The mineral CaF2 has a solubility of 2.1 x 10^-4 M. What is the Ksp of CaF2?
a. 1.85x10^-11
b. 9.3x10^-12
c. 3.7x10^-11
d. 8.8x10^-3
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. Post your work if you get stuck.
To find the Ksp (solubility product constant) of CaF2, we need to use the information about the solubility of CaF2. The solubility of CaF2 is given as 2.1 x 10^-4 M.
The solubility product constant, Ksp, is the product of the concentrations of the ions in a saturated solution of a compound. Since CaF2 dissociates into Ca2+ and 2F- ions, the Ksp expression for CaF2 can be written as:
Ksp = [Ca2+][F-]^2
Since CaF2 dissociates into one Ca2+ ion and two F- ions, we can substitute the solubility of CaF2 to calculate the concentration of the ions.
The concentration of Ca2+ ions is equal to the solubility of CaF2, as only one Ca2+ ion is produced per formula unit. Hence, the concentration of Ca2+ ions is 2.1 x 10^-4 M.
The concentration of F- ions is equal to twice the solubility of CaF2, as two F- ions are produced per formula unit. Therefore, the concentration of F- ions is (2.1 x 10^-4 M) x 2 = 4.2 x 10^-4 M.
Now, we can substitute these values into the Ksp expression:
Ksp = (2.1 x 10^-4 M)(4.2 x 10^-4 M)^2
Ksp = 3.7 x 10^-11 M^3
Therefore, the Ksp of CaF2 is approximately 3.7 x 10^-11 M^3.
The correct option is c. 3.7x10^-11.
To determine the Ksp (solubility product constant) of CaF2, we first need to understand the balanced chemical equation for its dissolution. The equation is as follows:
CaF2 (s) ⇌ Ca2+ (aq) + 2F- (aq)
The solubility of CaF2 is given as 2.1 x 10^-4 M. This means that at equilibrium, the concentration of Ca2+ ions in the solution is 2.1 x 10^-4 M, and the concentration of F- ions is twice that, 4.2 x 10^-4 M.
Since the stoichiometry of the balanced equation is 1:2 (1 CaF2 yields 1 Ca2+ ion and 2 F- ions), we can calculate the solubility product constant (Ksp) using the following equation:
Ksp = [Ca2+][F-]^2
Plugging in the values, we get:
Ksp = (2.1 x 10^-4 M) * (4.2 x 10^-4 M)^2
Ksp = 1.85 x 10^-11
Therefore, the Ksp of CaF2 is 1.85 x 10^-11.
The correct answer is option a. 1.85 x 10^-11.