ch4

Bubbles of gas form as antablet dissolves in

water. The is produced by thereaction of citric acid, and sodium bicarbonate, in the tablet.

NaHCO3,H3C6H5O7,

CO2

Alka-Seltzer®CO2

Chapter4Aqueous Reactions and SolutionStoichiometry

BLB04-112-151-V4 2/21/02 1:33 PM Page 112

4.1 General Properties of Aqueous Solutions4.2 Precipitation Reactions4.3 Acid-Base Reactions4.4 Oxidation-Reduction Reactions4.5 Concentrations of Solutions4.6 Solution Stoichiometry and Chemical Analysis

ALMOST TWO THIRDS of our planet is covered bywater, and water is the most abundant substance inour bodies. Because water is so common, we tend totake its unique chemical and physical properties forgranted. We will see repeatedly throughout this text, however, that water possesses many unusual properties essential to support lifeon Earth.

One of the most important properties of water is its ability to dissolve a widevariety of substances. The water in nature, therefore, whether it is the purestdrinking water from the tap or water from a clear mountain stream, invariablycontains a variety of dissolved substances. Solutions in which water is the dis-solving medium are called aqueous solutions.

Many of the chemical reactions that take place within us and around usinvolve substances dissolved in water. Nutrients dissolved in blood are carried toour cells, where they enter into reactions that help keep us alive. Automobileparts rust when they come into frequent contact with aqueous solutions that con-tain various dissolved substances. Spectacular limestone caves (Figure 4.1 ») areformed by the dissolving action of underground water containing carbon dioxide,

(aq):

[4.1]

We saw in Chapter 3 a few simple types of chemical reactions andhow they are described. In this chapter we continue to examine chemi-

cal reactions by focusing on aqueous solutions. A great dealof important chemistry occurs in aqueous solutions, and

we need to learn the vocabulary and concepts used todescribe and understand this chemistry. In addition,

we will extend the concepts of stoichiometry that welearned in Chapter 3 by considering how solution con-

centrations can be expressed and used.

CaCO3(s) + H2O(l) + CO2(aq) ¡ Ca(HCO3)2(aq)

CO2

113

» What’s Ahead «• We begin by examining the nature

of the substances dissolved inwater, whether they exist in wateras ions, molecules, or as some mix-ture of the two. This information isnecessary to understand the natureof reactants in aqueous solutions.

• Three major types of chemicalprocesses occur in aqueous solu-tion: precipitation reactions, acid-base reactions, andoxidation-reduction reactions.

• Precipitation reactions are those inwhich soluble reactants yield aninsoluble product.

• Acid-base reactions are those inwhich ions are transferredbetween reactants.

• Oxidation-reduction reactions arethose in which electrons are trans-ferred between reactants.

• Reactions between ions can be rep-resented by ionic equations thatshow, for example, how ions cancombine to form precipitates, orhow they are removed from thesolution or changed in some otherway.

• After examining the common typesof chemical reactions and how theyare recognized and described, weconsider how the concentrations ofsolutions can be expressed.

• We conclude the chapter by exam-ining how the concepts of stoi-chiometry and concentration canbe used to determine the amountsor concentrations of varioussubstances.

H+

BLB04-112-151-V4 2/21/02 1:34 PM Page 113

114 Chapter 4 Aqueous Reactions and Solution Stoichiometry

Á Figure 4.1 When dissolves inwater, the resulting solution is slightlyacidic. Limestone caves are formed bythe dissolving action of this acidicsolution acting on in thelimestone.

CaCO3

CO2

(a) (b) (c)

��

�

��

�

�

noions

fewions

manyions

Á Figure 4.2 A device for detecting ions in solution. The ability of a solution to conductelectricity depends on the number of ions it contains. (a) A nonelectrolyte solution does notcontain ions, and the bulb does not light. (b and c) An electrolyte solution contains ions toserve as charge carriers, causing the bulb to light. If the solution contains a small number ofions, the bulb will be only dimly lit, as in (b). If the solution contains a large number of ions,the bulb will be brightly lit, as in (c).

Pure water and aqueous solutions ofnonelectrolytes are poor conductorsof electricity.

4.1 General Properties of Aqueous Solutions

A solution is a homogeneous mixture of two or more substances. • (Section 1.2)The substance present in greater quantity is usually called the solvent. The othersubstances in the solution are known as the solutes; they are said to be dissolvedin the solvent. When a small amount of sodium chloride (NaCl) is dissolved ina large quantity of water, for example, the water is the solvent and the sodiumchloride is the solute.

Electrolytic Properties

Imagine preparing two aqueous solutions—one by dissolving a teaspoon of tablesalt (sodium chloride) in a cup of water and the other by dissolving a teaspoonof table sugar (sucrose) in a cup of water. Both solutions are clear and colorless.How do they differ? One way, which might not be immediately obvious, is intheir electrical conductivity: The salt solution is a good conductor of electricity,whereas the sugar solution is not.

Whether or not a solution conducts electricity can be determined by usinga device such as that shown in Figure 4.2 ¥. To light the bulb, an electric currentmust flow between the two electrodes that are immersed in the solution.Although water itself is a poor conductor of electricity, the presence of ions caus-es aqueous solutions to become good conductors. Ions carry electrical chargefrom one electrode to another, completing the electrical circuit. Thus, the con-ductivity of NaCl solutions indicates the presence of ions in the solution, andthe lack of conductivity of sucrose solutions indicates the absence of ions. WhenNaCl dissolves in water, the solution contains and ions, each surroundedby water molecules. When sucrose dissolves in water, the solutioncontains only neutral sucrose molecules surrounded by water molecules.

A substance (such as NaCl) whose aqueous solutions contain ions is calledan electrolyte. A substance (such as ) that does not form ions inC12H22O11

(C12H22O11)Cl-Na+

ANIMATIONElectrolytes and Nonelectrolytes

3-D MODELSodium Chloride, Sucrose

BLB04-112-151-V4 2/21/02 1:34 PM Page 114

4.1 General Properties of Aqueous Solutions 115

�

��

��

�

�

�

�

�

�

��

��

�

��

��

�

��

�

�

�

��

�

�

�

��

�

��

�

��

(b)(a)

Á Figure 4.3 (a) Dissolution of an ionic compound. When an ionic compound dissolvesin water, molecules separate, surround, and disperse the ions into the liquid. (b) Methanol, a molecular compound, dissolves without forming ions. Themethanol molecules can be found by looking for the black spheres, which represent carbonatoms. In both parts (a) and (b), the water molecules have been moved apart so the soluteparticles can be seen more clearly.

CH3OH,H2O

solution is called a nonelectrolyte. The difference between NaCl and arises largely because NaCl is ionic, whereas is molecular.

Ionic Compounds in Water

Recall from Section 2.7 and especially Figure 2.23 that solid NaCl consists of an order-ly arrangement of and ions. When NaCl dissolves in water, each ion sepa-rates from the solid structure and disperses throughout the solution as shown inFigure 4.3(a) Á. The ionic solid dissociates into its component ions as it dissolves.

Water is a very effective solvent for ionic compounds. Although water is anelectrically neutral molecule, one end of the molecule (the O atom) is rich in elec-trons and thus possesses a partial negative charge. The other end (the H atoms)has a partial positive charge. Positive ions (cations) are attracted by the negativeend of and negative ions (anions) are attracted by the positive end. As anionic compound dissolves, the ions become surrounded by molecules asshown in Figure 4.3(a). This process helps stabilize the ions in solution and pre-vents cations and anions from recombining. Furthermore, because the ions andtheir shells of surrounding water molecules are free to move about, the ionsbecome dispersed uniformly throughout the solution.

We can usually predict the nature of the ions present in a solution of an ioniccompound from the chemical name of the substance. Sodium sulfate for example, dissociates into sodium ions and sulfate ions Youmust remember the formulas and charges of common ions (Tables 2.4 and 2.5)to understand the forms in which ionic compounds exist in aqueous solution.

Molecular Compounds in Water

When a molecular compound dissolves in water, the solution usually consists ofintact molecules dispersed throughout the solution. Consequently, mostmolecular compounds are nonelectrolytes. For example, a solution of methanol

in water consists entirely of molecules dispersed throughoutthe water [Figure 4.3(b)].

CH3OH(CH3OH)

(SO4

2-).(Na+)(Na2SO4),

H2OH2O,

Cl-Na+

C12H22O11

C12H22O11

3-D MODELEthanol

ANIMATIONDissolution of NaCl in Water

BLB04-112-151-V4 2/21/02 1:34 PM Page 115

2�

2�

2�2�

�

�

�

�

�

��

�


The difference between strongelectrolytes and weak electrolytes isqualitative and somewhat arbitrary.

The symbol for equilibrium ( ) isnot to be confused with the double-headed arrow ( ). Students oftenincorrectly interchange thesesymbols.

4

L

There are, however, a few molecular substances whose aqueous solutionscontain ions. Most important of these are acid solutions. For example, whenHCl(g) dissolves in water to form hydrochloric acid, HCl(aq), it ionizes or breaksapart into and ions.

