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Chapter 4HIL04-124-167v4 1/13/04 12:09 PM 24

Chemical Reactions in Aqueous Solutions

125

CO N T E N T S4.1 Some Electrical Properties

of Aqueous Solutions4.2 Reactions of Acids and Bases4.3 Reactions that Form

Precipitates4.4 Reactions Involving

Oxidation and Reduction4.5 Applications of Oxidation

and Reduction4.6 Titrations

THREE-QUARTERS OF Earth’s surface is covered with water, but thisnatural water is not chemically pure. It contains dissolved substances.Solutions are formed as liquid water comes into contact with gasesin the atmosphere, materials in Earth’s solid crust, and materials released into the environment by human activities. The water in living organisms isalso in the form of aqueous solutions. In fact, water is such a good solvent for so manyionic and molecular substances that it has been called the universal solvent.

Reactions that take place in aqueous solutions can be grouped into a few basiccategories, all of which have important applications, as we will see in this chapter. Inexploring these reactions, we will continue to stress ideas from previous chapters.Later in the text, we will examine each reaction type in more detail.

4.1 Some Electrical Properties of Aqueous SolutionsThe outward appearance of an aqueous solution does not tell us what is present at themicroscopic level, that is, whether the solute particles are ions, molecules, or a mixtureof the two. Nevertheless, some of the earliest insights into the microscopic nature ofaqueous solutions came through macroscopic observations on the ability of solutionsto conduct electricity. To understand why, let’s first look at some significant early dis-coveries about electricity.

Static electricity, such as that produced by running a comb through your hair, hasbeen recognized since ancient times. By the end of the eighteenth century, two types ofelectric charge—positive and negative—had been identified, and the interactionsbetween positively and negatively charged objects were well understood (Figure 4.1).

At the beginning of the nineteenth century, it was discovered that electricity couldflow through metal wires. We now know that an electric current is a flow of chargedparticles. In solid and liquid metals, those charged particles are electrons. Metals aregood electrical conductors, that is, they transmit electric current with relatively little

> Many of the chemical reactions that we study take place in aqueous solution.Seawater is a solution of ions, principally and but it also contains dissolved

and other gases from the atmosphere.The fluids of living cells, such as our bloodcells, are also solutions of ions and other dissolved substances; their composition isquite close to that of seawater.The three types of reactions that we consider in thischapter all occur in the sea and in our bodies.

CO2

Cl-,Na+

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126 Chapter 4 � Chemical Reactions in Aqueous Solutions

(a) (b) (c) (d)

+ − + + − −

� FIGURE 4.1 Electrostatic forcesThe objects pictured here are plastic-foam peanuts, the type used as packing material. (a) The peanuts carry no electric charge, andtherefore no electrostatic force acts between them. (b) The peanuts are oppositely charged and therefore attract each other. One hasa positive charge; the other has a negative charge. In both (c) and (d), the peanuts carry like charges and therefore repel each other.

+ −

−+

CathodeAnode

+

−−

+++

−−

Electricity source

e− e−

� FIGURE 4.2 Conduction ofelectric current through a solutionThe electricity source directs electronsfrom the electrodes through the wiresfrom the anode to the cathode. In thesolution, cations are attracted to thecathode and anions areattracted to the anode This migra-tion of ions represents the flow of elec-tricity through the solution.

1+ 2.1- 21- 2,

1+ 2

resistance. The electrons in a metal wire can be set in motion in different ways—by anelectric generator in a power plant, for example, or by a chemical reaction in an elec-tric battery.

Molten (liquid) ionic compounds and aqueous solutions of ionic compounds arealso good electrical conductors, but in these cases ions are the charged particles thatflow. Michael Faraday (1791–1867) did much of the early work on the conduction ofelectricity through solutions, the topic of interest in this section. Faraday coined anumber of terms that are still used by chemists today:

• Electrodes are electrical conductors (such as wires or metal plates) partiallyimmersed in a solution and connected to a source of electricity. The anode isthe electrode connected to the positive pole of the source of electricity, and thecathode is the electrode connected to the negative pole.

• An ion is a carrier of electricity through a solution. (Ion is derived from Greek andmeans “wanderer.”) Anions have a negative charge and are attracted to theanode cations carry a positive charge and are attracted to the cathode

Figure 4.2 suggests how electricity passes through a solution. The external source ofelectricity—a battery—withdraws electrons from the positive anode and forces themonto the negative cathode. The ions in solution then migrate to the electrodes—cationsto the negatively charged cathode, anions to the positively charged anode—carryingelectricity through the solution and completing the electric circuit.

Arrhenius’s Theory of Electrolytic DissociationFaraday did not speculate about how the ions that conduct electricity are formed in asolution. Other scientists at the time thought that as it entered a solution via the elec-trodes, the electric current caused solute molecules to break apart, or dissociate, intocations and anions. In his doctoral dissertation in 1884, however, Svante Arrheniuspresented the hypothesis that certain substances, such as NaCl and HCl, dissociate intocations and anions when they dissolve in water. In other words, electricity does notproduce ions in an aqueous solution; rather, the ions that already exist in solutionallow electricity to flow. And, of course, if there are no ions, there is no electric cur-rent. We call a solute that produces enough ions to make a solution an electrical con-ductor an electrolyte. Arrhenius’s ideas are now called the theory of electrolyticdissociation.

Figure 4.3 illustrates a way to demonstrate the relative abilities of solutions toconduct electric current. We place two electrodes made of graphite (the material inpencil “lead”) in the solution to be tested and connect them by wires to a source ofelectric current. We also place an electric lightbulb in the circuit. Electrons can passfreely through the wires and the bulb, but they cannot pass through the solution.Unless the solution contains enough ions to carry electric charge, no electric currentpasses. Suppose that the solutions in the beakers are all 1 M in a solute. No matterwhat the solute is, we will always observe one of the following three possibilities.

1-2.1+21+2;1-2

Dissolution of NaCl in Wateranimation

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4.1 � Some Electrical Properties of Aqueous Solutions 127

*Ordinary tap water is not pure. It contains enough dissolved ionic compounds to conduct electricity to a limitedextent. This is why we must exercise great care when handling electrical equipment near pools of water.

(a)1 M CH3OH

NonelectrolyteSolute consistsof molecules;

no ions

(c)1 M CH3COOH(aq)

Weak electrolyteSolute consists

mostly of molecules;some ions:

CH3COO� H3O�+–

(b)1 M NaCl(aq)

Strong electrolyteSolute consists of ions:

Na� Cl�+

+

+

++

+

+

––

––

–

–

–

� FIGURE 4.3 Electrolytic properties of aqueous solutionsIn order for there to be an electric current through each solution, cations and anions must be present and must be free to migratebetween the two graphite electrodes. In the microscopic view of each solution, solvent molecules are not shown because we want tofocus on the solutes; the blue background suggests the presence of water as the solvent. In (a), there are no ions. In (b), all the soluteparticles are ions: and In (c), about one in every 200 molecules ionizes, producing an acetate ion, and a hydro-gen ion, which attaches itself to a water molecule, forming H3O+.

CH3COO-,Cl-.Na+

Figure 4.3a: The bulb does not light. This observation signifies that there areessentially no ions in the solution, certainly not nearly enough to carry a significantamount of charge through the solution. This is what we see when the liquid in thebeaker is either pure water* or an aqueous solution of a molecular substance that doesnot ionize, such as ethanol, or sucrose,

A nonelectrolyte is a solute that is present in solution almost exclusively as mole-cules. A solution of a nonelectrolyte is a nonconductor of electricity.

Figure 4.3b: The bulb lights brightly. This indicates that a large number of ions arepresent in the solution. As in NaCl(s), there are no NaCl molecules in NaCl(aq), butonly separate and ions. We say that NaCl(aq) is a strong electrolyte.

A strong electrolyte is a solute that is present in solution almost exclusively asions. A solution of a strong electrolyte is a good electrical conductor.

Cl-Na+

C12H22O11.CH3CH2OH,

Electrolytes and Nonelectrolytesanimation

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Later in the text, we will find thesebracket symbols to be indispensablefor representing molar concentrationsin algebraic equations.

Problem-Solving NoteFrom this point on in the text, we willnot routinely show the cancellation ofunits in calculations, although youshould continue to use them as an aidin problem-solving. However, we willretain cancel marks for emphasis incertain equations and derivations.

*Dissociation is complete in very dilute solutions of ionic substances. In more concentrated solutions this willlikely not be the case. For our purposes in this text, we will generally assume that dissociation is complete.

Hydrogen chloride is also a strong electrolyte in an aqueous solution. In this case,however, HCl is a molecular substance in the pure state. It forms and ionswhen dissolved in water.

Figure 4.3c: The bulb lights but only dimly. This is what we expect if a soluteexists partly in molecular form and partly in ionic form in solution. Acetic acid,

behaves this way in aqueous solution.

A weak electrolyte is a solute that is only partially ionized in solution. A solutionof a weak electrolyte is a poor conductor of electricity.

Note that the terms “strong” and “weak” refer to the extent to which an electrolyteproduces ions in solution. If an electrolyte exists almost exclusively as ions in solu-tion, it is a strong electrolyte; if much of it remains in molecular form, it is a weakelectrolyte. Thus, even though an extremely dilute solution of sodium chloride mightnot conduct electricity as well as a somewhat more concentrated solution of aceticacid, NaCl is still a strong electrolyte and is still a weak electrolyte. Thefollowing generalizations classify water-soluble solutes according to their electrolyticproperties:

• Essentially all soluble ionic compounds are strong electrolytes.

• Only a few molecular compounds (mainly a few acids, such as HCl) are strongelectrolytes.

• Most molecular compounds are either nonelectrolytes or weak electrolytes.

• Most organic compounds are molecular and nonelectrolytes; carboxylic acids anda class of organic compounds called amines (page 132) are weak electrolytes.

We can use these generalizations to decide how best to represent the solute in anaqueous solution. Because a solute that is a nonelectrolyte exists only in molecularform, we use its molecular formula. For example, we represent an aqueous solution ofethanol as When we dissolve a water-soluble ionic compound—astrong electrolyte—the cations and anions dissociate completely from one another andappear in solution as independent solute particles*. We can represent the dissociationof NaCl as

The above the arrow signifies that the dissociation occurs when NaCl(s) dis-solves in water, but is not a reactant in the usual sense. In many cases, we cancontinue to represent aqueous solutions of ionic compounds as we did in Chapter 3,that is, without indicating their dissociation, as in NaCl(aq). However, in other situa-tions the ionic form— —gives a better representation of the matterat hand, as we will see next.

Calculating Ion Concentrations in SolutionAt times, we may need to determine the concentrations of the individual ions in anaqueous solution of some ionic compound. Thus, because one mole each of ionsand ions appear for every mole of NaCl(s) dissolved, the ion concentrations in a0.010 M NaCl(aq) solution are 0.010 M and 0.010 M A commonconvention for representing molar concentrations uses a set of brackets enclosing thesolute symbol: and

Because the dissociation of produces two ions and oneion per formula unit, the ion concentrations in a solution are

per liter and 0.010 mol per liter. That is,and Example 4.1 illustrates an additional idea about ion concen-trations in solution: There is only one concentration for any given ion in a solution,even if the ion has more than one source. Additionally, the total ion concentration in asolution is the sum of the molarities of the individual ions.

[SO4

2-] = 0.010 M.[Na+] = 0.020 MSO4

2-Na+2 * 0.010 mol0.010 M Na2SO4

SO4

2-Na+Na2SO4

[Cl-] = 0.010 M Cl-(aq).Na+(aq)[Na+] = 0.010 M

Cl-(aq).Na+(aq)Cl-

Na+

Na+(aq) + Cl-(aq)

H2OH2O

NaCl(s) H2O99: Na+(aq) + Cl-(aq)

CH3CH2OH(aq).

CH3COOH

CH3COOH,

Cl-H+

Emphasis: When a strong elec-trolyte dissolves in water, the

solute species are ions. A solution ofsodium chloride could be said to containno “sodium chloride.” Rather, it containssodium ions and chloride ions. Studentsoften have a great deal of difficultyabsorbing and using this concept.

Strong and Weak Electrolytesmovie

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4.2 � Reactions of Acids and Bases 129

Example 4.1Calculate the molarity of each ion in an aqueous solution that is 0.00384 M and 0.00202 M NaCl. In addition, calculate thetotal ion concentration of the solution.

STRATEGY

First, note that there are three types of ions in the solution— and Of these three types, comes only from thedissolved comes only from the NaCl, and comes from both solutes. Thus, we can establish just fromthe molarity of and just from the molarity of NaCl(aq). For we will have to work with both molarities.Finally, we sum the molarities of the individual ions to obtain the total ion concentration.

SOLUTION

[Na+],[Cl-]Na2SO4(aq)[SO4

2-]Na+Cl-Na2SO4 ,SO4

2-Cl-.SO4

2-,Na+,

Na2SO4

The dissociation of the ionic solutes in the solution is repre-sented by the equations describing the dissolution of the two salts. Salt Cation Anion

Salt Cation Anion

NaCl(s) H2O99: Na+(aq) + Cl-(aq)

Na2SO4(s) H2O99: 2 Na+(aq) + SO4

2-(aq)

Because one mole of produces one mole ofthe molarity of the is the same as that

given for Na2SO4(aq).SO4

2-SO4

2-(aq),Na2SO4(s)

[SO4

2-] =0.00384 mol Na2SO4

1 L*

1 mol SO4

2-

1 mol Na2SO4= 0.00384 M

The situation is similar for The molarity of is the same as that of the NaCl(aq).

Cl-(aq)Cl-(aq).[Cl-] =

0.00202 mol NaCl

1 L*

1 mol Cl-

1 mol NaCl= 0.00202 M

The molarity of the is the sum of the two terms in theright column, one for each source of Na+.

Na+

= 0.00768 M + 0.00202 M = 0.00970 M

+ B0.00202 mol NaCl

1 L*

1 mol Na+

1 mol NaClR

[Na+] = B 0.00384 mol Na2SO4

1 L*

2 mol Na+

1 mol Na2SO4R

The sum of the individual ion molarities gives the total ionconcentration. = 0.00970 M + 0.00202 M + 0.00384 M = 0.01556 M

Total ion concentration = [Na+] + [Cl-] + [SO4

2-]

ASSESSMENT

As you become more familiar with calculating ion concentrations, you should be able to work a problem of this sort just by inspectingthe formulas of the solutes, thereby arriving directly at the conclusions; and

0.00970 M.

EXERCISE 4.1A

Seawater is essentially 0.438 M NaCl and 0.0512 M together with several other minor solutes. What are the molarities ofand in seawater?

EXERCISE 4.1B

Each year, oral rehydration therapy (ORT)—feeding a person an electrolyte solution—saves the lives of a million children worldwidewho become dehydrated from diarrhea. Each liter of the electrolyte solution contains 3.5 g sodium chloride, 1.5 g potassium chloride,2.9 g sodium citrate and 20.0 g glucose Calculate the molarity of each species present in the solution.(Hint: Sodium citrate is a strong electrolyte, and glucose is a nonelectrolyte.)

(C6H12O6).(Na3C6H5O7),

Cl-Mg2+,Na+,MgCl2 ,

=12 * 0.003842 + 0.00202=[Na+][Cl-] = 0.00202 M,[SO4

2-] = 0.00384 M,

4.2 Reactions of Acids and BasesWhen we first encountered Arrhenius’s theory of acids and bases in Chapter 2, wefocused on names and formulas. In Chapter 3, we used acids and/or bases as reac-tants in a few of the reactions. In this section, we will look at acid–base reactions inmore detail.

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� A space-filling model of sulfuricacid, a strong acid in its first ionizationand a weak acid in its second.


� A ball-and-stick model of aceticacid, a weak acid.CH3COOH,

*The simple hydrogen ion, does not exist in aqueous solutions. Instead, it is associated with several molecules; that is, it is present as where n is an integer. The most important of these hydrated hydro-gen ions has Its formula is and it is called the hydronium ion. We will use the abbreviation hereand switch to in Chapter 15.H3O+

H+H3O+,n = 1.H(H2O)n

+,H2OH+,

Strong and Weak AcidsWe will adopt a more general acid–base theory in Chapter 15. However, for now wewill continue to use the Arrhenius view that an acid is a substance that produces hydro-gen ions * in aqueous solution, and we will expand the definition of an acid in away that explains the ability of aqueous acid solutions to conduct electricity.

Acids that are completely ionized in water produce aqueous solutions that aregood electrical conductors. These acids are strong electrolytes and are called strongacids. As just one example, the equation representing the ionization of hydrogen chlo-ride in water to produce hydrochloric acid, a strong acid, is

We can therefore represent an aqueous solution of hydrochloric acid either as HCl(aq)or as and we can calculate the concentrations of cations and anionsas in Example 4.1. Specifically, the ionization equation tells us that 1 mol HCl(g) dis-sociates to yield 1 mol and 1 mol Thus, in 0.0010 M HCl,

Because essentially all the HCl molecules have ionized, we take the concentration ofHCl molecules to be zero:

The vast majority of acids are weak electrolytes; that is, only some of their mole-cules ionize in aqueous solution. The rest remain as intact molecules. These acids arecalled weak acids. The carboxylic acids (Section 2.9) containing one to four carbonatoms are soluble in water and are weak acids. In 1.0 M for example, onlyabout 0.5% of the molecules ionize. Because most weak acids remain largely in mole-cular form, we use a molecular formula, such as to represent them.We can write the following equation to represent the limited ionization of acetic acid:

Ionization:

Acetic acid Acetate ion

Taken by itself, this equation is misleading because it falsely implies that the ioniza-tion goes to completion. Ionization does not go to completion, however. The ions areable to recombine to form neutral molecules in a reverse reaction:

Recombination of ions:

The best way to represent the limited ionization of a weak acid is to write a singleequation with a double arrow

A double arrow in a chemical equation signifies that the reaction is reversible: Aforward reaction and a reverse reaction occur simultaneously. Instead of going to com-pletion, a reversible reaction reaches a state of equilibrium in which the concentrationsof reactants and products remain constant over time. We will discuss reversible reac-tions and the nature of equilibrium in more detail in Chapter 14, but for now, we havesimply noted the significance of the double arrow.

Some acids can produce two or more hydrogen ions per molecule of the acid. Sul-furic acid, and phosphoric acid, are two common examples. Sulfuricacid is interesting in that it is a strong acid in its first ionization and a weak acid in itssecond:

First ionization:

Second ionization: HSO4

-(aq) ∆ H+(aq) + SO4

2-(aq)

H2SO4(aq) ¡ H+(aq) + HSO4

-(aq)

H3PO4,H2SO4,

CH3COOH(aq) ∆ H+(aq) + CH3COO-(aq)

(∆ ):

H+(aq) + CH3COO-(aq) ¡ CH3COOH(aq)

CH3COOH(aq) ¡ H+(aq) + CH3COO-(aq)

CH3COOH(aq),

CH3COOH,

[HCl] = 0

[H+] = 0.0010 M and [Cl-] = 0.0010 M

Cl-(aq).H+(aq)

H+(aq) + Cl-(aq),

HCl(g) H2O99: H+(aq) + Cl-(aq)

(H+)

Introduction to Acids animation

Acetic Acid 3D model

Emphasis: It generally isn’t neces-sary to memorize a list of weak

acids. Remember the six strong acids(Table 4.1). Almost any other acidencountered will be weak.

Sulfuric Acid 3D model

The phrase “strong acid”oftenbrings to mind words such as

“corrosive.” However, a strong acid is sim-ply an acid that dissociates completely inwater. Some weak acids (notably HF) aremore corrosive and more dangerous touse than some strong acids.

