Fundamentals F9

[Katie 3H]

"M.9 Copper(II) nitrate reacts with sodium hydroxide to produce a precipitate of light blue copper(II) hydroxide. (a) Write the net ionic equation for the reaction. "
For this question, what is the difference between the net ionic equation and just a regular balanced equation? How do I write the net ionic equation for this question?

[Kelly Bui 2I]

A net ionic equation shows the compounds and elements that are directly involved in the chemical reaction. When you solve for a net ionic equation, you would remove the spectator ions (the ions that are present in the solution but are not involved in the chemical reaction). This is different from a regular balanced equation which would show all the elements and compounds involved.

To find the net ionic equation, you would need to know the charges of each of the atoms. Copper (II) nitrate is [Cu(NO3)2] which would react with sodium hydroxide (NaOH) to produce copper(II) hydroxide [Cu(OH)2]. This equation would also produce Na(NO3) if you use the criss-cross method.

Then, you would balance out the equation to: Cu(NO3)2 + 2NaOH --> Cu(OH)2 + 2Na(NO3).

After that, you separate the aqueous solutions into ions while the solid solutions are left as is:
Cu + 2(NO3) + 2Na +2(OH) --> Cu(OH)2 + 2Na + 2(NO3)

Finally, you would cross out the ions that are repeated on the reactant and product side (aka spectator ions):
Cu + 2(OH) --> Cu(OH)2