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16.4: Net Ionic Equations

Difficulty Level: At Grade Created by: CK-12
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Lesson Objectives

  • Write net ionic equations for double-replacement reactions that produce precipitates, gases, or molecular compounds.
  • Write net ionic equations for single-replacement reactions.
  • Use the solubility rules to predict precipitate formation.

Lesson Vocabulary

  • ionic equation
  • molecular equation
  • net ionic equation
  • spectator ion

Check Your Understanding

Recalling Prior Knowledge

  • What are the five types of chemical reactions?
  • What is dissociation?

Several types of reactions were introduced in the chapter Chemical Reactions: combination, decomposition, single-replacement, double-replacement, and combustion. Single-replacement and double-replacement reactions occur most frequently in aqueous solution. In this lesson, you will learn about various ways to depict these chemical reactions.

Aqueous Reactions

When ionic compounds are dissolved into water, the polar water molecules break apart the solid crystal lattice, resulting in the hydrated ions being evenly distributed through the water. As you have learned, this process is called dissociation, and it is the reason that ionic compounds tend to be strong electrolytes. When two different ionic compounds that have been dissolved in water are mixed, a chemical reaction may occur between certain pairs of the hydrated ions.

Consider the double-replacement reaction that occurs when a solution of sodium chloride is mixed with a solution of silver nitrate. The driving force behind this reaction is the formation of the silver chloride precipitate.

\begin{align*}\mathrm{NaCl}{(aq)}+\mathrm{AgNO}_{3}{(aq)} \rightarrow \mathrm{NaNO}_{3}{(aq)}+\mathrm{AgCl}{(s)}\end{align*}NaCl(aq)+AgNO3(aq)NaNO3(aq)+AgCl(s)

This is called a molecular equation. A molecular equation is an equation in which the formulas of the compounds are written as though all substances exist as molecules. However, there is a better way to show what is happening in this reaction. All of the aqueous compounds can be written as ions, because they are actually present in the water as dissociated ions.

\begin{align*}\mathrm{Na}^+{(aq)}+\text{Cl}^-{(aq)}+\text{Ag}^+{(aq)}+\text{NO}^-_{3}{(aq)} \rightarrow \text{Na}^+{(aq)}+\text{NO}^-_{3}{(aq)}+\text{AgCl}{(s)}\end{align*}Na+(aq)+Cl(aq)+Ag+(aq)+NO3(aq)Na+(aq)+NO3(aq)+AgCl(s)

This equation is called an ionic equation, an equation in which dissolved ionic compounds are shown as free ions.

If you look carefully at the last equation, you will notice that the sodium ion and the nitrate ion appear unchanged on both sides of the equation. When the two solutions are mixed, neither the Na+ nor the NO3 ions participate in the reaction. Although they are still present in the solution, they do not need to be included when describing the chemical reaction that occurs upon mixing.

\begin{align*}\cancel{\text{Na}^+{(aq)}}+\text{Cl}^-{(aq)}+\text{Ag}^+{(aq)}+\cancel{\text{NO}^-_{3}{(aq)}} \rightarrow \cancel{\text{Na}^+{(aq)}}+\cancel{\text{NO}^-_{3}{(aq)}}+\text{AgCl}{(s)}\end{align*}Na+(aq)+Cl(aq)+Ag+(aq)+NO3(aq)Na+(aq)+NO3(aq)+AgCl(s)

A spectator ion is an ion that does not take part in the chemical reaction and is found in solution both before and after the reaction. In the above reaction, the sodium ion and the nitrate ion are both spectator ions. The equation can now be written without the spectator ions.

\begin{align*}\text{Ag}^+{(aq)}+\text{Cl}^-{(aq)} \rightarrow \text{AgCl}{(s)}\end{align*}Ag+(aq)+Cl(aq)AgCl(s)

The net ionic equation is the chemical equation that shows only those elements, compounds, and ions that are directly involved in the chemical reaction. Notice that in writing the net ionic equation, the positively charged silver cation was written first on the reactant side, followed by the negatively charged chloride anion. This is somewhat customary because that is the order in which the ions must be written in the silver chloride product. However, it is not absolutely necessary to order the reactants in this way.

Net ionic equations must be balanced by both mass and charge. An equation that is balanced by mass has equal amounts of each element on both sides. Balancing by charge means that the total charge is the same on both sides of the equation. In the above equation, the overall charge is zero, or neutral, on both sides of the equation. As a general rule, if you balance the molecular equation properly, the net ionic equation will end up being balanced by both mass and charge.

Sample Problem 16.9: Writing and Balancing Net Ionic Equations

When aqueous solutions of copper(II) chloride and potassium phosphate are mixed, a precipitate of copper(II) phosphate is formed. Write a balanced net ionic equation for this reaction.

