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One man says that Calcium Fluoride solution can not exist, because it is only hydrated cations (N+) and anions (F-) surrounded by water molecules and that CaF2 can exist only in solid State, not in a solution. I think he is wrong because if the water contains Calcium and Fluoride ions dissolved, then it is a CaF2 solution the same way that if it contains Sodium and Chloride ions it is a saline solution. Who is right?

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    $\begingroup$ Welcome to ChemistrySE! Your question is quite unclear. What do you mean by "N+"? $\endgroup$ Commented Apr 23, 2018 at 1:29

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If you add any ionic compound into water, it will dissolve and dissociate into ions... up to a certain extent.

For $\ce{NaCl}$, pretty much all of it dissociates into $\ce{Na+}$ and $\ce{Cl-}$. We say that this compound is soluble.

However, only a little bit of $\ce{CaF2}$ will dissociate into ions. This number is so small that we say this compound is insoluble.

Still, a $\ce{CaF2}$ solution does exist. it is just in a really small concentration. So basically, both of you are wrong; a solution with this compound does exist, but it by no means acts like an $\ce{NaCl}$ solution.

If you want to know how much dissolves, google its $\ce{K_{sp}}$, or its solubility product. The $\ce{K_{sp}}$ of a compound $\ce{X_aY_b}$ is equal to $\ce{[X]^a[Y]^b}$. This way, you can calculate the amount of the compound dissolves and dissociates.

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I agree with the previous answer. Almost all group 1 metals are soluble in water and are considered spectator ions. Also compounds containing chloride, bromide, and iodide are almost always soluble. This explains why NaCl is soluble because Na is a group 1 metal and Cl is chloride. In the situation of CaF2, Calcium is a group 2 metal and is most of the time not soluble and fluoride is not part of the soluble halogen group. So, most likely CaF2 cannot exist in an aqueous solution because neither of the elements are soluble so CaF2 will always exist as a solid inside of an aqueous solution.

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