Given a saturated solution of CaF2, write the ionic equilibrium and use it to calculate the solubility of CaF2.

Question

My answers for the ionic equation and ksp values are
CaF2(s)<->Ca2+(aq)+ 2F-(aq)
Ksp= [Ca2+][F-]2
ksp=2.02*10^-4

it asks what will happen to the value of ksp when a solid NaF is added to the solution?

and what would be the new solubility of CaF2 (in a 0.100 mol/L NaF)

Answer 1

...........CaF2(s) ==> Ca^2+ + 2F^-
I..........solid........0.......0
C..........solid........x.......2x
E..........solid........x.......2x

Ksp = (Ca^2+)(F^-)^2
Substitute from the ICE chart and solve for x. The question asks for solubility of CaF2. You will calculate x = (Ca^2+) in moles/L but since 1 mol CaF2 dissolves to give 1 mol Ca&2+, then x is also solubility CaF2.

If NaF is added to the solution what happens? Remember Le Chatelier's Principle which says that a system at equilibrium will try to undo what we do to it. So if we add F^-, the reaction will try to get rid of it. How can it do that. Only one way. That's by shifting to the left so that the solubility of CaF2 is decreased.

If we make the (F^-) = 0.1 M, then
Ksp = (Ca^2+)(F^-)^2
Ksp = (x)(0.1).
Solver for x.

Answer 2

what is the name for CaF2 in chemistry?

Answer 3

You're a cheesehead