What is the net ionic equation of CuSO4 plus KOH?

Answer 1

First off, NaCO3 is unbalanced. It should be Na2CO3.

Na2CO3 --> 2 Na+ + CO32-

CuSO4 --> Cu2+ + SO42-

Together only Cu2+ and CO32- react with each other, Sodium and Sulfate don't form a precipitate and are called: tribune ions, ie. don't react and can be left out of the reaction equation (marked by hooked parenthesis [.+.] )

Cu2+ + [SO42-+ 2 Na+] + CO32- --> (CuCO3)precip. solid+ [SO42-+ 2 Na+]

Simpler:

Cu2+aq. + CO32-aq. --> (CuCO3)precip. solid

Answer 2

First Derive molocular equation using oxidation numbers

CuSO4+KOH-->Cu(OH)2+K2SO4

Then Balance CuSO4+2KOH-->Cu(OH)2+K2SO4

Then determine spectator ions by noting which elements do not change oxidation numbers

Remove your spectator ions and you have your net ionic equation

Answer 3

2NaOH+MgSO4 -------> Na2SO4+ Mg(OH)2

There you go ;)

Answer 4

You know, I have the same question and cant find the answer anywhere. Just my luck.

Answer 5

There is no reaction between these chemicals, so there is NO net ionic equation.

Answer 6

2OH^- + Cu^2+ --------> Cu(OH)2!

Answer 7

2I- + Cu+2 = CuI2 (s)

Answer 8

Any net ionic equation.