wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

K1 and K2 for dissociation of H2S are 4×103 and 1×105. Calculate sulphide ion concentration in 0.1 MH2S solution.


A

104

No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

108

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

105

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D

106

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C

105


H2SH+HS
K1=[H][HS][H2S]=4×103
[H]=Cα,[HS]=Cα,[H2S]=C(1α)
or 4×103=CαCαC(1α)=Cα2(1α)
4×103=0.1×α2(1α) (1α should not be neglected)
or α=0.18,
[H]=Cα=0.1×0.18=0.018M
[HS]=Cα=0.1×0.18=0.018M
[H2S]=C(1α)=0.1(10.18)=0.082M
Now, HS further dissociate to H and S2,
C1[HS]=0.018M
HS H+S2
1 0 0
(1α1)α1α1
K2=1×105=[H][S2]HS
because [H] already in solution =0.018 and thus, dissociation of [HS] further suppresses due to common ion effect and 1α1.
1×105=0.018×C1α1C1(1α1)=0.018×α1
α1=1×1050.018=5.55×104
[S2]=C1α1=0.018×5.55×104
=105


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Le Chateliers Principle
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon