Question

A solution contains `0.1MH_(2)S` and `0.3MHCl` .Calculate the conc.of `S^(2-)` and `HS^(-)` ions is solutions.Given `K_(a_(1))` and `K_(a_(2))` are `10^(-7)`and `1.3xx10^(-13)` respectively.

Answer 1

`H_(2)ShArr H^(+)+HS^(-),K_(a_(1))=10^(-7)`
`HS^(-)hArrH^(+)+S^(2-),K_(a_(2))=1.3xx10^(-13)`
`HClrarr H^(+)+Cl^(-)`
Due to coomon ion effect exerted by `H^(+)`of `HCl` ,the dissocation of `H_(2)S` are supposed and the `[H^(+)]` in solution is mainly due to `HCl`.
`K_(a_(1))=([H^(+)][HS^(-)])/([H_(2)S])`
`10^(-7)=([0.3][HS^(-)])/([0.1])[:.[H^(+)]` from `HCl=0.3` would have dissociated negligibly]
`[HS^(-)]=(10^(-7)xx0.1)/(0.3)=3.3xx10^(-8)M`
Further `K_(a_(2))=([H^(+)][S^(2-)])/([HS^(-)])rArr 1.3xx10^(-13)([0.3][s^(2-)])/(3.3xx10^(-8))`
`[S^(2=)]=(1.3xx10^(-13)xx3.3xx10^(-8))/(0.03)=1.43xx10^(-19)M`