An aqueous solution contains 0.10 (M) `H_2S` and 0.20 (M) HCl. If the equilibrium constants for the formation of `HS^-` from `H_2S` 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. `5xx10^(-8)`
B. `3xx10^(-20)`
C. `6xx10^(-21)`
D. `5xx10^(-19)`

Explanation

`H_2S `HS^(-1) `K_1xxK_2=1xx10^(-7)xx1.2xx10^(-13)`
`=([H^+][HS^-])/([H_2S])xx([H^+][S^(2-)])/([HS^-])`
or `([H^+][S^-])/([H_2S])=1.2xx10^(-20)`
Due to the common ion effect the degree of dissociation of a weak electrolyte such as `H_2S` weak electrolyte such as `H_2S` become very small and hence the equilibrium concentration of `H_2S` is taken same as that of initial concentration i.e, `[H_2S]=0.10(M)`. The `H^+` ion concentration is hence same as that of the HCl concentration i. `[H^+]=0.20(M)[H^+]=0.20(M)`
`:.((0.2)^2[S^(2-)])/(0.1)=1.2xx10^(-20)`
or `[S^(2-)]=3xx10^(-20)(M)=1.1xx10^(-9)(M)`