How would you calculate the solubility of #LaF_3# in grams per liter in the following solutions? a) 0.049 M #KF# solution b) 0.070 M #LaCl_3# solution
1 Answer
Here's what I got.
Explanation:
I'll show you how to solve part (a) and leave part (b) to you as practice.
Lanthanum trifluoride,
#"LaF"_ ( 3(s)) rightleftharpoons "La"_ ((aq))^(3+) + color(red)(3)"F"_ ((aq))^(-)#
The position of the equilibrium is given to you by the value of the solubility product constant,
More specifically, the smaller the
Now, you want to know how much lanthanum trifluoride can be dissolved in a solution that is
In other words, potassium fluoride dissociates completely in aqueous solution to form potassium cations,
#"KF" _ ((aq)) -> "K"_ ((aq))^(+) + "F"_ ((aq))^(-)#
Since every mole of potassium fluoride that dissociates form
#["F"^(-)] = 1 xx ["KF"] = "0.049 M"#
The presence of the fluoride anions will affect the position of the equilibrium in accordance with Le Chatelier's Principle.
More specifically, the equilibrium will shift to the left in an attempt to decrease the extra concentration of dissolved fluoride anions. As a result, the solubility of the salt will decrease
So, if you take
#"LaF"_ ( 3(s)) " "rightleftharpoons" " "La"_ ((aq))^(3+) " "+" " color(red)(3)"F"_ ((aq))^(-)#
By definition, the solubility product constant is equal to
#K_(sp) = ["La"^(3+)] * ["F"^(-)]^color(red)(3)#
In this case, you have
#K_(sp) = s * (0.049 + color(red)(3)s)^color(red)(3)" " " "color(orange)("(*)")#
This will be equivalent to
#K_(sp) = s * [0.049^3 + (3s)^3 + 3 * 0.049^2 * (3s) + 3 * 0.049 * (3s)^2]#
#K_(sp) = s * (0.00011765 + 27s^3 + 0.02161s + 1.323s^2)#
#K_(sp) = 27s^4 + 1.323s^3 + 0.02161s^2 + 0.00011765s#
This can be rearranged to give
#27s^4 + 1.323s^3 + 0.02161s^2 + 0.00011765s - K_(sp) = 0#
Since the problem doesn't provide you with a value for the
#K_(sp) = 2.0 * 10^(-19)#
Now, this quartic equation will produce four solutions, one positive, one negative, and two complex.
https://www.easycalculation.com/algebra/quartic-equation.php
Since you're looking for
#s = 1.7 * 10^(-15)#
Alternatively, you can use the fact that
#0.049 + 3s ~~ 0.049#
Equation
#K_(sp) = s * (0.049)^3#
This will get you
#s = K_(sp)/0.049^3 = (2.0 * 10^(-19))/(0.049^3) = 1.7 * 10^(-15)#
As you can see, the approximation holds.
This means that in a solution that is
To find the solubility in grams per liter, use the salt's molar mass
#1.8 * 10^(-15) color(red)(cancel(color(black)("moles LaF"_3)))/"L" * "195.9 g"/(1color(red)(cancel(color(black)("mole LaF"_3)))) = color(green)(|bar(ul(color(white)(a/a)color(black)(3.5 * 10^(-13)"g L"^(-1))color(white)(a/a)|)))#
I'll leave the answer rounded to two sig figs.
By comparison, the solubility of lanthanum trifluoride in pure water is
https://socratic.org/questions/what-s-the-solubility-in-grams-per-liter-of-laf3-in-pure-water