A solution contains 0.1 M `H_2S and 0.3 M HCl.` Calculate the conc. of `S^(2-) and HS^(-)` ions in solution. Given `K_(a_(1)) and K_(a_(2))" for "H_2S" are "10^(-7) and 1.3 xx 10^(-13)" respectively"`.

Explanation

`H_(2)S Leftrightarrow H^(+)+HS^(-)" "K_(a_(1))=10^(-7)`
`HS^(-) Leftrightarrow H^(+)+S^(2-) K_(a_(2))=1.3 xx 10^(-13)`
Due to common ion effect the dissociation of `H_2S` is suppressed and the `[H^+]` in solution is due to HCl
`therefore K_(a_(1))=([H^(+)][HS^(-1)])/([H_(2)S])`
`10^(-7)= ([0.3][HS^(-1)])/([0.1]) [? [H^(+)]" from HCl "=0.3]`
`therefore [HS^(-1)]=(10^(-7) xx 0.1)/(0.3)`
`=3.3 xx 106^(-8)M`
Further `K_(a_(2))=([H^(+)][S^(2-)])/([HS^(-)]) and K_(a_(1))=([H^(+)][HS^(-1)])/([H_(2)S]`
`therefore K_(a_(1)) xx K_(a_(2))=([H^(+)][S^(2-)])/([H_(2)S])`
`10^(-7) xx 1.3 xx 10^(-13)=([0.3]^(3)[S^(2-)])/([0.1])`
`therefore [S^(2-)]=(1.3 xx 10^(-20)xx 0.1)/(0.09)`
`=1.44 xx 10^(-20)M`