Question

An excess of solid CaF₂ is shaken with a 0.100 M KF solution until the solution is saturated with calcium fluoride. (Use Ksp = 3.2 x 10⁻¹¹ for CaF₂.)

Calculate the molar concentration of Ca₂⁺ in the saturated solution neglecting activity effects.

Answer 1

To find the molar concentration of Ca2+ in a saturated solution of CaF2 with a given Ksp of 3.2 x 10^-11, solve for 's' in the equation Ksp = 4s^3. The concentration of Ca2+ will be the cube root of (3.2 x 10^-11 / 4).

Step-by-step explanation:

The question asks to calculate the molar concentration of Ca2+ in a saturated solution of calcium fluoride (CaF2), using a provided solubility product constant (Ksp). The dissolution equation for CaF2 is:

CaF2 (s) → Ca2+ (aq) + 2F− (aq)

Given the Ksp = 3.2 x 10−11 for CaF2, and the fact that the solution is already saturated with KF, we can assert that the only source of fluoride ions is from the dissolving CaF2. The solubility of CaF2 (s) in KF can be represented by 's' for Ca2+ and '2s' for F−, where 's' is the solubility of CaF2 in moles per liter.

The Ksp equation for CaF2 is:
Ksp = [Ca2+][F−]2 = s(2s)2 = 4s3

Substituting the given Ksp into the equation:

3.2 x 10−11 = 4s3

Solving for 's', we find that 's' is the cube root of (3.2 x 10−11 / 4). The molar concentration of Ca2+ ions, represented by 's', will be the answer to the question.