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A solution contains `0.1M H_(2)S` and `0.3M HCI`. Calculate the conc.of `S^(2-)` and `HS^(-)` ions in solution. Given `K_(a_(1))` and `K_(a_(2))` for `H_(2)S` are `10^(-7)` and `1.3xx10^(-7)` respectively. |
Answer» `H_(2)S hArr H^(+)+HS^(-)` , `(K_(a_(1))=10^(-7))` `HS^(-) hArr H^(+) + S^(2-)` , `(K_(a_(2))=1.3xx10^(-13))` `HCI rarr H^(+)+CI^(-)` Due to common ion effect, the dissociation of `H_(2)S` is suppressed and the `[H^(+)]` in solution is due to HCI. `:. K_(a_(1))=([H^(+)][HS^(-)])/([H_(2)S])` `10^(-7)=([0.3][HS^(-)])/([0.1]) [.: [H^(+)]` form `HCI = 0.3]` `:. [HS^(-)]= (10^(-7)xx0.1)/(0.3)=3.3xx10^(-8)M` Further `K_(a_(1))=([H^(+)][S^(2-)])/([HS^(-)])` and `K_(a_(1))= ([H^(+)][HS^(-))]/([H_(2)S])` `:. K_(a_(1))xxK_(a_(2))= ([H^(+)]^(2)[S^(2-)])/([0.1])` `:. [S^(2-)]= (1.3xx10^(-20)xx0.1)/(0.09)` `=1.44xx10^(-20)M` |
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