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*In Arrhenius’s time, chemists generally believed that a substance must contain OH groups in order to be a base.Thus, was thought to be (ammonium hydroxide). Even though there is no compelling evidencefor the existence of molecules, this formula is still often seen as a representation of NH3(aq).NH4OH

NH4OHNH3(aq)

Phosphoric acid, on the other hand, is a weak acid in each of its three ionization steps.We will discuss sulfuric acid and phosphoric acid further in Chapter 15.

Can we look at a chemical formula and tell whether a substance is an acid? If it is,can we tell whether it is a strong or weak acid? We will tackle questions such as thesein our in-depth treatment of acids in Chapter 15, but even at this point we can get somepartial answers by applying the following ideas.

1. The molecular formula of an acid is generally written with ionizable H atomsappearing first. Thus, and are acids containing one, two,and three ionizable H atoms, respectively. Methane, has four H atoms, butthey are not ionizable, and therefore is not an acid. From its name, we knowthat acetic acid is an acid; and if we see its formula written as we knowthat it has one H atom that is ionizable and three H atoms that are not.

2. Structural and condensed structural formulas show where H atoms are found in amolecule; their locations usually help identify ionizable H atoms. The condensedstructural formula for acetic acid, indicates that three of the H atomsare bonded to the C atom. Just like the four H atoms in the three in the group are not ionizable. Only the H atom bonded to one of the O atoms in the car-boxylic acid group, is ionizable.

The simplest way to tell whether an acid is strong or weak is to note that therearen’t many strong acids; the most common ones are those in Table 4.1. Unless youhave information to the contrary, you can assume that any other acid is a weak acid.

Strong and Weak BasesBy the Arrhenius definition, a base is a substance that produces hydroxide ions,in aqueous solution. Moreover, because ionic compounds are strong electrolytes, ionichydroxides, such as NaOH, are strong bases.

Some compounds produce ions by reacting with water, not just by dissolv-ing in it. These substances are also bases. Gaseous ammonia dissolves in water, andthe formed reacts with water to produce some ions:

As in the case of acetic acid, most of the molecules remain nonionized at equi-librium, and ammonia is therefore a weak base.* Most molecular substances that actas bases are weak bases.

NH3

NH3(aq) + H2O(l) ∆ NH4

+(aq) + OH-(aq)

NH3(aq)

OH-

NaOH(s) H2O99: Na+(aq) + OH-(aq)

OH-,

¬COOH,

CH3CH4,CH3COOH,

HC2H3O2,CH4

CH4,H3PO4H2SO4,HNO3,

Table 4.1 Common Strong Acids and Strong Bases

Acids Bases

Binary Hydrogen Group 1A Group 2A Compounds Oxoacids hydroxides hydroxides

HCl LiOHHBr a NaOHHI KOH

RbOHCsOH

a is a strong acid in its first ionization step but weak in its second ionization step.H2SO4

Ba(OH)2

Sr(OH)2HClO4

Ca(OH)2H2SO4

Mg(OH)2HNO3

Students should be coached toremember that a COOH group

indicates a carboxylic acid.¬

Emphasis: This reaction ofammonia with water is quite

simple. A hydrogen ion is donated bythe water molecule and accepted by theammonia molecule.

Introduction to Bases movie

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Application NotePhenol red is familiar to people whomaintain a home swimming pool. It isthe indicator commonly used toensure that the pool water is keptabout neutral, that is, neither acidic nor basic.

*The complete-formula equation is often called a molecular equation, but this term is misleading because manyof the chemical formulas written in such an equation, for example, NaCl(aq), represent not molecules but formulaunits of ionic compounds.

The organic substances known as amines are molecular compounds in which oneor more of the H atoms in are replaced by a hydrocarbon group. In these twoamines, one of the H atoms has been replaced:

or or

Methylamine Ethylamine

The replacement of two and three H atoms, respectively, is seen in dimethylamineand trimethylamine

Amines containing one to four carbon atoms are water soluble, and like theyare weak bases. For example,

Methylamine Methylammonium ion

As with acids, we can often determine whether a substance is a base from its for-mula. For example, if the formula indicates an ionic compound containing metal ionsand ions, we expect it to be a strong base. NaOH and KOH are strong bases, andTable 4.1 lists several others. In contrast, methanol, is not an Arrhenius base.There are only nonmetals in the compound—no metals—and therefore is amolecular compound. The OH group is covalently bonded to the C atom, not presentas Similarly, acetic acid is not a base because the OH group in it isbonded to the C atom. Instead, acetic acid produces and is therefore an acid. Toidentify a weak base, you usually need to see a chemical equation for the ionizationreaction. However, you can identify many weak bases by using these facts:

There are only a few common strong bases (Table 4.1), and the most commonweak bases are ammonia and the amines.

Acid–Base Reactions: NeutralizationIn any reaction between an aqueous acid and base, the identifying characteristics ofthe acid and base (Section 2.8) cancel out, or neutralize, each other. As noted inChapter 2, this process is called neutralization, and one of the products of the processis a salt. Generally, we don’t see anything happen in a neutralization reaction at themacroscopic level because both the acid and base are usually colorless, as is the saltsolution formed from them. How then do we know when a solution is neutral, that is,neither acidic nor basic? As we mentioned in Section 2.8, one characteristic of acidsand bases is their ability to affect the color of litmus, a natural dye. A substance whosecolor is affected by the relative amounts of and ions in solution is called anacid–base indicator. For example, as shown in Figure 4.4, the indicator phenol red isyellow in acidic solutions, orange in neutral solutions, and red in basic solutions.

If we use conventional formulas for the acid and base, we can write what wemight call a complete-formula equation* for a neutralization reaction:

Acid Base Salt Water

However, this complete-formula equation is not always the best way to show whathappens in the neutralization. To show that the HCl(aq), a strong acid; NaOH(aq), astrong base; and NaCl(aq), a salt, exist in solution as ions, we can write the equation inionic form:

Acid Base Salt Water

('''')''''*('''')''''*(''')'''*

H2O(l)+Na+(aq) + Cl-(aq)¡Na+(aq) + OH-(aq)+H+(aq) + Cl-(aq)

HCl(aq) + NaOH(aq) ¡ NaCl(aq) + H2O(l)

OH-H+

H+(CH3COOH)OH-.

CH3OHCH3OH,

OH-

CH3NH2(aq) + H2O(l) ∆ CH3NH3

+(aq) + OH-(aq)

NH3

[(CH3)3N].[(CH3)2NH]

CH3CH2NH2

H¬

Hƒ

Cƒ

H

¬

Hƒ

Cƒ

H

¬

Hƒ

N

¬H

CH3NH2

H¬

Hƒ

Cƒ

H

¬

Hƒ

N

¬H

NH3

� A ball-and-stick model ofmethylamine, a weak base.CH3NH2 ,

Emphasis: Point out the similaritybetween the reaction of methy-

lamine with water and the reaction ofammonia with water.

Methylamine 3D model

Neutralization Reactions activity

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(a) (b) (c)

> FIGURE 4.4 Phenol red—anacid–base indicatorPhenol red is (a) yellow in an acidic solu-tion (0.10 M HCl), (b) orange in a neutralsolution (0.10 M NaCl), and (c) red in abasic solution (0.10 M NaOH).

QUESTION: What would happen to thecolor if ammonia were added to thebeaker on the right?

Moreover, we can eliminate spectator ions—those that just “look on” and appear inthe same form on the two sides of the ionic equation. Eliminating these ions reducesthe equation to a simple form called a net ionic equation. In the equation we justwrote, and are spectator ions, and the net ionic equation is

The net ionic equation conveys the essence of this neutralization: and ionscombine to form water. In general, net ionic equations represent the gist of a reaction. Ifthe spectator ions in a neutralization reaction are those of a soluble salt, they remain insolution. Evaporation of the water leaves the pure salt, such as NaCl (table salt).

Example 4.2Barium nitrate, used to produce a green color in fireworks, can be made by the reaction ofnitric acid with barium hydroxide. Write (a) a complete-formula equation, (b) an ionic equa-tion, and (c) a net ionic equation for this neutralization reaction.

SOLUTION

(a) Write chemical formulas for the substances involved in the reaction, and then balancethe equation:

Nitric acid Barium hydroxide Barium nitrate Water

(not balanced)

(balanced)

(b) Now, represent the strong electrolytes by the formulas of their ions and the nonelec-trolyte water by its molecular formula:

A strong acid A strong base A salt Water

(c) Cancel the spectator ions ( and ) in the ionic equation:

This gives the equation

or, more simply, the net ionic equation.

EXERCISE 4.2A

Calcium hydroxide is used to neutralize a waste stream of hydrochloric acid. Write (a) acomplete-formula equation, (b) an ionic equation, and (c) a net ionic equation for this neu-tralization reaction.

EXERCISE 4.2B

Write the (a) complete-formula equation, (b) ionic equation, and (c) net ionic equation forthe following neutralization reaction.

KHSO4(aq) + NaOH(aq) ¡ ?

H+(aq) + OH-(aq) ¡ H2O(l)

2 H+(aq) + 2 OH-(aq) ¡ 2 H2O(l)

¡ Ba2+(aq) + 2 NO3

-(aq) + 2 H2O(l)2 H+(aq) + 2 NO3

-(aq) + Ba2+(aq) + 2 OH-(aq)

NO3

-Ba2+

2 H2O(l)+Ba2+(aq) + 2 NO3

-(aq)¡Ba2+(aq) + 2 OH-(aq)+2 H+(aq) + 2 NO3

-(aq)

$'%'&$''''%''''&$''''%''''&$'''''%'''''&

2 HNO3(aq) + Ba(OH)2(aq) ¡ Ba(NO3)2(aq) + 2 H2O(l)

HNO3(aq) + Ba(OH)2(aq) ¡ Ba(NO3)2(aq) + H2O(l)

OH-H+H+(aq) + OH-(aq) ¡ H2O(l)

Cl-Na+ Emphasis: A net ionic equation isa general equation, and allows us

to make predictions. The equation at theleft tells us that any solution containingH will react with any solution contain-ing OH to form water.-

+

A common student error is to dis-sociate every formula in the

equation. Remind the student to firstidentify the strong electrolytes in theequation, and to dissociate only thosestrong electrolytes.

Another common student error isto assume that all acid-base reac-

tions have the same net ionic equation.Exercise 4.2B is a good way to demon-strate otherwise.

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Acid–Base Reactions: Additional ExamplesWhile some acid–base reactions can be described by the simple net ionic equationinvolving and other acid–base reactions have different netionic equations. For example, magnesium hydroxide, though only slightly soluble inwater, is a strong base because what little does dissolve is completely dissociated intoions. In the net ionic equation that shows how a slurry of in water neu-tralizes excess stomach acid, we should use the formula of solid magnesium hydroxideon the left. On the right side, hydroxide ion from has combined with toform leaving magnesium ion as

This equation helps us understand why magnesium hydroxide, which is only veryslightly soluble in pure water, dissolves readily in an acidic solution. Initially,combines with the relatively few ions present in the slurry, forming

However, the removed ions are immediately replaced by new ions,released as more dissolves. The replacement ions suffer the samefate as the original—combination with and removal as Rather quickly, allthe in the acid solution dissolves.

The net ionic equation for the neutralization of HCl(aq) with

which shows that from the acid combines directly with molecules, looksrather different from that of a strong acid and strong base,

and for two reasons:

• is a weak base, and we should therefore write its complete formula.

• The molecule contains no which means it cannot be satisfactorily de-scribed as an Arrhenius base.

In the same way that can combine with to form and with to formit can combine with certain other anions to produce either weak electrolytes or

nonelectrolytes. In a broad sense, these reactions are also acid–base reactions.In baking, carbon dioxide gas causes dough to rise. The process involves a reac-

tion between baking soda, and some acidic ingredient in the dough(Figure 4.5). Specifically, from a weak acid (let’s call it HA) reacts with the hydro-gen carbonate ions from the baking soda to form the weak acid carbonic acid,Carbonic acid is unstable and decomposes to and The equations forthese two reactions and the net ionic equation are

The reaction of carbonate ion with an acid also produces carbonic acidthat decomposes to and In similar reactions, sulfite ion andhydrogen sulfite ion react with an acid to produce unstable sulfurous acid

which decomposes to and Net ionic equations for a few gas-forming reactions are summarized in Table 4.2.

SO2(g).H2O(l)(H2SO3),(HSO3

-)(SO3

2-)CO2(g).H2O(l)(CO3

2-)

HA(aq) + HCO3

-(aq) ¡ H2CO3(aq ) ¡

Net: HA(aq) + HCO3

-(aq) ¡

H2CO3(aq) + A-(aq)

H2O(l) + CO2(g)

A-(aq) + H2O(l) + CO2(g)

CO2(g).H2O(l)H2CO3.

H+NaHCO3,

NH4

+,NH3H2OOH-H+

OH-,NH3

NH3

H+(aq) + OH-(aq) ¡ H2O(l)

NH3H+

H+(aq) + NH3(aq) ¡ NH4

+(aq)

NH3(aq),Mg(OH)2(s)

H2O(l).H+OH-Mg(OH)2(s)

OH-OH-H2O(l).Mg(OH)2OH-

H+(aq)

Mg(OH)2(s) + 2 H+(aq) ¡ Mg2+(aq) + 2 H2O(l)

Mg2+(aq):H2O(l),H+Mg(OH)2

Mg(OH)2(s)

H2O(l),OH-(aq),H+(aq),

� Water-insoluble hydroxides such as(milk of magnesia) and

are used as antacids. Usingionic hydroxides that are soluble inwater would be quite dangerous. Inhigh concentrations, is highlybasic and causes severe tissueburning and scarring.

OH-(aq)

Al(OH)3(s)Mg(OH)2(s)

� FIGURE 4.5 The leaveningaction of baking sodaWhen acidified, here with citric acid

from a lemon, baking sodareacts to produce carbonic

acid which decomposes to car-bon dioxide and water. The carbon diox-ide gas produces a lift in the dough being baked.

(H2CO3),(NaHCO3)(H3C6H5O7)

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Table 4.2 Some Common Gas-Forming Acid–Base Reactions

Anion Reaction with

S2- + 2 H+ ¡ H2S(g)S2-HS- + H+ ¡ H2S(g)HS-SO3

2- + 2 H+ ¡ SO2(g) + H2O(l)SO3

2-HSO3

- + H+ ¡ SO2(g) + H2O(l)HSO3

-CO3

2- + 2 H+ ¡ CO2(g) + H2O(l)CO3

2-HCO3

- + H+ ¡ CO2(g) + H2O(l)HCO3

-

H�

Example 4.3 A Conceptual ExampleExplain the observations illustrated in Figure 4.6.

ANALYSIS AND CONCLUSIONS

The bulb in Figure 4.6(a) is only dimly lit because acetic acid is a weak acid and therefore aweak electrolyte [recall Figure 4.3(c)]. The situation in (b) is similar because ammonia isa weak base and therefore also ionizes only slightly. When the two solutions are mixed,which is what has been done in (c), the ions from the readily combine with

molecules to form ions:

By removing ions from the solution, this reaction causes more mole-cules to ionize, producing more to react with more and so on. Soon all the

molecules ionize, and the neutralization goes to completion:

Weak acid Weak base Salt

The original weak acid and weak base are replaced by an aqueous solution of a salt—anionic compound and strong electrolyte. The solution is now a good electrical conductor, asseen in Figure 4.6(c).

EXERCISE 4.3A

In a situation similar to that in Figure 4.6, describe the observations you would expect tomake if the original solutions were and

EXERCISE 4.3B

Describe the observations you would expect in a situation similar to that in Figure 4.6 if youstart with a slurry of in water (milk of magnesia) and add vinegar [essentially1 M ] until all the dissolves and some vinegar remains inexcess.

Mg(OH)2(s)CH3COOH(aq)Mg(OH)2(s)

HNO3(aq).CH3NH2(aq)

('''''')''''''*

NH4

+(aq) + CH3COO-(aq)¡NH3(aq)+CH3COOH(aq)

CH3COOHNH3 ,H+

CH3COOHH+

H+(aq) + NH3(aq) ¡ NH4

+(aq)

NH4

+NH3

CH3COOHH+

(a) (b) (c)

� FIGURE 4.6 Change in electrical conductivity as a result of a chemical reaction(a) When the beaker contains a 1 M solution of acetic acid, the bulb in the electric circuit glows only very dimly.(b) When the beaker contains a 1 M solution of ammonia, the bulb again glows only dimly. (c) When the two solutions are inthe same beaker, the bulb glows brightly. What happens when the two solutions are mixed is described in Example 4.3.

NH3,CH3COOH,

There are many common sub-stances containing carbonates and

bicarbonates that can be shown to reactwith acids, even with weak acids. Black-board chalk, antacids containing calciumcarbonate, marble, eggshell, baking soda,and washing soda are some examples.

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4.3 Reactions that Form PrecipitatesEarlier in the chapter, we noted that many ionic compounds dissolve in water, but thereis a limit to how much will dissolve in a given quantity of water. For NaCl(aq) at25 °C, the limit is a concentration of about 5.47 M NaCl. Because the maximum soluteconcentration for NaCl is relatively high, we say that NaCl(s) is readily soluble inwater. If the maximum concentration of solute is less than about 0.01 M, we generallyrefer to this sparingly soluble solute as insoluble in water.

Combinations of certain cations and anions, then, yield ionic compounds that areinsoluble in water. If these ions are taken from separate sources and brought togetherin an aqueous solution, the insoluble ionic compound, a precipitate, comes out of thesolution and settles to the bottom of the container as a solid. We say that the insolublecompound precipitates from the solution, and a chemical reaction between ions thatproduces a precipitate is called a precipitation reaction. Figure 4.7 shows that a pre-cipitate of insoluble silver iodide forms when solutions of the soluble compounds sil-ver nitrate and potassium iodide are mixed.

Predicting Precipitation ReactionsLet’s now consider how we might have predicted that a precipitate of silver iodideshould form in the reaction illustrated in Figure 4.7. When asked to predict a chemicalreaction, you will have information about the reactants—the left side of an equation—and will need to provide information about the products—the right side of the equa-tion. You will also need to recognize when no reaction occurs, noted by writing “noreaction” on the right side of the equation.

In predicting precipitation reactions, it generally helps to write the equation in itsionic form, as in this equation for the addition of to KI(aq):

The only simple compounds that could form by combinations of these ions that aredifferent from the reactants are and AgI. A precipitation reaction will occuronly if a potential product is insoluble.

In short, to make predictions, we need to know which ionic compounds are solu-ble in water and which are not. We can look up solubility data in handbooks, but thatmay not be necessary. Memorizing solubilities is easier than it might first seembecause a lot of data for common ionic compounds can be summarized in a fewsolubility guidelines, which are listed in Table 4.3.

The guidelines indicate that all nitrates are soluble in water, and so we concludethat is soluble. Similarly, Table 4.3 tells us that all chlorides, bromides, andiodides are soluble except those of and and thus we conclude thatAgI is insoluble. We can complete the equation by showing the AgI precipitate as asolid and the as dissociated into ions.

Ag+(aq) + NO3

-(aq) + K+(aq) + I-(aq) ¡ AgI(s) + K+(aq) + NO3

-(aq)

KNO3

Hg2

2+,Ag+,Pb2+,KNO3

KNO3

Ag+(aq) + NO3

-(aq) + K+(aq) + I-(aq) ¡ ?

AgNO3(aq)

This solubility guideline is only qualita-tive and somewhat arbitrary.We dis-cuss solubility more quantitatively inChapter 16.

