Question

What is the molar solubility of `"CaF"_2` in water in terms of its `K_(sp)`?

Answer 1

`s = root(3)(K_(sp)/4)`

Explanation:

The molar solubility of an insoluble ionic compound tells you how many moles of said compound you can dissolve in one liter of water.

Insoluble ionic compounds do not dissociate completely in aqueous solution, which implies that an equilibrium is established between the undissolved solid and the dissolved ions.

The solubility product constant, `K_(sp)`, essentially tells you how far to the left this equilibrium lies.

In your case, calcium fluoride, `"CaF"_2`, is considered insoluble in aqueous solution. The small amounts of calcium fluoride that do dissociate will produce calcium cations, `"Ca"^(2+)`, and fluoride anions, `"F"^(-)`, in solution

`"CaF"_ (color(red)(2)(s)) rightleftharpoons "Ca"_ ((aq))^(2+) + color(red)(2)"F"_ ((aq))^(-)`

Notice that every mole of calcium fluoride that dissolves produces `1` mole of calcium cations and `color(red)(2)` moles of fluoride anions.

If you take `s` to be the concentration of calcium fluoride that dissociates, i.e. its molar solubility, you can use an ICE table to find the value of `s`

`"CaF"_ (color(red)(2)(s)) " "rightleftharpoons" " "Ca"_ ((aq))^(2+) " "+" " color(red)(2)"F"_ ((aq))^(-)`

`color(purple)("I")color(white)(aaaacolor(black)(-)aaaaaaaaaaaacolor(black)(0)aaaaaaaaaacolor(black)(0)`
`color(purple)("C")color(white)(aaaacolor(black)(-)aaaaaaaaaacolor(black)((+s))aaaaaacolor(black)((+color(red)(2)s))`
`color(purple)("E")color(white)(aaaacolor(black)(-)aaaaaaaaaaaacolor(black)(s)aaaaaaaaacolor(black)(color(red)(2)s)`

By definition, the solubility product constant will be

`K_(sp) = ["Ca"^(2+)] * ["F"^(-)]^color(red)(2)`

This will be equivalent to

`K_(sp) = s * (color(red)(2)s)^color(red)(2)`

`K_(sp) = 4s^3`

Therefore, the molar solubility of calcium fluoride in terms of its `K_(sp)` will be

`color(green)(|bar(ul(color(white)(a/a)color(black)(s = root(3)(K_(sp)/4))color(white)(a/a)|)))`