Solubility and Net Ionic Equations

Net ionic equations are useful in that they show only those chemical species participating in a chemical reaction. The key to being able to write net ionic equations is the ability to recognize monoatomic and polyatomic ions, and the solubility rules.

The following solubility rules predict the solubility of many ionic compound when applied in order:

 Exceptions 1. All Na+, K+, and NH4+ salts are soluble. 2. All NO3-, C2H3O2-, ClO3-, and ClO4- salts are soluble. 3. All Ag+, Pb2+, and Hg22+ salts are insoluble. 4. All Cl-, Br-, and I- salts are soluble. 5. All CO32-, O2-, S2-, OH-, PO43-, CrO42-, Cr2O72- and SO32- salts are insoluble. Group IIA S2- and Ba(OH)2 are soluble. 6. All SO42- salts are soluble. CaSO4, SrSO4, and BaSO4 are insoluble.

Let's first start with a complete chemical equation and see how the net ionic equation is derived. Take for example the reaction of lead(II) nitrate with sodium chloride to form lead(II) chloride and sodium nitrate, shown below:

Pb(NO3)2(aq) + 2 NaCl(aq) PbCl2(s) + 2 NaNO3(aq)
This complete equation may be rewritten in ionic form by using the solubility rules. Rule 2 confirms that lead(II) nitrate is soluble and therefore dissociated. Rule 1 says the same about NaCl. As products, sodium nitrate is predicted to be soluble (rules 1 and 2) and will be dissociated. The lead(II) chloride, however, is insoluble (rule 3). The above equation written in dissociated form is:
Pb2+(aq) +  2 NO3-(aq)  +  2 Na+(aq)  +  2 Cl-(aq) PbCl2(s)  +  2 Na+(aq)  +  2 NO3-(aq)
At this point, one may cancel out those ions which have not participated in the reaction. Notice how the nitrate ions and sodium ions remain unchanged on both sides of the reaction.
Pb2+(aq)  +  2 NO3-(aq)  +  2 Na+(aq)  +  2 Cl-(aq) PbCl2(s)  +  2 Na+(aq)  +  2 NO3-(aq)
What remains is the net ionic equation, showing only those chemical species participating in a chemical process:
Pb2+(aq) + 2 Cl-(aq) PbCl2(s)
After a while, you should be able to predict the net ionic equation given only the reactants. For instance, suppose you had to determine the net ionic equation resulting from the reaction of solutions of sodium sulfate and barium bromide:
BaBr2(aq) + Na2SO4(aq) ?
One way to tackle this problem is to examine what ions are found together in solution:  Ba2+, Br-, Na+, and SO42-. We know that barium ions and bromide ions are soluble together (rule 4), but will sodium ions or sulfate ions combine with barium ions to form an insoluble compound? Barium ions and sodium ions, both being positive in charge will repel each other, so no compound is expected to form between them. On the other hand, sulfate ions and barium ions would form barium sulfate (insoluble; rule 6). The sodium ions must therefore combine with bromide ions to form sodium bromide. According to the solubility rules, sodium bromide should be soluble (rule 1). Now we can write a complete balanced equation:
BaBr2(aq)  +  Na2SO4(aq) BaSO4(s)  +  2 NaBr(aq)
As before, the above equation can be rewritten showing the soluble species as ions in solution:
Ba2+(aq)  +  2 Br-(aq)  +  2 Na+(aq)  +  SO42-(aq) BaSO4(s)  +  2 Na+(aq)  +  2 Br-(aq)
Next, cross out any species which have not changed on both sides of the reactions; these are sometimes called spectator ions:
Ba2+(aq)  +  2 Br-(aq)  +  2 Na+(aq)  +  SO42-(aq) BaSO4(s)  +  2 Na+(aq)  +  2 Br-(aq)
What remains is the balanced, net ionic equation:
Ba2+(aq) + SO42-(aq) BaSO4(s)