I am not sure if i fully understand your question
Unfortunately, that shows. Question is about a redox reaction between copper ions and Mg.
but I believe you're just asked to find out how much Cu is in a sample. I'd just use CuSO4, this is soluble, then just add NaCl(s). The copper will crash out as CuCl2(s).
CuCl2 is pretty well soluble, so it won't crash out, at least not quantitatively.
You might be thinking of a redox reaction.
Mg(s) + HCl -------> Mg^2+(aq) +2Cl- + H2(g)
See above. While definitely some Mg will dissolve in the acid, that's not the reaction that the separation is based on.
My biggest concern was if CuO would disassociate. I didn't know the answer off the top of my head. I was/am preparing for an exam, i was pretty tired when i looked at this topic. That's silly of me to say CuCl2 is not soluble. should use a carbonate.
I wasn't sure if the original question was hypothetical or not. I looked in to all this. CuO has a ksp of 10^-20, really product favored so it won't disassociate, that was my main concern. Adding the HCl. Does cause it to dissolve, So yeah, I was wrong about the Water/acid solubility thing. I was thinking more along the line of solvents.
So now, for sure CuO does dissolve in HCl. So we're good. I was also concerned about if a redox reaction would occur with Mg(s) + Cu+
I looked up the reduction potentials for
Cu+ + e- -----> Cu(s) = 0.52V
Mg^2+ + 2e- ----> Mg(s) = -1.18V
Mg(s) + 2Cu+ -------> 2Cu(s) + Mg^2+ = 1.70V
Volts are positive so the reaction is spontaneous, it will occur understand standard conditions.
So the equation looks good, it will occur, everything is set.
So to answer the question directly. what happen to the Cu2O when you dissolve it into the HCl?
This is the reaction
CuO + 2HCl ----> 2Cu^2+ + H2O(l)This is how it works.
The key is this 2CuO(s) ------> Cu^2+ + O2(g), this reaction isn't very soluble.
But when you add HCl the O2 reacts with H+(from HCl). This causes the O2(g) concentration to go down. remember equilibrium? If the O2 goes down, more O2 will be produced, causing a greater disassociation of CuO(s) resulting in more Cu^2+ which means the solubility of CuO(s) is increasing.
If interested here are the details
1. 2CuO(s) ------> 2Cu^2+ + O2(g) ksp = 1.0e-20 (very tiny amount of products)
2. 4HCl --------> 4H+ + 4Cl- very soluble
3. O2(g) + 4H+ -------> 2H2O Ecell = 1.23 (will naturally occur(react))
All the above reactions happen, so we're good to go. If you combine all the reactions this is what you get.
2CuO(s) + 4HCl + O2(g) + 4H+ -------> 2Cu2+ + O2(g) +4H+ + 4Cl- + 2H2O
Cancel things out that are on both sides
2CuO(s) + 4HCl +
-------> 2Cu2+ +
+ 4Cl- + 2H2O
You end up with this reaction.
2CuO(s) + 4HCl -----> 2Cu2+ 4Cl- + 2H2O(l)
CuO(s) + 2HCl -----> Cu2+ 2Cl- + H2O(l)
Same as original.