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:

$$NaCl_{(aq)}+KBr_{(aq)} \longrightarrow NaBr_{(aq)}+KCl_{(aq)}$$

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:

$$ \begin{align} NaCl &\longrightarrow Na^+ + Cl^- \\[5pt] KBr &\longrightarrow K^+ +Br^- \end{align}$$

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:

$$ \begin{align} NaBr &\longrightarrow Na^+ + Br^- \\[5pt] KCl &\longrightarrow K^+ +Cl^- \end{align}$$

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.

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:

$$AgNO_{3 \; (aq)}+NaCl_{(aq)} \longrightarrow AgCl_{(s)}+NaNO_{3 \; (aq)}$$

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.

$$Ag^+ + \cancel{NO_3^-} + \cancel{Na^+} + Cl^- \longrightarrow AgCl_{(s)} + \cancel{Na^+} + \cancel{NO_3^-}$$

So the net ionic equation is:

$$Ag^+ + Cl^- \longrightarrow AgCl_{(s)}$$

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:

$$Mg(NO_3)_2 + Na_2CO_3 \longrightarrow MgCO_3 + NaNO_3$$

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:

$$Mg(NO_3)_{2 \; (aq)} + Na_2CO_{3 \; (aq)} \longrightarrow MgCO_{3 \; (s)}+NaNO_{3 \; (aq)}$$

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

$$Mg^{2+} + 2 \, NO_3^- + 2 Na^+ + CO_3^{2-} \longrightarrow MgCO_{3 \;(s)} + 2 \, Na^+ + 2 \, NO_3^-$$

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

$$Mg^{2+} + \cancel{2 \, NO_3^-} + \cancel{2 Na^+} + CO_3^{2-} \longrightarrow MgCO_{3 \;(s)} + \cancel{2 \, Na^+} + \cancel{2 \, NO_3^-}$$

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

$$Mg^{2+} + CO_3^{2-} \longrightarrow MgCO_{3 \; (s)}$$

Example 3

Strontium bromide and potassium sulfate react in aqueous solution to form strontium sulfate, which is insoluble (K_sp = 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:

$$SrBr_{2 \; (aq)} + K_2SO_{4 \; (aq)} \longrightarrow SrSO_{4 \; (s)} + KBr_{(aq)}$$

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:

$$Sr^{2+} + \cancel{2 \, Br^-} + \cancel{2 \, K^+} + SO_4^{2-} \longrightarrow SrSO_{4\; (s)} + \cancel{2 \, K^+} + \cancel{2 \, Br^-}$$

The net ionic equation is then

$$Sr^{2+} + SO_4^{2-} \longrightarrow SrSO_{4 \; (s)}$$

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:

$$Cr(NO_3)_{3 \; (aq)} + FeSO_{4 \; (aq)} \longrightarrow Cr_2(SO_4)_{3 \; (aq)} + Fe(NO_3)_{2 \; (aq)}$$

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

$$2 \, Cr(NO_3)_{3 \; (aq)} + 3 \, FeSO_{4 \; (aq)} \longrightarrow Cr_2(SO_4)_{3 \; (aq)} + 3 \, Fe(NO_3)_{2 \; (aq)}$$

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

$$2 \, Cr^{2+} + 6 \, NO_3^- + 3 \, Fe^{2+} + 3 \, SO_4^{2-} \longrightarrow 2 \, Cr^{2+} + 3 \, SO_4^{2-} + 3 \, Fe^{2+} + 6 \, NO_3^-$$

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.

$$\cancel{2 \, Cr^{2+}} + \cancel{6 \, NO_3^-} + \cancel{3 \, Fe^{2+}} + \cancel{3 \, SO_4^{2-}} \longrightarrow \cancel{2 \, Cr^{2+}} + \cancel{3 \, SO_4^{2-}} + \cancel{3 \, Fe^{2+}} + \cancel{6 \, NO_3^-}$$

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 (K_sp = 5 x 10^-5)

   Solution

   Ammonium sulfate is (NH₄)₂SO₄ and calcium hydroxide is Ca(OH)₂. The double displacement reaction that can take place is:

   $$(NH_4)SO_4 + Ca(OH)_2 \longrightarrow NH_4OH + CaSO_4$$

   Now while (NH₄)₂SO₄ and NH₄OH are soluble by the solubility rules, the K_sp values for Ca(OH)₂ and CaSO₄ are 5.02 × 10^-6 and 4.93 × 10^-5, respectively (calcium salts are often not soluble).

