An aqueous solution contains `0.10 M H_(2)S` and 0.20 M HCl. If the equilibrium constants for the formation of `HS^(–)` from `H_(2)S` is `1.0xx10^(–7)` and that of `S^(2-)` from `HS^(–)` ions is `1.2xx10^(–13)` then the concentration of `S^(2-)` ions in aqueous solution is

Options

A. `6xx10^(-21)`
B. `5xx10^(-19)`
C. `5xx10^(-8)`
D. `3xx10^(-20)`

Answer

D

Explanation

`HCl to H^(+) + CL^(-)`
0.2M 0.2 M
`H_(2)S to H^(+) + HS^(-) K_(1)=10^(-7)`
`HS^(-) to H^(+) + S^(2-) K_(2)=1.2xx10^(-13)`
`H_(2)S to 2H^(+) + S^(2-) K=K_(1).K_(2)=1.2xx10^(-20)`
`K+([H^(+)]^(2)[HS^(-)])/([H_(2)S]) [H^(+)]=0.2M, [H_(2)S=0.1M`
`1.2xx10^(-20)=((0.2)^(2)[S^(2-)]=3xx10^(-20)M`