A solution contains `0.1M H_(2)S` and `0.3M HCI`. Calculate the conc.of `S^(2-)` and `HS^(-)` ions in solution. Given `K_(a_(1))` and `K_(a_(2))` for `H_(2)S` are `10^(-7)` and `1.3xx10^(-7)` respectively.

Explanation

`H_(2)S hArr H^(+)+HS^(-)` , `(K_(a_(1))=10^(-7))`
`HS^(-) hArr H^(+) + S^(2-)` , `(K_(a_(2))=1.3xx10^(-13))`
`HCI rarr H^(+)+CI^(-)`
Due to common ion effect, the dissociation of `H_(2)S` is suppressed and the `[H^(+)]` in solution is due to HCI.
`:. K_(a_(1))=([H^(+)][HS^(-)])/([H_(2)S])`
`10^(-7)=([0.3][HS^(-)])/([0.1]) [.: [H^(+)]` form `HCI = 0.3]`
`:. [HS^(-)]= (10^(-7)xx0.1)/(0.3)=3.3xx10^(-8)M`
Further `K_(a_(1))=([H^(+)][S^(2-)])/([HS^(-)])` and `K_(a_(1))= ([H^(+)][HS^(-))]/([H_(2)S])`
`:. K_(a_(1))xxK_(a_(2))= ([H^(+)]^(2)[S^(2-)])/([0.1])`
`:. [S^(2-)]= (1.3xx10^(-20)xx0.1)/(0.09)`
`=1.44xx10^(-20)M`