Strong and Weak Electrolytes

There are two categories of electrolytes, strong electrolytes and weak electrolytes,which differ in the extent to which they conduct electricity. Strong electrolytesare those solutes that exist in solution completely or nearly completely as ions.Essentially all soluble ionic compounds (such as NaCl) and a few molecular com-pounds (such as HCl) are strong electrolytes. Weak electrolytes are those solutesthat exist in solution mostly in the form of molecules with only a small fractionin the form of ions. For example, in a solution of acetic acid most ofthe solute is present as molecules. Only a small fraction (about 1%) ofthe is present as and ions.

We must be careful not to confuse the extent to which an electrolyte dis-solves with whether it is strong or weak. For example, is extremelysoluble in water but is a weak electrolyte. on the other hand, is notvery soluble, but the amount of the substance that does dissolve dissociatesalmost completely, so is a strong electrolyte.

When a weak electrolyte such as acetic acid ionizes in solution, we write thereaction in the following manner:

[4.2]

The double arrow means that the reaction is significant in both directions. At anygiven moment some molecules are ionizing to form and At the same time, and ions are recombining to form Thebalance between these opposing processes determines the relative concentrations ofions and neutral molecules. This balance produces a state of chemical equilibriumthat varies from one weak electrolyte to another. Chemical equilibria are extremelyimportant, and we will devote Chapters 15–17 to examining them in detail.

Chemists use a double arrow to represent the ionization of weak electrolytesand a single arrow to represent the ionization of strong electrolytes. Because HClis a strong electrolyte, we write the equation for the ionization of HCl as follows:

[4.3]

The single arrow indicates that the and ions have no tendency to recom-bine in water to form HCl molecules.

In the sections ahead we will begin to look more closely at how we can use thecomposition of a compound to predict whether it is a strong electrolyte, weak elec-trolyte, or nonelectrolyte. For the moment, it is important only to remember thatsoluble ionic compounds are strong electrolytes. We identify ionic compounds as being onescomposed of metals and nonmetals [such as NaCl, and ], or com-pounds containing the ammonium ion, [such as and ].

SAMPLE EXERCISE 4.1 The diagram on the left represents an aqueous solution of one of the following com-pounds: KCl, or Which solution does it best represent?

Solution The diagram shows twice as many cations as anions, consistent with the for-mulation

PRACTICE EXERCISEIf you were to draw diagrams (such as that shown on the left) representing aqueous solu-tions of each of the following ionic compounds, how many anions would you show if thediagram contained six cations? (a) (b) (c) (d)Answers: (a) 6; (b) 12; (c) 2; (d) 9

Al2(SO4)3 .Na3PO4;Ca(NO3)2 ;NiSO4;

K2SO4.

K2SO4.MgCl2,

(NH4)2CO3NH4BrNH4

+Al(NO3)3FeSO4,

Cl-H+HCl(aq) ¡ H+(aq) + Cl-(aq)

HC2H3O2.C2H3O2

-H+C2H3O2

-.H+HC2H3O2

HC2H3O2(aq) ∆ H+(aq) + C2H3O2

-(aq)

Ba(OH)2

Ba(OH)2,HC2H3O2

C2H3O2

-(aq)H+(aq)HC2H3O2

HC2H3O2

(HC2H3O2)

Cl-(aq)H+(aq)

Universal indicator and aconductivity probe are used toexplore the relative acidity andconductivity of a series of aqueousacids. Bassam Z. Shakhashiri,“Conductivity and Extent ofDissociation of Acids in AqueousSolution,” Chemical Demonstrations:A Handbook for Teachers of Chemistry,Vol. 3 (The University of WisconsinPress, Madison, 1989) pp. 140–145.

CQ

MOVIEStrong and Weak Electrolytes

3-D MODELHCl, Acetic Acid

BLB04-112-151-V4 2/21/02 1:34 PM Page 116

4.2 Precipitation Reactions 117

Students explore a variety of ionicreactions that result in the formationof colored precipitates.L.R. Summerlin, Christie L. Borgford,and Julie B. Ealy, “Name ThatPrecipitate,” Chemical Demonstrations,A Sourcebook for Teachers, Vol. 2(American Chemical Society,Washington, DC, 1988) pp. 123–125.

Richard A. Kjonaas, “An Analogy forSolubility: Marbles and Magnets,”J. Chem. Educ., Vol. 61, 1984, 765.

4.2 Precipitation Reactions

Figure 4.4 ¥ shows two clear solutions being mixed, one containing lead nitrateand the other containing potassium iodide (KI). The reaction between

these two solutes produces an insoluble yellow product. Reactions that result inthe formation of an insoluble product are known as precipitation reactions. Aprecipitate is an insoluble solid formed by a reaction in solution. In Figure 4.4 theprecipitate is lead iodide a compound that has a very low solubility in water:

[4.4]

The other product of this reaction, potassium nitrate, remains in solution.Precipitation reactions occur when certain pairs of oppositely charged ions attract

each other so strongly that they form an insoluble ionic solid. To predict whether cer-tain combinations of ions form insoluble compounds, we must consider someguidelines or rules concerning the solubilities of common ionic compounds.

Pb(NO3)2(aq) + 2KI(aq) ¡ PbI2(s) + 2KNO3(aq)

(PbI2),

[Pb(NO3)2]

Pb(NO3)2(aq) PbI2(s) � 2KNO3(aq)2 KI(aq)

�

K�

I�

Pb2�

NO3�

Á Figure 4.4 The addition of a colorless solution of potassium iodide (KI) to a colorlesssolution of lead nitrate produces a yellow precipitate of lead iodide thatslowly settles to the bottom of the beaker.

(PbI2)[Pb(NO3)2]

MOVIEPrecipitation Reactions

BLB04-112-151-V4 2/21/02 1:34 PM Page 117


TABLE 4.1 Solubility Guidelines for Common Ionic Compounds in Water

Soluble Ionic Compounds Important Exceptions

Compounds containing NoneNoneCompounds of and Compounds of and Compounds of and Compounds of and

Insoluble Ionic Compounds Important Exceptions

Compounds containing Compounds of the alkali metal cations, and and

Compounds of and the alkali metal cations

Compounds of and the alkali metal cations

Compounds of the alkali metal cations, and and Ba2+Sr2+,Ca2+,

OH-

NH4

+PO4

3-

NH4

+CO3

2-Ba2+Sr2+,Ca2+,

NH4

+,S2-

Pb2+Hg2

2+,Ba2+,Sr2+,SO4

2-Pb2+Hg2

2+,Ag+,I-Pb2+Hg2

2+,Ag+,Br-Pb2+Hg2

2+,Ag+,Cl-C2H3O2

-NO3

-

Solubility Guidelines for Ionic Compounds

The solubility of a substance is the amount of that substance that can be dis-solved in a given quantity of solvent. Only mol of dissolves in aliter of water at In our discussions any substance with a solubility less than0.01 mol L will be referred to as insoluble. In those cases the attraction betweenthe oppositely charged ions in the solid is too great for the water molecules to sep-arate them to any significant extent, and the substance remains largely undis-solved.

Unfortunately, there are no rules based on simple physical properties suchas ionic charge to guide us in predicting whether a particular ionic compoundwill be soluble or not. Experimental observations, however, have led to guide-lines for predicting solubility for ionic compounds. For example, experimentsshow that all common ionic compounds that contain the nitrate anion, are soluble in water. Table 4.1 ¥ summarizes the solubility guidelines for commonionic compounds. The table is organized according to the anion in the compound,but it reveals many important facts about cations. Note that all common ioniccompounds of the alkali metal ions (group 1A of the periodic table) and of the ammoniumion are soluble in water.(NH4

+)

NO3

-,

> 25°C.PbI21.2 * 10-3

SAMPLE EXERCISE 4.2Classify the following ionic compounds as soluble or insoluble: (a) sodium carbonate

(b) lead sulfate

SolutionAnalyze: We are given the names and formulas of two ionic compounds and asked topredict whether they are soluble or insoluble in water.Plan: We can use Table 4.1 to answer the question. Thus, we need to focus on the anionin each compound because the table is organized by anions.Solve: (a) According to Table 4.1, most carbonates are insoluble, but carbonates of thealkali metal cations (such as sodium ion) are an exception to this rule and are soluble.Thus, is soluble in water.

(b) Table 4.1 indicates that although most sulfates are water soluble, the sulfate ofis an exception. Thus, is insoluble in water.

PRACTICE EXERCISEClassify the following compounds as soluble or insoluble: (a) cobalt(II) hydroxide;(b) barium nitrate; (c) ammonium phosphate.Answers: (a) insoluble; (b) soluble; (c) soluble

PbSO4

Pb2+Na2CO3

(PbSO4).(Na2CO3);

The solubility of a series of silver saltsand complexes is explored in thiscolorful demonstration. Lee R.Summerlin, Christie L. Borgford, andJulie B. Ealy, “Solubility of SomeSilver Compounds,” ChemicalDemonstrations, A Sourcebook forTeachers, Vol. 2 (American ChemicalSociety, Washington, DC, 1988)pp. 83–85.

