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