AP CHEMISTRY
posted by HELP PLEASE .
Calculate the solubility of LaF3 in grams per liter in the following solutions. (Ksp = 2. 1019.)
a) pure water
Ans: .00182g/L
b) 0.037 M KF solution
c) 0.055 M LaCl3 solution
NEED b&c
PLEASE HELP

Let x = solubility LaF3.
...........LaF3 ==> La^3+ + 3F^
............x.......x.......3x
...........KF ==> K^+ + F^
initial.0.037......0....0
change...0.037..0.037...0.037
equil.....0.....0.037...0.037
Ksp LaF3 = (La^3+)(F^)^3
From the above charts.
(La^3+) = x
(F^) = 3x from the LaF3 and 0.037 from the KF to make a total of 3x+0.037
Solve for x
c is done the same way except
(La^3+) = x from LaF3 + 0.0055 from LaCl3.
(F^) = 3x
Solve for x.