When you turn a balanced equation into a net ionic equation, you ignore all the ions that are spectator ions (aqueous on one side as well as the other). You only write the solid and the aqueous ions that form that solid.

So, what happens when you try to write a net ionic equation if ALL the ions are aqueous? Would you write down every ion in the reactants/products, or write nothing?

  • $\begingroup$ If all the ions remain aqueous, there is no reaction because nothing was produced. Everything remains in solution before and "after". $\endgroup$ – Shafter Sep 10 '14 at 5:37

If you have a balanced equation like the following:

$$\ce{NaCl_{(aq)} + KBr_{(aq)} -> NaBr_{(aq)} + KCl_{(aq)}}$$

and the solubility products for all the species are big enough that there's essentially no solid, there's really no distinction between the left and right sides of the equation. It's a balanced equation, but nothing has changed. We started with $\ce{Na+_{(aq)} + K+_{(aq)} + Br^{-}_{(aq)} + Cl^{-}_{(aq)}}$ in solution and we still have that on the right side. So there's really not a net ionic equation because there's no net change.


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