I started with copper (II) sulfate solution left over from a recrystallization. I added sodium carbonate (definitely an excess) and about 200 of water and got a lot of bubbling. I heated it, for no particular reason. Few hours later I filtered the solution, which now looked dingy gray. I got a deep blue solution with a pH of around 11, and what was in the filter was black...

The filtrate is still deep blue with no precipitate, I'm washing and drying the residue from the filter, but I'm still curious what I made. I should have gotten basic copper carbonate, I've done this before to get copper carbonate, I can't remember if I've used sodium carbonate or bicarbonate last times though. I didn't think it made a difference.

  • $\begingroup$ chemistry.stackexchange.com/questions/34605/… $\endgroup$ Apr 22, 2020 at 6:50
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    $\begingroup$ You made basic copper carbonate. $\endgroup$ Apr 22, 2020 at 7:17
  • $\begingroup$ I though copper carbonate was 1: insoluble in water and 2: green-blue powder. I have a dark blue solution and black power... I've done this before, I have copper carbonate from previous runs, idk this time is different. $\endgroup$ Apr 22, 2020 at 7:22
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    $\begingroup$ It is probably copper oxide, that you get by heating copper hydroxide and possibly basic copper carbonates as well. $\endgroup$
    – Poutnik
    Apr 22, 2020 at 7:34
  • $\begingroup$ Any idea why the filtrate is deep blue? That, to me, indicates dissolved copper but I shouldn't have anything soluble after adding the sodium carbonate, right? $\endgroup$ Apr 22, 2020 at 7:39

1 Answer 1


There are two surprising phenomena in this experiment.

First : $\ce{CuSO_4}$ reacts in a unusual way with $\ce{Na_2CO_3}$. The equation is : $$\ce{2 CuSO_4 + 2 Na_2CO_3 + H_2O -> Cu(OH)_2·CuCO_3 + CO_2 + 2 Na_2SO_4}$$ This forms some bubbles of $\ce{CO_2}$ as you have seen. And it produces a strange precipitate of the so-called basic copper carbonate, $\ce{Cu(OH)_2·CuCO_3}$, which is well known in geology as the mineral called malachite. This formula is sometimes written as $\ce{Cu_2CO_3(OH)_2}$. This is what you have obtained as your insoluble deposit.

Second : Basic Copper carbonate gets redissolved in a big excess of carbonate ions, producing a deep blue solution containing an anion which is probably $\ce{[Cu(CO_3)_2]^{2-}}$ This is what you have obtained in your blue filtrate.

  • $\begingroup$ Wow, thank you for the very specific answer! I'm wondering, is there a way to somehow recover the copper carbonate/anion from the filtrate? After filtering I've let it sit in a wide mouth jar covered with a coffee filter and there's some white precipitate at the bottom and fractal-tree looking crystals crawling up the sides. $\endgroup$ Apr 23, 2020 at 2:28
  • $\begingroup$ Your white precipitate may be sodium hydrogenocarbonate $\ce{NaHCO_3}$ which is not very soluble in water : $6.9$ g in $100$ mL, and less in $\ce{Na_2CO_3}$ solutions. This substance appears in $\ce{Na_2CO_3}$ solution exposed to air and/or to $\ce{CO_2}$ $\endgroup$
    – Maurice
    Apr 23, 2020 at 8:48
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    $\begingroup$ To recover copper carbonate, the excess carbonate must be destroyed, by adding an acid. The trouble is : first the excess carbonate ions will produce huge amounts of CO2. Then the anion $\ce{[Cu(CO_3)_2]^{2-}}$ will be decomposed and the malachite precipitate will soon appear again. But it is difficult to know with some precision the amount of acid necessary to obtain malachite quantitively, because if an excess of acid is added, the malachite will be dissolved and at the end you will obtain a $\ce{Cu^{2+}}$ solution like in the very beginning of your experiments. $\endgroup$
    – Maurice
    Apr 23, 2020 at 15:20
  • $\begingroup$ I guess I'll add sulfuric acid little by little with stirring. This really is just an experiment with leftover chemicals so... $\endgroup$ Apr 23, 2020 at 22:50

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