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–)?

[F⁻]² = 0.068 M

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

the electron configuration that you gave me has 16 electrons ergo it has 16 protons ergo it is the element sulfur.