This will probably bore everybody to death, but I don't know how to figure this out.
My intent was to carry out a simple double replacement between CuSO4 and NaOH to give Cu(OH)2 and Na2SO4. i placed 24.97g Copper II sulfate
pentahydrate (0.16 moles) in a beaker and added 100 mL dH2O. To fully dissolve the salt, I placed the solution on a hotplate and stirred periodicaly
as it heated.
In another beaker, I combined 8.00g (0.2 moles) NaOH with 10 mL dH2O. I was supposed to pour the basic solution into the copper sulfate soln w/
constant stirring, but I just poured a little in. a BLACK (not the expected light blue) precipitate formed. When I poured some more in, the expected
light blue copper hydroxide precipitate formed. On stirring, however, the mixture grew warmer and became a heavy black sludge.
Filtering gave a slightly cloudy, but basically clear filtrate, which I am evaporating now. It is yielding a whitish residue, but not enough to
account for a significant amount of the starting reactants.
The only departures I took from the specified procedure were to heat the blue CuSO4 soln to facillitate complete dissolution, and that I was not
initially stirring the soln as I added the NaOH. As the hydroxide and sulfate anions have charges of -1 and -2, respectively, and the sodium and
copper cations are likewise disparate in charge, I can't see the possibility if hetero anion or hetero cation salts such as Na(Cu)SO4 or Cu(OH)SO4, or
any other such combinations. There should only be light blue copper hydroxide and white sodium sulfate, (with a little unreacted CuSO4 remaining as
the NaOH was the limiting reagent).
Can anybody tell me what this black shit is?
[Edited on 27-9-2009 by birdman]entropy51 - 27-9-2009 at 13:14
Cupric oxide. You can dissolve it in acetic acid to make cupric acetate. Using cold solutions give the hydroxide. Heating turns the hydroxide to
oxide.
[Edited on 27-9-2009 by entropy51]birdman - 27-9-2009 at 13:31
Thank you!
I might have scrapped that lab and given up. I'm new at chemistry and get frustrated easily. Your answer reassures me that I am at least paying
close enough atttention to realize where I most likely went wrong. Without your help, however, I probably would have given up and moved onto the next
lab b4 ever reading enough about Cu(OH)2 to figure that out.entropy51 - 27-9-2009 at 13:43
Always read up on the chemicals and reactions you are using, preferably before the experiment.
If you have no chemistry book, find one online, or if really desperate, use Wikipedia, which lists copper(II) hydroxide, but I couldn't paste the link
for some reason.
[Edited on 27-9-2009 by entropy51]birdman - 27-9-2009 at 14:17
Yeah,
I went to wikipedia and read the copper II hydroxide entry after reading your initial reply. Thanks again!birdman - 27-9-2009 at 18:03
Any advice on recovering dry copper II hydroxide for purposes of stoichiometric determinations w/o its decomposition to the oxide? I'm not having much
luck with that and I'm almost out of copper sulfate. Just letting it sit results in cupric oxide and gentle heating just speeds that process up.
I'm supposed to dry the product to weigh it and determine the reaction's agreement with its theoretical stoichiometry and then combine it with
aqueous ammonia to make Schweizer's Reagent. I suspect the person who came up with this lab didn't try it himself first. I suspect I won't be able to
do this in an aerobic environment.not_important - 27-9-2009 at 19:08
Any advice on recovering dry copper II hydroxide...and then combine it with aqueous ammonia ...I suspect I won't be able to do this in an aerobic
environment.
Wash the Cu(OH)2 with water twice, then alcohol, then twice with ether or acetone - cold washes in every case.
The moist hydroxide reacts much more quickly with aqueous ammonia.
Cu(II) is stable in air under most conditions, you don't need to worry about working in an aerobic setting.
Schweizer's Reagent and Le Chatelier
trb456 - 14-2-2011 at 12:18
Sorry to revive this old thread, but I am doing this very experiment to make Schweizer's reagent.
Does anyone have a reference, or just knows, as to whether there is a big Le Chatelier effect with NH3 reacting with Cu(OH)2 to get Cu(NH3)4(OH)2 ? I
used an excess of aqueous NH3, and I noticed that a lot of unreacted Cu(OH)2 remained. But as I draw off Cu(NH3)4(OH)2, more is formed. I measured
pretty carefully at the start to get 0.1 mole of product, so I'm curious at to whether it will keep reacting until mostly NH4OH remains.
P.S. If I can get the Cu(NH3)4(OH)2 to evaporate, I'll try and post some crystal pictures!
