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CuCl2(Aq) + NaOH gone horribly wrong?


xeluc

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I figured that when I mixed these compounds that I would get a solution of NaCl and a Precipitate of Cu(OH)2. Instead, I have a blue solution, and a dark dark Grey Precipitate. Can anyone explain this? Undoubtably my CuCl2 is slightly contaminated, But only slightly. There is as much grey sediment as I put CuCl2, so I don't think this was due to contaminents.

 

(An Excess of NaOH was used.)

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When you use a very concentrated solution of NaOH, then part of the copper hydroxide redissolves again. Copper is somewhat amphoteric and in concentrated NaOH the so-called cuprate ion is formed:

 

Cu(OH)2 + 2OH(-) ----> CuO2(2-) + 2H2O

 

Here the copper hydroxide acts as acid!

 

The cuprate ion is royal blue, it is not the sky-blue of copper (II) ions in water, but a deeper blue, more like indigo.

 

Another effect that may plague you is that Cu(OH)2 is not very stable. Even at moderate heat, it decomposes and looses water, even when it is in water:

 

Cu(OH)2 ---> CuO + H2O

 

This may explain why you get a dark grey precipitate.

 

Just try the following to see if my hypothesis is right:

 

Add excess acid to a suspension of the dark grey stuff. If it dissolves and the liquid becomes light sky-blue on dilution, then it indeed is CuO.

 

The following web-page may help you. Read the part about copper in oxidation state +2 and focus on the precipitates of copper hydroxide and the cuprate ion. Is the color of your solution similar to the dilute cuprate solution on the webpage?

 

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

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Yeah, all the colors look the same. My blue solution WAS a lot darker. after leaving the precipitate to settle, it lightened up considerably. It is definatly not a Cu2+ solution, the color is much darker. I did this experiment once before and got what I thought was Copper Hydroxide. It does turn dark after heating. But this turned dark right away. There was no heating involved. So how did the Cu(OH)2 Decompose?

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Yeah, all the colors look the same. My blue solution WAS a lot darker. after leaving the precipitate to settle, it lightened up considerably. It is definatly not a Cu2+ solution, the color is much darker. I did this experiment once before and got what I thought was Copper Hydroxide. It does turn dark after heating. But this turned dark right away. There was no heating involved. So how did the Cu(OH)2 Decompose?

Even when a solution is luke-warm, the Cu(OH)2 can decompose. It might be that you prepared the solution of NaOH from solid NaOH and that some heat of hydration was left in the liquid. Your CuCl2 also may contain a considerable amount of HCl as impurity, which with the hydroxide releases quite some heat. So, there are quite some possible explanations for your observations. So, indeed, your precipitate most likely is (impure) CuO.

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So If I cool all the reactants than I should get Cu)OH)2... right? Well, I jsut bought me some HCl :) I will do some testing tomorrow. Right now I have to worry about what I'm going to do about my radiator that jsut decided to take a shit and spill antifreeze everywhere...

 

 

EDIT: Also, would you know how to turn CuO into Cu2O? Maybe a strong Oxidizing agent? Hope that doesn't sound stupid, It's a guess.

 

EDIT 2: just realized, I wanna go DOWN an oxidation state. Now I have no idea. Reducer? lol.....

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If you use cold reactants and the CuCl2 is not too acidic, then I expect you to get green copper hydroxochloride Cu(OH)Cl. Getting really pure Cu(OH)2 is hard from CuCl2, because chloride ions coprecipitate with the hydroxide.

 

If you want Cu2O, then try the following:

 

Crunch a tablet of vitamin C (not the multivitamin), dissolve in water and filter away the white crud. The clear solution contains vitamin C. Add this solution to a solution of NaOH.

Now add a solution of CuCl2 to this NaOH/vitamin C solution. You will get a green precipitate, which turns orange or yellow. Assure that you have excess vitamin C. The Cu2O produced in this way is very flocculent and not that easy to separate from the liquid.