BLB04-112-151-V4 2/21/02 1:34 PM Page 118

4.2 Precipitation Reactions 119

To predict whether a precipitate forms when we mix aqueous solutions oftwo strong electrolytes, we must (1) note the ions present in the reactants, (2) con-sider the possible combinations of the cations and anions, and (3) use Table 4.1to determine if any of these combinations is insoluble. For example, will a pre-cipitate form when solutions of and NaOH are mixed? Because

and NaOH are both soluble ionic compounds, they are both strongelectrolytes. Mixing and NaOH(aq) first produces a solution con-taining and ions. Will either of the cations interact witheither of the anions to form an insoluble compound? In addition to the reactants,the other possible interactions are with and with FromTable 4.1 we see that is insoluble and will thus form a precipitate.

however, is soluble, so and will remain in solution. The bal-anced equation for the precipitation reaction is

[4.5]

Exchange (Metathesis) Reactions

Notice in Equation 4.5 that the cations in the two reactants exchange anions—ends up with and ends up with The chemical formulas

of the products are based on the charges of the ions—two ions are need-ed to give a neutral compound with and one ion is needed to givea neutral compound with • (Section 2.7) It is only after the chemicalformulas of the products are determined that the equation can be balanced.

Reactions in which positive ions and negative ions appear to exchange part-ners conform to the following general equation:

[4.6]

Example:

Such reactions are known as exchange reactions, or metathesis reactions (meh-TATH-eh-sis, which is the Greek word for “to transpose”). Precipitation reac-tions conform to this pattern, as do many acid-base reactions, as we will see inSection 4.3.

SAMPLE EXERCISE 4.3(a) Predict the identity of the precipitate that forms when solutions of and are mixed. (b) Write the balanced chemical equation for the reaction.

SolutionAnalyze: We are given two ionic reactants and asked to predict the insoluble productthat they form.Plan: We need to write down the ions present in the reactants and to exchange theanions between the two cations. Once we have written the chemical formulas for theseproducts, we can use Table 4.1 to determine which is insoluble in water. Knowing theproducts also allows us to write the equation for the reaction.Solve: (a) The reactants contain and ions. If we exchange theanions, we will have and KCl. According to Table 4.1, most compounds of

are soluble but those of are not. Thus, is insoluble and will precip-itate from solution. KCl, on the other hand, is soluble.

(b) From part (a) we know the chemical formulas of the products, and KCl.The balanced equation with phase labels shown is

PRACTICE EXERCISE(a) What compound precipitates when solutions of and LiOH are mixed?(b) Write a balanced equation for the reaction. (c) Will a precipitate form when solu-tions of and KOH are mixed?Answers: (a) (b)(c) No (both possible products are water soluble)

3Li2SO4(aq);+2Fe(OH)3(s)Fe2(SO4)3(aq) + 6LiOH(aq) ¡Fe(OH)3;Ba(NO3)2

Fe21SO423

BaCl2(aq) + K2SO4(aq) ¡ BaSO4(s) + 2KCl(aq)

BaSO4

BaSO4Ba2+SO4

2-BaSO4

SO4

2-K+,Cl-,Ba2+,

K2SO4BaCl2

AgNO3(aq) + KCl(aq) ¡ AgCl(s) + KNO3(aq)

AX + BY ¡ AY + BX

Na+.NO3

-Mg2+,OH-

NO3

-.Na+OH-,Mg2+

Mg(NO3)2(aq) + 2NaOH(aq) ¡ Mg(OH)2(s) + 2NaNO3(aq)

NO3

-Na+NaNO3,Mg(OH)2

NO3

-.Na+OH-Mg2+

OH-Na+,NO3

-,Mg2+,Mg(NO3)2(aq)

Mg(NO3)2

Mg(NO3)2

One of the following is needed todrive a metathesis reaction: theformation of a precipitate, thegeneration of a gas, the productionof a weak electrolyte, or theproduction of a nonelectrolyte.

BLB04-112-151-V4 2/21/02 1:34 PM Page 119


Writing the net ionic equation makesit easier for students to focus on theions participating in a chemicalreaction. It is important to rememberthat, although they do not appear inthe net ionic equation, the spectatorions are still present in solution.

When writing net ionic equations,students often try to split polyatomicions into smaller units.

Ionic Equations

In writing chemical equations for reactions in aqueous solution, it is often use-ful to indicate explicitly whether the dissolved substances are present predom-inantly as ions or as molecules. Let’s reconsider the precipitation reaction between

and 2KI, shown previously in Figure 4.4:

An equation written in this fashion, showing the complete chemical formulas ofthe reactants and products, is called a molecular equation because it shows thechemical formulas of the reactants and products without indicating their ioniccharacter. Because KI, and are all soluble ionic compounds andtherefore strong electrolytes, we can write the chemical equation to indicateexplicitly the ions that are in the solution:

[4.7]

An equation written in this form, with all soluble strong electrolytes shown asions, is known as a complete ionic equation.

Notice that and appear on both sides of Equation 4.7. Ionsthat appear in identical forms among both the reactants and products of a com-plete ionic equation are called spectator ions. They are present but play no directrole in the reaction. When spectator ions are omitted from the equation (theycancel out like algebraic quantities), we are left with the net ionic equation:

[4.8]

A net ionic equation includes only the ions and molecules directly involvedin the reaction. Charge is conserved in reactions, so the sum of the charges ofthe ions must be the same on both sides of a balanced net ionic equation. In thiscase the charge of the cation and the two charges of the anions add togive zero, the charge of the electrically neutral product. If every ion in a completeionic equation is a spectator, then no reaction occurs.

Net ionic equations are widely used to illustrate the similarities betweenlarge numbers of reactions involving electrolytes. For example, Equation 4.8expresses the essential feature of the precipitation reaction between any strongelectrolyte containing and any strong electrolyte containing : The and ions combine to form a precipitate of Thus, a net ionic equationdemonstrates that more than one set of reactants can lead to the same net reac-tion. The complete equation, on the other hand, identifies the actual reactantsthat participate in a reaction.

Net ionic equations also point out that the chemical behavior of a strongelectrolyte solution is due to the various kinds of ions it contains. Aqueous solu-tions of KI and for example, share many chemical similarities becauseboth contain ions. Each kind of ion has its own chemical characteristics thatdiffer very much from those of its parent atom.

The following steps summarize the procedure for writing net ionic equations:

1. Write a balanced molecular equation for the reaction.

2. Rewrite the equation to show the ions that form in solution when each sol-uble strong electrolyte dissociates or ionizes into its component ions. Onlystrong electrolytes dissolved in aqueous solution are written in ionic form.

3. Identify and cancel spectator ions.

I-MgI2,

PbI2 .I-(aq)Pb2+(aq)I-Pb2+

1-2+

Pb2+(aq) + 2I-(aq) ¡ PbI2(s)

NO3

-(aq)K+1aq2

PbI2(s) + 2K+(aq) + 2NO3

-(aq)

Pb2+(aq) + 2NO3

-(aq) + 2K+(aq) + 2I-(aq) ¡

KNO3Pb(NO3)2 ,

Pb(NO3)2(aq) + 2KI(aq) ¡ PbI2(s) + 2KNO3(aq)

Pb(NO3)2

Betty J. Wruck, “Reinforcing NetIonic Equation Writing,” J. Chem.Educ., Vol. 73, 1996, 149–150.

ACTIVITYWriting a Net Ionic Equation

BLB04-112-151-V4 2/21/02 1:34 PM Page 120

Á Figure 4.5 Some common acids(left), and bases (right) that arehousehold products.

4.3 Acid-Base Reactions 121

HC2H3O2

HNO3

HCl

H

O

C

N

Cl

SAMPLE EXERCISE 4.4Write the net ionic equation for the precipitation reaction that occurs when solutionsof calcium chloride and sodium carbonate are mixed.

SolutionAnalyze: Our task is to write a net ionic equation for a precipitation reaction, given thenames of the reactants present in solution.Plan: We first need to write the chemical formulas of the reactants and products andto determine which product is insoluble. Then we write and balance the molecularequation. Next, we write each soluble strong electrolyte as separated ions to obtainthe complete ionic equation. Finally, we eliminate the spectator ions to obtain the netionic equation.Solve: Calcium chloride is composed of calcium ions, and chloride ions, hencean aqueous solution of the substance is Sodium carbonate is composed ofions and ions; hence an aqueous solution of the compound is In themolecular equations for precipitation reactions, the anions and cations appear to exchangepartners. Thus, we put and together to give and and togeth-er to give NaCl.According to the solubility guidelines in Table 4.1, is insoluble andNaCl is soluble. The balanced molecular equation is

In a complete ionic equation, only dissolved strong electrolytes (such as solubleionic compounds) are written as separate ions. As the (aq) designations remind us,

and NaCl are all dissolved in the solution. Furthermore, they are allstrong electrolytes. is an ionic compound, but it is not soluble. We do not writethe formula of any insoluble compound as its component ions. Thus, the completeionic equation is

and are spectator ions. Canceling them gives the following net ionic equation:

Check: We can check our result by confirming that both the elements and the electriccharge are balanced. Each side has 1 Ca, 1 C, and 3 O, and the net charge on each sideequals 0.Comment: If none of the ions in an ionic equation is removed from solution or changedin some way, then they all are spectator ions and a reaction does not occur.

PRACTICE EXERCISEWrite the net ionic equation for the precipitation reaction that occurs when aqueoussolutions of silver nitrate and potassium phosphate are mixed.Answer:

4.3 Acid-Base Reactions

Many acids and bases are industrial and household substances (Figure 4.5 »), andsome are important components of biological fluids. Hydrochloric acid, for exam-ple, is not only an important industrial chemical but also the main constituent of gas-tric juice in our stomach. Acids and bases also happen to be common electrolytes.

