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

Idid the calculation and i get 875,000,000 which can be written as 8.75 x 10^6?