Chemistry Solubility?

Question

I am having trouble understanding solubility.

Calculate the solubility of LaF3 in grams per liter in the following:

Pure water

0.025 M KF solution

0.150 M LaCl3 solution

Answer 1

According to http://www.northland.cc.mn.us/chemistry/solubility_products.htm

Ksp (LaF3) =4*10^-17

.. .. .. .. .. LaF3 <=> La+3 + 3F-
Initial .. .. .. .. .. .. .. .. .. L .. .. .. F
React .. .. .. .. .. .. .. .. .x .. .. .. 3x
At equil .. .. .. .. .. .. .. L-x .. .. F-3x

Ksp =[La+3][F-]^3= (L-x)(F-3x)^3

For pure water L=F=0, so
Ksp =x*(3x)^3 = 27x^4 => x= (Ksp/27)^(1/4) =>
x= ((4*10^-17)/27)^(1/4) = 3.49*10^-5 M

In the presence of 0.025 M KF, L=0 and F=0.025 so
Ksp= x(0.025-3x)^3=4*10^-17
Assume that 0.025 >> x so that 0.025-3x=0.025
Then x*0.025^3 =4*10^-17 =>

x=(4*10^-17)/0.025^3 =2.56 *10^-12 << 0.025 so our assumption is valid

In the presence of 0.150 LaCl3, L=0.15 and F=0 so
Ksp = (0.15-x)x^3 =4.10^-17
If 0.15 >> x so that 0.15-x=0.15, then
0.15x^3= 4*10^-17 =>

x= (((4*10^-17)/0.15))^(1/3) =6.44 *10^-6 M << 0.15 so our assumption is valid

Answer 2

Have you the Ksp of LaF3?
Can you give me?
I will answer at once