What is the molar solubility of CaF2 if o,1o mol of NaF are added to 1.0L of a saturated solution of CaF2? The Ksp for CaF2 is 4.0 x 10^11.
I get 4,0 x 10^9. Is this right?
4.0 x 10^-5
Surely you made a typo with Ksp. Surely it is 4.0 x 10^-11 and not +11.
S then is 4.0 x 10^-9 M (not +9)
To determine the molar solubility of CaF2 in this scenario, we need to use the concept of the common ion effect.
The common ion effect states that the solubility of a salt is reduced when a common ion is added to the solution. Here, the common ion is fluoride (F-), which is present in both NaF and CaF2.
First, let's write the balanced chemical equation for the dissociation of CaF2:
CaF2(s) ⇌ Ca2+(aq) + 2F-(aq)
The solubility product constant (Ksp) expression for CaF2 can be written as:
Ksp = [Ca2+][F-]^2
Since CaF2 is already in a saturated solution, we can assume that it is at equilibrium, meaning the concentrations of Ca2+ and F- ions from the dissociation are constant. Let's denote the molar solubility of CaF2 as 'x', which represents the concentration of Ca2+ and F- ions.
Using the given value of Ksp (4.0 x 10^11), we can set up the following equation:
4.0 x 10^11 = (x)(2x)^2
Simplifying, we have:
4.0 x 10^11 = 4x^3
Dividing both sides by 4, we get:
1.0 x 10^11 = x^3
Now, taking the cube root of both sides:
x ≈ 1.0 x 10^3
Therefore, the molar solubility of CaF2 is approximately 1.0 x 10^3 mol/L or 1000 mol/L.
Based on your calculation, you arrived at a molar solubility of 4.0 x 10^9 mol/L, which is not correct.