What's the solubility (in grams per liter) of #"LaF"_3# in pure water?

In order to solve this problem, you would need the value of the costante di solubilità del prodotto, #K_(sp)#, Per lanthanum trifluoride, #"LaF"_3#, which is usually given to you with the problem.

In this case, I'll pick

#K_(sp) = 2.0 * 10^(-19)#.

You could approach this problem by using an Tavolo ICE (more here: http://en.wikipedia.org/wiki/RICE_chart) to help you find the solubilità molare, #s#, of lanthanum trifluoride in aqueous solution.

Since you're dealing with an insolubile composto ionico, 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))^(-)#

#color(purple)("I") color(white)(aaaaacolor(black)(-)aaaaaaaaaaacolor(black)(0)aaaaaaaaaaacolor(black)(0)#
#color(purple)("C") color(white)(aaaacolor(black)(-)aaaaaaaaaacolor(black)((+s))aaaaaaacolor(black)((+color(red)(3)s))#
#color(purple)("E") color(white)(aaaacolor(black)(-)aaaaaaaaaaaacolor(black)(s)aaaaaaaaaaacolor(black)(color(red)(3)s)#

Inizialmente, il concentrazioni tutte lungo la #"La"^(3+)# e #"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.

Per definizione, the solubility product constant for this equilibrium will be

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

Questo sarà equivalente a

#K_(sp) = s * (color(red)(3)s)^color(red)(3)#

#2.0 * 10^(-19) = 27s^4#

Avrai così

#s = root(4)((2.0 * 10^(-19))/27) = 9.3 * 10^(-6)#

Dal #s# rappresenta la solubilità molare of the salt, i.e. how many moles of lanthanum trifluorice can be dissolved in a liter of water, Si avrà

#s = 9.3 * 10^(-6)"mol L"^(-)#

In order to express the solubility in grammi per litro, #"g L"^(-1)#, use lanthanum trifluoride's massa molare

#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))#