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A saturated solution of H_2S in water has concetration of approximately 0.10 M. What is the pH of this soluton and equilibrium concentrations of H_2S,HS^(-) andS^(2-)? Hydrogen sulphide is a diprotic acid and its dissociation constants are K_(a_1)=9.1xx10^(-8) ,K_(a_2)=1.3xx10^(-3) "mol L"^(-1) respectively. |
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Answer» `{:(,H_2S+H_2O hArr, H_3O^(+)+, HS^-),("Initial conc.", 0.10,0,0),("At equilibrium" , 0.10-x, x ,x):}` (x is amount of `H_2S` dissociated ) `K_(a_1)=([H_3O^+][HS^-])/([H_2S])=9.1xx10^(-8)` or `(x xx x)/(0.10-x)=9.1xx10^(-8)` Solving `x=9.5x10^(-5)` `therefore [H_3O^+]=[HS^-]=9.5xx10^(-5) "mol L"^(-1)` `pH=-log [H_3O^+]=-log (9.5xx10^(-5))` `=-log 9.5 +5 =-0.98+5=4.02` `[H_2S]=0.10-9.5xx10^(-5)`=0.10 M To calculate the concentration of `S^(2-)` ion , we are to consider the second dissociation : `{:(,HS^(-)+H_2O hArr, H_3O^(+)+,S^(2-)),("Initial conc.", 9.5xx10^(-5), 9.5xx10^(-5),0),("At equi.",9.5xx10^(-5)-x, 9.5xx10^(-5)+x ,x):}` `K_(a_2)=([H_3O^+][S^(2-)])/([HS^-])` `=((9.5xx10^(-5)+x)xx x)/(9.5xx10^(-5)-x)` Assuming x to be very very small : `9.5xx10^(-5)-x APPROX 9.5 xx10^(-5)` and `9.5xx10^(-5)+x` =`9.5xx10^(-5)` Solving `x=1.3xx10^(-13)` `therefore [S^(2-)]=1.3xx10^(-13)` Thus, `[H_3O^+]=9.5xx10^(-5)` M , `[HS^-]=9.5xx10^(-5)` M, `[S^(2-)]=1.3xx10^(-13)M , [H_2S]`=0.01 M , pH=4.02 |
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