What is the minimum concentration?

Question

An aqueous solution at room temperature is 0.25 M in both Ba²⁺ and Ca²⁺ ions. You want to separate these two ions by selective precipitation using carbonate. What is the minimum concentration of carbonate required to start the SECOND compound precipitating? (BaCO₃ Ksp = 2.58 × 10⁻⁹ M, CaCO₃ Ksp = 3.36 × 10⁻⁹ M).

Answer 1

A). Use the Law of Mass Action to use the Ksp values given.

1. From the two Ksp reactions, we use only the products cation metal and the concentration of the carbonate anion. The solid reactants are not used.

For BaCO3:

Ksp = 2.58 x 10^-9 = [Ba⁺²_(aq)] * [CO₃^-2_(aq)]

For CaCO3: 

Ksp = 3.36 x 10^-9 = [Ca⁺²_(aq)] * [CO₃^-2_(aq)]

B). Calculate the equilibrium concentrations of carbonate, [CO₃^-2].

For BaCO₃:

Ksp = 2.58 x 10^-9 = [Ba⁺²_(aq)] * [CO₃^-2_(aq)]

[CO₃^-2_(aq)]_BaCO3 = Ksp / [Ba⁺²_(aq)]

[CO₃^-2_(aq)]_BaCO3 = (2.58 x 10^-9) / (0.25 M) = 1.0 x 10^-8 M

For CaCO₃:

Ksp = 3.36 x 10^-9 = [Ca⁺²_(aq)] * [CO₃^-2_(aq)]

[CO3^-2_(aq)]_CaCO3 = Ksp / [Ca⁺²_(aq)]

[CO₃^-2_(aq)]_CaCO3 = (3.36 x 10^-9) / (0.25 M) = 1.3 x 10^-8 M

C). Analyze the results

The concentration of carbonate needed to precipitate BaCO3 is less than the concentration needed to precipitate CaCO₃. Therefore, BaCO3 will appear before CaCO₃.

So… which is the second compound to precipitate? CaCO₃. At what concentration of carbonate will it precipitate? 1.3 x 10-8 M.