Determine the molar solubility of baf2 in pure water. ksp for baf2 = 2.45 x 10-5. 1.83 x 10-2 m 1.23 x 10-5 m 2.90 x 10-2 m 4.95 x 10-3 m 6.13 x 10-6 m
The Molar solubility of baf2 in pure water is 1.83 x 10⁽₋₂⁾ if ksp for baf2 = 2.45 x 10⁻⁵. solubility product equilibrium reaction from the balance equation of reaction is: k₍sp₎ = [Ba⁺²] [F⁻]² using mole ratios from one to another, [Ba⁺²] = x and [F⁻]² = 2x k₍sp₎ = [Ba⁺²] [F⁻]² k₍sp₎ = [x][2x]² ksp = 2.45 x 10⁻⁵ then, 2.45 x 10⁻⁵ = [x][2x]² 4x³ = 2.45 x 10⁻⁵ x = ∛(2.45 x 10⁻⁵)/4 = 1.83 x 10⁻²m so, x is molar solubility which is 1.83 x 10⁻²m
