InterviewSolution
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InterviewSolution
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The dissociation constants for acetic acid and HCN at `25^(@)C` are `1.5xx10^(-5)` and `4.5xx10^(-10)` , respectively. The equilibrium constant for the equilibirum `CN^(-) + CH_(3)COOHhArr HCN + CH_(3)COO^(-)` would beA. `3.0xx10^(-5)`B. `3.0xx10^(-4)`C. `3.0xx10^(4)`D. `3.0xx10^(5)` |
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Answer» Correct Answer - C We are given `CH_(3)COOH hArr CH_(3)COO^(-)+H^(+), K_(1)=1.5xx10^(-5)` `HCN hArr H^(+)+CN^(-), K_(2)= 4.5xx10^(-10)` If we reverse the second equilibrium expression and add it to the first one, we get `CN^(-)+CH_(3)COOH hArr CH_(3)COO^(-)+HCN` `:. K_(eq)=K_(1)xx(1)/(K_(2))=(1.5xx10^(-5))/(4.5xx10^(-10))` `=0.33xx10^(5)` `=3.3xx10^(4)` Note that equilibrium constant expressions are multiplied when we add the expressions. When w reverse the expression, the new equilibrium constant is always the reciprocal of the previous one. |
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