MCQ
An aqueous solution ox $X$ 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 $II$
$P$. $\quad \underset{X}{\left( C _2 H _5\right)_3 N }+\underset{ Y }{ CH _3 COOH }$ $1.$ Conductivity decreases and then increases
$Q.$ $\quad \underset{X}{ KI (0.1 M )}+\underset{ Y }{ AgNO _3(0.01 M )}$ $2.$ Conductivity decreases and then does not change much
$R.$ $\quad \underset{X}{ CH _3 COOH }+\underset{ K }{ KOH }$ $3.$ Conductivity increases and then does not change much
$S$. $\quad \underset{X}{ NaOH }+\underset{Y}{ HI }$ $4.$ Conductivity does not change much and then increases

Codes: $ \quad P \quad Q \quad R \quad S $ 

  • $\quad 3 \quad 4 \quad 2 \quad 1 $
  • B
    $\quad 4 \quad 3 \quad 2 \quad 1 $
  • C
    $\quad 2 \quad 3 \quad 4 \quad 1 $
  • D
    $\quad 1 \quad 4 \quad 3 \quad 2 $

Answer

Correct option: A.
$\quad 3 \quad 4 \quad 2 \quad 1 $
a
$(P)$ $\left( C _2 H _5\right)_3 N + CH _3 Y COOH \longrightarrow CH _3 COO ^{-}( aq )+\left( C _2 H _5\right)_3 NH ^{+}( aq )$

As $CH _3 COOH$ is a weak acid, its conductivity is already less. On addition of weak base, acid-base reaction takes place and new ions are created. So conductivity increases.

$(Q)$ $KI (0.1 M )+ AgNO _3(0.01 M ) \longrightarrow AgI \downarrow$ (ppt) $+ KNO _3$ (aq).

As the only reaction taking place is precipitation of AgI and in place of $Ag ^{+}, K ^{+}$is coming in the solution, conductivity remain nearly constant and then increases.

$(R)$ $CH _3 COOH + KOH \longrightarrow CH _3 COOK ( aq )+ H _2 O$

$OH ^{-}( aq )$ is getting replaced by $CH _3 COO ^{-}$, which has poorer conductivity.

So conductivity dereases and then after the end point, due to common ion effect, no further creation of ions take place. So, conductivity remain nearly same.

$(S)$ $NaOH + HI \longrightarrow NaI ( aq )+ H _2 O$

As $H ^{+}$is getting replaced by $Na ^{+}$conductivity dereases and after end point, due to $OH ^{-}$, it increases.

So answer of 39 is : $(P) -(3) ;(Q)-(4) ;(R)-(2) ;(S)-(1)$. Answer is $(D)$.

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