# chemistry

the Ksp for CaF2 is 3.9*10-11. 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

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1. 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.

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