Question

# The Ksp of barium fluoride is 1.00 x 10–6. The Ksp of calcium fluoride is 3.90...

The Ksp of barium fluoride is 1.00 x 10–6. The Ksp of calcium fluoride is 3.90 x 10–11.

An aqueous solution of sodium fluoride is slowly added to a water sample that contains barium ion (4.45×10-2M ) and calcium ion (5.50×10-2M ). What is the remaining concentration of the first ion to precipitate when the second ion begins to precipitate?

BaF2 ⇒ Ba2+ + 2F-

Ksp = [Ba2+][F-]2
1.00 x 10–6=[4.45×10-2M][F-]2
[F-]2 = 0.225 x 10-4
[F-]= 0.4743x 10-2 M
= 4.743 x 10-3 M
Therefore, BaF2 will precipitate when the [F-] = 4.743 x 10-3 M

CaF2 ⇒ Ca2+ + 2F-
Ksp = [Ca2+][F-]2
3.90 x 10–11=[5.50×10-2M][F-]2
[F-]2 = 0.7091 x 10-9
[F-]= 2.663 x 10-5 M
Therefore, CaF2 will precipitate when the [F-] = 2.663 x 10-5 M

Since, [F-]CaF2 < [F-]BaF2; the CaF2 will precipitate first.
Therefore, the concentration of Ca2+ ion remaining at [F-] =4.743 x 10-3 M is ;

Ksp = [Ca2+][F-]2
3.90 x 10–11= [Ca2+] x(4.743 x 10-3)2 M
[Ca2+] = 0.1733 x 10–5 M
[Ca2+] = 1.733 x 10–6 M