Hi, I'm self-studying for the MCATs and I came across this question that isn't making sense to me. Any help is appreciated.
The Passage basically lists a series of reactions where the precipitates PbSO
4, PbI
2, and PbCO
3 are formed (in that order).
The question:A soluble form of Pb
2+ can be carefully added to a solution to sequentially precipitate and separate anions present in the solution. When Pb
2+ is added, what order will the following anions be precipitated?
The answer:The rxns described in the passage show that lead(II) is successively precipitated as PbSO
4, PbI
2, and PbCO
3. This sequence shows (assuming equal anion concentrations as must be done here) that PbCO
3 is less soluble than PbI
2, and PbI
2 is less soluble than PbSO
4. The order in which the anions precipitate Pb
2+ is: CO
32- then I
- then SO
42-.
My question:I don't understand where they get that the order of precipitation is CO
32- then I
- then SO
42-. Why isn't it SO
42- then I
- then CO
32- (because that's the order the precipitates were formed during the rxn series)??