[blueberries]

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 PbSO₄, PbI₂, and PbCO₃ are formed (in that order).

The question:
A soluble form of Pb²⁺ can be carefully added to a solution to sequentially precipitate and separate anions present in the solution. When Pb²⁺ 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 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-.

My question:
I don't understand where they get that the order of precipitation is CO₃^2- then I^- then SO₄^2-. Why isn't it SO₄^2- then I^- then CO₃^2- (because that's the order the precipitates were formed during the rxn series)??

[XGen]

If there is no prior background information given, such as solubility constants, it is very hard to judge relative solubility between compounds deemed "insoluble". I only know because of intuition.

Using solubility constants, it is a matter of taking square roots and cube roots.

[PIQgoogleme]

Sounds like a poorly worded question...

but I think "When Pb2+ is added, what order will the following anions be precipitated?" is their way of saying "which of the anions will yield the most precipitate?"

Since the passage it listed I^- as replacing SO₄^2-, and CO₃^2- as replacing I^-, we know that the carbonate is the favored precipitate, and thus it will form in greatest quantity.

[XGen]

I am confused because I fail to see how lead iodide precipitates before lead sulfate. After comparing the solubility constants, I see that lead sulfate is less soluble, and therefore should precipitate first. Perhaps this is an error?

[gertrudetrumpet]

The reaction is probably with lead ions in solution. Then sulfate is added. Then iodide is added. Then carbonate is added. They are not separate reactions I think, but in the same container.

[Borek]

The way I read it you have a solution containing all three anions and you slowly add lead, precipitating each anion separately. That would precipitate carbonate first, sulfate second, iodide third.

But I agree wording is confusing and not clear.

[blueberries]

There were no solubility constants given. I kind of agree with PIQgoogleme in that I think the question is poorly worded, but asking which anion will yield the most precipitate. No solubility constants were given.

Here is all the information given in the passage, if it helps:

A series of chemical rxns were carried out to study the chemistry of lead.

Rxn 1
Initially 15.0 mmL of 0.300 M Pb(NO₃)₂(aq) was mixed with 15.0 mL of .300 M Na₂SO₄(aq). All the Pb(NO₃)₂ reacted to form Compound A, a white precipitate. Compound A was removed by filtration.
(I established that Compound A is PbSO₄)

Rxn2
Next, 15.0 mL of 0.300 M KI(aq) was added to Compound A. The mixture was agitated and some of Compound A dissolved. In addition, a yellow precipitate of PbI₂(s) was formed.

Rxn3
The PbI₂(s) was separated and mixed with 15.0 mL of 0.300 M Na₂CO₃ (aq). A white precipitate of PbCO₃(s) formed. All of the PbI₂(s) was convered into PbCO₃(s).

Rxn4
The PbCO₃(s) was removed by filtration and a small sample gave off a gas when treated with dilute HCl.

[Borek]

which anion will yield the most precipitate

No, it doesn't ask about yields, but about order:

what order will the following anions be precipitated?

Without any further analysis from the third reaction it is obvious carbonate will precipitate before iodide. The only thing that is not clear at the first sight is when will the sulfate precipitate.

[blueberries]

Ok, but why is it clear that carbonate will precipitate first? I don't understand where the order is coming from.

[Borek]

All of the PbI₂(s) was convered into PbCO₃(s).

That clearly means carbonate has much lower solubility than iodide.

Ok, but why is it clear that carbonate will precipitate first?

First of iodide and carbonate, not necessarily of all three.

Actually it will precipitate first of all three as well, but I know if it after checking with K_sp values.

[blueberries]

Ok, I understand now why Carbonate is first, don't really understand sulfate as much, but I'm ok with that. It's just one practice question.

Thanks for all your help everyone!

[boobyrayray]

It's all in the wording. Pay close attention to how they hint what the solubility is. From A to PbI2, the passage says that only some of Compound A dissolved. From PbI2 to PbCO3, the passage says that all of the PbI2 was converted.

If all the PbI2 was converted to PbCO3, we know that Ksp is favoring the production of PbCO3, and therefore PbCO3 is less soluble and will favorably precipitate.

If only some of compound A dissolved to form PbI2, this is an indication that Ksp is not as favorable towards the production of PbI2. This reaction favors Compound A as a ppt over PbI2, meaning that PbI2 is more soluble.

In conclusion, CO3 2- is least soluble and will ppt first, followed by I- and SO4 2- respectively.