What concentration of the barium ion, ba2+, must be exceeded to precipitate baf2 from a solution that is 1.00×10−2 m in the fluoride ion, f−? ksp for barium fluoride is 2.45×10−5 . express your answer with the appropriate units?

Answer is: concentration of the barium ion is 0.245 M.

Chemical reaction: BaF₂ → Ba²⁺ + 2F⁻.

[F⁻] = 1.00·10⁻² M.

Ksp = 2.45·10⁻⁵.

Ksp = [Ba²⁺] · [F⁻]².

[Ba²⁺] = Ksp ÷ [F⁻]².

[Ba²⁺] = 2.45·10⁻⁵ ÷ (1.00·10⁻² M)².

[Ba²⁺] = 0.245 M.

it would be a gas formation