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Hi SDNers,
I have a question about the first passage in the PS section of AAMC 3. In this passage, a series of chemical reactions are carried out to study the chemistry of lead.
In reaction 1, PbSO4(s) is produced. In reaction 2, PbI2(s) is formed. In reaction 3, PbI2(s) is converted into PbCO3(s).
Question 4 asks: "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, in what order will the following anions be precipitated?"
A. SO4(2-) then I-
B. CO3(2-) then I-
C. SO4(2-) then CO3(2-)
D. I- then CO3(2-)
The correct answer is B. AAMC says that "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 B is in the correct order, and answers A, C, and D are all in the opposite order. Thus, answer B is the best answer.
How was I supposed to infer from the passage that that is the answer? Is it because they made the solids and then dissolved them to create the following solid in that order. Let me know how I should have been able to figure this out
Thank you for all your help!
I have a question about the first passage in the PS section of AAMC 3. In this passage, a series of chemical reactions are carried out to study the chemistry of lead.
In reaction 1, PbSO4(s) is produced. In reaction 2, PbI2(s) is formed. In reaction 3, PbI2(s) is converted into PbCO3(s).
Question 4 asks: "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, in what order will the following anions be precipitated?"
A. SO4(2-) then I-
B. CO3(2-) then I-
C. SO4(2-) then CO3(2-)
D. I- then CO3(2-)
The correct answer is B. AAMC says that "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 B is in the correct order, and answers A, C, and D are all in the opposite order. Thus, answer B is the best answer.
How was I supposed to infer from the passage that that is the answer? Is it because they made the solids and then dissolved them to create the following solid in that order. Let me know how I should have been able to figure this out
Thank you for all your help!
