The mineral CaF2 has a solubility of 2.1 x 10^-4 M. What is the Ksp of CaF2?
pls Bob if you're here!
CaF2 ==> Ca^+2 + 2F^-
Ksp = (Ca^+2)(F^-)^2
You know solubility CaF2 = 2.1 x 10^-4; therefore, (Ca^+2) = 2.1 x 10^-4. Since (F^-) is twice that, (F^-) = 4.2 x 10^-4. Plug those values into the Ksp expression I wrote above and solve for Ksp.
................CaF2 ==> Ca^2+ + 2F^-
I.................solid...........0............0
C............... solid...........x.............2x
E.................solid...........x............2x
Ksp = (Ca^2+)(F^-)^2 = (x)(2x)^2 = 4x^3
The problems tells you the value of x. Plug and chug.
Note: You can plug in differently depending upon how you want to do it.
Ksp = (Ca^2+)(F^-)^2 = (2.1E-4)(4.2E-4)^2 = ?
You should get the same number either way. Try it.
thank you omg thank you
To calculate the solubility product constant (Ksp) of CaF2, we need to know the molar concentrations of calcium ions (Ca2+) and fluoride ions (F-) in a saturated solution of CaF2. The molar solubility of CaF2 is given as 2.1 x 10^-4 M, which means that at equilibrium, the concentration of Ca2+ ions and F- ions in the solution is both equal to 2.1 x 10^-4 M.
The balanced chemical equation for the dissolution of CaF2 is:
CaF2 (s) ⇌ Ca2+ (aq) + 2F- (aq)
Since the stoichiometry of the balanced equation is 1:1, the concentration of Ca2+ is also 2.1 x 10^-4 M, and the concentration of F- is 2 x (2.1 x 10^-4 M) = 4.2 x 10^-4 M.
Now, we can use the concentrations of Ca2+ and F- ions to calculate the Ksp value.
Ksp = [Ca2+][F-]^2
Ksp = (2.1 x 10^-4 M) * (4.2 x 10^-4 M)^2
Ksp = 1.47 x 10^-11
So, the Ksp value of CaF2 is 1.47 x 10^-11.
Note: The value may vary depending on the temperature and the source of the solubility data.