When you revolve a well balanced equation right into a net ionic equation, you overlook all the ions that are spectator ions (aqueous on one side as well as the other). You just write the solid and the aqueous ions that create that solid.

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So, what happens once you attempt to compose a net ionic equation if ALL the ions are aqueous? Would you compose down eextremely ion in the reactants/products, or write nothing? If you have actually a balanced equation prefer the following:

$$ceNaCl_(aq) + KBr_(aq) -> NaBr_(aq) + KCl_(aq)$$

and also the solubility products for all the species are huge enough that there"s basically no solid, there"s really no distinction in between the left and also best sides of the equation. It"s a well balanced equation, however nopoint has adjusted. We began with $ceNa+_(aq) + K+_(aq) + Br^-_(aq) + Cl^-_(aq)$ in solution and we still have actually that on the right side. So there"s really not a net ionic equation because there"s no net readjust. Thanks for contributing a solution to Chemisattempt Stack Exchange!

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