An aqueous solution contains 0.10 M `H_2S` and 0.20 M HCl. If the equilibrium constants for the formation of HS from HS is `1.0 xx 10^(-7)` and that of `S^(2-)?` from `HS^(-)` ions is `1.2 xx 10^(-7)` then the concentration of `S^(2-)` ions in aqueous solution is

Options

A. ` 5 xx 10^(-8)`
B. `3 xx 10^(-20)`
C. `6xx 10^(-21)`
D. `5xx 10^(-19)`

Answer

B

Explanation

Given : (i) ` H_(2) S hArr HS^(-) + H^(+),`
`K_(a_(1)) = 1.0 xx 10^(-7)`
(ii) `HS^(-) hArr S^(2-) + H^(+), K_(a_(2))= 1.2 xx 10^(-13)`
For the reaction , ` H_(2)S hArr 2 H^(+) + S^(2-)`
(obtained by adding (i) and (ii) )
`K_(a) = K_(a_(1)) xx K_(a_(2))`
` :. (1.0 xx 10^(-7)) (1.2 xx 10^(-13))= ([H^(+)]^(2) [S^(2-)])/([H_(2)S]) `
In presence of HCl, `[H^(+)]` are obtained mainly from HCl. Hence ,
` [H^(+)] = 0.20 M, [H_(2) S]= 0.1 M `
`:. 1.2 xx 10^(-20) = ((0.2 )^(2)[S^(2-)])/0.1`
or `[S^(2-)] = 3 xx 10^(20)`