November 27, 2024, 05:38:12 PM
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Topic: What fraction of all lead species will be present as Pb^2+ at equilibrium?  (Read 5033 times)

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Offline bucky91

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A water sample contains 1.0 x 10^-4 M of Pb(II) and 0.75 M Cl-. Assuming that no reactions other than that between lead and chlorine occur, what fraction of all of the lead species will Pb^2+ be at equilibrium?

Please help...I've been stuck for ages and have no idea what to do :S

Offline sjb

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A water sample contains 1.0 x 10^-4 M of Pb(II) and 0.75 M Cl-. Assuming that no reactions other than that between lead and chlorine occur, what fraction of all of the lead species will Pb^2+ be at equilibrium?

Please help...I've been stuck for ages and have no idea what to do :S

Solubility product?

Offline bucky91

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...?

Offline Arkcon

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This problem will require you to use the solubility product, from a table in your book or elsewhere, or given to you in another part of this problem.  Look here, there are some solved practice problems:  http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch18/ksp.php
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

Offline bucky91

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Thanks!

This is what I've done so far...no idea if it's right or wrong though :s

Pb2+ + Cl- <---> PbCl+

PbCl+ + Cl- <---> PbCl2

PbCl2 + Cl- <---> PbCl3-

PbCl3- + Cl- <---> PbCl42-

For the 1st equation: K1 = [PbCl+]
                                    [Pb2+][Cl-]

The questions says that: [Pb2+] = 1.0 x 10-4 M
                                     [Cl-] = 0.75 M

This site: http://chemed.chem.wisc.edu/chempaths/GenChem-Textbook/The-Solubility-Product-622.html  gives the solubility product for PbCl2 as: 1.6 × 10-5. So would this be K2? How do I find K1 then? Or am I totally off track here...

I have such a hard time with all this K stuff :(



Offline sjb

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Thanks!

This is what I've done so far...no idea if it's right or wrong though :s

Pb2+ + Cl- <---> PbCl+

PbCl+ + Cl- <---> PbCl2

PbCl2 + Cl- <---> PbCl3-

PbCl3- + Cl- <---> PbCl42-

For the 1st equation: K1 = [PbCl+]
                                    [Pb2+][Cl-]

The questions says that: [Pb2+] = 1.0 x 10-4 M
                                     [Cl-] = 0.75 M

This site: http://chemed.chem.wisc.edu/chempaths/GenChem-Textbook/The-Solubility-Product-622.html  gives the solubility product for PbCl2 as: 1.6 × 10-5. So would this be K2? How do I find K1 then? Or am I totally off track here...

I have such a hard time with all this K stuff :(

No, Ksp refers to Pb2+(aq) + 2Cl-(aq)  ::equil:: PbCl2(s); so a combination of K1 and K2

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