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QUESTION

# The mass solubility of CaF2 in water is 0.016 g/L in water. What is the molar concentration of fluoride ions in a saturated solution of CaF2? And what is the value of Ksp for CaF2?

Here's what I got.

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 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))##