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 (4.50×10-2M ) and calcium ion (3.75×10-2M ). What is the remaining concentration of the first ion to precipitate when the second ion begins to precipitate?

Answer 1

BaF2 (s) --> Ba+2 + 2 F-

ionic product = [Ba+2] [F-]^2

precipitation will start when ionic product = Ksp

so

4.5 x 10-2 x [F-]^2 = 1 x 10-6

[F-] = 4.714 x 1o^-3 M

so

4.714 x 10-3 M F- is needed for BaF2 to start precipitating

similarly

for CaF2

CaF2 --> Ca+2 + 2 F-

3.9 x 10-11 = 3.75 x 10-2 x [F-]^2

[F-] = 3.2249 x 10-5 M

so

we can see that

CaF2 will start precipitating first and then BaF2

now

when BaF2 starts to precipitate , [F-] = 4.714 x 10-3 M

we need to use this value of [F-] to calculate the concentration of the other ion Ca+2

now

[Ca+2] [F-]^2 = 3.9 x 10-11

[Ca+2] [4.714 x 10-3]^2 = 3.9 x 10-11

[Ca+2] = 1.755 x 10-6 M

so

the remaining concentration of the first ion , Ca+2 is 1.755 x 10-6 M