   The dissociation reactions are

   $$ \begin{align} (NH_4)_2SO_4 &\longrightarrow 2 \, NH_4^+ + SO_4^{2-} \\[5pt] Ca(OH)_2 &\longrightarrow Ca^{2+} + 2 \, OH^- \; \; \color{#E90F89}{\text{insoluble}} \end{align}$$

   The ionic reaction is

   $$\cancel{2 \, NH_4^+} + SO_4^{2-} + Ca(OH)_2 \longrightarrow \cancel{2 \, NH_4^+} + 2 \, OH^- + CaSO_{4 \; (s)}$$

   Cancelling the ammonium ions tha appear on both sides of this balanced equation gives us the net ionic equation:

   $$SO_4^{2-} + Ca(OH)_2 \longrightarrow 2 \, OH^- + CaSO_{4 \; (s)}$$




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.

   Solution

   The overall reaction is

   $$Pb(NO_3)_2 + MgI_2 \longrightarrow PbI_2 + Mg(NO_3)_2$$

   The solubilities are: Pb(NO₃)₂ and MgI₂ are soluble in water. PbI₂ is insoluble, but Mg(NO₂)₂, like most nitrates, is soluble.

   The ionic equation is

   $$Pb^{2+} + \cancel{2 \, NO_3^-} + \cancel{Mg^{2+}} + 2 \, I^- \longrightarrow PbI_{2 \; (s)} + \cancel{Mg^{2+}} + \cancel{2 \, NO_3^-}$$

   Cancelling what occurs on both sides gives the net ionic equation:

   $$Pb^{2+} + 2 \, I^- \longrightarrow PbI_{2 \; (s)}$$




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.

   Solution

   Reaction:

   $$CaCl_{2 \; (aq)} + K_2CO_{3 \; (aq)} \longrightarrow 2 \, KCl_{(aq)}+CaCO_{3 \; (s)}$$

   Ionic reaction:

   $$Ca^{2+} + \cancel{2 \, Cl^-} + \cancel{2 \, K^+} + CO_3^{2-} \longrightarrow \cancel{2 \, K^+} + \cancel{2 \, Cl^-} + CaCO_{3 \; (s)}$$

   Net ionic reaction:

   $$Ca^{2+} + CO_3^{2-} \longrightarrow CaCO_{3 \; (s)}$$




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.

   Solution

   Reaction:

   $$2(NH_4)_3PO_{4 \; (aq)} + 3 \, ZnCl_{2 \; (aq)} \longrightarrow 6 \, NH_4Cl_{(aq)}+Zn_3(PO_4)_{2 \; (s)}$$

   Ionic reaction:

   $$\cancel{6 \, NH_4^+} + 8\, PO_3^{3-} + 3 \, Zn^{2+} + \cancel{6 \, Cl^-} \longrightarrow \cancel{6 \, NH_4^+} + \cancel{6 Cl^-} + Zn_3(PO_4)_{2 \; (s)}$$

   Net ionic reaction:

   $$8 \, PO_3^{3-} + 3 \, Zn^{2+} \longrightarrow Zn_3(PO_4)_{2 \; (s)}$$




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.

   Solution

   Reaction (single-displacement):

   $$3 \, Mg_{(s)} + 2 \, FeCl_{3 \; (aq)} \longrightarrow 3 \, MgCl_{2 \; (aq)} + 2 \, Fe_{(s)}$$

   Ionic reaction:

   $$3 \, Mg_{(s)}+2 \, Fe^{3+}+\cancel{6 \, Cl^-} \longrightarrow 3 \, Mg^{2+}+\cancel{6 \, Cl^-} + 2 \, Fe_{(s)}$$

   Net ionic reaction:

   $$3 \, Mg_{(s)}+2 \, Fe_{(aq)}^{3+} \longrightarrow 3 \, Mg_{(aq)}^{2+} + 2 \, Fe_{(s)}$$