Net Ionic Equations and Determining When Double and Single Displacement Reactions Occur

We have discussed how to tell if a double displacement reaction occurs or not, and how to decide if a single displacement reaction occurs or not. In both cases, you can answer the “will it occur?” question without thinking about what is going on at the molecular and ionic level, but delving into that will lead us to a deeper understanding of what is happening and why. In both cases, we can use a net ionic equation to help us understand. 

Net Ionic Equations of Double Displacement Reactions

We’ll look at two different examples of double displacement reactions to see how net ionic equations are created and what they can tell us about when double displacement reactions occur and when they don’t.

Example 1: A solution of barium nitrate reacts with a solution of sodium phosphate

The balanced reaction would be written this way:

3 Ba(NO3)2 + 2 Na3PO4 → Ba3(PO4)2 + 6 NaNO3 

Let’s look up all of the compounds in this equation on the solubility table. We find that Barium phosphate is insoluble, while the rest are soluble. That would mean that we would say that this reaction DOES occur. 

Now, remembering that when ionic compounds dissolve, they break up into their ions, we can rewrite this equation like this:

3 Ba+2(aq) + 6 NO3-1(aq) + 6 Na+1(aq) + 2 PO4-3(aq) → Ba3(PO4)2 (s) + 6 Na+1(aq) + 6 NO3-1(aq) 

This version of the reaction  is called the Complete Ionic Equation.

Note that this has kept the number of ions consistent (The 3 Ba(NO3)2 produces 3 Ba ions and 6 nitrate ions). Also notice that the barium phosphate is still together since it is insoluble and does not break up into ions.

Now, we can go one step further. On the left side of the reaction, there are 6 nitrate ions swimming around on their own. The same 6 nitrate ions are still there on the right side. The same can be said of the sodium ions. Since their situation did not change during the course of the reaction, they are called “spectator ions” (they just watch, rather than participate).

Since they don’t influence anything, we can remove them from our reaction, leaving this:

3 Ba+2(aq) + 2 PO4-3(aq) → Ba3(PO4)2 (s)

This is called the Net Ionic Equation. This version of the reaction shows what is really going on. In a very real sense, the sodium and nitrate ions are irrelevant. Stated differently, this reaction would appear exactly the same if we replaced the sodium ions with potassium ions and the nitrates with acetates.

Example 2: A solution of copper II nitrate reacts with a solution of sodium sulfate

The balanced reaction is:

Cu(NO3)2 + Na2SO4 → CuSO4 + 2 NaNO3

All four of these compounds are soluble, so we would predict that this reaction does NOT occur. 

To see why, we remember that when ionic compounds dissolve they break up into ions, so the complete ionic equation therefore looks like this:

Cu+2(aq) + 2 NO3-1(aq) + 2 Na+1(aq) + SO4-2(aq) → Cu+2(aq) + SO4-2(aq) + 2 Na+1(aq) + 2 NO3-1(aq)

If you look at this reaction, it should quickly become clear that EVERYTHING is a spectator. That is, none of the ions change their situation at all in this reaction. If we eliminated the spectators, we would be left with this:

→ 

Not really anything. Hmmmm.

The Point:

So, what do we mean when we ask “Will a Reaction Occur?” What we mean, simply, is does the net ionic equation show a change?

Net Ionic Equations for Single Displacement Reactions

Now let’s see how net ionic equations differ for single displacement reactions compared to double displacement reactions. We’ll use a simple example: Copper reacts with a solution of silver nitrate:

Cu + 2 AgNO3 → Cu(NO3)2 + 2 Ag

Both silver nitrate and copper II nitrate are soluble, and therefore break up into ions when they dissolve. That  gives us this complete ionic equation:

Cu(s) + 2 Ag+1(aq)  + NO3-1(aq) → Cu+2(aq) + 2 NO3-1(aq) + 2 Ag(s)

As with the double displacement reaction we can find spectator ions, but in this case there is only ONE spectator (nitrate) rather than 2. If we eliminate those, we get the net ionic equation:

Cu(s) + 2 Ag+1(aq) → Cu+2(aq) + 2 Ag(s)

Looking at this net ionic equation, it should become clear why, when we consider which single displacement reactions occur and which do not, we ONLY look at the elements that are involved, not the other ion, since it is just a spectator. In other words, our example would have looked exactly the same if we had used silver chlorate rather than silver nitrate.