# Which ions are accounted for in total and net ionic equations?

There are a number of things I don't understand about ionic equations.

First off, when you do net ionic equations, is it correct that you're only focusing on precipitates? The way I understood it, you put "no reaction" if there are no solid or gaseous products. To clarify, if all of the products of the reaction are aqueous, we look at it as if no reaction had taken place.

Then I went and looked in my book, and that made it seem like all products that weren't covalent were spectator ions, i.e., covalent molecules were the only ones you note. Now I'm even more confused.

• Welcome to Chemistry.SE! You are not the first one who has struggled with spectator ions, complete and net ionic equations. Please feel free to use the search function of this site. In addition, this video on net ionic equations might be helpful. – Klaus-Dieter Warzecha Jan 22 '14 at 6:21

After reading into the matter (as a non-native English speaker, I've never heard of total and net ionic equations) here goes my small little guide to ionic equations:

## Total Ionic Equations

A total ionic equation is where you start with a balanced equation and write all the ionic compounds in their ionic form. For example, in the following reaction $$\ce{2 Na3PO4 + 3 CaCl2 -> 6 NaCl + Ca3(PO4)2 v},$$ one of the products precipitates (noted by the downwards pointing arrow). So we will not write it in ionic form for the total ionic equation, which is the following: $$\ce{6 Na+ + 2PO4^3- + 3Ca^2+ + 6 Cl- -> 6 Na+ + 6 Cl- + Ca3(PO4)2 v}.$$

## Net Ionic Equations

Net ionic equations can be derived from total ionic equations: We just eliminate the compounds that appear both on the right and on the left side of the reaction arrow. Thus we get for the above total ionic equation: $$\ce{2 PO4^3- + 3 Ca^2+ -> Ca3(PO4)2 v}$$

The ions that were eliminated are termed spectator ions, because they do not participate in any notable chemical reaction.

If something is still unclear, maybe have a go at Kristy M. Bailey's website (Oklahoma City Community College), which explains it pretty neatly.