Point: We are adding excess KI, which is a strong electrolyte. The concentration of K⁺ and I⁻ ions in the solution increases significantly, causing a sharp increase in conductivity. Conclusion: The conductivity does not change much, then increases. This matches List-II, description 4.

R. $\mathop {C{H_3}COOH}\limits_X$ + $\mathop {KOH}\limits_Y$ This is the titration of a strong base (potassium hydroxide, Y) with a weak acid (acetic acid, X).

1. Initial Solution (Y): The initial solution is aqueous KOH, a strong base, fully dissociated into K⁺ and OH⁻ ions. Due to the very high mobility of OH⁻ ions, the initial conductivity is high.
2. During Titration (adding X): The neutralization reaction is: $KOH + CH₃COOH → CH₃COOK + H₂O$ The net ionic equation is: $OH⁻ + CH₃COOH → CH₃COO⁻ + H₂O$ The highly mobile OH⁻ ions are consumed and replaced by the much less mobile acetate ions (CH₃COO⁻). This leads to a significant decrease in conductivity.
3. After Equivalence Point: We are adding excess CH₃COOH, which is a weak acid and dissociates very little. The presence of the common ion CH₃COO⁻ from the salt further suppresses its dissociation. Thus, the concentration of ions does not increase significantly, and the conductivity does not change much. Conclusion: The conductivity decreases, then does not change much. This matches List-II, description 2.

S. $\mathop {NaOH}\limits_X$ + $\mathop {HI}\limits_Y$ This is the titration of a strong acid (hydroiodic acid, Y) with a strong base (sodium hydroxide, X).

1. Initial Solution (Y): The initial solution is aqueous HI, a strong acid, fully dissociated into H⁺ and I⁻ ions. Due to the exceptionally high mobility of H⁺ ions, the initial conductivity is very high.
2. During Titration (adding X): The neutralization reaction is: $HI + NaOH → NaI + H₂O$ The net ionic equation is: $H⁺ + OH⁻ → H₂O$ The highly mobile H⁺ ions are consumed and replaced by the much less mobile Na⁺ ions. This leads to a sharp decrease in conductivity. The conductivity is minimum at the equivalence point.
3. After Equivalence Point: We are adding excess NaOH, which is a strong base and fully dissociates into Na⁺ and highly mobile OH⁻ ions. The concentration of ions, particularly the fast-moving OH⁻, increases, causing a sharp increase in conductivity. Conclusion: The conductivity decreases, then increases. This matches List-II, description 1.

Summary of Matches:

• P → 3
• Q → 4
• R → 2
• S → 1

This corresponds to the code P - 3; Q - 4; R - 2; S - 1.