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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asked by anon
  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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