Acids

Acids are substances that ionize in aqueous solutions to form hydrogen ions,thereby increasing the concentration of ions. Because a hydrogen atomconsists of a proton and an electron, is simply a proton. Thus, acids are oftencalled proton donors. Molecular models of three common acids, HCl, and are shown in the margin.

Molecules of different acids can ionize to form different numbers of ions.Both HCl and are monoprotic acids, which yield one per molecule ofH+HNO3

H+HC2H3O2,

HNO3,H+

H+(aq)

3Ag+(aq) + PO4

3-(aq) ¡ Ag3PO4(s)

Ca2+(aq) + CO3

2-(aq) ¡ CaCO3(s)

Na+Cl-

CaCO3(s) + 2Na+(aq) + 2Cl-(aq)Ca2+(aq) + 2Cl-(aq) + 2Na+(aq) + CO3

2-(aq) ¡

CaCO3

Na2CO3,CaCl2 ,

CaCl2(aq) + Na2CO3(aq) ¡ CaCO3(s) + 2NaCl(aq)

CaCO3

Cl-Na+CaCO3CO3

2-Ca2+

Na2CO3(aq).CO3

2-Na+CaCl2(aq).

Cl-;Ca2+,

BLB04-112-151-V4 2/21/02 1:34 PM Page 121


Strong Acids Strong Bases

Hydrochloric, HCl Group 1A metal hydroxides (LiOH, NaOH, KOH, RbOH, CsOH)

Hydrobromic, HBr Heavy group 2A metal hydroxides

Hydroiodic, HI

Chloric,

Perchloric,

Nitric,

Sulfuric, H2SO4

HNO3

HClO4

HClO3

Ba(OH)2]Sr(OH)2 ,[Ca(OH)2 ,

TABLE 4.2 Common Strong Acids and Bases

acid. Sulfuric acid, is a diprotic acid, one that yields two per moleculeof acid. The ionization of and other diprotic acids occurs in two steps:

[4.9]

[4.10]

Although is a strong electrolyte, only the first ionization is complete.Thus, aqueous solutions of sulfuric acid contain a mixture of

and

Bases

Bases are substances that accept (react with) ions. Bases produce hydroxideions when they dissolve in water. Ionic hydroxide compounds such asNaOH, KOH, and are among the most common bases. When dissolvedin water, they dissociate into their component ions, introducing ions intothe solution.

Compounds that do not contain ions can also be bases. For example,ammonia is a common base. When added to water, it accepts an ionfrom the water molecule and thereby produces an ion (Figure 4.6 «):

[4.11]

Because only a small fraction of the (about 1%) forms and ions, ammonia is a weak electrolyte.

Strong and Weak Acids and Bases

Acids and bases that are strong electrolytes (completely ionized in solution) arecalled strong acids and strong bases. Those that are weak electrolytes (partlyionized) are called weak acids and weak bases. Strong acids are more reactivethan weak acids when the reactivity depends only on the concentration of The reactivity of an acid, however, can depend on the anion as well as on For example, hydrofluoric acid (HF) is a weak acid (only partly ionized in aque-ous solution), but it is very reactive and vigorously attacks many substances,including glass. This reactivity is due to the combined action of and

Table 4.2 ¥ lists the common strong acids and bases. You should commit theseto memory. As you examine this table, notice first that some of the most commonacids, such as HCl, and are strong. Second, three of the strongacids result from combining a hydrogen atom and a halogen atom. (HF, however,is a weak acid.) Third, the list of strong acids is very short. Most acids are weak.Fourth, the only common strong bases are the hydroxides of and

(the alkali metals, group 1A) and the hydroxides of and Ba2+Sr2+,Ca2+,Cs+Rb+,K+,Na+,Li+,

H2SO4,HNO3,

F-(aq).H+(aq)

H+(aq).H+(aq).

OH-NH4

+NH3

NH3(aq) + H2O(l) ∆ NH4

+(aq) + OH-(aq)

OH-H+1NH32

OH-

OH-Ca(OH)2

(OH-)H+

SO4

2-(aq).HSO4

-(aq),H+(aq),

H2SO4

HSO4

-(aq) ∆ H+(aq) + SO4

2-(aq)

H2SO4(aq) ¡ H+(aq) + HSO4

-(aq)

H2SO4

H+H2SO4,

H2O OH�

NH3 NH4�

Á Figure 4.6 An molecule actsas a proton donor (acid), and as aproton acceptor (base). Only a fractionof the reacts with is aweak electrolyte.

NH3H2O;NH3

NH3

H2O

H. van Lubeck, “Significance,Concentration Calculations, Weakand Strong Acids,” J. Chem. Educ.,Vol. 60, 1983, 189.

Other definitions of acids and basesare discussed in Chapter 16.

ANIMATIONIntroduction to Aqueous Acids,Introduction to Aqueous Bases

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CQ


�

��

�

�

� �

��

� �

�

�

�

�

�

�

�

�

��

�

� �

�

�

HX HY HZ

Strong Electrolyte Weak Electrolyte Nonelectrolyte

Ionic All None NoneMolecular Strong acids Weak acids

(see Table 4.2) Weak bases All other compounds(NH3)(H Á )

TABLE 4.3 Summary of the Electrolytic Behavior of Common Soluble Ionic and Molecular Compounds

(the heavy alkaline earths, group 2A). These are the common soluble metal hydrox-ides. Most other metal hydroxides are insoluble in water. The most common weakbase is which reacts with water to form ions (Equation 4.11).

SAMPLE EXERCISE 4.5The following diagrams represent aqueous solutions of three acids (HX, HY, and HZ)with water molecules omitted for clarity. Rank them from strongest to weakest.

OH-NH3,

Solution The strongest acid is the one with the most ions and fewest undissociat-ed acid molecules in solution. Hence, the order is HY is a strongacid because it is totally ionized (no HY molecules in solution), whereas both HX andHZ are weak acids, whose solutions consist of a mixture of molecules and ions.

PRACTICE EXERCISEImagine a diagram showing 10 ions and 10 ions. If this solution were mixedwith the one pictured above for HY, what would the diagram look like that represents thesolution after any possible reaction? ( ions will react with ions to form )Answer: The final diagram would show 10 ions, 2 ions, 8 ions, and

molecules.

Identifying Strong and Weak Electrolytes

If we remember the common strong acids and bases (Table 4.2) and also rememberthat is a weak base, we can make reasonable predictions about the electrolyt-ic behavior of a great number of water-soluble substances. Table 4.3 ¥ summarizesour observations about electrolytes. To classify a soluble substance as a strong elec-trolyte, weak electrolyte, or nonelectrolyte, we simply work our way down andacross this table. We first ask ourselves whether the substance is ionic or molecu-lar. If it is ionic, it is a strong electrolyte. If it is molecular, we ask whether it is anacid. (Does it have H first in the chemical formula?) If it is an acid, we rely on thememorized list from Table 4.2 to determine whether it is a strong or weak elec-trolyte. If an acid is not listed in Table 4.2, it is probably a weak electrolyte. Forexample, and are not listed in Table 4.2 and are weakacids. is the only weak base that we consider in this chapter. (There are com-pounds called amines that are related to and are also molecular bases, but wewill not consider them until Chapter 16.) Finally, any molecular substance that weencounter in this chapter that is not an acid or is probably a nonelectrolyte.NH3

NH3

NH3

HC7H5O2H2SO3,H3PO4,

NH3

8 H2OY-OH-Na+

H2O.OH-H+

OH-Na+

HY 7 HZ 7 HX.H+

Albert Kowalak, “When is a StrongElectrolyte Strong?” J. Chem. Educ.,Vol. 65, 1988 , 607.

John J. Fortman “Pictorial AnalogiesX: Solutions of Electrolytes,” J. Chem.Educ., Vol. 71, 1994, 27–28.

The pH of a variety of householdchemicals is determined usingindicators and pH meters. Bassam Z.Shakhashiri, “Food is Usually Acidic,Cleaners Are Usually Basic,“ ChemicalDemonstrations: A Handbook forTeachers of Chemistry, Vol. 3 (TheUniversity of Wisconsin Press,Madison, 1989) pp. 65–69.

ACTIVITYStrong Acids

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Á Figure 4.7 The acid-base indicatorbromthymol blue is blue in basic solutionand yellow in acidic solution. The leftflask shows the indicator in the presenceof a base, aqueous ammonia (herelabeled as ammonium hydroxide). Theright flask shows the indicator in thepresence of hydrochloric acid, HCl.


Salts may be thought of as ioniccompounds that are neither acidsnor bases.

* Tasting chemical solutions is not a good practice. However, we have all had acids such as ascorbicacid (vitamin C), acetylsalicylic acid (aspirin), and citric acid (in citrus fruits) in our mouths, and weare familiar with their characteristic sour taste. Soaps, which are basic, have the characteristic bittertaste of bases.

SAMPLE EXERCISE 4.6Classify each of the following dissolved substances as a strong electrolyte, weak elec-trolyte, or nonelectrolyte: (ethanol), (formic acid),KOH.

