4.02: Precipitation Reactions — libretexts.dev mirror

Learning Objectives

To identify a precipitation reaction and predict solubilities.

Exchange (Double-Displacement) Reactions

A precipitation reaction is a reaction that yields an insoluble product—a precipitate—when two solutions are mixed. We described a precipitation reaction in which a colorless solution of silver nitrate was mixed with a yellow-orange solution of potassium dichromate to give a reddish precipitate of silver dichromate:

$\begin{matrix} (4.2.1) & AgNO_{3} (aq) + K_{2} Cr_{2} O_{7} (aq)  Ag_{2} Cr_{2} O_{7} (s) + KNO_{3} (aq) \end{matrix}$

This unbalanced equation has the general form of an exchange reaction:

$\begin{matrix} (4.2.2) & \overset{soluble}{\overset{⏞}{AC}} + \overset{soluble}{\overset{⏞}{BD}} \to \underset{insoluble}{\underset{⏟}{AD}} + \overset{soluble}{\overset{⏞}{BC}} \end{matrix}$

The solubility and insoluble annotations are specific to the reaction in Equation $4.2.1$ and not characteristic of all exchange reactions (e.g., both products can be soluble or insoluble). Precipitation reactions are a subclass of exchange reactions that occur between ionic compounds when one of the products is insoluble. Because both components of each compound change partners, such reactions are sometimes called double-displacement reactions. Two important uses of precipitation reactions are to isolate metals that have been extracted from their ores and to recover precious metals for recycling.

Video $4.2.1$ : Mixing Potassium Chromate and Silver Nitrate together to initiate a precipitation reaction (Equation $4.2.1$ ).

While full chemical equations show the identities of the reactants and the products and give the stoichiometries of the reactions, they are less effective at describing what is actually occurring in solution. In contrast, equations that show only the hydrated species focus our attention on the chemistry that is taking place and allow us to see similarities between reactions that might not otherwise be apparent.

Let’s consider the reaction of silver nitrate with potassium dichromate above. When aqueous solutions of silver nitrate and potassium dichromate are mixed, silver dichromate forms as a red solid. The overall balanced chemical equation for the reaction shows each reactant and product as undissociated, electrically neutral compounds:

$\begin{matrix} (4.2.3) & 2 AgNO_{3} (aq) + K_{2} Cr_{2} O_{7} (aq) \to Ag_{2} Cr_{2} O_{7} (s) + 2 KNO_{3} (aq) \end{matrix}$

Although Equation $4.2.3$ gives the identity of the reactants and the products, it does not show the identities of the actual species in solution. Because ionic substances such as $AgNO_{3}$ and $K_{2} Cr_{2} O_{7}$ are strong electrolytes (i.e., they dissociate completely in aqueous solution to form ions). In contrast, because $Ag_{2} Cr_{2} O_{7}$ is not very soluble, it separates from the solution as a solid. To find out what is actually occurring in solution, it is more informative to write the reaction as a complete ionic equation showing which ions and molecules are hydrated and which are present in other forms and phases:

$\begin{matrix} (4.2.4) & 2 Ag^{+} (aq) + 2 NO_{3}^{-} (aq) + 2 K^{+} (aq) + Cr_{2} O_{7}^{2 -} (aq)  Ag_{2} Cr_{2} O_{7} (s) + 2 K^{+} (aq) + 2 NO_{3}^{-} (aq) \end{matrix}$

Note that $K^{+} (aq)$ and $NO_{3}^{-} (aq)$ ions are present on both sides of Equation $4.2.4$ and their coefficients are the same on both sides. These ions are called spectator ions because they do not participate in the actual reaction. Canceling the spectator ions gives the net ionic equation, which shows only those species that participate in the chemical reaction:

$\begin{matrix} (4.2.5) & 2 A g^{+} (a q) + C r_{2} O_{7}^{2 -} (a q) \to A g_{2} C r_{2} O_{7} (s) \end{matrix}$

Both mass and charge must be conserved in chemical reactions because the numbers of electrons and protons do not change. For charge to be conserved, the sum of the charges of the ions multiplied by their coefficients must be the same on both sides of the equation. In Equation $4.2.5$ , the charge on the left side is 2(+1) + 1(−2) = 0, which is the same as the charge of a neutral $Ag_{2} Cr_{2} O_{7}$ formula unit on the right side.

