Question

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?

Answer 1

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)

Step-by-step 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