Question

Question #f53d2

Answer 1

Here's what I got.

Explanation:

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))`