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(A) $(P)$ $(C_2H_5)_3N + CH_3COOH \longrightarrow (C_2H_5)_3NH^+ + CH_3COO^-$. Since both reactants are weak,the formation of ions leads to an increas

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$3 \quad 4 \quad 2 \quad 1$

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(A) $(P)$ $(C_2H_5)_3N + CH_3COOH \longrightarrow (C_2H_5)_3NH^+ + CH_3COO^-$. Since both reactants are weak,the formation of ions leads to an increase in conductivity initially. After the equivalence point,the addition of excess weak acid does not significantly change the ion concentration,so it remains nearly constant. Thus,$P-3$.$(Q)$ $KI + AgNO_3 \longrightarrow AgI(s) + KNO_3$. Here,$Ag^+$ ions are replaced by $K^+$ ions in the solution. Since the mobility of $K^+$ is comparable to $Ag^+$,the conductivity does not change much initially. After the equivalence point,excess $AgNO_3$ adds $Ag^+$ and $NO_3^-$ ions,causing conductivity to increase. Thus,$Q-4$.$(R)$ $CH_3COOH + KOH \longrightarrow CH_3COOK + H_2O$. $OH^-$ ions (high mobility) are replaced by $CH_3COO^-$ ions (low mobility),causing conductivity to decrease. After the equivalence point,excess $KOH$ adds $K^+$ and $OH^-$ ions,but the effect is minimal due to the common ion effect or dilution. Thus,$R-2$.$(S)$ $NaOH + HI \longrightarrow NaI + H_2O$. $H^+$ ions (very high mobility) are replaced by $Na^+$ ions (low mobility),causing conductivity to decrease. After the equivalence point,excess $HI$ adds $H^+$ and $I^-$ ions,causing conductivity to increase. Thus,$S-1$.The correct matching is $P-3, Q-4, R-2, S-1$,which corresponds to option $(A)$.

List-$I$	List-$II$
$P$. $\underset{X}{(C_2H_5)_3N} + \underset{Y}{CH_3COOH}$	$1$. Conductivity decreases and then increases
$Q$. $\underset{X}{KI (0.1 \ M)} + \underset{Y}{AgNO_3 (0.01 \ M)}$	$2$. Conductivity decreases and then does not change much
$R$. $\underset{X}{CH_3COOH} + \underset{Y}{KOH}$	$3$. Conductivity increases and then does not change much
$S$. $\underset{X}{NaOH} + \underset{Y}{HI}$	$4$. Conductivity does not change much and then increases

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