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I need to prepare several solutions with different concentrations of $\ce{F-}$, in order to carry out some investigations on fluoridated water. Would this be achievable by dissolving calcium fluoride, $\ce{CaF2}$, in water? Wikipedia writes that $\ce{CaF2}$ is insoluble in water, so I am not sure.

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Calcium fluoride has a $K_\mathrm{sp}=3.9\times 10^{-11}$. That means a saturated solution of calcium fluoride will have a fluoride concentration of $4.28\times 10^{-4}\mathrm{\ M}$, which, assuming the density of this dilute solution is approximately that of water, is $\mathrm{8.13\ mg/L}$ fluoride.

Fluoridated water has fluoride concentration ranging from $\mathrm{0.5\ mg/L}$ to $\mathrm{1.5\ mg/L}$, so your saturated calcium fluoride solution is right in that range. If you need to test lower concentrations, dilute your solution. If you need to test higher concentrations, you are out of luck. You cannot make more concentrated fluoride solutions with calcium fluoride.

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