CaF2 has a Ksp = 3.8*10^-11 at 25°C. What is the [F-] in a saturated solution of CaF2 at 25°C?

Question

jfraser · Answer

What is the equation for the dissociation of CaF2?

anonymous · Answer

$$CaF_{2(s)} \rightarrow Ca ^{2+} + 2F ^{-}$$ we know that there are 2 mols of F- for every one mole of CaF2 and since it's a solid it's omitted from the solubility expression therefore, 

$$Ksp = [Ca ^{2+}][2F^-]^2$$ 

Since we know the Ksp, we can solve for molarity of F- by inserting x's.$$Ksp = (x)(2x)^2 = (x)(4x^2)= 4x^3 $$

therefore: $$x = \sqrt[3]{Ksp/4}$$ where$$x=[F^-]$$