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# 17.8: Reactions Between Ions in Solutions

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## Lesson Objectives

The student will:

• use a solubility chart or solubility rules to determine whether or not a given substance is soluble in water.
• use a solubility chart or solubility rules to determine if a precipitate is likely when specified solutions are combined.
• write molecular, ionic, and net ionic equations.
• identify spectator ions in ionic equations.
• explain what is meant by a precipitation reaction and give an example of one.
• identify the spectator ions in any precipitation reaction.
• write the balanced complete ionic equation and the net ionic equation for any precipitation reaction.

## Vocabulary

• gravimetric analysis
• ionic equation
• net ionic equation
• precipitate
• precipitation reaction
• spectator ions

## Introduction

Many ionic compounds are said to be soluble in water, while others are said to be insoluble. However, no ionic compound is completely insoluble in water. Every ionic compound dissociates into its ions to some extent when placed in water. In fact, the solubility of ionic compounds ranges across a full spectrum from as little solubility as $1 \times 10^{-100}$ moles/liter to $20$ moles/liter. When solutes are dissolved in a solution, the solution is transparent so the dissolved solute particles cannot be visually detected. If undissolved particles are present in a liquid, they form a cloudy barrier to light passing through the liquid and hence their presence can be detected visually. Eventually, the un-dissolved particles will settle to the bottom of the container, making it more apparent that an un-dissolved solid is present.

When ionic solutions are mixed together, it is often possible to form an insoluble ionic compound even though both original compounds were soluble. For example, both silver nitrate and sodium chloride are soluble compounds. In $1.0 \ \mathrm{M}$ solutions of these substances, the compounds would completely dissociate. One solution would contain $\mathrm{Ag}^+$ and $\mathrm{NO}_3^-$ ions, and the other solution would contain $\mathrm{Na}^+$ and $\mathrm{Cl}^-$ ions. When these two solutions are poured together, all of these ions move around in the solution and come into contact with each other. When a silver ion combines with a chloride ion, they form an insoluble solid particle that will not dissolve. Therefore, when the two solutions are mixed, a cloudy, non-transparent substance forms that will eventually settle to the bottom of the container, as illustrated below. When a non-soluble substance is formed in a solution, it is called a precipitate. A precipitation reaction is a reaction in which soluble ions in separate solutions are mixed together to form an insoluble compound that settles out of the solution.

## Product Precipitates

Once a solid substance has been separated into its ions, the ions are then available for reactions. When a compound is in the solid state, the ions are held with electrostatic attractions. It must first be dissolved in solution before the ions can move freely. Take for example the reaction between sodium chloride and silver nitrate. Both of these compounds are available commonly in the solid form. First, both solids are dissolved in water, as seen below:

$\mathrm{AgNO}_{3(s)} \rightarrow \mathrm{Ag}^+_{(aq)} + \mathrm{NO}_{3(aq)}^-$
$\mathrm{NaCl}_{(s)} \rightarrow \mathrm{Na}^+_{(aq)} + \mathrm{Cl}^-_{(aq)}$

Once the solid dissolves to separate into its ions in solution, these ions are available to react together in the following chemical reaction:

$\mathrm{AgNO}_{3(aq)} + \mathrm{NaCl}_{(aq)} \rightarrow \mathrm{AgCl}_{(s)} + \mathrm{NaNO}_{3(aq)}$

Make note that the $\mathrm{AgNO}_{3(aq)}$ and $\mathrm{NaCl}_{(aq)}$ reactants show the ions are in solution. In other words, $\mathrm{AgNO}_{3(aq)}$ is equivalent to $\mathrm{Ag}^+_{(aq)} + \mathrm{NO}^-_{3(aq)}$, and $\mathrm{NaCl}_{(aq)}$ is equivalent to $\mathrm{Na}^+_{(aq)} + \mathrm{Cl}^-_{(aq)}$. The equation represented above is a double displacement reaction, which means the cations exchange anions in the reactants to form the products. The precipitate that will form in this case is silver chloride.

The same reactions can be seen when substances undergo ionization. Remember that ionization forms ions in solution. For example, look at the equations below for the ionization of sulfuric acid and the dissolution of sodium hydroxide. Notice how they both end up with ions in solution.

