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!
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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