Calculate the molar solubility of barium fluoride, baf2, in each liquid or solution. for barium fluoride, ksp =2.45 x 10-5.

BaF₂ when it dissolves, dissociates as follows;
BaF₂ --> Ba²⁺ + 2F⁻
Molar solubility is the number of moles that can be dissolved in 1 L of solution. If molar solubility of BaF₂ is x, then molar solubility of Ba²⁺ is x and solubility of F⁻ is 2x.
ksp = [Ba²⁺][F⁻]²
ksp = (x)(2x)²
2.45 x 10⁻⁵ = 4x³
x³ = 0.6125 x 10⁻⁵
x = 0.0183 mol/L is molar solubility of BaF₂

The answer should be vaporization