What is the molar solubility of #"CaF"_2# in water in terms of its #K_(sp)#?
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,
In your case, calcium fluoride,
#"CaF"_ (color(red)(2)(s)) rightleftharpoons "Ca"_ ((aq))^(2+) + color(red)(2)"F"_ ((aq))^(-)#
Notice that every mole of calcium fluoride that dissolves produces
If you take
#"CaF"_ (color(red)(2)(s)) " "rightleftharpoons" " "Ca"_ ((aq))^(2+) " "+" " color(red)(2)"F"_ ((aq))^(-)#
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
#color(green)(|bar(ul(color(white)(a/a)color(black)(s = root(3)(K_(sp)/4))color(white)(a/a)|)))#