+

++

+

++

+ +

+

+

+

__

__

__

_

_ +_ +_++ ++_

_

_ _+ +_ _

_

� FIGURE 4.7 The precipitation of silver iodide, Agl(s)When clear, colorless silver nitrate solu-tion is added to clear, colorless potassiumiodide solution, the product is a yellowprecipitate of silver iodide. The micro-scopic view shows a droplet of solution about to strike the surface of theKI solution. Notice that the only sub-stance that forms is AgI (represented bythe cluster of gray and blue ions); the and (and excess ) remain as ionsin solution. This means that the net reac-tion is between silver ions and iodideions: I-(aq) ¡ AgI(s)+Ag+(aq)

I-NO3

-K+

AgNO3

Table 4.3 General Guidelines for the Water Solubilities of Common Ionic Compounds

Almost all nitrates, acetates, perchlorates, group 1A metal salts, and ammonium salts areSOLUBLE.

Most chlorides, bromides, and iodides are SOLUBLE. Exceptions: those of and

Most sulfates are SOLUBLE. Exceptions: those of and ( isslightly soluble).

Most carbonates, hydroxides, phosphates, and sulfides are INSOLUBLE. Exceptions:ammonium and group 1A metal salts of any of those anions are soluble; hydroxides and sulfides of and are slightly to moderately soluble.Ba2+Sr2+,Ca2+,

CaSO4Hg2

2+Pb2+,Ba2+,Sr2+,

Hg2

2+.Ag+,Pb2+,

Precipitation Reactions movie

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4.3 � Reactions that Form Precipitates 137

Because our interest is usually in the net ionic equation, we can eliminate the specta-tor ions and write

This equation tells us that, regardless of their sources, if and are placed in thesame solution, they will form a precipitate of AgI(s).

Example 4.4Predict whether a precipitation reaction will occur in each of the following cases. If so, writea net ionic equation for the reaction.

(a) (c)(b)

STRATEGY

We must consider the simple compounds that could be formed from the combinations ofions present. Then we can use the solubility guidelines to determine whether any potentialproducts are insoluble compounds. If one or more of these potential products are insoluble,a reaction will occur. If all potential products are soluble, there will be no reaction.

SOLUTION

(a) The initial reactants are in aqueous solution. We can begin by writing the equation inionic form:

Then we can identify the possible products: NaCl and From Table 4.3, we seethat sodium compounds (group 1A) are soluble, as are most sulfates, including that ofmagnesium. Thus, we conclude

(no reaction)

(b) We can represent the reactants in solution by an ionic equation:

Here the possible products are and CuS. According to the solubility guidelines,all nitrates are soluble, but most sulfides, including CuS, are not. Thus, the reaction is

The and are spectator ions and can be canceled, yielding the net ionicequation

(c) The ionic equation is

The potential products are KCl and Potassium compounds (group 1A) aresoluble, and among carbonates, only those of the group 1A metals are soluble. Becausezinc is not in group 1A, we expect a precipitate of and an aqueous solutioncontaining and ions. The net ionic equation is

EXERCISE 4.4A

Predict whether a reaction will occur in each of the following cases. If so, write a net ionicequation for the reaction.

(a) (c)(b)

EXERCISE 4.4B

In each of the following cases, predict whether a reaction will occur, and, if so, write thenet ionic equation for the reaction.

(a) (c)(b) Mg(OH)2(s) + NaOH(aq) ¡ ?

NaHCO3(aq) + Ca(OH)2(aq) ¡ ?ZnSO4(aq) + BaS(aq) ¡ ?

FeCl3(aq) + Na2S(aq) ¡ ?

Sr(NO3)2(aq) + Na2SO4(aq) ¡ ?MgSO4(aq) + KOH(aq) ¡ ?

CO3

2-(aq) + Zn2+(aq) ¡ ZnCO3(s)

Cl-K+ZnCO3(s)

ZnCO3 .

2 K+(aq) + CO3

2-(aq) + Zn2+(aq) + 2 Cl-(aq) ¡ ?

S2-(aq) + Cu2+(aq) ¡ CuS(s)

NO3

-NH4

+¡ CuS(s) + 2 NH4

+(aq) + 2 NO3

-(aq)

2 NH4

+(aq) + S2-(aq) + Cu2+(aq) + 2 NO3

-(aq)

NH4NO3

2 NH4

+(aq) + S2-(aq) + Cu2+(aq) + 2 NO3

-(aq) ¡ ?

Na2SO4(aq) + MgCl2(aq) ¡ no precipitate

MgSO4 .

2 Na+(aq) + SO4

2-(aq) + Mg2+(aq) + 2 Cl-(aq) ¡ ?

(NH4)2S(aq) + Cu(NO3)2(aq) ¡ ?

K2CO3(aq) + ZnCl2(aq) ¡ ?Na2SO4(aq) + MgCl2(aq) ¡ ?

I-Ag+Ag+(aq) + I-(aq) ¡ AgI(s)

A precipitate like AgI does notdissociate to any significant

extent, which is why it is a precipitate. Ifit dissociated, it would dissolve in doing so.

A common student error encoun-tered in problems like

Example 4.4(a) is to write the formulas ofthe products as MgSO4 and Na2Cl2.Remind the student to write correct for-mulas for the compounds, rather thansimply swapping cation for cation.

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Example 4.5 A Conceptual ExampleFigure 4.8 shows that the dropwise addition of to produces a precipi-tate. What is the precipitate?


The reactants, and are both soluble in water. That all ammonium compoundsare soluble means the precipitate is not likely to contain It must therefore contain

But what is the anion? Recall that is a weak base and that it produces ionsin aqueous solution:

From the solubility guidelines, we expect to combine with to forminsoluble

To get the net ionic equation for the precipitation reaction, we need to multiply Equation (a)by 3 to get three ions and add the resulting equation to Equation (b):

The cancellation slashes show that the hydroxide ions formed in the ionization of areconsumed in the precipitation of and that no free appears in the netionic equation.

EXERCISE 4.5A

Suppose that after all the is precipitated, a large quantity of HCl(aq) is addedto the beaker in Figure 4.8. Describe what you would expect to see, and write a net ionicequation for this change.

EXERCISE 4.5B

Potassium palmitate is a typical water-soluble soap. It is formed by the neutralization ofpalmitic acid, with potassium hydroxide. When calcium chloride isadded to an aqueous solution of potassium palmitate, a gray precipitate is observed. Whatis the likely precipitate? Write ionic and net ionic equations for its formation.

Some Applications of Precipitation ReactionsIn many cases, precipitation reactions are an attractive method of preparing chemicalsubstances because they generally can be done with simple equipment and give a highpercent yield of product. Table 4.4 lists a few industrially important precipitationreactions.

In other cases, precipitation reactions are used in chemical analysis. Analysis of asample of matter to find out what it contains, but with no concern for how much, iscalled a qualitative analysis. For example, as shown in Figure 4.9, a sample of munic-ipal water typically becomes cloudy when is added to it, signaling thatchloride ion is probably present in the water.

Within the precision of the measurements used, the actual yield of a precipitationreaction is often equal to the theoretical yield. In this case, the reaction can be used fora quantitative analysis. For example, if the source of the chloride ion in a water-soluble material is NaCl, we can determine the actual quantity of NaCl present, asshown in Example 4.6.

Cl-(from sample under analysis) + Ag+(aq) ¡ AgCl(s)

AgNO3(aq)

CH3(CH2)14COOH,

Fe(OH)3(s)

OH-(aq)Fe(OH)3(s)NH3

3 NH3(aq) + 3 H2O(l) ∆ Fe3+(aq) + 3 OH-(aq) ¡

Net: Fe3+(aq) + 3 NH3(aq) + 3 H2O(l) ¡

3 NH4

+(aq) + 3 OH-(aq)

Fe(OH)3(s)

3 NH4

+(aq) + Fe(OH)3(s)

OH-

(b) Fe3+(aq) + 3 OH-(aq) ¡ Fe(OH)3(s)

Fe(OH)3(s):Fe3+(aq)OH-(aq)

(a) NH3(aq) + H2O(l) ∆ NH4

+(aq) + OH-(aq)

OH-NH3Fe3+.NH4

+.FeCl3 ,NH3

FeCl3(aq)NH3(aq)

� FIGURE 4.8 Predicting theproduct of a precipitation reactionAddition of to pro-duces a precipitate.

FeCl3(aq)NH3(aq)

� FIGURE 4.9 Chloride ion in tap waterWhen a drop or two of isadded to tap water, the solution turnscloudy. Eventually, a trace of white pre-cipitate settles to the bottom of the beaker.Most likely the precipitate is AgCl(s),indicating the presence of in the water.

QUESTION: What other ions in tap watermight be responsible for this precipitate?

Cl-

AgNO3(aq)

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4.3 � Reactions that Form Precipitates 139

Example 4.6One cup (about 240 g) of a certain clear chicken broth yields 4.302 g AgCl when excess

is added to it. Assuming that all the is derived from NaCl, what is the massof NaCl in the sample of broth?

STRATEGY

In this problem, chloride ions derived from NaCl are precipitated as AgCl. From the givenmass of AgCl, we can determine the number of moles of chloride ions initially present. Fromthe moles of we can calculate first the moles of NaCl and then the mass of NaCl.

SOLUTION

The equation for the precipitation reaction is

The stoichiometric equivalence from which we derive a stoichiometric factor is

and the series of conversions required in the stoichiometric calculation is

In the usual manner, we can combine these conversions into a single setup:

4.302 g AgCl ×1 mol AgCl

143.32 g AgCl? g NaCl =

1.754 g NaCl=

The answer:

(?) the unit

We want (?)and the unit

g NaCl.

Stoichiometric factorconverts mol AgCl

to mol Cl–.

Inverse of molarmass converts fromg AgCl to mol AgCl.

This is the massof precipitate

obtained.

Factor toconvert mol Cl–

to mol NaCl.

Molar massconverts mol NaCl

to g NaCl.

× ×

×1 mol Cl−

1 mol AgCl

1 mol NaCl

1 mol Cl−58.443 g NaCl

1 mol NaCl

g AgCl ¡ mol AgCl ¡ mol Cl- ¡ mol NaCl ¡ g NaCl

1 mol Cl-(aq) � 1 mol AgCl(s)

Ag+(aq) + Cl-(aq) ¡ AgCl(s)

Cl-,

Cl-AgNO3(aq)

Table 4.4 Some Precipitation Reactions of Practical Importance

Reaction in Aqueous Solution Application

Water purification. (The gelatinous precipitate carriesdown suspended matter.)

Removal of phosphates from wastewater in sewage treatment.

Precipitation of magnesium ion from seawater. (First step in the Dow process for extractingmagnesium from seawater.)

Preparation of AgBr for use in photographic film.

Production of lithopone, a mixture used as a whitepigment in both water paints and oil paints.

Preparation of calcium hydrogen phosphate dihydrate,used as a polishing agent in toothpastes.

H3PO4(aq) + Ca(OH)2(aq) ¡ CaHPO4# 2 H2O(s)

Zn2+(aq) + SO4

2-(aq) + Ba2+(aq) + S2-(aq) ¡ ZnS(s) + BaSO4(s)

Ag+(aq) + Br-(aq) ¡ AgBr(s)

Mg2+(aq) + 2 OH-(aq) ¡ Mg(OH)2(s)

Al3+(aq) + PO4

3-(aq) ¡ AlPO4(s)

Al3+(aq) + 3 OH-(aq) ¡ Al(OH)3(s)

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ASSESSMENT

Note that the mass of the broth (about 240 g) does not enter into this calculation. We areinterested only in the mass of NaCl present. On the other hand, if we had been required tofind the percentage of NaCl in the broth, the mass of the broth would have been required[% NaCl 0.7% NaCl].

EXERCISE 4.6A

What is the mass percent NaCl in a mixture of sodium chloride and sodium nitrate if a0.9056-g sample of the mixture yields 0.9372 g AgCl(s) when allowed to react with excess

EXERCISE 4.6B

Consider the seawater sample described in Exercise 4.1A. How many grams of precipitatewould you expect to get by adding (a) an excess of to 225 mL of the seawater,(b) an excess of NaOH(aq) to 5.00 L of the seawater? What are the precipitates?

4.4 Reactions Involving Oxidation and ReductionThe third category of chemical reactions that we consider in this chapter, oxidation–reduction reactions, is perhaps the largest of all. It includes all combustion processes,most metabolic reactions in living organisms, extraction of metals from their ores,manufacture of countless chemicals, and many of the reactions occurring in our nat-ural environment. Chemists often refer to oxidation–reduction reactions by reversingthe words and shortening them to redox reactions.

The term oxidation was originally used to describe reactions in which a substancecombines with oxygen. The opposite process, the removal of oxygen, was calledreduction. To encompass a wide range of reactions, however, we need much broaderdefinitions of oxidation and reduction. To assist us in formulating these more compre-hensive definitions, let’s first consider the concept of oxidation numbers.

Oxidation NumbersAs an aid to understanding, we have tried, whenever possible, to relate observationsat the macroscopic level to behavior at the molecular level. Oxidation numbers,however, don’t precisely reflect anything at the microscopic or molecular level.They are an arbitrary construction that chemists find useful when dealing with oxi-dation and reduction.

Oxidation numbers are easier to illustrate than to define, but after we have consid-ered some practical examples, you should be able to relate them to this definition: Anoxidation number represents either the actual charge on a monatomic ion or ahypothetical charge assigned to an atom in a molecule or in a polyatomic ion.

Consider the formation of sodium chloride. Each Na atom loses one electron, andeach Cl atom gains one electron. The compound is made up of and ions. Wesay that Na has an oxidation number of and Cl has an oxidation number of Inthe ionic compound chlorine also has the oxidation number existing as

ions. The oxidation number of calcium, however, is it is present as ions.The total of the oxidation numbers of the atoms (ions) in a formula unit of is

In the formation of a molecule, no electrons are transferred; instead they areshared. We can, however, arbitrarily assign oxidation numbers as if electrons weretransferred. For example, in the molecule we assign each H atom an oxidationnumber of If we also require that the total of the oxidation numbers for the threeatoms in the molecule be zero, then we must assign the O atom an oxidation number of

because In the molecule, the H atoms are identical and must have the same oxidation

number. If we require the sum of these oxidation numbers to be zero, then the oxida-tion number of each H atom must also be zero.

H2

21+12 - 2 = 0.-2,

+1.H2O,

+2 + 21-12 = 0.CaCl2

Ca2++2;Cl--1,CaCl2 ,

-1.+1Cl-Na+

AgNO3(aq)

AgNO3(aq)?

11.754 g NaCl>240 g broth2 * 100% LL

The related term oxidation state refersto the state, or condition, correspond-ing to a given oxidation number. Forexample, ion is an oxidation stateof chlorine that has an oxidation num-ber of The terms oxidation numberand oxidation state are often usedinterchangeably.

-1.

Cl-

Oxidation number is sometimescolloquially described as “the

charge an atom would have if it were an ion.”

The compound CaCl2 is ionic, sothe oxidation numbers are simply

the charges on the ions. The compoundH2O is not ionic, so we must arbitrarilyassign values for the oxidation numbers.

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4.4 � Reactions Involving Oxidation and Reduction 141

Because all the atoms in a molecule ofan element are alike, each Cl atom in

and each S atom in has an oxida-tion number of 0.

S8Cl2

The principal exception to Rule 5occurs when H is bonded to a metal, asit is in compounds called metalhydrides. In these cases, H has an oxi-dation number of Examples areNaH and CaH2 .

-1.

The principal exceptions to Rule 6occur when O atoms are bonded toone another, as in peroxides (for exam-ple, in the oxidation number of Ois ) and in superoxides (for example,in the oxidation number of O is ).-1>2

KO2

-1H2O2

From these examples, you can see that we must assign oxidation numbers system-atically. We can deal with the great majority of compounds with the following rules.Important exceptions are listed in notes in the margin. The rules are listed by priority;if two rules contradict each other, use the one with the higher priority. This generallytakes care of exceptions. For each rule, we have provided some examples. Additionalexamples are illustrated in Example 4.7.

1. For the atoms in a neutral species—an isolated atom, a molecule, or a formulaunit—the sum of all the oxidation numbers is 0.Examples: The oxidation number of an uncombined Fe atom is 0. The sum of theoxidation numbers of all the atoms in and is 0. The sum ofthe oxidation numbers of the ions in is 0.

2. For the atoms in an ion, the sum of the oxidation numbers is equal to the charge onthe ion.Examples: The oxidation number of Cr in the ion is The sum of the oxi-dation numbers in is and the sum in is

3. In compounds, the group 1A metals all have an oxidation number of and thegroup 2A metals all have an oxidation number ofExamples: The oxidation number of Na in is and that of Ca in

is

4. In compounds, the oxidation number of fluorine isExamples: The oxidation number of F is in HF, and

5. In compounds, hydrogen has an oxidation number ofExamples: The oxidation number of H is in HCl, and

6. In most compounds, oxygen has an oxidation number ofExamples: The oxidation number of O is in CO, and

7. In binary compounds with metals, group 7A elements have an oxidation numberof group 6A elements have an oxidation number of and group 5A ele-ments have an oxidation number ofExamples: The oxidation number of Br is in that of S is in and that of N is in

Example 4.7What are the oxidation numbers assigned to the atoms of each element in

(a) (b) (c) (d) (e)

STRATEGY

In applying the rules for determining oxidation numbers, we must adhere to the priorityorder in which they are listed above. Note, however, that except in the case of uncombinedelements, we cannot apply Rule 1 first.

SOLUTION

(a) The oxidation number of K is (Rule 3). The oxidation number of O is (Rule 6), and the total for four O atoms is For these two elements, the total is

The oxidation number of the Cl atom in this ternary compound mustbe to give a total of zero for all atoms in the formula unit(Rule 1).

(b) The oxidation number of O is (Rule 6), and the total for seven O atoms is Thetotal of the oxidation numbers in this ion must be (Rule 2). Therefore the total ofthe oxidation numbers of two Cr atoms is and that of one Cr atom is

(c) Keeping in mind that the total for the formula unit must be 0 (Rule 1) and that the oxi-dation number of Ca is (Rule 3), the oxidation number of H must be rather thanits usual (Rule 5). Thus, for the sum of the oxidation numbers is+2 + 12 * -12 = 0.

CaH2+1-1+2

+6.+12,-2

-14.-2

1+1 - 8 + 7 = 02+7,+1 - 8 = -7.

-8.-2+1

Fe3O4Na2O2CaH2Cr2O7

2-KClO4

Mg3N2.-3Na2S,-2CaBr2,-1

-3.-2,-1,

ClO4

-.C6H12O6,CH3OH,-2

-2.

CH4.NH3,H2O,+1+1.

SO2F2.ClF3,-1-1.

+2.Ca3(PO4)2

+1Na2SO4

+2.+1

+1.NH4

+-3,PO4

3-+3.Cr3+

MgBr2

C6H12O6S8,Cl2 ,

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� FIGURE 4.10 The thermitereaction


(d) The oxidation number of Na is (Rule 3), and for the two Na atoms, The totalfor the formula unit must be 0 (Rule 1). Even though the oxidation number of O is usu-ally (Rule 6), here it must be so that the total for the two O atoms is Rule 3takes priority over Rule 6.

(e) The oxidation number of O is (Rule 6). For four O atoms, the total is The totalfor the formula unit must be 0 (Rule 1). The total for three Fe atoms must be and foreach Fe atom,

ASSESSMENT

Remember that the sum of the oxidation numbers of all the atoms present must add up toequal the total charge on the molecular or ionic formula. Notice how Rules 1 and 2 wereused at one point or another in each part of this example.

