chemistry
posted by anon on .
the Ksp for CaF2 is 3.9*1011. assuming that calcium flouride dissociates completely upon dissolving and that there are no other important equilibria affecting the solubility, calculate the solubility in pure water, of calcium floride in m/L and in g/L

CaF2 ==> Ca^+2 + 2F^
Ksp = (Ca^+)(F^)^2
Set up an ICE chart, substitute into Ksp, and solve.
Let S = solubility CaF2 in M (moles/L)
Then S is the concn of Ca^+2
and 2S is concn F^
3.9 x 10^11 = (S)(2S)^2
Solve for S for moles/L
g = moles x molar mass for g/L.