InterviewSolution
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InterviewSolution
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A reaction `CaF_(2)hArrCa^(2+)+2F^(-)` is at equilibrium. If the concentration of `Ca^(2+)` is increased four times, what will be the change in `F^(-)` concentration as compared to the initial concentration of `F^(-)`?A. `(1)/(4)` timesB. `(1)/(2)` timesC. `4` timesD. `2` times |
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Answer» Correct Answer - B The dissociation constant `K` of the reaction ` =([ca^(2+)][F^(-)]^(2))/([CaF_(2)])` When the concentration of `Ca^(2+)` increases and becomes four times, the conentration of `F^(-)` must decrease by the same value. This is necessary in order to maintain `K` as constant. Since the `F^(-)` concentration is raised to the power `2` in the expression, the concentration of fluoride ion must decrease by `1//2`. |
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