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Topic: order of solubility without using ksp  (Read 1743 times)

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Offline erhernandez 02

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order of solubility without using ksp
« on: October 22, 2013, 11:34:59 PM »
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?


Offline Borek

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Re: order of solubility without using ksp
« Reply #1 on: October 23, 2013, 03:03:01 AM »
PbSO4 is mixed with KI to produce PbI2 precipitate.

Which means in the presence of both SO42- and I- iodide would precipitate first, doesn't it? (that is, assuming similar concentrations of both ions, not differing by many orders of magnitude).
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Offline AWK

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Re: order of solubility without using ksp
« Reply #2 on: October 23, 2013, 06:36:02 AM »
Quote
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).
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