[Soverdu]

Hi everyody,

hope someone can help me identify some ions in this solution:

Mixing HCl(aq) with a nitrate salt gives HNO3(aq) in an equalibrium:

H₃O⁺(aq)+2NO₃^-(aq)+NO₂(aq)  3HNO₃(l)

Adding copper to this solution produces a brownish gas NO₂ and when the reaction dies out the solution is completely black. My question is what is this black stuff?

Cu(s) + 4HNO₃(aq)  Cu(NO₃)₂(aq) + 2H₂O(l) +2NO₂(g)

The Cu(NO₃)₂(aq) produced should be blue. And if i take a few drops from the solution and mix with water it turns out blue, so what is going on here?

Appreciate your help here,
Soverdu

[billnotgatez]

with a nitrate salt

Just wondering what the nitrate salt was that you started with

[Radu]

http://en.wikipedia.org/wiki/Copper(II)_chloride
As you see, theoretically in a concentrated Cu²⁺ we have colors ranging from light green to light blue. Maybe your solution is too concentrated, and your black a dark concentrated blue.
    Or, what if NO₂ remains in a concentrated solution and acts as a ligand, leading to brown species?

[Soverdu]

Sorry for the slow follow up on this thread

with a nitrate salt

Just wondering what the nitrate salt was that you started with

I used KNO3 but I know the same color forms with NaNO3

http://en.wikipedia.org/wiki/Copper(II)_chloride
As you see, theoretically in a concentrated Cu²⁺ we have colors ranging from light green to light blue. Maybe your solution is too concentrated, and your black a dark concentrated blue.
    Or, what if NO₂ remains in a concentrated solution and acts as a ligand, leading to brown species?

Short after the beginning of the reaction the solution is green while the black product slowly forms around the cupper. If the solution is stirred the black product disappears until it starts forming around the cupper again. Hours later the solution is completely black.

You might be right that the black color could be due a high conc. of  CuCl2 but I think the ligand theory sounds interesting.

Is it possible that [Cu(NO₂)₆]^4- could have formed?

[billnotgatez]

@Soverdu

So you are saying

Your first reaction is
KNO₃(aq) + HCl(aq)  HNO₃(aq) + KCl(aq)

Your second reaction is
Cu(s) + 4HNO₃(aq) +4KCl(aq) + 4KNO₃(aq) + 4HCl(aq)  Cu(NO₃)₂(aq) + 2H₂O(l) +2NO₂(g) + 4KCl(aq) + 4KNO₃(aq) + 4HCl(aq) 

[Borek]

Your second reaction is
Cu(s) + 4HNO₃(aq) +4KCl(aq) + 4KNO₃(aq) + 4HCl(aq)  Cu(NO₃)₂(aq) + 2H₂O(l) +2NO₂(g) + 4KCl(aq) + 4KNO₃(aq) + 4HCl(aq)

Bill, that's wrong. You can't have the same substance between reactants and products.

[billnotgatez]

@Borek
You are correct.
I was trying to make a point and trying to make sure of the situation in short hand

I should have written the question this way

@Soverdu
So you are saying
Assuming your first equation is correct and is in equilibrium
Your first reaction is
KNO₃(aq) + HCl(aq)  HNO₃(aq) + KCl(aq)

So now you have a reaction vessel containing
KNO₃(aq)
HCl(aq)
HNO₃(aq)
KCl(aq)

To which you add
Cu(s)

and so far you believe that the following reaction happens
Cu(s) + 4HNO₃(aq)  Cu(NO₃)₂(aq) + 2H₂O(l) + 2NO₂(g)

Therefore you believe that you have created the or still have present in the reaction vessel
KNO₃(aq)
HCl(aq)
HNO₃(aq)                  [possibly totally reacted and not present]
KCl(aq)
Cu(s)                        [possibly totally reacted and not present]
Cu(NO₃)₂(aq)
H₂O(l)                     [as part of the reaction]
NO₂(g)

And you are wondering what might cause the solution to be black.
Is this the situation?

[Borek]

Bill, listing these substances doesn't make much sense - they should be listed as ions and all reactions should be net ionic.

[Soverdu]

@ billnotgatez

Yes that's the overall situation.

Right after the reaction has stopped the solution completely black and there's a light green solid in it.

After letting the dead reaction mixture sit for a few days it has turned into a greenish blue color and the solid about the same color. I think the solid might be CuCl₂ · (dihydrate?)

[billnotgatez]

@Soverdu
So you think there is another reaction happening like
CU(s) + 2 HCl(aq) = CUCl₂(aq) + H₂(g)
and eventually you get
CuCl₂·2H₂O

From WIKI
http://en.wikipedia.org/wiki/Copper%28II%29_chloride

Copper(II) chloride is the chemical compound with the formula CuCl₂. This is a light brown solid, which slowly absorbs moisture to form a blue-green dihydrate.

So now you believe that you have created the or still have present in the reaction vessel
KNO₃(aq)
HCl(aq)                             [possibly totally reacted and not present]
HNO₃(aq)                  [possibly totally reacted and not present]
KCl(aq)
Cu(s)                        [possibly totally reacted and not present]
Cu(NO₃)₂(aq)
H₂O(l)                     [as part of the reaction]
NO₂(g)
CUCl₂(aq)                    [later becoming CuCl₂·2H₂O]
H₂(g)

And you believe that is the cause for the solution to be black color that turns into greenish blue color
Is this the situation?


@Borek
You are correct again.
I should be depicting the ions
Even with expanding my post I am still dealing with the question too informally.

[Soverdu]

Yes I think there's a good chance that it is the CuCl₂ that makes the solution black.

So you think there is another reaction happening like
CU(s) + 2 HCl(aq) = CUCl₂(aq) + H₂(g)

I don't think this is the reaction since HCl is unable to oxidize copper.

Rather copper is oxidized by HNO₃ and copper(ll) reacts with Cl^- to form CuCl₂

Cu(s) + 4HNO3(aq)  Cu(NO3)2(aq) + 2H2O(l) +2NO2(g)

Cu(NO3)2(aq)  Cu²⁺(aq) + 2NO₃^-(aq)

Cu²⁺(aq) + 2Cl^-(aq)  CuCl₂(aq)

CuCl₂ has lower solubility than Cu(NO3)2(aq) so some of it precipitate.

Does this sound plausibly?

[billnotgatez]

I don't think this is the reaction since HCl is unable to oxidize copper.

Good catch!

@Borek

Bill, listing these substances doesn't make much sense - they should be listed as ions and all reactions should be net ionic.

Check @Soverdu previous post - It appears that @Soverdu does it far better than me.