Question

What [F⁻] is required to reduce [Ca²⁺] to 1.0 × 10⁻⁴ M by precipitation of CaF₂?

a) 1.0 × 10⁻⁴ M
b) 2.0 × 10⁻⁴ M
c) 4.0 × 10⁻⁴ M
d) 5.0 × 10⁻⁴ M

Answer 1

The fluoride ion concentration needed to reduce the calcium ion concentration to 1.0 × 10−4 M by precipitation of CaF2 is closest to 2.0 × 10−4 M, using the solubility product constant of 3.45 × 10−11 for CaF2.

Step-by-step explanation:

To find the fluoride ion concentration ([F−]) that is required to reduce the calcium ion concentration ([Ca2+]) to 1.0 × 10−4 M by precipitation of CaF2, we can use the solubility product constant (Ksp).

The solubility product of calcium fluoride (CaF2) is provided as 3.45 × 10−11. The dissolution equation of CaF2 is CaF2 ⇌ Ca2+ + 2 F−, and so the Ksp equation can be written as:

Ksp = [Ca2+][F−]2

By substituting the known values, we have:

3.45 × 10−11 = (1.0 × 10−4 M)([F−])2

From this equation, we can solve for [F−] as follows:

[F−] = √(3.45 × 10−11 / 1.0 × 10−4 M) = 1.86 × 10−4 M

Thus, the closest answer to the calculated [F−] is 2.0 × 10−4 M (answer choice b).