By eliminating the spectator ions, we can focus on the chemistry that takes place in a solution. For example, the overall chemical equation for the reaction between silver fluoride and ammonium dichromate is as follows:

$\begin{matrix} (4.2.6) & 2 A g F (a q) + {(N H_{4})}_{2} C r_{2} O_{7} (a q) \to A g_{2} C r_{2} O_{7} (s) + 2 N H_{4} F (a q) \end{matrix}$

The complete ionic equation for this reaction is as follows:

$\begin{matrix} (4.2.7) & 2 Ag^{+} (aq) + 2 F^{-} (aq) + 2 NH_{4}^{+} (aq) + Cr_{2} O_{7}^{2 -} (aq) \to Ag_{2} Cr_{2} O_{7} (s) + 2 NH_{4}^{+} (aq) + 2 F^{-} (aq) \end{matrix}$

Because two $NH_{4}^{+} (aq)$ and two $F^{-} (aq)$ ions appear on both sides of Equation $4.2.7$ , they are spectator ions. They can therefore be canceled to give the net ionic equation (Equation $4.2.8$ ), which is identical to Equation $4.2.5$ :

$\begin{matrix} (4.2.8) & 2 Ag^{+} (aq) + Cr_{2} O_{7}^{2 -} (aq)  Ag_{2} Cr_{2} O_{7} (s) \end{matrix}$

If we look at net ionic equations, it becomes apparent that many different combinations of reactants can result in the same net chemical reaction. For example, we can predict that silver fluoride could be replaced by silver nitrate in the preceding reaction without affecting the outcome of the reaction.

Determining the Products for Precipitation Reactions: Determining the Products for Precipitation Reactions, YouTube(opens in new window) [youtu.be]

Example $4.2.1$ : Balancing Precipitation Equations

Write the overall chemical equation, the complete ionic equation, and the net ionic equation for the reaction of aqueous barium nitrate with aqueous sodium phosphate to give solid barium phosphate and a solution of sodium nitrate.

Given: reactants and products

Asked for: overall, complete ionic, and net ionic equations

Strategy:

Write and balance the overall chemical equation. Write all the soluble reactants and products in their dissociated form to give the complete ionic equation; then cancel species that appear on both sides of the complete ionic equation to give the net ionic equation.

Solution:

From the information given, we can write the unbalanced chemical equation for the reaction:

$\begin{matrix} Ba (NO_{3})_{2} (aq) + Na_{3} PO_{4} (aq)  Ba_{3} (PO_{4})_{2} (s) + NaNO_{3} (aq) \end{matrix}$

Because the product is Ba₃(PO₄)₂, which contains three Ba²⁺ ions and two PO₄³⁻ ions per formula unit, we can balance the equation by inspection:

$\begin{matrix} 3 Ba (NO_{3})_{2} (aq) + 2 Na_{3} PO_{4} (aq)  Ba_{3} (PO_{4})_{2} (s) + 6 NaNO_{3} (aq) \end{matrix}$

This is the overall balanced chemical equation for the reaction, showing the reactants and products in their undissociated form. To obtain the complete ionic equation, we write each soluble reactant and product in dissociated form:

$\begin{matrix} 3 Ba^{2 +} (aq) + 6 NO_{3}^{-} (aq) + 6 Na^{+} (aq) + 2 PO_{4}^{3 -} (aq) \to Ba_{3} (PO_{4})_{2} (s) + 6 Na^{+} (aq) + 6 NO_{3}^{-} (aq) \end{matrix}$

The six NO₃⁻(aq) ions and the six Na⁺(aq) ions that appear on both sides of the equation are spectator ions that can be canceled to give the net ionic equation:

$\begin{matrix} 3 Ba^{2 +} (aq) + 2 PO_{4}^{3 -} (aq)  Ba_{3} (PO_{4})_{2} (s) \end{matrix}$

Exercise $4.2.1$ : Mixing Silver Fluoride with Sodium Phosphate

Write the overall chemical equation, the complete ionic equation, and the net ionic equation for the reaction of aqueous silver fluoride with aqueous sodium phosphate to give solid silver phosphate and a solution of sodium fluoride.

Answer

overall chemical equation:

$\begin{matrix} 3 AgF (aq) + Na_{3} PO_{4} (aq)  Ag_{3} PO_{4} (s) + 3 NaF (aq) \end{matrix}$

complete ionic equation:

$\begin{matrix} 3 Ag^{+} (aq) + 3 F^{-} (aq) + 3 Na^{+} (aq) + PO_{4}^{3 -} (aq)  Ag_{3} PO_{4} (s) + 3 Na^{+} (aq) + 3 F^{-} (aq) \end{matrix}$

net ionic equation:

$\begin{matrix} 3 Ag^{+} (aq) + PO_{4}^{3 -} (aq)  Ag_{3} PO_{4} (s) \end{matrix}$

So far, we have always indicated whether a reaction will occur when solutions are mixed and, if so, what products will form. As you advance in chemistry, however, you will need to predict the results of mixing solutions of compounds, anticipate what kind of reaction (if any) will occur, and predict the identities of the products. Students tend to think that this means they are supposed to “just know” what will happen when two substances are mixed. Nothing could be further from the truth: an infinite number of chemical reactions is possible, and neither you nor anyone else could possibly memorize them all. Instead, you must begin by identifying the various reactions that could occur and then assessing which is the most probable (or least improbable) outcome.

