Ionic EquilibriumHard
Question
Which is/are correct statements :
(a) When 100 ml of 0.1 M NaCN solution is titrated with 0.1 M HCl solution the variation of pH of solution with volume of HCl added will be (as shown in figure):
(b) Variation of degree of dissociation a with concentration for a weak electrolyte at a particular temperature is best represented by (as shown in figure) :
(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) When 100 ml of 0.1 M NaCN solution is titrated with 0.1 M HCl solution the variation of pH of solution with volume of HCl added will be (as shown in figure):
(b) Variation of degree of dissociation a with concentration for a weak electrolyte at a particular temperature is best represented by (as shown in figure) :
(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.
Options
A.a & c
B.b & c
C.a, b & c
D.b only
Solution
(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
Ka =
when α << 1 then α =
as C increases ⇒ αdecreases
as C is tending to zero ⇒ α will be unity
(c) At 1/4th neutralisation
CH3COOH + NaOH → CH3COONa + H2O
pH = pKa + log
= pKa + log 
At 3/4th neutralisation
pH = pKa + log 3
so difference in pH = ᐃ(pH) = log 3 - log
= 2 log 3
(b) For a weak electrolyte
Ka =
when α << 1 then α =
as C increases ⇒ αdecreases
as C is tending to zero ⇒ α will be unity
(c) At 1/4th neutralisation
CH3COOH + NaOH → CH3COONa + H2O
pH = pKa + log
= pKa + log At 3/4th neutralisation
pH = pKa + log 3
so difference in pH = ᐃ(pH) = log 3 - log
= 2 log 3Create a free account to view solution
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