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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
A saturated solution of H2S in water has concentration of approximately 0.10 M. What is the pH of this solution and equilibrium concentrations of H2S, HS-and S2-? Hydrogen sulphide is a diprotic acid and its dissociation constants are Ka1 = 9.1 x 10-8, Ka2 = 1.3 x 10-13 mol L-1 respectively.H2S 2H+ + S2-0.1-x                         2x       xk=4x^3/(0.1-x)k=ka1*ka2solving for x we get[S2-]=6.6*10^-8however the answer is 1.3 x 10-13 Mwhere am i going wrong

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5 years ago

```							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-] = 0Let theamount ofH2S dissociated be x. Therefore,Atequilibrium:[H2S] = 0.10 – x[H3O+] = x[HS-] = xHence,Ka1= [H3O+] [HS-] / [H2S]= 9.1 x 10-8x2/ (0.10 – x) = 9.1 x 10-8assuming x << 0.10x2/ 0.10 = 9.1 x 10-8x2= 9.1 x 10-8x 0.1 = 9.1 x 10-9x = 9.5 x 10-5therefore,[H3O+] = [HS-] = 9.5 x 10-5mol L-1pH = – 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 MTocalculatetheconcentrationof 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-] = 0Let theamount ofH2S dissociated be x. Therefore,Atequilibrium:[HS-] = 9.5 x 10-5– x[H3O+] = 9.5 x 10-5+ x[S2-] = xHence,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 10-5and9.5 x 10-5+ x = 9.5 x 10-51.3 x 10-13= 9.5 x 10-5x (x) / 9.5 x 10-5X = 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 10-13MpH = 4.02
```
5 years ago
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