What's the solubility (in grams per liter) of #"LaF"_3# in pure water?
1 Answer
Explanation:
In order to solve this problem, you would need the value of the solubility product constant,
In this case, I'll pick
#K_(sp) = 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,
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))^(-)#
Initially, the concentrations of the
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)#
This will be equivalent to
#K_(sp) = s * (color(red)(3)s)^color(red)(3)#
#2.0 * 10^(-19) = 27s^4#
You will thus have
#s = root(4)((2.0 * 10^(-19))/27) = 9.3 * 10^(-6)#
Since
#s = 9.3 * 10^(-6)"mol L"^(-)#
In order to express the solubility in grams per liter,
#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))#