Step 1: Plan the problem.

Write and balance the molecular equation first, making sure that all formulas are correct. Then write the ionic equation, showing all aqueous substances as ions. Carry through any coefficients. Finally, eliminate spectator ions and write the net ionic equation.

Step 2: Solve.

Molecular equation:

\begin{align*}\text{3CuCl}_{2}{(aq)}+2\text{K}_3\text{PO}_{4}{(aq)} \rightarrow 6\text{KCl}{(aq)}+\text{Cu}_3(\text{PO}_4)_{2}{(s)}\end{align*}3CuCl2(aq)+2K3PO4(aq)6KCl(aq)+Cu3(PO4)2(s)

Ionic equation:

\begin{align*}\text{3Cu}^{2+}{(aq)}+6\text{Cl}^-{(aq)}+6\text{K}^+{(aq)}+2\text{PO}^{3-}_{4}{(aq)} \rightarrow 6\text{K}^+{(aq)}+6\text{Cl}^-{(aq)}+\text{Cu}_3(\text{PO}_4)_{2}{(s)}\end{align*}3Cu2+(aq)+6Cl(aq)+6K+(aq)+2PO34(aq)6K+(aq)+6Cl(aq)+Cu3(PO4)2(s)

Notice that the balancing is carried through when writing the dissociated ions. For example, there are six chloride ions on the reactant side because the coefficient of 3 is multiplied by the subscript of 2 in the copper(II) chloride formula. The spectator ions are K+ and Cl and can be eliminated.

Net ionic equation:

\begin{align*}\text{3Cu}^{2+}{(aq)}+2\text{PO}^{3-}_{4}{(aq)} \rightarrow \text{Cu}_3(\text{PO}_4)_{2}{(s)}\end{align*}3Cu2+(aq)+2PO34(aq)Cu3(PO4)2(s)

Step 3: Think about your result.

For a precipitation reaction, the net ionic equation always shows the two ions that come together to form the precipitate. The equation is balanced by mass and charge.

Practice Problem
  1. Write the net ionic equation for the reaction of calcium nitrate with lithium hydroxide. The products are aqueous lithium nitrate and a calcium hydroxide precipitate.

Some other double-replacement reactions do not produce a precipitate as one of the products. The production of a gas and/or a molecular compound such as water may also drive the reaction. For example, consider the reaction of a solution of sodium carbonate with a solution of hydrochloric acid (HCl). The products of the reaction are aqueous sodium chloride, carbon dioxide, and water. The balanced molecular equation is:

\begin{align*}\text{Na}_2\text{CO}_{3}{(aq)}+2\text{HCl}{(aq)} \rightarrow 2\text{NaCl}{(aq)}+\text{CO}_{2}{(g)}+\text{H}_2\text{O}{(l)}\end{align*}Na2CO3(aq)+2HCl(aq)2NaCl(aq)+CO2(g)+H2O(l)

The ionic equation is:

\begin{align*}\text{2Na}^+{(aq)}+\text{CO}^{2-}_{3}{(aq)}+2\text{H}^+{(aq)}+2\text{Cl}^-{(aq)} \rightarrow 2\text{Na}^+{(aq)}+2\text{Cl}^-{(aq)}+\text{CO}_{2}{(g)}+\text{H}_2\text{O}{(l)}\end{align*}2Na+(aq)+CO23(aq)+2H+(aq)+2Cl(aq)2Na+(aq)+2Cl(aq)+CO2(g)+H2O(l)

The sodium and chloride ions are spectator ions, making the final net ionic equation:

\begin{align*}\text{2H}^+{(aq)}+\text{CO}^{2-}_{3}{(aq)} \rightarrow \text{CO}_{2}{(g)}+\text{H}_2\text{O}{(l)}\end{align*}2H+(aq)+CO23(aq)CO2(g)+H2O(l)

You will obtain the correct net ionic equation for any reaction as long as you follow the steps in the examples.

A single-replacement reaction is one in which a pure, neutral element replaces another element in a compound. A neutral element would not carry a charge, so it will not be a spectator ion. The example below shows the reaction of solid magnesium metal with aqueous silver nitrate to form aqueous magnesium nitrate and silver metal.

