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 (Λ₀) values (in mS m² mol⁻¹) for different ions in aqueous solutions are given below:

Ions	$A{g}^{+}$	$K^{+}$	$N{a}^{+}$	$H^{+}$	$N{O}_3^{-}$	$C{l}^{-}$	$S{O}_4^{2-}$	$O{H}^{-}$	$C{H}_{3}CO{O}^{-}$
${\Lambda }_0$	6.2	7.4	5.0	35.0	7.2	7.6	16.0	19.9	4.1

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	List-II
(P) Titrate: KCl       Titrant: AgNO3	(1)
(Q) Titrate: AgNO3        Titrant: KCl	(2)
(R) Titrate: NaOH       Titrant: HCl	(3)
(S) Titrate: NaOH      Titrant: CH3COOH	(4)
	(5)

The correct option is P-3, Q-4, R-2, S-5.

Analysis of the Limiting Ionic Conductivity (${\Lambda }_0$):
Based on the provided table:
• $H^{+}$ = 35.0 mS m² mol⁻¹
• $O{H}^{-}$ = 19.9 mS m² mol⁻¹
• $C{l}^{-}$ = 7.6 mS m² mol⁻¹
• $K^{+}$ = 7.4 mS m² mol⁻¹
• $N{O}_3^{-}$ = 7.2 mS m² mol⁻¹
• $A{g}^{+}$ = 6.2 mS m² mol⁻¹
• $N{a}^{+}$ = 5.0 mS m² mol⁻¹
• $C{H}_3$ = 4.1 mS m² mol⁻¹

Step-by-Step Matching of the Titrations:

1. Combination (P): Titrate = KCl, Titrant = AgNO₃
• Reaction: $KCl+AgN{O}_3\to AgCl\left(s\right)+KN{O}_3$
• Before equivalence: Highly mobile $C{l}^{-}$ ions (${\Lambda }_0=7.6$) are precipitated as solid AgCl and replaced by slightly less mobile $N{O}_3^{-}$ ions (${\Lambda }_0=7.2$). The concentration of $K^{+}$ remains unchanged. Consequently, the conductance decreases very slightly.
• After equivalence: Excess $AgN{O}_3$ is added, contributing mobile $A{g}^{+}$ (6.2) and $N{O}_3^{-}$ (7.2) ions, which causes a steady increase in conductance.
• This matches the graph where the conductance decreases slightly and then increases, corresponding to (3) (Image 2).

2. Combination (Q): Titrate = AgNO₃, Titrant = KCl
• Reaction: $AgN{O}_3+KCl\to AgCl\left(s\right)+KN{O}_3$
• Before equivalence: $A{g}^{+}$ ions (${\Lambda }_0=6.2$) are precipitated and replaced by slightly more mobile $K^{+}$ ions (${\Lambda }_0=7.4$). The concentration of $N{O}_3^{-}$ remains unchanged. Therefore, the conductance increases slightly.
• After equivalence: Excess KCl is added, introducing highly mobile $K^{+}$ (7.4) and $C{l}^{-}$ (7.6) ions, which results in a steeper increase in conductance.
• This matches the graph where conductance increases slowly and then increases with a steeper slope, corresponding to (4) (Image 3).

3. Combination (R): Titrate = NaOH, Titrant = HCl
• Reaction: $NaOH+HCl\to NaCl+H_{2}O$
• Before equivalence: The highly conducting $O{H}^{-}$ ions (${\Lambda }_0=19.9$) are replaced by less conducting $C{l}^{-}$ ions (${\Lambda }_0=7.6$). Thus, the conductance decreases sharply.
• After equivalence: Excess HCl is added, introducing highly conducting $H^{+}$ ions (${\Lambda }_0=35.0$) and $C{l}^{-}$ (7.6) ions, causing the conductance to rise steeply.
• This matches the classic V-shaped curve, corresponding to (2) (Image 1).

4. Combination (S): Titrate = NaOH, Titrant = CH₃COOH
• Reaction: $NaOH+C{H}_3$
• Before equivalence: Highly mobile $O{H}^{-}$ ions (${\Lambda }_0=19.9$) are replaced by poorly conducting acetate ions, $C{H}_3$ (${\Lambda }_0=4.1$). This results in a sharp decrease in conductance.
• After equivalence: Excess weak acid $C{H}_3$ is added. Because of the common ion effect from the acetate salt, the dissociation of this weak acid is heavily suppressed. Consequently, the conductance remains nearly constant.
• This matches the graph where the conductance decreases and then levels off, corresponding to (5) (Image 4).

Conclusion:
Matching the pairs yields:
• P → 3
• Q → 4
• R → 2
• S → 5