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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))