SolutionAnalyze: We are given several chemical formulas and asked to classify each substanceas a strong electrolyte, weak electrolyte, or nonelectrolyte.Plan: The approach we take is outlined in Table 4.3. We can predict whether a sub-stance is ionic or molecular, based on its composition. As we saw in Section 2.7, mostionic compounds we encounter in this text are composed of both a metal and a non-metal, whereas most molecular compounds are composed only of nonmetals.Solve: Two compounds fit the criteria for ionic compounds: and KOH. Both arestrong electrolytes. The three remaining compounds are molecular. Two, and

are acids. Nitric acid, is a common strong acid (a strong electrolyte),as shown in Table 4.2. Because most acids are weak acids, our best guess would bethat is a weak acid (weak electrolyte). This is correct. The remaining molecu-lar compound, is neither an acid nor a base, so it is a nonelectrolyte.Comment: Although has an OH group, it is not a metal hydroxide; thus, itis not a base. Rather, it is a member of a class of organic compounds that have bonds and which are known as alcohols. • (Section 2.9)

PRACTICE EXERCISEConsider solutions in which 0.1 mol of each of the following compounds is dissolvedin 1 L of water: (calcium nitrate), (glucose), (sodiumacetate), and (acetic acid). Rank the solutions in order of increasing electri-cal conductivity, based on the fact that the greater the number of ions in solution, thegreater the conductivity.Answer: (nonelectrolyte) (weak electrolyte, existing mainly in theform of molecules with few ions) (strong electrolyte that provides two ions,

and (strong electrolyte that provides three ions, and

Neutralization Reactions and Salts

Solutions of acids and bases have very different properties. Acids have a sourtaste, whereas bases have a bitter taste.* Acids can change the colors of certaindyes in a specific way that differs from the effect of a base (Figure 4.7 «). The dyeknown as litmus, for example, is changed from blue to red by an acid, and fromred to blue by a base. In addition, acidic and basic solutions differ in chemicalproperties in several important ways that we will explore in this chapter and inlater chapters.

When a solution of an acid and that of a base are mixed, a neutralizationreaction occurs. The products of the reaction have none of the characteristic prop-erties of either the acidic or the basic solutions. For example, when hydrochloricacid is mixed with a solution of sodium hydroxide, the following reaction occurs:

[4.12](acid) (base) (water) (salt)

Water and table salt, NaCl, are the products of the reaction. By analogy to thisreaction, the term salt has come to mean any ionic compound whose cationcomes from a base (for example, from NaOH) and whose anion comes froman acid (for example, from HCl). In general, a neutralization reaction betweenan acid and a metal hydroxide produces water and a salt.

Cl-Na+

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

2NO3

-2 Ca2+C2H3O2

- ) 6 Ca(NO3)2Na+6 NaC2H3O2

6 HC2H3O2C6H12O6

HC2H3O2

NaC2H3O2C6H12O6Ca(NO3)2

C ¬ OHC2H5OH

C2H5OH,HCHO2

HNO3,HCHO2,HNO3

CaCl2

HCHO2C2H5OHHNO3,CaCl2 ,

BLB04-112-151-V4 2/21/02 1:34 PM Page 124


(a) (b) (c)

Á Figure 4.8 (a) Milk of magnesia is a suspension of magnesium hydroxide, in water. (b) The magnesium hydroxide dissolves upon the addition of hydrochloric acid,HCl(aq). (c) The final clear solution contains soluble shown in Equation 4.15.MgCl2(aq),

Mg(OH)2(s),

Because HCl, NaOH, and NaCl are all soluble strong electrolytes, the com-plete ionic equation associated with Equation 4.12 is

[4.13]

Therefore, the net ionic equation is

[4.14]

Equation 4.14 summarizes the essential feature of the neutralization reactionbetween any strong acid and any strong base: and ions combineto form

Figure 4.8 Á shows the reaction between hydrochloric acid and another base,which is insoluble in water. A milky white suspension of

called milk of magnesia is seen dissolving as the neutralization reaction occurs:

Molecular equation: [4.15]

Net ionic equation: [4.16]

Notice that the ions (this time in a solid reactant) and ions combine toform Because the ions exchange partners, neutralization reactions betweenacids and metal hydroxides are also metathesis reactions.

SAMPLE EXERCISE 4.7(a) Write a balanced complete chemical equation for the reaction between aqueoussolutions of acetic acid and barium hydroxide (b) Write the netionic equation for this reaction.

SolutionAnalyze: We are given the chemical formulas for an acid and a base and asked to writea balanced chemical equation and then a net ionic equation for their neutralizationreaction.Plan: As Equation 4.12 and the italicized statement that follows it indicate, neutral-ization reactions form two products, and a salt. We examine the cation of the baseand the anion of the acid to determine the composition of the salt.

H2O

[Ba(OH)2].(HC2H3O2)

H2O.H+OH-

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

Mg(OH)2(s) + 2HCl(aq) ¡ MgCl2(aq) + 2H2O(l)

Mg(OH)2Mg1OH22 ,

H2O.OH-(aq)H+(aq)

H+(aq) + OH-(aq) ¡ H2O(l)

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

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Solve: (a) The salt will contain the cation of the base and the anion of the acidThus, the formula of the salt is According to the solubility

guidelines in Table 4.1, this compound is soluble. The unbalanced equation for theneutralization reaction is

To balance the equation, we must provide two molecules of to furnish thetwo ions and to supply the two ions needed to combine with the two

ions of the base. The balanced equation is

(b) To write the ionic equation, we must determine whether or not each com-pound in aqueous solution is a strong electrolyte. is a weak electrolyte (weakacid), is a strong electrolyte, and is also a strong electrolyte.Thus, the complete ionic equation is

Eliminating the spectator ions gives

Simplifying the coefficients gives the net ionic equation.

Check: We can determine whether the molecular equation is correctly balanced bycounting the number of atoms of each kind on both sides of the arrow. (There are 10 H,6 O, 4 C, and 1 Ba on each side.) However, it is often easier to check equations by count-ing groups: There are groups as well as 1 Ba, and 4 additional H atoms and2 additional O atoms on each side of the equation. The net ionic equation checks outbecause the numbers of each kind of element and the net charge are the same on bothsides of the equation.

PRACTICE EXERCISE(a) Write a balanced equation for the reaction of carbonic acid and potassiumhydroxide (KOH). (b) Write the net ionic equation for this reaction.Answers: (a) (b)

( is a weak electrolyte, whereas KOHand are strong electrolytes.)

Acid-Base Reactions with Gas Formation

There are many bases besides that react with to form molecular com-pounds. Two of these that you might encounter in the laboratory are the sulfideion and the carbonate ion. Both of these anions react with acids to form gases thathave low solubilities in water. Hydrogen sulfide the substance that givesrotten eggs their foul odor, forms when an acid such as HCl(aq) reacts with ametal sulfide such as :



Carbonates and bicarbonates react with acids to form gas. Reaction ofor with an acid first gives carbonic acid For example,

when hydrochloric acid is added to sodium bicarbonate, the following reactionoccurs:

[4.19]HCl(aq) + NaHCO3(aq) ¡ NaCl(aq) + H2CO3(aq)

1H2CO32.HCO3

-CO3

2-CO2

2H+(aq) + S2-(aq) ¡ H2S(g)

2HCl(aq) + Na2S(aq) ¡ H2S(g) + 2NaCl(aq)

Na2S

(H2S),

H+OH-

K2CO3

H2CO32H2O(l) + CO3

2-(aq);2OH-(aq) ¡+H2CO3(aq)H2CO3(aq) + 2KOH(aq) ¡ 2H2O(l) + K2CO3(aq);

(H2CO3)

2C2H3O2

HC2H3O2(aq) + OH-(aq) ¡ H2O(l) + C2H3O2

-(aq)

2HC2H3O2(aq) + 2OH-(aq) ¡ 2H2O(l) + 2C2H3O2

-(aq)

2HC2H3O2(aq) + Ba2+(aq) + 2OH-(aq) ¡ 2H2O(l) + Ba2+(aq) + 2C2H3O2

-(aq)

Ba(C2H3O2)2Ba1OH22HC2H3O2

2HC2H3O2(aq) + Ba(OH)2(aq) ¡ 2H2O(l) + Ba(C2H3O2)2(aq)

OH-H+C2H3O2

-HC2H3O2

HC2H3O2(aq) + Ba(OH)2(aq) ¡ H2O(l) + Ba(C2H3O2)2(aq)

Ba(C2H3O2)2 .(C2H3O2

-).(Ba2+)

Bassam Z. Shakhashiri, “Fizzing andFoaming: Reactions of Acids withCarbonates,“ ChemicalDemonstrations: A Handbook forTeachers of Chemistry, Vol. 3 (TheUniversity of Wisconsin Press,Madison, 1989) pp. 96–99.

The acidic or basic nature ofsolutions of gases is investigated,John J. Fortman and Katherine M.Stubbs, “Demonstrations with RedCabbage Indicator,”J. Chem. Educ.,Vol. 69, 1992, 66–67.

An example of a reaction involvingtwo solids ( and is demonstrated. L.R. Summerlin,Christie L. Borgford, and Julie B. Ealy,“A Hand-Held Reaction: Productionof Ammonia Gas,” ChemicalDemonstrations, A Sourcebook forTeachers, Vol. 2 (American ChemicalSociety, Washington, DC, 1988) p. 37.

