# Question f53d2

Nov 22, 2015

Here's what I got.

#### Explanation:

Your strategy here is to convert the solubility of calcium fluoride, ${\text{CaF}}_{2}$, from grams per liter to moles per liter, then use an ICE table to determine the ${K}_{s p}$ 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, ${\text{Ca}}^{2 +}$, and fluoride anions, ${\text{F}}^{-}$, according to the following equilibrium reaction

${\text{CaF"_text(2(s]) rightleftharpoons "Ca"_text((aq])^(2+) + color(red)(2)"F}}_{\textrm{\left(a q\right]}}^{-}$

Since every mole of calcium fluoride produces $\textcolor{red}{2}$ moles of fluoride anions, you can say that the molarity of the fluoride anions in a saturated calcium fluoride solution will be

$\left[{\text{F"^(-)] = color(red)(2) xx ["CaF}}_{2}\right]$

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, ${K}_{s p}$, use an ICE table

${\text{ ""CaF"_text(2(s]) " "rightleftharpoons" " "Ca"_text((aq])^(2+) " "+" " color(red)(2)"F}}_{\textrm{\left(a q\right]}}^{-}$

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

By definition, ${K}_{s p}$ is equal to

${K}_{s p} = s \cdot {\left(\textcolor{red}{2} s\right)}^{\textcolor{red}{2}} = s \cdot 4 {s}^{2} = 4 {s}^{3}$

Here $s$ represents the molar solubility of calcium fluoride. This means that ${K}_{s p}$ will be equal to

${K}_{s p} = 4 \cdot {0.000205}^{3} = \textcolor{g r e e n}{3.4 \cdot {10}^{- 11}}$