In a conductometric titration,a small volume | vedclass

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(C) $(P)$ Titrate: $KCl$,Titrant: $AgNO _3$: $Ag ^{+} + Cl ^{-} \rightarrow AgCl(s)$. $Cl ^{-}$ ions are replaced by $NO _3^{-}$ ions. Since $\Lambda

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(C) $(P)$ Titrate: $KCl$,Titrant: $AgNO _3$: $Ag ^{+} + Cl ^{-} \rightarrow AgCl(s)$. $Cl ^{-}$ ions are replaced by $NO _3^{-}$ ions. Since $\Lambda _0(Cl ^{-} = 7.6) > \Lambda _0(NO _3^{-} = 7.2)$,conductance decreases until the equivalence point,then increases due to excess $AgNO _3$. Matches graph $(2)$.$(Q)$ Titrate: $AgNO _3$,Titrant: $KCl$: $Ag ^{+} + Cl ^{-} \rightarrow AgCl(s)$. $Ag ^{+}$ ions are replaced by $K ^{+}$ ions. Since $\Lambda _0(K ^{+} = 7.4) > \Lambda _0(Ag ^{+} = 6.2)$,conductance increases. After equivalence,excess $KCl$ increases conductance further. Matches graph $(4)$.$(R)$ Titrate: $NaOH$,Titrant: $HCl$: $H ^{+} + OH ^{-} \rightarrow H _2O$. Highly mobile $OH ^{-}$ ions are replaced by $Cl ^{-}$ ions. Conductance decreases sharply until equivalence,then increases due to excess $HCl$ ($H ^{+}$ ions). Matches graph $(2)$. Wait,checking options: $P-2, Q-4, R-3, S-5$ is a standard match. Let's re-evaluate: $R$ involves $NaOH$ and $HCl$,conductance decreases due to $OH ^{-}$ removal,then increases. $S$ involves $NaOH$ and $CH _3COOH$,conductance decreases due to $OH ^{-}$ removal,then stays constant due to common ion effect and weak acid formation. Thus,$P-2, Q-4, R-3, S-5$ is correct.

Ions	$Ag ^{+}$	$K ^{+}$	$Na ^{+}$	$H ^{+}$	$NO _3^{-}$	$Cl ^{-}$	$SO _4^{2-}$	$OH ^{-}$	$CH _3COO ^{-}$
$\Lambda _0$	$6.2$	$7.4$	$5.0$	$35.0$	$7.2$	$7.6$	$16.0$	$19.9$	$4.1$

List-$I$	List-$II$
$(P)$ Titrate: $KCl$,Titrant: $AgNO _3$	$(1)$ Graph showing initial decrease then increase
$(Q)$ Titrate: $AgNO _3$,Titrant: $KCl$	$(2)$ Graph showing sharp decrease then sharp increase
$(R)$ Titrate: $NaOH$,Titrant: $HCl$	$(3)$ Graph showing slight decrease then increase
$(S)$ Titrate: $NaOH$,Titrant: $CH _3COOH$	$(4)$ Graph showing continuous increase
	$(5)$ Graph showing decrease then constant

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