A saturated solution of H 2 S in water has concentration of approximately 0.10 M. What is the pH of this solution and equilibrium concentrations of H 2 S, HS - and S 2- ? Hydrogen sulphide is a diprotic acid and its dissociation constants are K a1 = 9.1 x 10 -8 , K a2 = 1.3 x 10 -13 mol L -1 respectively. H 2 S 2H+ + S2- 0.1-x 2x x k=4x^3/(0.1-x) k=ka1*ka2 solving for x we get [S2-]=6.6*10^-8 however the answer is 1.3 x 10 -13 M where am i going wrong

							The seconddissociationconstant is very small than the firstdissociationconstant and therefore, theconcentrationof H3O+is obtainedonlyfrom the firstdissociationconstant:
H2S + H2O < -------------> H3O++ HS-
Initialconcentration:
[H2S] = 0.10
[H3O+] = 0
[HS-] = 0
Let theamount ofH2S dissociated be x. Therefore,
Atequilibrium:
[H2S] = 0.10 – x
[H3O+] = x
[HS-] = x
Hence,
Ka1= [H3O+] [HS-] / [H2S]
= 9.1 x 10-8
x2/ (0.10 – x) = 9.1 x 10-8
assuming x << 0.10
x2/ 0.10 = 9.1 x 10-8
x2= 9.1 x 10-8x 0.1 = 9.1 x 10-9
x = 9.5 x 10-5
therefore,
[H3O+] = [HS-] = 9.5 x 10-5mol L-1
pH = – log [H3O+] = – log (9.5 x 10-5)
= – log 9.5 + 5 = -0.98 + 5 = 4.02
[H2S] = 0.10 – 9.5 x 10-5= 0.10 M
Tocalculatetheconcentrationof S2-ion, we have to consider the seconddissociation:
HS-+ H2O < -------> H3O++ S2-
Initialconcentration:
[HS-] = 9.5 x 10-5
[H3O+] = 9.5 x 10-5
[S2-] = 0
Let theamount ofH2S dissociated be x. Therefore,
Atequilibrium:
[HS-] = 9.5 x 10-5– x
[H3O+] = 9.5 x 10-5+ x
[S2-] = x
Hence,
Ka2= [H3O+] [S2-] / [HS-]
= (9.5 x 10-5+ x) (x) / (9.5 x 10-5– x)
Assuming x to be very small:
9.5 x 10-5– x ≈ 9.5 x 10-5and
9.5 x 10-5+ x = 9.5 x 10-5
1.3 x 10-13= 9.5 x 10-5x (x) / 9.5 x 10-5
X = 1.3 x 10-13
[S2-] = 1.3 x 10-13thus,
[H3O+] = 9.5 x 10-5M
[HS-] = 9.5 x 10-5M
[S2-] = 1.3 x 10-13M
pH = 4.02