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A certain weak acid has a dissociation constant `1.0xx10^(-4)`. The equilibrium constant for its reaction with a strong base is :A. `1.0xx10^(-4)`B. `1.0xx10^(-10)`C. `1xx10^(-10)`D. `1.0xx10^(-14)` |
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Answer» Correct Answer - C `HAhArrH^(+)+A^(-)` `K_(a)=([H^(+)][A^(-)])/([HA])` (`i`) Also `HA+B^(+)+OH^(-) rarr B^(+)+A^(-)+H_(2)O` `K_(eq.)=([H_(2)O][A^(-)])/([HA][OH])` (`ii`) By (`i`) and (`ii`), `=(K_(eq))/(K_(a))=(1)/([H^(+)][OH^(-)])=(1)/(K_(w))` `:. K_(eq)=(K_(a))/(K_(w))=(10^(-4))/(10^(-14))=10^(-10)` |
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