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 `1.0 xx 10^(-7)` and that of `S^(2-)` from `HS^-` ions is `1.2 xx 10^(-13)` then the concentration of `S^(2-)` ions in aqueous solution is :

Options

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

Answer

B

Explanation

` therefore Ka(iii) = ka(i) xx ka(ii)`
` = (1 xx 10^(-7) (1.2 xx 10^(-13))`
` = 1.2 xx 10^(-20)`
(iii) of ka (iii) = ([H^+][S^(2-)])/([H_2S])`
Where , `[H_2S] = 0.1 M, [H^+] = 0.2M = [HCl]`
` therefore 1.2 xx 10^(-20) = (0.2)^2 (S^(2-) ) // 0.1`
`therefore [S^(2-)] = (0.1 xx 1.2 xx 10^(-20))/((0.2)^2)`
` = 0.3 xx 10^(-20) = 3 xx 10^(-21)`