K1 and K2 for dissociation of H2S are 4×10−3 and 1×10−5. Calculate sulphide ion concentration in 0.1 MH2S solution.
10−5
H2S⇌H⊕+HS⊖
K1=[H⊕][HS⊕][H2S]=4×10−3
[H⊕]=Cα,[HS⊖]=Cα,[H2S]=C(1−α)
or 4×10−3=Cα⋅CαC(1−α)=Cα2(1−α)
4×10−3=0.1×α2(1−α) (1−α 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(1−α)=0.1(1−0.18)=0.082M
Now, HS⊖ further dissociate to H⊖ and S2−,
C1[HS⊖]=0.018M
HS⊖ H⊕+S2− ⇌
1 0 0
(1α1)α1α1
∴K2=1×10−5=[H⊕][S2−]HS⊖
because [H⊕] already in solution =0.018 and thus, dissociation of [HS⊖] further suppresses due to common ion effect and 1−α≈1.
∴1×10−5=0.018×C1α1C1(1−α1)=0.018×α1
∴α1=1×10−50.018=5.55×10−4
∴[S2−]=C1α1=0.018×5.55×10−4
=10−5