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I have a solution of CuCl2 and with HCl, and I want to seperate CuCl2 from the HCl. I cannot boil or distill the solution, since I don't have a fume hood or access to a safe enough space outdoors. So I need to find a way to get the CuCl2 out of solution, or at least get rid of the HCl in order to boil off the water safely. I can filter out percipitates, if that helps in any way.

Any ideas?

Edit: I reacted small amounts with H2O2 and bubbling air through, and an excess of copper. This worked, but takes far too long. I have access to copper oxides (likely aged), but there was a discussion in the comments about solubility.

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    $\begingroup$ If you remember my correct answer (use copper metal in the presence of air and even weak acid like dilute HCl). it will work being electrochemcial in nature. This sources confirms that CuO will NOT dissolve in dilute HCl which is what you likely have. To quote: pubs.acs.org/doi/10.1021/acsreagents.4106 to quote: "The monograph also details the following specifications and corresponding tests for verifying that a substance meets ACS Reagent Grade specifications including: Assay, Insoluble in Dilute Hydrochloric Acid,." . So, if the HCl is dilute stirring every hour is more like hourly forever! $\endgroup$
    – AJKOER
    Commented Feb 1, 2022 at 15:58
  • $\begingroup$ If CuO, as the source claims, becomes insoluble at some HCl concentration level, how can there not be an associated HCl concentration issue? Note, as the electro-chemistry proceeds regardless of concentration as long as there is some HCl, I would argue that such an approach is the only possible path. But, I have further refined my answer and have presented it for input/review elsewhere. Upon completion of that process, I may provide a search phrase here for those actually seeking a workable solution and interested in advancing their range of chemistry (important for me, working on patents). $\endgroup$
    – AJKOER
    Commented Feb 1, 2022 at 20:52
  • $\begingroup$ @AJOKER I did react a small batch with copper metal and bubbling air through the solution, and yielded the desired product. A byproduct of this reaction was insoluble CuCl, which I am able to reuse in future runs. However, the reaction is very slow. Is there a way I can accelerate the reaction? And could I ask you to quote the relevant parts of your source? I am unable to access it, since I am am a physics student, and our department has no license. Thank you. I would love to accept your answer, but I am unable to find it. $\endgroup$
    – AeonSiraus
    Commented Feb 11, 2022 at 3:02
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    $\begingroup$ @AJKOER Please be careful when accusing other participants of particular conduct, here regarding their voting, which I remind you is strictly anonymous unless they leave a comment directed to you indicating that they have voted in a particular way. Note that accusations on your part could be construed as harassment. Others are at liberty to vote as they please on other posts, as are you. $\endgroup$
    – Buck Thorn ♦
    Commented Feb 12, 2022 at 12:51
  • $\begingroup$ There is a reputed assertion in my deleted answer that it is 'speculative'. My answer actually explains the chemistry around the prep of basic copper chloride per an existing patent and cited, with reaction equations no less, in Wikipedia here en.wikipedia.org/wiki/Dicopper_chloride_trihydroxide as a 'commercial process'. The only speculation exists for those not acquainted with the associated science. $\endgroup$
    – AJKOER
    Commented Feb 13, 2022 at 18:07

1 Answer 1

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Apparently your solution contain $\ce{CuCl2}$ and $\ce{HCl}$. If you want to save only $\ce{CuCl2}$ you should first add an excess of black copper oxide $\ce{CuO}$ and stir at least one hour. The following reaction will consume unwanted $\ce{HCl}$ : $$\ce{CuO + 2 HCl -> CuCl2 + H2O}$$ Then filtrate the solution to get rid of the $\ce{CuO}$ excess. The obtained blue solution contains no $\ce{HCl}$ any more. You may evaporate it to $≤ 100°$C. You should obtain by cooling to room temperature a blue-green hydrated copper chloride $\ce{CuCl2·2H2O}$. But this hydrated copper chloride is not easy to use, because it is deliquescent : it will soon get dissolved in the water attracted from the humidity of the air. It is much better to heat it to a temperature > $110$°C. Heated to between $110$°C and $498$°C, $\ce{CuCl2}$ solutions produce, after evaporation, an anhydrous copper chloride $\ce{CuCl2}$, which is a brownish yellow powder. And this powder is easier to use, because it is only slightly hygroscopic. It slowly attracts water but will not produce a viscous liquid like $\ce{CuCl2· 2H2O}$.

These two copper chlorides are both very soluble in water. At $0$° C, $100$ mL water dissolves $\ce{71 g CuCl2}$ and $\ce{110 g CuCl2·2H2O}$.

$\ce{CuCl2}$ melts at $498$°C, and is decomposed into $\ce{CuCl + Cl2}$ at $993$°C.

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  • $\begingroup$ Maurice: Do you really expect, given the limited resources outlined, access to Cupric Oxide ? If this is, indeed, correct, this answer is not useful. Please do not downgrade my both useful, correct and interesting answer. $\endgroup$
    – AJKOER
    Commented Feb 1, 2022 at 3:57
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    $\begingroup$ Neutralization of CuCl2 by NaOH and boiling solution with Cu(OH)2 provides CuO easily. $\endgroup$
    – Poutnik
    Commented Feb 1, 2022 at 6:48
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    $\begingroup$ No, do not add NaOH, as it can destroy all (or part) of your CuCl2 leaving an insoluble Cu(OH)2 and even result in a clear solution of NaCl and water. Any amount of NaOH also introduces a non-removal Sodium ion presence. And why use NaOH as one more likely has Baking Soda (NaHCO3) ? My answer (edited for a clear reference to Patina) introduces no contaminants and occurs naturally. $\endgroup$
    – AJKOER
    Commented Feb 1, 2022 at 11:27
  • $\begingroup$ @AJKOER It was - thought obviously - meant to just a part of the solution, not the whole one. Sure, NaHCO3 would work. // Another way is adding metallic copper, as you similarly pointed out, and carefully adding hydrogen peroxide. // Your answer diverges a lot - let others to evalute its qualities. $\endgroup$
    – Poutnik
    Commented Feb 1, 2022 at 12:11
  • $\begingroup$ Thanks, I have re-arranged my answer for more clarity as I progress in levels of chemistry around copper. $\endgroup$
    – AJKOER
    Commented Feb 1, 2022 at 13:12

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