$\mathrm{H}_2\mathrm{SO}_{4(aq)} \rightarrow 2 \ \mathrm{H}^+_{(aq)} + \mathrm{SO}^{2-}_{4(aq)}$
$\mathrm{NaOH}_{(aq)} \rightarrow \mathrm{Na}^+_{(aq)} + \mathrm{OH}^-_{(aq)}$

When sulfuric acid reacts with sodium hydroxide, we have a double displacement reaction where the cations exchange anions, as seen below:

$\mathrm{H}_2\mathrm{SO}_{4(aq)} + 2 \ \mathrm{NaOH}_{(aq)} \rightarrow 2 \ \mathrm{H}_2\mathrm{O}_{(l)} + \mathrm{Na}_2\mathrm{SO}_{4(aq)}$

Notice that liquid water is produced, not a solid. The second product is an aqueous ionic solution containing the ions $\mathrm{Na}^+_{(aq)}$ and $\mathrm{SO}^{2-}_{4(aq)}$ (illustrated below).

## Solubility Charts and Solubility Rules

How do you know if a precipitate is produced in a double replacement reaction? Not all reactions of this type will produce a precipitate. If you were to mix a solution of table salt ($\mathrm{NaCl}_{(aq)}$) and Epsom salts ($\mathrm{MgSO}_{4(aq)}$) in water, you would not get a precipitate. The reaction is seen below:

$2 \ \mathrm{NaCl}_{(aq)} + \mathrm{MgSO}_{4(aq)} \rightarrow \mathrm{Na}_2\mathrm{SO}_{4(aq)} + \mathrm{MgCl}_{2(aq)}$

Scientists use a set of solubility rules or a solubility chart to determine whether or not a precipitate will form. Table below represents the solubility chart for the most common cations and anions found in ionic solids.

Solubility Chart
$\mathrm{C}_2\mathrm{H}_3\mathrm{O}_2^-$ $\mathrm{Br}^-$ $\mathrm{CO}_3^{2-}$ $\mathrm{Cl}^-$ $\mathrm{ClO}_3^-$ $\mathrm{CrO}_4^{2-}$ $\mathrm{I}^-$ $\mathrm{NO}_3^-$ $\mathrm{OH}^-$ $\mathrm{O}^{2-}$ $\mathrm{PO}_4^{3-}$ $\mathrm{SO}_4^{2-}$ $\mathrm{S}^{2-}$
$\mathrm{Ag}^+$ ii I I I S ii I S * ii I ii I
$\mathrm{Al}^{3+}$ S S * S S * S S I I I S *
$\mathrm{Ba}^{2+}$ S S ii S S I S S S S I I *
$\mathrm{Ca}^{2+}$ S S ii S S ii S S ii ii ii ii ii
$\mathrm{Cu}^{2+}$ S S * S S * * S I I I S I
$\mathrm{Fe}^{2+}$ S S ii S S * S S I I I S I
$\mathrm{Fe}^{3+}$ S S * S S I S S I I ii ii *
$\mathrm{Hg}_2^{2+}$ S S * S S ii ii S I ii I I I
$\mathrm{K}^+$ S S S S S S S S S S S S S
$\mathrm{Mg}^{2+}$ S S ii S S ii S S I I ii S *
$\mathrm{Mn}^{2+}$ S S ii S S * S S I I ii S I
$\mathrm{Na}^+$ S S S S S S S S S S S S S
$\mathrm{NH}_4^+$ S S S S S S S S S S S S S
$\mathrm{Pb}^{2+}$ S I I I S I I S ii ii I ii I
$\mathrm{Sn}^{2+}$ * S * S S I S S I I I S I
$\mathrm{Sn}^{4+}$ S S * * S ii * S ii I * S I
$\mathrm{Sr}^{2+}$ S S ii S S ii S S S S I ii S
$\mathrm{Zn}^{2+}$ S S ii S S ii S S I ii I S I

$\text{S} = \text{soluble in water}, \text{I} = \text{insoluble in water}, \text{ii} = \text{partially soluble in water}, * = \text{unknown or does not exist}.$

Let’s now see how we use the solubility chart to determine if two compounds will form a precipitate when they react. If we had a reaction between sodium bromide and silver nitrate, we know that this is a reaction between two compounds and therefore is a double replacement reaction. How do we know the states of the products formed? The reaction is seen below.

$\mathrm{NaBr}_{(aq)} + \mathrm{AgNO}_{3(aq)} \rightarrow \mathrm{NaNO}_{3(?)} + \mathrm{AgBr}_{(?)}$

Look at the solubility chart and see if you can predict if the reaction will produce any precipitates. If you look across the row for sodium ion, all sodium compounds are soluble (S), therefore you can fill in this part of the equation:

$\mathrm{NaBr}_{(aq)} + \mathrm{AgNO}_{3(aq)} \rightarrow \mathrm{NaNO}_{3(aq)} + \mathrm{AgBr}_{(?)}$

If you look across the row for silver, under bromide ion, you find an I for insoluble.

