The first ionization constant of `H_(2)S` is `9.1xx10^(-8)`. Calculate the concentration of `HS^(Θ)` ion in its `0.1 M` solution. How will this concentration be affected if the solution is `0.1 M` in `HCl` also? If the second dissociation constant if `H_(2)S` is `1.2xx10^(-13)`, calculate the concentration of `S^(2-)` under both conditions.

Explanation

`underset(C(1-alpha))(H_(2)S)hArr underset(Calpha)(H^(+))+underset(C alpha)(HS^(-))`
I. `K_(a_(1)) = 9.1xx10^(-8) = ([H^(+)][HS^(-)])/([H_(2)S])`
`("Given: [H_(2)S]= 0.1=C)`
`9.1xx10^(-8)= (C^(2)alpha^(2))/(C(1- alpha))= Calpha^(2) , (.: 1-alpha ~~1)`
`9.1xx10^(-8)=0.1xxalpha^(2)`
`:. alpha= 9.54xx10^(-4)`
`:. [HS^(-)]= Calpha= 0.1xx9.54xx10^(-4)`
`=9.54xx10^(-5)M`
II. If `[H^(+)]= 0.1`, then
`K_(a_(1))=([H^(+)][HS^(-)])/([H_(2)S])`
`9.1xx10^(-8)= (0.1xxCalpha)/(C )=0.1xxalpha`
`:. alpha = 9.1xx10^(-7)`
`:. [HS^(-)]=Calpha = 0.1xx9.1xx10^(-7)`
`= 9.1xx10^(-8)M`
`[S^(2-)]` in Case:
`underset((9.5xx10^(-5)-x))(HS^(-))hArr underset((9.54xx10^(-5)+x))(H^(+))+underset(x)(S^(2-))`
`K_(a_(2))=([H^(+)][S^(2-)])/([HS^(-)])`
`1.2xx10^(-13)=((9.5xx10^(-5)+x).x)/((9.5xx10^(-5)-x))`
or `x= K_(a_(2))`
(x is very small due to common ion effect)
or `[S^(2-)]= 1.2xx10^(-13)M`
`[S^(2-)]` in II Case:
`underset(C(1-alpha))(H_(2)S)hArr underset(0.1)(2H^(+))+underset(Calpha)(S^(2-))`
`K_(a_(1))xxK_(a_(2))=([H^(+)]^(2)[S^(2-)])/([H_(2)S])=((0.1)^(2)xxCalpha)/(C(1-alpha))`
`:. alpha=(9.1xx10^(-8)xx1.2xx10^(-13))/(10^(-2))`
`= 10.92xx10^(-19)`
`:. [S^(2-)]=Calpha=0.1xx10.92xx10^(-19)`
`= 1.092xx10^(-19)= 1.092xx10^(-10)M`