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. | (C2H5)3N + CH3COOH X Y | 1. | Conductivity decreases and then increases |
Q. | KI (0.1M) + AgNO3(0.01M) X Y | 2. | Conductivity decreases and then does not change much |
R. | CH3COOH + KOH X Y | 3. | Conductivity increases and then does not change much |
S. | NaOH + HI X Y | 4. | Conductivity does not change much and then increases |
Codes
P Q R S
Text Solution
Verified by ExpertsA
(P) 
+ 
⎯→ CH 3 COO – (aq) + (C 2 H 5 ) 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) K I (0.1 M) + AgNO 3 (0.01 M) ⎯→ Ag I ↓ (ppt) + KNO 3 (aq).
As the only reaction taking place is precipitation of Ag I and in place of Ag + , K + is coming in the solution, conductivity remain nearly constant and then increases.
(R) CH 3 COOH + KOH ⎯→ 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 + H I ⎯→ Na I (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) – ; (Q) – ; (R) – ; (S) – . Answer is .
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