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

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To write the net ionic equation for the reaction between Cu(NH4)2(SO4)2 x 6H2O and NaOH, identify the cations and anions involved. The key ions are Cu²⁺, NH4⁺, SO4²⁻, Na⁺, and OH⁻. A precipitate is expected to form, likely involving Cu²⁺ and OH⁻, leading to the formation of Cu(OH)2. The final net ionic equation will focus on the ions that participate in the formation of this precipitate, excluding spectator ions. Understanding the dissociation of the compounds is crucial for accurately representing the reaction.
Selophane
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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
 
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Selophane said:
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.
 
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!
 

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