Question

(a)When 100 ml of 0.1M NaCN solution is titrated with 0.1 M HCl solution the variation of pH of solution with volume of HCl added will be :

(b)Variation of degree of dissociation `alpha` with concentration for a weak electrolyte at a particular temperature is best represented by :

( c)0.1 M acetic acid solution is titrated against 0.1 M NaOH solution. The difference in pH between 1/4 and 3/4 stages of neutralization of acid will be 2 log 3.
A. T,F,T
B. F,F,F
C. T,T,T
D. F,T,F

Answer 1

Correct Answer - C
(a)Initially pH will decrease fast, then slowly due to buffer formation and then will decrease fast as buffer action diminishes.
(b) For a weak electrolyte
`K_a=(Calpha^2)/((1-alpha))` when `alphaltlt 1` then `alpha=sqrt(k_a/C)`
as C increases `implies` `alpha` decreases
as C is tending to zero `implies` `alpha` will be unity
( c) At `1//4^(th)` neutralisation
`CH_3COOH+NaOHtoCH_3COONa+H_2O`
`(0.1xx3/4) " " (0.1xx1/4)`
`pH=pK_a+"log" ([CH_3COO^(-)])/([CH_3COOH])=pK_a+ "log" (1/3)`
At `3//4^(th)` neutralisation
`pH=pK_a+"log" 3`
so difference in pH=`Delta(pH)="log"3-"log"1/3=2"log"3`