Saturday
July 26, 2014

Homework Help: Chemistry

Posted by Frederique on Tuesday, November 7, 2006 at 10:16pm.

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?

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