The most important step in analyzing an unknown reaction is to write down all the species—whether molecules or dissociated ions—that are actually present in the solution (not forgetting the solvent itself) so that you can assess which species are most likely to react with one another. The easiest way to make that kind of prediction is to attempt to place the reaction into one of several familiar classifications, refinements of the five general kinds of reactions (acid–base, exchange, condensation, cleavage, and oxidation–reduction reactions). In the sections that follow, we discuss three of the most important kinds of reactions that occur in aqueous solutions: precipitation reactions (also known as exchange reactions), acid–base reactions, and oxidation–reduction reactions.

Predicting Solubilities

Table $4.2.1$ gives guidelines for predicting the solubility of a wide variety of ionic compounds. To determine whether a precipitation reaction will occur, we identify each species in the solution and then refer to Table $4.2.1$ to see which, if any, combination(s) of cation and anion are likely to produce an insoluble salt. In doing so, it is important to recognize that soluble and insoluble are relative terms that span a wide range of actual solubilities. We will discuss solubilities in more detail later, where you will learn that very small amounts of the constituent ions remain in solution even after precipitation of an “insoluble” salt. For our purposes, however, we will assume that precipitation of an insoluble salt is complete.

Table $4.2.1$ : Guidelines for Predicting the Solubility of Ionic Compounds in Water
	Soluble		Exceptions
Rule 1	most salts that contain an alkali metal (Li⁺, Na⁺, K⁺, Rb⁺, and Cs⁺) and ammonium (NH₄⁺)
Rule 2	most salts that contain the nitrate (NO₃⁻) anion
Rule 3	most salts of anions derived from monocarboxylic acids (e.g., CH₃CO₂⁻)	but not	silver acetate and salts of long-chain carboxylates
Rule 4	most chloride, bromide, and iodide salts	but not	salts of metal ions located on the lower right side of the periodic table (e.g., Cu⁺, Ag⁺, Pb²⁺, and Hg₂²⁺).
	Insoluble		Exceptions
Rule 5	most salts that contain the hydroxide (OH⁻) and sulfide (S²⁻) anions	but not	salts of the alkali metals (group 1), the heavier alkaline earths (Ca²⁺, Sr²⁺, and Ba²⁺ in group 2), and the NH₄⁺ ion.
Rule 6	most carbonate (CO₃²⁻) and phosphate (PO₄³⁻) salts	but not	salts of the alkali metals or the NH₄⁺ ion.
Rule 7	most sulfate (SO₄²⁻) salts that contain main group cations with a charge ≥ +2	but not	salts of +1 cations, Mg²⁺, and dipositive transition metal cations (e.g., Ni²⁺)

Just as important as predicting the product of a reaction is knowing when a chemical reaction will not occur. Simply mixing solutions of two different chemical substances does not guarantee that a reaction will take place. For example, if 500 mL of a 1.0 M aqueous NaCl solution is mixed with 500 mL of a 1.0 M aqueous KBr solution, the final solution has a volume of 1.00 L and contains 0.50 M Na⁺(aq), 0.50 M Cl⁻(aq), 0.50 M K⁺(aq), and 0.50 M Br⁻(aq). As you will see in the following sections, none of these species reacts with any of the others. When these solutions are mixed, the only effect is to dilute each solution with the other (Figure $4.2.1$ ).

Diagrams showing three flasks with illustrations of particles inside, labeled 'Molarity' and 'Mole' on the sides. — Figure $4.2.1$ : The Effect of Mixing Aqueous KBr and NaCl Solutions. Because no net reaction occurs, the only effect is to dilute each solution with the other. (Water molecules are omitted from molecular views of the solutions for clarity.)

Example $4.2.2$

Using the information in Table $4.2.1$ , predict what will happen in each case involving strong electrolytes. Write the net ionic equation for any reaction that occurs.

Aqueous solutions of barium chloride and lithium sulfate are mixed.
Aqueous solutions of rubidium hydroxide and cobalt(II) chloride are mixed.
Aqueous solutions of strontium bromide and aluminum nitrate are mixed.
Solid lead(II) acetate is added to an aqueous solution of ammonium iodide.

Given: reactants

Asked for: reaction and net ionic equation

Strategy:

Identify the ions present in solution and write the products of each possible exchange reaction.
Refer to Table $4.2.1$ to determine which, if any, of the products is insoluble and will therefore form a precipitate. If a precipitate forms, write the net ionic equation for the reaction.

