Question

An aqueous solution of sodium fluoride is slowly added to a water sample that contains barium ion (4.15×10-2 M ) and calcium ion (3.55×10-2 M ). The Ksp of barium fluoride is 1.00x10–6. The Ksp of calcium fluoride is 3.90x10–11. What is the remaining concentration of the first ion to precipitate when the second ion begins to precipitate?

Answer 1

BaF2 <-> Ba+2 + 2F-

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

10^-6 = (4.15*10^-2)(F-)^2

[F-] = ((10^-6) /(4.15*10^-2))^0.5 = 0.004908 M for BaF2 = 4.9*10^-3

CaF2 <-> Ca+2 + 2F-

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

3.9*10^-11 = (3.55*10^-2)(F-)^2

[F-] = ((3.9*10^-11) /(3.55*10^-2))^0.5 = 0.000033 M for CaF2 = 3.3*10^-5

Cleraly, CaF2 precipitates first

then

[F-] left when Ba+2 preciptiate

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

3.9*10^-11 = [Ba+2] * (0.000033)^2

[Ba+2] = (3.9*10^-11) / ( 0.000033^2)

[Ba+2] = 0.035812 M