anonymous
  • anonymous
CaF2 has a Ksp = 3.8*10^-11 at 25°C. What is the [F-] in a saturated solution of CaF2 at 25°C?
Chemistry
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
JFraser
  • JFraser
What is the equation for the dissociation of CaF2?
anonymous
  • anonymous
\[CaF_{2(s)} \rightarrow Ca ^{2+} + 2F ^{-}\] 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\[x=[F^-]\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.