Question

A solution contains 1.53×10^-2 M calcium acetate and 1.53×10^-2 M barium nitrate. Solid potassium fluoride is added slowly to this mixture. What is the concentration of calcium ion when barium ion begins to precipitate? Solubility product constant data is found in the Chemistry References.

[Ca²⁺] = _____M

Answer 1

[CaAcetate] = 1.53*10^-3

[Ba(NO3)2] = 1.53*10^-2

Then KF(s) is added

find Ca+2 and Ba+2 when Ba+2 preciptites

important note: Ksp values vary a lot depending on the reference you use, therefore, please prodvide Ksp values for precie, or 100% accuracy.

Ksp BaF2 1.0×10–6

Ksp CaF2 5.3×10–9

First, identify the Concentration of F- required in order to start precipitateion of Ba+2 ions

so

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

10^-6 = (1.53*10^-2)(F-)^2

[F-] = ((10^-6) / (1.53*10^-2))^0.5 = 0.008084 M

this is the concentration in which Ba+2 will be preciptiate

now...

Get Ca+2 concentration for this case:

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

Assume all Ca+2 preciptiate initially ( this is for math)

initially:

[Ca+2] = 0

[F-] = 0.008084

in equilibrium

[Ca+2] = 0 + x

[F-] = 0.008084 - 2x

substitute in Ksp

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

5.3*10^-9 = x ( 0.008084 - 2x) ^2

solve for x

assume 0.008084 - 2x is too small so 0.008084

5.3*10^-9 = x ( 0.008084)^2

x = 5.3*10^-9 /(( 0.008084)^2))

x = 0.00008110044 M

so

[Ca+2] = 0.0000811004 M