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^(-13)` respectively.

Options

A. `1.44 xx 10^(-19), 3.3 xx 10^(-7)`
B. `1.44 xx 10^(-20), 3.3 xx 10^(-8)`
C. `1.44 xx 10^(-22), 3.3 xx 10^(-3)`
D. `1.44 xx 10^(-18), 3.3 xx 10^(-5)`

Answer

B

Explanation

`H_(2)S hArr H^(+) + HS^(-), k_(a_(1)) = 10^(-7)`
`HS^(-) hArr H^(+) + S^(2-), K_(a_(2)) = 1.3 xx 10^(-13)`
`HCl rarr H^(+) + Cl^(-)`
due to common ion effect the dissociation of `H_(2)S` is suppressed and the `[H^(+)]` in solution is due to
`HCl, k_(a_(1)) = ([H^(+)][HS^(-)])/([H_(2)S])`
`10^(-7) = ([0.3][HS^(-)])/([0.1]) [:'[H^(+)] `from `HCl = 0.3`]
`[HS^(-)] = (10^(-7) xx 0.1)/(0.3) = 3.3 xx 10^(-8) M`
`k_(a_(2)) = ([H^(+)][S^(-2)])/([HS^(-)])`
`1.3 xx 10^(-13) = ([0.3][S^(-2)])/(3.3 xx 10^(-8))`
further `[S^(-2)] = 1.3 xx 10^(-13) xx 3.3 xx 10^(-8)//0.03 = 1.43 xx 10^(-20) M`