Question #f53d2

Your strategy here is to convert the solubility of calcium fluoride, #"CaF"_2#, from grams per liter to moles per liter, then use an ICE table to determine the #K_(sp)# of the compound.

So, to convert calcium fluoride's solubility from grams per liter to moles per liter, you need to use the molar mass of the compound

#0.016color(red)(cancel(color(black)("g")))/"L" * "1 mole CaF"_2/(78.07color(red)(cancel(color(black)("g")))) = "0.000205 moles CaF"_2#

Now, the little calcium fluoride that dissolves in aqueous solution will produce calciumcations, #"Ca"^(2+)#, and fluoride anions, #"F"^(-)#, according to the following equilibrium reaction

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

Since every mole of calcium fluoride produces #color(red)(2)# moles of fluoride anions, you can say that the molarity of the fluoride anions in a saturated calcium fluoride solution will be

#["F"^(-)] = color(red)(2) xx ["CaF"_2]#

In this case, you will get

#"F"^(-) = 2 xx "0.000205 M" = color(green)("0.00041 M")#

To determine the value of the solubility product constant, #K_(sp)#, use an ICE table

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

#color(purple)("I")" " " " " " - " " " " " " " " " " "0 " " " " " " " " "0#
#color(purple)("C")" " " " " "- " " " " " " " " "(+s)" " " " " " " "(+color(red)(2)s)#
#color(purple)("E")" " " " " "- " " " " " " " " " "s" " " " " " " "color(red)(2)s#

By definition, #K_(sp)# is equal to

#K_(sp) = s * (color(red)(2)s)^color(red)(2) = s * 4s^2 = 4s^3#

Here #s# represents the molar solubility of calcium fluoride. This means that #K_(sp)# will be equal to

#K_(sp) = 4 * 0.000205^3 = color(green)(3.4 * 10^(-11))#