Answer is: concentration
of the fluoride ion is 0.0368 M.
Chemical reaction: CaF₂(s) → Ca²⁺(aq) + 2F⁻(aq). Ksp = 2.0·10⁻⁴ M.
Ksp = [Ca²⁺] ·
[F⁻]². [Ca²⁺] = x. [F⁻] = 2[Ca²⁺] = 2x. Ksp = x · (2x)². 2.0·10⁻⁴ = 4x³. x = [F⁻] = ∛(2.0·10⁻⁴ ÷ 4). [F⁻] = 0.0368 M.