Usually, fractional oxidation numbers, as seen in part (e) signify an average. The com-pound is actually Two of the Fe atoms have oxidation numbers of and one has an oxidation number of The average is

EXERCISE 4.7A

Assign an oxidation number to each atom in

(a) (b) (c) (d) (e) (f) (g)

EXERCISE 4.7B

Assign an oxidation number to each underlined atom:

(a) (b) (c) (d) (e) (f) (g)

Identifying Oxidation–Reduction ReactionsThe spectacular reaction pictured in Figure 4.10, called the thermite reaction, is usedto produce liquid iron for welding large iron objects:

Even by the limited definitions we gave at the start of this section, we can call this anoxidation–reduction reaction. The reactant Al is oxidized to aluminum atomsgain oxygen atoms. The reactant is reduced to Fe; iron(III) oxide loses oxy-gen atoms.

Now consider how we can use oxidation numbers to identify an oxidation–reduction reaction. In the equation for the thermite reaction, the oxidation numbers ofthe Al, Fe, and O atoms are assigned according to the conventions just established andwritten as small numbers above the chemical symbols:

In the thermite reaction, the oxidation number of Al atoms (red) increases from 0 toand the oxidation number of Fe atoms (blue) decreases from to 0, illustrating

this oxidation-number view of oxidation–reduction:

In an oxidation–reduction reaction, the oxidation number of one or more elementsincreases—an oxidation process—and the oxidation number of one or more ele-ments decreases—a reduction.

The reaction pictured in Figure 4.11 differs strikingly from the thermite reaction,but the expanded definition identifies this as an oxidation–reduction reaction, as wecan see from the oxidation numbers noted in the following equation:

Mg(s) is oxidized to Mg2+(aq), and Cu2+(aq) is reduced to Cu(s).

Mg(s) + Cu2+(aq)

0 +2 +2Mg2+(aq) + Cu(s)

0

+3+3,

2 Al(s) � Fe2O3(s)

0 �32 Fe(l) � Al2O3(s)

0�2 �3 �2

Fe2O3

Al2O3;

2 Al(s) + Fe2O3(s) ¡ 2 Fe(l) + Al2O3(s)

C2O4

2-NO2

+,C3O2 ,S4O6

2-,P3O10

5-,CHCl3 ,HSbF6 ,

CsO2ClO2

-,NaMnO4 ,HAsO4

2-,CH3F,P4 ,Al2O3 ,

+8>3.=13 + 3 + 22>3+2.+3,Fe2O3

# FeO.Fe3O4

+8>3.+8,

-8.-2

-2.-1-2

+2.+1

Problem-Solving NoteThe compounds and the polyatomic ion demonstrate how the oxidation num-ber of carbon can vary in organic com-pounds.

C2O4

2-C3O2 ,CHCl3 ,CH3F,

Thermite Reaction movie

HIL04-124-167v4 1/13/04 12:10 PM Page 142


Cu2+

Cu2+

Mg2+

Mg2+

Mg2+

Mg2+

Cu2+

Cu2+

Cu

Cu CuCu2+2e

(a) (b) (c)

� FIGURE 4.11 An oxidation–reduction reaction(a) A coil of magnesium ribbon immersed in a solution of (b) A microscopic view shows electronsbeing transferred from metallic magnesium to a ion, which is thereby converted to an atom of Cu metal. Thetransfer of electrons also produces a ion in solution. (For clarity, sulfate ions have been left out of thismicroscopic view.) (c) After a few hours, all of the has been displaced from the solution, leaving a depositof red-brown copper metal, some unreacted magnesium, and clear, colorless

QUESTION: Does the mass of the coil increase, decrease, or remain the same during the course of this oxidation–reduction reaction? Why?

MgSO4(aq).Cu2+

Mg2+Cu2+

CuSO4(aq).

The simple reaction shown in Figure 4.11 suggests the most fundamentaldefinition of oxidation–reduction reactions. When you look closely at part (b) of thefigure, you see that two things happen simultaneously:

Oxidation:

Reduction:

The oxidation number of Mg increases from 0 to —an oxidation—as the Mg atomloses two electrons. The ion gains the two electrons, and its oxidation numberdecreases from to 0—a reduction. Because electrons, particles of matter, can beneither created nor destroyed, oxidation and reduction must always occur together.This fundamental definition of oxidation and reduction reflects these ideas:

An oxidation–reduction reaction consists of two processes that occur simultane-ously. In one process, called oxidation, electrons are lost, and in the other, calledreduction, they are gained.

Oxidation–Reduction EquationsPermanganate ion, in an acidic aqueous solution, oxidizes ion to and is itself reduced to ion. Let’s see if we can write an equation describing thisreaction. From the description of the reaction, we see that and shouldboth appear on the left side and and on the right:

(incomplete)

Can you see why we have labeled this equation incomplete? There are four Oatoms on the left that are not accounted for on the right. Recall, however, that thedescription of the reaction also indicates the presence of both and (“an acidicaqueous solution”). If we include these species in the equation, specifically, with

on the right to balance the four O atoms in and on the left to bal-ance eight H atoms in the we obtain

(not balanced)MnO4

-(aq) + Fe2+(aq) + 8 H+(aq) ¡ Mn2+(aq) + Fe3+(aq) + 4 H2O(l)

4 H2O,8 H+MnO4

-4 H2O

H2OH+

MnO4

-(aq) + Fe2+(aq) ¡ Mn2+(aq) + Fe3+(aq)

Fe3+Mn2+Fe2+MnO4

-Mn2+

Fe3+Fe2+MnO4

-,

+2Cu2+

+2

Cu2+(aq) + 2e- ¡ Cu(s)

Mg(s) ¡ Mg2+(aq) + 2e-

Oxidation-Reduction Reactions—Part 1 animation

The reaction in Figure 4.11 iseasy to carry out and is rapid

enough for the blue colorchange to be apparent within afew minutes. A second demon-stration using a coil of copperwire in silver nitrate solutionshows the gradual appearance ofblue . The two demon-strations accompany oneanother nicely.

Cu2+(aq)

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Are you wondering why we have labeled this equation not balanced? Clearly it is bal-anced for numbers of atoms—1 Mn, 4 O, 1 Fe, and 8 H on each side. The importantrequirement that the equation fails to meet is that electric charge must be balanced.Electric charge cannot be created or destroyed in a chemical reaction. As the equationstands, there is a net charge of on the left and only on the right. As usual, wecan, by trial and error, change coefficients in the equation to achieve a balance. Thus,we get the desired charge balance by using the coefficient 5 for both and The net charges on the two sides of the equation are now equal; both are

(balanced)

Unfortunately, the trial-and-error method of achieving charge balance often doesnot work. What we need is a systematic method of balancing redox equations that ismore likely to succeed. We present such a method based on the transfer of electrons inredox reactions in Chapter 18, where we need the method for other purposes as well.For the present, your main tasks will be to

• Identify oxidation–reduction reactions.

• Balance certain simple redox equations by inspection.

• Recognize, in all cases, whether a redox equation is properly balanced.

Oxidizing and Reducing AgentsAn inorganic chemistry treatise lists dinitrogen tetroxide, as a “fairly strongoxidizing agent” and hydrazine, as a “powerful reducing agent.” Such termsdescribe the way substances participate in redox reactions, and we will find it helpfulto understand their meanings.

In an oxidation–reduction reaction, the substance that is oxidized—because itcauses some other substance to be reduced—is called a reducing agent. Similarly, thesubstance that is reduced is called an oxidizing agent because it causes another sub-stance to be oxidized. The financial practice of lending or borrowing money providesan analogy. In such a transaction, there must be both a lending agent and a borrower inorder for the transaction to occur. Furthermore, the amount of money gained by theborrower must equal the amount of money given up by the lender. Likewise, both areducing agent and an oxidizing agent are needed for a redox reaction, and the numberof electrons obtained by the oxidizing agent must equal the number given up by thereducing agent.

Based on our description of oxidizing and reducing agents, we might well predictthat nitrogen tetroxide and hydrazine should react with each other. They do indeed.The reaction between these two substances, which is accompanied by the release oflarge quantities of heat, is the basis of a rocket propulsion system:

In the reaction, is reduced to (oxidation number of N decreases from to 0), making the oxidizing agent, and is oxidized to (oxidation num-ber of N increases from to 0), making the reducing agent. Although thechanges in oxidation numbers occur in N atoms, we do not call the atoms themselvesthe oxidizing or reducing agents. The compounds in which these atoms are found( and ) are the oxidizing and reducing agents, respectively.

Oxidation Numbers of NonmetalsSome compounds and ions that contain the nonmetallic elements nitrogen, sulfur,or chlorine are listed in Figure 4.12. They are arranged, in order of decreasing oxi-dation number of the nonmetal atoms, in columns that correspond to the periodictable. We can use this figure to illustrate some additional ideas about oxidizing andreducing agents:

N2H4N2O4

N2H4-2N2N2H4N2O4

+4N2N2O4

N2O4(l) + 2 N2H4(l) ¡ 3 N2(g) + 4 H2O(g)

N2H4,N2O4,

MnO4

-(aq) + 5 Fe2+(aq) + 8 H+ ¡ Mn2+(aq) + 5 Fe3+(aq) + 4 H2O(l)

17+ :Fe3+.Fe2+

5+9+

Problem-Solving NoteFor an algebraic derivation, let theunknown coefficient of and be x.The net charge on the left is

and that on the right is

x = 5

-x = -5

2x - 3x = 2 - 8 + 1

2x + 8 - 1 = 3x + 2

3x + 2:2x + 8 - 1,

Fe3+Fe2+

Oxidation-Reduction Reactions—Part 2 animation

HIL04-124-167v4 1/13/04 12:10 PM Page 144


• The maximum oxidation number for a nonmetal atom is equal to the atom’s groupnumber in the periodic table: for group 5A atoms, for group 6A, and for group 7A. Oxygen and fluorine are exceptions (recall rules 4 and 6 for assign-ing oxidation numbers).

• The minimum oxidation number for a nonmetal atom is equal to the group numberminus eight: for group 5A atoms, for group 6A, and for group 7A.

• Species in which a nonmetal atom has its maximum oxidation number are invari-ably oxidizing agents because the oxidation number of the nonmetal atom canonly decrease in a redox reaction. Thus, in a redox reaction, can only be anoxidizing agent.

• Species in which a nonmetal atom has its minimum oxidation number are reduc-ing agents. Thus, in a redox reaction, can only be a reducing agent.

• In principle, species in which a nonmetal atom has an intermediate oxidationnumber can be either an oxidizing or a reducing agent, depending on the particu-lar reaction. Generally, one role or the other is more common. For example,with the oxidation number for N, is almost always an oxidizing agent, and

with the oxidation number for N, is almost always a reducing agent.Even though it is high on the oxidation-number scale for sulfur, usuallyacts as a reducing agent. Conversely, though low on the oxidation numberscale for chlorine, is generally an oxidizing agent.

• A redox reaction in which the same substance is both an oxidizing and a reducingagent is called a disproportionation reaction. Thus, when disproportion-ates, half the is oxidized to and half is reduced to

Metals As Reducing AgentsIn all common metallic compounds, the metal atom has a positive oxidation number.Elemental metals, of course, have atoms with oxidation number 0, their lowest com-mon oxidation state. Metals are reducing agents, but their strengths as reducing agentsvary widely. The atoms of some metals, such as those of groups 1A and 2A, lose elec-trons easily, which means the atoms are readily oxidized to metal cations and are

2 H2O2(aq) ¡ 2 H2O(l) + O2(g)

H2O(l):O2(g)H2O2

H2O2

Cl2 ,SO3

2--2N2H4,

+4N2O4,

H2S

NO3

-

-1-2-3

+7+6+5

Group 7AGroup 6AGroup 5A

NO3− +5

N2O4 +4

NO2− +3

NO +2

N2O +1

N2 0

NH2OH −1

N2H4 −2

NH3 −3

SO42− +6

S2O62− +5

SO32− +4

S2O42− +3

S2O32− +2

S2Cl2 +1

S8 0

H2S2 −1

H2S −2

ClO4− +7

Cl2O6 +6

ClO3− +5

ClO2 +4

ClO2− +3

+2

+1

0

ClO−

Cl2

Cl− −1> FIGURE 4.12 Oxidationnumbers in some nitrogen-, sulfur-,and chlorine-containing speciesThe species in red can act only as oxidiz-ing agents; those in blue can act only asreducing agents. Those in between can actas either, depending on the reaction.

Redox Chemistry of Tin and Zincmovie

HIL04-124-167v4 1/13/04 12:10 PM Page 145

K

Ca

Na

Mg

Al

Cr

Zn

Fe

Cd

Ni

Sn

Pb

Cu

Ag

Hg

Au

Powerful

Strength as areducing agent

Strong

Good

Fair

Poor

Very poor

H2

� FIGURE 4.13 Activity series ofsome metals


therefore powerful reducing agents. Other metals, such as silver and gold, are oxidizedwith great difficulty. They are exceptionally poor reducing agents. Figure 4.13 listssome common metals in a sequence called the activity series of the metals. As a gen-eral rule,

A metal will displace from solution the ions of any metal that lies below it in theactivity series.

For example, with the activity series we could have predicted the reaction pictured inFigure 4.11:

Mg lies above Cu in the activity series. Therefore Mg(s), a good reducing agent,reduces to Cu(s), and the Mg(s) is itself oxidized to Because takes the place of in the solution and Cu replaces Mg as the solid, we say thatmagnesium displaces copper(II) ion from solution.

We can also use the activity series to predict with confidence that if we add Ag(s)rather than Mg(s) to there is no reaction:

Because it lies below Cu in the activity series, silver is unable to reduce to Cu(s).

The usefulness of the activity series of the metals is greatly increased by includingbecause doing this permits us to say that any metal above hydrogen in the series

can react with an acid to produce For example,

Any metal below hydrogen cannot react with an acid to produce For example,

Finally, there are a few special circumstances in which a metal that lies below hydro-gen in the activity series may still undergo reaction in an acidic solution. One suchcase is described in Example 4.8.

Example 4.8 A Conceptual ExampleExplain the difference in what happens when a copper-clad penny is immersed in(a) hydrochloric acid and (b) nitric acid, as shown in Figure 4.14.

Ag(s) + H+(aq) ¡ no reaction

H2(g).

2 Al(s) + 6 H+(aq) ¡ 2 Al3+(aq) + 3 H2(g)

H2(g).H2,

Cu2+(aq),

Ag(s) + Cu2+(aq) ¡ no reaction

Cu2+(aq),

Cu2+Mg2+Mg2+(aq).Cu2+

Mg(s) + Cu2+(aq) ¡ Mg2+(aq) + Cu(s)

� FIGURE 4.14 The action of HCl(aq) and on copperA copper-clad penny does not react with hydrochloric acid (left), but it does reactwith concentrated nitric acid, producing red-brown fumes and a green-blue solu-tion (right).

HNO3(aq)


(a) Because copper lies below hydrogen in the activity series of the metals, Cu(s) cannotreduce to and be oxidized to Looking at it the other way,is not a strong enough oxidizing agent to oxidize Cu(s) to Chloride ion inCu2+(aq).

H+Cu2+(aq).H2(g)H+(aq)

Formation of Silver Crystalsmovie

It might be pointed out to thestudent that gold is an obviously

poor reducing agent. That is why itremains shiny for centuries, even underwater—it is among the poorest of reduc-ing agents and oxidizes only with thegreatest difficulty.

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HCl(aq) can only be a reducing agent. As neither nor can oxidize Cu, weexpect no reaction between Cu(s) and HCl(aq).

(b) In contrast, we clearly observe a reaction between copper and nitric acid. The Cu(s) isoxidized to (blue color). From the HCl analysis, we know that is notstrong enough as an oxidizing agent to oxidize Cu(s) to and we know that

is an oxidizing agent here because the N atom has its highest possible oxi-dation number, Therefore, copper is being oxidized to by Figure 4.12 suggests that might be reduced to any one of several products.We have no way at this point of predicting what that product might be, but the red-brown gas proves to be nitrogen dioxide, The oxidation number of nitrogen in

is the same as in that is, Using this information, we can write a netionic equation:

(not balanced)

Finally, the equation is balanced by inspection.

EXERCISE 4.8A

The oxidizing agent potassium dichromate, reacts one way when heated withhydrochloric acid and another way when heated with nitric acid. With one of the acids, agas is evolved and the solution color changes from red-orange to green. With the otheracid, the original red-orange color remains unchanged; that is, no reaction occurs. Explainthis difference in behavior.

EXERCISE 4.8B

Since 1982, U.S. pennies have been made of zinc with a thin copper coating. If the edge ofa new cent is notched with a knife and dropped in hydrochloric acid overnight, only a hol-low shell of copper remains the next morning. How would the resulting solution differfrom the two solutions in Figure 4.14?

K2Cr2O7(s)

Cu(s) + 4 H+(aq) + 2 NO3

-(aq) ¡ Cu2+(aq) + 2 NO2(g) + 2 H2O(l)

Cu(s) + H+(aq) + NO3

-(aq) ¡ Cu2+(aq) + NO2(g) + H2O(l)

+4.N2O4 ,NO2

NO2 .

NO3

-(aq)NO3

-(aq).Cu2+(aq)+5.NO3

-(aq)Cu2+(aq),

H+(aq)Cu2+(aq)

Cl-H+

Glassmaking and glassblowing are among the oldest of artforms. Modern museums display glass bottles and vasesthat are thousands of years old. Stained-glass windows

constructed in the Middle Ages still decorate many churchesthroughout Europe. In these windows, artisans have made use ofintricate combination of colored glass to generate their mosaics.The color of many forms of glass is directly related to the oxida-tion states of impurities found throughout the glass matrix.

Ordinary (soda-lime) glass is made from sand sodium carbonate (“soda”), and calcium oxide (“lime”). Addi-tives confer other properties, including color. Often the sand

(SiO2),

Oxidation–Reduction and the Color of Glasscontains a small proportion of iron, usually in the oxidationstate, which imparts a pale tan color. During processing the ironattains the oxidation state. Look at a piece of window glassend-on, or at a “clear” glass bottle, and you can see the greencolor that is characteristic of iron(II) ion.

Colorless glass is more difficult to make than “bottle-green” glass. Small batches of colorless optical glass are madeusing sand that is essentially free of impurities. For largebatches, manganese(IV) oxide can be added. The oxi-dizes the iron back to Fe(III) ion. The manganese is converted tothe oxidation state, which is pale pink. The right combina-tion of Mn(II) and Fe(III) ions makes the glass nearly colorless.

Another redox reaction occurs when manganese-containingglass is exposed to intense sunlight. Ultraviolet light is energeticenough to convert manganese gradually from the to the oxidation state, which is purple. The manganese is locked intothe structure of the glass, so it does not come into contact withother species (as it could in an aqueous solution) and thereforecannot be reduced back to a lower oxidation state. After years ofexposure to sunlight, the colorless manganese-containing glassappears purple. Brief exposure to energetic radiation such asgamma rays (Chapter 19) can cause the same transition.

+7+7+2

+2

MnO2

+2

+3

> This composite imageshows a glass electricalinsulator before irradiationwith gamma rays (left) andafter (right), illustrating theoxidation of Mn(II) to Mn(VII).

The demonstration described inExercise 4.8B is easy to carry out,

and the students often find it amusing.Rinse the copper shell thoroughly withwater before passing it around forexamination!

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4.5 Applications of Oxidation and ReductionAs we noted at the beginning of the previous section, redox reactions have manyimportant applications. Let’s look at a few of these.

In Everyday LifeOxygen is undoubtedly the most important oxidizing agent in all aspects of life. Weuse it to oxidize fuels for heating our homes and propelling our automobiles. Oxygencorrodes metals by oxidizing them to cations, as in the rusting of iron. Oxygen even“burns” the foods we eat, to release the energy we need for life.

