Question

The molar solubility of barium fluoride in a 0.216 M ammonium fluoride solution is

M.

Answer 1

Concentration of Ammonium fluoride = NH4F = 0.216M

NH4F ------------- NH4+ + F-

0.216M                           0.216M

BaF2-------------------------- Ba+2   + 2 F-

                                         S             2S+0.216

Ksp = [Ba+2][F-]^2

Ksp= sx(2s+0.216)^2

Ksp = s(0.216)^2     [ (2s)^2 is neglected]

Ksp of BaF2= 2.45x10^-5

2.45x10^-5 = s(0.216)^2

S= 2.45x10^-5/0.04666

S= 52.5x10^-5

S= 5.25x10^-4 M

Solubility of BaF2= 5.25x10^4M