Question

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?

Answer 1

BaF₂ ⇒ Ba²⁺ + 2F^-

K_sp = [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^-
K_sp = [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 ;

K_sp = [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