Question

An aqueous solution of X is added slowly to an aqueous solution of Y as shown in List I. The variation in conductivity of these reactions is given in List II. Match List I with List II and select the correct answer using the code given below the lists:

List I		List II
P	$\underset{(X)}{(C_2{H}_5{)}_{3}N}+\underset{(Y)}{C{H}_{3}COOH}$	1	Conductivity decreases and then increases.
Q	$\underset{(X)}{KI(0.1M)}+\underset{(Y)}{AgN{O}_3(0.01M)}$	2	Conductivity decreases and then does not change much.
R	$\underset{(X)}{C{H}_{3}COOH}+\underset{(Y)}{\mathrm{KOH}}$	3	Conductivity increases and then does not change much.
S	$\underset{(X)}{\mathrm{NaOH}}+\underset{(Y)}{\mathrm{HI}}$	4	Conductivity does not change much and then increases.

• A. $(P)-(3),(Q)-(4),(R)-(2),(S)-(1)$
• B. $(P)-(4),(Q)-(3),(R)-(2),(S)-(1)$
• C. $(P)-(2),(Q)-(3),(R)-(4),(S)-(1)$
• D. $(P)-(1),(Q)-(4),(R)-(3),(S)-(2)$

Answer 1

Answer: (A)

$P:\underset{(X)}{(C_2{H}_5{)}_{3}N}+\underset{(Y)}{C{H}_{3}COOH}\to C{H}_{3}CO{O}_{(aq)}^{-}+(C_2{H}_5{)}_{3}N{H}^{+}$
Initially conductivity increases due to ion formation after that it becomes practically constant because X alone cannot form ions.
$Q:\underset{(X)}{KI(0.1M}+\underset{(Y)}{AgN{O}_3(0.01M}\to AgI\downarrow +KN{O}_3$
Number of ions in the solution remains constant until all the $AgN{O}_3$ precipitated as $AgI$. Thereafter conductance increases due to increase in number of ions.
$R:\underset{(X)}{C{H}_{3}COOH}+\underset{(Y)}{\mathrm{KOH}}\to C{H}_{3}COOK+H_{2}O$
Initially conductivity decreases due to the decrease in the number of $O{H}^{-}$ ions thereafter it slowly increases due to increase in number of $H^{+}$ ions.
$S:\underset{(X)}{\mathrm{NaOH}}+\underset{(Y)}{\mathrm{HI}}\to NaI+H_{2}O$
Initially conductivity decreases due to decrease in $H^{+}$ ions and then increases due to increase in $O{H}^{-}$ ions.