What's the solubility (in grams per liter) of "LaF"_3LaF3 in pure water?

1 Answer
Stefan V. · Meave60
Dec 22, 2014

1.8 * 10^(-3) "g L"^(-1)1.8⋅10−3g L−1

Explanation:

In order to solve this problem, you would need the value of the solubility product constant, K_(sp)Ksp, for lanthanum trifluoride, "LaF"_3LaF3, which is usually given to you with the problem.

In this case, I'll pick

K_(sp) = 2.0 * 10^(-19)Ksp=2.0⋅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, ss, 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.

" " "LaF"_ (color(red)(3)(s)) " "rightleftharpoons" " "La"_ ((aq))^(3+) " "+" " color(red)(3)"F"_ ((aq))^(-) LaF3(s) ⇌ La3+(aq) + 3F−(aq)

color(purple)("I") color(white)(aaaaacolor(black)(-)aaaaaaaaaaacolor(black)(0)aaaaaaaaaaacolor(black)(0)Iaaaaa−aaaaaaaaaaa0aaaaaaaaaaa0
color(purple)("C") color(white)(aaaacolor(black)(-)aaaaaaaaaacolor(black)((+s))aaaaaaacolor(black)((+color(red)(3)s))Caaaa−aaaaaaaaaa(+s)aaaaaaa(+3s)
color(purple)("E") color(white)(aaaacolor(black)(-)aaaaaaaaaaaacolor(black)(s)aaaaaaaaaaacolor(black)(color(red)(3)s)Eaaaa−aaaaaaaaaaaasaaaaaaaaaaa3s

Initially, the concentrations of the "La"^(3+)La3+ and "F"^(-)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_(sp) = ["La"^(3+)] * ["F"^(-)]^color(red)(3)Ksp=[La3+]⋅[F−]3

This will be equivalent to

K_(sp) = s * (color(red)(3)s)^color(red)(3)Ksp=s⋅(3s)3

2.0 * 10^(-19) = 27s^42.0⋅10−19=27s4

You will thus have

s = root(4)((2.0 * 10^(-19))/27) = 9.3 * 10^(-6)s=4√2.0⋅10−1927=9.3⋅10−6

Since ss 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 * 10^(-6)"mol L"^(-)s=9.3⋅10−6mol L−

In order to express the solubility in grams per liter, "g L"^(-1)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))

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