 

Another way of getting Cu2O (the red stuff) is dissolving Cu(OH)2 in excess solution of NaOH, to which tartaric acid is added. This gives a deep blue solution (google Fehling's reagent for more info). Add some glycerin or dextrose to this solution and let stand. After a few hours you have a nice compact precipitate of Cu2O, which easily is separated from the liquid.

 

Another nice experiment is making the dark blue tetrammine copper (II) complex, Cu(NH3)4(2+). This can be made by adding a solution of CuCl2 to a solution of ammonia. You need a large excess of ammonia.

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Thanks a lot. I added my grey stuff to some diluted HCl. At frist nothing happened. Then it rapidly cleared up into a crystal clear liquid. After doign this some more, the liquid turned bluish green. After boiling it turned green. Definatly Copper (II) Chloride.

 

Would Copper (I) Oxide Be turned into Copper (I) Chloride on addition of HCl or would the HCl further oxidize Cu+ to Cu2+

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Would Copper (I) Oxide Be turned into Copper (I) Chloride on addition of HCl or would the HCl further oxidize Cu+ to Cu2+

Suppose that HCl would oxidize Cu(+) to Cu(2+), how do you think that would happen with hydrochloric acid as the oxidizer? Think of this and try to find a possible equation. By thinking about this and attempting to find a reaction-equation, you'll gain a lot of insight. When you come back with an answer, I'll praise you or I'll correct you :D, depending on your smartness ;).

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rrrrrr If you have ONLY HCl and Cu+, I don't really see HOW it could happen, but that happens a lot, which is why I ask these questions... ;) I threw the equation together.. I found out that H+ isn't going to be taking any more electrons. Cl- Obviously won't. The way I see it, noone wants Cu+'s electron. :-( So, my guess is that this would not happen. It could be that some other complex is formed as a side reaction that would oxidize it, of course this is jsut a wild guess.. ;) I do beleive H2O2 would do this though, if not, maybe in an acidic solution... So anyway I guess I'm ready for my... Correction ;)

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Well in fact, time to praise :) . Your reasoning is very good. Indeed, nothing in the mix wants the electron, so the Cu(+) is not oxidized.

If you take red Cu2O and add HCl, then you see it turn white and CuCl (not CuCl2 !) is formed. In concentrated HCl you'll see it dissolve to the colorless complex ion CuCl2(-).

 

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

 

HOWEVER, there is one big pitfall. Oxygen from the air is capable of oxidizing Cu(+) very well and it certainly does.

 

Cu(+) is not present as a simple ion in hydrochloric acid, but it is present as the colorless CuCl2(-). When oxygen is present, then the Cu(+) in the CuCl2(-) complex ion is oxidized to Cu(2+).

 

What happens next is incredibly complex and I believe even the world-leading chemists do not fully understand what happens then. The only thing I can say about this is that a mixed-valency complex is formed. A compound, containing copper (I), copper (II) and chloride is formed. This compound has a VERY intense color.

 

A proposed formula for this compound is ClCu(μ-Cl)CuCl, where the copper atoms have oxidation state somewhere between +1 and +2.

 

If you have concentrated HCl, H2O2 and Cu-metal and CuCl2, then you can do most of the experiments, I have performed a year ago and which I put on my website. I invite you to repeat some of these and to enter the world of mixed-valency copper chemistry ;) . Be surprised, I mean it :) !

 

http://woelen.scheikunde.net/science/chem/riddles/copperI+copperII/index.html

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Thanks a lot, your awsome

 

The amount of knowledge I have Amassed in such a short time is really amazing, and most of it is thanks to people like you and YT. :)

 

ACTUALLY, It is VERY funny that you should mention that, I have also witnessed your dark compound. :) I didn't have the knowledge I now possess and didn't try to think about it, but now I realize that it's really weird for it to turn so dark when Cu+ ions are colorless... As for an explanation... looks liek your all over it.. I have no idea what else it could be.

 

EDIT: Also, in your experiment where you mixed your brown complex with NaOH... Well I believe that your yellow Precipitate may be Cu2O. It can be anywhere from yellow to red. I have made some that was yellow/ orange. It behaves exactly like the CuO I made when disolved in HCl, Although I do remember it disolving easier. So, I'm jsut offering a hypothesis here, I'm sure you have done much more thinkign about this though.

