Question

What is the molar solubility of BaF2 in .033M KF (aq)? The BaF2 Ksp is 1.8*10^-7....

What is the molar solubility of BaF2 in .033M KF (aq)? The BaF2 Ksp is 1.8*10^-7.

I got 5.5*10^-6 which is incorrect. Please explain!

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Answer #1

BaF2 <---------------------> Ba+2 + 2F-

                                         S             2S ----------------> in absence of common ion

                                        S              2S+ 0.033 ------------> in presence of common ion

                                        S              0.033    --------------> (0.033 >>> S because KF is strong electrolyte)

Ksp = [Ba+2][F-]^2

1.8 x 10^-7 = (S) x (0.033)^2

S = 1.65 x 10^-4 M

molar solubility = 1.65 x 10^-4 M

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