A solution contains ` 0.10 M H_2S and 0.3 HCl` Calculate the concentration of `S^(2-) and HS^(-)` ions in the solution . For `H_2S , K_(a_1) = 1 xx 10^(-7) and K_(a_2) = 1.3 xx 10^(-13)`

Explanation

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