A common household oxidizing agent is hydrogen peroxide, usually foundin the form of an aqueous solution containing 3% An advantage of hydrogenperoxide over other oxidizing agents is that in most reactions it is converted to water,an innocuous product. The 3% hydrogen peroxide solution is used in medicine asan antiseptic to treat minor cuts and abrasions. An enzyme in blood catalyzes thedecomposition of hydrogen peroxide which, as noted on page 145, is a disproportion-ation reaction:

The escaping bubbles of oxygen gas help to carry dirt and germs out of a wound.Benzoyl peroxide is a powerful oxidizing agent that has long been

used at 5% and 10% concentrations for treating acne. In addition to its antibacterialaction, benzoyl peroxide acts as a skin irritant, causing old skin to flake off and bereplaced by newer, fresher-looking skin. However, when used on areas exposed to sun-light, benzoyl peroxide is thought to promote skin cancer.

Chlorine and its compounds are commonly encountered as oxidizing agents indaily life. Elemental is used to kill microorganisms, both in drinking water andin wastewater treatment. Swimming pools are usually disinfected by “chlorination.” Inlarge pools, the chlorine is often introduced from steel cylinders in which it is stored asa liquid. The chlorine disproportionates in water to form HCl and hypochlorous acid,HOCl, which is the actual sanitizing agent:

The HCl(aq) soon makes the pool water too acidic, and it must be neutralized byadding a base such as sodium carbonate.

Small swimming pools are sometimes chlorinated with sodium hypochlorite,NaOCl, formed by dissolving in NaOH(aq). The added NaOH causes these poolsto become too basic, and hydrochloric acid is added to neutralize the excess basicity.Calcium hypochlorite, is used to disinfect clothing and bedding in hospitalsand nursing homes.

Nearly any oxidizing agent can be used as a bleach to remove unwanted colorfrom fabrics, hair, or other materials. However, some are too expensive, some harmfabrics, some produce undesirable products, and some are simply unsafe.

In Foods and NutritionIn food chemistry, the substances known as antioxidants are reducing agents. Ascorbicacid (vitamin C), which is soluble in water, is thought to retard potentially damagingoxidation of living cells. Tocopherol (vitamin E) is a fat-soluble antioxidant. In thebody, it is thought to act by scavenging harmful by-products of metabolism, such asthe highly reactive molecular fragments called free radicals. In foods, vitamin E actsto prevent fats from being oxidized and thus becoming rancid.

When it acts as an antioxidant, vitamin C is oxidized to dehydroascorbicacid, For example, when nitrite ions from foods get into the bloodstream,they oxidize iron in hemoglobin, destroying its ability to carry oxygen. In the stomach,however, ascorbic acid reduces nitrite ion to NO(g):

Ascorbic acid Dehydroascorbic(vitamin C) acid

C6H8O6(aq) + 2 H+(aq) + 2 NO2

-(aq) ¡ C6H6O6(aq) + 2 H2O(l) + 2 NO(g)

C6H6O6.(C6H8O6)

Ca(OCl)2 ,

Cl2

Cl2(g) + H2O(l) ¡ H+(aq) + Cl-(aq) + HOCl(aq)

Cl2

(C6H5COO)2

2 H2O2(aq) ¡ 2 H2O(l) + O2(g)

H2O2.H2O2,

� Swimming pool “chlorine” used totreat small home pools is an alkalinesolution of sodium hypochlorite,formed by the reaction of chlorine gaswith aqueous sodium hydroxide:

¡ NaOCl(aq) + NaCl(aq) + H2O(l)

Cl2(g) + 2 NaOH(aq)

� Hydrogen peroxide “bubbles” whenpoured over a cut because of theproduction of gaseous oxygen.

HIL04-124-167v4 1/13/04 12:10 PM Page 148

4.5 � Applications of Oxidation and Reduction 149

Many colored organic materials are made up of large molecules that contain alternate double and single carbon-to-carbonbonds. An example is lycopene, the compound that gives tomatoes their bright red color:

The brackets and subscript 2 mean that the formula within the brackets is doubled. Think of the structure shown here as thecomplete formula folded over from right to left; the right-hand bracket is like a hinge. When you open up the hinge, you getthe complete formula. Chemists often use space-saving devices like this to represent complex formulas. The interesting feature ofthe lycopene molecule is the series of alternating double and single bonds that are shown in color. This feature is called aconjugated system, and such systems often confer the property of color to organic compounds.

Bleaching agents oxidize colored molecules such as lycopene by destroying the conjugated system. Thus, lycopene is oxidizedby aqueous hypochlorite ions to colorless organic compounds, and the hypochlorite ion is reduced to chloride ion:

Hypochlorite bleaches are safe and effective for cotton and linen fabrics because they donot oxidize the cellulose that makes up these fabrics. However, hypochlorites do oxidize theprotein and protein-like molecules that make up wool, silk, and nylon and therefore should notbe used on those fabrics. Chlorine dioxide and sodium chlorite are equallygood oxidizing agents that do less damage to fabrics than do chlorine and hypochlorites.

Some other bleaching agents are hydrogen peroxide, “oxygen” bleaches containingsodium percarbonate (often represented as to indicate an associationbetween and ), sodium perborate and a variety of chlorine-containing organic compounds that release in water.

Stain removal is not nearly so simple a process as bleaching. A few stain removers areoxidizing agents or reducing agents; others have quite different chemical natures. Nearly allstains require rather specific stain removers.

Hydrogen peroxide in cold water removes bloodstains from cotton and linen fabrics.Potassium permanganate can be used to remove most stains from white fabrics (except rayon).The purple permanganate stain then can be removed in a redox reaction with oxalic acid:

Iodine, used as a disinfectant, often stains clothing it contacts. The iodine stain is readilyremoved in a redox reaction with sodium thiosulfate:

I2 + 2 S2O3

2-(aq) ¡ 2 I-(aq) + S4O6

2-(aq)

5 H2C2O4(aq) + 2 MnO4

-(aq) + 6 H+(aq) ¡ 2 Mn2+(aq) + 8 H2O + 10 CO2(g)

Cl2

(NaBO2# H2O2),H2O2Na2CO3

2 Na2CO3# 3 H2O2

(NaClO2)(ClO2)

C40H56 + OCl-(aq) ¡ Cl-(aq) + colorless organic products

C

CH3

CH3 CH2CH2CH C

CH3

CH CH CH C

CH3

CH CH CH C

CH3

CH CH2

Oxidation–Reduction in Bleaching and in Stain Removal

Green plants carry out the redox reaction that makes possible almost all life onEarth. They do this through a process called photosynthesis, in which carbon dioxideand water are converted to glucose, a simple sugar. The synthesis of glucose requires avariety of proteins called enzymes and a green pigment called chlorophyll that con-verts sunlight into chemical energy. The overall change that occurs is

In this reaction, is reduced to glucose and is oxidized to oxygen gas. Otherreactions convert the simple sugar to more complex carbohydrates and to plant pro-teins and oils.

In IndustryThe most widely used oxidizing agent in industrial processes, and certainly the cheap-est and least objectionable environmentally, is oxygen itself. In the first step of theconversion of iron to steel, high-pressure oxygen gas is blown over molten impure iron

H2OCO2

6 CO2 + 6 H2O ¡ C6H12O6 + 6 O2

� Pure water (left) has littleability to remove a dried tomatosauce stain. Sodium hypochlorite,NaOCl(aq) (right), easily bleachesthe stain away by oxidizing thecolored pigments of the sauce tocolorless products.

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� An oxyacetylene torch being usedto weld air-conditioning equipment in Malaysia.


Application NoteOzone, is a strong oxidizing agentthat finds significant use in waterpurification. Gaseous chlorine dioxide,

was used in December 2001 todestroy possible anthrax contamina-tion in the Hart Senate Office Buildingin Washington, DC.

ClO2 ,

O3 ,

Application NoteTungsten has a high electrical conduc-tivity and the highest melting point ofall the metals (3410 °C).These proper-ties make it ideal for use as the light-emitting filaments in electriclightbulbs.

(pig iron). This process burns off carbon and sulfur as gaseous oxides. The oxygenalso converts the elements Si, P, and Mn to their solid oxides.

Oxygen is used to oxidize hydrogen, or acetylene, in torches for weld-ing and cutting metals. The heat released in such reactions provides the high tempera-tures needed to melt metals. In the oxyacetylene torch, for example, the reaction is

Another group of important industrial oxidizing agents includes chlorine gas andchlorine compounds in which the chlorine atoms have positive oxidation numbers. Asnoted above, chlorine gas and solutions containing hypochlorite ion, are used inwater-treatment plants to kill pathogenic (disease-causing) microorganisms, and theseagents are also used in the paper and textile industries for bleaching.

The principal industrial reducing agents are carbon and hydrogen. Carbon is oftenused in a solid form known as coke, which is produced by driving off the volatile mat-ter from coal. In the blast furnace method for manufacturing iron from iron ore, theactual reducing agent is carbon monoxide, which is produced from coke:

The CO(g) then reduces the iron ions in the ore to elemental iron:

Hydrogen is used as a reducing agent in smaller-scale processes and in caseswhere a metal might react with carbon to form an objectionable metal carbide. As anexample, passing a stream of over at 1200 °C produces tungsten metal:

In Organic ChemistryDichromate ion, is a widely used oxidizing agent in many areas of chemistry.In organic chemistry, it is used to oxidize alcohols (ROH). The product of the reactiondepends on the location of the OH group in the alcohol molecule, on the relative pro-portions of alcohol and dichromate ion, and on reaction conditions. If the OH group ison a terminal carbon atom and the product is distilled off as formed, the product is analdehyde. The characteristic functional group of an aldehyde is

It is often written on one line as CHO, in which case the carbon–oxygen double bondis understood. As an example, consider the reaction

Ethanol AcetaldehydeBoiling point 78 °C Boiling point 21 °C

Because it has a lower boiling point, the acetaldehyde can be boiled off from the reac-tion mixture, leaving the ethanol behind. If the acetaldehyde is not removed as itforms, it is further oxidized to acetic acid. In this case, the overall reaction is

From the oxidation number of Cr in and in we see thatis reduced in this reaction. This indicates that ethanol is oxidized to acetalde-

hyde in the first reaction and to acetic acid in the second. The average oxidation num-bers of the C atoms are in ethanol, in acetaldehyde, and 0 in acetic acid. Theoxidations described here are pictured in Figure 4.15.

When the OH group of an alcohol is bonded to an interior carbon atom, the productof oxidation of the alcohol is called a ketone, featuring the functional group known asthe carbonyl group with two alkyl groups attached to the carbonyl carbon atom:

C

O

-1-2

Cr2O7

2-1+32,Cr3+1+62Cr2O7

2-

3 CH3CH2OH + 2 Cr2O7

2- + 16 H+ ¡ 3 CH3COOH + 4 Cr3+ + 11 H2O

3 CH3CH2OH + Cr2O7

2- + 8 H+ ¡ 3 CH3CHO + 2 Cr3+ + 7 H2O

C

O

H

Cr2O7

2-,

WO3(s) + 3 H2(g) ¡ W(s) + 3 H2O(g)

WO3H2(g)

Fe2O3(s) + 3 CO(g) ¡ 2 Fe(l) + 3 CO2(g)

C(s) + CO2(g) ¡ 2 CO(g)

C(s) + O2(g) ¡ CO2(g)

OCl-,

2 C2H2(g) + 5 O2(g) ¡ 4 CO2(g) + 2 H2O(g)

C2H2,H2,

Application NoteWhen a person drinks an alcoholicbeverage, enzymes in the liver causethe ethanol to be oxidized to acetalde-hyde. If the person drinks moderately,the acetaldehyde is further oxidized toacetic acid and then to carbon dioxideand water.The liver can handle about 1oz of ethanol an hour, the quantity inone average drink. If a person drinksmore than that, the acetaldehyde con-centration builds up and the individualgets intoxicated. Acetaldehyde isthought to be responsible for many ofthe harmful effects of ethanol, such ashangovers and fetal alcohol syndromein babies born to women who drinkheavily while pregnant.

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4.6 � Titrations 151

(a) (b) (c)

> FIGURE 4.15 The oxidation ofethanol by dichromate ion in acidicsolution(a) Orange about to beadded to colorless thathas been acidified with (b) Theethanol solution becomes colored becauseof the added (c) After thereaction, the solution turns pale violet,signifying that the is gone andthat is now present.Cr3+(aq)

Cr2O7

2-

Cr2O7

2-(aq).

H2SO4.CH3CH2OH(aq)

K2Cr2O7(aq)

The simplest ketone, derived from 2-propanol, is the common solvent acetone,used in applications as varied as varnishes and lacquers, rubber cement,

and nail polish remover:

As we have just seen, aldehydes and ketones can be formed by the oxidation ofalcohols. Conversely, aldehydes and ketones can be reduced to alcohols. Reduction ofthe carbonyl group is important in living organisms. For example, in anaerobic exer-cise (exercise in which the supply of oxygen is limited), pyruvic acid is reduced to lac-tic acid in the muscles:

(Pyruvic acid is both a carboxylic acid and a ketone; only the ketone group isreduced.) The buildup of lactic acid during vigorous exercise is responsible in largepart for the muscle pain that we experience.

4.6 TitrationsIn Section 4.2, we described the neutralization of HCl(aq) with NaOH(aq). By whatprocess is neutralization achieved? In a procedure called titration, one reactant iscarefully added to another reactant until the two have combined in their exact stoi-chiometric proportions. Although titrations may differ in the nature of the chemicalreaction that occurs, all titrations have this in common: Both reactants are fully con-sumed at the end of the titration.

The objective of a titration is usually to find the number of moles or grams, theconcentration, or the percentage of the analyte (the substance we are looking for) in asample. This is usually done by measuring the precise volume and concentration ofa titrant (the solution being added) needed to react completely with the analyte. Stoi-chiometric relationships are then used to determine the quantity of analyte presentfrom the number of moles of titrant added. Let’s look at somecommon types of titrations.

Acid–Base TitrationsThe most frequently encountered titration is one in which one reactant is an acid and the other is a base. An example of an acid–base titration is pictured in Figure 4.16. The central piece of equipment in a titration is a buret, a long

1volume * molarity2

Pyruvic acidC

O O

CH3 C OHLactic acid

CH

OH O

CH3 C OHreduction

3 CH3CCH33 CH3CHCH3 2 Cr3+ ++ 7 H2O Acetone

2-PropanoneIsopropyl alcohol

2-Propanol

OH O

8 H+Cr2O72− ++

(CH3)2CO,

Different titrations have beencombined into a single section in

this edition, in part because of the greatsimilarities between different types oftitrations. Emphasis of the similaritiesmay help the student understand alltypes of titrations.

Acid-Base Titration animation

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(a) (b) (c)

� FIGURE 4.16 The technique of titrationThe steps in the technique corresponding to the photographs (a), (b), and (c) are described in the text.

QUESTION: Adding water to increase the volume of the analyte solution before beginning the titration changes the concentrationof the analyte solution in the flask but does not affect the volume of titrant needed to reach the equivalence point. Why not?

graduated tube with a stopcock (valve) at one end. The buret is filled with thetitrant, and the solution containing the analyte is placed in a small flask. A fewdrops of an indicator are usually added to the flask. The indicator is chosen tosignal, through its change in color, when the equivalence point in the titration has been reached. The equivalence point in acid–base reactions occurs when the titrant completely neutralizes the analyte. Consider the titration pictured inFigure 4.16:

(a) A precisely measured volume of HCl(aq) of unknown concentration is deliveredinto the flask (with a device called a pipet). This is the analyte. A few drops of theacid–base indicator phenolphthalein are added. Phenolphthalein is colorless inacidic solution and pink in basic solution. Thus, the acidic solution in the flask isinitially colorless.

(b) NaOH(aq) from the buret is slowly added to the flask. At first, the HCl is in excessand the NaOH is the limiting reactant. After it is stirred, the solution remainscolorless, indicating that it is still acidic.

(c) When just enough NaOH has been added to react with all the HCl, the amountsof NaOH and HCl have been brought to their stoichiometric proportions,and the equivalence point has been reached. A tiny amount (less than one drop) of excess NaOH(aq) makes the solution basic, and we see a color change to pink; this color change marks the endpoint. At the endpoint, the titra-tion is stopped and the volume of NaOH(aq) solution delivered from the buret is recorded. From this information, the concentration of the HCl(aq) solutioncan be determined.

From this example, it should be apparent that a successful titration requires anindicator that changes color at (or very closely after) the equivalence point. We willdiscuss how to choose such an indicator in Chapter 15.

Acid–base titrations see widespread use. Swimming pool water, wastewater, coaland coal ash, protein in foods, some pollutants, and raw industrial ingredients can allbe analyzed by acid–base titrations. Happily, calculations involved in these titrationsare not really new calculations. They are simply different applications of the solutionstoichiometry calculations we already know.

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Example 4.9What volume (mL) of 0.2010 M NaOH is required to neutralize 20.00 mL of 0.1030 M HCl in an acid–base titration?

STRATEGY

We need to do four things to solve this problem. In general, other titration problems can be solved with a similar approach.

Step 1: Write an equation describing the neutralization and obtain a stoichiometric factor relating moles of NaOH and moles of HCl.

Step 2: Determine how many moles of HCl are to be neutralized.

Step 3: Find the number of moles of NaOH required in the neutralization.

Step 4: Determine the volume of the solution containing this number of moles of NaOH.

SOLUTION

Step 1: First write an equation for the neutralizationreaction.

NaOH(aq) + HCl(aq) ¡ NaCl(aq) + H2O(l)

From this equation, we relate the amount ofthe titrant, NaOH(aq), to a given amount ofthe analyte, HCl(aq), and write the appro-priate stoichiometric factor as a ratio.

1 mol NaOH

1 mol HCl

Step 2: The number of moles of HCl to be titrated isthe product of the volume, in liters, and themolarity of the HCl(aq). = 0.002060 mol HCl

? mol HCl = 0.02000 L HCl(aq) *0.1030 mol HCl

1 L HCl(aq)

Step 3: Now, we use the stoichiometric factor toconvert moles of HCl to moles of NaOH re-quired for the titration. = 0.002060 mol NaOH

? mol NaOH = 0.002060 mol HCl *1 mol NaOH

1 mol HCl

Step 4: Finally, we use the inverse of the molarity ofNaOH as a conversion factor to find the vol-ume of the NaOH(aq) and convert fromliters to milliliters.

= 10.25 ml NaOH(aq)

? mL NaOH(aq) = 0.002060 mol NaOH *1 L NaOH(aq)

0.2010 mol NaOH*

1000 mL

1 L

As in most stoichiometric calculations, we can combine these steps into a single setup:

ASSESSMENT

The molarity of the NaOH(aq) is just about twice that of the HCl(aq), and the combining mole ratio of the acid and base is Thissuggests that the volume of NaOH(aq) required should be just about one-half that of the HCl(aq), and it is: 10.25 mL of the NaOH(aq)neutralizes 20.00 mL of the HCl(aq).

1 : 1.

We want (?)and the unit

mL NaOH(aq).

0.02000 L HCl(aq)0.1030 mol HCl

1 L HCl(aq)? mL NaOH(aq) =

10.25 mL NaOH(aq)=

1 mol NaOH1 mol HCl

Molarity convertsL HCl(aq)

to mol HCI.

The answer:

(?) the unit

Stoichiometric factorconverts mol HCI

to mol NaOH.

We start withthe volume (in L) of

HCI(aq) to be titrated.

Inverse of molarityconverts moles NaOHto liters NaOH(aq).

ConvertsL NaOH(aq) tomL NaOH(aq).

