# AP CHEMISTRY

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Calculate the solubility of LaF3 in grams per liter in the following solutions. (Ksp = 2. 10-19.)
a) pure water
Ans: .00182g/L

b) 0.037 M KF solution

c) 0.055 M LaCl3 solution

NEED b&c

• AP CHEMISTRY -

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.

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