Question

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

Answer 1

What is the equation for the dissociation of CaF2?

Answer 2

\[CaF_{2(s)} \rightarrow Ca ^{2+} + 2F ^{-}\] 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\[x=[F^-]\]