Consider a solution that is 2.5×10−2 M in Fe2+ and 1.1×10−2 M in Mg2+. (Ksp for FeCO3 is 3.07×10−11 and Ksp for MgCO3 is 6.82×10−6.) If potassium carbonate is used to selectively precipitate one of the cations while leaving the other cation in solution, which cation will precipitate first?

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

i believe it is 40–40–20%

explanation: