# What's the solubility (in grams per liter) of "LaF"_3 in pure water?

Dec 22, 2014

$1.8 \cdot {10}^{- 3} {\text{g L}}^{- 1}$

#### Explanation:

In order to solve this problem, you would need the value of the solubility product constant, ${K}_{s p}$, for lanthanum trifluoride, ${\text{LaF}}_{3}$, which is usually given to you with the problem.

In this case, I'll pick

${K}_{s p} = 2.0 \cdot {10}^{- 19}$.

You could approach this problem by using an ICE table (more here: http://en.wikipedia.org/wiki/RICE_chart) to help you find the molar solubility, $s$, of lanthanum trifluoride in aqueous solution.

Since you're dealing with an insoluble ionic compound, an equilibrium will be established between the undissolved solid and the dissolved ions.

${\text{ " "LaF"_ (color(red)(3)(s)) " "rightleftharpoons" " "La"_ ((aq))^(3+) " "+" " color(red)(3)"F}}_{\left(a q\right)}^{-}$

color(purple)("I") color(white)(aaaaacolor(black)(-)aaaaaaaaaaacolor(black)(0)aaaaaaaaaaacolor(black)(0)
color(purple)("C") color(white)(aaaacolor(black)(-)aaaaaaaaaacolor(black)((+s))aaaaaaacolor(black)((+color(red)(3)s))
color(purple)("E") color(white)(aaaacolor(black)(-)aaaaaaaaaaaacolor(black)(s)aaaaaaaaaaacolor(black)(color(red)(3)s)

Initially, the concentrations of the ${\text{La}}^{3 +}$ and ${\text{F}}^{-}$ ions are equal to zero - the solid was not yet placed in water.

Keep in mind that the solid's concentration is presumed to be either unknown or constant, which is why it's not relevant here.

By definition, the solubility product constant for this equilibrium will be

${K}_{s p} = {\left[{\text{La"^(3+)] * ["F}}^{-}\right]}^{\textcolor{red}{3}}$

This will be equivalent to

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

$2.0 \cdot {10}^{- 19} = 27 {s}^{4}$

You will thus have

$s = \sqrt[4]{\frac{2.0 \cdot {10}^{- 19}}{27}} = 9.3 \cdot {10}^{- 6}$

Since $s$ represents the molar solubility of the salt, i.e. how many moles of lanthanum trifluorice can be dissolved in a liter of water, you will have

$s = 9.3 \cdot {10}^{- 6} {\text{mol L}}^{-}$

In order to express the solubility in grams per liter, ${\text{g L}}^{- 1}$, use lanthanum trifluoride's molar mass

9.3 * 10^(-6) color(red)(cancel(color(black)("mol")))/"L" * "195.9 g"/(1color(red)(cancel(color(black)("mol")))) = color(green)(1.8 * 10^(-3) "g L"^(-1))