Sciencemadness Discussion Board

Copper Hydroxide Problem

TheNerdyFarmer - 6-11-2016 at 19:15

I have been attempting the seemingly simple synthesis of Copper hydroxide but it keeps failing. I always end up getting a black or dark green precipitate when I do it. The reaction seems pretty temperature sensitive because when it is warm it turns black and when room temperature it seems to turn blue then to a dark green.

What I assumed at first was the copper sulfate I was using (root killer). So I went through the whole process of crystalizing a bunch out but it still failed.

Then I tried to cool my reactants down because it was a pretty exothermic reaction. So then the copper hydroxide started blue but ended up a dark greenish black. I then thought that it was oxidizing to copper oxide (seeing as copper hydroxide turns to copper oxide over time) but couldn't think of a reason why.

Could someone please help show me what I'm doing wrong.

DraconicAcid - 6-11-2016 at 20:15

If you're just adding sodium hydroxide to a solution of copper(II) sulphate, you're probably getting a basic sulphate.

Neme - 6-11-2016 at 21:29

Nile red's video: https://www.youtube.com/watch?v=YniCGw-rqmU
If I remember correctly, he had same issue so he used other route via some NH4OH.

Fulmen - 7-11-2016 at 02:23

Edit:

Sorry, I was under the impression that Cu(OH)2 didn't exist as a pure substance. Wikipedia seems to disagree.

My best advice is to use cold reagents and not an excess of lye, this promotes the conversion into the oxide.

[Edited on 7-11-16 by Fulmen]

TheNerdyFarmer - 7-11-2016 at 03:56

Thanks for the replies. I think I'm gonna try nile red's method and see if that works for me. The book I was going by says to simply make your solutions and then dump them in with each other. Anyway thanks for the help. :)

byko3y - 7-11-2016 at 06:08

You need either ammonia or glycerol - both serve to complex the copper and suppress its dehydration. You can't make a good quality Cu(OH)2 in a pure water. Neither it is sold in a pure form: 32733 Copper(II) hydroxide, tech. 94%, stab.
PS: soluble phosphate can also be used for stabilization, most likely it is the one that is used nowadays.

[Edited on 7-11-2016 by byko3y]

woelen - 7-11-2016 at 06:54

Glycerol does not sound like a good alternative to me. It forms highly soluble complexes which do not precipitate and besides that, there will be slow internal oxidation/reduction of the complex, resulting in red copper(I) oxide and some oxidized glycerol.

With ammonia you could do this, but be sure to use excess copper sulfate. If you use excess ammonia, then you get the highly soluble royal blue cuprammine complex of copper(II).

Copper(II) hydroxide is a very unstable compound and very easily loses water:
Cu(OH)2 --> CuO + H2O

Even when stored under water it easily loses water. This causes the change from blue to black when you try to heat or dry the copper(II) hydroxide. I have seen procedures for preparation and isolation of high purity copper(II) hydroxide, but these involved the use of vacuum for drying. The process requires the use of cold dilute solutions of copper sulfate and cold dilute ammonia (or dilute sodium hydroxide) and allowing the material to settle at the bottom. These solutions should be free of carbon dioxide or carbonate.
You then can rinse with a lot of water and then the vacuum filtered slurry must be allowed to dry under vacuum in the cold. Drying in air does not work. It will absorb CO2 from air and become green/cyan instead of pure blue, due to formation of a basic carbonate.

wg48 - 7-11-2016 at 07:51

I have never noticed Cu(2)OH2 decomposing though I have only seen it on the way to other reactions.

The following link confirms its instablity and to me its exotic mechanisms.

http://www.sciencedirect.com/science/article/pii/S1293255803...

"Copper hydroxide Cu(OH)2 is metastable. It easily transforms into copper oxide CuO more stable, either in the solid state by a thermal dehydration or at room temperature, in aqueous basic solutions. In the solid state, the transformation is performed at a relatively low temperature, 423 K. It is a topotactic or a pseudomorphic transformation owing to clear relationships between axes of the two solids, in the three directions. The reacting process is described and the corresponding vectorial relations between crystal parameters are proposed. It is not the same case in aqueous basic solutions. Copper hydroxide gives rise to oxide through the formation of a complex anion, Cu(OH)42−, by a reconstructive transformation involving a dissolution reaction followed by a precipitation."

I had to look up topotactic and pseudomorphic LOL

DraconicAcid - 7-11-2016 at 08:53

In one of our labs, we have our students make Cu(OH)2 (or some variant thereof) and then decompose it to the oxide. Excess hydroxide causes it to decompose much more readily (if you use exactly 1:2 CuSO4:NaOH, it stays blue no matter how long it's left in boiling water).

I have read that to get a filterable, light-blue Cu(OH)2, you have to add dilute ammonia, without adding enough to get the complex ion.

byko3y - 7-11-2016 at 09:09

woelen, I've being storing my Cu(OH)2 prepared with 1% glycerol as stabilizer for 6 month already, and it's still the same blue solid as it was when just prepared (I might have used slightly more glycerol probably, that's why it's more stable than the one with 1% stabilizer).
I see no way for a solid copper hydroxide to oxidize glycerol at room temperature without some kind of catalyst (radical initiator, additional inert solvent as a reaction medium, etc).

woelen - 7-11-2016 at 13:01

I have quite different experience with glycerol in combination with copper(II) salts in alkaline environments. However, I used more concentrated solutions in my experiments, with excess base. In such solutions, the copper(II) forms a soluble deep blue complex, which slowly decomposes, giving brick-red copper(I) oxide.

ave369 - 10-11-2016 at 08:26

I thought Cu(OH)2 only exists as a wet goop, and decomposes to CuO if you try to dry it.

byko3y - 10-11-2016 at 09:35

Already discussed above - your assumption is not completely correct.