# Double replacement reactions

Definition and examples of double replacement reactions. Predicting and balancing neutralization and precipitation reactions.

## What is a double replacement reaction?

Double replacement reactions—also called double displacement, exchange, or metathesis reactions—occur when parts of two ionic compounds are exchanged, making two new compounds. The overall pattern of a double replacement reaction looks like this:
$\greenD{\text {A}^+\text{B}^-} + \maroonD{\text{C}^+\text{D}^-} \rightarrow \greenD{\text {A}^+}\maroonD{\text {D}^- }+ \maroonD{\text{C}^+}\greenD{\text{B}^-}$
You can think of the reaction as swapping the cations or the anions, but not swapping both since you would end up with the same substances you started with. The solvent for a double replacement reaction is usually water, and the reactants and products are usually ionic compounds—but they can also be acids or bases.
Here is an example of a double replacement reaction:
$\greenD{\text{BaCl}_2} (aq) +\maroonD{ \text{Na}_2 \text{SO}_4} (aq)\rightarrow \greenD{\text{Ba}}\maroonD{\text{SO}_4} (s)+ 2\maroonD{\text{Na}}\greenD{\text{Cl} }(aq)$
In this example, the cations are $\greenD{\text{Ba}^{2+}}$ and $\maroonD{\text{Na}^+}$, and the anions are $\greenD{\text{Cl}^-}$ and $\maroonD{\text{SO}_4^{2-}}$. If we swap the anions, or cations, we get as our products $\text{BaSO}_4$ and $\text{NaCl}$.
Once we swap the anion/cation partners to get the products, we also need to make sure that the product compounds are neutral since the reactant compounds are neutral. This means checking that the charges for the anion(s) and cation(s) cancel out within a compound. In $\text{BaSO}_4$, the 2+ charge on the cation $\text{Ba}^{2+}$ cancels out the 2- charge on the $\text{SO}_4^{2-}$ to give an overall neutral compound.
Also, we have to check that the subscripts in the products are the lowest possible integer value. For example, in the above reaction we know that one of the products is $2\text{NaCl}$ and not $\text{Na}_2 \text{Cl}_2$ because even though both compounds are neutral, we can divide both subscripts in $\text{Na}_2 \text{Cl}_2$by two to get $\text{NaCl}$.
If you are having trouble remembering the charges on different ions, you can review by watching the video on common polyatomic ions.

## Precipitation and neutralization reactions

Identifying double replacement reactions is usually fairly straightforward once you can recognize the pattern. Predicting whether the reaction will occur can be trickier; it helps to be able to recognize some common types of double replacement reactions. In this article we will be discussing precipitation reactions and neutralization reactions.
A precipitation reaction is when two aqueous ionic compounds form a new ionic compound that is not soluble in water. One example is the reaction between lead (II) nitrate and potassium iodide. Both compounds are white solids that can be dissolved in water to make clear, colorless solutions. When you combine the two clear solutions, you get the following reaction:
$\text{Pb(NO}_3)_2 (aq) + 2\text{KI} (aq)\rightarrow 2\text{KNO}_3 (aq)+ \goldE{\text{PbI}_2}(s)$
We made a beautiful golden solid from two clear solutions! In real life, your reaction flask might look something like the picture below.
A cloud of solid yellow lead (II) iodide forms when clear solutions of lead (II) nitrate and potassium iodide are combined.
Combining aqueous solutions of lead (II) nitrate and potassium iodide results in the formation of insoluble lead (II) iodide, a yellow solid. Image credit: PRHaney on Wikimedia Commons, CC-BY-SA 3.0
The insoluble product compound is called the precipitate. The solvent and soluble components of the reaction are called the supernatant or supernate. We can use solubility rules to predict whether a precipitation reaction will take place. The formation of a solid precipitate is the driving force that makes the reaction proceed in the forward direction.
The formation of the precipitate lowers the concentration of the insoluble product in solution, which drives the reaction further in the forward direction. This can be understood in more detail using Le Châtelier’s principle.
Concept check: What is in our supernatant?
Our supernatant is everything that is not the precipitate, so it is made up of our solvent $\text{H}_2 \text O$ and $\text{KNO}_3 (aq)$, which can also be written as the dissociated ions $\text K^+ (aq)$ and $\text {NO}_3^-(aq)$.
Neutralization reactions are a type of double replacement reaction that occurs between an acid and a base. The following is an example of a neutralization reaction:
$\blueD{\text H} \text F (aq)+\text{Na}\blueD{\text{OH}} (aq)\rightarrow \blueD{\text H_2 \text O} + \text{NaF}(aq)$
$acid + base \rightarrow \text H_2 \text O + salt$
An aqueous neutralization reaction generally produces water and a new ionic compound, also called a salt. The hardest part of identifying a neutralization reaction is recognizing that you have an acid and a base for the reactants. Once you know you have a neutralization reaction, you can generally predict the reaction will occur in the forward direction as long as you have a strong acid and/or a strong base as a reactant.
If you would like to know more about acid-base reactions, you can watch this video on the acid-base properties of salts.
A fun neutralization reaction that you may have tried is the combination of baking soda—sodium bicarbonate, $\text{NaHCO}_3$—and vinegar—mostly water with acetic acid, $\text {CH}_3 \text{COOH} (aq)$)—which produces carbonic acid—$\text {H}_2 \text{CO}_3$—and sodium acetate—$\text{NaCH}_3 \text{COO}$. If you have tried this reaction at home, you probably remember a lot of fizzing because the neutralization reaction is accompanied by a gas-producing reaction, where the carbonic acid decomposes into carbon dioxide gas—bubbles!—and water.
Note that double replacement reactions can be written as molecular, complete ionic, or net ionic equations. In this article we are only writing out the molecular equation, but you probably want to be familiar with writing the other forms of the equation as well.

