An aqueous solution contains0.10 M H_(2)S" is "1.0 xx 10^(-7)" and that of " S^(2-) "ion from " HS^(-) "ions is " 1.2 xx 10^(-3) ," then the concentration of " S^(2-) " ions in aqueous solution is "

An aqueous solution contains0.10 M H_(2)S" is "1.0 xx 10^(-7)" and tha

1.	An aqueous solution contains0.10 M H_(2)S" is "1.0 xx 10^(-7)" and that of " S^(2-) "ion from " HS^(-) "ions is " 1.2 xx 10^(-3) ," then the concentration of " S^(2-) " ions in aqueous solution is "
Answer» ` 5 xx 10^(-8)` `3 xx 10^(-20)` `6xx 10^(-21)` `5xx 10^(-19)` Solution :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)`