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

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SUMMARY

The net ionic equation for the reaction between Cu(NH4)2(SO4)2 x 6H2O and NaOH involves the dissociation of the reactants into their constituent ions. The relevant ions include Cu2+, NH4+, SO42-, Na+, and OH-. The formation of a precipitate indicates that Cu(OH)2 is likely produced in this reaction, as it is insoluble in water. The complete net ionic equation can be derived by identifying the ions that participate in the formation of the precipitate.

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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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