Based on data given and using a formula wt of CaF2 = 78g/mol;
Molar solubility CaF2 = (7E-2/78)M = 8.9E-4M. Ksp = [Ca][F]^2 =(8.9E-4)(8.9E-4)^2 = 7E-10 which does not match published Ksp-values for CaF2. Never the less, In presence of 0.50M CaCl2...
CaF2 <=> Ca+2 + 2F^-
Ceq: --- 0.50M 2x
Ksp = [Ca][F]^2 = (0.50)(2x)^2 = (0.50)(4x^2) = 2x^2 = 7E-10 => x = Solubility in the presence of 0.50M CaCl2 = [(7E-10/2)]^1/2 = 1.9E-5M
Comparing to solubility in water S(HOH)=8.9E-4M > S(Ca)=1.9E-5M. This is consistent with common ion effect.
the solubility of CaF in water is 7*10^-2 gm/l. find its solubility in 0.5m CaCl2 soln
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