[interminable]

In lab we synthesized copper (II) acetate starting from copper (II) sulfate pentahydrate. First ammonia was added, followed by sodium hydroxide.

I've been working on the balanced equations, and I understand we're forming a Cu(OH)2 precipitate twice, (we dissolved it with excess ammonia the first time and reformed it with the NaOH).

We dissolve the Cu(OH)2 in 10% acetic acid solution to obtain the acetate product. I'm just wondering why dissolving the copper (II) sulfate directly in the acetic acid and skipping all the other steps wouldn't work.

[Schrödinger]

If you were to add acetic acid directly, then the reaction (supposing that they react) would be :

CuSO₄ + CH₃COOH  (CH₃COO)₂Cu + H₂SO₄

But as you can see, a weak acid (acetic acid) is displacing a stronger acid (sulphuric acid) from its salt. That's not possible.

On the other hand, when you add NaOH, the acid being displaced is H₂O, which is certainly weaker than acetic acid. (wirte the equation)


P.S : I recently learned on this forum that there might be rare exceptions to this rule. http://www.chemicalforums.com/index.php?topic=40518.0

[interminable]

Thanks, the answer seems obvious now, haha.
 
Cu(OH)2 + CH3COOH --> Cu(CH3COO)2 + H2O

This is what you were referring to, right?

[Schrödinger]

Yes