Question

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

Answer 1

`1.4 * 10^(-11)"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_(sp)`. You'll find this listed as

`K_(sp) = 2.0 * 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

`"LaF"_text(3(s]) rightleftharpoons "La"_text((aq])^(3+) + color(red)(3)"F"_text((aq])^(-)`

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, `"KF"`.

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

`"KF"_text((aq]) -> "K"_text((aq])^(+) + "F"_text((aq])^(-)`

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

`" ""LaF"_text(3(s]) " "rightleftharpoons" " "La"_text((aq])^(3+) " "+" " color(red)(3)"F"_text((aq])^(-)`

`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_(sp) = ["La"^(3+)] * ["F"^(-)]^color(red)(3)`

In your case, you will have

`K_(sp) = x * (1.4 * 10^(-2) + color(red)(3)x)^color(red)(3)`

Now, because `K_(sp)` is so small, you can use the approximation

`1.4 * 10^(-2) + color(red)(3)x ~~ 1.4 * 10^(-2)`

This will get you

`2 * 10^(-19) = x * (1.4 * 10^(-2))^color(red)(3)`

`x = (2 * 10^(-19))/(2.744 * 10^(-6)) = 7.29 * 10^(-14)`

The molar solubility of lanthanum fluoride in a solution that contains `1.4 * 10^(-2)"M"` potassium fluoride will be equal to `7.3 * 10^(-14)"M"`.

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

`7.3 * 10^(-13)color(red)(cancel(color(black)("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)color(white)(a/a)|)))`

The answer is rounded to two sig figs.