1 L NaOH(aq)0.2010 mol NaOH

1000 mL NaOH(aq)1 L NaOH(aq)

× ×

× ×

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EXERCISE 4.9A

What volume of 0.01060 M HBr(aq) is required to neutralize 25.00 mL of 0.01580

EXERCISE 4.9B

A 2.000-g sample of a sulfuric acid solution that is 96.5% by mass is dissolved in a quantity of water and titrated. What vol-ume of 0.3580 M KOH(aq) is required for the titration? Assume that at the equivalence point the solution in the flask is K2SO4(aq).

H2SO4

2 HBr(aq) + Ba(OH)2(aq) ¡ BaBr2(aq) + 2 H2O(l)

M Ba(OH)2

Application NoteVinegar can contain from 4 to 10%acetic acid.The exact amount of aceticacid in a particular vinegar can bedetermined by titration with a base ofknown concentration.

Example 4.10A 10.00-mL sample of an aqueous solution of calcium hydroxide is neutralized by 23.30 mLof What is the molarity of the calcium hydroxide solution?

STRATEGY

To determine the molarity of the we need to determine how many moles ofare consumed in the titration and divide this number by the volume of the sample

(0.01000 L, or 10.00 mL). To determine the number of moles of we can proceedthrough much of the problem as we did in Example 4.9.

SOLUTION

Let’s start with a balanced equation for this reaction:

Nitric acid Calcium hydroxide Calcium nitrate Water

We can determine the number of moles of through a series of conversions:

We have just calculated the number of moles of found in a 10.00-mL (0.01000 L)sample. Now we can use the definition of molarity to write

EXERCISE 4.10A

What volume of 0.550 M NaOH(aq) is required to neutralize a 10.00-mL sample of vine-gar that is 4.12% by mass acetic acid, Assume that the vinegar has a densityof

EXERCISE 4.10B

A sample of battery acid is to be analyzed for its sulfuric acid content. A 1.00-mL samplehas a mass of 1.239 g. This 1.00-mL sample is diluted to 250.0 mL with water, and10.00 mL of the diluted acid is neutralized by 32.44 mL of 0.00986 M NaOH. What is themass percent in the battery acid?H2SO4

1.01 g>mL.CH3COOH?

Molarity =2.330 * 10-4 mol Ca(OH)2(aq)

0.01000 L= 0.02330 M Ca(OH)2

Ca(OH)2

2.330 10−4 mol Ca(OH)2=

The answer:

(?) the unit

We want (?)and the unitmol Ca(OH)2.

Converts mL to L

Molarityconverts L acidto mol HNO3.

Stoichiometricfactor convertsmol HNO3 tomol Ca(OH)2.

mL HNO3required intitration.

23.30 mL HNO3 ×1 L

1000 mL

0.02000 mol HNO3

1 L

1 mol Ca(OH)2

2 mol HNO3? mol Ca(OH)2 =

×

× ×

L HNO3(aq) mol HNO3 mol Ca(OH)2mL HNO3(aq)

Ca(OH)2

2 HNO3(aq) + Ca(OH)2(aq) ¡ Ca(NO3)2(aq) + 2 H2O(l)

Ca(OH)2 ,Ca(OH)2

Ca(OH)2(aq),

0.02000 M HNO3(aq).Emphasis: Students should beencouraged to write complete

units such as “moles NaOH”or “mL HCl.”Incomplete labeling as simply “moles”and “mL”can lead to endless confusion.

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Precipitation TitrationsA reaction that forms a precipitate may also be the basis for a titration. The most com-mon precipitation titrations involve an aqueous solution of silver ion, either as thetitrant or as the analyte. For example, in one type of precipitation titration the amountof chloride ion in an unknown solution is found by titrating the sample with silvernitrate solution. Insoluble silver chloride is formed during the titration. Potassiumchromate can be used as the indicator; red-brown silver chromate forms only after allof the chloride ions have precipitated from solution. The endpoint of this titration isindicated by the first appearance of this colored precipitate, as shown in Figure 4.17.

Example 4.11Suppose a 0.4096-g sample from a box of commercial table salt is dissolved in water and requires 49.57 mL of 0.1387 M to completely precipitate the chloride ion. If the chloride ion present in solution comes only from the sodium chloride, find the mass ofNaCl in the sample. Is commercial table salt pure sodium chloride?

STRATEGY

As with other solution stoichiometry problems, we can determine first the number of moles of NaCl and then the NaCl mass from thevolume and molarity of the the appropriate stoichiometric factor from the balanced equation for the precipitation reaction,and the molar mass of NaCl.

SOLUTION

AgNO3(aq),

AgNO3(aq)

We start with the balanced equation for the pre-cipitation reaction.

AgNO3(aq) + NaCl(aq) ¡ NaNO3(aq) + AgCl(s)

An outline of the problem illustrates our strategyfor calculating the mass of NaCl in the sample ofcommercial table salt.

¡ mol NaCl ¡ g NaClmL AgNO3(aq) ¡ L AgNO3(aq) ¡ mol AgNO3

In a first step, we calculate the moles of from the given volume and molarity.

AgNO3

= 6.875 * 10-3 mol AgNO3

? mol AgNO3 = 49.57 mL AgNO3(aq) *1 L

1000 mL*

0.1387 mol AgNO3

1 L AgNO3(aq)

In the second step, the mass of the analyte (NaCl)is calculated from the stoichiometric ratio ofNaCl to and the molar mass of NaCl.AgNO3

= 0.4018 g NaCl

? g NaCl = 6.875 * 10-3 mol AgNO3 *1 mol NaCl

1 mol AgNO3*

58.44 g NaCl

1 mol NaCl

ASSESSMENT

The calculated mass of the NaCl is less than the mass of the sample of table salt, as it should be because the sample is not pure NaCl.Also, notice that we did not need to use the mass of table salt in our calculation, although we would need that information if we wantedto find the mass percent of NaCl in the table salt.

> FIGURE 4.17A precipitation titration(Left) Silver nitrate solution in the buretreacts with chloride ion in the flask, pro-ducing a precipitate of AgCl (Center).The indicator is yellow (Right)At the endpoint, when the is con-sumed, we see the formation of a brick-red precipitate of Ag2CrO4.

Cl-CrO4

2-.

(a) (b) (c)

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Note that we could have combined the two steps into a single setup in which we would not have had to record the intermediateresult: The final answer proves to be the same by either method; 0.4018 g NaCl.

EXERCISE 4.11A

A solution containing a water-soluble salt of the radioactive element thorium can be titrated with oxalic acid solution to form insolu-ble thorium(IV) oxalate. If 25.00 mL of a solution of thorium(IV) ion requires 19.63 mL of 0.02500 M for complete precip-itation, find the molar concentration of thorium(IV) ion in the unknown solution.

EXERCISE 4.11B

A mixture of 0.1015 g of NaCl and 0.1324 g KCl is dissolved in water and titrated with 0.1500 M What volume of thewill be needed to reach the endpoint?AgNO3(aq)

AgNO3(aq).

H2C2O4

6.875 * 10-3 mol AgNO3 .

Problem-Solving NoteIn Examples 4.11 and 4.12, the mass of asample is given. A common error intitration calculations is to begin with thismass rather than a volume and molarity.Additionally,note that in Example 4.11this mass is not needed,and in Example4.12 the sample is not a pure substance.

Redox TitrationsIn a redox titration, one reactant—often the titrant—is an oxidizing agent and the otherreactant is a reducing agent. Permanganate ion, usually from is one of themost commonly used oxidizing agents in the chemical laboratory and makes an excel-lent titrant. When is used as a titrant, a separate indicator often is notrequired because even a slight excess of will produce a visible pink colorin the analyte solution when the endpoint of the titration is reached.

In a titration for determining the percent iron in an iron ore sample,can be used to oxidize to in an acidic solution:

The stoichiometric equivalence between the reactants is

The titration is illustrated in Figure 4.18, and sample data are presented in Exam-ple 4.12.

5 mol Fe2+ � 1 mol MnO4-

5 Fe2+(aq) + MnO4

-(aq) + 8 H+(aq) ¡ 5 Fe3+(aq) + Mn2+(aq) + 4 H2O(l)

Fe3+Fe2+MnO4

-(aq)

MnO4

-(aq)MnO4

-(aq)

KMnO4,

(a) (b) (c)

N FIGURE 4.18 A redox titrationusing permanganate ion as theoxidizing agent(a) The acidic solution in the flask (clear)contains an unknown amount of and the buret contains of aknown concentration (dark purple). (b) Asthe titrant is added to the solution,the intense color of each drop quickly dis-appears as reacts with (c) When all the has been oxidizedto the next drop of pro-duces a lasting pink (very light purple)color in the solution.

KMnO4(aq)Fe3+,Fe2+

Fe2+.MnO4

-

Fe2+

KMnO4(aq)Fe2+,

Example 4.12A 0.2865-g sample of an iron ore is dissolved in acid, and the iron is converted entirely to

To titrate the resulting solution, 0.02645 L of 0.02250 M is required.What is the mass percent of iron in the ore?

STRATEGY

We can work this problem in two parts. The first requires finding the number of grams of Fein the unknown solution from the titration data. This calculation is similar to ones we have

KMnO4(aq)Fe2+(aq).

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done for acid–base and precipitation titrations. The second part is a straightforward calcula-tion of a percentage, using the sample mass of 0.2865 g.

SOLUTION

The purpose of each factor in the calculation involving titration data is indicated in the fol-lowing outline:

Finally, the mass percent of iron is

ASSESSMENT

Because iron is only one component of iron ore, the mass of iron in the ore sample must beless than 0.2865 g and the percent iron must be less than 100%. These facts can be used toassess the plausibility of the results obtained in the calculation.

EXERCISE 4.12A

Suppose the titration in Example 4.12 was carried out with 0.02250 M ratherthan What volume of the would be required? The net ionicequation is

EXERCISE 4.12B

A 20.00-mL sample of is required to titrate 0.2378 g sodium oxalate in anacidic solution. How many milliliters of this same are required to titrate a25.00-mL sample of 0.1010 M in an acidic solution?

¡ 2 Mn2+(aq) + 8 H2O(l) + 10 CO2(g)

2 MnO4

-(aq) + 16 H+(aq) + 5 C2O4

2-(aq)

FeSO4

KMnO4(aq)KMnO4(aq)

6 Fe2+(aq) + Cr2O7

2-(aq) + 14 H+(aq) ¡ 6 Fe3+(aq) + 2 Cr3+(aq) + 7 H2O(l)

K2Cr2O7(aq)KMnO4(aq).K2Cr2O7(aq)

% Fe =0.1662 g Fe

0.2865 g iron ore* 100% = 58.01% Fe

0.02645 L × 0.02250 mol KMnO4

1 L? g Fe =

0.1662 g Fe=

The answer:

(?) the unit

We want(?) and theunit g Fe.

Stoichiometricfactor convertsmol MnO4

– tomol Fe2+.

Molarityconverts

L of titrantto mol KMnO4.

Factor toconvert

mol KMnO4to mol MnO4

–.

The volumeof titrant.

Factor toconvert mol Fe2+

to mol Fe.

Molar massconverts molFe to g Fe.

× ×1 mol Fe1 mol Fe2+

55.847 g Fe1 mol Fe

× 1 mol MnO4−

1 mol KMnO4× 5 mol Fe2+

1 mol MnO4−

Cumulative ExampleSodium nitrite is used in the production of fabric dyes, as a meat preservative, as a bleach for fibers, and in photography. It is preparedby passing nitrogen monoxide gas and oxygen gas into an aqueous solution of sodium carbonate. Carbon dioxide gas is the other prod-uct of the reaction. (a) Write a balanced equation for the reaction. (b) What mass of sodium nitrite should be produced in the reactionof 748 g of with the other reactants in excess? (c) In another preparation, the reactants are 225 mL of 1.50 M 22.1 g nitrogen monoxide, and excess What mass of sodium nitrite should be produced if the reaction has a yield of 95.1%?

STRATEGY

For part (a), we can write the formulas for reactants and products (Sections 2.6 and 2.7), and then construct and balance the equation(Section 3.7). Using the balanced equation, we can solve part (b), starting with grams of and ending with grams of (Section 3.8) Part (c) can be broken down into three steps: Convert volume of to moles using molarity (Section 3.11); solve asNa2CO3

NaNO2Na2CO3

O2 .Na2CO3(aq),Na2CO3 ,

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a limiting reactant problem (Section 3.9) for the mass of produced from NO and use that theoretical yield of productalong with percentage yield (Section 3.10) to find the actual yield.

SOLUTION

Na2CO3 ;NaNO2

(a) First, we write formulas for thereactants and products, then placethe formulas in their proper placesin the equation.

Reactants Products

NO + O2 + Na2CO3 ¡ NaNO2 + CO2

To balance, begin with an elementthat appears in only one compoundon each side (Na, for example).

NO + O2 + Na2CO3 ¡ 2 NaNO2 + CO2

Then balance the N atoms. 2 NO + O2 + Na2CO3 ¡ 2 NaNO2 + CO2

Next balance the O atoms. 2 NO + 12 O2 + Na2CO3 ¡ 2 NaNO2 + CO2

Finally, multiply through by 2 to re-move the fraction.

4 NO + O2 + 2 Na2CO3 ¡ 4 NaNO2 + 2 CO2

(b) A series of conversions is requiredfor this stoichiometric calculation.

g Na2CO3 ¡ mol Na2CO3 ¡ mol NaNO2 ¡ g NaNO2

We can perform the three stepseither individually or in a singlesetup as shown here. = 974 g NaNO2

? g NaNO2 = 748 g Na2CO3 *1 mol Na2CO3

106.0 g Na2CO3*

4 mol NaNO2

2 mol Na2CO3*

69.00 g NaNO2

1 mol NaNO2

(c) Convert volume to moles ofNa2CO3 . 225 mL *

1 L

1000 mL*

1.50 mol Na2CO3

1 L= 0.338 mol Na2CO3

Determine which is the limiting re-actant by comparing the yield fromthe with that from the NO.Na2CO3

22.1 g NO *1 mol NO

30.01 g NO*

4 mol NaNO2

4 mol NO= 0.736 mol NaNO2

0.338 mol Na2CO3 *4 mol NaNO2

2 mol Na2CO3= 0.676 mol NaNO2

Calculate the theoretical yield basedon the limiting reactant, Na2CO3 . 0.676 mol NaNO2 *

69.00 g NaNO2

1 mol NaNO2= 46.6 g NaNO2

Use the percentage yield (95.1%) tocalculate the actual or expected yield.

Actual yield (g) = 95.1% * 46.6 g NaNO2>100% = 44.3 g NaNO2

95.1% =actual yield (g)

46.6 g NaNO2* 100%

ASSESSMENT

The balanced equation has four N atoms, twelve O atoms, four Na atoms, and two C atoms on each side. Because the molar mass ofis roughly half that of and because the mole ratio between and is the mass of we get

in part (b) should be comparable to the mass of used. In part (c), because the yields of product from the two reactants are sosimilar, it is difficult to determine whether the limiting reactant is correct except by close inspection of the setup. However, we can notethat the actual yield of is indeed less than the theoretical yield, as it should be.NaNO2

Na2CO3

NaNO22 : 1,Na2CO3NaNO2Na2CO3NaNO2

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the reducing agent is oxidized. In adisproportionation reaction, thesame substance acts as both oxidizingagent and reducing agent. Strong oxi-dizing agents include a few nonmetalsand some species having atoms withhigh oxidation numbers. Strong re-ducing agents include the active met-als and some species having atomswith low oxidation numbers. Theactivity series of the metals ranksmetals in order of their strength as re-ducing agents. It can be used to pre-dict reactions between a metal and other metal ions in solution.

4.5 Applications of Oxidation and Reduction—In everyday life, peroxides and hypochlorites areencountered as oxidizing agents. In industry,oxygen gas, chlorine, and chlorine-containingcompounds are used as oxidizing agents. Antiox-idants such as vitamin C are reducing agents thatcan scavenge reactive free radicals. Photosynthe-sis is an important redox reaction. Redox reac-tions are used in organic chemistry to oxidize alcohols to aldehydesand ketones.

4.6 Titrations—A titration is a type of chemical analysis in whichtwo reactants are combined in exact stoichiometric proportions. In atitration, the volume of a solution of known concentration (the titrant)is added to a solution containing the analyte or sought-for substance.The equivalence point occurs when the reactants are present in stoi-chiometric proportions. Often an indicator is used to render theequivalence point visible through a color change. That color changeindicates the endpoint of the titration, Titrations can involveacid–base reactions, precipitation reactions, or redox reactions.

Assessment Goals 159

Concept Review with Key Terms

4.1 Some Electrical Properties of Aqueous Solutions—Solubleionic compounds are completely dissociated into ions in aqueous solu-tion and are therefore strong electrolytes. A few water-soluble molec-ular compounds are completely ionized in aqueous solution and arealso strong electrolytes. Most molecular compounds exist in solutioneither as molecules (nonelectrolytes) or as a mixture of molecules andions (weak electrolytes).

4.2 Reactions of Acids and Bases—A fewacids are strong acids; these acids are strongelectrolytes. However, most acids are weakacids (weak electrolytes). The common strongbases are water-soluble ionic hydroxides.Weak bases, like the weak acids, are molecu-lar compounds that exist as a mixture of molecules and ions in aque-ous solution. Many common weak bases are related to ammonia.

Neutralization reactions between acids and bases are conve-niently represented by ionic equations and net ionic equations. Netionic equations include only those ions that undergo a chemical reac-tion in solution; spectator ions are eliminated. The neutralizationreaction of an acid and a base produces water and an ionic compoundcalled a salt. The color of an indicator can be used to determinewhether a solution is acidic, basic, or neutral.

4.3 Reactions that Form Precipitates—Another important type of reaction in solutionis one in which ions combine to form an insol-uble solid—a precipitate. Solubility guide-lines can often be used to predict precipitationreactions (review Table 4.3). Qualitative andquantitative chemical analyses and industrialprocesses frequently make use of precipitationreactions.

4.4 Reactions Involving Oxidation andReduction—The oxidation number is thecharge on a monatomic ion; for species other than monatomic ions, itis a hypothetical charge on an atom, assigned by a set of rules.Oxidation is an increase in oxidation number accompanied by loss ofelectrons. Reduction is a decrease in oxidation number accompaniedby gain of electrons. The two processes occur simultaneously in aredox reaction. In a redox reaction, the oxidizing agent is reduced and

+

++

+

+

+––

––

––

KCaNaMgAlCrZnFeCdNiSnPb

CuAgHgAu

Powerful

Strength as a reducing agent

Strong

Good

Fair

Poor

Very poor

H2

Assessment Goals

When you have mastered the material in this chapter, you will be able to

• Identify strong electrolytes, weak electrolytes, and nonelectrolytes.

• Calculate ion concentrations in solutions of strong electrolytes.

• Classify a substance as an acid or base.

• Describe and identify strong and weak acids and strong and weakbases.

• Describe neutralization reactions.

• Identify spectator ions in a solution. Write a net ionic equation for areaction in solution.

• Cite and use the solubility guidelines in Table 4.3.

• Predict whether precipitation will occur when certain ionic com-pounds are present in the same solution.

• Solve stoichiometry problems based on precipitation reactions.• Assign oxidation numbers to elements in compounds or ions.• Recognize a redox reaction, and determine whether the equation for

the reaction is balanced.• Define and recognize oxidizing and reducing agents.• Use the activity series to predict the products of redox reactions in-

volving metals and metal ions.• Describe some oxidation–reduction reactions of practical

significance.• Describe how a titration is performed, and carry out calculations re-

lated to titrations.

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Self-Assessment Questions

1. The best electrical conductor of the following aqueous solutionsis (a) 0.10 M NaCl; (b) 0.10 M (c) 0.15 M

(d) 0.10 M

2. Identify each of the following substances as either a strong acid,a weak acid, a strong base, a weak base, or a salt.

(a) (d)

(b) HCl (e)

(c) LiOH (f)

3. Which of the following solutions has the highest and which hasthe lowest concentration of

(a) (c)

(b) (d)

4. Which of the following solutions has the highest total concen-tration of ions?

(a) (c)

(b) 0.030 M KCl (d)

5. Which of the following aqueous solutions is the best electricalconductor? Explain.

(a) 0.08 M NaCl (c)

(b) (d)

6. The greatest will be found in which of the following aque-ous solutions? (a) (b) (c) 0.10M NaOH, (d)

7. According to the Arrhenius theory, (a) are all hydrogen-containingcompounds acids? (b) Are all compounds containing OH groupsbases? Explain.

8. In each of the following pairs of mixtures, reaction occurs in onemixture but not the other. Explain why this is so, and write a netionic equation for the reaction that occurs.

(a)

or

(b)

or

9. When treated with dilute HCl(aq), which compound produces agaseous product? (a) ZnO, (b) (c) (d)

10. When ammonium carbonate is added to a zinc chloride solution,zinc ions will precipitate from solution. Write the (a) complete-formula equation, (b) ionic equation, and (c) net ionic equationfor this reaction.

11. Only one of the following compounds is insoluble in water.Which one must it be? Explain.

(a) (c)

(b) (d)

12. Which of the following compounds reacts to precipitate from an aqueous solution of Write an equation for the reaction.

(a) (c)

(b) NaI (d) NaNO3

Na2CO3Na2S

MgCl2 ?Mg2+

PbSO4ZnCl2

CuSO4Ba(NO3)2

Na2SO4.BaSO3,NaNO3,

ZnCl2(aq) + KOH(aq)

ZnCl2(aq) + MgSO4(aq)

NaOH(aq) + CH3NH2(aq)

HCl(aq) + CH3CH2NH2(aq)

0.10 M CH3NH2.0.10 M H2CO3,0.10 M HNO3,

[H+]

2.0 M C6H12O61.0 M CH3CH2OH

0.10 M CH3COOH

0.025 M K2SO4

0.040 M Al(NO3)30.012 M Al2(SO4)3

0.050 M Mg(NO3)20.040 M Ca(NO3)2

0.040 M Al(NO3)30.10 M KNO3

NO3

-?

CH3CH2COOH

NH4NO3

CH3NH2H3PO4

CH2OHCHOHCH2OH.CH3COOH;CH3CH2OH;

13. What simple chemical test can you perform to determine whethera particular barium compound is or

14. What is the usual oxidation number of hydrogen atoms in com-pounds? What is that of oxygen atoms in compounds? What aresome exceptions?

15. What happens to the oxidation number of one of its elementswhen a compound is oxidized, and when it is reduced?

16. Of the following metals, all will react with HCl(aq) except(a) Ca, (b) Cu, (c) Fe, (d) Zn.

17. In the reaction,which statement is true? (a) Cu is the

oxidizing agent; (b) is the oxidizing agent; (c) is the ox-idizing agent; (d) is reduced.

18. Indicate the oxidation number of the underlined atom in each ofthe following.

(a) (f)

(b) (g)

(c) (h)

(d) (i)

(e) (j)

19. Use the conventions on page 141 to determine the oxidationnumbers of the carbon atoms in the following organic compounds.

(a) (d)

(b) (e)

(c)

20. In the reaction

is the oxidized or reduced or neither? Explain.

21. A reaction occurs in one of these mixtures but not the other. Ex-plain why this is so, and write a net ionic equation for the reac-tion that occurs.

or

22. Both magnesium and aluminum react with an acidic solution toproduce hydrogen. Why is it that only one of the followingequations correctly describes the reaction?

and

23. What is the equivalence point in an acid–base titration? What isthe function of the indicator in the titration? What is the rela-tionship between an equivalence point and an endpoint?

24. The complete neutralization of 10.00 mL of 0.100 M requires (a) 10.00 mL of 0.100 M NaOH; (b) 100.0 mL of0.0200 M NaOH; (c) 5.00 mL of 0.100 M KOH; or(d) 20.00 mL of 0.100 M Ba(OH)2.

H2SO4(aq)

Al(s) + 2H+(aq) ¡ Al3+(aq) + H2(g)

Mg(s) + 2H+(aq) ¡ Mg2+(aq) + H2(g)

Au(s) + HCl(aq)Zn(s) + CH3COOH(aq)

H2SO4(aq)

CuSO4(aq) + 2 H2O(l) + SO2(g)Cu(s) + 2 H2SO4(aq) ¡

C2H2O2

C2H2O4CH2O2

C2H6OC2H6

NH2OHPH4

+S4O6

2-TeF6

P2O7

4-K2Se

SrTiO3ClO2

-CaRuO3Cr

SO4

2-H+SO2

2 H2O(l) + SO2(g),+Cu2+(aq)SO4

2-(aq) ¡+Cu(s) + 4 H+(aq)

BaCO3(s)BaSO4(s)

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Problems 161

Problems

Types of Electrolytes25. Identify each substance as a strong electrolyte, a weak elec-

trolyte, or a nonelectrolyte.

(a) HBr (d) HCOOH

(b) KCl (e) NaOH

(c) HI (f)26. Identify each substance as either a strong acid, a weak acid,

a strong base, a weak base, or a salt.

(a) (d)(b) KOH (e)(c) (f) CH3CH2NH2BaCl2

NH4I

CH3CH2COOHNa2SO4

Ca(NO3)2

27. Which, if any, of the following potassium compounds when dis-solved in water could cause the bulb in the apparatus shown inFigure 4.3 to glow brightly: the sulfate, the hydroxide, the car-bonate, the chloride, the acetate? Explain.

28. A solution of a substance causes the bulb in the apparatus shownin Figure 4.3 to glow weakly. Is the substance necessarily aweak electrolyte? Explain.

Concentrations in Aqueous Solutions29. Determine the molarity of each of the following.

(a) and in 0.647 M

(b) and in 0.035 M

(c) and in 1.07 M

30. Determine the molarity of each of the following.

(a) and in 0.231 M

(b) and in 0.850 M

(c) and in 0.2000 M

31. A solution is 0.0554 M NaCl and 0.0145 M What areand in this solution?

32. A solution is 0.015 M each in LiCl, and What is the molarity of each ion in this solution?

33. In what volume of solution must 16.11 g of be dissolvedto make a solution that is 0.1000 M in chloride ion?

34. In what volume of solution must 31.7 g of oxalic acid dihydrate,be dissolved to make a solution that is

0.0859 M in oxalic acid?

35. Without doing detailed calculations, place the following solu-tions in order from highest to lowest 0.10 M 0.040 M 0.047 M

36. Without doing detailed calculations, place the following solu-tions in order from highest to lowest total concentration of ions:0.030 M KCl, 0.025 M 0.040 M

37. An aqueous solution is prepared by dissolving 18.3 gin water and diluting to 285 mL of solution.

What is in this solution?

38. What is the total concentration of ions in the solution in Prob-lem 37?

[SO4

2-]MgSO4

# 7 H2O

Fe(NO3)3K2SO4,

Ca(NO3)2 .Al(NO3)3 ,KNO3,[NO3

-]:

H2C2O4# 2 H2O,

MgCl2

AlCl3 .Li2SO4,MgI2 ,

[SO4

2-][Cl-],[Na+],Na2SO4.

Fe(NH4)2(SO4)2SO4

2-Fe2+Mg(CH3COO)2CH3COO-Mg2+

KHCO3HCO3

-K+

Al2(SO4)3SO4

2-Al3+CaI2I-Ca2+

LiNO3NO3

-Li+39. The components of seawater are sometimes expressed in mil-

ligrams per liter Use the description of seawater givenin Exercise 4.1A to determine the chloride ion content of seawa-ter in mg seawater.

40. A unit commonly used to describe low concentrations of asolute is ppm (parts per million, meaning, for example, gramssolute per million grams of solution). If the concentration ofchloride ion in a municipal water supply is given as 30.6 ppm

what is the molarity of in the water? (Assume the den-sity of the water is )

41. What volume of 0.0250 M should be diluted to 250.0 mLto obtain a solution with

42. A solution is prepared by mixing 100.0 mL 0.438 M NaCl,100.0 mL 0.0512 M and 250.0 mL of water. What are

and in the resulting solution?

43. Without doing detailed calculations, place these solutions inorder from lowest to highest

(a) 0.21 M RbCl

(b) 0.45 M

(c) 1.20 moles of dissolved in 2.00 L solution

(d) a solution that is 0.15 M KCl and 0.35 M NaCl

44. Without doing detailed calculations, determine which of the fol-lowing contains the greatest mass of the element nitrogen.

(a) 1.00 L of 0.0020 M aluminum nitrate

(b) 500 mL of an ammonium nitrate solution containing 80 mg

(c) 100 mL of a 0.10% by mass magnesium nitrate solution1d = 1.00 g>mL2.

N>L

MgCl2

FeCl3# 9 H2O

[Cl-].

[Cl-][Mg2+],[Na+],MgCl2 ,

[Cl-] = 0.0135 M?MgCl2

1.00g>mL.Cl-Cl-,

Cl->L

(mg>L).

Acid–Base Reactions45. Write equations to show the ionization of the following aqueous

acids and bases.

(a) HBr (d)(b) LiOH (e)(c) HF (f) HCOOH

(CH3)2NH

HIO3

46. Write equations to show the ionization of the following aqueousacids and bases.

(a) (d)(b) (e)(c) (f) HClO4Ba(OH)2

HSO4

-CH3(CH2)2COOH(aq)

CH3CH2NH2HNO2

HIL04-124-167v4 1/13/04 12:11 PM Page 161


47. Without performing detailed calculations, place the followingaqueous solutions in order from lowest to highest concentrationof ion.

(a) 0.10 M HCl (c) 0.10 M

(b) 0.10 M (d) 0.15 M

48. Consider the four solutions in Problem 47. Might the order beaffected if (a) the HCl was changed to 0.11 M; (b) the waschanged to 0.050 M; (c) the was changed to0.090 M? Explain.

49. Which of the following have the same net ionic reaction asHCl(aq) and NaOH(aq), (a) and (b) and (c) and HBr(aq)?Write a net ionic equation for the reaction that occurs ineach case.

NH3(aq)HClO4(aq).Ba(OH)2(aq)HNO3(aq).CH3COOH(aq)

CH3COOHNH3

NH3H2SO4

CH3COOH

H+

50. Strontium iodide can be made by the reaction of solid strontiumcarbonate with hydroiodic acid. Write the (a) complete-formulaequation, (b) ionic equation, and (c) net ionic equation for this reaction.

51. Lime deposits on brass faucets are mostly Though usu-ally white, the deposits may be slightly green from copper in thebrass [recall Figure 2.1(b)]. The deposits may be removed bysoaking the faucet in hydrochloric acid. (The commercial prod-uct is often called muriatic acid.) Write a net ionic equation forthe reaction that occurs.

52. A paste of sodium hydrogen carbonate (sodium bicarbonate)and water can be used to relieve the pain of an ant bite. The irri-tant in the ant bite is formic acid (HCOOH). Write a net ionicequation for the reaction that occurs.

CaCO3.

Ionic Equations53. Each equation represents the mixing of two aqueous solutions.

Complete each as a net ionic equation. If no reaction occurs,write NR.

(a)

(b)

(c)

(d)

(e)

(f)

54. Each equation represents the mixing of two aqueous solutions.Complete each as a net ionic equation. If no reaction occurs,write NR.

(a)

(b)

(c)

(d)

(e)

(f) CH3CH2COOH(aq) + Ba2+(aq) + 2 OH-(aq) ¡

Na+(aq) + OH-(aq) + Mg2+(aq) + 2 Cl-(aq) ¡

K+(aq) + OH-(aq) + Na+(aq) + HSO4

-(aq) ¡

2 Na+(aq) + SO4

2-(aq) + Cu2+(aq) + 2 Cl-(aq) ¡

+ SO4

2-(aq) ¡Pb2+(aq) + 2 NO3

-(aq) + Mg2+(aq)

Ba2+(aq) + 2 Cl-(aq) + 2 Na+(aq) + CO3

2-(aq) ¡

K+(aq) + HSO4

-(aq) + Na+(aq) + OH-(aq) ¡

Ba2+(aq) + 2 OH-(aq) + H+(aq) + I-(aq) ¡

H+(aq) + Cl-(aq) + CH3COOH(aq) ¡

Cr3+(aq) + 3 Cl-(aq) + Li+(aq) + OH-(aq) ¡

Mg2+(aq) + 2 Br-(aq) + Zn2+(aq) + SO4

2-(aq) ¡

K+(aq) + I-(aq) + Pb2+(aq) + 2 NO3

-(aq) ¡

55. Predict whether a reaction is likely to occur in each of the fol-lowing cases. If so, write a net ionic equation for the reaction.

(a)

(b)

(c)

(d)

(e)

56. Predict whether a reaction is likely to occur in each of the fol-lowing cases. If so, write a net ionic equation for the reaction.

(a)

(b)

(c)

(d)

(e) NaOH(aq) + Mg(NO3)2(aq) ¡

MgBr2(aq) + ZnSO4(aq) ¡

NH3(aq) + H2SO4(aq) ¡

Cr(OH)3(s) + HBr(aq) ¡

BaS(aq) + CuSO4(aq) ¡

KBr(aq) + Zn(NO3)2(aq) ¡

CuSO4(aq) + Na2CO3(aq) ¡

CH3COOH(aq) + H2SO4(aq) ¡

HCOOH(aq) + NH3(aq) ¡

Mg(OH)2(s) + HI(aq) ¡

Solubility Guidelines and Precipitation Reactions57. Classify the following as being soluble or insoluble in water:

(a) (c)

(b) (d)

58. Which of the following compounds reacts to precipitate from an aqueous solution of Write an equation for the reaction.

(a) (c)

(b) KBr (d)

59. You suspect that a certain white powder is either orYou add dilute HCl(aq) and obtain a clear solution

as shown on the right. Does the test indicate what the powder is?If not, what test would you perform instead? Explain.

Mg(OH)2(s).MgSO4(s)

NaNO3

Na2CO3Na2SO4

FeCl2 ?Fe2+

PbSCuCl2

CuSO4Sr(NO3)2

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Problems 163

60. You suspect that a particular solution is either orYou add dilute KOH(aq) and obtain the result

shown. Does the test indicate what the solution is? If not, whattest would you perform instead? Explain.

Cu(NO3)2(aq).CuCl2(aq) 61. When aqueous solutions of copper(II) nitrate and potassium car-

bonate are mixed, a precipitate forms. Write the net ionic equa-tion for this reaction.

62. When aqueous solutions of iron(III) chloride and sodium sulfideare mixed, a precipitate forms. Write the net ionic equation forthis reaction.

63. You suspect that a particular unlabeled aqueous solution is oneof the following: or Ex-plain how you can use precipitation reactions on small test sam-ples of the solution to determine its identity. You have availablethe variety of aqueous solutions usually found in a generalchemistry laboratory.

64. The addition of to an aqueous solution containing asingle unknown ionic compound as a solute produces a whiteprecipitate. List three compounds that the unknown might beand three that it cannot be. Explain your choices.

MgCl2(g)

Ba(NO3)2(aq).NH3(aq)Na2SO4(aq),

Oxidation–Reduction Reactions65. Indicate whether the first-named substance in each change un-

dergoes an oxidation, a reduction, or neither. Explain yourreasoning.

(a) Blue changes to green when exposedto air.

(b) Yellow changes to orange whenacidified.

(c) Dinitrogen pentoxide produces nitric acid when it reactswith water.

66. Indicate whether the first-named substance in each change un-dergoes an oxidation, a reduction, or neither. Explain yourreasoning.

(a) Sulfur trioxide gas produces sulfuric acid when passed intowater.

(b) Nitrogen dioxide converts to dinitrogen tetroxide whencooled.

(c) Carbon monoxide is converted to methane in the presence ofhydrogen.

67. Balance the following redox equations by inspection.

(a)

(b)

(c)

(d)

(e)

68. Balance the following redox equations by inspection, except inany case where the reaction is not possible.

(a)

(b)

(c) SeO3

2- + I- + H+ ¡ Se + I2 + H2O

Ca(ClO)2 + HCl ¡ CaCl2 + H2O + Cl2

CH4 + NO2 ¡ N2 + CO2 + H2O

IO4

- + I- + H+ ¡ I2 + H2O

Ag + H+ + NO3

- ¡ Ag+ + H2O + NO

CH4 + NO ¡ N2 + CO2 + H2O

NO + H2 ¡ NH3 + H2O

HCl + O2 ¡ Cl2 + H2O

K2Cr2O7(aq)K2CrO4(aq)

CrCl3(aq)CrCl2(aq)

(d)

(e)

69. Identify the oxidizing and reducing agents in Problem 67.

70. Identify the oxidizing and reducing agents in Problem 68.

71. The reactions represented by the following equations cannotoccur as written. Why?

(a)

(b)

72. The reactions represented by the following equations cannotoccur as written. Why?

(a)

(b)

73. Use the activity series of the metals to predict chemical reac-tions in the following cases. Write a plausible balanced equationfor each reaction that does occur and NR for those that do not.

(a)

(b)

(c)

(d)

74. An unknown metal M gives the following results in some labo-ratory tests. Use these data to find the approximate location ofM in the activity series on page 146.

M(s) + Zn2+(aq) ¡ no reaction

2 Al(s) + 3 M2+(aq) ¡ 2 Al3+(aq) + 3 M(s)

M(s) + Fe2+(aq) ¡ M2+(aq) + Fe(s)

M(s) + Cu2+(aq) ¡ M2+(aq) + Cu(s)

M(s) + 2 H+(aq) ¡ M2+(aq) + H2(g)

Au(s) + H+(aq) ¡Fe(s) + Ag+(aq) ¡Cu(s) + Zn2+(aq) ¡Zn(s) + H+(aq) ¡

NH3(g) + H2O(g) ¡ N2H4(g) + NO2(g)

¡ H2O(l) + Cl-(aq)O3(g) + ClO3

-(aq) + OH-(aq)

Fe2S3(s) + H2O(l) ¡ Fe(OH)3(s) + S(s)

PbO + V3+ + H2O ¡ PbO2 + VO2+ + H+

Zn + Cr2O7

2- + H+ ¡ Zn2+ + Cr3+ + H2O

Fe2+ + NO3

- + H+ ¡ Fe3+ + H2O + NO

Titrations75. How many milliliters of 0.0195 M HCl are required to titrate

(a) 25.00 mL of 0.0365 M KOH(aq), (b) 10.00 mL of 0.0116 M(c) 20.00 mL of 0.0225 M NH3(aq)?Ca(OH)2(aq),

76. How many milliliters of 0.0108 M are required totitrate (a) 20.00 mL of 0.0265 M (b) 25.00 mL of0.0213 M HCl(aq), (c) 10.00 mL of 0.0868 M CH3COOH(aq)?

H2SO4(aq),Ba(OH)2(aq)

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77. Vinegar is an aqueous solution of acetic acid,A 10.00-mL sample of a particular vinegar requires 31.45 mL of0.2560 M KOH for its titration. What is the molarity of aceticacid in the vinegar?

78. Most window cleaners are aqueous solutions of ammonia. A 10.00-mL sample of a particular window cleaner requires39.95 mL of 0.1008 M HCl for its titration. What is the molarityof ammonia in the window cleaner?

79. A tablet of a dietary supplement containing calcium carbonate isfound to neutralize 38.8 mL of 0.251 M HCl, forming calciumchloride, water, and Calculate the number of milligrams of(a) and (b) in the tablet.

80. A 5.00% NaOH solution by mass has a density of What is the minimum molarity of an HCl(aq) solution that canbe used to titrate a 5.00-mL sample of the NaOH(aq) if the titra-tion is to be accomplished without having to refill a 50.00-mLburet used in the titration?

