Question

The solubility of lithium fluoride, LiF, is 1.6 g/L, or 6.2 x 10â2 M.

Answer 1

Answer:

a. LiF(s) ⇄ Li⁺(aq) + F⁻(aq)

b. [Li⁺] = [F⁻] = 6.2 x 10⁻² M

c. Ksp = [Li⁺] [F⁻]

d. Ksp = 3.8 × 10⁻³

Explanation:

The solubility (S) of lithium fluoride, LiF, is 1.6 g/L, or 6.2 x 10⁻² M.

a. The balanced solubility equilibrium equation for LiF is:

LiF(s) ⇄ Li⁺(aq) + F⁻(aq)

b. We will make an ICE chart.

        LiF(s) ⇄ Li⁺(aq) + F⁻(aq)

I                       0           0

C                     +S         +S

E                       S           S

Then, [Li⁺] = [F⁻] = S = 6.2 x 10⁻² M

c. The solubility product constant, Ksp​, is the equilibrium constant for a solid substance dissolving in an aqueous solution.

Ksp = [Li⁺] [F⁻]

d.

Ksp = [Li⁺] [F⁻] = (6.2 x 10⁻²)² = 3.8 × 10⁻³