# 4.14: Types of Chemical Reactions - Single and Double Replacement Reactions


Learning Objectives

• Recognize chemical reactions as single-replacement reactions and double-replacement reactions.
• Use the periodic table, an activity series, or solubility rules to predict whether single-replacement reactions or double-replacement reactions will occur.

You were introduced to chemical reactions in module 3. Many chemical reactions can be classified as belonging to a particular type of reaction which enables you to predict products given the reactants. You have already learned how to predict the products and write the balanced equations for the complete combustion of hydrocarbons.  In this module you will be introduced to two other classes of reactions that take place in aqueous solutions.

A single-replacement reaction is a chemical reaction in which one element is substituted for another element in a compound, generating a new element and a new compound as products. For example,

$\ce{2HCl(aq) + Zn(s) → ZnCl2(aq) + H2(g)}\nonumber$

is an example of a single-replacement reaction. The hydrogen atoms in $$\ce{HCl}$$ are replaced by $$\ce{Zn}$$ atoms, and in the process a new element-hydrogen-is formed. Another example of a single-replacement reaction is

$\ce{2NaCl(aq) + F2(g) → 2NaF(s) + Cl2(g)}\nonumber$

Here the negatively charged ion changes from chloride to fluoride. A typical characteristic of a single-replacement reaction is that there is one element as a reactant and another element as a product.

A double-replacement reaction occurs when parts of two ionic compounds are exchanged, making two new compounds. A characteristic of a double-replacement equation is that there are two compounds as reactants and two different compounds as products. An example is

$\ce{CuCl2(aq) + 2AgNO3(aq) → Cu(NO3)2(aq) + 2AgCl(s)}\nonumber$

There are two equivalent ways of considering a double-replacement equation: either the cations are swapped, or the anions are swapped. (You cannot swap both; you would end up with the same substances you started with.) Either perspective should allow you to predict the proper products, as long as you pair a cation with an anion and not a cation with a cation or an anion with an anion.

Example $$\PageIndex{3}$$

Predict the products of this double-replacement equation:

$\ce{BaCl2 + Na2SO4 → }\nonumber$

Solution

Thinking about the reaction as either switching the cations or switching the anions, we would expect the products to be BaSO4 and NaCl.

Exercise $$\PageIndex{3}$$

Predict the products of this double-replacement equation:

$\ce{KBr + AgNO3 → }\nonumber$

KNO3 and AgBr

Predicting whether a double-replacement reaction occurs is somewhat more difficult than predicting a single-replacement reaction. However, there is one type of double-replacement reaction that we can predict: the precipitation reaction. A precipitation reaction occurs when two ionic compounds are dissolved in water and form a new ionic compound that does not dissolve; this new compound falls out of solution as a solid precipitate. The formation of a solid precipitate is the driving force that makes the reaction proceed.

To judge whether double-replacement reactions will occur, we need to know what kinds of ionic compounds form precipitates. For this, we use solubility rules, which are general statements that predict which ionic compounds dissolve (are soluble) and which do not (are not soluble or insoluble). Table $$\PageIndex{1}$$ lists some general solubility rules. We need to consider each ionic compound (both the reactants and the possible products) in light of the solubility rules. If a compound is soluble, we use the (aq) label with it, indicating it dissolves. If a compound is not soluble, we use the (s) label with it and assume that it will precipitate out of solution. If everything is soluble, then no reaction will be expected.

Table $$\PageIndex{1}$$: Some Useful Solubility Rules (soluble)
These compounds generally dissolve in water (are soluble): Exceptions:
All compounds of Li+, Na+, K+, Rb+, Cs+, and NH4+ None
All compounds of NO3 and C2H3O2 None
Compounds of Cl, Br, I Ag+, Hg22+, Pb2+
Compounds of SO42 Hg22+, Pb2+, Sr2+, Ba2+
Table $$\PageIndex{2}$$: Some Useful Solubility Rules (insoluble)
These compounds generally do not dissolve in water (are insoluble): Exceptions:
Compounds of CO32− and PO43− Compounds of Li+, Na+, K+, Rb+, Cs+, and NH4+
Compounds of OH Compounds of Li+, Na+, K+, Rb+, Cs+, NH4+, Sr2+, and Ba2+

For example, consider the possible double-replacement reaction between Na2SO4 and SrCl2. The solubility rules say that all ionic sodium compounds are soluble and all ionic chloride compounds are soluble except for Ag+, Hg22+, and Pb2+, which are not being considered here. Therefore, Na2SO4 and SrCl2 are both soluble. The possible double-replacement reaction products are NaCl and SrSO4. Are these soluble? NaCl is (by the same rule we just quoted), but what about SrSO4? Compounds of the sulfate ion are generally soluble, but Sr2+ is an exception: we expect it to be insoluble-a precipitate. Therefore, we expect a reaction to occur, and the balanced chemical equation would be

$\ce{Na2SO4(aq) + SrCl2(aq) → 2NaCl(aq) + SrSO4(s)}\nonumber$

You would expect to see a visual change corresponding to SrSO4 precipitating out of solution (Figure $$\PageIndex{2}$$).

Example $$\PageIndex{4}$$:

Will a double-replacement reaction occur? If so, identify the products.

1. Ca(NO3)2 + KBr → ?
2. NaOH + FeCl2 → ?

Solution

1. According to the solubility rules, both Ca(NO3)2 and KBr are soluble. Now we consider what the double-replacement products would be by switching the cations (or the anions)-namely, CaBr2 and KNO3. However, the solubility rules predict that these two substances would also be soluble, so no precipitate would form. Thus, we predict no reaction in this case.
2. According to the solubility rules, both NaOH and FeCl2 are expected to be soluble. If we assume that a double-replacement reaction may occur, we need to consider the possible products, which would be NaCl and Fe(OH)2. NaCl is soluble, but, according to the solubility rules, Fe(OH)2 is not. Therefore, a reaction would occur, and Fe(OH)2(s) would precipitate out of solution. The balanced chemical equation is $\ce{2NaOH(aq) + FeCl2(aq) → 2NaCl(aq) + Fe(OH)2(s)}\nonumber$

Exercise $$\PageIndex{4}$$

$\ce{Sr(NO3)2 + KCl → }\nonumber$