Question

You dissolve a small amount of your copper product in a few mL of water. When you added eight drops of 6 M HCl to the copper complex, you observe that the bright blue color fades and no precipitate forms. Write a net ionic equation for any reaction taking place. Is this compound resistant to attack by acid? How can you tell?

If it is helpful, the reaction involved in the experiment is:

Cu(H₂O)₄²⁺ (aq) + SO₄^2- (aq) + 4NH₃ (aq) --> Cu(NH₃)₄ SO₄ H₂O (s) + 3H₂O

Answer 1

The reaction between NH3(aq) and Cu2+(aq) happens in two stages and NH3(aq) behaves in a different way in each stage.

(1) First, when a small amount of NH3(aq) is added:

NH3(aq) is a weak base and dissociates in water to generate OH- ions.

NH3 + H2O ←→ NH4+ + OH-

These OH- ions then deprotonate the H2O ligands in the hexaaqua complex to form the hydroxide, which is a pale blue precipitate.

so if we are adding HCl , what it is doing is consuming the OH-

[Cu(H2O)6]2+(aq) + 2OH-(aq)→ [Cu(OH)2(H2O)4](s) + 2H2O (l)

Notice that when the hydroxide is formed, the overall charge goes from 2+ to 0. Since the hydroxide carries no charge, it is unable to engage in ion-dipole interactions with water, hence comes out of the solution as a precipitate. but now hydroxide is used so no precipitate is formed.

(2) Next, when more NH3(aq) is added:

NH3 here behaves as a ligand and binds directly to the Cu2+ centre, forming [Cu(NH3)4(H2O)2](aq) which is a deep blue complex.

[Cu(H2O)6]2+(aq) + 4NH3(aq)→ [Cu(NH3)4(H2O)2]2+(aq) + 4H2O (l)

Here we get a deep blue solution because the complex ion carries a 2+ charge which enables it to engage in ion-dipole interactions with water hence exists as a soluble complex.