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

Calculation of `[HS^(-)]` in 0.1 `M H_(2)S` solution
Let the degree of dissociation of `H_(2)S =alpha`

`k_(a1) =[[H^(+)][HS^(-)]]/[[H_(2)S]] =[[alpha][alpha]]/[0.1 -alpha]] =[[alpha][alpha]]/(0.1)`
`9.1 xx10^(-8) =[[alpha][alpha]]/(0.1) " or [alpha]^(2) =9.1 xx 10^(-9) =91 xx 10^(-10)`
`alpha =(91 xx 10^(-10))^(1//2) =9.54 xx 10^(-5) " or " [HS^(-)] = 9.54 xx 10^(-5)`
(II). Calculation of `[HS^(-)]` in 0.1 M HCI solution
When `H_(2)S` solution is mixed with 0.1 M HCI solution, the dissociation may be represented as
`H_(2)S hArr H^(+) + HS^(-) , HCI to H^(+) +Cr^(-)`
`[H^(+)]` due to dissciation of HCI (strong acid) =0.1 M
Let `[H^(+)]` due to dissociation of `H_(2)S` (weak acid) =x M
Total concentration of `H^(+)` ions i.e., `[H^(+)] =0.1 +x ~~ 0.1 M`
(x being very small can be neglected)
`[HS^(-)]` in solution =x M
Concentration of `H_(2)S` left undissociated i.e., `[H_(2)S] = 0.1 - x ~~ 0.1 M`
`K_(a1) =[[H^(+)][HS^(-)]]/[[H_(2)S]] " or " 9.1 xx 10^(-8) = ((0.1) xx(X))/((0.1))`
(III). Calculation of `[S^(2-)]` in the obsence of HCI
`H_(2)S hArr H^(+) + HS^(-) , K_(a1) =9.1 xx 10^(-8)`
`HS^(-) hArr H^(+) +S^(2-) , K_(a2) =1.2 xx 10^(-13)`
In order to calculate `K_(a)` for the overall reaction add the two equations.
`k_(a) =K_(a1) xx K_(a2) =(9.1 xx 10^(-8)) xx (1.2 xx 10^(-13)) = 1.092 xx 10^(-20)`
Dissociation of `H_(2)S` may be represented as :
`H_(2)S hArr 2H^(+) +S^(2-)`
`0.1 -y ~~ 0.1 " "2y " "y`
`k_(a) =[[H^(+)][S^(2-)]]/[[H_(2)S]] " or " 1.092 xx 10^(-20) = (4y^(2) xx Y)/(0.1)`
`4y^(3) =1.092 xx 10^(-21) " or " y=((1.092 xx10^(-21))/(4))^(1/3) = (273 xx 10^(-24))^(1//3)`
`y= 6.49 xx 10^(-8) M~~ 6.5 xx 10^(-8) M`
(IV). Calculation of `[S^(2-)]` in the presence of 0.1 M HCI
Let `[S^(2-)]` due to dissociation of `H_(2)S = Z M`
Dissociation of `H_(2)S` may be represented as :
`H_(2)S hArr 2H^(+) + S^(2-)`
`0.1 -Z ~~ 0.1 " "2Z " "Z`
Total concentration of `H^(+)` ions i.e., `[H^(+)] =0.1 + 2Z ~~ 0.1 M`
`[S^(2-)]` in solution =Z
`K_(a) =[[H^(+)]^(2)[S^(2-)]]/[[H_(2)S]] " or " 1.092 xx 10^(-20) =((0.1)^(2) xx Z)/(0.1)`
`Z= (1.092 xx 10^(-20) xx 0.1)/((0.1)) = 1.092 xx 10^(-19) M`