Question

Calculate the molar solubility of CaF₂ in a solution containing 0.250 M Ca(NO₃)₂. Ksp = 1.46 x 10⁻¹⁰

a. 1.46 x 10⁻⁵ M
b. 2.92 x 10⁻⁶ M
c. 5.84 x 10⁻⁶ M
d. 1.17 x 10⁻⁵ M

Answer 1

To calculate the molar solubility of CaF₂ in a solution containing 0.250 M Ca(NO₃)₂, we use the solubility product constant (Ksp) and the stoichiometry of the dissolution equation. The molar solubility of CaF₂ in the solution is 1.46 x 10⁻⁵ M.

Step-by-step explanation:

To calculate the molar solubility of CaF2 in a solution containing 0.250 M Ca(NO3)2, we need to use the solubility product constant (Ksp) and the stoichiometry of the dissolution equation.

First, we can set up the equation:

Ksp = [Ca2+][F-]2 = (s)(2s)2 = 4s3

Since the concentration of Ca2+ in the solution is 0.250 M, we can substitute this value into the equation and solve for s:

Ksp = 4s3

1.46 x 10-10 = 4s3

s = 1.46 x 10-5 M

Therefore, the molar solubility of CaF2 in the solution is 1.46 x 10-5 M, so the answer is a. 1.46 x 10-5 M.