Balanced molecular equation:

\begin{align*}\text{Mg}{(s)}+2\text{AgNO}_{3}{(aq)} \rightarrow \text{Mg(NO}_{3}{)}_{2}{(aq)}+2\text{Ag}{(s)}\end{align*}Mg(s)+2AgNO3(aq)Mg(NO3)2(aq)+2Ag(s)

Ionic equation:

\begin{align*}\text{Mg}{(s)}+2\text{Ag}^+{(aq)}+2\text{NO}^-_{3}{(aq)} \rightarrow \text{Mg}^{2+}{(aq)}+2\text{NO}^-_{3}{(aq)}+2\text{Ag}{(s)}\end{align*}Mg(s)+2Ag+(aq)+2NO3(aq)Mg2+(aq)+2NO3(aq)+2Ag(s)

The only spectator ion is the nitrate ion, so the net ionic equation is:

\begin{align*}\text{Mg}{(s)}+2\text{Ag}^+{(aq)} \rightarrow \text{Mg}^{2+}{(aq)}+2\text{Ag}{(s)}\end{align*}Mg(s)+2Ag+(aq)Mg2+(aq)+2Ag(s)

Notice that the overall charge on both sides of the equation is now +2, instead of zero like it was in the previous examples. This is typical for a single-replacement reaction. Because both sides of the reaction carry the same total charge, it is still balanced. This type of single-replacement reaction is called a metal replacement. Other common categories of single-replacement reactions are hydrogen replacement and halogen replacement. These were discussed in the chapter Chemical Reactions.

Predicting Precipitates

Some combinations of aqueous reactants result in the formation of a solid precipitate as a product. However, some combinations will not produce such a product. If solutions of sodium nitrate and ammonium chloride are mixed, no reaction occurs. One could write a molecular equation showing a double-replacement reaction, but both products, sodium chloride and ammonium nitrate, are soluble and would remain in the solution as ions. Every ion is a spectator ion, so there is no net ionic equation.

It is useful to be able to predict when a precipitate will form from a given mixture of ions. To do so, you can use a set of guidelines called the solubility rules (Table below).

Solubility Rules for Ionic Compounds in Water
Solubility Ionic Compound
Soluble Compounds containing the alkali metal ions (Li+, Na+, K+, Rb+, Cs+) or the ammonium ion (NH4+)
Soluble Compounds containing the nitrate ion (NO3), acetate ion (CH3COO), chlorate ion (ClO3), or bicarbonate ion (HCO3)
Mostly soluble

Compounds containing the chloride ion (Cl), bromide ion (Br), or iodide ion (I)

Exceptions are those compounds that also contain silver (Ag+), mercury(I) (Hg22+), or lead(II) (Pb2+)

Mostly soluble

Compounds containing the sulfate ion (SO42−)

Exceptions are the sulfate salts of silver (Ag+), calcium (Ca2+), strontium (Sr2+), barium (Ba2+), mercury(I) (Hg22+), or lead(II) (Pb2+) ions

Mostly insoluble

Compounds containing the carbonate ion (CO32−), phosphate ion (PO43−), chromate ion (CrO42−), sulfide ion (S2−), or silicate ion (SiO32-)

Exceptions are those compounds that also contain the alkali metals or ammonium

Mostly insoluble

Compounds containing the hydroxide ion (OH)

Exceptions are hydroxide salts of the alkali metals and the barium ion (Ba2+)

As an example on how to use the solubility rules, predict if a precipitate will form when solutions of cesium bromide and lead(II) nitrate are mixed.

\begin{align*}\text{Cs}^+{(aq)}+\text{Br}^-{(aq)}+\text{Pb}^{2+}{(aq)}+2\text{NO}^-_{3}{(aq)} \rightarrow \ ?\end{align*}Cs+(aq)+Br(aq)+Pb2+(aq)+2NO3(aq) ?

The potential precipitates from a double-replacement reaction are cesium nitrate and lead(II) bromide. According to the solubility rules table, cesium nitrate is soluble because all compounds containing the nitrate ion, as well as all compounds containing the alkali metal ions, are soluble. Most compounds containing the bromide ion are soluble, but lead(II) is an exception. Therefore, the cesium and nitrate ions are spectator ions and the lead(II) bromide is a precipitate. The balanced net ionic reaction is:

\begin{align*}\text{Pb}^{2+}{(aq)}+2\text{Br}^-{(aq)} \rightarrow \text{PbBr}_{2}{(s)}\end{align*}Pb2+(aq)+2Br(aq)PbBr2(s)

Lesson Summary

  • Single-replacement and double-replacement reactions often place in aqueous solution, in which dissociated ionic compounds are more accurately represented as free ions.
  • Spectator ions do not participate directly in the chemical reaction. An equation that shows only the substances in the reaction is called a net ionic equation. Net ionic equations must be balanced by both mass and charge.
  • The solubility rules can be used to predict whether a precipitate is produced from a given mixture of ions.