Ca(OH)2(s)NH4Cl(s)

Bassam Z. Shakhashiri,“Determination of NeutralizingCapacity of Antacids,” ChemicalDemonstrations: A Handbook forTeachers of Chemistry, Vol. 3 (TheUniversity of Wisconsin Press,Madison, 1989) pp. 162–166.

An antacid, milk of magnesia, ismixed with acid in thisdemonstration. Lee R. Summerlin,Christie L. Borgford, and Julie B. Ealy,“Milk of Magnesia Versus Acid,”Chemical Demonstrations, ASourcebook for Teachers, Vol. 2(American Chemical Society,Washington, DC, 1988) p. 173.

BLB04-112-151-V4 2/21/02 1:34 PM Page 126


Carbonic acid is unstable; if present in solution in sufficient concentrations, itdecomposes to form which escapes from the solution as a gas.

[4.20]

The decomposition of produces bubbles of gas, as shown inFigure 4.9 ». The overall reaction is summarized by the following equations:

Molecular equation:[4.21]

Net ionic equation:[4.22]

Both and are used as acid neutralizers in acid spills. Thebicarbonate or carbonate salt is added until the fizzing due to the formation of

stops. Sometimes sodium bicarbonate is used as an antacid to soothe anupset stomach. In that case the reacts with stomach acid to form The fizz when tablets are added to water is due to the reac-tion of sodium bicarbonate and citric acid.

Alka-Seltzer®CO2(g).HCO3

-CO2(g)

Na2CO3NaHCO3

H+(aq) + HCO3

-(aq) ¡ H2O(l) + CO2(g)

HCl(aq) + NaHCO3(aq) ¡ NaCl(aq) + H2O(l) + CO2(g)

CO2H2CO3

H2CO3(aq) ¡ H2O(l) + CO2(g)

CO2,

Chemistry at Work Antacids

Á Figure 4.10 Antacids and acid inhibitors are common,over-the-counter drugs. Tagamet and Pepcid are acidinhibitors, whereas the other products are antacids.

AC®HB®

Commercial Name Acid-Neutralizing Agents

and Milk of Magnesia

and and

CaCO3Tums®NaAl(OH)2CO3Rolaids®

Al(OH)3Mg(OH)2Mylanta®Al(OH)3Mg(OH)2Maalox®

Mg(OH)2

CaCO3Mg(OH)2Di-Gel®Al(OH)3Amphojel®NaHCO3Alka-Seltzer®

TABLE 4.4 Some Common Antacids

The stomach secretes acids to help digest foods. These acids,which include hydrochloric acid, contain about 0.1 mol of per liter of solution. The stomach and digestive tract are nor-mally protected from the corrosive effects of stomach acid by amucosal lining. Holes can develop in this lining, however,allowing the acid to attack the underlying tissue, causingpainful damage. These holes, known as ulcers, can be caused bythe secretion of excess acids or by a weakness in the digestivelining. Recent studies indicate, however, that many ulcers are

H+caused by bacterial infection. Between 10 and 20% of Ameri-cans suffer from ulcers at some point in their lives, and manyothers experience occasional indigestion or heartburn due todigestive acids entering the esophagus.

We can address the problem of excess stomach acid in twosimple ways: (1) removing the excess acid, or (2) decreasing theproduction of acid. Those substances that remove excess acidare called antacids, whereas those that decrease the productionof acid are called acid inhibitors. Figure 4.10 « shows severalcommon, over-the-counter drugs of both types.

Antacids are simple bases that neutralize digestive acids.Their ability to neutralize acids is due to the hydroxide, car-bonate, or bicarbonate ions they contain. Table 4.4 ¥ lists theactive ingredients in some antacids.

The newer generation of antiulcer drugs, such asand are acid inhibitors. They act on acid-

producing cells in the lining of the stomach. Formulations thatcontrol acid in this way are now available as over-the-counterdrugs.

Zantac®,Tagamet®

Á Figure 4.9 Carbonates react withacids to form carbon dioxide gas. Here

(white solid) reacts withhydrochloric acid; the bubbles containCO2 .

NaHCO3

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Á Figure 4.11 Corrosion at theterminals of a battery, caused by attack ofthe metal by sulfuric acid from the battery.

Substanceoxidized(loseselectron)

Substancereduced(gainselectron)

e�

Á Figure 4.12 Oxidation is the lossof electrons by a substance; reduction isthe gain of electrons by a substance.Oxidation of one substance is alwaysaccompanied by reduction of another.


When one substance is oxidized,another substance must be reduced.

Students may be familiar with twogood mnemonics for redoxreactions: (1) LEO the lion says GER:Lose electrons oxidation, gainelectrons reduction, and (2) OIL RIG:Oxidation involves loss of electrons,reduction involves gain of electrons.

Gion Calzaferri, “OxidationNumbers,” J. Chem. Educ., Vol. 76,1999, 362–363.

4.4 Oxidation-Reduction Reactions

In precipitation reactions cations and anions come together to form an insolubleionic compound. In neutralization reactions ions and ions come togeth-er to form molecules. Now let’s consider a third important kind of reactionin which electrons are transferred between reactants. Such reactions are calledoxidation-reduction, or redox, reactions.

Oxidation and Reduction

The corrosion of iron (rusting) and of other metals, such as the corrosion of theterminals of an automobile battery, are familiar processes. What we call corrosionis the conversion of a metal into a metal compound by a reaction between themetal and some substance in its environment. Rusting involves the reaction ofoxygen with iron in the presence of water. The corrosion shown in Figure 4.11 «results from the reaction of battery acid with the metal clamp.

When a metal corrodes, it loses electrons and forms cations. For example,calcium is vigorously attacked by acids to form calcium ions :

[4.23]

When an atom, ion, or molecule has become more positively charged (that is,when it has lost electrons), we say that it has been oxidized. Loss of electrons by asubstance is called oxidation. Thus, Ca, which has no net charge, is oxidized (under-goes oxidation) in Equation 4.23, forming

The term oxidation is used because the first reactions of this sort to be stud-ied thoroughly were reactions with oxygen. Many metals react directly with in air to form metal oxides. In these reactions the metal loses electrons to oxygen,forming an ionic compound of the metal ion and oxide ion. For example, whencalcium metal is exposed to air, the bright metallic surface of the metal tarnish-es as CaO forms:

[4.24]

As Ca is oxidized in Equation 4.24, oxygen is transformed from neutral to two ions. When an atom, ion, or molecule has become more negativelycharged (gained electrons), we say that it is reduced. Gain of electrons by a sub-stance is called reduction. When one reactant loses electrons, another reactantmust gain them; the oxidation of one substance is always accompanied by thereduction of another as electrons are transferred between them, as shown inFigure 4.12 «.

Oxidation Numbers

Before we can properly identify an oxidation-reduction reaction, we must havesome way of keeping track of the electrons gained by the substance reduced andthose lost by the substance oxidized. The concept of oxidation numbers (alsocalled oxidation states) was devised as a simple way of keeping track of electronsin reactions. The oxidation number of an atom in a substance is the actual chargeof the atom if it is a monoatomic ion; otherwise, it is the hypothetical chargeassigned to the atom using a set of rules. Oxidation occurs when there is anincrease in oxidation number, whereas reduction occurs when there is a decreasein oxidation number.

We use the following rules for assigning oxidation numbers:

1. For an atom in its elemental form the oxidation number is always zero. Thus,each H atom in the molecule has an oxidation number of 0, and each Patom in the molecule has an oxidation number of 0.P4

H2

O2-O2

2Ca(s) + O2(g) ¡ 2CaO(s)

O2

Ca2+.

Ca(s) + 2H+(aq) ¡ Ca2+(aq) + H2(g)

(Ca2+)

(H2SO4)

H2OOH-H+

ANIMATIONOxidation-Reduction Reactions: Part I, Oxidation-ReductionReactions: Part II

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4.4 Oxidation-Reduction Reactions 129

Lee R. Summerlin and James L. Ealy,Jr., “Oxidation States of Manganese:

, , , and ,” Chemical Demonstrations,

A Sourcebook for Teachers, Vol. 1(American Chemical Society,Washington, DC, 1988)pp. 133–134.

Mn2+Mn4+Mn6+Mn7+

2. For any monoatomic ion the oxidation number equals the charge on the ion. Thus,has an oxidation number of has an oxidation state of and so

forth. The alkali metal ions (group 1A) always have a charge, and there-fore the alkali metals always have an oxidation number of in their com-pounds. Similarly, the alkaline earth metals (group 2A) are always andaluminum (group 3A) is always in their compounds. (In writing oxida-tion numbers, we will write the sign before the number to distinguish themfrom the actual electronic charges, which we write with the number first.)

3. Nonmetals usually have negative oxidation numbers, although they cansometimes be positive:(a) The oxidation number of oxygen is usually in both ionic and molecu-

lar compounds. The major exception is in compounds called perox-ides, which contain the ion, giving each oxygen an oxidationnumber of

(b) The oxidation number of hydrogen is when bonded to nonmetals andwhen bonded to metals.

