I'm currently in AP chemistry at my high school, which has proved to be a fascinating course.  My class just recently finished our work in a laboratory experiment in which we were responsible for discovering the identities of ten unknown compounds.  My lab group and I were fortunate enough to finish a couple days early, giving us some extra time.  During this extra time we discovered some interesting reactions that I know I have a few questions about. ?   Our curiosity lead us to combine HCl together with CuSO4, which yielded a vibrant lime green color.  We believed the products that were produced from this reaction to be acidic, so in a means of disposal, it seemed logical to combine the products of our reaction with baking soda.  When an excess of baking soda is present the reaction seems to neutralize, as it gives off gas, but it also shifts back to a blue colored solution, which is quickly absorbed by the excess baking soda.  However, this new reaction is endothermic!  Leading me to become confused.  Initially I believed that I knew what was going on, yet now I find myself lost.
   Here's what we had:
   We put a volume of 6 M HCl into a beaker (I think it was about 20 mL).  Into this we slowly added drops of .2 M CuSO4 until it turned a vibrant fluorescent green.  This was really cool.
   From what we thought the following should occur (but after the oddities that we've observed we seek clarification):
      2 HCl (aq) + CuSO4 (aq) = CuCl2 (aq) + H2SO4 (aq)
   Theoretically, we then had a solution of CuCl2 and H2SO4, which we though was acidic.  Then we attempted to neutralize:
   CuCl2 (aq) + CuSO4 (aq) + NaHCO3 (s) = ?  (We don't know what the combination of these three yields; thus, this is were we’re stuck!)  All we know is that when we performed this in the lab, we found that the rxn was endothermic, as our temperature dropped from 21.0?C to 8.10?C.  It also became a solution of robin's egg blue.
   We were thinking that multiple reactions are occurring at the same time, yet we are unsure of what is going on.  We also thought that maybe something is occurring with the charge of the copper in the reaction, perhaps it goes from Cu2+ to Cu+ and vice versa.  Something might be happening with the color of the solution in relationship with the oxidation state of the copper, yet again, we’re not sure...
   Does anyone know what we have created as final products, and how the rxn of those three compounds works?  Also, I was thinking since it went back to blue, maybe this shows an example of Le Chatelier's principle?  I'm not sure.  Help please!  Thanks!!!

2 HCl (aq) + CuSO4 (aq) = CuCl2 (aq) + H2SO4

First of all: use only net ionic reactions in this case, as some of the ions are only spectators. SO₄^2- for sure.

First step is probably Cu²⁺ reacting with Cl^- to form not a CuCl₂ as you suggest (although that's not completely incorrect). Copper ions in water are hydrated (they form complex with water particles being ligands). In the presence of Cl^- some of the water particles are replaced by Cl^- anions thus the color of the solution changes (color observed is a color of copper ions complexed; note that anhydrous CuSO₄ is white).

On addition of NaHCO₃ it first neutralizes H⁺ that was left in solution, then it probably rises pH to the moment when colloidal copper hydroxide precipitates. It is colloidal so it will not just fall down the baker, but will rather form blue "snots" in the solution.

I don't suppose any readox reaction is going on here, as there is no reducing reagent present.

Wilco did many similar experiments and I believe he may add something if he will see the thread.

Borek, do you think that the pH of the solution would affect the preference of copper cation for Cl- or SO42-?

In other words is CuCl2 or CuSO4 favoured in acidic or basic solutions? (I know these will also be hydrated).

Would the solution change on the addition of excess NaCl? (as a source of excess Cl-)

I don't get you.

As long as the solution has pH low enough (no Cu(OH)₂) precipitation it doesn't contain CuSO₄ nor CuCl₂ - it contains Cu²⁺ (complexed by water and/or Cl^-), SO₄^2- and Cl^-. Thus it is not possible to tell which salt is favoured, as there is no salt in the solution, ions only.

Once you add excess of Cl^- (regardless of the form - be it NaCl solution, or HCl solution), you will observe higher concentrations of chloride complexes. I don't have creation constant tables here with me at the moment, but I can check them later if you want.

there must be factors which affect whether the Cu cation has a greater affinity for H2O, SO4, or Cl ligands. THe complexes must form to some degree in solution otherwise there would be no changes in colour.

OK, now I get you

I will never describe the solutions in terms you are referring too. Every reaction - be it dissociation, complexation, precipitation - has some equilibrium constant. This constant doesn't change with pH. However, pH can change relative concentrations of ions, thus shifting equilibrium.

In this particular case concentration of Cl^- will be virtually not affected by the pH changes. SO₄^2- can get easily protonated (pKa₂=2) to HSO₄^- and as such ill be less and less active as complexing agent when the pH goes down (no idea what is the value of complex creation constant for Cu²⁺/SO₄^2-, probably much lower than for Cu²⁺/Cl^- pair). WHen pH goes up above 4 both concentration of Cl^- and SO^42- is not changing, so as long the Cu(OH)₂ doesn't start to precipitate nothing happens.

Chemistry of water/chloride complexes can be pretty interesting - especially in case of chromium there are things happening that are not easily explained at first sight

CuCl₄^2- is a green complex in solution.

It is formed in the presence of excess chloride anions

Cu²⁺ + 4Cl^- <-> CuCl₄^2- dH = -ve

The solution is acidic because excess HCl was added to it. Addition of bicarbonate neutralises the H+ and this results in effervescence of carbon dioxide.

When you add excess bicarbonate, you not only remove the H+, but also precipitate the copper ions in the form of Cu(OH)₂. This is the blue precipitate you get in your solution.

The consequent removal of Cu²⁺ shifts the equilibrium backwards. The backward reaction is endothermic. The extent of the shift of the equilibrium is so big that the heat liberated in the process of effervescence is not sufficient to compensate the drop in temperature of the solution.