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^(-)`?

Options

A. `1//4` of the initial value
B. `1//2` of the initial value
C. `2` times of the initial value
D. none of these

Explanation

`K_(c)=[Ca^(2+)][F^(-)]^(2)`, If `[Ca^(2+)]=4xx[Ca^(2+)]`,
To have `K_(c)` constant `[F^(-)]` should be `([F^(-)])/(2)`.