Given a saturated solution of CaF2, write the ionic equilibrium and use it to calculate the solubility of CaF2.
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)
...........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.
what is the name for CaF2 in chemistry?
You're a cheesehead
To determine what will happen to the value of Ksp when a solid NaF is added to the solution, we need to consider the reaction that occurs between NaF and CaF2.
NaF(s) → Na+(aq) + F-(aq)
When NaF is added, it dissociates into Na+ and F-. The F- ions will be in excess and can react with the Ca2+ ions from CaF2 to form more CaF2. However, since the reaction is a double-displacement reaction, it will not affect the concentration of Ca2+ and F- ions in the solution. Therefore, the value of Ksp will remain the same.
Now, to calculate the new solubility of CaF2 in a 0.100 mol/L NaF solution, we need to consider the common-ion effect. The F- ions from NaF and CaF2 will both contribute to the concentration of F- in the solution.
Let's assume the solubility of CaF2 is x mol/L. Since CaF2 dissociates to form Ca2+ and 2F-, the F- concentration will be 2x mol/L.
Considering the reaction: CaF2(s) ↔ Ca2+(aq) + 2F-(aq)
The F- concentration from NaF is 0.100 mol/L, and the F- concentration from CaF2 is 2x mol/L. Therefore, the total F- concentration in the solution is 0.100 + 2x mol/L.
Using the Ksp expression for CaF2: Ksp = [Ca2+][F-]2 = [Ca2+](0.100 + 2x)2
Since the value of Ksp is constant, we can now solve for x to find the new solubility of CaF2.