Ok, I haven't done this in awhile, this is taking me back a couple years, but its good to go back into the past. :+)
Ok lets write out the equation...
CaF2 = Calcium fluoride.
Important thing here to remember is your table of elements and where stuff located on there, we noticed that Ca is on the the 2nd row, which is going to have a 2+ charge, and F (fluoride) is going to be all the way at the end and going to have a -1 charge. The goal of the combined elements is to get to the Noble Gas status, basically how can you get this to zero, well if you have 2 F = (2 x (-1) = -2 and 1 Ca = 2+ so we have 2+ - 2 = 0 and wooo its zero, so it should work that way.
So, write out the full equation and balance it...
CaF2 --> Ca + 2F
Ok, it looks balance, 1 Ca on both sides and 2 F on the other, it looks good so far.
Now, for this We put them into x equations, ooo make them both X sort of like, like this...
(x)(2x)2 = 3.9 x 10-11
4x3 = 3.9 x 10-11
Divide both sides by 4 =
x3 = 9.75x10-12
Cube root of both sides =
x = 2.13x10-4 molar.
Ok now, find the total weight of CaF2 = Ca = 40.078g + 2(19.00g) F = 78.078g
So, now we multiply 2.13x10
-4 molar x 78.078g/mol =
0.166 grams/per L
Which that should be your final answer for solubility of CaF2 Whew a very long one, with lots of other stuff to know, sort of a long step process, but the end result is always a good rewards. :+)
If you need any more help with this question feel free to PM me :+)