AAMC #3 Passage 1 question

[mangos]

there is this one question in AAMC #3 that i don't understand. its in the first passage. here is the answer

 

The reactions described in the passage show that lead(II) is successively precipitated as PbSO4,

PbI2, and PbCO3. This sequence shows (assuming equal anion concentrations, as must be done

here) that PbCO3 is less soluble than PbI2, and PbI2 is less soluble than PbSO4. The order in which

the anions precipitate Pb2+ is: CO3

2- then I- then SO4

2-. When this sequence is applied to the

question, answer choice B is in the correct order, and answers A, C, and D are all in the opposite

order. Thus, answer choice B is the best answer.

 

what i dont' get is if PbSO4 precipiates first, then how is it more soluable?

 

thanks