(c) The oxidation number of fluorine is in all compounds. The otherhalogens have an oxidation number of in most binary compounds.When combined with oxygen, as in oxyanions, however, they have pos-itive oxidation states.

4. The sum of the oxidation numbers of all atoms in a neutral compound is zero. Thesum of the oxidation numbers in a polyatomic ion equals the charge of the ion. Forexample, in the hydronium ion, the oxidation number of each hy-drogen is and that of oxygen is Thus the sum of the oxidation num-bers is which equals the net charge of the ion. This ruleis very useful in obtaining the oxidation number of one atom in a compoundor ion if you know the oxidation numbers of the other atoms, as illustratedin Sample Exercise 4.8.

SAMPLE EXERCISE 4.8Determine the oxidation state of sulfur in each of the following: (a) (b) (c)

(d) (e)

Solution (a) When bonded to a nonmetal, hydrogen has an oxidation number of (rule 3b). Because the molecule is neutral, the sum of the oxidation numbers mustequal zero (rule 4). Letting x equal the oxidation number of S, we have Thus, S has an oxidation number of

(b) Because this is an elemental form of sulfur, the oxidation number of S is 0(rule 1).

(c) Because this is a binary compound, we expect chlorine to have an oxidationnumber of (rule 3c). The sum of the oxidation numbers must equal zero (rule 4). Let-ting x equal the oxidation number of S, we have Consequently, theoxidation number of S must be

(d) Sodium, an alkali metal, always has an oxidation number of in its com-pounds (rule 2). Oxygen has a common oxidation state of (rule 3a). Letting x equalthe oxidation number of S, we have Therefore, the oxidationnumber of S in this compound is

(e) The oxidation state of O is (rule 3a). The sum of the oxidation numbersequals the net charge of the ion (rule 4). Thus we have From this relation we conclude that the oxidation number of S in this ion is

These examples illustrate that the oxidation number of a given element dependson the compound in which it occurs. The oxidation numbers of sulfur, as seen in theseexamples, range from to

PRACTICE EXERCISEWhat is the oxidation state of the boldfaced element in each of the following: (a)(b) NaH; (c) (d) (e)Answers: (a) (b) (c) (d) (e) -1+4;+6;-1;+5;

BaO2 ?SnBr4;Cr2O7

2-;P2O5;

+6.-2

+6.x + 41-22 = -2.SO4

2--2,-2

+4.21+12 + x + 31-22 = 0.

-2+1

+2.x + 21-12 = 0.

-1

-2.21+12 + x = 0.

H2S+1

SO4

2-.Na2SO3;SCl2 ;S8 ;H2S;

31+12 + 1-22 = +1,-2.+1

H3O+,

-1-1

-1+1-1.

O2

2-

-2

+3+2,

+11+

-2,+1, S2-K+

ACTIVITYOxidation Numbers

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⁄⁄

Oxidation of Metals by Acids and Salts

There are many kinds of redox reactions. For example, combustion reactions areredox reactions because elemental oxygen is converted to compounds of oxy-gen. • (Section 3.2) In this chapter we consider the redox reactions betweenmetals and either acids or salts. In Chapter 20 we will examine more complexkinds of redox reactions.

The reaction of a metal with either an acid or a metal salt conforms to the fol-lowing general pattern:

[4.25]

Examples:

These reactions are called displacement reactions because the ion in solution isdisplaced or replaced through oxidation of an element.

Many metals undergo displacement reactions with acids, producing saltsand hydrogen gas. For example, magnesium metal reacts with hydrochloric acidto form magnesium chloride and hydrogen gas (Figure 4.13 «). To show thatoxidation and reduction have occurred, the oxidation number for each atom isshown below the chemical equation for this reaction:

[4.26]

Notice that the oxidation number of Mg changes from 0 to The increase inthe oxidation number indicates that the atom has lost electrons and has thereforebeen oxidized. The ion of the acid decreases in oxidation number from to0, indicating that this ion has gained electrons and has therefore been reduced.The oxidation number of the ion remains and it is a spectator ion in thereaction. The net ionic equation is as follows:

[4.27]

Metals can also be oxidized by aqueous solutions of various salts. Iron metal,for example, is oxidized to by aqueous solutions of such as



The oxidation of Fe to form in this reaction is accompanied by the reductionof to Ni. Remember: Whenever one substance is oxidized, some other substancemust be reduced.

Ni2+Fe2+

Fe(s) + Ni2+(aq) ¡ Fe2+(aq) + Ni(s)

Fe(s) + Ni(NO3)2(aq) ¡ Fe(NO3)2(aq) + Ni(s)

Ni(NO3)2(aq):Ni2+,Fe2+

Mg(s) + 2H+(aq) ¡ Mg2+(aq) + H2(g)

-1,Cl-

+1H+

+2.

0-1+2-1+10

Mg(s) + 2HCl(aq) ¡ MgCl2(aq) + H2(g),

Mn(s) + Pb(NO3)2(aq) ¡ Mn(NO3)2(aq) + Pb(s)

Zn(s) + 2HBr(aq) ¡ ZnBr2(aq) + H2(g)

A + BX ¡ AX + B

Á Figure 4.13 Many metals, such asthe magnesium shown here, react withacids to form hydrogen gas. The bubblesare due to the hydrogen gas.


Lee R. Summerlin, Christie L.Borgford, and Julie B. Ealy,“Producing Hydrogen Gas fromCalcium Metal,” ChemicalDemonstrations, A Sourcebook forTeachers, Vol. 2 (American ChemicalSociety, Washington, DC, 1988)pp. 51–52.

An overhead projector demonstationemploying hydrogen gas formation.Lee R. Summerlin and James L. Ealy,Jr., “Activity Series for Some Metals,”Chemical Demonstrations,A Sourcebook for Teachers, Vol. 1(American Chemical Society,Washington, DC, 1988) p. 150.

2Al(s) + 6HBr(aq) ¡ 2AlBr3(aq) + 3H2(g)

SAMPLE EXERCISE 4.9Write the balanced molecular and net ionic equations for the reaction of aluminum with hydrobromic acid.

SolutionAnalyze: We must write the equation for the redox reaction between a metal and an acid.Plan: Metals react with acids to form salts and gas. To write the balanced equation, we must write the chemical formulas forthe two reactants and then determine the formula of the salt. The salt is composed of the cation formed by the metal and the anionof the acid.Solve: The formulas of the given reac-tants are Al and HBr. The cationformed by Al is and the anionfrom hydrobromic acid is Thus,the salt formed in the reaction is Writing the reactants and products andthen balancing the equation gives

AlBr3.Br-.

Al3+,

H2

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Both HBr and are soluble strongelectrolytes. Thus, the complete ionicequation is

Because is a spectator ion, the netionic equation is

Comment: The substance oxidized is the aluminum metal because its oxidation state changes from 0 to in the cation, therebyincreasing in oxidation number. The is reduced because its oxidation state changes from to 0 in

PRACTICE EXERCISE(a) Write the balanced molecular and net ionic equations for the reaction between magnesium and cobalt(II) sulfate. (b) What isoxidized and what is reduced in the reaction?Answers: (a) (b) Mg is oxidized and

is reduced.Co2+Mg(s) + Co2+(aq) ¡ Mg2+(aq) + Co(s);Mg(s) + CoSO4(aq) ¡ MgSO4(aq) + Co(s);

H2.+1H++3

Br-

AlBr3


Eas

e of

oxi

dat

ion

incr

ease

s

TABLE 4.5 Activity Series of Metals in Aqueous Solution

Metal Oxidation Reaction

Lithium Li(s) Li�(aq) � e�

Potassium K(s) K�(aq) � e�

Barium Ba(s) Ba2�(aq) � 2e�

Calcium Ca(s) Ca2�(aq) � 2e�

Sodium Na(s) Na�(aq) � e�

Magnesium Mg(s) Mg2�(aq) � 2e�

Aluminum Al(s) Al3�(aq) � 3e�

Manganese Mn(s) Mn2�(aq) � 2e�

Zinc Zn(s) Zn2�(aq) � 2e�

Chromium Cr(s) Cr3�(aq) � 3e�

Iron Fe(s) Fe2�(aq) � 2e�

Cobalt Co(s) Co2�(aq) � 2e�

Nickel Ni(s) Ni2�(aq) � 2e�

Tin Sn(s) Sn2�(aq) � 2e�

Lead Pb(s) Pb2�(aq) � 2e�

Copper Cu(s) Cu2�(aq) � 2e�

Silver Ag(s) Ag�(aq) � e�

Mercury Hg(l) Hg2�(aq) � 2e�

Platinum Pt(s) Pt2�(aq) � 2e�

Gold Au(s) Au3�(aq) � 3e�

Hydrogen H2(g) 2H�(aq) � 2e�

Bassam Z. Shakhashiri, “An ActivitySeries: Zinc, Copper, and Silver HalfCells,“ Chemical Demonstrations: AHandbook for Teachers of Chemistry,Vol. 4 (The University of WisconsinPress, Madison, 1992) pp. 101–106.

The zinc core of copper-coatedpennies reacts with acid to formpennies that float in thisdemonstration. Lee R. Summerlin,Christie L. Borgford, and Julie B. Ealy,“Floating Pennies,” ChemicalDemonstrations, A Sourcebook forTeachers, Vol. 2 (American ChemicalSociety, Washington, DC, 1988) p. 63.

