- #1
steckona
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This question is from a practice MCAT and I am very confused about the logic in the answer they provide. I would really appreciate any help.
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, in which order will the following anions be precipitated?"
Ans:
"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 Pb^2+ is: CO3^-2 then I- then SO4^-2"
- I see that PbCO3 must be the least soluble because it precipitates last but then how does it precipitate Pb^2+ first?
Thank-you so much!
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, in which order will the following anions be precipitated?"
Ans:
"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 Pb^2+ is: CO3^-2 then I- then SO4^-2"
- I see that PbCO3 must be the least soluble because it precipitates last but then how does it precipitate Pb^2+ first?
Thank-you so much!