 

Also, lol... I had to disolve a decent amount of CuO in order to obtain visible Cu2+ ions. See in one of your experiments, you take CuCl2 + HCl + copper wire and throw it in a test tube. It becomes dark (Cu+ and Cu2+) But when it is stoppered, it goes clear. so there must be some reduction going on here.

 

I'm going to attempt something to further my understanding of this...

 

[math]

\ce{Cu2+ + Cu -> 2Cu+}

[/math]

 

So, is the Copper acting as a reducer? I hope so, it all makes sense to me at least..

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So, is the Copper acting as a reducer? I hope so, it all makes sense to me at least..

Indeed, the copper acts as reductor and it reduces the Cu(2+).

 

So, in my test tube experiment you have green Cu(2+) (not as free ion, but complexed, also called coordinated, to chloride) and this reacts with the copper metal. A simplified equation is given below.

 

Cu(2+) + Cu ---> 2Cu(+)

 

The Cu(+) gives the dark compound with Cu(2+).

 

When the test tube is not stoppered, then the liquid remains dark and finally becomes green again. The Cu(+) is VERY sensitive to oxidation by oxygen from the air, so any Cu(+) formed is oxidized to Cu(2+) by oxygen. When the test tube is stoppered, then no new oxygen can enter the tube and then finally it becomes (almost) colorless, because all Cu(2+) is reduced by the copper metal. As soon as the test tube is opened again, oxygen can go in and the Cu(+) is oxidized to Cu(2+) and the brown complex with copper (I) and copper (II) is formed.

 

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

 

Now about the complexes. The story I told above is just the principle of the redox reaction between copper (II) and metallic copper (0). In reality, copper exists in the form of anionic species at high concentrations of HCl.

 

Copper (II) exists as green CuCl4(2-) and copper (I) exists as colorless CuCl2(-). So, complex ions with chloride attached to copper are formed. These ions are not basically different than e.g. SO4(2-), where oxo-groups are attached to a central sulphur core.

 

So, the real reaction when copper wire is added is the following:

 

CuCl4(2-) + Cu ---> 2 CuCl2(-)

 

The CuCl2(-) reacts with CuCl4(2-), forming the deep brown complex. The mechanism behind this is amazingly complex and not fully understood (at least not by me, but I'm afraid by no one).

 

 

The chloride ions also have a very important role in this. If you have a solution of CuSO4 in 30% H2SO4 and you add copper metal, then nothing happens. You will not get copper (I). Chloride stabilizes the copper (I) complex.

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so, if i have oxygen and copper all the time, will the Cu2+ ions continue to disolve the copper? I put some copper in a 20 ounce bottle with some HCl and Copper Oxide (should disolve into the Cu2+ ion) and nothing really change much throughout the day. I added some H2O2 later to try and get more Cu2+ into solution. After wards, the solution is a little darker green, when the wire is no in HCl, I can see the little bubbles of liquid turn dark, and redisolve when placed ni the HCl. So can I continue to make CuCl2 in this way until not more Cl- Ions are available? (it IS kinda slow though.)

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Basically, yes... When you have HCl with some Cu(2+) in it, then when oxygen from the air is allowed to reach the mix, you can continue dissolving copper, turning that into copper (I), which is turned to copper (II) by the oxygen from the air. The limiting reagent is HCl. The net reaction is

 

2Cu + O2 + 4H(+) + 8Cl(-) ---> 2CuCl4(2-) + 2H2O

 

As you can see, the reaction requires H(+) and Cl(-). Long before all HCl is used up, the reaction becomes so slow that in practice it hardly continues.

 

But, the equation, given above, only is a net reaction. In reality, the reaction goes through the copper (I) complex, as you pointed out already. So, the liquid becomes very dark. When the copper metal is removed, though, the liquid finally will become green, because all copper (I) is converted to copper (II) by oxidation with oxygen from the air.

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