The concentration of calcium ion in a town's supply of drinking water is 0.0020 M. (This water is referred to as hard water because it contains such a large concentration of Ca21.) Suppose the water is to be fluoridated by the addition of NaF for the purpose of reducing tooth decay. What is the maximum concentration of fluoride ion that can be achieved in the water before precipitation of CaF2 begins? Will the water supply attain the level of fluoride ion recommended by the U. S. Public Health Service, about 5 * 1025 M (1 mg fluorine per liter) ?

1.41x10⁻⁴ M

It will not attain the level recommended.

Explanation:

When the NaF is added to the water, it will dissociate, forming the ions Na⁺ and F⁻. The fluoride ion may react with the Ca⁺² ions to form the salt CaF₂, which can dissociate too. The dissociation, or solubilization, is characterized by the solubility product (Kps), which is an equilibrium constant. For CaF₂, Kps = 4.0x10⁻¹¹, so:

CaF₂ (s) ⇄ Ca⁺² + 2F⁻ (aq)

The solids are not put in the equilibrium expression, so:

Kps = [Ca⁺²] * [F⁻]²

4.0x10⁻¹¹ = 0.0020*[F⁻]²

[F⁻]² = 2.0x10⁻⁸

[F⁻] = √2.0x10⁻⁸

[F⁻] = 1.41x10⁻⁴ M

Which is the maximum concentration to the solution be saturated, after that, the solid will precipitated. Because the concentration of fluoride ion is lower than the recommended, it will not attain.