What is the molar solubility of barium fluoride (baf2) in water? the solubility-product constant for baf2 is 1.7 * 10-6 at 25∘c?

Molar solubility is the number of moles that are dissolved in 1 L solution.

when BaF₂ dissolves it dissociates into the following ions

BaF₂ - - > Ba²⁺ + 2F⁻

if the molar solubility of BaF₂ is X, then molar solubility of Ba²⁺ is X and F⁻ is 2x

then the formula for the solubility product constant - ksp is;

ksp = [Ba²⁺][F⁻]²

ksp = X * (2X) ²

ksp = 4X³

since ksp = 1.7 x 10⁻⁶

4X³ = 1.7 x 10⁻⁶

X = 0.0075 M

molar solubility of BaF₂ is 0.0075 M