81. Without doing detailed calculations, determine which of the fol-lowing popular antacids is able to neutralize more stomach acid[dilute HCl(aq)] if equal masses are compared: Alka-Seltzer®

(sodium hydrogen carbonate) or Tums® (calcium carbonate).

82. Without doing detailed calculations, determine which of the fol-lowing is more effective in reducing the acidity of a home swim-ming pool if equal masses are compared: caustic soda (sodiumhydroxide) or soda ash (sodium carbonate).

83. Which of the following points in the titration of 10.00 mL of1.00 M with 0.500 M NaOH would produce a solu-tion with the “molecular view” like the figure? After the volumeof 0.5000 M NaOH added is (a) 0.00 mL, (b) 5.00 mL,(c) 20.00 mL, (d) 22.00 mL, (e) 30.00 mL. Explain.

= Na+

= OH−

= CH3COOH

= CH3COO−

CH3COOH

1.054 g>mL.

Ca2+CaCO3

CO2.

CH3COOH. 84. Refer to Problem 83. In a similar manner to the figure shown,draw a sketch to represent the titration mixture after 10.00 mL of0.5000 M NaOH has been added.

85. How many milliliters of 0.02091 M would be needed totitrate (a) 25.00 mL of 0.1235 M KI, (b) 40.00 mL of 0.01944 M

(c) 0.0323 g of

86. How many milliliters of 0.1467 M would be neededto titrate each of the samples given in Problem 85?

87. Titration of 0.2558 g of that has been dissolved in waterrequires 41.60 mL of a solution of Write the bal-anced net ionic equation for the reaction, and calculate the molarconcentration of the solution.

88. A sample of NaCl weighing 1.4477 g is dissolved in water anddiluted to 250.0 mL. What volume of this solution will beneeded to titrate 25.00 mL of 0.1000 M

89. How many milliliters of 0.1050 M are required forthe titration of (a) 20.00 mL of 0.3252 M and (b) a1.065-g sample of The balanced equations are:

90. For the reactions described in Problem 89, what is the molarityof if (a) 22.55 mL is required to titrate10.00 mL of 0.2434 M and (b) if a 567.4 mg sampleof requires 31.61 mL for its titration?

91. The concentration of can be determined by titrationwith in basic solution. The balanced equation is

A 25.00-mL sample of requires 34.77 mL of0.05876 M for its titration. What is the molarity of the

92. The concentration of is to be determined by titra-tion against A 0.1156-g sample of requires27.08 mL of the for its titration. The balanced equa-tion is given below. What is the molarity of the

¡ 10 H3AsO4 + 4 Mn2+5 As2O3 + 4 MnO4

- + 9 H2O + 12 H+

KMnO4(aq)?KMnO4(aq)

As2O3(s)As2O3(s).KMnO4(aq)

Mn2+(aq)?KMnO4(aq)

Mn2+(aq)

3 Mn2+ + 2 MnO4

- + 4 OH- ¡ 5 MnO2(s) + 2 H2O

MnO4

-(aq)Mn2+(aq)

KMnO4(aq)KNO2

FeSO4(aq),KMnO4(aq)KMnO4(aq)

5 NO2

- + 2 MnO4

- + 6 H+ ¡ 2 Mn2+ + 5 NO3

- + 3 H2O

5 Fe2+ + MnO4

- + 8 H+ ¡ 5 Fe3+ + Mn2+ + 4 H2O

KNO2 ?Fe2+(aq),

KMnO4(aq)

AgNO3(aq)?

Ba(ClO4)2

Ba(ClO4)2 .Na2SO4

Pb(NO3)2

Na2CO3 ?FeCl3 ,

AgNO3

Additional Problems

97. A white solid is known to be either orAn aqueous solution prepared from the solid yields

a white precipitate when treated with What must thesolid be?

98. What reagent solution (including pure water) would you use toseparate the cations in the following pairs, that is, with onecation appearing in solution and the other in a precipitate?

(a) and

(b) and

(c) and

(d) and

(e) and BaSO4(s)Mg(OH)2(s)

CuCO3(s)PbSO4(s)

KNO3(s)AgNO3(s)

Na2CO3(s)MgCO3(s)

NaCl(s)BaCl2(s)

Ba(NO3)2 .Mg(NO3)2 .

MgSO4,MgCl2 ,Problems marked with an * may be more challenging than others.

93. A sample of human blood serum has a density of and contains 18.9 mg of and 365 mg of per 100 mL. Cal-culate the molar concentrations of and in the sample.

94. Would nitric acid and acetic acid be equally effective and usefulfor removing lime deposits on a brass faucet? (See Problem 51.)Explain.

95. A sample of ordinary table salt is 98.8% NaCl and 1.2% by mass. What is if 6.85 g of this mixture is dissolved in500.0 mL of an aqueous solution?

96. A solution is 0.0240 M KI and 0.0146 M What volume ofwater should be added to 100.0 mL of this solution to produce asolution with [I-] = 0.0500 M?

MgI2.

[Cl-]MgCl2

Cl-K+Cl-K+

1.022 g>mL

HIL04-124-167v4 1/13/04 12:11 PM Page 164

109. Methanol can reduce chlorate ion to chlorine dioxidein an acidic solution. The methanol is oxidized to carbon diox-ide. What volume of methanol in milli-liters, is needed to produce 125 kg The balancedequation for the reaction is

110. What volume of 0.185 M must be added to 235 mL of0.206 M KCl to produce a solution with a concentration of0.250 M

111. The exact concentration of an aqueous solution of oxalic acid(HOOCCOOH, that is, ) is determined by an acid–basetitration. Then the oxalic acid solution is used to determine theconcentration of by a redox titration in acidic solu-tion. The titration of 25.00-mL samples of the oxalic acidsolution requires 32.15 mL of 0.1050 M NaOH and 28.12 mL ofthe What is the molarity of the (Hint:The balanced equation can be derived from Problem 104)

112. A piece of marble (assume it to be pure ) reacts with2.00 L of 2.52 M HCl. After dissolution of the marble, a10.00-mL sample of the remaining HCl(aq) is withdrawn, addedto some water, and titrated with 24.87 mL of 0.9987 M NaOH.What must have been the mass of the piece of marble? Com-ment on the precision of this method; that is, how many signifi-cant figures are justified in the result?

113. A sample of impure potassium hydroxide consists of 92.25%KOH, 2.25% and 5.50% by mass. A 1.250-g sam-ple of this solid is allowed to react with 25.00 mL of 1.840 MHCl. The excess HCl is neutralized with 1.050 M KOH, and thesolution is then evaporated to dryness. What solid residue is ob-tained, and what is its mass?

114. Feldspars are a group of rock-forming silicate minerals thatmake up over half of Earth’s crust. Two elements commonlyfound in feldspars are sodium and potassium. In the analysis offeldspars, the percentages of sodium and potassium are fre-quently given as % and % (see Chapter 3, Prob-lem 128). The sodium and potassium content of a 0.7500-gsample of a feldspar mineral is obtained as a 0.2250 g mixture ofNaCl(s) and KCl(s). The chloride mixture is dissolved in waterand requires 22.61 mL of 0.1525 M for completeprecipitation of the chloride ion as AgCl(s). Determine the masspercent of and in the mineral.

115. For use as a fuel for generating electrical power, natural gasmust be freed of sulfur impurities to comply with environmentalstandards. The sulfur content of a 4.476-g sample of natural gaswas determined by burning the sample in excess oxygen. Thegases produced were bubbled through a 3% solution of which oxidized the to sulfuric acid. Then 25.00 mL of0.00923 M NaOH, an amount in excess of that needed to neu-tralize the sulfuric acid, was added to the solution. The excessNaOH was titrated, requiring 13.33 mL of 0.01007 M HCl forthe titration. Calculate the percentage of sulfur in the sample.

SO2

H2O2,

K2ONa2O

AgNO3(aq)

K2ONa2O

H2OK2CO3,

CaCO3

KMnO4(aq)?KMnO4(aq).

KMnO4(aq)

H2C2O4

Cl-?

MgCl2

6 ClO2 + CO2 + 5 H2OCH3OH + 6 H+ + 6 ClO3

- ¡

ClO2(g)?1d = 0.791 g>mL2,

(CH3OH)

Additional Problems 165

99. A railroad tank car carrying of concentrated sulfu-ric acid derails and spills its load. The acid is 93.2% andhas a density of How many kilograms of sodiumcarbonate (soda ash) are needed to neutralize the acid? (Hint:What is the neutralization reaction?)

100. To 125 mL of 1.05 M is added 75 mL of 4.5 MHCl(aq). Then the solution is evaporated to dryness. What massof NaCl(s) is obtained?

101. A 15,000-gallon home swimming pool is disinfected by thedaily addition of 0.50 gal of a “chlorine” solution—NaOCl inNaOH(aq). To maintain the proper acidity in the pool, the basiccomponents in the chlorine solution must be neutralized. By ex-periment, it is found that about 220 mL of an HCl(aq) solutionthat is 31.4% HCl by mass is required to neu-tralize 0.50 gal of the chlorine solution. What is the ofthe chlorine solution?

102. What is the molarity of in the solution formed by mixing25.10 mL of 0.2455 M NaOH and 35.05 mL of 0.1524 M

103. A 0.235-g sample of a solid that is 92.5% NaOH and 7.5%requires 45.6 mL of an HCl(aq) solution for its titra-

tion. What is the molarity of the HCl(aq)?

104. To titrate a 5.00-mL sample of a saturated aqueous solution ofsodium oxalate, requires 25.82 mL of 0.02140 M

How many grams of would be present in250.0 mL of the saturated solution? The balanced equation is

105. The compound has half its Mn atoms with oxida-tion number and half with oxidation number What is theoxidation number of the neodymium?

106. Biochemists sometimes describe reduction as the gain of Hatoms. Use examples from the text to show that this definitionconforms to that based on oxidation numbers.

107. A 25.00-mL sample of 0.1996 M solution requires48.97 mL of 0.3000 M for titration. The is convertedto in the titration reaction. What is the oxidation numberfor vanadium ion after the titration is complete?

108. A stock buffer solution is prepared by dissolving 31.5 g ofand 282 mL of concentrated ammonia in water and di-

luting to 1.00 liter. The concentrated ammonia solutionis 17.0 M Ten milliliters of the buffer

solution is added to 40.0 mL of This mixture istitrated with 0.01000 M ethylenediamine tetraacetate ion, sym-bolized here as

It takes 38.26 mL of the solution to reach the equivalencepoint. Use the data given, as needed, to calculate (a) the mass of

in the (b) the concentration of chloride ionin the final solution after titration. Assume that the volumes ofsolution may be added to obtain the total volume.

MgCl2(aq),MgCl2

Y4-

Mg2+ + Y4- ¡ MgY2-

Y4-.

MgCl2(aq).NH3.1d = 0.899 g>mL2

NH4Cl

Ce3+Ce4+Ce4+

V2+

+4.+3NdCaMn2O6

2 Mn2+(aq) + 8 H2O(l) + 10 CO2(g)

5 C2O4

2-(aq) + 2 MnO4

-(aq) + 16 H+(aq) ¡

Na2C2O4KMnO4(aq).Na2C2O4,

Ca(OH)2

HNO3 ?

OH-11 gal = 3.785 L2

[OH-]1d = 1.16 g>mL2

Na2CO3(aq)

1.84 g>mL.H2SO4

1.5 * 103 L

*

*

*

*

*

*

*

HIL04-124-167v4 1/13/04 12:11 PM Page 165


Apply Your Knowledge

) and an average depth of 4.45 m? (d) What massof oxygen would be consumed by the decay of the amount ofalgal protoplasm in (c), assuming the decay is the reverse of theabove reaction? (e) If the algae die when the dissolved oxygenconcentration reaches 15.0 mg would the algal decaydeplete the dissolved oxygen in the pond? (f) If phosphates fromsewage entering the lake raise the concentration to

what mass of algal protoplasm would be formed,assuming that the is the limiting reactant?

121. [Laboratory] Although Figure 4.6 gives a dramatic picture,chemists use precise measurements to determine how well a so-lution conducts electric current and how the conductancechanges as the result of a chemical reaction. For our purposes,we do not need the details of the method used; we just need tonote the ideas that follow.

The conductivity of a solution depends on the total concen-tration of ions.

Ions differ in their individual abilities to carry electriccurrent.

and are much better electrical conductors than otherions.

Shown here are four idealized graphs of the electrical conduc-tance of a solution during the course of a titration. For example,(a) represents the titration of HCl(aq) with NaOH(aq)

The conductance starts high because is greatest at the start. The conductance decreases during the titration because reacts with to form the nonelectrolyte and isreplaced by which does not conduct current nearly as well.The conductance falls to a minimum at the equivalence point,where the solution is simply NaCl(aq). Beyond the equivalence

Na+,H+H2O,OH-

H+[H+]

Con

duct

ance

mL titrant

Con

duct

ance

mL titrant(a) HCl(aq) titrated with NaOH(aq) (b) ?

Con

duct

ance

mL titrant

Con

duct

ance

mL titrant(c) ? (d) ?

H+ + Cl- + Na+ + OH- ¡ Na+ + Cl- + H2O

OH-H+

HPO4

2-110 mg>L,

HPO4

2-

O2>L,

1 hm = 100 m116. [Biochemical] The Kjeldahl analysis is used to determine theprotein content of foods. The food analyzed is heated with sul-furic acid in the presence of a catalyst. The carbon in the proteinis converted to the hydrogen to water, and the nitrogento The mixture is then treated with concentratedNaOH(aq) and heated to drive off The is neu-tralized by an excess of a standard acid, and the acid presentafter the neutralization is titrated with a standard base. From thetitration data and the known mass of the sample, the % N can becalculated. Based on the fact that the average percentage of ni-trogen in most plant and animal protein is about 16%, the per-cent protein is obtained by multiplying the % N by the factor6.25. A 2.500-g sample of meat is subjected to Kjeldahl analy-sis. The liberated is absorbed in 50.00 mL of

The excess acid requires 19.90 mL of 0.5510 MNaOH for its complete neutralization. A separate 25.00-mLsample of the requires 22.65 mL of the 0.5510 MNaOH for its titration. What is the percent protein in the meat?

117. [Laboratory] Back titration may be performed when a reactionis too slow for direct titration. An antacid tablet weighing 0.7023grams and containing calcium carbonate plus starch andsweeteners was added to 50.00 mL of 0.3000 M HCl

in a flask. After the reaction was complete,the excess HCl in the flask required 28.06 mL of 0.1200 MNaOH for titration to the equivalence point.Use the data given, as needed, to calculate the mass of inthe tablet.

118. [Laboratory] In a Volhard precipitation titration, a piece ofsterling silver alloy weighing 0.5039 grams was dissolved in10.0 mL of concentrated (15 M) nitric acid. The solution was di-luted to 60.0 mL with water, and 5.00 mL of 0.100 M Fe was added as an indicator. The contents of the flask were titratedwith 43.56 mL of 0.1005 M KSCN to reach the red endpoint.Write the net ionic equation for the titration reaction (the prod-uct of the titration reaction was AgSCN precipitate), and use thedata given as needed to calculate the percentage of silver in the alloy.

119. [Environmental] Incineration of a chlorine-containing toxicwaste such as a polychlorinated biphenyl (PCB) produces and HCl.

Balance the equation for this combustion reaction. Comment onthe advantages and disadvantages of incineration as a method ofdisposal of such wastes.

120. [Environmental] Algal protoplasm is a complex mixture ofmany substances, but its composition can be represented by a“formula,” The protoplasm is produced dur-ing photosynthesis by a process that we can represent by an“equation.”

(a) Balance the equation. (b) Which is the limiting nutrient,or in a lake that has the following concentra-

tions: and (c) Whatmass of algal protoplasm would be produced in a pond wih an area of 2.55 hectares (1 hectare 1 square hectometer;=

65.0 mg>L?HPO4

2-,434 mg>L;NO3

-,HPO4

2-,NO3

-

+ energy ¡ C106H263O110N16P + O2

CO2 + H2O + H+ + NO3

- + HPO4

2- + trace elements

C106H263O110N16P.

C12H4Cl6 + O2 ¡ CO2 + HCl + H2O

CO2

(NO3)3

CaCO3

1d = 1.01 g>mL2

1d = 1.03 g>mL2

H2SO4(aq)

H2SO4(aq).NH3(g)

NH3(g)NH3(g).(NH4)2SO4.

CO2(g),

*

*

*

*

HIL04-124-167v4 1/13/04 12:11 PM Page 166

e-Media Problems 167

point, the conductance again rises because of the accumulationof excess which is a very good electrical conductor.

Match each of the following titrations to the appropriateremaining graph: (b), (c), or (d). In the manner used for graph(a), explain the general shape of each graph.

122. [Laboratory] To be most effective as an antiseptic solution,should contain 3.0% by mass. The following

experiment is done to determine if a particular hydrogen per-oxide solution is at full strength. A 10.00-mL sample of theaqueous solution is treated with an excess of KI(aq). The liber-

H2O2H2O2(aq)

Ba(OH)2 with (NH4)2SO4 as the titrant

CH3COOH(aq) with NH3 (aq) as the titrant

NH3(aq) with HCl (aq) as the titrant

OH-,ated forms the triiodide ion, and the triiodide-ion solu-tion requires 28.91 mL of 0.1522 M for its titration.Is the at full strength? (Assume that the density ofthe is .)

(not balanced)

(not balanced)

123. [Laboratory] In the discussion of redox titrations, it was statedthat the titrant is usually the oxidizing agent. In such a titration,the analyte must first be converted entirely to a single low oxi-dation state. Once the analyte is so converted, it is usuallytitrated immediately and as quickly as possible. Suggest practi-cal reasons for this procedure.

I3

-(aq) + S2O3

2-(aq) ¡ S4O6

2-(aq) + I-(aq)

H2O2(aq) + H+(aq) + I-(aq) ¡ H2O(l) + I3

-(aq)

1.00 g>mLH2O2(aq)H2O2(aq)

Na2S2O3

I3

-,I2

e-Media Problems

The activities described in these problems can be found in the e-MediaActivities and Interactive Student Tutorial (IST) modules of theCompanion Website, http://chem.prenhall.com/hillpetrucci.

124. View the Electrolytes and Nonelectrolytes and Dissolution ofNaCl in Water animations (Section 4–1). Describe on an atom-ic scale the difference between solid sodium chloride and anaqueous solution of sodium chloride. What property of the solution is responsible for its being categorized as an electrolyte?

125. From the Strong and Weak Electrolytes movie (Section 4–1),what can you deduce from the experimental observation (bright-ness of the lightbulb) about the relative fraction of ions in solu-tion for the hydrogen chloride and acetic acid solutions?

126. In the Precipitation Reactions movie (Section 4–3), what fac-tors will influence the amount of precipitate formed in the test

tube for each of the two reactions shown? Write a balancedchemical equation for each case, and discuss the properties ofthe solutions involved in the two reactions.

127. Consider the reaction that is depicted in the Formation of Sil-ver Crystals movie (Section 4–5). Predict the effect of changingthe wire placed in the solution from copper to zinc. Wouldchanging the wire from copper to gold produce a similar effect?Describe what would be observed if a solution of lead nitratewere instead added to the beaker containing the copper wire.

128. For the reaction shown in the Acid–Base Titrations animation(Section 4–6), calculate the concentration of each species foundin solution after 20.0 mL of the 0.100 M sodium hydroxide so-lution has been added to the acidic solution being titrated. Howdoes this differ from the solution concentrations after 60.0 mLof the sodium hydroxide solution has been added?

*

HIL04-124-167v4 1/13/04 12:11 PM Page 167

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