Lesson Review Questions

Reviewing Concepts

  1. Substances that are in which state(s) in a molecular equation are broken down into ions to make an ionic equation?
  2. One or more of three possible types of products are generally formed in a double-replacement reaction. What are they?
  3. What happens to a spectator ion during a chemical reaction?
  4. Which statement below is true concerning a net ionic equation?
    1. The overall charge on both sides of the equation must be zero.
    2. The overall charge on both sides of the equation must be equal.
  5. Use the solubility rules to determine whether each of the following compounds is soluble or insoluble in water.
    1. (NH4)3PO4
    2. CaBr2
    3. AgBr
    4. Li2SO4
    5. Mn(OH)2
    6. SrCO3
    7. Pt(NO3)2
    8. Fe2(CrO4)3


  1. Write a balanced net ionic equation for the reactions represented by each of the following unbalanced molecular equations:
    1. \begin{align*}\text{Na}_3\text{PO}_4{(aq)}+\text{Fe(NO}_3)_3{(aq)} \rightarrow \text{NaNO}_3{(aq)}+\text{FePO}_4{(s)}\end{align*}Na3PO4(aq)+Fe(NO3)3(aq)NaNO3(aq)+FePO4(s)
    2. \begin{align*}\text{HCl}{(aq)}+\text{Mg(OH)}_2{(s)} \rightarrow \text{MgCl}_2{(aq)}+\text{H}_2\text{O}{(l)}\end{align*}HCl(aq)+Mg(OH)2(s)MgCl2(aq)+H2O(l)
    3. \begin{align*}\text{Na}{(s)}+\text{H}_2\text{O}{(l)} \rightarrow \text{NaOH}{(aq)}+\text{H}_{2}{(g)}\end{align*}Na(s)+H2O(l)NaOH(aq)+H2(g)
    4. \begin{align*}\text{K}_2\text{S}{(aq)}+\text{HBr}{(aq)} \rightarrow \text{KBr}{(aq)}+\text{H}_2\text{S}{(g)}\end{align*}K2S(aq)+HBr(aq)KBr(aq)+H2S(g)
  2. Will a precipitate form when aqueous solutions of the following salts are mixed? If so, write the formula of the precipitate.
    1. CaCl2 and Mg(NO3)2
    2. K2CrO4 and MgI2
    3. NaCH3COO and Ni(ClO3)2
    4. Ba(OH)2 and Al2(SO4)3
  3. Finish the molecular equations below, balance them, and then write a corresponding net ionic equation. You will need to use the solubility rules to determine whether any of the products will precipitate.
    1. \begin{align*}\text{Na}_2\text{CO}_3{(aq)}+\text{ZnCl}_2{(aq)} \rightarrow\end{align*}Na2CO3(aq)+ZnCl2(aq)
    2. \begin{align*}\text{NH}_4\text{Cl}{(aq)}+\text{Pb(NO}_3)_2{(aq)} \rightarrow\end{align*}NH4Cl(aq)+Pb(NO3)2(aq)
    3. \begin{align*}\text{Al}{(s)}+\text{HI}{(aq)} \rightarrow\end{align*}
    4. \begin{align*}\text{HCl}{(aq)}+\text{Ba(OH)}_2{(aq)} \rightarrow\end{align*}
    5. \begin{align*}\text{Zn}{(s)}+\text{Fe(NO}_3)_3{(aq)} \rightarrow\end{align*}
    6. \begin{align*}\text{Cl}_2{(g)}+\text{KBr}{(aq)} \rightarrow\end{align*}
  4. Write a balanced net ionic equation from each of the word equations below.
    1. Aqueous cobalt(III) chloride reacts with aqueous ammonium sulfate to produce aqueous ammonium chloride and solid cobalt(III) sulfate.
    2. Aluminum metal reacts with a solution of copper(II) acetate to produce aqueous aluminum acetate and copper metal.
    3. Nickel metal reacts with sulfuric acid (H2SO4) to produce aqueous nickel(II) sulfate and hydrogen gas.
  5. Write a balanced net ionic equation for each of the following situations. You will need to determine the products first.
    1. Aqueous solutions of potassium hydroxide and chromium(III) chloride are mixed.
    2. Magnesium metal is dipped into a solution of lead(II) nitrate.
    3. Solid iron(III) hydroxide is added to a solution of hydrochloric acid.
    4. Fluorine gas is bubbled into a solution of sodium iodide.

Further Reading / Supplemental Links

Points to Consider

Heat and energy are important concepts in chemistry, as most chemical reactions are accompanied by a transfer of energy.

  • What is heat and how does a transfer of heat occur?
  • How do different substances respond to an input or loss of heat?

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Date Created:
Mar 01, 2013
Last Modified:
Aug 16, 2016
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