How do I write a net ionic equation for an aqueous double displacement reaction?

1. Identify the ions. Oxalic acid contains H⁺ and C₂O₄^2-ions; lead(II) nitrate contains Pb²⁺ and NO₃^- ions.
2. Swap the ions. The H⁺ goes with the NO₃^- to form HNO₃ on the product side. The C₂O₄^2-goes with the Pb²⁺ to form PbC₂O₄^2-on the product side.
3. Classify the products. At least one of the products must be either a) a weak electrolyte or b) an insoluble salt, or no reaction occurs. In your reaction, the lead(II) oxalate is an insoluble salt, and the nitric acid is a soluble strong acid.

1. Balance the equation.
   H₂C₂O₄(aq) + Pb(NO₃)₂(aq) PbC₂O₄(s) + 2 HNO₃(aq)
2. Break all strong electrolytes into ions. In your reaction, oxalic acid is NOT a strong electrolyte. You'll have to carry it over into the ionic equation unchanged. You also can't consider the lead(II) oxalate a strong electrolyte because it remains a solid:
   H₂C₂O₄(aq) + Pb²⁺(aq) + 2 NO₃^-(aq) PbC₂O₄(s) + 2 H⁺(aq) + 2 NO₃^-(aq)
   This equation is called a "complete ionic equation" or "total ionic equation" because all the strong electrolytes are written as ions.
3. Eliminate spectator ions. These are ions that appear on both sides of the equation (and so, are unchanged by the reaction). The nitrate ion is the only spectator in this reaction.
   H₂C₂O₄(aq) + Pb²⁺(aq) PbC₂O₄(s) + 2 H⁺(aq)
   This last equation is the net ionic equation.

Author: Fred Senese senese@antoine.frostburg.edu