Answer:
This question is incomplete, here's the complete 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)
Explanation:
Part A
Fe2+ will precipitate first as solubility product of FeCO3 is lesser than solubility product of MgCO3.
Part B
FeCO3\= Fe2+ + CO32-
Ksp = [Fe2+][CO32-] = 3.07 x 10-11
3.07 x 10-11 = (2.1 x 10-2)[CO32-]
[CO32-] = 1.5 x 10-9 M to precipitate the Fe2+ ions
Part C
MgCO3\= Mg2+ + CO32-
Ksp = [Mg2+][CO32-] = 3.07 x 10-11
6.82 x 10-6 = (1.6 x 10-2)[CO32-]
[CO32-] = 4.3 x 10-4 M to precipitate the Mg2+ ions
Ksp = [Fe2+][CO32-] = 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
