Sciencemadness Discussion Board

Identifying reaction product

Kagutsuchi - 23-6-2015 at 09:46

I've dissolved some copper in nitric acid but there was a little nitric acid left in the solution. I added some ammonia to form a complex with the Cu2+ ions, but accidently I added some Na2CO3. Some light blue solid material formed in the solution. Does anyone have an idea what could it be?

Loptr - 23-6-2015 at 09:56

The precipitate is likely basic copper carbonate, CuCO3.

Cu(NO3)2 + Na2CO3 -> CuCO3 + 2 NaNO3

It would also help if you provided the amounts of reactants used.

[Edited on 23-6-2015 by Loptr]

diggafromdover - 23-6-2015 at 10:00

Copper Carbonate? Copper Carbonate is not water soluble and as such would precipitated. It's a one way reaction.

Kagutsuchi - 23-6-2015 at 11:36

I used about 5-6 grams of Cu(NO3)2, 20 ml of circa 60% nitric acid and 15 grams of Na2CO3.

MrHomeScientist - 23-6-2015 at 12:10

By some rough stoichiometry calculations, 15g of carbonate is just about enough to neutralize 20mL of your acid. That plus the ammonia you added (you didn't specify how much) means your solution is almost certainly pH >= 7, allowing some copper carbonate to fall out. It's about the only possibility for the blue solid, really.

Also, basic copper carbonate is Cu<sub>2</sub>(OH)<sub>2</sub>CO<sub>3</sub>. CuCO<sub>3</sub> by itself doesn't really exist (or at least, it's extremely hard to make).

Loptr - 23-6-2015 at 13:34

Ah, that's right. CuCO3 reacts with water to form basic copper carbonate.

blogfast25 - 23-6-2015 at 13:50

Quote: Originally posted by Loptr  
Ah, that's right. CuCO3 reacts with water to form basic copper carbonate.


Hmmm.... in a sense:

2 Cu<sup>+</sup>(aq) + 2 CO<sub>3</sub><sup>2-</sup> (aq) + H<sub>2</sub>O(l) === > Cu<sub>2</sub>(OH)<sub>2</sub>CO<sub>3</sub>(s) + CO<sub>2</sub>(g)

Copper basic carbonate is kind of a double salt: CuCO3.Cu(OH)2


[Edited on 23-6-2015 by blogfast25]

Kagutsuchi - 23-6-2015 at 19:18

Thanks guys, it really seems that way. The ammonia was about 20 ml 20% ammonia but I think you were right about the case. Later today I'll do some analytical testing but I did not have time for that yesterday :)