Question

Zinc fluoride (ZnF2) dissociates according to the following equation: ZnF2(s) ⇌ Zn2+(aq) + 2F–(aq) Ksp of this equilibrium is 3.0 × 10–2. If the concentration of zinc ions (Zn2+) in a solution is measured to be 4.4 × 10–1 M, what is the concentration of the fluoride ions (F–)?

Answer 1

[F⁻]² = 0.068 M

Step-by-step explanation:

ZnF₂(s) ⇌ Zn²⁺(aq) + 2F⁻(aq)

At equilibrium the Kps can be calculated by the following equation:

Kps = [Zn²⁺][F⁻]²

Therefore, the concentration of F⁻¹ can be calculated:

[F⁻]² = Kps/ [Zn²⁺]

[F⁻]² = = 3.0 x 10⁻² / 4.4 x 10⁻¹M

[F⁻]² = 0.068 M