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

The solubilità molare di insolubile ionic compound tells you how many talpe of said compound you can dissolve in un litro d'acqua.

Insolubile composti ionici do not dissociate completely in aqueous solution, which implies that an equilibrium is established between the non disciolto solid and the sciolto ioni.

The costante di solubilità del prodotto, #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+)#e anioni di fluoro, #"F"^(-)#, in solution

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

Notare che ogni talpa of calcium fluoride that dissolves produces #1# Talpa of calcium cations and #color(red)(2)# talpe di anioni di fluoro.

Se prendi #s# essere la concentrazione of calcium fluoride that dissocia, i.e. its solubilità molare, you can use an ICE table per trovare il valore di #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)#

Questo sarà equivalente a

#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)# sarà

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