A reaction ${\mathrm{CaF}}_2 \rightleftharpoons {\mathrm{Ca}}^{+2} + 2{\mathrm{F}}^{-}$  is at equilibrium. If the concentration of ${\mathrm{Ca}}^{2+}$ is increased four times, what will be the change in ${\mathrm{F}}^{-}$ concentration as compared to the initial concentration of ${\mathrm{F}}^{-}$ ?

The dissociation constant K of the reaction $= \frac{\left[{\mathrm{Ca}}^{2+}\right] {\left[{\mathrm{F}}^{-}\right]}^2 }{\left[{\mathrm{CaF}}_2\right] }$

When the concentration of ${\mathrm{Ca}}^{2+}$ increases and becomes four times, the concentration of ${\mathrm{F}}^{-}$ must decrease by the same value. This is necessary in order to maintain K as constant. Since the ${\mathrm{F}}^{-}$ concentration is raised to the power 2 in the expression, the concentration of fluoride ion must decrease by 1/2.