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Topic: Chem Lab ID of unknowns  (Read 7652 times)

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Offline ldixon

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Chem Lab ID of unknowns
« on: October 06, 2006, 10:19:00 PM »
I had to find 5 unknown solutions in a recent chem lab and now I am doing the equations, but im not sure about two of them:

Deep blue percipitate
NH3 (aq) + CuSO4 (aq) -------> Cu(NH3)4 SO4 (s)

brown heavy percipitate
4NaI (aq) + 2CuSO4 (aq) -------> 2CuI (s) + I2 (s) + 2Na2SO4 (aq)

Are these reactions correct?  I got percipitates for both of them.

Offline ldixon

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Re: Chem Lab ID of unknowns
« Reply #1 on: October 09, 2006, 09:35:37 PM »
can anyone help me with this one? 

Offline Borek

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Re: Chem Lab ID of unknowns
« Reply #2 on: October 10, 2006, 02:44:15 AM »
First is wrong. Cu(NH3)42+ is what is produced when your precipitate dissolves in excess ammonia.
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Offline AWK

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Re: Chem Lab ID of unknowns
« Reply #3 on: October 10, 2006, 03:56:54 AM »
I had to find 5 unknown solutions in a recent chem lab and now I am doing the equations, but im not sure about two of them:

Deep blue percipitate
NH3 (aq) + CuSO4 (aq) -------> Cu(NH3)4 SO4 (s)

brown heavy percipitate
4NaI (aq) + 2CuSO4 (aq) -------> 2CuI (s) + I2 (s) + 2Na2SO4 (aq)

Are these reactions correct?  I got percipitates for both of them.
Stoichiometry of reaction is correct. In the first case you got a deep blue solution. In the second case - iodine forms a brown precipitate, CuI is almost white.
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Offline ldixon

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Re: Chem Lab ID of unknowns
« Reply #4 on: October 10, 2006, 12:25:29 PM »
Borek would the first one just be:

NH3 (aq) + CuSO4 (aq) -------> Cu(NH3)2 (aq) + SO4 (aq)

Is the second one correct?

Offline Borek

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Re: Chem Lab ID of unknowns
« Reply #5 on: October 10, 2006, 01:53:24 PM »
Write correct charges. 4 not 2 ammonia molecules.
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Offline ldixon

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Re: Chem Lab ID of unknowns
« Reply #6 on: October 10, 2006, 02:37:51 PM »
So it would be:
4NH3(aq) + CuSO4(aq) --------> Cu(NH3)4(aq) + SO4(aq)

Would this be a percipitate? 

Offline Borek

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Re: Chem Lab ID of unknowns
« Reply #7 on: October 10, 2006, 03:05:38 PM »
Charges are still wrong. Besides, CuSO4 was dissociated from the start, SO42- is just a spectator.
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Offline ldixon

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Re: Chem Lab ID of unknowns
« Reply #8 on: October 10, 2006, 03:20:13 PM »
Ok by charges you mean?
                                                           
4NH3-1(aq) + Cu+2SO4-2(aq) --------> Cu+4(NH3-1)4(aq) + SO4-2(aq)

Yes I understand that SO42- is a spectator but im not sure if this is a percipitate?

Or since SO4-2(aq) is a spectator it would just be:
4NH3-1(aq) + Cu+2(aq) --------> Cu+4(NH3-1)4(aq)


Offline Dan

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Re: Chem Lab ID of unknowns
« Reply #9 on: October 10, 2006, 03:26:48 PM »
Ammonia, (NH3) is neutral, and Cu+4 is wrong

remember, charges must balance
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Offline Borek

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Re: Chem Lab ID of unknowns
« Reply #10 on: October 10, 2006, 03:37:03 PM »
Precipitate is copper hydroxide, it gets dissolved when you add more ammonia. CuSO4 is well soluble. Sulfates of copper/ammonia complexes are well soluble as well.
« Last Edit: October 10, 2006, 03:45:08 PM by Borek »
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Offline ldixon

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Re: Chem Lab ID of unknowns
« Reply #11 on: October 10, 2006, 03:55:18 PM »
Is there a way to write an equation for this whole reaction?

If the charge on (NH3) is neutral then Cu would also be neutral?

Offline Borek

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Re: Chem Lab ID of unknowns
« Reply #12 on: October 10, 2006, 04:04:49 PM »
No, copper cation has a charge, as well as sulfate anion. Note: copper sulfate molecule is neutral, so the copper cation must have charge opposite to sulfate.
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