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Something weird I've noticed wiht Cu(OH)2


xeluc
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So I threw some CuCl2 solution in NaOH solution. I got what I wanted. However, something odd has happened. I add enough CuCl2 until no more Precipitate forms. Then I stir the stalagtite-looking things of Cu(OH)2 and the precipitate turns A LOT lighter. The addition of more NaOH darkens the precipitate. I'm really not sure of any other precipitate that could be formed and it is unlikely that when you stir the Cu(OH)2 it spontaniously reacts, but how then can you explain more NaOH darkening the precipitate to near it original darkness? Any impurities are very minimal.

 

One thing I thought of was partical size. Cu2O can be lots of different colors depending on partical size, but I don't think that stirring would create small enough particals? Also, that doesn't explain the darkening after adding NaOH solution. What sort of reaction is going on here?

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Ok, I tried the reaction and got similar results. I add drops of CuCl2 (aq) to NaOH (aq) and the blue precipitate forms, and after a while the CuCl2 doesn't react anymore. However, I swished the test tube around and the precipitate dissolved, leaving a green solution (lighter than the CuCl2 soln). So I add some more NaOH, and I get a black precipitate.

 

What is going on?

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The black prec. is the hydroxide losing water and becoming Oxide. This happens from heat produced when you disolve NaOH in water (Basicly ANY heat, but this is usually the culprit This stuff is finicky and reverts to CuO very easily). When I said darken, I meant that it turns back to the darkish blue prec. that I started with, not to black, I understand that process. I do appreciate you conducting this experiemnt however.

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Cuprate ions are only formed at VERY high concentration of NaOH, so I do not think it is that. I'll try the experiment as well. It might be that a basic chloride is precipitated and that that gives rise to the special colors observed. If more NaOH is added, an ion exchange could occur (chloride in the precipitate, replaced by hydroxide ions). More to be said on this, when the experiment is done :).

 

EDIT: I did the experiment as follows:

 

Take a solution of NaOH and add solution of CuCl2, while stirring/shaking after each few drops. At first, a nice bright blue precipitate is obtained, fairly dark. At a certain point, however, the color becomes lighter. The precipitate becomes lighter green/blue.

 

Then, I added NaOH solution again, while shaking all the time. This results in darkening and shifting from green/blue to bright blue again, until a certain point is reached. From that point on, the color does not shift anymore.

 

I explain this as follows:

 

Initially, there is NaOH in excess, and addition of the CuCl2-solution causes formation of bright blue Cu(OH)2. At a certain point, when all NaOH is used up, then addition of more CuCl2 causes formation of basic copper chloride, something like CuCl(OH), or more general [math]CuCl_{2-x}(OH)_x[/math].

 

When lateron, more NaOH is added again, then the basic copper chloride looses its Cl(-) ions again and Cu(OH)2 is formed again.

 

---------------------------------------------------------------

 

Now, to you observations. I think that your liquid was very uneven, with unused NaOH below and unused CuCl2 at the top. At the top, there is excess CuCl2 and then you see the basic copper chloride. But, when you shake, then the unused NaOH reacts with the excess and basic copper chloride, forming Cu(OH)2.

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surely when all the chlorine rects with the sodium to form sodium chloride, it becomes a non reactant and can more or less be writen out the equasion entirely, leaving sodium hydroxide and copper hydroxide as the only 2 "Players" in the game?

from this point the only possible reaction would be to make the Cuprate of sodium and water.

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No, the chloride (not chlorine) certainly plays an important role. It cannot be cancelled out with sodium ion. Chloride forms complexes with copper.

 

Just try dissolving some CuSO4 in plain water and dissolve in a solution of NaCl (fairly concentrated NaCl). You'll definitely see a difference!

 

Next, when you add NaOH, then a complicated precipitate is formed, which can be regarded as hydrated copper hydroxide/chloride. This is not the same as a mix of copper hydroxide and copper chloride, but it is a so-called berthollide compound.

 

Berthollide --> indefinite stoichiometry, no precise formula can be given, only a formula like I give above with an uncertain parameter x.

Daltonide --> definite stoichiometry (e.g. CuCl2, CuSO4, many other compounds).

 

The formation of berthollide compounds is an annoying thing in chemistry, but unfortunately it is more common than purely daltonide compounds.

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only because sodium sulphate is also quite a stable compound.

but in this, the Cl as NaCl should be the more stable of all the possibles?

 

here`s how I see it, 2NaOH +CuCl2 = 2NaCl + Cu(OH)2

and then it`s done.

 

further addition of NaOH to the Cu(OH)2 will serve to form the Cuprate surely!?

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Basically I agree with your net reaction, but the underlying mechanism is MUCH more complex. I do not think that if you add carefully stoichiometrically computed amounts of NaOH and CuCl2 to each other, that you end up with a precipitate of Cu(OH)2 and a solution of NaCl, but that you end up with a berthollide compound CuOH/Cl and still NaOH in solution.

 

This experiment does not have anything to do with stability of NaCl or Na2SO4. These compounds simply do not exist in aqueous solution, you only have ions Na(+) and ions Cl(-). The Na(+) ions just are spectator ions, but the Cl(-) ions are not. They do very complicated things in the reaction.

 

Only when there is quite some excess NaOH, the chloride-content of the precipitate will become very low and one can speak of Cu(OH)2.

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  • 3 weeks later...

Cu(OH)2 is bright blue. If you use an excess amount of CuSO4, then you'll get pale blue basic copper sulfate, Cu(OH)x(SO4)y, with 0.5x+y=1.

 

On my website I have pictures of the precipitate:

 

http://woelen.scheikunde.net/science/chem/solutions/cu.html

 

Look at the section for oxidation state +2. The bright fairly dark material is Cu(OH)2, the lighter material is basic copper sulfate.

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