[gaubonganvung]

In one experiment they ask to use an given amount of unknown-copper contain compound  as a sample and put it in the HCl solution. After dissolving it into the solution another amount of Mg turning is added to the mixture to let the reaction take place in order to isolate the Cu ion by turning it into Cu(s). My question is that if the sample is Cu2O, what happen to the Cu2O when you dissolve it into the HCl? Will there be another reaction that take place 1st in which Cu2O will disassociate and forming CuCl before the Cu react with Mg and forming Cu(s)? If so is CuCl insoluble or soluble in that mixture of solution? and can the reaction involving Mg take place with CuCl?

Since the rule is that Cl-,I- and Br- is insoluble when combine with Ag+,Pb2+,Mercury (I), and Cu 1+,if one compound is insoluble in aqueous solution does that mean that it will also insoluble in acidic solution? or they are only not soluble in water solution?
 I am sorry for such long question I am quite bad at summarizing stuff.Thank you guys

[Borek]

CuCl is very weakly soluble, so yes, you can expect it to precipitate - that is, after addition of hydrochloric acid CuCl will replace Cu₂O.

But my bet is that you will given Cu(II) compound.

[gaubonganvung]

CuCl is very weakly soluble, so yes, you can expect it to precipitate - that is, after addition of hydrochloric acid CuCl will replace Cu₂O.

But my bet is that you will given Cu(II) compound.

Thanks Borek for you answer:) We were given Cu(II) however it was one of the question in the activity that ask what if Cu2O were used as well in the sample of unknown compound which make me quite confuse between acid solution and aqueous solution. I just wonder if I will be able to convert the amount of Cu2O in the sample to Cu(s) using Mg turning if it will form insoluble salt with the CL- ions. I also wonder when some salt is insoluble does it mean that they will not soluble in very diluted chemical solution as well?

[Borek]

There are no "insoluble" substances, everything is soluble - to some extent. In the case of salts we use so called solubility product to predict concentration of ions in saturated solutions.

And yes, CuCl will be reduced by Mg, just slowly (because what is getting reduced is the dissolved Cu⁺, and its concentration is low, reaction is slow).

[asmcriminal]

In one experiment they ask to use an given amount of unknown-copper contain compound  as a sample and put it in the HCl solution. After dissolving it into the solution another amount of Mg turning is added to the mixture to let the reaction take place in order to isolate the Cu ion by turning it into Cu(s). My question is that if the sample is Cu2O, what happen to the Cu2O when you dissolve it into the HCl? Will there be another reaction that take place 1st in which Cu2O will disassociate and forming CuCl before the Cu react with Mg and forming Cu(s)? If so is CuCl insoluble or soluble in that mixture of solution? and can the reaction involving Mg take place with CuCl?

Since the rule is that Cl-,I- and Br- is insoluble when combine with Ag+,Pb2+,Mercury (I), and Cu 1+,if one compound is insoluble in aqueous solution does that mean that it will also insoluble in acidic solution? or they are only not soluble in water solution?
 I am sorry for such long question I am quite bad at summarizing stuff.Thank you guys

To get an accurate answer to your question, you need to look up reduction potentials, deltaG. I am not going to look it up for you, but i am doing some ochem homework(not sure if it applies) I did the reaction from an o-chem perspective and I ended up with CuCl(s) and CuOH(s).

If this is correct, then you can't figure out how much Cu you have, because CuCl and CuOH have different densities.

I am not sure if i fully understand your question, 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). Now you can do your math.

To answer your question about HCl. Aqueous means H2O. That's your solvent. When you see something like NaCl(aq) this means NaCl in WATER.  If something isn't soluble in water, it's not going to be soluable in HCl because HCl is in WATER. The solvent in both cases is Water.

You might be thinking of a redox reaction.

Mg(s) + HCl -------> Mg^2+(aq) +2Cl- + H2(g)

Notice on the product side Mg is a solid. on the reactants side it's a cation. You can't see cations in water. So it appears it dissolved, but it didn't. There was an electron transfer between the H's and the Mg that took place. 

[Borek]

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.

[asmcriminal]

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.
------------------------------------------------------------

To gaubonganvung
 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
Reactions
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 + O2(g) + 4H+ -------> 2Cu2+ + O2(g) + 4H+ + 4Cl- + 2H2O

You end up with this reaction.
2CuO(s) + 4HCl -----> 2Cu2+ 4Cl- + 2H2O(l)

Balance it
 CuO(s) + 2HCl -----> Cu2+ 2Cl- + H2O(l) 
Same as original.

[Borek]

what happen to the Cu2O when you dissolve it into the HCl?

This is the reaction
CuO + 2HCl ----> 2Cu^2+ + H2O(l)

You do understand that CuO and Cu₂O are two different compounds?

Not that it changes much - in both cases copper is reduced by metallic magnesium, just Cu²⁺ is reduced much faster (because of the higher solubility).

[asmcriminal]

what happen to the Cu2O when you dissolve it into the HCl?

This is the reaction
CuO + 2HCl ----> 2Cu^2+ + H2O(l)

You do understand that CuO and Cu₂O are two different compounds?

Not that it changes much - in both cases copper is reduced by metallic magnesium, just Cu²⁺ is reduced much faster (because of the higher solubility).

Misread, have ADD. But obviously as you stated the point still stands.