Question

A solution contains 1.39×10^-2 M barium acetate and 1.28×10^-2 M calcium nitrate.
Solid ammonium sulfite is added slowly to this mixture.

What is the concentration of calcium ion when barium ion begins to precipitate?
[Ca²⁺] =  M

Answer 1

First, identify Ksp:

Ksp of BaSO3 = 5.0×10-¹⁰

Ksp of CaSO3 = 6.8×10^–8

now..

BaSO3 <-< Ba+2 + SO3-2

Ksp = [Ba+2][SO3-2]

CaSO3 <-> Ca+2 + SO3-2

Ksp = [Ca+2][SO3-2]

calculate solubility:

5.0×10-¹⁰ = S*S

S = sqrt(5*10^-10) = 2.2360*10^-5 M of BaSOs...

so BASOs preicpitates first

calculate SO3-2

Ksp = [Ba+2][SO3-2]

(5*10^-10) = (1.39*10^-3)*S

S = (5*10^-10)/ (1.39*10^-3) = 3.5971*10^-7

then, for Ca+2:

CaSO3 :

Ksp = [Ca+2][SO3-2]

6.8*10^-8 = *[CA+2][3.5971*10^-7]

Ca+2 = (6.8*10^-8)/(3.5971*10^-7) = 0.1890