# What is the solubility of lanthanum trichloride in grams per liter of solution at 25^@"C" ?

Mar 28, 2016

$1.4 \cdot {10}^{- 11} {\text{g L}}^{- 1}$

#### Explanation:

In order to be able to solve this problem, you need to know the value of lanthanum trifluoride's solubility product constant, ${K}_{s p}$. You'll find this listed as

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

So, lanthanum trifluoride is *insoluble8 in aqueous solution, which means that when you place it water, an equilibrium is established between the dissolved ions and the undissolved solid

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

Now, the thing to keep in mind here is that you're interested in finding the solubility of lanthanum trifluoride in a solution that contains aqueous potassium fluoride, $\text{KF}$.

Potassium fluoride is a soluble ionic compound that dissociates completely in aqueous solution to form potassium cations and fluoride anions

${\text{KF"_text((aq]) -> "K"_text((aq])^(+) + "F}}_{\textrm{\left(a q\right]}}^{-}$

The presence of these fluoride anions will affect the dissociation equilibrium of lanthanum trifluoride, i.e. less solid will dissolve because of the additional fluoride anions present in solution - this is known as the common ion effect.

Potassium fluoride dissociates in a $1 : 1$ mole ratio with the fluoride anions,which means that you can take the initial concentration of fluoride anions to be

["F"^(-)]_0 = 1.4 * 10^(-2)"M"

Set up an ICE table to help you find the molar solubility of the compound

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

color(purple)("I")" " " "-" " " " " " " " " " " " "0" " " " " " "1.4 * 10^(-2)
color(purple)("C")" " " "-" " " " " " " " " "(+x)" " " " "(+color(red)(3)x)
color(purple)("E")" " " "-" " " " " " " " " " " "x" " " " "1.4 * 10^(-2) + color(red)(3)x

By definition, the solubility product constant will be equal to

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

In your case, you will have

${K}_{s p} = x \cdot {\left(1.4 \cdot {10}^{- 2} + \textcolor{red}{3} x\right)}^{\textcolor{red}{3}}$

Now, because ${K}_{s p}$ is so small, you can use the approximation

$1.4 \cdot {10}^{- 2} + \textcolor{red}{3} x \approx 1.4 \cdot {10}^{- 2}$

This will get you

$2 \cdot {10}^{- 19} = x \cdot {\left(1.4 \cdot {10}^{- 2}\right)}^{\textcolor{red}{3}}$

$x = \frac{2 \cdot {10}^{- 19}}{2.744 \cdot {10}^{- 6}} = 7.29 \cdot {10}^{- 14}$

The molar solubility of lanthanum fluoride in a solution that contains $1.4 \cdot {10}^{- 2} \text{M}$ potassium fluoride will be equal to $7.3 \cdot {10}^{- 14} \text{M}$.

To get the solubility in grams per liter, use lanthanum trifluoride's molar mass

$7.3 \cdot {10}^{- 13} \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{mol")))"L"^(-1) * "195.9 g"/(1color(red)(cancel(color(black)("mole")))) = color(green)(|bar(ul(color(white)(a/a)1.4 * 10^(-11)"g L}}^{- 1} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to two sig figs.