Question

a solution is made by dissolving cadmium fluoride (CdF2 Ksp =6.44×10 −3 ) in 100.0 mL of water until excess solid is present. Solution 2 is prepared by dissolving lithium fluoride (LiF, Ksp= 1.84 x 10 −3 ) in 200.0 mL of water until excess solid is present. Which solution has a higher fluoride ion concentration? what is the molar concentration of F-1 in the CdF2 solution? what is the molar concentration of F-1 in the LiF solution?

Answer 1

Answer:

Answers are in the explanation

Explanation:

Ksp of CdF₂ is:

CdF₂(s) ⇄ Cd²⁺(aq) + 2F⁻(aq)

Ksp = 6.44x10⁻³ = [Cd²⁺] [F⁻]²

When an excess of solid is present, the solution is saturated, the molarity of Cd²⁺ is X and F⁻ 2X:

6.44x10⁻³ = [X] [2X]²

6.44x10⁻³ = 4X³

X = 0.1172M

[F⁻] = 0.2344M

Ksp of LiF is:

LiF(s) ⇄ Li⁺(aq) + F⁻(aq)

Ksp = 1.84x10⁻³ = [Li⁺] [F⁻]

When an excess of solid is present, the solution is saturated, the molarity of Li⁺ and F⁻ is XX:

1.84x10⁻³ = [X] [X]

1.84x10⁻³ = X²

X = 0.0429

[F⁻] = 0.0429M

The solution of CdF₂ has the higher fluoride ion concentration