Question

Consider a solution that is 2.1×10⁻² M in Fe2+ and 1.6×10⁻² M in Mg2+.

Part A

If potassium carbonate is used to selectively precipitate one of the cations while leaving the other cation in solution, which cation will precipitate first? ANSWER: Fe 2+

Part B

What minimum concentration of K2CO3 is required to cause the precipitation of the cation that precipitates first? ANSWER: [K2CO3] = 1.5×10⁻⁹ M

Part C

What is the remaining concentration of the cation that precipitates first, when the other cation just begins to precipitate? (ANSWER IS NOT .021 or 2.0E-6)

Answer 1

Part A

Fe²⁺ will precipitate first as solubility product of FeCO₃ is lower than solubility product of MgCO₃.

Part B

FeCO₃ Fe²⁺ + CO₃^2-
K_sp = [Fe²⁺][CO₃^2-] = 3.07 x 10^-11
3.07 x 10^-11 = (2.1 x 10^-2)[CO₃^2-]
[CO₃^2-] = 1.5 x 10^-9 M to precipitate the Fe²⁺ ions

Part C

MgCO₃ Mg²⁺ + CO₃^2-
K_sp = [Mg²⁺][CO₃^2-] = 3.07 x 10^-11
6.82 x 10^-6 = (1.6 x 10^-2)[CO₃^2-]
[CO₃^2-] = 4.3 x 10^-4 M to precipitate the Mg²⁺ ions

K_sp = [Fe²⁺][CO₃^2-] = 3.07 x 10^-11
3.07 x 10^-11 = [Fe²⁺](4.3 x 10^-4)
[Fe²⁺] = 7.1 x 10^-8 M when the Mg²⁺ ions precipitates