[erhernandez 02]

PbSO4 is mixed with KI to produce PbI2 precipitate.

Solid PbI2 is mixed with Na2CO3. The products should be PbCO3 and NaI (as free floating I believe).

The question is, in what order will precipitate form with Pb?

With ksp values I can tell which is more insoluble and most likely to form a precipitate. But the problem doesn't give ksp values. Ihe answer to the question is CO3-, I-, SO4-2.

All the poly atomic ions are insoluble with lead, but how can one tell the order of solubility?

I believe it's because when mixed and broken into separate ions, precipitate will form from the most insoluble, hence the order...but sounds a bit too simple or over simplified. Thoughts?

[Borek]

PbSO4 is mixed with KI to produce PbI2 precipitate.

Which means in the presence of both SO₄^2- and I^- iodide would precipitate first, doesn't it? (that is, assuming similar concentrations of both ions, not differing by many orders of magnitude).

[AWK]

PbSO₄ is mixed with KI to produce PbI₂ precipitate.

Solid PbI₂ is mixed with Na₂CO₃. The products should be PbCO₃ and NaI

From the information above the conclusion is correct.
But this information is not exactly correct. Reactions of solutions with solids (PbSO₄ is also solid) are very slow and need much time for equilibrating. You can only observe a traces of PbI₂ in the first reaction because the color of iodide is very intensive but it does not mean that solubility of lead iodide is lower than PbSO₄.
From the same point of view the second reaction proceeds though the extent of reaction is invisible for a very long time (PbCO₃ is colourless).