Therefore we can complete the equation:

$\mathrm{NaBr}_{(aq)} + \mathrm{AgNO}_{3(aq)} \rightarrow \mathrm{NaNO}_{3(aq)} + \mathrm{AgBr}_{(s)}$

Let’s try another one. Take the reaction between ammonium phosphate and lead acetate.

$2 \ (\mathrm{NH}_4)_3\mathrm{PO}_{4(aq)} + 3 \ \mathrm{Pb(C}_2\mathrm{H}_3\mathrm{O}_2)_{2(aq)} \rightarrow 6 \ \mathrm{NH}_4\mathrm{C}_2\mathrm{H}_3\mathrm{O}_{2(aq)} + \mathrm{Pb}_3(\mathrm{PO}_4)_{2(s)}$

How did we know that the lead(II) phosphate would precipitate from solution? Why did the ammonium acetate not precipitate? If we follow the ammonium row in the solubility table across to the acetate column, we find an “S” at the intersection. This indicates that this compound is soluble. If we follow the lead row across to the phosphate column, we find an “I” at the intersection. This indicates that this compound is insoluble and therefore a precipitate of this compound will form.

Rather than using a solubility chart, some scientists simply use a set of solubility rules. Solubility rules are summaries of information about which ionic compounds are soluble in aqueous solutions and which are not. The rules are as follows:

1. All group 1 metals and ammonium compounds are soluble.
2. All nitrates, chlorates, and bicarbonates are soluble.
3. Halides are soluble except for $\mathrm{Ag}^+$, $\mathrm{Hg}_2^{2+}$, and $\mathrm{Pb}^{2+}$.
4. Sulfates are soluble except for $\mathrm{Ag}^+$, $\mathrm{Ba}^{2+}$, $\mathrm{Ca}^{2+}$, $\mathrm{Hg}_2^{2+}$, $\mathrm{Sr}^{2+}$, and $\mathrm{Pb}^{2+}$.
5. Carbonates, chromates, phosphates, and sulfides are insoluble except those from rule #1.
6. Hydroxides are insoluble except for those in rule #1 and $\mathrm{Ba}^{2+}$.

It is important to remember that this is a priority set of rules. What this means is that Rule #1 is first. All group 1 metals and ammonium compounds are always soluble. For example, even though sulfide compounds are rarely soluble for any cation (rule #5), they will be soluble with group 1 metal ions or with ammonium ions (rule #1). It also does not matter whether you use the set of rules or the solubility chart. They both provide the same information; the chart is easier to read for some, while the rules are easier to remember for others.

Example:

Complete the following reactions. Use the solubility table to predict whether precipitates will form in each of the reactions.

1. $\mathrm{Pb(NO}_3)_{2(aq)} + \mathrm{KI}_{(aq)} \rightarrow$
2. $\mathrm{BaCl}_{2(aq)} + \mathrm{Na}_2\mathrm{SO}_{4(aq)} \rightarrow$

Solution:

1. $\mathrm{Pb(NO}_3)_{2(aq)} + 2 \ \mathrm{KI}_{(aq)} \rightarrow \mathrm{PbI}_{2(s)} + 2 \ \mathrm{KNO}_{3(aq)}$
2. $\mathrm{BaCl}_{2(aq)} + \mathrm{Na}_2\mathrm{SO}_{4(aq)}\rightarrow \mathrm{BaSO}_{4(s)} + 2 \ \mathrm{NaCl}_{(aq)}$

Suppose that a chemist has a solution that contains both $\mathrm{Pb}^{2+}$ and $\mathrm{Zn}^{2+}$ ions. If these two ions were dissolved in the solution as nitrates, then the only anion present is the nitrate ion. If the chemist added some $\mathrm{NaCl}$ to the solution, the zinc ions would remain in solution because $\mathrm{ZnCl}_2$ is soluble, but the lead ions and the chloride ions would form the precipitate $\mathrm{PbCl}_2$. If this mixture is poured through a piece of filter paper, the dissolved zinc ions would pass through the filter paper with the solution but the solid $\mathrm{PbCl}_2$ would be filtered out (illustrated below). Therefore, the chemist would have separated the zinc ions (now in the solution) and the lead ions (now in the filter paper). This process for separating and identifying ions by selective precipitation and filtration is known as gravimetric analysis.

If you are called upon to determine a process for separating ions from each other, you should look in the solubility table to determine a reagent that will form a precipitate with one of the ions but not with the other.

## Ionic Equations

Remember earlier we said that $\mathrm{NaCl}_{(aq)}$ should be visualized as $\mathrm{Na}^+_{(aq)}$ and $\mathrm{Cl}^-_{(aq)}$, and $\mathrm{AgNO}_{3(aq)}$ as $\mathrm{Ag}^+_{(aq)}$ and $\mathrm{NO}^-_{3(aq)}$. This is so we can write chemical reactions as molecular or formula equations. The equation below is an example of a formula equation.

$\mathrm{NaCl}_{(aq)} + \mathrm{AgNO}_{3(aq)} \rightarrow \mathrm{NaNO}_{3(aq)} + \mathrm{AgCl}_{(s)}$

In an ionic equation, the separated ions are written in the chemical equation. Let’s rewrite the equation above as a total ionic equation, which is a better representation of a double replacement reaction.

$\mathrm{Na}^+_{(aq)} + \mathrm{Cl}^-_{(aq)} + \mathrm{Ag}^+_{(aq)} + \mathrm{NO}^-_{3(aq)} \rightarrow \mathrm{Na}^+_{(aq)} + \mathrm{NO}^-_{3(aq)} + \mathrm{AgCl}_{(s)}$

Example:

Write the ionic equation for each of the following.

1. $\mathrm{BaCl}_{2(aq)} + \mathrm{Na}_2\mathrm{SO}_{4(aq)} \rightarrow 2 \ \mathrm{NaCl}_{(aq)} + \mathrm{BaSO}_{4(s)}$
2. $2 \ \mathrm{K}_3\mathrm{PO}_{4(aq)} + 3 \ \mathrm{Ca(NO}_3)_{2(aq)} \rightarrow 6 \ \mathrm{KNO}_{3(aq)} + \mathrm{Ca}_3(\mathrm{PO}_4)_{2(s)}$

Solution:

1. $\mathrm{Ba}^{2+}_{(aq)} + 2 \ \mathrm{Cl}^-_{(aq)} + 2 \ \mathrm{Na}^+_{(aq)} + \mathrm{SO}^{2-}_{4(aq)} \rightarrow 2 \ \mathrm{Na}^+_{(aq)} + 2 \ \mathrm{Cl}^-_{(aq)} + \mathrm{BaSO}_{4(s)}$
2. $6 \ \mathrm{K}^+_{(aq)} + 2 \ \mathrm{PO}^{3-}_{4(aq)} + 3 \ \mathrm{Ca}^{2+}_{(aq)} + 6 \ \mathrm{NO}^-_{3(aq)} \rightarrow 6 \ \mathrm{K}^+_{(aq)} + 6 \ \mathrm{NO}^-_{3(aq)} + \mathrm{Ca}_3(\mathrm{PO}_4)_{2(s)}$

## Net Ionic Equations

If you look at the equation below, what do you notice is the same on both sides of the equation?

$\mathrm{Na}^+_{(aq)} + \mathrm{Cl}^-_{(aq)} + \mathrm{Ag}^+_{(aq)} + \mathrm{NO}^-_{3(aq)} \rightarrow \mathrm{Na}^+_{(aq)} + \mathrm{NO}^-_{3(aq)} + \mathrm{AgCl}_{(s)}$

Do you see that $\mathrm{Na}^+_{(aq)}$ and $\mathrm{NO}^-_{3(aq)}$ appear on both sides of this equation in the same form? These ions, because they appear on both sides of the equation and in the same form, are called spectator ions. A spectator ion is an ion in the ionic equation that appears in the same form on both sides of the equation, indicating they do not participate in the overall reaction. Therefore $\mathrm{Na}^+_{(aq)}$ and $\mathrm{NO}^-_{3(aq)}$ are spectator ions for this reaction. By removing the spectator ions, the overall reaction will become:

$\mathrm{Ag}^+_{(aq)} + \mathrm{Cl}^-_{(aq)} \rightarrow \mathrm{AgCl}_{(s)}$

Without the spectator ions, we see only the ions that are responsible for forming the solid silver chloride. This equation, represented above, is the net ionic equation. The net ionic equation is the overall equation that results when spectator ions are removed from the ionic equation. The net ionic equation gives us all of the essential information we need: what ions we need to form our solid. In this case, it is irrelevant whether we had sodium chloride or potassium chloride, as what is important is that the chloride ion and silver ion are present.

Example:

Write the net ionic equation for each of the following. Name the spectator ions.

