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The first and second dissociation constant of an acid `H_(2)A` are `1.0xx10^(-5)` and `5.0xx10^(-10)` repectively. The overall dissociation constant of the acid will beA. `5.0xx10^(-5)`B. `5.0xx10^(15)`C. `5.0xx10^(-15)`D. `0.2xx10^(5)` |
Answer» Correct Answer - C `H_(2)AhArrHA^(-)+H^(+)` `K_(1) = ([HA^(-)][H^(+)])/([H_(2)A])` ……(1) `HA^(-)hArr H^(+)+A^(2-)` `K_(2) = ([H^(+)][A^(2-)])/([HA^(-)])` ……(2) For the reaction, `H_(2)A hArr 2H^(+)+A^(2-)` `K = ([HA^(-)][H^(+)])/([H_(2)A])xx([H^(+)][A^(2-)])/([HA^(-)])` `= ([H^(+)]^(2-)[A^(2-)])/([H_(2)A])= K_(1)xxK_(2)` `1xx10^(-5)xx5xx10^(-10)` `=5xx10^(-15)` |
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