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When etching circuit boards, a somewhat commonly used substitute for (expensive and messy) ferric chloride is a mixture of an acid, a source of chloride ions, and hydrogen peroxide. If I understand the chemistry correctly, the acidic environment allows the copper to react with the chlorine ions to make $\ce{CuCl2}$, which will more slowly react with the copper again to make $\ce{2 CuCl}$. Adding oxygen to the solution (from the peroxide or just by aerating it) makes $\ce{CuO}$ and $\ce{CuCl2}$ again, allowing the process to repeat.

I meant to mix $\ce{H2O2}$, acetic acid and table salt, but accidentally used citric acid instead of acetic acid. After realizing my mistake, I added scraps of aluminum foil to the mixture to displace the copper, and render the mixture safer to dispose of. I went to filter out the copper and dispose of the rest, when I found that there was a layer of some kind of gelatinous substance coating the exposed surfaces in my container. Does anyone have any idea of what it could be, and how I might safely dispose it?

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    $\begingroup$ Initial guess: some aluminate. Not sure, though. $\endgroup$ – Jan Feb 2 '16 at 23:44
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1) You should not use acetic acid (or any organic acid) in this setup. It potentially may form some (even if minuscule) amount of chloracetic acid which is really hazardous. Furthermore, copper acetate (and other carboxilates) has low solubility. Go for either genuine $\ce{HCl}$ or for $\ce{H2SO4}$ used in lead accumulators.

2) Most likely it is aluminium hydroxide. If the substance is bluish, it may be also copper hydroxide.

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  • $\begingroup$ Regarding chloroacetic acid, I was under the impression that it is only dangerous on contact, and that any amount present when I finally go to dispose of it would be easily neutralized by baking soda. Regarding my gelatinous substance, it does seem to be a mixture of aluminum hydroxide and sodium citrate. $\endgroup$ – P... Feb 3 '16 at 18:20
  • $\begingroup$ Oh, and don't forget the Na from the salt. The intended overall reaction is 4 Cu + 4 NaCl + 4 CH3COOH + O2 -> 4 CuCl + 4 CH3COONa + 2 H2O $\endgroup$ – P... Feb 3 '16 at 18:26

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