Stripping chemical equations of "spectators"


Consider an ionic reaction taking place in aqueous solution. Often mixing two clear ionic solutions results in another clear solution. How can we tell if a reaction takes place? Which ions are reacting?

Here's an example of mixing two ionic solutions in which nothing noticeable really happens:

NaCl (aq) + KBr (aq) → NaBr (aq) + KCl (aq)

When these two transparent solutions of soluble ions are mixed, a third transparent solution of soluble ions results.

If we look more closely at the two ionic compounds in the solutions that are mixed, we see that each is soluble, and dissolves according to these reactions:

NaCl → Na+ + Cl-

KBr → K+ + Br-

The only reaction that could happen in this case is a double displacement reaction that produces NaBr and KCl. Each of these, however, is also a soluble salt, dissociating into the same ionic parts:

NaBr → Na+ + Br-

KCl → K+ + Cl-

The resulting mixture of ions is the same in both pre- and post-mix sets of compounds. Both contain Na+, Cl-, K+ and Br- ions, and no insoluble compounds can result from any combination of these ions. No real reaction has occurred here, just dissolving and mixing of ions in water.

Now consider a different kind of ionic reaction, a double-displacement reaction in which one of the products of the swapping of ions results in an insoluble compound which mostly precipitates as a solid. Here's an example.

consider two solutions of soluble lead nitrate [ Pb(NO3)2 ] and soluble potassium iodate (KIO3). The dissociation reactions are

Pb(NO3)2 → Pb2+ + 2 NO3-

KIO3 → K+ + IO3-

Now when the solutions are mixed, this reaction takes place:

Pb2+(aq) + 2 IO3-(aq) → Pb(IO3)2 (s)

The figure below illustrates the process. Notice that Pb(IO3)2 (s) is insoluble and precipitates from the solution. Clearly, in this case, a reaction has occurred.


When these soluble salts are mixed, the shuffling of cations and anions produces an insoluble salt, Pb(IO3)2, which precipitates from the clear solution as a yellow solid.


Now in this reaction, we can see that of the four kinds of ions produced by the dissociation of the original ionic compounds, only two were involved in the reaction, Pb2+ and IO3-. The other two ions, K+ and NO3-, we call "spectator ions."

In this case, the net ionic reaction, the reaction that only shows ions actually involved in forming a new product, is:


Pb2+(aq) + 2 IO3-(aq) → Pb(IO3)2 (s)


In this section we'll look at how we can easily arrive at the net ionic reaction for any ionic process. We'll do it by working examples.


Example 1

When solutions silver nitrate (AgNO3) and sodium chloride (NaCl) are mixed, solid (insoluble) silver chloride precipitates from the solution. Write the net ionic equation for this reaction. Which are the spectator ions?

Solution: We begin by writing the double-displacement reaction. It helps to include the phases in such a reaction:

Now break all soluble ionic compounds on both sides into their constituent ions. Then cancel any ions or compounds that appear on both sides of the reaction arrow. These are spectators, and aren't actually involved in the reaction.

So the net ionic equation is:

,

where the nitrate and sodium ions are spectators.


Example 2

Aqueous solutions of magnesium nitrate [ Mg(NO3)2 ] and of sodium carbonate ( Na2CO3 ) are combined, resulting in a possible double displacement reaction. Determine what happens after the solutions are mixed, and write the net ionic equation that describes it.

Solution: The double-displacement reaction that occurs is:

Now we ought to keep track of the solubilities of these compounds. According to the solubility rules, most nitrate compounds are soluble, so Mg(NO3)2 is soluble. The same is true for Na2CO3, as most sodium salts are soluble, even though carbonates can be problematic. It follows, then, that NaNO3 is soluble. MgCO3, on the other hand, is a fairly insoluble salt, with a solubility product constant of about 7 x 10-6 M2, so in this solution, it's reasonable to expect that it precipitates. We can modify our double-displacement reaction to this:

Now we can break the aqueous (soluble) compounds into their constituent ions:

and cancel the ions that appear on both sides of the equation, algebraically. Those are the spectator ions.

Finally, the net ionic equation is that of the formation of MgCO3:


Example 3

Strontium bromide and potassium sulfate react in aqueous solution to form strontium sulfate, which is insoluble (Ksp = 3.44 x 10-7), and aqueous potassium bromide. Write a net ionic equation for this reaction.

Solution: First, we need to find chemical formulae for each of these compounds. This is a good skill to review if you're out of practice.

Strontium (being in the second column of the periodic table) forms a +2 ion, so it will need two Br- ions to form the neutral compound SrBr2. The sulfate ion is a common ion that you should memorize; its charge is -2, so potassium (K+) sulfate is K2SO4. Potassium bromide is formed from +1 and -1 ions, respectively: KBr. Finally, the +2 and -2 charges of the strontium cation and the sulfate anion form a neutral compound in a 1:1 ratio: SrSO4.

Our first guess at the double-displacement reaction gives:

The states of each compound were given, so they're included. Now we break each ionic compound into its constituent ions and cross out any ions that appear on both sides of the equation:

The net ionic equation is then


Example 4 – no reaction

An aqueous solution of chromium (III) nitrate is added to an aqueous solution of iron (II) sulfate. Write the net ionic equation for any reaction that occurs.

Solution: First, chromium (III) nitrate is Cr(NO3)3 and iron (II) sulfate is FeSO4. The products of a double displacement reaction would be chromium (III) sulfate, Cr2(SO4)3, and iron (II) nitrate, Fe(NO3)2. The simple double-displacement reaction is:

Be careful here. Don't forget to balance the equation:

Now if we break all of these soluble compounds into their constituent ions, we get the overall ionic equation:

Now notice that all of the ions on the product side are represented on the reactant side. That is, all of the ions are spectator ions, so there is no reaction here.

The result of this mixture is not a chemical reaction, just a simple co-dilution of ions.


Practice problems



  1. An aqueous solution of ammonium sulfate is mixed with an aqueous solution of calcium hydroxide of equal concentration. Write a double-displacement reaction for this procedure, then write the net ionic equation. Calcium sulfate is a sparingly soluble salt (Ksp = 5 x 10-5)

  2. Lead (II) nitrate and magnesium iodide are mixed in aqueous solution. Write the net ionic equation that results from the resulting reaction or rearrangement, if there is one.

  3. Solutions of calcium chloride (soluble) and potassium carbonate (most potassium salts are soluble) are mixed. Write the net ionic equation for the resulting reaction.

  4. Aqueous solutions of ammonium phosphate and zinc chloride are mixed. Write a balanced equation for the reaction that could occur, including state information. If there is a reaction, write the net ionic equation.

  5. Some magnesium (Mg) metal is added to a solution of iron (III) chloride. Write a balanced reaction, including states (s, l, g, aq) for the process that occurs. Write a net ionic equation for the reaction.

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