wiz-icon
MyQuestionIcon
MyQuestionIcon
18
You visited us 18 times! Enjoying our articles? Unlock Full Access!
Question

Two weak bases AOH and BOH having equal concentrations of 0.5 M are present in a solution. Calculate the ratio, [A+][B+] at equilibrium.
Dissociation constants for AOH and BOH are 1.1×108 and 1.7×109 respectively.

A
3.53
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
0.66
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
6.47
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
9.45
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 6.47

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)
Since α1 and α2 are very small in comparison to unity for weak bases.So 1α11 and 1α21.
Kb1=(C1α1+C2α2)(C1α1)C1(1α1)

Kb1C1=(C1α1+C2α2)(C1α1)
Similarly,
Kb2C2=(C1α1+C2α2)(C2α2)
Now,
Kb1C1Kb2C2=(C1α1+C2α2)(C1α1)(C1α1+C2α2)(C2α2)Kb1C1Kb2C2=(C1α1)(C2α2)=[A+][B+][A+][B+]=Kb1C1Kb2C2=1.1×108×0.51.7×109×0.5[A+][B+]=6.47

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Degree of Dissociation
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon