Reactions
Beakers 4
Chemical with precipitates
of lead(II) iodide and
mercury(II) iodide.
C
hemical reactions are the heart of chemistry. Some reactions, such as those
accompanying a forest re or the explosion of dynamite, are quite dramatic.
Others are much less obvious, although all chemical reactions must involve
detectable change. A chemical reaction involves a change from reactant substances
to product substances, and the product substances will have physical and chemical properties different from those of the reactants.
Figure 4.1 shows an experimenter adding a colorless solution of potassium
iodide, KI, to a colorless solution of lead(II) nitrate, Pb(NO3)2. What you see is the formation of a cloud of bright yellow crystals where the two solutions have come into
contact, clear evidence of a chemical reaction. The bright yellow crystals are lead(II)
iodide, PbI2, one of the reaction products. We call a solid that forms during a chemical reaction in solution a precipitate; the reaction is a precipitation reaction.
In this chapter, we will discuss the major types of chemical reactions, including
precipitation reactions. Some of the most important reactions we will describe involve
ions in aqueous (water) solution. Therefore, we will rst look at these ions and see how
we represent by chemical equations the reactions involving ions in aqueous solution.
Some questions we will answer are: What is the evidence for ions in solution?
How do we write chemical equations for reactions involving ions? How can we classify and describe the many reactions we observe so that we can begin to understand
them? What is the quantitative description of solutions and reactions in solution?
CONTENTS
Ions in Aqueous Solution
4.1 Ionic Theory of Solutions and
Solubility Rules
4.2 Molecular and Ionic Equations
Types of Chemical Reactions
4.3
4.4
4.5
4.6
Precipitation Reactions
AcidBase Reactions
OxidationReduction Reactions
Balancing Simple
OxidationReduction Equations
Working with Solutions
4.7 Molar Concentration
4.8 Diluting Solutions
Quantitative Analysis
4.9 Gravimetric Analysis
4.10 Volumetric Analysis
123
124
CHAPTER 4
Chemical Reactions
Figure 4.1
Reaction of potassium iodide
solution and lead(II) nitrate
solution
The reactant solutions are
colorless, but one of the
products, lead(II) iodide, forms
as a yellow precipitate.
Ions in Aqueous Solution
You probably have heard that you should not operate electrical equipment while standing in water. And you may have read a murder mystery in which the victim was electrocuted when an electrical appliance accidentally fell into his or her bath water.
Actually, if the water were truly pure, the person would be safe from electrocution,
because pure water is a nonconductor of electricity. Bath water or water as it ows
from the faucet, however, is a solution of water with small amounts of dissolved substances in it, and these dissolved substances make the solution an electrical conductor. This allows an electric current to ow from an electrical appliance to the human
body. Let us look at the nature of such solutions.
4.1 Ionic Theory of Solutions and Solubility Rules
Arrhenius submitted his ionic
theory as part of his doctoral
dissertation to the faculty at
Uppsala, Sweden, in 1884. It
was not well received and he
barely passed. In 1903,
however, he was awarded the
Nobel Prize in chemistry for
this theory.
Chemists began studying the electrical behavior of substances in the early nineteenth
century, and they knew that you could make pure water electrically conducting by
dissolving certain substances in it. In 1884, the young Swedish chemist Svante Arrhenius proposed the ionic theory of solutions to account for this conductivity. He said
that certain substances produce freely moving ions when they dissolve in water, and
these ions conduct an electric current in an aqueous solution.
Suppose you dissolve sodium chloride, NaCl, in water. From our discussion in
Section 2.6, you may remember that sodium chloride is an ionic solid consisting of
sodium ions, Na , and chloride ions, Cl , held in a regular, xed array. When you dissolve solid sodium chloride in water, the Na and Cl ions go into solution as freely
moving ions. Now suppose you dip electric wires that are connected to the poles of a
battery into a solution of sodium chloride. The wire that connects to the positive pole
of the battery attracts the negatively charged chloride ions in solution, because of their
opposite charges. Similarly, the wire connected to the negative pole of the battery
attracts the positively charged sodium ions in solution (Figure 4.2). Thus, the ions in
the solution begin to move, and these moving charges form the electric current in the
solution. (In a wire, it is moving electrons that constitute the electric current.)
4.1
Ionic Theory of Solutions and Solubility Rules
125
Now consider pure water. Water consists of molecules, each of which is electrically neutral. Since each molecule carries no net electric charge, it carries no overall
electric charge when it moves. Thus, pure water is a nonconductor of electricity.
In summary, although water is itself nonconducting, it has the ability to dissolve
various substances, some of which go into solution as freely moving ions. An aqueous solution of ions is electrically conducting.
Battery
+
Cl
Na+
Figure 4.2
Motion of ions in solution
Ions are in xed positions in a
crystal. During the solution
process, however, ions leave the
crystal and become freely
moving. Note that Na ions
(small gray spheres) are attracted
to the negative wire, whereas Cl
ions (large green spheres) are
attracted to the positive wire.
Electrolytes and Nonelectrolytes
We can divide the substances that dissolve in water into two broad classes, electrolytes
and nonelectrolytes. An electrolyte is a substance that dissolves in water to give an
electrically conducting solution. Sodium chloride, table salt, is an example of an electrolyte. When most ionic substances dissolve in water, ions that were in xed sites in
the crystalline solid go into the surrounding aqueous solution, where they are free to
move about. The resulting solution is conducting because the moving ions form an
electric current. Thus, in general, ionic solids that dissolve in water are electrolytes.
Not all electrolytes are ionic substances. Certain molecular substances dissolve
in water to form ions. The resulting solution is electrically conducting, and so we say
that the molecular substance is an electrolyte. An example is hydrogen chloride gas,
HCl(g), which is a molecular substance. Hydrogen chloride gas dissolves in water,
giving HCl(aq), which in turn produces hydrogen ions, H , and chloride ions, Cl ,
in aqueous solution. (The solution of H and Cl ions is called hydrochloric acid.)
HO
HCl(aq) 2 H (aq)
Cl ( aq)
We will look more closely at molecular electrolytes, such as HCl, at the end of this
section.
A nonelectrolyte is a substance that dissolves in water to give a nonconducting
or very poorly conducting solution. A common example is sucrose, C12H22O11, which
is ordinary table sugar. Another example is methanol, CH3OH, a compound used in
car window washer solution. Both of these are molecular substances. The solution
process occurs because molecules of the substance mix with molecules of water. Molecules are electrically neutral and cannot carry an electric current, so the solution is
electrically nonconducting.
Observing the Electrical Conductivity of a Solution
Demonstration of conductivity of
liquid versus frozen solution:
Summerlin et al., Chemical
Demonstrations, Vol. 2
(Washington, D.C.: Am. Chem.
Soc., 1987), no. 71, p. 132.
Video no. 3 (Series A),Electrical
Conductivity of Aqueous
Solutions.
Figure 4.3 shows a simple apparatus that allows you to observe the ability of a solution to conduct an electric current. The apparatus has two electrodes, here they are
at metal plates, dipping into the solution in a beaker. One electrode connects directly
to a battery through a wire. The other electrode connects by a wire to a light bulb
that connects with another wire to the other side of the battery. For an electrical current to ow from the battery, there must be a complete circuit, which allows the current to ow from the positive pole of the battery through the circuit to the negative
pole of the battery. To have a complete circuit, the solution in the beaker must conduct electricity, as the wires do. If the solution is conducting, the circuit is complete
and the bulb lights. If the solution is nonconducting, the circuit is incomplete and the
bulb does not light.
The beaker shown on the left side of Figure 4.3 contains pure water. Because the
bulb is not lit, we conclude that pure water is a nonconductor (or very poor conductor) of electricity, which is what we expect from our earlier discussion. The beaker
126
CHAPTER 4
Chemical Reactions
Figure 4.3
Testing the electrical
conductivity of a solution
Left: Pure water does not
conduct; therefore the bulb does
not light. Right: A solution of
sodium chloride allows the
current to pass through it, and
the bulb lights.
Video Series C,Conductivities of
Aqueous Solutions.
shown on the right side of Figure 4.3 contains a solution of sodium chloride in water.
In this case, the bulb burns brightly, showing that the solution is a very good conductor of electricity, due to the movement of ions in the solution.
How brightly the bulb lights tells you whether the solution is a very good conductor (contains a strong electrolyte) or only a moderately good conductor (contains
a weak electrolyte). The solution of sodium chloride burns brightly, so we conclude
that sodium chloride is a strong electrolyte. Let us look more closely at such strong
and weak electrolytes.
Strong and Weak Electrolytes
When electrolytes dissolve in water they produce ions, but to varying extents. A
strong electrolyte is an electrolyte that exists in solution almost entirely as ions.
Most ionic solids that dissolve in water do so by going into the solution almost
completely as ions, so they are strong electrolytes. An example is sodium chloride.
We can represent the dissolution of sodium chloride in water by the following equation:
HO
NaCl(s) 2 Na (aq)
Video no. 23 (Series D),
Conductivity Apparatus.
Cl ( aq)
A weak electrolyte is an electrolyte that dissolves in water to give a relatively
small percentage of ions. These are generally molecular substances. Ammonia, NH3,
is an example. Pure ammonia is a gas that readily dissolves in water and goes into
solution as ammonia molecules, NH3(aq). When you buy ammonia in the grocery
store, you are buying an aqueous solution of ammonia. Ammonia molecules react with
water to form ammonium ions, NH4 , and hydroxide ions, OH .
NH3(aq)
H2O(l ) NH4 (aq)
OH (aq)
However, these ions, NH4
OH , react with each other to give back ammonia molecules and water molecules.
NH4 (aq)
OH (aq) NH3(aq)
H2O(l )
Both reactions, the original one and its reverse, occur constantly and simultaneously.
We denote this situation by writing a single equation with a double arrow:
NH3(aq)
H2O(l ) B NH4 (aq)
A
OH (aq)
4.1
Ionic Theory of Solutions and Solubility Rules
127
Figure 4.4
Comparing strong and weak
electrolytes
The apparatus is similar to that in
Figure 4.3, but this time strong
and weak electrolytes are
compared.The solution on the
left is of HCl (a strong electrolyte)
and that on the right is of NH3
(a weak electrolyte). Note how
much more brightly the bulb on
the left burns compared with
that on the right.
As a result of this forward and reverse reaction, just a small percentage of the NH3
molecules (about 3%) have reacted at any given moment to form ions. Thus, ammonia is a weak electrolyte. The electrical conductivities of a strong and a weak electrolyte are contrasted in Figure 4.4.
Most soluble molecular substances are either nonelectrolytes or weak electrolytes.
An exception is hydrogen chloride gas, HCl(g), which dissolves in water to produce
hydrogen ions and chloride ions. We represent its reaction with water by an equation
with a single arrow:
HCl(aq) H (aq)
Cl (aq)
Since hydrogen chloride dissolves to give almost entirely ions, hydrogen chloride (or
hydrochloric acid) is a strong electrolyte.
Solubility Rules
It is clear from the preceding discussion that substances vary widely in their solubility, or ability to dissolve, in water. Some compounds, such as sodium chloride and
ethyl alcohol (CH3CH2OH), dissolve readily and are said to be soluble. Others, such
as calcium carbonate (which occurs naturally as limestone and marble) and benzene
(C6H6), have quite limited solubilities and are thus said to be insoluble.
Soluble ionic compounds form solutions that contain many ions and therefore are
strong electrolytes. To predict the solubility of ionic compounds, chemists have developed solubility rules. Table 4.1 lists eight solubility rules for ionic compounds. These
rules apply to most of the common ionic compounds that we will discuss in this
course. Example 4.1 illustrates how to use the rules.
EXAMPLE 4.1
Using the Solubility Rules
Determine whether the following compounds are soluble or insoluble in water.
a. Hg2Cl2
PROBLEM STRATEGY
b. KI
c. lead(II) nitrate
You should refer to Table 4.1 for the solubility rules of ionic compounds to see
which compounds are soluble in water. The soluble ionic compounds are those that
readily dissolve in water. The insoluble compounds hardly dissolve at all.
128
CHAPTER 4
Chemical Reactions
SOLUTION
a. According to Rule 3 in Table 4.1, most compounds that contain chloride, Cl ,
are soluble. However, Hg2Cl2 is listed as one of the exceptions to this rule so
it does not dissolve in water. Therefore, Hg2Cl2 is not soluble in water.
b. According to both Rule 1, Group IA compounds are soluble, and Rule 3, most
iodides are soluble, KI is expected to be soluble. Therefore, KI is soluble in
water.
c. According to Rule 2, compounds containing nitrates, NO3 , are soluble. Since
there are no exceptions to this rule, lead(II) nitrate, Pb(NO3)2, is soluble in
water.
EXERCISE 4.1
Determine whether the following compounds are soluble or insoluble in water.
a. NaBr
b. Ba(OH)2
c. calcium carbonate
See Problems 4.23 and 4.24.
Let us summarize the main points in this section. Compounds that dissolve in
water are soluble; those that dissolve little, or not at all, are insoluble. Soluble substances are either electrolytes or nonelectrolytes. Nonelectrolytes form nonconducting aqueous solutions because they dissolve completely as molecules. Electrolytes form electrically conducting solutions in water because they dissolve to
give ions in solution. Electrolytes can be strong or weak. Almost all soluble ionic
substances are strong electrolytes. Soluble molecular substances usually are nonelectrolytes or weak electrolytes; the latter solution consists primarily of molecules,
but has a small percentage of ions. Ammonia, NH3, is an example of a molecular
substance that is a weak electrolyte. A few molecular substances (such as HCl) dissolve almost entirely as ions in the solution and are therefore strong electrolytes.
The solubility rules can be used to predict the solubility of ionic compounds in
water.
Table 4.1
Solubility Rules for Ionic Compounds
Rule
Applies to
Statement
Exceptions
1
Li , Na , K , NH4
2
3
C2H3O2 , NO3
Cl , Br , I
4
SO42
Group IA and ammonium
compounds are soluble.
Acetates and nitrates are soluble.
Most chlorides, bromides, and
iodides are soluble.
Most sulfates are soluble.
5
6
7
8
CO32
PO43
S2
OH
Most carbonates are insoluble.
Most phosphates are insoluble.
Most suldes are insoluble.
Most hydroxides are insoluble.
AgCl, Hg2Cl2, PbCl2, AgBr, HgBr2, Hg2Br2,
PbBr2, AgI, HgI2, Hg2I2, PbI2
CaSO4, SrSO4, BaSO4, Ag2SO4, Hg2SO4,
PbSO4
Group IA carbonates, (NH4)2CO3
Group IA phosphates, (NH4)3PO4
Group IA suldes, (NH4)2S
Group IA hydroxides, Ca(OH)2, Sr(OH)2,
Ba(OH)2
4.2
CONCEPT CHECK 4.1
Molecular and Ionic Equations
129
LiI(s) and CH3OH(l) are each introduced into a separate beaker containing
water. Using the drawings shown here, label each beaker with the appropriate
compound and indicate whether you would expect each substance to be a strong
electrolyte, weak electrolyte, or nonelectrolyte.
+
4.2 Molecular and Ionic Equations
We use chemical equations to help us describe chemical reactions. For a reaction
involving ions, we have a choice of chemical equations, depending on the kind of
information we want to convey. We can represent such a reaction by a molecular
equation, a complete ionic equation, or a net ionic equation.
To illustrate these different kinds of equations, consider the preparation of precipitated calcium carbonate, CaCO3. This white, ne powdery compound is used as
a paper ller to brighten and retain ink, as an antacid (as in the trade-named Tums),
and as a mild abrasive in toothpastes. One way to prepare this compound is to react
calcium hydroxide, Ca(OH)2, with sodium carbonate, Na2CO3. Let us look at the different ways to write the equation for this reaction.
Molecular Equations
You could write the equation for this reaction as follows:
Ca(OH)2(aq)
Na2CO3(aq) CaCO3(s)
2NaOH(aq)
We call this a molecular equation, which is a chemical equation in which the reactants and products are written as if they were molecular substances, even though they
may actually exist in solution as ions. The molecular equation is useful because it is
130
CHAPTER 4
Chemical Reactions
explicit about what the reactant solutions are and what products you obtain. The equation says that you add aqueous solutions of calcium hydroxide and sodium carbonate
to the reaction vessel. As a result of the reaction, the insoluble, white calcium carbonate solid forms in the solution; that is, calcium carbonate precipitates. After you
remove the precipitate, you are left with a solution of sodium hydroxide. The molecular equation closely describes what you actually do in the laboratory or in an industrial process.
Complete Ionic Equations
Although a molecular equation is useful in describing the actual reactant and product substances, it does not tell you what is happening at the level of ions. That is,
it does not give you an ionic-theory interpretation of the reaction. Because this kind
of information is useful, you often need to rewrite the molecular equation as an ionic
equation.
Again, consider the reaction of calcium hydroxide, Ca(OH)2, and sodium carbonate, Na2CO3. Both are soluble ionic substances and therefore strong electrolytes;
when they dissolve in water, they go into solution as ions. Each formula unit of
Ca(OH)2 forms one Ca2 ion and two OH ions in solution. If you want to emphasize that the solution contains freely moving ions, it would be better to write Ca2 (aq)
2OH (aq) in place of Ca(OH)2(aq). Similarly, each formula unit of Na2CO3 forms
two Na ions and one CO32 ion in solution, and you would emphasize this by writing 2Na (aq)
CO32 (aq) in place of Na2CO3(aq). The reactant side of the equation becomes
Ca2 (aq)
2OH (aq)
2Na (aq)
CO32 (aq)
Thus, the reaction mixture begins as a solution of four different kinds of ions.
Now let us look at the product side of the equation. One product is the precipitate CaCO3(s). According to the solubility rules, this is an insoluble ionic compound
so it will exist in water as a solid. We leave the formula as CaCO3(s) to convey this
information in the equation. On the other hand, NaOH is a soluble ionic substance
and therefore a strong electrolyte; it dissolves in aqueous solution to give the freely
moving ions, which we denote by writing Na (aq) OH (aq). The complete equation is
Ca2 (aq)
2OH (aq)
2Na (aq)
CO32 (aq)
CaCO3(s)
2Na (aq)
2OH (aq)
The purpose of such a complete ionic equation is to represent each substance by
its predominant form in the reaction mixture. For example, if the substance is a soluble ionic compound, it probably dissolves as individual ions (so it is a strong electrolyte). In a complete ionic equation, you represent the compound as separate ions.
If the substance is a weak electrolyte, it is present in solution primarily as molecules,
so you represent it by its molecular formula. If the substance is an insoluble ionic
compound, you represent it by the formula of the compound, not by the formulas of
the separate ions in solution.
Thus, a complete ionic equation is a chemical equation in which strong electrolytes (such as soluble ionic compounds) are written as separate ions in the solution. You represent other reactants and products by the formulas of the compounds,
indicating any soluble substance by (aq) after its formula and any insoluble solid substance by (s) after its formula.
4.2
Molecular and Ionic Equations
131
Net Ionic Equations
In the complete ionic equation representing the reaction of calcium hydroxide and
sodium carbonate, some ions (OH and Na ) appear on both sides of the equation.
This means that nothing happens to these ions as the reaction occurs. They are called
spectator ions. A spectator ion is an ion in an ionic equation that does not take part
in the reaction. You can cancel such ions from both sides to express the essential reaction that occurs.
Ca2 (aq)
2OH (aq)
2Na (aq)
CO32 (aq)
CaCO3(s)
2Na (aq)
2OH (aq)
The resulting equation is
Ca2 (aq)
CO32 (aq) CaCO3(s)
This is the net ionic equation, an ionic equation from which spectator ions have been
canceled. It shows that the reaction that actually occurs at the ionic level is between
calcium ions and carbonate ions to form solid calcium carbonate.
From the net ionic equation, you can see that mixing any solution of calcium ion
with any solution of carbonate ion will give you this same reaction. For example, the
strong electrolyte calcium nitrate, Ca(NO3), dissolves readily in water to provide a
source of calcium ions. (According to the solubility rules, calcium nitrate is soluble,
so it goes into solution as Ca2 and NO3 ions.) Similarly, the strong electrolyte
potassium carbonate, K2CO3, dissolves readily in water to provide a source of carbonate ions. (According to the solubility rules, potassium carbonate is soluble, so it
goes into solution as K and CO32 ions.) When you mix solutions of these two compounds, you obtain a solution of calcium ions and carbonate ions, which react to form
the insoluble calcium carbonate. The other product is potassium nitrate, a soluble ionic
compound, and therefore a strong electrolyte. The molecular equation representing the
reaction is
Ca(NO3)2(aq)
K2CO3(aq) CaCO3(s)
2KNO3(aq)
You obtain the complete ionic equation from this molecular equation by rewriting
each of the soluble ionic compounds as ions, but retaining the formula for the precipitate CaCO3(s):
Ca2 (aq)
2NO3 (aq)
2K (aq)
CO32 (aq)
CaCO3(s)
2K (aq)
2NO3 (aq)
The net ionic equation is
Figure 4.5
Limestone formations
It is believed that most limestone
formed as a precipitate of
calcium carbonate (and other
carbonates) from seawater.The
photograph shows limestone
formations at Bryce Point, Bryce
Canyon National Park, Utah. More
than sixty million years ago, this
area was covered by seawater.
Ca2 (aq)
CO32 (aq) CaCO3(s)
Note that the net ionic equation is identical to the one obtained from the reaction of
Ca(OH)2 and Na2CO3. The essential reaction is the same whether you mix solutions
of calcium hydroxide and sodium carbonate or solutions of calcium nitrate and potassium carbonate.
The value of the net ionic equation is its generality. For example, seawater contains Ca2 and CO32 ions from various sources. Whatever the sources of these ions,
you expect them to react to form a precipitate of calcium carbonate. In seawater, this
precipitate results in sediments of calcium carbonate, which eventually form limestone
(Figure 4.5).
132
CHAPTER 4
Chemical Reactions
EXAMPLE 4.2
An alternate example is
provided in the Instructors
Resource Manual.
Writing Net Ionic Equations
Write a net ionic equation for each of the following molecular equations.
a. 2HClO4(aq)
Ca(OH)2(aq) Ca(ClO4)2(aq)
2H2O(l )
Perchloric acid, HClO4, is a strong electrolyte, forming H and ClO4 ions in
solution. Ca(ClO4)2 is a soluble ionic compound.
b. HC2H3O2(aq)
NaOH(aq) NaC2H3O2(aq)
H2O(l )
Acetic acid, HC2H3O2, is a molecular substance and a weak electrolyte.
PROBLEM STRATEGY
You will need to convert the molecular equation to the complete ionic equation,
then cancel spectator ions to obtain the net ionic equation. For each ionic compound in the reaction, use the solubility rules to determine if the compound will
be soluble (in the solution as ions) or insoluble (present as an undissolved solid).
For the complete ionic equation, represent all of the strong electrolytes by their
separate ions in solution, adding (aq) after the formula of each. Retain the formulas of the other compounds. An ionic compound should have (aq) after its formula
if it is soluble or (s) if it is insoluble.
SOLUTION
a. According to the solubility rules presented in Table 4.1 and the problem statement, Ca(OH)2 and Ca(ClO4)2 are soluble ionic compounds so they are strong electrolytes. The problem statement notes that HClO4 is also a strong electrolyte. You
write each strong electrolyte in the form of separate ions. Water, H2O, is a nonelectrolyte (or very weak electrolyte), so you retain its molecular formula. The
complete ionic equation is
2H (aq)
2ClO4 (aq)
Ca2 (aq)
2OH (aq)
Ca2 (aq)
2ClO4 (aq)
2H2O(l )
After canceling spectator ions and dividing by 2, you get the following net ionic
equation:
H (aq)
OH (aq) H2O(l)
b. According to the solubility rules, NaOH and NaC2H3O2 are soluble ionic compounds, so they are strong electrolytes. The problem statement notes that HC2H3O2
is a weak electrolyte, which you write by its molecular formula. Water, H2O, is a
nonelectrolyte, so you retain its molecular formula also. The complete ionic equation is
HC2H3O2(aq)
Na (aq)
OH (aq) Na (aq)
C2H3O2 (aq)
H2O(l )
and the net ionic equation is
HC2H3O2(aq)
EXERCISE 4.2
OH (aq) C2H3O2 (aq)
H2O(l )
Write complete ionic and net ionic equations for each of the following molecular
equations.
a. 2HNO3(aq)
Mg(OH)2(s) 2H2O(l)
Nitric acid, HNO3, is a strong electrolyte.
b. Pb(NO3)2(aq)
Na2SO4(aq) PbSO4(s)
Mg(NO3)2(aq)
2NaNO3(aq)
See Problems 4.27 and 4.28.
4.3
Precipitation Reactions
133
Types of Chemical Reactions
Demonstration of precipitation:
Summerlin and Ealy, Chemical
Demonstrations, Vol. 1
(Washington, D.C.: Am. Chem.
Soc., 1985), no. 1214, pp. 2325;
Summerlin et al., Chemical
Demonstrations, Vol. 2
(Washington, D.C.: Am. Chem.
Soc., 1987), no. 6566,
pp. 121125.
Among the several million known substances, many millions of chemical reactions
are possible. Beginning students are often bewildered by the possibilities. How can I
know when two substances will react when they are mixed? How can I predict the
products? Although it is not possible to give completely general answers to these questions, it is possible to make sense of chemical reactions. Most of the reactions we will
study belong to one of three types:
1. Precipitation reactions. In these reactions, you mix solutions of two ionic substances and a solid ionic substance (a precipitate) forms.
2. Acidbase reactions. An acid substance reacts with a substance called a base.
Such reactions involve the transfer of a proton between reactants.
3. Oxidationreduction reactions. These involve the transfer of electrons between
reactants.
We will look at each of these types of reactions. By the time you nish this chapter, you should feel much more comfortable with the descriptions of chemical reactions that you will encounter in this course.
4.3 Precipitation Reactions
Video no. 4 (Series A),
Precipitation Reactions.
Video Series D,Chemical
Casserole,The Ammonia
Fountain, and Fun with
Limewater.
In the previous section, we used a precipitation reaction to illustrate how to convert
a molecular equation to an ionic equation. A precipitation reaction occurs in aqueous
solution because one product is insoluble. A precipitate is an insoluble solid compound formed during a chemical reaction in solution. To predict whether a precipitate will form when you mix two solutions of ionic compounds, you need to know
whether any of the potential products that might form are insoluble or not. This is
another application of the solubility rules (Section 4.1).
Predicting Precipitation Reactions
Now let us see how you would go about predicting whether a precipitation reaction
will occur. Suppose you mix together solutions of magnesium chloride, MgCl2, and
silver nitrate, AgNO3. You can write the potential reactants as follows:
MgCl2
Demonstration of metathesis
reaction between solid Pb(NO3)2
and solid KI: Summerlin and Ealy,
Chemical Demonstrations, Vol. 1
(Washington, D.C.: Am. Chem.
Soc., 1985), no. 75, p. 121.
Demonstration of variation of
conductivity during a metathesis
reaction (decrease of ions):
Summerlin et al., Chemical
Demonstrations, Vol. 2
(Washington, D.C.: Am. Chem.
Soc., 1987), no. 70, pp. 130131.
AgNO3
How can you tell if a reaction will occur, and if it does, what products to expect?
When you write a precipitation reaction as a molecular equation, the reaction has
the form of an exchange reaction. An exchange (or metathesis) reaction is a reaction between compounds that, when written as a molecular equation, appears to
involve the exchange of parts between the two reactants. In a precipitation reaction,
the anions exchange between the two cations (or vice versa).
Let us momentarily assume that a reaction does occur between magnesium chloride and silver nitrate. If you exchange the anions, you get silver chloride and magnesium nitrate. Once you gure out the formulas of these potential products, you can
write the molecular equation. The formulas are AgCl and Mg(NO3)2. (If you do not
recall how to write the formula of an ionic compound, given the ions, refer back to
Example 2.3.) The balanced equation, assuming there is a reaction, is
MgCl2
2AgNO3 2AgCl
Mg(NO3)2
134
CHAPTER 4
Chemical Reactions
The reactants MgCl2 and
AgNO3 must be added in
correct amounts; otherwise,
the excess reactant will remain
along with the product
Mg(NO3)2.
Let us verify that MgCl2 and AgNO3 are soluble and then check the solubilities
of the products. Rule 3 in Table 4.1 says that chlorides are soluble, with certain exceptions, which do not include magnesium chloride. Thus, we predict that magnesium
chloride is soluble. Rule 2 indicates that nitrates are soluble, so AgNO3 is soluble as
well. The potential products are silver chloride and magnesium nitrate. According to
Rule 3, silver chloride is one of the exceptions to the general solubility of chlorides.
Therefore, we predict that the silver chloride is insoluble. Magnesium nitrate is soluble according to Rule 2.
Now we can append the appropriate phase labels to the compounds in the preceding equation.
MgCl2(aq)
2AgNO3(aq) 2AgCl(s)
Mg(NO3)2(aq)
We predict that reaction occurs because silver chloride is insoluble and precipitates
from the reaction mixture. Figure 4.6 shows the formation of the white silver chloride from this reaction. If you separate the precipitate from the solution by pouring it
through lter paper, the solution that passes through (the ltrate) contains magnesium
nitrate, which you could obtain by evaporating the water. The molecular equation is
a summary of the actual reactants and products in the reaction.
To see the reaction that occurs on an ionic level, you need to rewrite the molecular
equation as a net ionic equation. You rst write the strong electrolytes (here soluble
ionic compounds) in the form of ions, leaving the formula of the precipitate unchanged.
NO3
Mg2+
Mg2 (aq)
2Cl (aq)
2Ag (aq)
2NO3 (aq)
2AgCl(s)
Mg2 (aq)
2NO3 (aq)
After canceling spectator ions and reducing the coefcients to the smallest whole
numbers, you obtain the net ionic equation:
Ag+
Cl
Figure 4.6
Reaction of magnesium
chloride and silver nitrate
Magnesium chloride solution is
added to a beaker of silver nitrate
solution. A white precipitate of
silver chloride forms.
EXAMPLE 4.3
Two alternate examples are
provided in the Instructors
Resource Manual.
Ag (aq)
Cl (aq) AgCl(s)
This equation represents the essential reaction that occurs: Ag ions and Cl ions in
aqueous solution react to form solid silver chloride.
If silver chloride were soluble, a reaction would not have occurred. When you
rst mix solutions of MgCl2(aq) and AgNO3(aq), you obtain a solution of four
ions: Mg2 (aq), Cl (aq), Ag (aq), and NO3 (aq). If no precipitate formed, you
would end up simply with a solution of these four ions. However, because Ag
and Cl react to give the precipitate AgCl, the effect is to remove these ions from
the reaction mixture as an insoluble compound and leave behind a solution of
Mg(NO3)2(aq).
Deciding Whether Precipitation Occurs
For each of the following, decide whether a precipitation reaction occurs. If it does,
write the balanced molecular equation, and then the net ionic equation. If no reaction occurs, write the compounds followed by an arrow and then NR (no reaction).
a. Aqueous solutions of sodium chloride and iron(II) nitrate are mixed.
b. Aqueous solutions of aluminum sulfate and sodium hydroxide are mixed.
PROBLEM STRATEGY
Start by writing the formulas of the compounds that are mixed. (If you have trouble with this, see Examples 2.3 and 2.4.) Then, assuming momentarily that the
compounds do react, write the exchange reaction. Make sure that you write the
correct formulas of the products. Using the solubility rules, append phase labels to
4.3
135
Precipitation Reactions
each formula in the equation. If one of the products is a precipitate, reaction occurs;
otherwise, no reaction occurs.
SOLUTION
a. The formulas of the compounds are NaCl and Fe(NO3)2. Exchanging anions,
you get sodium nitrate, NaNO3, and iron(II) chloride, FeCl2. The equation for the
exchange reaction is
NaCl
Fe(NO3)2 NaNO3
FeCl2
(not balanced)
Referring to Table 4.1, note that NaCl and NaNO3 are soluble (Rule 1). Also,
iron(II) nitrate is soluble (Rule 2), and iron(II) chloride is soluble. (Rule 3 says
that chlorides are soluble with some exceptions, none of which include FeCl2.)
Since there is no precipitate, no reaction occurs. You obtain simply an aqueous
solution of the four different ions (Na , Cl , Fe2 , and NO3 ). For the answer,
we write
NaCl(aq)
Fe(NO3)2(aq) NR
b. The formulas of the compounds are Al2(SO4)3 and NaOH. Exchanging anions,
you get aluminum hydroxide, Al(OH)3, and sodium sulfate, Na2SO4. The equation
for the exchange reaction is
Al2(SO4)3
NaOH Al(OH)3
Na2SO4
(not balanced)
From Table 4.1, you see that Al2(SO4)3 is soluble (Rule 4), NaOH and Na2SO4 are
soluble (Rule 1), and Al(OH)3 is insoluble (Rule 8). Thus, aluminum hydroxide
precipitates. The balanced molecular equation with phase labels is
Al2(SO4)3(aq)
6NaOH(aq) 2Al(OH)3(s)
3Na2SO4(aq)
To get the net ionic equation, you write the strong electrolytes (here, soluble ionic
compounds) as ions in aqueous solution and cancel spectator ions.
2Al3 (aq)
3SO42 (aq)
6Na (aq)
6OH (aq)
2Al(OH)3(s)
6Na (aq)
3SO42 (aq)
The net ionic equation is
Al3 (aq)
3OH (aq) Al(OH)3(s)
Thus, aluminum ion reacts with hydroxide ion to precipitate aluminum hydroxide.
EXERCISE 4.3
You mix aqueous solutions of sodium iodide and lead(II) acetate. If a reaction
occurs, write the balanced molecular equation and the net ionic equation. If no
reaction occurs, indicate this by writing the formulas of the compounds and an
arrow followed by NR.
See Problems 4.31, 4.32, 4.33, and 4.34.
CONCEPT CHECK 4.2
Your lab partner tells you that she mixed two solutions that
contain ions. You analyze the solution and nd that it contains
the ions and precipitate shown in the beaker.
a. Write the molecular equation for the reaction.
b. Write the complete ionic equation for the reaction.
c. Write the net ionic equation for the reaction.
Na+(aq)
C2H3O2 (aq)
SrSO4(s)
136
CHAPTER 4
Chemical Reactions
4.4 AcidBase Reactions
Demonstration of acidbase
indicators: Bassam Z. Shakhashiri,
Chemical Demonstrations, Vol. 3
(Madison: University of Wisconsin
Press, 1989), no. 8.18.3,
pp. 3349; Summerlin and Ealy,
Chemical Demonstrations, Vol. 1
(Washington, D.C.: Am. Chem.
Soc., 1985), no. 2225, pp. 3740.
Acids and bases are some of the most important electrolytes. You can recognize acids
and bases by some simple properties. Acids have a sour taste. Solutions of bases, on
the other hand, have a bitter taste and a soapy feel. (Of course, you should never taste
laboratory chemicals.) Some examples of acids are acetic acid, present in vinegar; citric acid, a constituent of lemon juice; and hydrochloric acid, found in the digestive
uid of the stomach. An example of a base is aqueous ammonia, often used as a
household cleaner. Table 4.2 lists further examples; see also Figure 4.7.
Another simple property of acids and bases is their ability to cause color
changes in certain dyes. An acidbase indicator is a dye used to distinguish
between acidic and basic solutions by means of the color changes it undergoes in
these solutions. Such dyes are common in natural materials. The amber color of tea,
for example, is lightened by the addition of lemon juice (citric acid). Red cabbage
juice changes to green then yellow when a base is added (Figure 4.8). The green
and yellow colors change back to red when an acid is added. Litmus is a common
laboratory acidbase indicator. This dye, produced from certain species of lichens,
turns red in acidic solution and blue in basic solution. Phenolphthalein (fee nol
thay leen), another laboratory acidbase indicator, is colorless in acidic solution
and pink in basic solution.
An important chemical characteristic of acids and bases is the way they react with
one another. To understand these acidbase reactions, we need to have precise denitions of the terms acid and base.
Figure 4.7
Household acids and bases
Shown are a variety of household
products that are either acids or
bases.
Denitions of Acid and Base
When Arrhenius developed his ionic theory of solutions, he also gave the classic definitions of acids and bases. According to Arrhenius, an acid is a substance that pro-
Table 4.2
Common Acids and Bases
Name
Acids:
Acetic acid
Acetylsalicylic acid
Ascorbic acid
Citric acid
Hydrochloric acid
Sulfuric acid
Bases:
Ammonia
Calcium hydroxide
Magnesium hydroxide
Sodium hydroxide
Formula
Remarks
HC2H3O2
HC9H7O4
H2C6H6O6
H3C6H5O7
HCl
H2SO4
Found in vinegar
Aspirin
Vitamin C
Found in lemon juice
Found in gastric juice (digestive uid in stomach)
Battery acid
NH3
Ca(OH)2
Mg(OH)2
NaOH
Aqueous solution used as a household cleaner
Slaked lime (used in mortar for building construction)
Milk of magnesia (antacid and laxative)
Drain cleaners, oven cleaners
4.4
AcidBase Reactions
137
Figure 4.8
Red cabbage juice as an
acidbase indicator
Left: Preparation of red cabbage
juice.The beaker (green solution)
contains red cabbage juice and
sodium bicarbonate (baking
soda). Right: Red cabbage juice
has been added from a pipet to
solutions in the beakers.These
solutions vary in acidity from
highly acidic on the left to highly
basic on the right.
Demonstration of properties of
acids and bases: Bassam
Z. Shakhashiri, Chemical
Demonstrations, Vol. 3 (Madison:
University of Wisconsin Press,
1989), no. 8.5, p. 58.
duces hydrogen ions, H , when it dissolves in water. An example is nitric acid, HNO3,
a molecular substance that dissolves in water to give H and NO3 .
Video no. 37 (Series D),
Red Cabbage Juice.
Demonstration of indicators from
plants: Bassam Z. Shakhashiri,
Chemical Demonstrations, Vol. 3
(Madison: University of Wisconsin
Press, 1989), no. 8.4, pp. 5057.
Solutions of ammonia are
sometimes called ammonium
hydroxide and given the
formula NH4OH(aq), based on
analogy with NaOH. No NH4OH
molecule or compound has
ever been found, however.
HO
HNO3(aq) 2 H (aq)
NO3 ( aq)
An Arrhenius base is a substance that produces hydroxide ions, OH , when it
dissolves in water. For example, sodium hydroxide is a base.
HO
NaOH(s) 2 Na (aq)
OH ( aq)
The molecular substance ammonia, NH3, is also a base because it yields hydroxide
ions when it reacts with water.
NH3(aq)
H2O(l ) B NH4 (aq)
A
OH (aq)
+
+
Although the Arrhenius concept of acids and bases is useful, it is somewhat limited. For example, it tends to single out the OH ion as the source of base character,
when other ions or molecules can play a similar role. In 1923, Johannes N. Brnsted
and Thomas M. Lowry independently noted that many reactions involve nothing more
than the transfer of a proton (H ) between reactants, and they realized that they could
use this idea to expand the denitions of acids and bases to describe a large class of
chemical reactions. In this view, acidbase reactions are proton-transfer reactions.
Consider the preceding reaction. It involves the transfer of a proton from the water
molecule, H2O, to the ammonia molecule, NH3.
H
NH3(aq)
The BrnstedLowry concept is
treated in detail in Chapter 16.
Animation: Proton transfer.
A
H2O(l ) B NH4 (aq)
OH (aq)
Once the proton, H , has left the H2O molecule, it leaves behind an OH ion. When
the H adds to NH3, an NH4 ion results. H2O is said to donate a proton to NH3,
and NH3 is said to accept a proton from H2O.
Brnsted and Lowry defined an acid as the species (molecule or ion) that
donates a proton to another species in a proton-transfer reaction. They dened a
138
CHAPTER 4
Chemical Reactions
base as the species (molecule or ion) that accepts a proton in a proton-transfer reaction. In the reaction of ammonia with water, the H2O molecule is the acid, because
it donates a proton. The NH3 molecule is a base, because it accepts a proton.
H
Demonstration of the
conductivity of strong and weak
acids: Bassam Z. Shakhashiri,
Chemical Demonstrations, Vol. 3
(Madison: University of Wisconsin
Press, 1989), no. 8.21, pp. 140145.
NH3(aq)
base
A
H2O(l ) B NH4 (aq)
The dissolution of nitric acid, HNO3, in water is actually a proton-transfer reaction, although the following equation, which we used as an illustration of an Arrhenius acid, does not spell that out.
HO
HNO3(aq) 2 H (aq)
The H (aq) ion has variable
structure in solution. In solids,
there is evidence for
[H(H2O)n] , where n is 1, 2, 3, 4,
or 6.
OH (aq)
acid
NO3 (aq)
To see that this is a proton-transfer reaction, we need to clarify the structure of the
hydrogen ion, H (aq). This ion consists of a proton (H ) in association with water
molecules, which is what (aq) means. This is not a weak association, however,
because the proton (or hydrogen nucleus) would be expected to attract electrons
strongly to itself. In fact, the H (aq) ion might be better thought of as a proton
chemically bonded to a water molecule to give the H3O ion, with other water
molecules less strongly associated with this ion, which we represent by the phaselabeled formula H3O (aq). Written in this form, we usually call this the hydronium ion. It is important to understand that the hydrogen ion, H (aq), and the
hydronium ion, H3O (aq), represent precisely the same physical ion. For simplicity, we often write the formula for this ion as H (aq), but when we want to be
explicit about the proton-transfer aspect of a reaction, we write the formula as
H3O (aq).
Now let us rewrite the preceding equation by replacing H (aq) by H3O (aq). To
maintain a balanced equation, we will also need to add H2O(l) to the left side.
HNO3(aq)
H2O(l) NO3 (aq)
H3O (aq)
+
+
Note that the reaction involves simply a transfer of a proton (H ) from HNO3 to H2O:
H
HNO3(aq)
acid
We will discuss the
BrnstedLowry concept
of acids and bases more
thoroughly in Chapter 16.
H2O(l ) NO3 (aq)
H3O (aq)
base
The HNO3 molecule is an acid (proton donor) and H2O is a base (proton acceptor). Note that H2O may function as an acid or a base, depending on the other
reactant.
The Arrhenius denitions and those of Brnsted and Lowry are essentially equivalent for aqueous solutions, although their points of view are different. For instance,
sodium hydroxide and ammonia are bases in the Arrhenius view because they increase
the percentage of OH ion in the aqueous solution. They are bases in the Brnsted
Lowry view because they provide species (OH from the strong electrolyte sodium
hydroxide and NH3 from ammonia) that can accept protons.
4.4
AcidBase Reactions
139
Table 4.3
Common Strong Acids and Bases
Strong Acids
=
=
=
=
H3O+
Cl
F
HF
Strong Bases
HClO4
H2SO4
HI
HBr
HCl
HNO3
LiOH
NaOH
KOH
Ca(OH)2
Sr(OH)2
Ba(OH)2
Strong and Weak Acids and Bases
A
Acids and bases are classied as strong or weak, depending on whether they are strong
or weak electrolytes. A strong acid is an acid that ionizes completely in water; it is
a strong electrolyte. Hydrochloric acid, HCl(aq), and nitric acid, HNO3(aq), are examples of strong acids. Using the hydronium ion notation, we write the respective equations as follows:
HCl(aq)
H2O(l ) H3O (aq)
Cl (aq)
HNO3(aq)
H2O(l ) H3O (aq)
NO3 (aq)
Table 4.3 lists six common strong acids. Most of the other acids we will discuss are
weak acids.
A weak acid is an acid that only partly ionizes in water; it is a weak electrolyte.
Examples of weak acids are hydrocyanic acid, HCN(aq), and hydrouoric acid, HF(aq).
These molecules react with water to produce a small percentage of ions in solution.
HCN(aq)
B
Figure 4.9
Molecular views comparing
the strong acid HCl and the
weak acid HF in water (H2O
molecules are omitted for
clarity)
Top: Due to its complete reaction
with water, the strong acid HCl
exists as the hydronium ion,
H3O (aq), and Cl (aq) in aqueous
solution. Because of this complete
reaction, there are essentially no
HCl(aq) molecules in the solution.
Bottom: Due to the reaction with
only very few of the water molecules, the weak acid HF(aq) exists
in aqueous solution largely as
HF(aq) molecules with very little
H3O (aq) and F (aq) being
produced.
H2O(l) B H3O (aq)
A
CN (aq)
HF(aq)
H2O(l) B H3O (aq)
A
F (aq)
Figure 4.9 presents molecular views of the strong acid HCl(aq) and the weak acid
HF(aq). Note how both the strong and weak acid undergo the same reaction with
water (proton donors); however, in the case of the weak acid such as HF, only a small
portion of the acid molecules actually undergoes reaction leaving the majority of the
acid molecules unreacted. On the other hand, strong acids like HCl and HNO3 undergo
complete reaction with water. The result is that weak acids can produce little H3O
in aqueous solution, whereas strong acids can produce large amounts. Hence, when
chemists refer to a weak acid, they are often thinking about a substance that produces
a limited amount of H3O (aq), whereas when chemists refer to a strong acid, they
are thinking about a substance that can produce large amounts of H3O (aq).
A strong base is a base that is present in aqueous solution entirely as ions, one
of which is OH ; it is a strong electrolyte. The ionic compound sodium hydroxide,
NaOH, is an example of a strong base. It dissolves in water as Na and OH .
HO
NaOH(s) 2 Na (aq)
OH (aq)
The hydroxides of Groups IA and IIA elements, except for beryllium hydroxide, are
strong bases (see Table 4.3).
140
CHAPTER 4
Chemical Reactions
In many cases, when chemists
are thinking about strong
versus weak bases in aqueous
solution, they consider a weak
base to be a substance that
can produce only limited
amounts of OH (aq) and a
strong base to be a substance
that can produce large
amounts of OH (aq).
EXAMPLE 4.4
An alternate example is
provided in the Instructors
Resource Manual.
SOLUTION
A weak base is a base that is only partly ionized in water; it is a weak electrolyte. Ammonia, NH3, is an example.
NH3(aq)
H2O(l) B NH4 (aq)
A
OH (aq)
You will nd it important to be able to identify an acid or base as strong or
weak. When you write an ionic equation, you represent strong acids and bases by
the ions they form and weak acids and bases by the formulas of the compounds.
The next example will give you some practice identifying acids and bases as strong
or weak.
Classifying Acids and Bases as Strong or Weak
Identify each of the following compounds as a strong or weak acid or base:
a. LiOH b. HC2H3O2 c. HBr d. HNO2
Refer to Table 4.3 for the common strong acids and bases. You can assume that
other acids and bases are weak.
a. As noted in Table 4.3, LiOH is a strong base.
b. Acetic acid, HC2H3O2, is not one of the strong acids listed in Table 4.3; therefore, we assume HC2H3O2 is a weak acid.
c. As noted in Table 4.3, HBr is a strong acid.
d. Nitrous acid, HNO2, is not one of the strong acids listed in Table 4.3; therefore,
we assume HNO2 is a weak acid.
EXERCISE 4.4
Label each of the following as a strong or weak acid or base:
a. H3PO4 b. HClO c. HClO4 d. Sr(OH)2
See Problems 4.35 and 4.36.
Neutralization Reactions
Demonstration of neutralization
of acid by milk of magnesia:
Summerlin et al., Chemical
Demonstrations, Vol. 2
(Washington, D.C.: Am. Chem.
Soc., 1987), no. 90, p. 171.
One of the chemical properties of acids and bases is that they neutralize one another. A neutralization reaction is a reaction of an acid and a base that results in
an ionic compound and possibly water. When a base is added to an acid solution,
the acid is said to be neutralized. The ionic compound that is a product of a neutralization reaction is called a salt. Most ionic compounds other than hydroxides
and oxides are salts. Salts can be obtained from neutralization reactions such as:
2HCl(aq)
acid
HCN(aq)
acid
Ca(OH)2(aq) CaCl2(aq)
base
KOH(aq) KCN(aq)
base
2H2O(l )
salt
H2O(l )
salt
The salt formed in a neutralization reaction consists of cations obtained from the
base and anions obtained from the acid. In the rst example, the base is Ca(OH)2,
which supplies Ca2 cations; the acid is HCl, which supplies Cl anions. The salt
contains Ca2 and Cl ions (CaCl2).
We wrote these reactions as molecular equations. Written in this form, the
equations make explicit the reactant compounds and the salts produced. However, to
discuss the essential reactions that occur, you need to write the net ionic equations.
4.4
AcidBase Reactions
141
You will see that the net ionic equation for each acidbase reaction involves the
transfer of a proton.
The rst reaction involves a strong acid, HCl(aq), and a strong base,
Ca(OH)2(aq). Both Ca(OH)2 and the product CaCl2, being soluble ionic compounds, are strong electrolytes. Writing these strong electrolytes in the form of ions
gives the following complete ionic equation:
2H (aq)
2Cl (aq)
Ca2 (aq)
2OH (aq) Ca2 (aq)
2Cl (aq)
2H2O(l )
Canceling spectator ions and dividing by 2 gives the net ionic equation:
H
H (aq)
OH (aq) H2O(l )
Note the transfer of a proton from the hydrogen ion (or hydronium ion, H3O ) to
the hydroxide ion.
The second reaction involves HCN(aq), a weak acid, and KOH(aq), a strong
base; the product is KCN, a strong electrolyte. The net ionic equation is:
H
OH (aq) CN (aq)
HCN(aq)
H2O(l )
Note the proton transfer, characteristic of an acidbase reaction.
In each of these examples, hydroxide ions latch strongly onto protons to form
water. Because water is a very stable substance, it effectively provides the driving
force of the reaction.
Although water is one of the products in most neutralization reactions, the reaction of an acid with the base ammonia provides a prominent exception. Consider
the reaction of sulfuric acid with ammonia:
H2SO4(aq)
acid
2NH3(aq) (NH4)2SO4(aq)
base
salt
The net ionic equation is
H
H (aq)
NH3(aq) NH4 (aq)
Note the proton transfer, a hallmark of an acidbase reaction. In this case, NH3 molecules latch onto protons and form relatively stable NH4 ions.
EXAMPLE 4.5
An alternate example is
provided in the Instructors
Resource Manual.
PROBLEM STRATEGY
Writing an Equation for a Neutralization
Write the molecular equation and then the net ionic equation for the neutralization
of nitrous acid, HNO2, by sodium hydroxide, NaOH, both in aqueous solution. Use
an arrow with H over it to show the proton transfer.
Start by writing the acid and base reactants and the salt product. Recall that the
salt consists of cations from the base and anions from the acid. Note also that water
is frequently a product; if it is, you will not be able to balance the equation without it. Once you have the balanced molecular equation, write the complete ionic
equation. Do that by representing any strong electrolytes by the formulas of their
ions. Finally, write the net ionic equation by canceling any spectator ions, and from
it note the proton transfer.
(continued)
142
CHAPTER 4
Chemical Reactions
SOLUTION
The salt consists of the cation from the base (Na ) and the anion from the acid
(NO2 ); its formula is NaNO2. You will need to add H2O as a product to complete and balance the molecular equation:
NaOH(aq) NaNO2(aq)
HNO2(aq)
H2O(l ) (molecular equation)
Note that NaOH (a strong base) and NaNO2 (a soluble ionic substance) are strong
electrolytes; HNO2 is a weak electrolyte (it is not one of the strong acids in
Table 4.3). You write both NaOH and NaNO2 in the form of ions. The complete
ionic equation is
HNO2(aq)
Na (aq)
OH (aq) Na (aq)
NO2 (aq)
H2O(l )
The net ionic equation is
OH (aq) NO2 (aq)
HNO2(aq)
H2O(l )
Note that a proton is transferred from HNO2 to the OH ion to yield the products:
H
HNO2(aq)
EXERCISE 4.5
OH (aq)
NO2 (aq)
H2O(l )
(net ionic equation)
Write the molecular equation and the net ionic equation for the neutralization of
hydrocyanic acid, HCN, by lithium hydroxide, LiOH, both in aqueous solution.
See Problems 4.37, 4.38, 4.39, and 4.40.
Acids such as HCl and HNO3 that have only one acidic hydrogen atom per acid
molecule are called monoprotic acids. A polyprotic acid is an acid that yields two
or more acidic hydrogens per molecule. Phosphoric acid is an example of a triprotic acid. By reacting this acid with different amounts of a base, you can obtain a
series of salts:
H3PO4(aq) NaOH(aq) NaH2PO4(aq) H2O(l )
Na2HPO4(aq) 2H2O(l )
H3PO4(aq) 2NaOH(aq)
H3PO4(aq) 3NaOH(aq) Na3PO4(aq) 3H2O(l )
Salts such as NaH2PO4 and Na2HPO4 that have acidic hydrogen atoms and can
undergo neutralization with bases are called acid salts.
EXERCISE 4.6
Video no. 36 (Series D),
Bicarb./Hydrochloric Acid
Volcano.
Demonstration of reaction of
acids with carbonates: Bassam
Z. Shakhashiri, Chemical
Demonstrations, Vol. 3 (Madison:
University of Wisconsin Press,
1989), no. 8.11, pp. 9699.
Write molecular and net ionic equations for the successive neutralizations of each
of the acidic hydrogens of sulfuric acid with potassium hydroxide. (That is, write
equations for the reaction of sulfuric acid with KOH to give the acid salt and for
the reaction of the acid salt with more KOH to give potassium sulfate.)
See Problems 4.41, 4.42, 4.43, and 4.44.
AcidBase Reactions with Gas Formation
Certain salts, notably carbonates, sultes, and suldes, react with acids to form a
gaseous product. Consider the reaction of sodium carbonate with hydrochloric acid.
The molecular equation for the reaction is
Na2CO3(aq)
carbonate
2HCl(aq) 2NaCl(aq)
acid
salt
H2O(l )
CO2( g)
4.4
Figure 4.10
Reaction of a carbonate with
an acid
Baking soda (sodium hydrogen
carbonate) reacts with acetic acid
in vinegar to evolve bubbles of
carbon dioxide.
AcidBase Reactions
143
Here, a carbonate (sodium carbonate) reacts with an acid (hydrochloric acid) to
produce a salt (sodium chloride), water, and carbon dioxide gas. A similar reaction
is shown in Figure 4.10, in which baking soda (sodium hydrogen carbonate) reacts
with the acetic acid in vinegar. Note the bubbles of carbon dioxide gas that evolve
during the reaction. This reaction of a carbonate with an acid is the basis of a simple test for carbonate minerals. When you treat a carbonate mineral or rock, such
as limestone, with hydrochloric acid, the material zzes, as bubbles of odorless
carbon dioxide form.
It is useful to consider the preceding reaction as an exchange, or metathesis,
reaction. Recall that in an exchange reaction, you obtain the products from the
reactants by exchanging the anions (or the cations) between the two reactants. In
this case, we interchange the carbonate ion with the chloride ion and we obtain the
following:
2HCl(aq) 2NaCl(aq)
Na2CO3(aq)
H2CO3(aq)
The last product shown in the equation is carbonic acid, H2CO3. (You obtain the formula of carbonic acid by noting that you need to associate two H ions with the carbonate ion, CO32 , to obtain a neutral compound.) Carbonic acid is unstable and
decomposes to water and carbon dioxide gas. The overall result is the equation we
wrote earlier:
Na2CO3(aq)
2HCl(aq) 2NaCl(aq)
H2O(l ) CO2( g)
H2CO3(aq)
The net ionic equation for this reaction is
CO32 (aq)
2H (aq) H2O(l )
CO2(g)
The carbonate ion reacts with hydrogen ion from the acid. If you write the hydrogen
ion as H3O (hydronium ion), you can see also that the reaction involves a proton
transfer:
2H
CO32 (aq)
CO32 , S2 , and SO32 are
bases. Thus the net equations
involve a basic anion with H ,
an acid.
2H3O (aq) H2CO3(aq)
2H2O(l ) 3H2O(l )
CO2( g)
From the broader perspective of the BrnstedLowry view, this is an acidbase reaction.
Sultes behave similarly to carbonates. When a sulte, such as sodium sulte
(Na2SO3), reacts with an acid, sulfur dioxide (SO2) is a product. Suldes, such as
sodium sulde (Na2S), react with acids to produce hydrogen sulde gas. The reaction can be viewed as a simple exchange reaction. The reactions are summarized
in Table 4.4.
Table 4.4
Some Ionic Compounds That Evolve Gases When Treated with Acids
Ionic Compound
Gas
Example
Carbonate (CO32 )
Sulte (SO32 )
Sulde (S2 )
CO2
SO2
H2S
Na2CO3 2HCl 2NaCl
Na2SO3 2HCl 2NaCl
Na2S H2SO4 Na2SO4
H2O
H2O
H2S
CO2
SO2
144
CHAPTER 4
Chemical Reactions
EXAMPLE 4.6
An alternate example is
provided in the Instructors
Resource Manual.
SOLUTION
Experiments in General
Chemistry, 4B,Ionic Reactions
in Aqueous Solutions.
Writing an Equation for a Reaction with Gas Formation
Write the molecular equation and the net ionic equation for the reaction of zinc
sulde with hydrochloric acid.
First write the reactants, noting from Table 4.1 that most suldes are insoluble
(except those of the IA metals and (NH4)2S).
ZnS(s)
HCl(aq)
If you look at this as an exchange reaction, the products are ZnCl2 and H2S. Since
chlorides are soluble (with some exceptions, not including ZnCl2), we write the
following balanced equation.
ZnS(s)
2HCl(aq) ZnCl2(aq)
H2S(g)
The complete ionic equation is
ZnS(s)
2H (aq)
2Cl (aq) Zn2 (aq)
2Cl (aq)
H2S( g)
and the net ionic equation is
ZnS(s)
EXERCISE 4.7
2H (aq) Zn2 (aq)
H2S(g)
Write the molecular equation and the net ionic equation for the reaction of calcium
carbonate with nitric acid.
See Problems 4.45, 4.46, 4.47, and 4.48.
CONCEPT CHECK 4.3
At times, we want to generalize the formula of certain important chemical substances; acids and bases fall into this category. Given the following reactions,
try to identify the acids, bases, and some examples of what the general symbols (M and A ) represent.
a. MOH(s) M (aq)
b. HA(aq)
c. H2A(aq)
OH (aq)
H2O(l) B H3O (aq)
A
2H2O(l) B H3O (aq)
A
A (aq)
HA (aq)
d. For parts a through c, come up with real examples for M and A.
4.5 OxidationReduction Reactions
Video no. 24 (Series D),
The Gold Penny.
In the two preceding sections, we described precipitation reactions (reactions producing a precipitate) and acidbase reactions (reactions involving proton transfer).
Here we discuss the third major class of reactions, oxidationreduction reactions,
which are reactions involving a transfer of electrons from one species to another.
As a simple example of an oxidationreduction reaction, let us look at what happens when you dip an iron nail into a blue solution of copper(II) sulfate (Figure 4.11).
What you see is that the iron nail becomes coated with a reddish-brown tinge of metallic copper. The molecular equation for this reaction is
Fe(s)
CuSO4(aq) FeSO4(aq)
Cu(s)
The net ionic equation is
Fe(s)
Cu2 (aq) Fe2 (aq)
Cu(s)
4.5
OxidationReduction Reactions
145
Figure 4.11
Reaction of iron with Cu2 (aq)
Left: Iron nail and copper(II) sulfate solution, which has a blue
color. Center: Fe reacts with
Cu2 (aq) to yield Fe2 (aq) and
Cu(s). In the molecular view water
and the sulfate anion have been
omitted. Right: The copper metal
plates out on the nail.
Cu2+
Fe2+
Cu
Fe
Video no. 27 (Series D),
Complexes of Vanadium.
The electron-transfer aspect of the reaction is apparent from this equation. Note that
each iron atom in the metal loses two electrons to form an iron(II) ion, and each
copper(II) ion gains two electrons to form a copper atom in the metal. The net effect
is that two electrons are transferred from each iron atom in the metal to each copper(II) ion.
The concept of oxidation numbers was developed as a simple way of keeping
track of electrons in a reaction. Using oxidation numbers, you can determine whether
or not electrons have been transferred from one atom to another. If electrons have
been transferred, an oxidationreduction reaction has occurred.
Oxidation Numbers
Ca
O
We dene the oxidation number (or oxidation state) of an atom in a substance as
the actual charge of the atom if it exists as a monatomic ion, or a hypothetical charge
assigned to the atom in the substance by simple rules. An oxidationreduction reaction is one in which one or more atoms change oxidation number, implying that there
has been a transfer of electrons.
Consider the combustion of calcium metal in oxygen gas (Figure 4.12).
2Ca(s)
O2
Ca2+
Figure 4.12
The burning of calcium
metal in oxygen
The burning calcium emits a
red-orange ame.
O2(g) 2CaO(s)
This is an oxidationreduction reaction. To see this, you assign oxidation numbers to
the atoms in the equation and then note that the atoms change oxidation number during the reaction.
Since the oxidation number of an atom in an element is always zero, Ca and O
in O2 have oxidation numbers of zero. Another rule follows from the denition of
oxidation number: The oxidation number of an atom that exists in a substance as a
monatomic ion equals the charge on that ion. So the oxidation number of Ca in CaO
is 2 (the charge on Ca2 ), and the oxidation number of O in CaO is 2 (the charge
146
CHAPTER 4
Chemical Reactions
on O2 ). To emphasize these oxidation numbers in an equation, we will write them
above the atomic symbols in the formulas.
0
2Ca(s)
Figure 4.13
The burning of calcium metal
in chlorine
The reaction appears similar to
the burning of calcium in oxygen.
0
2
2
O2( g) 2CaO(s)
From this, you see that the Ca and O atoms change in oxidation number during the
reaction. In effect, each calcium atom in the metal loses two electrons to form Ca2
ions, and each oxygen atom in O2 gains two electrons to form O2 ions. The net
result is a transfer of electrons from calcium to oxygen, so this reaction is an oxidationreduction reaction. In other words, an oxidationreduction reaction (or redox reaction) is a reaction in which electrons are transferred between species or in which
atoms change oxidation number.
Note that calcium has gained in oxidation number from 0 to 2. (Each calcium
atom loses two electrons.) We say that calcium has been oxidized. Oxygen, on the other
hand, has decreased in oxidation number from 0 to 2. (Each oxygen atom gains two
electrons.) We say that oxygen has been reduced. An oxidationreduction reaction always
involves both oxidation (the loss of electrons) and reduction (the gain of electrons).
Formerly, the term oxidation meant reaction with oxygen. The current denition greatly enlarges the meaning of this term. Consider the reaction of calcium metal
with chlorine gas (Figure 4.13); the reaction looks similar to the burning of calcium
in oxygen. The chemical equation is
0
Ca(s)
0
2
1
Cl2( g) CaCl2(s)
In this reaction, the calcium atom is oxidized, because it increases in oxidation number (from 0 to 2, as in the previous equation). Chlorine is reduced; it decreases in
oxidation number from 0 to 1. This is clearly an oxidationreduction reaction that
does not involve oxygen.
Oxidation-Number Rules
So far, we have used two rules for obtaining oxidation numbers: (1) the oxidation
number of an atom in an element is zero, and (2) the oxidation number of an atom
in a monatomic ion equals the charge on the ion. These and several other rules for
assigning oxidation numbers are given in Table 4.5.
In molecular substances, we use these rules to give the approximate charges on
the atoms. Consider the molecule SO2. Oxygen atoms tend to attract electrons, pulling
them from other atoms (sulfur in the case of SO2). As a result, an oxygen atom in
SO2 takes on a negative charge relative to the sulfur atom. The magnitude of the
charge on an oxygen atom in a molecule is not a full 2 charge as in the O2 ion.
However, it is convenient to assign an oxidation number of 2 to oxygen in SO2 (and
in most other compounds of oxygen) to help us express the approximate charge distribution in the molecule. Rule 3 in Table 4.5 says that an oxygen atom has an oxidation number of 2 in most of its compounds.
Rules 4 and 5 are similar in that they tell you what to expect for the oxidation
number of certain elements in their compounds. Rule 4, for instance, says that hydrogen has an oxidation number of 1 in most of its compounds.
Rule 6 states that the sum of the oxidation numbers of the atoms in a compound
is zero. This rule follows from the interpretation of oxidation numbers as (hypothetical) charges on the atoms. Because any compound is electrically neutral, the sum of
the charges on its atoms must be zero. This rule is easily extended to ions: the sum
OxidationReduction Reactions
4.5
147
Table 4.5
Rules for Assigning Oxidation Numbers
Rule
Applies to
Statement
1
2
Elements
Monatomic ions
3
Oxygen
4
Hydrogen
5
Halogens
6
Compounds and ions
The oxidation number of an atom in an element is zero.
The oxidation number of an atom in a monatomic ion
equals the charge on the ion.
The oxidation number of oxygen is 2 in most of its
compounds. (An exception is O in H2O2 and other
peroxides, where the oxidation number is 1.)
The oxidation number of hydrogen is 1 in most of its
compounds. (The oxidation number of hydrogen is
1 in binary compounds with a metal, such as CaH2.)
The oxidation number of uorine is 1 in all of its
compounds. Each of the other halogens (Cl, Br, I)
has an oxidation number of 1 in binary
compounds, except when the other element is
another halogen above it in the periodic table or
the other element is oxygen.
The sum of the oxidation numbers of the atoms in a
compound is zero. The sum of the oxidation numbers
of the atoms in a polyatomic ion equals the charge on
the ion.
of the oxidation numbers (hypothetical charges) of the atoms in a polyatomic ion
equals the charge on the ion.
You can use Rule 6 to obtain the oxidation number of one atom in a compound
or ion, if you know the oxidation numbers of the other atoms in the compound or
ion. Consider the SO2 molecule. According to Rule 6,
(Oxidation number of S)
2
(oxidation number of O)
0
or
(Oxidation number of S)
2
( 2)
0.
Therefore,
Oxidation number of S (in SO2)
2
( 2)
4
The next example illustrates how to use the rules in Table 4.5 to assign oxidation
numbers.
EXAMPLE 4.7
Assigning Oxidation Numbers
Use the rules from Table 4.5 to obtain the oxidation number of the chlorine atom
in each of the following: (a) HClO4 (perchloric acid), (b) ClO3 (chlorate ion).
PROBLEM STRATEGY
Two alternate examples are
provided in the Instructors
Resource Manual.
In each case, write the expression for the sum of the oxidation numbers, equating
this to zero for a compound or to the charge for an ion (Rule 6). Now, use Rules 2
to 5 to substitute oxidation numbers for particular atoms, such as 2 for oxygen and
1 for hydrogen, and solve for the unknown oxidation number (Cl in this example).
(continued)
148
CHAPTER 4
Chemical Reactions
SOLUTION
a. For perchloric acid, Rule 6 gives the equation
(Oxidation number of H)
(oxidation number of Cl)
(oxidation number of O)
0
4
Using Rules 3 and 4, you obtain
( 1)
(oxidation number of Cl)
4
( 2)
0
Therefore,
Oxidation number of Cl (in HClO4)
( 1)
4
( 2)
7
(oxidation number of O)
1
b. For the chlorate ion, Rule 6 gives the equation
(Oxidation number of Cl)
3
Using Rule 3, you obtain
(Oxidation number of Cl)
3
( 2)
1
Therefore,
Oxidation number of Cl (in ClO3 )
ANSWER CHECK
EXERCISE 4.8
1
3
( 2)
5
As most compounds do not have elements with very large positive or very large
negative oxidation numbers, you should always be on the alert for a possible
assignment mistake when you nd oxidation states greater than 6 or less than
4. (From this example you see that a 7 oxidation state is possible; however, it
only occurs in a limited number of cases.)
Obtain the oxidation numbers of the atoms in each of the following: (a) potassium
dichromate, K2Cr2O7, (b) permanganate ion, MnO4 .
See Problems 4.49, 4.50, 4.51, and 4.52.
Describing OxidationReduction Reactions
We use special terminology to describe oxidationreduction reactions. To illustrate
this, we will look again at the reaction of iron with copper(II) sulfate. The net ionic
equation is
0
2
2
Cu2 ( aq) Fe2 ( aq)
Fe(s)
0
Cu(s)
We can write this reaction in terms of two half-reactions. A half-reaction is one of
two parts of an oxidationreduction reaction, one part of which involves a loss of
electrons (or increase of oxidation number) and the other a gain of electrons (or
decrease of oxidation number). The half-reactions for the preceding equation are
0
2
Fe(s) Fe2 ( aq)
2
Cu2 ( aq)
2e
(electrons lost by Fe)
0
2e
Cu(s)
(electrons gained by Cu2 )
Oxidation is the half-reaction in which there is a loss of electrons by a species
(or an increase of oxidation number of an atom). Reduction is the half-reaction
in which there is a gain of electrons by a species (or a decrease in the oxidation
number of an atom). Thus, the equation Fe(s) Fe2 (aq)
2e represents the
oxidation half-reaction, and the equation Cu2 (aq)
2e Cu(s) represents
the reduction half-reaction.
4.5
OxidationReduction Reactions
149
Recall that a species that is oxidized loses electrons (or contains an atom that
increases in oxidation number) and a species that is reduced gains electrons (or contains an atom that decreases in oxidation number). An oxidizing agent is a species
that oxidizes another species; it is itself reduced. Similarly, a reducing agent is a
species that reduces another species; it is itself oxidized. In our example reaction, the
copper(II) ion is the oxidizing agent, whereas iron metal is the reducing agent.
The relationships among these terms are shown in the following diagram for the
reaction of iron with copper(II) ion.
oxidation
0
2
reducing
agent
0
2
Cu2 ( aq) Fe2 ( aq)
Fe(s)
Cu(s)
oxidizing
agent
Video Series C,Zinc and Iodine.
reduction
Video no. 19 (Series A),
Ammonium Dichromate
Volcano.
Video Series C,Ammonium
Dichromate Volcano.
Some Common OxidationReduction Reactions
Many of the oxidationreduction reactions can be described as:
1. Combination reactions
2. Decomposition reactions
3. Displacement reactions
4. Combustion reactions
We will describe examples of each of these in this section.
Figure 4.14
Combination reaction
Left: Sodium metal and chlorine
gas. Right: The spectacular combination reaction of sodium and
chlorine.
Combination Reactions A combination reaction is a reaction in which two
substances combine to form a third substance. Note that not all combination reactions
are oxidationreduction reactions. However, the simplest cases are those in which two
elements react to form a compound; these are clearly oxidationreduction reactions.
In Chapter 2, we discussed the reaction of sodium metal and chlorine gas, which is a
redox reaction (Figure 4.14).
2Na(s)
Cl2(g) 2NaCl(s)
Na
Cl2
NaCl
150
CHAPTER 4
Chemical Reactions
Antimony and chlorine also combine in a ery reaction.
2Sb
3Cl2 2SbCl3
Some combination reactions involve compounds as reactants and are not oxidation
reduction reactions. For example,
CaO(s)
SO2(g) CaSO3(s)
(If you check the oxidation numbers, you will see that this is not an oxidation
reduction reaction.)
Decomposition Reactions A decomposition reaction is a reaction in which a single compound reacts to give two or more substances. Often these reactions occur when
the temperature is raised. In Chapter 1, we described the decomposition of mercury(II)
oxide into its elements when the compound is heated (Figure 4.15). This is an
oxidationreduction reaction.
2HgO(s) 2Hg(l )
O2( g)
Another example is the preparation of oxygen by heating potassium chlorate with
manganese(IV) oxide as a catalyst.
O2
2KClO3(s) nO 2KCl(s)
M
Hg2+
2
3O2( g)
In this reaction, a compound decomposes into another compound and an element; it
also is an oxidationreduction reaction.
Not all decomposition reactions are of the oxidationreduction type. For example, calcium carbonate at high temperatures decomposes into calcium oxide and carbon dioxide.
CaCO3(s) CaO(s)
O
Hg
Figure 4.15
Decomposition reaction
The decomposition reaction
of mercury(II) oxide into its
elements, mercury and oxygen.
Demonstration of metal/metal
salt displacement reactions:
Summerlin and Ealy, Chemical
Demonstrations, Vol. 1
(Washington, D.C.: Am. Chem.
Soc., 1985), no. 96, p. 151.
CO2( g)
Is there a change in oxidation numbers? If not, this conrms that this is not an
oxidationreduction reaction.
Displacement Reactions A displacement reaction (also called a single-replacement
reaction) is a reaction in which an element reacts with a compound, displacing an element from it. Since these reactions involve an element and one of its compounds, these
must be oxidationreduction reactions. An example is the reaction that occurs when you
dip a copper metal strip into a solution of silver nitrate.
Cu(s)
2AgNO3(aq) Cu(NO3)2(aq)
2Ag(s)
From the molecular equation, it appears that copper displaces silver in silver nitrate,
producing crystals of silver metal and a greenish-blue solution of copper(II) nitrate.
The net ionic equation, however, shows that the reaction involves the transfer of electrons from copper metal to silver ion:
Cu(s)
2Ag (aq) Cu2 (aq)
2Ag(s)
When you dip a zinc metal strip into an acid solution, bubbles of hydrogen form
on the metal and escape from the solution (Figure 4.16).
Zn(s)
2HCl(aq) ZnCl2(aq)
H2(g)
4.5
OxidationReduction Reactions
151
Zinc displaces hydrogen in the acid, producing zinc chloride solution and hydrogen
gas. The net ionic equation is
Zn(s)
H2(g)
Whether a reaction occurs between a given element and a monatomic ion depends
on the relative ease with which the two species gain or lose electrons. Table 4.6 shows
the activity series of the elements, a listing of the elements in decreasing order of
their ease of losing electrons during reactions in aqueous solution. The metals listed
at the top are the strongest reducing agents (they lose electrons easily); those at the
bottom, the weakest. A free element reacts with the monatomic ion of another element if the free element is above the other element in the activity series.
Consider this reaction:
H2O
2K(s)
2H (aq) 2K (aq)
H2(g)
You would expect this reaction to proceed as written, because potassium metal (K) is
well above hydrogen in the activity series. In fact, potassium metal reacts violently
with water, which contains only a very small percentage of H ions. Imagine the reaction of potassium metal with a strong acid like HCl!
Cl
H+
Table 4.6
H2
React vigorously with acidic
solutions to give H2
Figure 4.16
Displacement reaction
Displacement reaction of zinc
metal and hydrochloric acid.
Hydrogen gas formed in the
reaction bubbles from the metal
surface that dips into the acid.
React with acids to give H2
Video Series C,Reactions of
Metals with Acids.
Demonstration of a displacement
reaction: Summerlin et al.,
Chemical Demonstrations, Vol. 2
(Washington, D.C.: Am. Chem.
Soc., 1987), no. 105, p. 107.
Video no. 8 (Series D),Sodium
and Potassium in Water.
Do not react with acids to give H2*
AGGBGGC AGGBGGC
Activity AGG
GBG
GGC
Cl
AGGGGGBGGGGGC Series of the Elements
Zn2+
Li
K
Ba
Ca
Na
Mg
Al
Zn
Cr
Fe
Cd
Co
Ni
Sn
Pb
H2
Cu
Hg
Ag
Au
React vigorously with liquid
water to give H2
React slowly with liquid water but
readily with steam to give H2
AG
GBG
GC
Zn
2H (aq) Zn2 (aq)
*Cu, Hg, and Ag react with HNO3 but do not produce H2. In these reactions, the metal is oxidized to the
metal ion, and NO3 ion is reduced to NO2 or other nitrogen species.
152
CHAPTER 4
Chemical Reactions
Figure 4.17
The combustion of iron wool
Iron reacts with oxygen in the air
to produce iron(III) oxide, Fe2O3.
The reaction is similar to the
rusting of iron but is much faster.
Figure 4.18
Fe
Rusting of iron wool
The temperature rises
perceptibly during the rusting,
showing that heat is released
by the reaction just as if the iron
were being burned (see Figure
4.17). Note the temperature
increase of the thermometer
wrapped with moist iron wool.
O
Fe3+
O2
A combustion reaction is a reaction in which a substance reacts with oxygen, usually with the rapid release of heat to produce a
ame. The products include one or more oxides. Oxygen changes oxidation number from 0 to 2, so combustions are oxidationreduction reactions.
Organic compounds usually burn in oxygen or air to yield carbon dioxide. If the
compound contains hydrogen (as most do), water is also a product. For instance,
butane (C4H10) burns in air as follows:
Combustion Reactions
2C4H10(g)
Video no. 3 (Series B),Does Steel
Wool Burn?
13O2(g) 8CO2(g)
10H2O(g)
Many metals burn in air as well. Although chunks of iron do not burn readily in
air, iron wool, which consists of ne strands of iron, does (Figure 4.17). The increased
surface area of the metal in iron wool allows oxygen from air to react quickly with it.
4Fe(s)
3O2(g) 2Fe2O3(s)
The rusting of iron wool is a similar reaction, although slower (Figure 4.18).
4.6 Balancing Simple OxidationReduction Equations
Balancing equations was
discussed in Section 2.10.
Oxidationreduction reactions can often be quite difcult to balance.
Some are
so complex in fact that chemists have written computer programs to accomplish the
task. In this section, we will develop a method for balancing simple oxidation
4.6
Balancing Simple OxidationReduction Equations
153
reduction reactions that can later be generalized for far more complex reactions. One
of the advantages of using this technique for even simple reactions is that you focus
on what makes oxidationreduction reactions different from other reaction types. See
Chapter 20 for a comprehensive treatment of this topic.
At rst glance, the equation representing the reaction of zinc metal with silver(I)
ions in solution might appear to be balanced.
Zn(s)
Ag (aq) Zn2 (aq)
Ag(s)
However, because a balanced chemical equation must have a charge balance as well
as a mass balance, this equation is not balanced: it has a total charge of 1 for the
reactants and 2 for the products. Let us apply the half-reaction method for balancing
this equation.
Half-Reaction Method Applied to Simple
OxidationReduction Equations
The half-reaction method consists of rst separating the equation into two halfreactions, one for oxidation, the other for reduction. You balance each half-reaction,
then combine them to obtain a balanced oxidationreduction reaction. Here is an illustration of the process. First we identify the species being oxidized and reduced and
assign the appropriate oxidation states.
0
1
Zn(s)
2
Zn2 ( aq)
Ag ( aq)
0
Ag(s)
Next, write the half-reactions in an unbalanced form.
Zn Zn2
(oxidation)
Ag Ag
(reduction)
Next, balance the charge in each equation by adding electrons to the more positive
side to create balanced half-reactions. Following this procedure, the balanced halfreactions are:
Zn Zn2
Ag
2e
(oxidation half-reaction)
e Ag
(reduction half-reaction)
Note that the number of electrons that Zn loses during the oxidation process (two)
exceeds the number of electrons gained by Ag during the reduction (one). Since,
according to the reduction half-reaction, each Ag is capable of gaining only one
electron, we need to double the amount of Ag in order for it to accept all of the
electrons produced by Zn during oxidation. To meet this goal and obtain the balanced oxidationreduction reaction, we multiply each half-reaction by a factor
(integer) so that when we add them together, the electrons cancel. We multiply
the first equation by 1 (the number of electrons in the second half-reaction) and
multiply the second equation by 2 (the number of electrons in the first halfreaction).
1
2 (Ag
Zn 2Ag
(Zn Zn2
e Ag)
2e Zn2
2e )
2Ag
2e
154
CHAPTER 4
Chemical Reactions
The electrons cancel, which nally yields the balanced oxidationreduction equation:
2Ag (aq) Zn2 (aq)
Zn(s)
2Ag(s)
Example 4.8 further illustrates this technique.
EXAMPLE 4.8
Balancing Simple OxidationReduction Reactions by the
Half-Reaction Method
Consider a more difcult problem, the combination (oxidationreduction) reaction
of magnesium metal and nitrogen gas:
N2(g) Mg3N2(s)
Mg(s)
Apply the half-reaction method to balance this equation.
PROBLEM STRATEGY
SOLUTION
Start by identifying the species undergoing oxidation and reduction and assigning
oxidation numbers. Then write the two balanced half-reactions, keeping in mind
that you add the electrons to the more positive side of the reaction. Next, multiply each of the half-reactions by a whole number that will cancel the electrons on
each side of the equation. Finally, add the half-reactions together to yield the balanced equation.
Identify the oxidation states of the elements:
0
0
Mg(s)
2
3
N2(g) Mg3N2(s)
In this problem, a molecular compound, nitrogen (N2), is undergoing reduction.
When a species undergoing reduction or oxidation is a molecule, write the formula
of the molecule (do not split it up) in the half-reaction. Also, make sure that both
the mass and the charge are balanced. (Note the 6e required to balance the charge
due to the 2N3 .)
Mg Mg2
N2
6e 2N
2e
3
(balanced oxidation half-reaction)
(balanced reduction half-reaction)
We now need to multiply each half-reaction by a factor that will cancel the
electrons.
1
3
(N2
3Mg
N2
(Mg Mg2
6e 2N3 )
6e 3Mg2
2e )
2N3
6e
Therefore, the balanced combination (oxidationreduction) reaction is
3Mg
N2 3Mg2
2N3
From inspecting the coefcients in this reaction, looking at the original equation,
and knowing that Mg2 and N3 will combine to form an ionic compound
(Mg3N2), we can rewrite the equation in the following form:
3Mg(s)
EXERCISE 4.9
N2(g) Mg3N2(s)
Use the half-reaction method to balance the equation Ca(s)
Cl2(g) CaCl2(s).
See Problems 4.59 and 4.60.
4.7
Molar Concentration
155
Working with Solutions
The majority of the chemical reactions discussed in this chapter take place in solution.
This is because the reaction between two solid reactants often proceeds very slowly or
not at all. In a solid, the molecules or ions in a crystal tend to occupy approximately xed
positions, so that the chance of two molecules or ions coming together to react is small.
In liquid solutions, reactant molecules are free to move throughout the liquid; therefore,
reaction is much faster. When you run reactions in liquid solutions, it is convenient to
dispense the amounts of reactants by measuring out volumes of reactant solutions. In the
next two sections, we will discuss calculations involved in making up solutions, and in
Section 4.10 we will describe stoichiometric calculations involving such solutions.
4.7 Molar Concentration
When we dissolve a substance in a liquid, we call the substance the solute and the
liquid the solvent. Consider ammonia solutions. Ammonia gas dissolves readily in
water, and aqueous ammonia solutions are often used in the laboratory. In such solutions, ammonia gas is the solute and water is the solvent.
The general term concentration refers to the quantity of solute in a standard quantity of solution. Qualitatively, we say that a solution is dilute when the solute concentration is low and concentrated when the solute concentration is high. Usually
these terms are used in a comparative sense and do not refer to a specic concentration. We say that one solution is more dilute, or less concentrated, than another. However, for commercially available solutions, the term concentrated refers to the maximum, or near maximum, concentration available. For example, concentrated aqueous
ammonia contains about 28% NH3 by mass.
In this example, we expressed the concentration quantitatively by giving the mass
percentage of solutethat is, the mass of solute in 100 g of solution. However, we
need a unit of concentration that is convenient for dispensing reactants in solution,
such as one that species moles of solute per solution volume.
Molar concentration, or molarity (M ), is dened as the moles of solute dissolved in one liter (cubic decimeter) of solution.
Molarity (M )
moles of solute
liters of solution
An aqueous solution that is 0.15 M NH3 (read this as 0.15 molar NH3) contains 0.15
mol NH3 per liter of solution. If you want to prepare a solution that is, for example,
0.200 M CuSO4, you place 0.200 mol CuSO4 in a 1.000-L volumetric ask, or a proportional amount in a ask of a different size (Figure 4.19). You add a small quantity
of water to dissolve the CuSO4. Then you ll the ask with additional water to the
mark on the neck and mix the solution. The following example shows how to calculate the molarity of a solution given the mass of solute and the volume of solution.
EXAMPLE 4.9
An alternate example is
provided in the Instructors
Resource Manual.
Calculating Molarity from Mass and Volume
A sample of NaNO3 weighing 0.38 g is placed in a 50.0-mL volumetric ask. The
ask is then lled with water to the mark on the neck, dissolving the solid. What
is the molarity of the resulting solution?
(continued)
156
CHAPTER 4
Chemical Reactions
Figure 4.19
Preparing a 0.200 M CuSO4 solution
Left: 0.0500 mol CuSO4 5H2O (12.48 g) is weighed on a platform balance. Center: The
copper(II) sulfate pentahydrate is transferred carefully to the volumetric ask. Right: Water is
added to bring the solution level to the mark on the neck of the 250-mL volumetric ask.The
molarity is 0.0500 mol/0.250 L 0.200 M.
PROBLEM STRATEGY
To calculate the molarity, you need the moles of solute. Therefore, you rst convert grams NaNO3 to moles. The molarity equals the moles of solute divided by
the liters of solution.
SOLUTION
You nd that 0.38 g NaNO3 is 4.47 10 3 mol NaNO3; the last signicant gure is
underlined. The volume of solution is 50.0 mL, or 50.0 10 3 L, so the molarity is
Molarity
ANSWER CHECK
EXERCISE 4.10
4.47 10 3 mol NaNO3
_
50.0 10 3 L soln
0.089 M NaNO3
Although very dilute solutions are possible, there is a limit as to how concentrated
solutions can be. Therefore, any answer that leads to solution concentrations that
are in excess of 20 M should be suspect.
A sample of sodium chloride, NaCl, weighing 0.0678 g is placed in a 25.0-mL volumetric ask. Enough water is added to dissolve the NaCl, and then the ask is
lled to the mark with water and carefully shaken to mix the contents. What is the
molarity of the resulting solution?
See Problems 4.61, 4.62, 4.63, and 4.64.
The advantage of molarity as a concentration unit is that the amount of solute is
related to the volume of solution. Rather than having to weigh out a specied mass of
substance, you can instead measure out a denite volume of solution of the substance,
which is usually easier. As the following example illustrates, molarity can be used as
a factor for converting from moles of solute to liters of solution, and vice versa.
EXAMPLE 4.10
An alternate example is
provided in the Instructors
Resource Manual.
Using Molarity as a Conversion Factor
An experiment calls for the addition to a reaction vessel of 0.184 g of sodium
hydroxide, NaOH, in aqueous solution. How many milliliters of 0.150 M NaOH
should be added?
4.8
PROBLEM STRATEGY
Diluting Solutions
157
You rst need to convert grams NaOH to moles NaOH, because molarity relates
moles of solute to volume of solution. Then, you convert moles NaOH to liters of
solution, using the molarity as a conversion factor. Here, 0.150 M means that 1 L
of solution contains 0.150 moles of solute, so the conversion factor is:
1 L soln
0.150 mol NaOH
Converts
mol NaOH to L soln
SOLUTION
Here is the calculation. (The molar mass of NaOH is 40.0 g/mol.)
0.184 g NaOH
1 mol NaOH
40.0 g NaOH
1 L soln
0.150 mol NaOH
3.07
10
2
L soln (or 30.7 mL)
You need to add 30.7 mL of 0.150 M NaOH solution to the reaction vessel.
EXERCISE 4.11
How many milliliters of 0.163 M NaCl are required to give 0.0958 g of sodium
chloride?
See Problems 4.65, 4.66, 4.67, and 4.68.
EXERCISE 4.12
How many moles of sodium chloride should be put in a 50.0-mL volumetric ask
to give a 0.15 M NaCl solution when the ask is lled to the mark with water?
How many grams of NaCl is this?
See Problems 4.69, 4.70, 4.71, and 4.72.
4.8 Diluting Solutions
Experiments in General
Chemistry, 4A,Conductivity
of Aqueous Solutions.
Commercially available aqueous ammonia (28.0% NH3) is 14.8 M NH3. Suppose,
however, that you want a solution that is 1.00 M NH3. You need to dilute the concentrated solution with a denite quantity of water. For this purpose, you must know
the relationship between the molarity of the solution before dilution (the initial molarity) and that after dilution (the nal molarity).
To obtain this relationship, rst recall the equation dening molarity:
Molarity
moles of solute
liters of solution
You can rearrange this to give
Moles of solute
molarity
liters of solution
The product of molarity and the volume (in liters) gives the moles of solute in the
solution. Writing Mi for the initial molar concentration and Vi for the initial volume
of solution, you get
Moles of solute
Mi
Vi
When the solution is diluted by adding more water, the concentration and volume
change to Mf (the nal molar concentration) and Vf (the nal volume), and the moles
of solute equals
Moles of solute
Mf
Vf
158
CHAPTER 4
Chemical Reactions
Figure 4.20
Molecular view of the
dilution process
Top: A molecular view of a solution
of Cl2 dissolved in water. Bottom:
The solution after performing a
dilution by adding water. Note
how the number of moles of Cl2 in
the container does not change
when perfoming the dilution, only
the concentration changes. In this
particular case, the concentration
of Cl2 drops to half of the starting
concentration because the volume
was doubled.
Add water
(solvent)
Figure 4.21
Preparing a solution by diluting
a concentrated one
A volume of concentrated
ammonia, similar to the one
in the beaker, has been added to
the volumetric ask. Here water
is being added from the plastic
squeeze bottle to bring the
volume up to the mark on
the ask.
EXAMPLE 4.11
An alternate example is
provided in the Instructors
Resource Manual.
SOLUTION
Because the moles of solute has not changed during the dilution (Figure 4.20),
Mi
Vi
Mf
Vf
(Note: You can use any volume units, but both Vi and Vf must be in the same units.)
The next example illustrates how you can use this formula to nd the volume of
a concentrated solution needed to prepare a given volume of dilute solution.
Diluting a Solution
You are given a solution of 14.8 M NH3. How many milliliters of this solution do
you require to give 100.0 mL of 1.00 M NH3 when diluted (Figure 4.21)?
You know the nal volume (100.0 mL), nal concentration (1.00 M), and initial
concentration (14.8 M). You write the dilution formula and rearrange it to give the
initial volume.
MiV
i
Mf Vf
V
i
Mf Vf
Mi
Now you substitute the known values into the right side of the equation.
V
i
1.00 M 100.0 mL
14.8 M
6.76 mL
4.9
ANSWER CHECK
EXERCISE 4.13
Gravimetric Analysis
159
When performing a dilution, the volume of the more concentrated solution should
always be less than the volume of the nal solution as it is in the photo. Related
to this concept is the fact that the initial concentration of a solution is always
greater than the nal concentration after dilution. These two concepts will always
allow you to check the reasonableness of your calculated quantities when using the
dilution equation.
You have a solution that is 1.5 M H2SO4 (sulfuric acid). How many milliliters of
this acid do you need to prepare 100.0 mL of 0.18 M H2SO4?
See Problems 4.73 and 4.74.
CONCEPT CHECK 4.4
Consider the following beakers. Each contains a solution of the hypothetical
atom X.
A
B
C
D
a. Arrange the beakers in order of increasing concentration of X.
b. Without adding or removing X, what specic things could you do to make
the concentrations of X equal in each beaker? (Hint: Think about dilutions.)
Quantitative Analysis
Analytical chemistry deals with the determination of composition of materialsthat
is, the analysis of materials. The materials that one might analyze include air, water,
food, hair, body uids, pharmaceutical preparations, and so forth. The analysis of materials is divided into qualitative and quantitative analysis. Qualitative analysis involves
the identication of substances or species present in a material. For instance, you might
determine that a sample of water contains lead(II) ion. Quantitative analysis, which
we will discuss in the last sections of this chapter, involves the determination of the
amount of a substance or species present in a material. In a quantitative analysis, you
might determine that the amount of lead(II) ion in a sample of water is 0.067 mg/L.
4.9 Gravimetric Analysis
Gravimetric analysis is a type of quantitative analysis in which the amount of a
species in a material is determined by converting the species to a product that can
be isolated completely and weighed. Precipitation reactions are frequently used in
gravimetric analyses. In these reactions, you determine the amount of an ionic species
by precipitating it from solution. The precipitate, or solid formed in the reaction, is
160
CHAPTER 4
Chemical Reactions
Figure 4.22
Gravimetric analysis for
barium ion
Left: A solution of potassium
chromate (yellow) is poured
down a stirring rod into a solution
containing an unknown amount
of barium ion, Ba2 . The yellow
precipitate that forms is barium
chromate, BaCrO4. Right: The
solution is ltered by pouring it
into a crucible containing a
porous glass partition. Afterward,
the crucible is heated to dry the
barium chromate. By weighing
the crucible before and afterward,
you can determine the mass of
precipitate.
then ltered from the solution, dried, and weighed. The advantages of a gravimetric
analysis are its simplicity (at least in theory) and its accuracy. The chief disadvantage
is that it requires meticulous, time-consuming work. Because of this, whenever possible, chemists use modern instrumental methods.
As an example of a gravimetric analysis, consider the problem of determining the
amount of lead in a sample of drinking water. Lead, if it occurs in the water, probably exists as the lead(II) ion, Pb2 . Lead(II) sulfate is a very insoluble compound of
lead(II) ion. When sodium sulfate, Na2SO4, is added to a solution containing Pb2 ,
lead(II) sulfate precipitates (that is, PbSO4 comes out of the solution as a ne, crystalline solid). If you assume that the lead is present in solution as lead(II) nitrate, you
can write the following equation for the reaction:
Na2SO4(aq)
Pb(NO3)2(aq) 2NaNO3(aq)
PbSO4(s)
You can separate the white precipitate of lead(II) sulfate from the solution by ltration. Then you dry and weigh the precipitate. Figure 4.22 shows the laboratory setup
used in a similar analysis.
EXAMPLE 4.12
An alternate example is
provided in the Instructors
Resource Manual.
Determining the Amount of a Species by Gravimetric Analysis
A 1.000-L sample of polluted water was analyzed for lead(II) ion, Pb2 , by adding
an excess of sodium sulfate to it. The mass of lead(II) sulfate that precipitated was
229.8 mg. What is the mass of lead in a liter of the water? Give the answer as milligrams of lead per liter of solution.
PROBLEM STRATEGY
All of the lead in the water solution is precipitated as lead(II) sulfate, PbSO4. If
you determine the percentage of lead in PbSO4, you can calculate the quantity of
lead in the water sample.
SOLUTION
Following Example 3.7, you obtain the mass percentage of Pb in PbSO4 by dividing the molar mass of Pb by the molar mass of PbSO4, then multiplying by 100%:
% Pb
207.2 g/mol
303.3 g/mol
100%
68.32%
4.10 Volumetric Analysis
161
Therefore, the 1.000-L sample of water contains
Amount Pb in sample
229.8 mg PbSO4
0.6832
157.0 mg Pb
The water sample contains 157.0 mg Pb per liter.
EXERCISE 4.14
You are given a sample of limestone, which is mostly CaCO3, to determine the
mass percentage of Ca in the rock. You dissolve the limestone in hydrochloric acid,
which gives a solution of calcium chloride. Then you precipitate the calcium ion
in solution by adding sodium oxalate, Na2C2O4. The precipitate is calcium oxalate,
CaC2O4. You nd that a sample of limestone weighing 128.3 mg gives 140.2 mg
of CaC2O4. What is the mass percentage of calcium in the limestone?
See Problems 4.77 and 4.78.
4.10 Volumetric Analysis
As you saw earlier, you can use molarity as a conversion factor, and in this way you
can calculate the volume of solution that is equivalent to a given mass of solute (see
Example 4.10). This means that you can replace mass measurements in solution reactions by volume measurements. In the next example, we look at the volumes of solutions involved in a given reaction.
EXAMPLE 4.13
An alternate example is
provided in the Instructors
Resource Manual.
PROBLEM STRATEGY
SOLUTION
Video no. 6 (Series A),
AcidBase Titration.
Calculating the Volume of Reactant Solution Needed
Consider the reaction of sulfuric acid, H2SO4, with sodium hydroxide, NaOH.
H2SO4(aq)
2NaOH(aq) 2H2O(l )
Na2SO4(aq)
Suppose a beaker contains 35.0 mL of 0.175 M H2SO4. How many milliliters of
0.250 M NaOH must be added to react completely with the sulfuric acid?
You convert from 35.0 mL (or 35.0
10 3 L) H2SO4 solution to moles H2SO4
(using the molarity of H2SO4), then to moles NaOH (from the chemical equation).
Finally, you convert this to volume of NaOH solution (using the molarity of
NaOH).
The calculation is as follows:
35.0
10
3
0.175 mol H2SO4 2 mol NaOH
1 L H2SO4 soln
1 mol H2SO4
1 L NaOH soln
4.90 10 2 L NaOH soln (or 49.0 mL NaOH soln)
0.250 mol NaOH
L H2SO4 soln
Thus, 35.0 mL of 0.175 M sulfuric acid solution reacts with exactly 49.0 mL of
0.250 M sodium hydroxide solution.
ANSWER CHECK
Whenever you perform a titration calculation, be sure that you have taken into
account the stoichiometry of the reaction between the acid and base (use the balanced chemical equation). In this case, two moles of NaOH are required to neutralize each mole of acid. Furthermore, when performing titration calculations, do
not be tempted to apply the dilution equation to solve the problem. If you were to
take such an approach here, you would arrive at an incorrect result since the dilution equation fails to take into account the stoichiometry of the reaction.
(continued)
162
CHAPTER 4
Chemical Reactions
EXERCISE 4.15
Nickel sulfate, NiSO4, reacts with sodium phosphate, Na3PO4, to give a pale
yellow-green precipitate of nickel phosphate, Ni3(PO4)2, and a solution of sodium
sulfate, Na2SO4.
3NiSO4(aq)
2Na3PO4(aq) Ni3(PO4)2(s)
3Na2SO4(aq)
How many milliliters of 0.375 M NiSO4 will react with 45.7 mL of 0.265 M
Na3PO4?
See Problems 4.83 and 4.84.
An important method for determining the amount of a particular substance is based
on measuring the volume of reactant solution. Suppose substance A reacts in solution
with substance B. If you know the volume and concentration of a solution of B that just
reacts with substance A in a sample, you can determine the amount of A. Titration is a
procedure for determining the amount of substance A by adding a carefully measured
volume of a solution with known concentration of B until the reaction of A and B is just
complete. Volumetric analysis is a method of analysis based on titration.
Figure 4.23 shows a ask containing hydrochloric acid with an unknown amount
of HCl being titrated with sodium hydroxide solution, NaOH, of known molarity. The
reaction is
NaOH(aq)
An indicator is a substance
that undergoes a color change
when a reaction approaches
completion. See Section 4.4.
Figure 4.23
Titration of an unknown
amount of HCl with NaOH
Left: The ask contains HCl and a
few drops of phenolphthalein
indicator; the buret contains
0.207 M NaOH (the buret reading
is 44.97 mL). Center: NaOH was
added to the solution in the ask
until a persistent faint pink color
was reached, marking the
endpoint of the titration (the
buret reading is 49.44 mL).
The amount of HCl can be
determined from the volume
of NaOH used (4.47 mL); see
Example 4.14. Right: The addition
of several drops of NaOH solution
beyond the endpoint gives a
deep pink color.
HCl(aq) NaCl(aq)
H2O(l )
To the HCl solution are added a few drops of phenolphthalein indicator.
Phenolphthalein is colorless in the hydrochloric acid but turns pink at the completion of
the reaction of NaOH with HCl. Sodium hydroxide with a concentration of 0.207 M
is contained in a buret, a glass tube graduated to measure the volume of liquid delivered from the stopcock. The solution in the buret is added to the HCl in the ask until
the phenolphthalein just changes from colorless to pink. At this point, the reaction is
complete and the volume of NaOH that reacts with the HCl is read from the buret.
This volume is then used to obtain the mass of HCl in the original solution.
163
A Checklist for Review
EXAMPLE 4.14
Calculating the Quantity of Substance in a Titrated Solution
A ask contains a solution with an unknown amount of HCl. This solution is
titrated with 0.207 M NaOH. It takes 4.47 mL NaOH to complete the reaction.
What is the mass of the HCl?
An alternate example is
provided in the Instructors
Resource Manual.
PROBLEM STRATEGY
SOLUTION
Experiments in General
Chemistry, 4C,How Much
Acetic Acid Is in Vinegar?
You convert the volume of NaOH (4.47 10 3 L NaOH solution) to moles NaOH
(from the molarity of NaOH). Then you convert moles NaOH to moles HCl (from
the chemical equation). Finally, you convert moles HCl to grams HCl.
The calculation is as follows:
4.47
10
3
L NaOH soln
0.207 mol NaOH
1 L NaOH soln
1 mol HCl
1 mol NaOH
36.5 g HCl
1 mol HCl
0.0338
EXERCISE 4.16
A 5.00-g sample of vinegar is titrated with 0.108 M NaOH. If the vinegar requires
39.1 mL of the NaOH solution for complete reaction, what is the mass percentage
of acetic acid, HC2H3O2, in the vinegar? The reaction is
HC2H3O2(aq)
NaOH(aq) NaC2H3O2(aq)
H2O(l )
See Problems 4.85 and 4.86.
CONCEPT CHECK 4.5
Consider three asks, each containing 0.10 mol of acid. You need to learn something about the acids in each of the asks, so you perform titration using an
NaOH solution. Here are the results of the experiment:
Flask A
10 mL of NaOH required for neutralization
Flask B
20 mL of NaOH required for neutralization
Flask C
30 mL of NaOH required for neutralization
a. What have you learned about each of these acids from performing the experiment?
b. Could you use the results of this experiment to determine the concentration
of the NaOH? If not, what assumption about the molecular formulas of the
acids would allow you to make the concentration determination?
A Checklist for Review
Important Terms
electrolyte (4.1)
nonelectrolyte (4.1)
strong electrolyte (4.1)
weak electrolyte (4.1)
molecular equation (4.2)
complete ionic
equation (4.2)
spectator ion (4.2)
net ionic equation (4.2)
precipitate (4.3)
exchange (metathesis)
reaction (4.3)
acidbase indicator (4.4)
acid (Arrhenius) (4.4)
base (Arrhenius) (4.4)
acid (BrnstedLowry) (4.4)
base (BrnstedLowry) (4.4)
strong acid (4.4)
weak acid (4.4)
strong base (4.4)
weak base (4.4)
neutralization
reaction (4.4)
salt (4.4)
polyprotic acid (4.4)
oxidation number
(oxidation state) (4.5)
oxidationreduction
reaction (redox
reaction) (4.5)
half-reaction (4.5)
oxidation (4.5)
164
CHAPTER 4
Chemical Reactions
reduction (4.5)
oxidizing agent (4.5)
reducing agent (4.5)
combination reaction (4.5)
decomposition
reaction (4.5)
displacement reaction
(single-replacement
reaction) (4.5)
combustion reaction (4.5)
molar concentration
(molarity) (M) (4.7)
quantitative analysis (p. 159)
gravimetric analysis (4.9)
titration (4.10)
volumetric analysis (4.10)
Key Equations
Molarity (M )
moles of solute
liters of solution
Mi
Vi
Mf
Vf
Summary of Facts and Concepts
Reactions often involve ions in aqueous solution. Many of the
compounds in such reactions are electrolytes, which are substances that dissolve in water to give ions. Electrolytes that
exist in solution almost entirely as ions are called strong electrolytes. Electrolytes that dissolve in water to give a relatively
small percentage of ions are called weak electrolytes. The solubility rules can be used to predict the extent to which an ionic
compound will dissolve in water. Most soluble ionic compounds are strong electrolytes.
We can represent a reaction involving ions in one of three
different ways, depending on what information we want to
convey. A molecular equation is one in which substances are
written as if they were molecular, even though they are ionic.
This equation closely describes what you actually do in the laboratory. However, this equation does not describe what is happening at the level of ions and molecules. For that purpose, we
rewrite the molecular equation as a complete ionic equation by
replacing the formulas for strong electrolytes by their ion formulas. If you cancel spectator ions from the complete ionic
equation, you obtain the net ionic equation.
Most of the important reactions we consider in this course
can be divided into three major classes: (1) precipitation reactions, (2) acidbase reactions, and (3) oxidationreduction reactions. A precipitation reaction occurs in aqueous solution because one product is insoluble. You can decide whether two
ionic compounds will result in a precipitation reaction, if you
know from solubility rules that one of the potential products is
insoluble.
Acids are substances that yield hydrogen ions in aqueous
solution or donate protons. Bases are substances that yield hydroxide ions in aqueous solution or accept protons. These
acidbase reactions are proton-transfer reactions. In this chapter, we covered neutralization reactions (reactions of acids and
bases to yield salts) and reactions of certain salts with acids to
yield a gas.
Oxidationreduction reactions are reactions involving a
transfer of electrons from one species to another or a change in
the oxidation number of atoms. The concept of oxidation numbers helps us describe this type of reaction. The atom that increases in oxidation number is said to undergo oxidation; the
atom that decreases in oxidation number is said to undergo
reduction. Oxidation and reduction must occur together in a
reaction. Many oxidationreduction reactions fall into the following categories: combination reactions, decomposition reactions, displacement reactions, and combustion reactions.
Oxidationreduction reactions can be balanced by the halfreaction method.
Molar concentration, or molarity, is the moles of solute in
a liter of solution. Knowing the molarity allows you to calculate the amount of solute in any volume of solution. Because
the moles of solute are constant during the dilution of a solution, you can determine to what volume to dilute a concentrated solution to give one of desired molarity.
Quantitative analysis involves the determination of the
amount of a species in a material. In gravimetric analysis, you
determine the amount of a species by converting it to a product
you can weigh. In volumetric analysis, you determine the
amount of a species by titration. Titration is a method of chemical analysis in which you measure the volume of solution of
known molarity that reacts with a compound of unknown
amount. You determine the amount of the compound from this
volume of solution.
Operational Skills
1. Using the solubility rules Given the formula of an ionic
compound, predict its solubility in water. (EXAMPLE 4.1)
2. Writing net ionic equations Given a molecular equation,
write the corresponding net ionic equation. (EXAMPLE 4.2)
3. Deciding whether precipitation occurs Using solubility
rules, decide whether two soluble ionic compounds react to
form a precipitate. If they do, write the net ionic equation.
(EXAMPLE 4.3)
Conceptual Problems
4. Classifying acids and bases as strong or weak Given the
formula of an acid or a base, classify it as strong or weak.
(EXAMPLE 4.4)
5. Writing an equation for a neutralization Given an acid
and a base, write the molecular equation and then the net ionic
equation for the neutralization reaction. (EXAMPLE 4.5)
6. Writing an equation for a reaction with gas formation
Given the reaction between a carbonate, sulte, or sulde
and an acid, write the molecular and the net ionic equations.
(EXAMPLE 4.6)
7. Assigning oxidation numbers Given the formula of a
simple compound or ion, obtain the oxidation numbers of
the atoms, using the rules for assigning oxidation numbers.
(EXAMPLE 4.7)
8. Balancing simple oxidationreduction reactions by the
half-reaction method Given an oxidationreduction reaction, balance it. (EXAMPLE 4.8)
9. Calculating molarity from mass and volume Given the
mass of the solute and the volume of the solution, calculate
the molarity. (EXAMPLE 4.9)
165
10. Using molarity as a conversion factor Given the volume
and molarity of a solution, calculate the amount of solute.
Or, given the amount of solute and the molarity of a solution, calculate the volume. (EXAMPLE 4.10)
11. Diluting a solution Calculate the volume of solution of
known molarity required to make a specied volume of solution with different molarity. (EXAMPLE 4.11)
12. Determining the amount of a species by gravimetric
analysis Given the amount of a precipitate in a gravimetric analysis, calculate the amount of a related species.
(EXAMPLE 4.12)
13. Calculating the volume of reactant solution needed
Given the chemical equation, calculate the volume of solution of known molarity of one substance that just reacts
with a given volume of solution of another substance.
(EXAMPLE 4.13)
14. Calculating the quantity of substance in a titrated solution Calculate the mass of one substance that reacts with a
given volume of solution of known molarity of another substance. (EXAMPLE 4.14)
Review Questions
4.1 Explain why some electrolyte solutions are strongly conducting, whereas others are weakly conducting.
4.2 Dene the terms strong electrolyte and weak electrolyte.
Give an example of each.
4.3 Explain the terms soluble and insoluble. Use the solubility rules to write the formula of an insoluble ionic compound.
4.4 What are the advantages and disadvantages of using a
molecular equation to represent an ionic reaction?
4.5 What is a spectator ion? Illustrate with a complete ionic
reaction.
4.6 What is a net ionic equation? What is the value in using
a net ionic equation? Give an example.
4.7 What are the major types of chemical reactions? Give a
brief description and an example of each.
4.8 Describe in words how you would prepare pure crystalline AgCl and NaNO3 from solid AgNO3 and solid NaCl.
4.9 Give an example of a neutralization reaction. Label the
acid, base, and salt.
4.10 Give an example of a polyprotic acid and write equations for the successive neutralizations of the acidic hydrogen
atoms of the acid molecule to produce a series of salts.
4.11 Why must oxidation and reduction occur together in a
reaction?
4.12 Give an example of a displacement reaction. What is
the oxidizing agent? What is the reducing agent?
4.13 Why is the product of molar concentration and volume
constant for a dilution problem?
4.14 Describe how the amount of sodium hydroxide in a
mixture can be determined by titration with hydrochloric acid
of known molarity.
Conceptual Problems
4.15 You need to perform gravimetric analysis of a water
sample in order to determine the amount of Ag present.
a. List three aqueous solutions that would be suitable for mixing with the sample to perform the analysis.
b. Would adding KNO3(aq) allow you to perform the analysis?
c. Assume you have performed the analysis and the silver
solid that formed is moderately soluble. How might this affect your analysis results?
4.16 In this problem you need to draw two pictures of solutions in beakers at different points in time. Time zero (t 0)
will be the hypothetical instant at which the reactants dissolve
in the solution (if they dissolve) before they react. Time after
166
CHAPTER 4
Chemical Reactions
mixing (t 0) will be the time required to allow sufcient interaction of the materials. For now, we assume that insoluble
solids have no ions in solution and do not worry about representing the stoichiometric amounts of the dissolved ions. Here
is an example: Solid NaCl and solid AgNO3 are added to a
beaker containing 250 mL of water.
Na+(aq)
Cl (aq)
Ag+(aq)
Na+(aq)
NO3 (aq)
NO3 (aq)
AgCl(s)
t=0
t>0
Note that we are not showing the H2O, and we are representing
only the ions and solids in solution. Using the same conditions
as the example (adding the solids to H2O), draw pictures of the
following:
a. solid lead(II) nitrate and solid ammonium chloride at t 0
and t 0
b. FeS(s) and NaNO3(s) at t 0 and t 0
c. solid lithium iodide and solid sodium carbonate at t 0 and
t0
4.17 You come across a beaker that contains water, aqueous
ammonium acetate, and a precipitate of calcium phosphate.
a. Write the balanced molecular equation for a reaction between
two solutions containing ions that could produce this solution.
b. Write the complete ionic equation for the reaction in part a.
c. Write the net ionic equation for the reaction in part a.
4.18 Three acid samples are prepared for titration by 0.01 M
NaOH:
1. Sample 1 is prepared by dissolving 0.01 mol of HCl in
50 mL of water.
2. Sample 2 is prepared by dissolving 0.01 mol of HCl in
60 mL of water.
3. Sample 3 is prepared by dissolving 0.01 mol of HCl in
70 mL of water.
a. Without performing a formal calculation, compare the concentrations of the three acid samples (rank them from highest to lowest).
b. When performing the titration, which sample, if any, will
require the largest volume of the 0.01 M NaOH for neutralization?
4.19 Would you expect a precipitation reaction between an
ionic compound that is an electrolyte and an ionic compound
that is a nonelectrolyte? Justify your answer.
4.20 Equal quantities of the hypothetical strong acid HX,
weak acid HA, and weak base BZ, are each added to a separate
beaker of water, producing the solutions depicted in the drawings. In the drawings, the relative amounts of each substance
present in the solution (neglecting the water) are shown. Identify the acid or base that was used to produce each of the solutions (HX, HA, or BZ).
= H3O+
= OH
A
B
C
4.21 Try and answer the following questions without using a
calculator.
a. A solution is made by mixing 1.0 L of 0.5 M NaCl and 0.5 L of
1.0 M CaCl2. Which ion is at the highest concentration in the
solution?
b. Another solution is made by mixing 0.50 L of 1.0 M KBr and
0.50 L of 1.0 M K3PO4. What is the concentration of each ion
in the solution?
4.22 If one mole of the following compounds were each placed
into separate beakers containing the same amount of water, rank
the Cl (aq) concentrations from highest to lowest (some may be
equivalent): KCl, AlCl3, PbCl2, NaCl, HCl, NH3, KOH, and HCN.
Practice Problems
Solubility Rules
4.23 Using solubility rules, predict the solubility in water of
the following ionic compounds.
a. PbS
c. Na2CO3
b. AgNO3
d. CaI2
4.24 Using solubility rules, predict the solubility in water of
the following ionic compounds.
c. NH4Br
a. Al(OH)3
d. NaOH
b. Li3P
Practice Problems
4.25 Using solubility rules, decide whether the following
ionic solids are soluble or insoluble in water. If they are soluble, indicate what ions you would expect to be present in solution.
a. AgBr
c. Ca3(PO4)2
b. Li2SO4
d. Na2CO3
4.26 Using solubility rules, decide whether the following
ionic solids are soluble or insoluble in water. If they are soluble, indicate what ions you would expect to be present in solution.
c. PbSO4
a. (NH4)2SO4
b. BaCO3
d. Ca(NO3)2
Ionic Equations
4.27 Write net ionic equations for the following molecular
equations. HBr is a strong electrolyte.
a. HBr(aq) KOH(aq) KBr(aq) H2O(l)
b. AgNO3(aq) NaBr(aq) AgBr(s) NaNO3(aq)
c. CaS(aq) 2HBr(aq) CaBr2(aq) H2S(g)
d. NaOH(aq) NH4Br(aq)
NaBr(aq) NH3(g) H2O(l)
4.28 Write net ionic equations for the following molecular
equations. HBr is a strong electrolyte.
a. HBr(aq) NH3(aq) NH4Br(aq)
b. 2HBr(aq) Ba(OH)2(aq) 2H2O(l) BaBr2(aq)
c. Pb(NO3)2(aq) 2NaBr(aq) PbBr2(s) 2NaNO3(aq)
d. MgCO3(s) H2SO4(aq)
MgSO4(aq) H2O(l) CO2(g)
4.29 Lead(II) nitrate solution and sodium sulfate solution are
mixed. Crystals of lead(II) sulfate come out of solution, leaving a solution of sodium nitrate. Write the molecular equation
and the net ionic equation for the reaction.
4.30 Potassium carbonate solution reacts with aqueous hydrobromic acid to give a solution of potassium bromide, carbon dioxide gas, and water. Write the molecular equation and
the net ionic equation for the reaction.
Precipitation
4.31 Write the molecular equation and the net ionic equation
for each of the following aqueous reactions. If no reaction occurs, write NR after the arrow.
a. FeSO4 NaCl
b. Na2CO3 MgBr2
c. MgSO4 NaOH
d. NiCl2 NaBr
4.32 Write the molecular equation and the net ionic equation
for each of the following aqueous reactions. If no reaction occurs, write NR after the arrow.
a. AgNO3 NaI
b. Ba(NO3)2 K2SO4
167
c. Mg(NO3)2 K2SO4
d. CaCl2 Al(NO3)3
4.33 For each of the following, write molecular and net ionic
equations for any precipitation reaction that occurs. If no reaction occurs, indicate this.
a. Solutions of barium nitrate and lithium sulfate are mixed.
b. Solutions of sodium bromide and calcium nitrate are mixed.
c. Solutions of aluminum sulfate and sodium hydroxide are
mixed.
d. Solutions of calcium bromide and sodium phosphate are
mixed.
4.34 For each of the following, write molecular and net ionic
equations for any precipitation reaction that occurs. If no reaction occurs, indicate this.
a. Zinc chloride and sodium sulde are dissolved in water.
b. Sodium sulde and calcium chloride are dissolved in water.
c. Magnesium sulfate and potassium iodide are dissolved in
water.
d. Magnesium sulfate and potassium carbonate are dissolved
in water.
Strong and Weak Acids and Bases
4.35 Classify each of the following as a strong or weak acid
or base.
a. HF
b. KOH
c. HClO4
d. HIO
4.36 Classify each of the following as a strong or weak acid
or base.
b. HCNO
c. Sr(OH)2
d. HI
a. NH3
Neutralization Reactions
4.37 Complete and balance each of the following molecular
equations (in aqueous solution); include phase labels. Then, for
each, write the net ionic equation.
a. NaOH HNO3
b. HCl Ba(OH)2
c. HC2H3O2 Ca(OH)2
d. NH3 HNO3
4.38 Complete and balance each of the following molecular
equations (in aqueous solution); include phase labels. Then, for
each, write the net ionic equation.
a. Al(OH)3 HCl
b. HBr Sr(OH)2
c. Ba(OH)2 HC2H3O2
d. HNO3 KOH
4.39 For each of the following, write the molecular equation,
including phase labels. Then write the net ionic equation. Note
that the salts formed in these reactions are soluble.
a. the neutralization of hydrobromic acid with calcium hydroxide solution
b. the reaction of solid aluminum hydroxide with nitric acid
168
CHAPTER 4
Chemical Reactions
c. the reaction of aqueous hydrogen cyanide with calcium hydroxide solution
d. the neutralization of lithium hydroxide solution by aqueous
hydrogen cyanide
4.40 For each of the following, write the molecular equation,
including phase labels. Then write the net ionic equation. Note
that the salts formed in these reactions are soluble.
a. the neutralization of lithium hydroxide solution by aqueous
perchloric acid
b. the reaction of barium hydroxide solution and aqueous nitrous acid
c. the reaction of sodium hydroxide solution and aqueous nitrous acid
d. the neutralization of aqueous hydrogen cyanide by aqueous
strontium hydroxide
4.41 Complete the right side of each of the following molecular equations. Then write the net ionic equations. Assume all
salts formed are soluble. Acid salts are possible.
a. 2KOH(aq) H3PO4(aq)
b. 3H2SO4(aq) 2Al(OH)3(s)
c. 2HC2H3O2(aq) Ca(OH)2(aq)
d. H2SO3(aq) NaOH(aq)
4.42 Complete the right side of each of the following molecular equations. Then write the net ionic equations. Assume all
salts formed are soluble. Acid salts are possible.
a. Ca(OH)2(aq) 2H2SO4(aq)
b. 2H3PO4(aq) Ca(OH)2(aq)
c. NaOH(aq) H2SO4(aq)
d. Sr(OH)2(aq) 2H2CO3(aq)
4.43 Write molecular and net ionic equations for the successive neutralizations of each acidic hydrogen of sulfurous acid
by aqueous calcium hydroxide. CaSO3 is insoluble; the acid
salt is soluble.
4.44 Write molecular and net ionic equations for the successive neutralizations of each acidic hydrogen of phosphoric acid
by calcium hydroxide solution. Ca3(PO4)2 is insoluble; assume
that the acid salts are soluble.
Reactions Evolving a Gas
4.45 The following reactions occur in aqueous solution.
Complete and balance the molecular equations using phase labels. Then write the net ionic equations.
a. CaS HBr
b. MgCO3 HNO3
c. K2SO3 H2SO4
4.46 The following reactions occur in aqueous solution.
Complete and balance the molecular equations using phase labels. Then write the net ionic equations.
a. BaCO3 HNO3
b. K2S HCl
c. CaSO3(s) HI
4.47 Write the molecular equation and the net ionic equation
for the reaction of solid iron(II) sulde and hydrochloric acid.
Add phase labels.
4.48 Write the molecular equation and the net ionic equation
for the reaction of solid barium carbonate and hydrogen bromide in aqueous solution. Add phase labels.
Oxidation Numbers
4.49 Obtain the oxidation number for the element noted in
each of the following.
b. Nb in NbO2
a. Ga in Ga2O3
c. Br in KBrO4
d. Mn in K2MnO4
4.50 Obtain the oxidation number for the element noted in
each of the following.
b. Hg in Hg2Cl2
a. Cr in CrO3
c. Ga in Ga(OH)3
d. P in Na3PO4
4.51 Obtain the oxidation number for the element noted in
each of the following.
b. I in IO3
a. N in NH2
c. Al in Al(OH)4
d. Cl in HClO4
4.52 Obtain the oxidation number for the element noted in
each of the following.
b. Cr in CrO42
a. N in NO2
2
c. Zn in Zn(OH)4
d. As in H2AsO3
4.53 Determine the oxidation numbers of all the elements in
each of the following compounds. (Hint: Look at the ions
present.)
b. Fe2(CrO4)3
a. Mn(ClO3)2
c. HgCr2O7
d. Co3(PO4)2
4.54 Determine the oxidation numbers of all the elements in
each of the following compounds. (Hint: Look at the ions
present.)
b. Cr2(SO4)3
a. Hg2(BrO3)2
c. CoSeO4
d. Pb(OH)2
Describing OxidationReduction Reactions
4.55 In the following reactions, label the oxidizing agent and
the reducing agent.
a. P4(s) 5O2(g) P4O10(s)
b. Co(s) Cl2(g) CoCl2(s)
4.56 In the following reactions, label the oxidizing agent and
the reducing agent.
a. ZnO(s) C(s) Zn(g) CO(g)
b. 8Fe(s) S8(s) 8FeS(s)
4.57 In the following reactions, label the oxidizing agent and
the reducing agent.
a. 2Al(s) 3F2(g) 2AlF3(s)
b. Hg2 (aq) NO2 (aq) H2O(l)
Hg(s) 2H (aq) NO3 (aq)
Practice Problems
4.58 In the following reactions, label the oxidizing agent and
the reducing agent.
a. Fe2O3(s) 3CO(g) 2Fe(s) 3CO2(g)
b. PbS(s) 4H2O2(aq) PbSO4(s) 4H2O(l)
Balancing OxidationReduction Reactions
4.59 Balance the following oxidationreduction reactions
by the half-reaction method.
a. CuCl2(aq) Al(s) AlCl3(aq) Cu(s)
b. Cr3 (aq) Zn(s) Cr(s) Zn2 (aq)
4.60 Balance the following oxidationreduction reactions
by the half-reaction method.
a. FeI3(aq) Mg(s) Fe(s) MgI2(aq)
b. H2(g) Ag (aq) Ag(s) H (aq)
Molarity
4.61 A sample of 0.0512 mol of iron(III) chloride, FeCl3,
was dissolved in water to give 25.0 mL of solution. What is the
molarity of the solution?
4.62 A 50.0-mL volume of AgNO3 solution contains 0.0285
mol AgNO3 (silver nitrate). What is the molarity of the solution?
4.63 An aqueous solution is made from 0.798 g of potassium
permanganate, KMnO4. If the volume of solution is 50.0 mL,
what is the molarity of KMnO4 in the solution?
4.64 A sample of oxalic acid, H2C2O4, weighing 1.192 g is
placed in a 100.0-mL volumetric ask, which is then lled to
the mark with water. What is the molarity of the solution?
4.65 What volume of 0.120 M CuSO4 is required to give
0.150 mol of copper(II) sulfate, CuSO4?
4.66 How many milliliters of 0.126 M HClO4 (perchloric
acid) are required to give 0.102 mol HClO4?
4.67 An experiment calls for 0.0353 g of potassium hydroxide, KOH. How many milliliters of 0.0176 M KOH are
required?
4.68 What is the volume (in milliliters) of 0.215 M H2SO4
(sulfuric acid) containing 0.949 g H2SO4?
4.69 Heme, obtained from red blood cells, binds oxygen, O2.
How many moles of heme are there in 150 mL of 0.0019 M
heme solution?
4.70 Insulin is a hormone that controls the use of glucose in
the body. How many moles of insulin are required to make up
28 mL of 0.0048 M insulin solution?
4.71 How many grams of sodium dichromate, Na2Cr2O7,
should be added to a 100.0-mL volumetric ask to prepare 0.025
M Na2Cr2O7 when the ask is lled to the mark with water?
169
4.72 Describe how you would prepare 2.50
102 mL of
0.20 M Na2SO4. What mass (in grams) of sodium sulfate,
Na2SO4, is needed?
4.73 You wish to prepare 0.12 M HNO3 from a stock solution of nitric acid that is 15.8 M. How many milliliters of the
stock solution do you require to make up 1.00 L of 0.12 M
HNO3?
4.74 A chemist wants to prepare 0.50 M HCl. Commercial
hydrochloric acid is 12.4 M. How many milliliters of the commercial acid does the chemist require to make up 1.50 L of the
dilute acid?
4.75 A 3.50 g sample of KCl is dissolved in 10.0 mL of water.
The resulting solution is then added to 60.0 mL of a 0.500 M
CaCl2(aq) solution. Assuming that the volumes are additive, calculate the concentrations of each ion present in the nal solution.
4.76 Calculate the concentrations of each ion present in a
solution that results from mixing 50.0 mL of a 0.20 M
NaClO3(aq) solution with 25.0 mL of a 0.20 M Na2SO4(aq).
Assume that the volumes are additive.
Gravimetric Analysis
4.77 A chemist added an excess of sodium sulfate to a solution of a soluble barium compound to precipitate all of the barium ion as barium sulfate, BaSO4. How many grams of barium
ion are in a 458-mg sample of the barium compound if a solution of the sample gave 513 mg BaSO4 precipitate? What is the
mass percentage of barium in the compound?
4.78 A soluble iodide was dissolved in water. Then an excess
of silver nitrate, AgNO3, was added to precipitate all of the iodide ion as silver iodide, AgI. If 1.545 g of the soluble iodide
gave 2.185 g of silver iodide, how many grams of iodine are in
the sample of soluble iodide? What is the mass percentage of
iodine, I, in the compound?
4.79 Copper has compounds with copper(I) ion or copper(II)
ion. A compound of copper and chlorine was treated with a solution of silver nitrate, AgNO3, to convert the chloride ion in the
compound to a precipitate of AgCl. A 59.40-mg sample of the
copper compound gave 86.00 mg AgCl.
a. Calculate the percentage of chlorine in the copper compound.
b. Decide whether the formula of the compound is CuCl or
CuCl2.
4.80 Gold has compounds containing gold(I) ion or gold(III)
ion. A compound of gold and chlorine was treated with a solution
of silver nitrate, AgNO3, to convert the chloride ion in the compound to a precipitate of AgCl. A 162.7-mg sample of the gold
compound gave 100.3 mg AgCl.
a. Calculate the percentage of the chlorine in the gold compound.
b. Decide whether the formula of the compound is AuCl or
AuCl3.
170
CHAPTER 4
Chemical Reactions
4.81 A compound of iron and chlorine is soluble in water. An
excess of silver nitrate was added to precipitate the chloride ion
as silver chloride. If a 134.8-mg sample of the compound gave
304.8 mg AgCl, what is the formula of the compound?
4.82 A 1.345-g sample of a compound of barium and oxygen
was dissolved in hydrochloric acid to give a solution of barium
ion, which was then precipitated with an excess of potassium
chromate to give 2.012 g of barium chromate, BaCrO4. What is
the formula of the compound?
Volumetric Analysis
H2SO4(aq)
2HNO3(aq)
Na2CO3(aq)
2NaNO3(aq)
H2O(l )
CO2(g)
4.84 A ask contains 49.8 mL of 0.150 M Ca(OH)2 (calcium
hydroxide). How many milliliters of 0.350 M Na2CO3 (sodium
carbonate) are required to react completely with the calcium
hydroxide in the following reaction?
Na2CO3(aq)
Ca(OH)2(aq) CaCO3(s)
2NaOH(aq)
4.85 How many milliliters of 0.150 M H2SO4 (sulfuric acid)
are required to react with 8.20 g of sodium hydrogen carbonate, NaHCO3, according to the following equation?
2H2O(l )
2CO2(g)
4.86 How many milliliters of 0.250 M KMnO4 are needed to
react with 3.36 g of iron(II) sulfate, FeSO4? The reaction is as
follows:
10FeSO4(aq) 2KMnO4(aq) 8H2SO4(aq)
5Fe2(SO4)3(aq) 2MnSO4(aq) K2SO4(aq) 8H2O(l )
4.87 A solution of hydrogen peroxide, H2O2, is titrated with
a solution of potassium permanganate, KMnO4. The reaction is
5H2O2(aq)
4.83 What volume of 0.250 M HNO3 (nitric acid) reacts with
44.8 mL of 0.150 M Na2CO3 (sodium carbonate) in the following reaction?
2NaHCO3(aq)
Na2SO4(aq)
2KMnO4(aq) 3H2SO4(aq)
5O2(g) 2MnSO4(aq) K2SO4(aq)
8H2O(l )
It requires 51.7 mL of 0.145 M KMnO4 to titrate 20.0 g of the
solution of hydrogen peroxide. What is the mass percentage of
H2O2 in the solution?
4.88 A 3.33-g sample of iron ore is transformed to a solution
of iron(II) sulfate, FeSO4, and this solution is titrated with
0.150 M K2Cr2O7 (potassium dichromate). If it requires 43.7
mL of potassium dichromate solution to titrate the iron(II) sulfate solution, what is the percentage of iron in the ore? The reaction is
6FeSO4(aq) K2Cr2O7(aq) 7H2SO4(aq)
3Fe2(SO4)3(aq) Cr2(SO4)3(aq) 7H2O(l ) K2SO4(aq)
General Problems
4.89 Magnesium metal reacts with hydrobromic acid to produce hydrogen gas and a solution of magnesium bromide.
Write the molecular equation for this reaction. Then write the
corresponding net ionic equation.
4.90 Aluminum metal reacts with perchloric acid to produce
hydrogen gas and a solution of aluminum perchlorate. Write
the molecular equation for this reaction. Then write the
corresponding net ionic equation.
4.91 Nickel(II) sulfate solution reacts with lithium hydroxide
solution to produce a precipitate of nickel(II) hydroxide and a
solution of lithium sulfate. Write the molecular equation for this
reaction. Then write the corresponding net ionic equation.
4.92 Potassium sulfate solution reacts with barium bromide
solution to produce a precipitate of barium sulfate and a solution of potassium bromide. Write the molecular equation for
this reaction. Then write the corresponding net ionic equation.
4.93 Decide whether a reaction occurs for each of the following. If it does not, write NR after the arrow. If it does, write the
balanced molecular equation; then write the net ionic equation.
a.
b.
c.
d.
LiOH HCN
Li2CO3 HNO3
LiCl AgNO3
LiCl MgSO4
4.94 Decide whether a reaction occurs for each of the following. If it does not, write NR after the arrow. If it does,
write the balanced molecular equation; then write the net
ionic equation.
a. Al(OH)3 HNO3
b. FeS HClO4
c. CaCl2 NaNO3
d. MgSO4 Ba(NO3)2
4.95 Complete and balance each of the following molecular
equations, including phase labels, if a reaction occurs. Then
write the net ionic equation. If no reaction occurs, write NR after the arrow.
a. Sr(OH)2 HC2H3O2
b. NH4I CsCl
c. NaNO3 CsCl
d. NH4I AgNO3
General Problems
4.96 Complete and balance each of the following molecular
equations, including phase labels, if a reaction occurs. Then
write the net ionic equation. If no reaction occurs, write NR
after the arrow.
a. HClO4 BaCO3
b. H2CO3 Sr(OH)2
c. K3PO4 MgCl2
d. FeSO4 MgCl2
4.97 Describe in words how you would do each of the following preparations. Then give the molecular equation for
each preparation.
a. CuCl2(s) from CuSO4(s)
b. Ca(C2H3O2)2(s) from CaCO3(s)
c. NaNO3(s) from Na2SO3(s)
d. MgCl2(s) from Mg(OH)2(s)
4.98 Describe in words how you would do each of the following preparations. Then give the molecular equation for
each preparation.
a. MgCl2(s) from MgCO3(s)
b. NaNO3(s) from NaCl(s)
c. Al(OH)3(s) from Al(NO3)3(s)
d. HCl(aq) from H2SO4(aq)
4.99 Classify each of the following reactions as a combination reaction, decomposition reaction, displacement reaction,
or combustion reaction.
a. When they are heated, ammonium dichromate crystals,
(NH4)2Cr2O7, decompose to give nitrogen, water vapor, and
solid chromium(III) oxide, Cr2O3.
b. When aqueous ammonium nitrite, NH4NO2, is heated, it
gives nitrogen and water vapor.
c. When gaseous ammonia, NH3, reacts with hydrogen chloride gas, HCl, ne crystals of ammonium chloride, NH4Cl,
are formed.
d. Aluminum added to an aqueous solution of sulfuric acid,
H2SO4, forms a solution of aluminum sulfate, Al2(SO4)3.
Hydrogen gas is released.
4.100 Classify each of the following reactions as a combination reaction, decomposition reaction, displacement reaction,
or combustion reaction.
a. When solid calcium oxide, CaO, is exposed to gaseous sulfur trioxide, SO3, solid calcium sulfate, CaSO4, is formed.
b. Calcium metal (solid) reacts with water to produce a solution of calcium hydroxide, Ca(OH)2, and hydrogen gas.
c. When solid sodium hydrogen sulte, NaHSO3, is heated,
solid sodium sulte, Na2SO3, sulfur dioxide gas, SO2, and
water vapor are formed.
d. Magnesium reacts with bromine to give magnesium bromide, MgBr2.
4.101 Consider the reaction of all pairs of the following
compounds in water solution: Ba(OH)2, Pb(NO3)2, H2SO4,
NaNO3, MgSO4.
171
a. Which pair (or pairs) forms one insoluble compound and
one soluble compound (not water)?
b. Which pair (or pairs) forms two insoluble compounds?
c. Which pair (or pairs) forms one insoluble compound and
water?
4.102 Consider the reaction of all pairs of the following
compounds in water solution: Sr(OH)2, AgNO3, H3PO4,
KNO3, CuSO4.
a. Which pair (or pairs) forms one insoluble compound and
one soluble compound (not water)?
b. Which pair (or pairs) forms two insoluble compounds?
c. Which pair (or pairs) forms one insoluble compound and
water?
4.103 An aqueous solution contains 4.50 g of calcium chloride, CaCl2, per liter. What is the molarity of CaCl2? When calcium chloride dissolves in water, the calcium ions, Ca2 , and
chloride ions, Cl , in the crystal go into the solution. What is
the molarity of each ion in the solution?
4.104 An aqueous solution contains 3.45 g of iron(III) sulfate, Fe2(SO4)3, per liter. What is the molarity of Fe2(SO4)3?
When the compound dissolves in water, the Fe3 ions and
SO42 ions in the crystal go into the solution. What is the molar concentration of each ion in the solution?
4.105 A stock solution of potassium dichromate, K2Cr2O7,
is made by dissolving 89.3 g of the compound in 1.00 L of solution. How many milliliters of this solution are required to
prepare 1.00 L of 0.100 M K2Cr2O7?
4.106 A 71.2-g sample of oxalic acid, H2C2O4, was dissolved in 1.00 L of solution. How would you prepare 1.00 L of
0.150 M H2C2O4 from this solution?
4.107 A solution contains 6.00% (by mass) NaBr (sodium
bromide). The density of the solution is 1.046 g/cm3. What is
the molarity of NaBr?
4.108 An aqueous solution contains 4.00% NH3 (ammonia)
by mass. The density of the aqueous ammonia is 0.979 g/mL.
What is the molarity of NH3 in the solution?
4.109 A barium mineral was dissolved in hydrochloric acid to
give a solution of barium ion. An excess of potassium sulfate was
added to 50.0 mL of the solution, and 1.128 g of barium sulfate
precipitate formed. Assume that the original solution was barium
chloride. What was the molarity of BaCl2 in this solution?
4.110 Bone was dissolved in hydrochloric acid, giving 50.0
mL of solution containing calcium chloride, CaCl2. To precipitate the calcium ion from the resulting solution, an excess of
potassium oxalate was added. The precipitate of calcium oxalate, CaC2O4, weighed 1.437 g. What was the molarity of
CaCl2 in the solution?
4.111 You have a sample of a rat poison whose active ingredient is thallium(I) sulfate. You analyze this sample for the
172
CHAPTER 4
Chemical Reactions
mass percentage of active ingredient by adding potassium iodide to precipitate yellow thallium(I) iodide. If the sample of
rat poison weighed 759.0 mg and you obtained 212.2 mg of the
dry precipitate, what is the mass percentage of the thallium(I)
sulfate in the rat poison?
4.112 An antacid tablet has calcium carbonate as the active
ingredient; other ingredients include a starch binder. You dissolve the tablet in hydrochloric acid and lter off insoluble material. You add potassium oxalate to the ltrate (containing calcium ion) to precipitate calcium oxalate. If a tablet weighing
0.680 g gave 0.6332 g of calcium oxalate, what is the mass
percentage of active ingredient in the tablet?
4.113 A sample of CuSO4 5H2O was heated to 110C,
where it lost water and gave another hydrate of copper(II) ion
that contains 32.50% Cu. A 98.77-mg sample of this new hydrate gave 116.66 mg of barium sulfate precipitate when
treated with a barium nitrate solution. What is the formula of
the new hydrate?
4.114 A sample of CuSO4 5H2O was heated to 100C,
where it lost water and gave another hydrate of copper(II) ion
that contained 29.76% Cu. An 85.42-mg sample of this new
hydrate gave 93.33 mg of barium sulfate precipitate when
treated with a barium nitrate solution. What is the formula of
the new hydrate?
4.115 A water-soluble compound of gold and chlorine is
treated with silver nitrate to convert the chlorine completely to
silver chloride, AgCl. In an experiment, 328 mg of the compound gave 464 mg of silver chloride. Calculate the percentage
of Cl in the compound. What is its empirical formula?
4.116 A solution of scandium chloride was treated with silver nitrate. The chlorine in the scandium compound was con-
verted to silver chloride, AgCl. A 58.9-mg sample of scandium
chloride gave 167.4 mg of silver chloride. What are the mass
percentages of Sc and Cl in scandium chloride? What is its empirical formula?
4.117 A 0.608-g sample of fertilizer contained nitrogen as
ammonium sulfate, (NH4)2SO4. It was analyzed for nitrogen
by heating with sodium hydroxide.
2NaOH(aq)
Na2SO4(aq)
(NH4)2SO4(s)
2H2O(l )
2NH3(g)
The ammonia was collected in 46.3 mL of 0.213 M HCl (hydrochloric acid), with which it reacted.
NH3(g)
HCl(aq) NH4Cl(aq)
This solution was titrated for excess hydrochloric acid with
44.3 mL of 0.128 M NaOH.
NaOH(aq)
HCl(aq) NaCl(aq)
H2O(l )
What is the percentage of nitrogen in the fertilizer?
4.118 An antacid tablet contains sodium hydrogen carbonate,
NaHCO3, and inert ingredients. A 0.500-g sample of powdered
tablet was mixed with 50.0 mL of 0.190 M HCl (hydrochloric
acid). The mixture was allowed to stand until it reacted.
NaHCO3(s)
HCl(aq) NaCl(aq)
H2O(l )
CO2(g)
The excess hydrochloric acid was titrated with 47.1 mL of
0.128 M NaOH (sodium hydroxide).
HCl(aq)
NaOH(aq) NaCl(aq)
H2O(l )
What is the percentage of sodium hydrogen carbonate in the
antacid?
Cumulative-Skills Problems
4.119 Lead(II) nitrate reacts with cesium sulfate in an aqueous precipitation reaction. What are the formulas of lead(II) nitrate and cesium sulfate? Write the molecular equation and net
ionic equation for the reaction. What are the names of the
products? Give the molecular equation for another reaction
that produces the same precipitate.
4.120 Silver nitrate reacts with strontium chloride in an
aqueous precipitation reaction. What are the formulas of silver
nitrate and strontium chloride? Write the molecular equation
and net ionic equation for the reaction. What are the names of
the products? Give the molecular equation for another reaction
that produces the same precipitate.
4.121 Elemental bromine is the source of bromine compounds.
The element is produced from certain brine solutions that occur
naturally. These brines are essentially solutions of calcium bromide that, when treated with chlorine gas, yield bromine in a dis-
placement reaction. What are the molecular equation and net
ionic equation for the reaction? A solution containing 40.0 g of
calcium bromide requires 14.2 g of chlorine to react completely
with it, and 22.2 g of calcium chloride is produced in addition to
whatever bromine is obtained. How many grams of calcium bromide are required to produce 10.0 pounds of bromine?
4.122 Barium carbonate is the source of barium compounds.
It is produced in an aqueous precipitation reaction from barium
sulde and sodium carbonate. (Barium sulde is a soluble compound obtained by heating the mineral barite, which is barium
sulfate, with carbon.) What are the molecular equation and net
ionic equation for the precipitation reaction? A solution containing 33.9 g of barium sulde requires 21.2 g of sodium carbonate
to react completely with it, and 15.6 g of sodium sulde is produced in addition to whatever barium carbonate is obtained.
How many grams of barium sulde are required to produce 10.0
tons of barium carbonate? (One ton equals 2000 pounds.)
173
Cumulative-Skills Problems
4.123 Mercury(II) nitrate is treated with hydrogen sulde,
H2S, forming a precipitate and a solution. Write the molecular
equation and the net ionic equation for the reaction. An acid is
formed; is it strong or weak? Name each of the products. If
81.15 g of mercury(II) nitrate and 8.52 g of hydrogen sulde
are mixed in 550.0 g of water to form 58.16 g of precipitate,
what is the mass of the solution after the reaction?
4.124 Mercury(II) nitrate is treated with hydrogen sulde,
H2S, forming a precipitate and a solution. Write the molecular
equation and the net ionic equation for the reaction. An acid is
formed; is it strong or weak? Name each of the products. If
65.65 g of mercury(II) nitrate and 4.26 g of hydrogen sulde
are mixed in 490.0 g of water to form 54.16 g of precipitate,
what is the mass of the solution after the reaction?
4.125 Iron forms a sulfide with the approximate formula
Fe7S8. Assume that the oxidation state of sulfur is 2, and
that iron atoms exist in both 2 and 3 oxidation states.
What is the ratio of Fe(II) atoms to Fe(III) atoms in this
compound?
4.126 A transition metal X forms an oxide of formula X2O3.
It is found that only 50% of X atoms in this compound are in
the 3 oxidation state. The only other stable oxidation states of
X are 2 and 5. What percentage of X atoms are in the 2
oxidation state in this compound?
4.127 What volume of a solution of ethanol, C2H6O, that is
94.0% ethanol by mass contains 0.200 mol C2H6O? The density of the solution is 0.807 g/mL.
4.128 What volume of a solution of ethylene glycol,
C2H6O2, that is 56.0% ethylene glycol by mass contains 0.350
mol C2H6O2? The density of the solution is 1.072 g/mL.
4.129 A 10.0-mL sample of potassium iodide solution was analyzed by adding an excess of silver nitrate solution to produce
silver iodide crystals, which were ltered from the solution.
KI(aq)
AgNO3(aq) KNO3(aq)
AgI(s)
If 2.183 g of silver iodide was obtained, what was the molarity
of the original KI solution?
4.130 A 25.0-mL sample of sodium sulfate solution was analyzed by adding an excess of barium chloride solution to produce barium sulfate crystals, which were ltered from the
solution.
Na2SO4(aq)
BaCl2(aq) 2NaCl(aq)
BaSO4(s)
If 5.719 g of barium sulfate was obtained, what was the molarity of the original Na2SO4 solution?
4.131 A metal, M, was converted to the sulfate, M2(SO4)3.
Then a solution of the sulfate was treated with barium chloride
to give barium sulfate crystals, which were ltered off.
M2(SO4)3(aq)
3BaCl2(aq) 2MCl3(aq)
3BaSO4(s)
If 1.200 g of the metal gave 6.026 g of barium sulfate, what is
the atomic weight of the metal? What is the metal?
4.132 A metal, M, was converted to the chloride, MCl2. Then
a solution of the chloride was treated with silver nitrate to give
silver chloride crystals, which were ltered from the solution.
2AgNO3(aq) M(NO3)2(aq)
MCl2(aq)
2AgCl(s)
If 2.434 g of the metal gave 7.964 g of silver chloride, what is
the atomic weight of the metal? What is the metal?
4.133 Phosphoric acid is prepared by dissolving phosphorus(V) oxide, P4O10, in water. What is the balanced equation
for this reaction? How many grams of P4O10 are required to
make 1.50 L of aqueous solution containing 5.00% phosphoric
acid by mass? The density of the solution is 1.025 g/mL.
4.134 Iron(III) chloride can be prepared by reacting iron
metal with chlorine. What is the balanced equation for this reaction? How many grams of iron are required to make 2.50 L
of aqueous solution containing 9.00% iron(III) chloride by
mass? The density of the solution is 1.067 g/mL.
4.135 An alloy of aluminum and magnesium was treated with
sodium hydroxide solution, in which only aluminum reacts.
2Al(s)
2NaOH(aq)
6H2O(l )
2NaAl(OH)4(aq)
3H2(g)
If a sample of alloy weighing 1.118 g gave 0.1068 g of hydrogen, what is the percentage of aluminum in the alloy?
4.136 An alloy of iron and carbon was treated with sulfuric
acid, in which only iron reacts.
2Fe(s)
3H2SO4(aq) Fe2(SO4)3(aq)
3H2(g)
If a sample of alloy weighing 2.358 g gave 0.1352 g of hydrogen, what is the percentage of iron in the alloy?
4.137 Determine the volume of sulfuric acid solution needed
to prepare 37.4 g of aluminum sulfate, Al2(SO4)3, by the reaction
2Al(s)
3H2SO4(aq) Al2(SO4)3(aq)
3H2(g)
The sulfuric acid solution, whose density is 1.104 g/mL, contains 15.0% H2SO4 by mass.
4.138 Determine the volume of sodium hydroxide solution
needed to prepare 26.2 g sodium phosphate, Na3PO4, by the reaction
3NaOH(aq)
H3PO4(aq) Na3PO4(aq)
3H2O(l )
The sodium hydroxide solution, whose density is 1.133 g/mL,
contains 12.0% NaOH by mass.
4.139 The active ingredients of an antacid tablet contained
only magnesium hydroxide and aluminum hydroxide. Complete neutralization of a sample of the active ingredients
174
CHAPTER 4
Chemical Reactions
required 48.5 mL of 0.187 M hydrochloric acid. The chloride
salts from this neutralization were obtained by evaporation of
the ltrate from the titration; they weighed 0.4200 g. What was
the percentage by mass of magnesium hydroxide in the active
ingredients of the antacid tablet?
4.140 The active ingredients in an antacid tablet contained
only calcium carbonate and magnesium carbonate. Complete
reaction of a sample of the active ingredients required 41.33
mL of 0.08750 M hydrochloric acid. The chloride salts from
the reaction were obtained by evaporation of the ltrate from
this titration; they weighed 0.1900 g. What was the percentage
by mass of the calcium carbonate in the active ingredients of
the antacid tablet?
Media Activities
Key: The media activities are designed to direct you to
electronic resources that will help you master important concepts or serve as a reference. All resources needed to complete
these activities are available on both the student CD and the
student web site (select Ebbing/Gammon General Chemistry
from www.college.hmco.com/chemistry), with the exception
of the ACE practice tests and the Molecule Library, which are
available on the student web site.
4.141 Open the Precipitation Reactions Understanding Concepts activity and go through the introduction and example.
a. What does the blue ball in the animation represent?
b. What ion is missing from the animation?
c. Consider the reaction between four formula units of
Pb(NO3)2(aq) and three formula units of CaCl2(aq). After
the reaction is complete, how many and which type of ions
remain in solution?
4.142 Open the Exercises in the Precipitation Reactions Understanding Concepts activity and work at least two problems.
a. For each of the problems that you work, write the net ionic
equation.
b. Write the net ionic equation for the reaction of PbNO3(aq)
and CaCl2(aq).
4.143 Open the Dissolution of a Solid in a Liquid Visualization.
a. If the process depicted in the animation were allowed to
continue, describe what you would eventually see.
b. If the crystal were MgCl2 instead of NaCl, how would the
animation compare?
c. If the crystal were PbCl2, how would the animation compare?
4.144 Open the Conductivities of Aqueous Solutions Visualization and play the demonstration illustrating the differences
in conductivity between strong, weak, and nonelectrolytes in
water.
a. Determine which solution was the strongest electrolyte and
explain why this is the case.
b. Based on the conductivity results, would you classify the
amino acid alanine as a strong or weak acid?
4.145 Watch the Zinc and Iodine Visualization to observe a
vigorous oxidationreduction reaction. The product of the reaction is zinc iodide.
a. Write and balance the equation using the half-reaction
method outlined in Section 4.6.
b. Identify the oxidizing agent and reducing agent in the reaction.
4.146 For review, take the ACE practice test for Chapter 4
on the student web site.
174
CHAPTER 4
Chemical Reactions
required 48.5 mL of 0.187 M hydrochloric acid. The chloride
salts from this neutralization were obtained by evaporation of
the ltrate from the titration; they weighed 0.4200 g. What was
the percentage by mass of magnesium hydroxide in the active
ingredients of the antacid tablet?
4.140 The active ingredients in an antacid tablet contained
only calcium carbonate and magnesium carbonate. Complete
reaction of a sample of the active ingredients required 41.33
mL of 0.08750 M hydrochloric acid. The chloride salts from
the reaction were obtained by evaporation of the ltrate from
this titration; they weighed 0.1900 g. What was the percentage
by mass of the calcium carbonate in the active ingredients of
the antacid tablet?
Media Activities
Key: The media activities are designed to direct you to
electronic resources that will help you master important concepts or serve as a reference. All resources needed to complete
these activities are available on both the student CD and the
student web site (select Ebbing/Gammon General Chemistry
from www.college.hmco.com/chemistry), with the exception
of the ACE practice tests and the Molecule Library, which are
available on the student web site.
4.141 Open the Precipitation Reactions Understanding Concepts activity and go through the introduction and example.
a. What does the blue ball in the animation represent?
b. What ion is missing from the animation?
c. Consider the reaction between four formula units of
Pb(NO3)2(aq) and three formula units of CaCl2(aq). After
the reaction is complete, how many and which type of ions
remain in solution?
4.142 Open the Exercises in the Precipitation Reactions Understanding Concepts activity and work at least two problems.
a. For each of the problems that you work, write the net ionic
equation.
b. Write the net ionic equation for the reaction of PbNO3(aq)
and CaCl2(aq).
4.143 Open the Dissolution of a Solid in a Liquid Visualization.
a. If the process depicted in the animation were allowed to
continue, describe what you would eventually see.
b. If the crystal were MgCl2 instead of NaCl, how would the
animation compare?
c. If the crystal were PbCl2, how would the animation compare?
4.144 Open the Conductivities of Aqueous Solutions Visualization and play the demonstration illustrating the differences
in conductivity between strong, weak, and nonelectrolytes in
water.
a. Determine which solution was the strongest electrolyte and
explain why this is the case.
b. Based on the conductivity results, would you classify the
amino acid alanine as a strong or weak acid?
4.145 Watch the Zinc and Iodine Visualization to observe a
vigorous oxidationreduction reaction. The product of the reaction is zinc iodide.
a. Write and balance the equation using the half-reaction
method outlined in Section 4.6.
b. Identify the oxidizing agent and reducing agent in the reaction.
4.146 For review, take the ACE practice test for Chapter 4
on the student web site.