How do you write a net ionic equation for Copper II sulfate that reacts Sodium Hydroxide?

Answer 1

First you have to write the entire equation (including states):

CuSO4(aq)+ 2NaOH(aq) -> Cu(OH)2(s) + Na2SO4(aq)

Now you have to write an ionic equation which involves splitting up each compound, that is in aqueous form, into it's parts and writing the charge of each part:

Cu(aq)2+ + SO4(aq)2- + 2Na(aq)1+ + 2OH(aq)-1 -> Cu(OH)2(s) + SO4(aq)2- + 2Na(aq)1+

Now to write the net ionic equation, you "cancel" out any thing that is the exact same on both sides of the equation (same state, number, charge). That leaves us with the following:

SO4(aq)2- + 2OH(aq)-1 -> Cu(OH)2(s)

This is the net ionic equation.

Answer 2

To write the net ionic equation for Copper II sulfate reacting with Sodium Hydroxide, first write the balanced molecular equation: CuSO4 + 2NaOH → Cu(OH)2 + Na2SO4. Then, write the total ionic equation by splitting all reactants and products into their respective ions. Finally, eliminate spectator ions (ions that appear on both sides of the equation) to get the net ionic equation: Cu2+ + 2OH- → Cu(OH)2.