My question is basically to find out what these reactants yield:
CuCl2(aq) + NH3(aq) --> ?
my best guess here is...
CuCl2(aq) + 6NH3(aq)-->[Cu(NH3)4(H2O)2]2+ + N2(g) + 2HCl
Im pretty sure about the [Cu(NH3)4(H2O)2]+2 part but not sure about the rest.
The primary product of this reaction is the copper(II) ammine complex ion, Cu(NH3)42+ and you make the remainder balance. I believe the ammine complex has a coordination number of 4 and not 6 although I'm not up on the latest research. I would do it like this.
CuCl2(aq) + 6NH3(aq) + 2HOH(l) ==>Cu(NH3)4^+2(aq) + 2Cl^-(aq) + 2NH4^+(aq) + 2 OH^-(aq).
There is absolutely no reason to believe N2(g) will be formed (and it isn't). If I find something to indicate a coordination number of six I will post another note here.
I later reviewed the work the reaction that is done is a series of copper related reactions, in a flow chart given it shows that the reactants CuCl2 and NH3 yield a product [Cu(NH3)4(H20)2]2+(aq) complex but i was wondering that clorine ions are present in the solution but how are the remaining nitrogen ions and hydrogen ions? are they floating around in this solution?
Any help I greatly appreciate.
I searched on the Internet and found what I believe to be a reliable source (from the UK) that confirms the coordination number of 6 with four NH3 and 2H2O; therefore, the complex ion [Cu(NH3)4(H2O)2]^2+ is entirely appropriate. The CuCl2 is present in aqueous solution as Cu^+2 and Cl^- and both of those stay as ions through the process; therefore, the Cl^- appears on the product side. There are no nirogen ions and few, if any, H^+ (especially in an NH3 solution). Actually, there are no Cu^+2 ions at the beginning either. Since this is an aqueous solution, I am sure the Cu^+2 is present as [Cu(H2O)6]^+2. If we start with that the equation is easier to write.
[Cu(H2O)6]Cl2 + 4NH3 ==>[Cu(NH3)4(H2O)2]^+2 + 2Cl^- + 4H2O.
Getting back to the original equation, if we write it like this,
CuCl2(aq) + 4NH3(aq) + 2H2O(l) ==> [Cu(NH3)4(H2O)2]^+2(aq) + 2Cl^-(aq)
we stick closer to what you are to start with and end with. Everything balances and you have no "nitrogen ions" to be concerned about.
Technically, I don't like to see NH3(aq) in solution without noting that it forms NH3 + HOH ==> NH4^+ + OH^-. IF you had H^+ that you asked about they would be neutralized with OH^- from the NH3 reaction. None of the ammonia will dissociate into a "nitrogen ion." I guess the short answer to your question, " but i was wondering that clorine ions are present in the solution but how are the remaining nitrogen ions and hydrogen ions? are they floating around in this solution?" is
CuCl2(aq) + 4NH3(aq) + 2H2O)l)==>[Cu(NH3)4(H2O)2]^+2(aq) + 2Cl^-(aq), which is the equation I wrote above, has no hydrogen ions or nitrogen ions to be conerned about.
Thanks for using the Jiskha boards. I hope this clears up any misunderstandings. Please follow up with anything you don't understand.
will it be
NH3 (aq) = NH4OH
NH4OH + CuCl2 --> Cu(NH3)Cl2 + H2O ?
I'm doing a lab on equilibriums with this equation, well actually we have to find the equation, but he did say that [Cu(H2O)6]+2 and [Cu(NH3)4(H20)2]+2 would be involved like you said, but some where in between these two he said there is some Cu(OH)2, which results from adding a little NH4OH to the CuCl2, uppon addin't more though it finally becomes what you had as your product, any idea what that middle step might be?
NH3(aq) --> NH4OH
NH4OH + CuCl2 --> Cu(OH)2 + NH4Cl
First a massive precipitate forms (the copper hydroxide), however as more ammonia is added the equilibrium shifts
to the diaquatetrammine copper(II) complex.
Your best guess:
CuCl2(aq) + 6NH3(aq) -->[Cu(NH3)4(H2O)2]2+ + N2(g) + 2HCl
I'd think that this is a more probable reaction:
CuCl2(aq) + 4NH3(aq)-->[Cu(NH3)4(H2O)2]2+ Cl2 , try smelling the products :)
(if you have the means to make them)
Based on the information you provided, it seems like there is an intermediate step involving the formation of Cu(OH)2. Here's the possible reaction pathway:
1. CuCl2(aq) + 2NH4OH(aq) → Cu(OH)2(s) + 2NH4Cl(aq)
In this step, CuCl2 reacts with NH4OH to form Cu(OH)2 (a precipitate) and NH4Cl.
2. Cu(OH)2(s) + 4NH3(aq) → Cu(NH3)4(OH)2(s) + 2H2O(l)
Cu(OH)2 reacts with NH3 to form a complex ion Cu(NH3)4(OH)2 (a precipitate) and water.
3. Cu(NH3)4(OH)2(s) + HCl(aq) → [Cu(NH3)4(H2O)2]2+(aq) + 2Cl^-(aq)
Cu(NH3)4(OH)2 reacts with HCl to form the desired complex ion [Cu(NH3)4(H2O)2]2+ and chloride ions.
Overall, the reaction can be summarized as:
CuCl2(aq) + 6NH3(aq) + 2NH4OH(aq) + HCl(aq) → [Cu(NH3)4(H2O)2]2+(aq) + 2Cl^-(aq) + 2NH4Cl(aq) + 2H2O(l)
I think the answer is just simply like this:
NH3(aq) --> NH4OH
NH4OH + CuCl2 --> Cu(OH)2 + NH4Cl
well this sucks. i have to write a lab report on this. i know the above reacton forms a blue soilution, and cu(no3)2 is blue in water...hmmm...
There is no answer because both ionic compounds are aqeous or soluble.
:D lol
CuCl2 + 4NH3 --> [Cu(NH3)4]Cl2