What are the concentration of `H_(3) O^(+), HS^(-) ` and `s^(2-)` is 0.01M `H_(2) S ` solution, given
`H_(2) S + H_(2) hArr H_(3) O^(+) + HS^(-) , K_(a_(1)) = 1.1 xx 10^(-7)`
`HS^(-) + H_(2) O hArr H_(3) O^(+) + S^(-) , K_(a_(2)) = 1.0 xx 10^(- 34 )`

Explanation

By ostwald dilution law, `[H_(3) O^(+)]` and `[HS^(-)]` due to first step ionization are `[H_(3) O^(+)]= [HS^(-)] = sqrt( Ka_(1) C )`
`= sqrt(1.1 xx 10^(-7) xx 0.1) = 1 xx 10^(-4) M `
For second step ionization `HS^(-) + H_(2) O rarr H_(3) O^(+) + S^(2-)`
Since `K_(a_(2)) lt lt lt K_(a)` hence ionization is very-very small and `[HS^(-)]` remains constant
`K_(a_(2)) = ([ H_(3) O^(+)][S^(-2)])/( [HS^(-)])`
`:. 1. 0 xx 10^(-14) = ( 1.0 xx 10^(-14) [ S^(2-)])/(1.0 xx 10^(-4))`
`:. [S^(2-) ] = 1.0 xx 10^(-14) M `