1. $\mathrm{Ba}^{2+}_{(aq)} + 2 \ \mathrm{Cl}^-_{(aq)} + 2 \ \mathrm{Na}^+_{(aq)} + \mathrm{SO}^{2-}_{4(aq)} \rightarrow 2 \ \mathrm{Na}^+_{(aq)} + 2 \ \mathrm{Cl}^-_{(aq)} + \mathrm{BaSO}_{4(s)}$
2. $6 \ \mathrm{K}^+_{(aq)} + 2 \ \mathrm{PO}^{3-}_{4(aq)} + 3 \ \mathrm{Ca}^{2+}_{(aq)} + 6 \ \mathrm{NO}^-_{3(aq)} \rightarrow 6 \ \mathrm{K}^+_{(aq)} + 6 \ \mathrm{NO}^-_{3(aq)} + \mathrm{Ca}_3(\mathrm{PO}_4)_{2(s)}$

Solution:

1. $\mathrm{Ba}^{2+}_{(aq)} + 2 \ \mathrm{Cl}^-_{(aq)} + 2 \ \mathrm{Na}^+_{(aq)} + \mathrm{SO}^{2-}_{4(aq)} \rightarrow 2 \ \mathrm{Na}^+_{(aq)} + 2 \ \mathrm{Cl}^-_{(aq)} + \mathrm{BaSO}_{4(s)}$
Net ionic equation: $\mathrm{Ba}^{2+}_{(aq)} + \mathrm{SO}^{2-}_{4(aq)} \rightarrow \mathrm{BaSO}_{4(s)}$
Spectator ions: $\mathrm{Cl}^-_{(aq)}$ and $\mathrm{Na}^+_{(aq)}$
1. $6 \ \mathrm{K}^+_{(aq)} + 2 \ \mathrm{PO}^{3-}_{4(aq)} + 3 \ \mathrm{Ca}^{2+}_{(aq)} + 6 \ \mathrm{NO}^-_{3(aq)} \rightarrow 6 \ \mathrm{K}^+_{(aq)} + 6 \ \mathrm{NO}^-_{3(aq)} + \mathrm{Ca}_3(\mathrm{PO}_4)_{2(s)}$
Net ionic equation: $2 \ \mathrm{PO}^{3-}_{4(aq)} + 3 \ \mathrm{Ca}^{2+}_{(aq)} \rightarrow \mathrm{Ca}_3(\mathrm{PO}_4)_{2(s)}$
Spectator ions: $\mathrm{K}^+_{(aq)}$ and $\mathrm{NO}^-_{3(aq)}$

## Lesson Summary

• A solubility chart is a grid showing the possible combinations of cations and anions and their solubilities in water. It is used to determine whether a precipitate is formed in a chemical reaction.
• Solubility rules are summaries of information about which ionic compounds are soluble in aqueous solutions and which are not.
• A total ionic equation is one in which all of the ions in a reaction are represented.
• A net ionic equation is one in which only the ions that produce the precipitate are represented.

This web site video shows a series of precipitation reactions.

This video is a ChemStudy film called “Molecular Spectroscopy.” The film is somewhat dated but the information is accurate.

## Review Questions

1. What is more valuable to use for determining solubility: a solubility chart or a set of solubility rules?
2. If you were told to visualize $\mathrm{Cu(NO}_3)_{2(aq)}$, what might this mean to you?

Use the solubility rules to determine the following solubilities in water.

1. Which of the following compounds is soluble in water?
1. $\mathrm{PbCl}_2$
2. $\mathrm{Hg}_2\mathrm{Cl}_2$
3. $(\mathrm{NH}_4)_2\mathrm{SO}_4$
4. $\mathrm{MgCO}_3$
5. $\mathrm{AgNO}_3$
6. $\mathrm{MgCl}_2$
7. $\mathrm{KOH}$
8. $\mathrm{PbSO}_4$
2. When only the ions that produce a precipitate are shown for a chemical equation, what type of reaction exists?
1. spectator equation
2. molecular equation
3. ionic equation
4. net ionic equation
3. Complete the following reactions:
1. $\mathrm{Na}_2\mathrm{S}_{(aq)} + \mathrm{ZnCl}_{2(aq)} \rightarrow$
2. $(\mathrm{NH}_4)_2\mathrm{CO}_{3(aq)} + \mathrm{CaCl}_{2(aq)} \rightarrow$
4. Write the ionic equations for the balanced molecular equations from question 7.
5. Write the net ionic equations for the ionic equations from question 8.
6. Identify the spectator ions for the ionic equations from question 8.

All images, unless otherwise stated, are created by the CK-12 Foundation and are under the Creative Commons license CC-BY-NC-SA.

Feb 23, 2012

Aug 01, 2014