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Question

Two weak bases AOH and BOH having equal concentrations of 1 M are present in solution at 25oC having dissociation constants as 1×1010 and 3×1010 respectively. Calculate the ratio α1α2

Here α1 is the degree of dissociation of AOH
α2 is degree of dissociation of BOH

A
1.5
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B
1
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C
0.5
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D
0.33
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Solution

The correct option is D 0.33
C1=C2=1 MKb1=1×1010 MKb2=3×1010 M
For two bases :

Weak Bases WB1+WB2 :
Mixture of two weak bases BOH and AOH:

For AOH :

AOH(aq)OH(aq) + A+(aq)

at t=0 C1 0 0

at t=teq C1C1α1 (C1α1+C2α2) C1α1

For BOH :

BOH(aq)OH(aq) + B+(aq)

at t=0 C2 0 0

at t=teq C2C2α2 (C1α1+C2α2) C2α2


Dissociation constant for AOH Kb1 :
Kb1=(C1α1+C2α2)(C1α1)C1(1α1)
Dissociation constant for BOH Kb2 :
Kb2=(C1α1+C2α2)(C2α2)C2(1α2)

C1Kb1+C2Kb2 = (C1α1+C2α2)2
[OH] = C1α1+C2α2 = C1Kb1+C2Kb2
[OH] = C1Kb1+C2Kb2
[OH]=1(1×1010+3×1010)
[OH]=2×105 M
Since , both are weak acids, so α1 and α2 are very small in comparison to unity for weak monoprotic acids. So, 1α11 and 1α21.

So:

Kb1×C1=(C1α1+C2α2)(C1α1)
Kb2×C2=(C1α1+C2α2)(C2α2)
Dividing both the above equations, we get:
Kb1Kb2=α1α2

α1α2=1×10103×1010α1α2=0.33


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