Writing Net Ionic Equation for Cu(NH4)2(SO4)2 x 6H2O + NaOH Reaction

[Selophane]

I'm trying to figure out how to go about writing the net ionic equation for :

$$Cu(NH_{4})_{2}(SO_{4})_{2} x 6H_{2}O + NaOH$$

incase that doesn't show, Cu(NH4)2(SO4)2 x 6H2O + NaOH... (thats a hydrate there)
I believe a precipitate is suppose to form, but i have no clue how to go about this one where it has 3 ions in it, Cu, NH4, and SO4... help is much appreciated!

Thanks,
Chris

 

[geoffjb]

I'm trying to figure out how to go about writing the net ionic equation for :

$$Cu(NH_{4})_{2}(SO_{4})_{2} x 6H_{2}O + NaOH$$

incase that doesn't show, Cu(NH4)2(SO4)2 x 6H2O + NaOH... (thats a hydrate there)
I believe a precipitate is suppose to form, but i have no clue how to go about this one where it has 3 ions in it, Cu, NH4, and SO4... help is much appreciated!

Thanks,
Chris

What are the cations? What are the anions? Break the equation up into them, and you should see your answer.

 

[Selophane]

hmm, ok, i think this is correct then?

Cu2+ 2NH4+ 2SO42- 6H+ 6OH- + Na+ OH-

not sure what ones form the products here tho...
thanks for the response!