## Example: predicting and balancing a double replacement reaction

Let’s take a look at an example where we don't know the products:
${\text{H}_2 \text{SO}_4} (aq) +{\text{Ba} \text{(OH)}_2} (aq) \rightarrow$
First, we can identify the cations and anions that will get swapped. The cations are $\text{H}^+$ and $\text{Ba}^{2+}$, and the anions are $\text{SO}_4^{2-}$ and $\text{OH}^-$. Swapping anions gives the products $\text{H}_2 \text O$ and $\text{BaSO}_4$:
${\text{H}_2 \text{SO}_4} (aq) + {\text{Ba} \text{(OH)}_2} (aq) \rightarrow \text{H}_2 \text O + \text{BaSO}_4$
We can see that our double replacement reaction is also a neutralization reaction since we are reacting sulfuric acid, a strong acid, with barium hydroxide, a strong base. What is the state of the product barium sulfate? If we check our solubility rules, we see that barium sulfate is insoluble and should precipitate out of solution. That means our reaction is a precipitation reaction, too! We can also include that information in our equation by adding the symbol $\blueD{(s)}$ after the $\text{BaSO}_4$.
We aren’t quite done yet, though. Our reaction is not balanced since we have unequal numbers of hydrogens and oxygens on both sides of the arrow. We can fix this by multiplying $\text{H}_2 \text O$ by $\redD{2}$ to give our final, balanced molecular equation:
$\text{H}_2 \text{SO}_4 (aq) + \text{Ba} \text{(OH)}_2 (aq) \rightarrow \redD{2}\text{H}_2 \text O + \text{BaSO}_4 \blueD{(s)}$

## Summary

Double replacement reactions have two ionic compounds that are exchanging anions or cations. Precipitation reactions and neutralization reactions are two common types of double replacement reactions. Precipitation reactions produce an insoluble product from two aqueous reactants, and you can identify a precipitation reaction using solubility rules. Neutralization reactions occur when the reactants are an acid and a base, and neutralization reactions are usually favorable as long as the reaction involves a strong acid and/or a strong base.

Kotz, J. C., P. M. Treichel, J. R. Townsend, and D. A. Treichel. "Precipitation Reactions." In Chemistry and Chemical Reactivity Instructor's Edition, 110-11. 9th ed. Stamford, CT: Cengage Learning, 2015.

## Try it!

#### Problem 1

For the following reactants, what are the products of the double displacement reaction?
$\text{FeCl}_3(aq) + \text{Ba(OH)}_2(aq) \rightarrow$
We have the cations $\text{Fe}^{3+}$ and $\text{Ba}^{2+}$, and the anions are $\text{Cl}^-$ and $\text{OH}^-$.
If we swap the anions, we get the products $\text{Fe(OH)}_3$ and $\text{BaCl}_2$. Based on solubility rules, $\text{Fe(OH)}_3$ is going to precipitate since most metal hydroxides are insoluble.
Therefore, $\text{Fe(OH)}_3 (s) + \text{BaCl}_2 (aq)$ is the correct answer.