Question

The solubility of lithium fluoride, LiF, is 1.6 g/L, or 6.2 x 10â2 M.
a. Write the balanced solubility equilibrium equation for LiF.
b. Determine the molar concentration of the lithium ion and the fluoride ion.
c. Write the Ksp expression for the reaction.
d. Calculate Ksp for lithium fluoride.

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