Solution:

A Because barium chloride and lithium sulfate are strong electrolytes, each dissociates completely in water to give a solution that contains the constituent anions and cations. Mixing the two solutions initially gives an aqueous solution that contains Ba²⁺, Cl⁻, Li⁺, and SO₄²⁻ ions. The only possible exchange reaction is to form LiCl and BaSO₄:

Mathematical equation with terms involving logarithms and variables, arranged in a circular format.

B We now need to decide whether either of these products is insoluble. Table $4.2.1$ shows that LiCl is soluble in water (rules 1 and 4), but BaSO₄ is not soluble in water (rule 5). Thus BaSO₄ will precipitate according to the net ionic equation

$\begin{matrix} B a^{2 +} (a q) + S O_{4}^{2 -} (a q) \to B a S O_{4} (s) \end{matrix}$

Although soluble barium salts are toxic, BaSO₄ is so insoluble that it can be used to diagnose stomach and intestinal problems without being absorbed into tissues. An outline of the digestive organs appears on x-rays of patients who have been given a “barium milkshake” or a “barium enema”—a suspension of very fine BaSO₄ particles in water.

X-ray image showing the lower digestive tract, including the colon and part of the spine. — An x-ray of the digestive organs of a patient who has swallowed a “barium milkshake.” A barium milkshake is a suspension of very fine BaSO₄ particles in water; the high atomic mass of barium makes it opaque to x-rays. from Wikipedia.

A Rubidium hydroxide and cobalt(II) chloride are strong electrolytes, so when aqueous solutions of these compounds are mixed, the resulting solution initially contains Rb⁺, OH⁻, Co²⁺, and Cl⁻ ions. The possible products of an exchange reaction are rubidium chloride and cobalt(II) hydroxide):

B According to Table $4.2.1$ , RbCl is soluble (rules 1 and 4), but Co(OH)₂ is not soluble (rule 5). Hence Co(OH)₂ will precipitate according to the following net ionic equation:

$C o^{2 +} (a q) + 2 O H^{-} (a q) \to C o {(O H)}_{2} (s)$
A When aqueous solutions of strontium bromide and aluminum nitrate are mixed, we initially obtain a solution that contains Sr²⁺, Br⁻, Al³⁺, and NO₃⁻ ions. The two possible products from an exchange reaction are aluminum bromide and strontium nitrate:

Mathematical equation involving variables x, y, and z, implying a relationship among them.

B According to Table $4.2.1$ , both AlBr₃ (rule 4) and Sr(NO₃)₂ (rule 2) are soluble. Thus no net reaction will occur.

A According to Table $4.2.1$ , lead acetate is soluble (rule 3). Thus solid lead acetate dissolves in water to give Pb²⁺ and CH₃CO₂⁻ ions. Because the solution also contains NH₄⁺ and I⁻ ions, the possible products of an exchange reaction are ammonium acetate and lead(II) iodide:

B According to Table $4.2.1$ , ammonium acetate is soluble (rules 1 and 3), but PbI₂ is insoluble (rule 4). Thus Pb(C₂H₃O₂)₂ will dissolve, and PbI₂ will precipitate. The net ionic equation is as follows:

$P b^{2 +} (a q) + 2 I^{-} (a q) \to P b I_{2} (s)$

Exercise $4.2.2$

Using the information in Table $4.2.1$ , predict what will happen in each case involving strong electrolytes. Write the net ionic equation for any reaction that occurs.

An aqueous solution of strontium hydroxide is added to an aqueous solution of iron(II) chloride.
Solid potassium phosphate is added to an aqueous solution of mercury(II) perchlorate.
Solid sodium fluoride is added to an aqueous solution of ammonium formate.
Aqueous solutions of calcium bromide and cesium carbonate are mixed.

Answer a: $F e^{2 +} (a q) + 2 O H^{-} (a q) \to F e {(O H)}_{2} (s)$
Answer b: $2 P O_{4}^{3 -} (a q) + 3 H g^{2 +} (a q) \to H g_{3} {(P O_{4})}_{2} (s)$
Answer c: $N a F (s)$ dissolves; no net reaction
Answer d: $C a^{2 +} (a q) + C O_{3}^{2 -} (a q) \to C a C O_{3} (s)$

Predicting the Solubility of Ionic Compounds: Predicting the Solubility of Ionic Compounds, YouTube(opens in new window) [youtu.be] (opens in new window)

Learning Objectives

Exchange (Double-Displacement) Reactions

Example 4.2.1: Balancing Precipitation Equations

Strategy:

Solution:

Exercise 4.2.1: Mixing Silver Fluoride with Sodium Phosphate

Predicting Solubilities

Example 4.2.2

Strategy:

Solution:

Exercise 4.2.2

Example $4.2.1$ : Balancing Precipitation Equations

Exercise $4.2.1$ : Mixing Silver Fluoride with Sodium Phosphate

Example $4.2.2$

Exercise $4.2.2$