A saturated solution of CaF2 (of MW 78.0
g/mol) contains 0.0167 grams per liter. What is the solubility product constant of CaF2?
..........CaF2 --> Ca^2+ + 2F^-
I.........solid....0........0
C.........solid....x........2x
E.........solid....x........2x
The easy way to do this is to convert g/L to mols/L and go from there.
mols/L = 0.0167/molar mass CaF2.
Then Ksp = (Ca^2+)(F^-)^2
Ksp = (x)(2x)^2 = 4x^3. You know what x is from above, solve for Ksp.
To find the solubility product constant (Ksp) of CaF2, we need to know the concentrations of Ca2+ and F- ions in the saturated solution.
The balanced chemical equation for the dissolution of CaF2 is as follows:
CaF2 (s) ↔ Ca2+ (aq) + 2F- (aq)
From the equation, we see that for every one mole of CaF2 that dissolves, one mole of Ca2+ and two moles of F- ions are formed.
Given that the molar mass of CaF2 is 78.0 g/mol, we can calculate the number of moles (n) of CaF2 dissolved per liter of solution:
n = mass/molar mass
n = 0.0167 g/78.0 g/mol
n = 0.000214 mol/L
Since every mole of CaF2 produces one mole of Ca2+ ions, the concentration of Ca2+ ions in the saturated solution is also 0.000214 mol/L.
Since every mole of CaF2 produces two moles of F- ions, the concentration of F- ions in the saturated solution is double that of the Ca2+ ions:
[Ca2+] = 0.000214 mol/L
[F-] = 2 × [Ca2+] = 2 × 0.000214 mol/L = 0.000428 mol/L
Now we can calculate the solubility product constant (Ksp) using the concentrations of Ca2+ and F- ions:
Ksp = [Ca2+][sp = (0.000214 mol/L)(0.000428 mol/L)
Ksp = 9.1592 x 10^-8
Therefore, the solubility product constant (Ksp) of CaF2 is 9.1592 x 10^-8.
To find the solubility product constant (Ksp) of CaF2, we need to use the given information about the mass of CaF2 in a saturated solution.
First, let's convert the mass of CaF2 to moles. We can do this by dividing the given mass (0.0167 grams) by the molar mass of CaF2 (78.0 g/mol):
moles = mass / molar mass
moles = 0.0167 g / 78.0 g/mol
Next, we need to convert the moles of CaF2 to moles per liter (Molarity). Since the solution is saturated, the amount of CaF2 is the maximum that can be dissolved in the given volume.
So, the molarity (M) of CaF2 in the saturated solution is equal to the moles of CaF2 divided by the volume (in liters) of the solution:
Molarity (M) = moles / volume (in liters)
Molarity (M) = (0.0167 g / 78.0 g/mol) / 1 L
Molarity (M) = 0.000214 M
Finally, the solubility product constant (Ksp) of CaF2 can be calculated by using the molarity of CaF2 in the saturated solution. The solubility product constant expression for CaF2 is:
CaF2 ⇌ Ca2+ + 2F-
Ksp = [Ca2+][F-]^2
Since CaF2 dissociates into Ca2+ and 2F-, we can substitute the molarity of CaF2 (0.000214 M) into the Ksp expression:
Ksp = (0.000214 M)(2 * 0.000214 M)^2
Ksp ≈ 3.378 x 10^-9
Therefore, the solubility product constant of CaF2 is approximately 3.378 x 10^-9.