The Activity Series

Can we predict whether a certain metal will be oxidized either by an acid or bya particular salt? This question is of practical importance as well as chemicalinterest. According to Equation 4.28, for example, it would be unwise to store asolution of nickel nitrate in an iron container because the solution would dis-solve the container. When a metal is oxidized, it appears to be eaten away as itreacts to form various compounds. Extensive oxidation can lead to the failure ofmetal machinery parts or the deterioration of metal structures.

Different metals vary in the ease with which they are oxidized. Zn is oxi-dized by aqueous solutions of for example, but Ag is not. Zn, therefore,loses electrons more readily than Ag; that is, Zn is easier to oxidize than Ag.

A list of metals arranged in order of decreasing ease of oxidation is called anactivity series. Table 4.5 ¥ gives the activity series in aqueous solution for manyof the most common metals. Hydrogen is also included in the table. The metalsat the top of the table, such as the alkali metals and the alkaline earth metals, aremost easily oxidized; that is, they react most readily to form compounds. They

Cu2+,

2Al(s) + 6H+(aq) ¡ 2Al3+(aq) + 3H2(g)

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

Hydrogen gas is collected as aproduct of the reaction of aluminumwith either HCl or NaOH. Lee R.Summerlin, Christie L. Borgford, andJulie B. Ealy, “ Making Hydrogen Gasfrom an Acid and a Base,” ChemicalDemonstrations, A Sourcebook forTeachers, Vol. 2 (American ChemicalSociety, Washington, DC, 1988)pp. 33–34.

Elements that lie near the top of thelist in Table 4.5 are referred to asactive metals.

R. Lipkin, “What Makes Gold Such aNoble Metal?” Science News, July 22,1995, 62.

MOVIEOxidation-Reduction Chemistry of Tin and Zinc

ACTIVITYPrecipitation, Redox, andNeutralization Reactions

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2AgNO3(aq) 2Ag(s)Cu(s) Cu(NO3)2(aq)

� �

Á Figure 4.14 When copper metal is placed in a solution of silver nitrate (a), a redoxreaction occurs, forming silver metal and a blue solution of copper(II) nitrate (b and c).

are called the active metals. The metals at the bottom of the activity series, suchas the transition elements from groups 8B and 1B, are very stable and form com-pounds less readily. These metals, which are used to make coins and jewelry, arecalled noble metals because of their low reactivity.

The activity series can be used to predict the outcome of reactions betweenmetals and either metal salts or acids. Any metal on the list can be oxidized by theions of elements below it. For example, copper is above silver in the series. Thus,copper metal will be oxidized by silver ions, as pictured in Figure 4.14 ¥:

[4.30]

The oxidation of copper to copper ions is accompanied by the reduction of sil-ver ions to silver metal. The silver metal is evident on the surface of the copperwires in Figure 4.14(b) and (c). The copper(II) nitrate produces a blue color inthe solution, which is most evident in part (c).

Only those metals above hydrogen in the activity series are able to react withacids to form For example, Ni reacts with HCl(aq) to form

[4.31]

Because elements below hydrogen in the activity series are not oxidized byCu does not react with HCl(aq). Interestingly, copper does react with nitric

acid, as shown previously in Figure 1.11. This reaction, however, is not a simpleoxidation of Cu by the ions of the acid. Instead, the metal is oxidized to by the nitrate ion of the acid, accompanied by the formation of brown nitrogendioxide,

[4.32]

What substance is reduced as copper is oxidized in Equation 4.32? In this casethe results from the reduction of We will examine reactions of thistype in more detail in Chapter 20.

NO3

-.NO2

Cu(s) + 4HNO3(aq) ¡ Cu(NO3)2(aq) + 2H2O(l) + 2NO2(g)

NO2(g):

Cu2+H+

H+,

Ni(s) + 2HCl(aq) ¡ NiCl2(aq) + H2(g)

H2 :H2.

Cu(s) + 2Ag+(aq) ¡ Cu2+(aq) + 2Ag(s)

MOVIEFormation of Silver Crystals

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A Closer Look The Aura of Goldof gold amounts to about (2000 tons). By contrast,over kg (16 million tons) of aluminum are producedannually. Gold is used mainly in jewelry (73%), coins (10%), andelectronics (9%). Its use in electronics relies on its excellent con-ductivity and its corrosion resistance. Gold is used, for example,to plate contacts in electrical switches, relays, and connections. Atypical telephone contains 33 gold-plated contacts.Gold is also used in computers and other microelectronic deviceswhere fine gold wire is used to link components.

Besides its value for jewelry, currency, and electronics, goldis also important in the health professions. Because of its resist-ance to corrosion by acids and other substances found in sali-va, gold is an ideal metal for dental crowns and caps, whichaccounts for about 3% of the annual use of the element. Thepure metal is too soft to use in dentistry, so it is combined withother metals to form alloys.

Touch-Tone®

1.5 * 10101.8 * 106 kgGold has been known since the earliest records of human exis-

tence. Throughout history people have cherished gold, havefought for it, and have died for it.

The physical and chemical properties of gold serve to makeit a special metal. First, its intrinsic beauty and rarity make itprecious. Second, gold is soft and can be easily formed intoartistic objects, jewelry, and coins (Figure 4.15 »). Third, gold isone of the least active metals (Table 4.5). It is not oxidized in airand does not react with water. It is unreactive toward basic solu-tions and nearly all acidic solutions. As a result, gold can befound in nature as a pure element rather than combined withoxygen or other elements, which accounts for its earlydiscovery.

Many of the early studies of the reactions of gold arosefrom the practice of alchemy, in which people attempted to turncheap metals, such as lead, into gold. Alchemists discoveredthat gold can be dissolved in a mixture of concentratedhydrochloric and nitric acids, known as aqua regia (“royalwater”). The action of nitric acid on gold is similar to that oncopper (Equation 4.32) in that the nitrate ion, rather than oxidizes the metal to The ions interact with toform highly stable ions. The net ionic equation for thereaction of gold with aqua regia is

All the gold ever mined would easily fit in a cube 19 m on aside and weighing about kg (125,000 tons). More than90% of this amount has been produced since the beginning of theCalifornia gold rush of 1848. Each year, worldwide production

1.1 * 108

AuCl4

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

Au(s) + NO3

-(aq) + 4H+(aq) + 4Cl-(aq) ¡

AuCl4

-Au3+Cl-Au3+.

H+,

3 : 1

SAMPLE EXERCISE 4.10Will an aqueous solution of iron(II) chloride oxidize magnesium metal? If so, writethe balanced molecular and net ionic equations for the reaction.

SolutionAnalyze: We are given two substances—an aqueous salt, and a metal, Mg—and asked if they react with each other.Plan: A reaction will occur if Mg is above in the activity series, Table 4.5. If the reac-tion occurs, the ion in will be reduced to Fe, and the elemental Mg will beoxidized to Solve: Because Mg is above Fe in the table, the reaction will occur. To write the formulafor the salt that is produced in the reaction, we must remember the charges on com-mon ions. Magnesium is always present in compounds as the chloride ion is The magnesium salt formed in the reaction is

Both and are soluble strong electrolytes and can be written in ionic form.then, is a spectator ion in the reaction. The net ionic equation is

The net ionic equation shows that Mg is oxidized and is reduced in this reaction.

PRACTICE EXERCISEWhich of the following metals will be oxidized by Zn, Cu, Fe?Answer: Zn and Fe

Pb(NO3)2 :

Fe2+

Mg(s) + Fe2+(aq) ¡ Mg2+(aq) + Fe(s)

Cl-,MgCl2FeCl2

Mg(s) + FeCl2(aq) ¡ MgCl2(aq) + Fe(s)

MgCl2 :Cl-.Mg2+;

Mg2+.FeCl2Fe2+

Fe2+

FeCl2 ,

« Figure 4.15 Portrait ofPharaoh Tutankhamun(1346–1337 B.C.) made fromgold and precious stones.From the inner coffin of thetomb of Tutankhamun.

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• If metathesis cannot occur, can the reactants possiblyengage in an oxidation-reduction reaction? This requiresthat there be both a reactant that can be oxidized and onethat can be reduced.

By asking questions such as these, you should be able topredict what might happen during the reaction. You might notalways be correct, but if you keep your wits about you, you willnot be far off. As you gain experience with chemical reactions,you will begin to look for reactants that might not be immedi-ately obvious, such as water from the solution or oxygen fromthe atmosphere.

One of the greatest tools available to us in chemistry isexperimentation. If you perform an experiment in which twosolutions are mixed, you can make observations that help youunderstand what is happening. For example, using the infor-mation in Table 4.1 to predict whether a precipitate will formis not nearly as exciting as actually seeing the precipitateform, as in Figure 4.4. Careful observations in the laboratoryportion of the course will make your lecture material easierto master.

Strategies in Chemistry Analyzing Chemical ReactionsIn this chapter you have been introduced to a great number ofchemical reactions. A major difficulty that students face in tryingto master material of this sort is gaining a “feel” for what happenswhen chemicals are allowed to react. In fact, you might marvel atthe ease with which your professor or teaching assistant can fig-ure out the results of a chemical reaction. One of our goals in thistextbook is to help you become more adept at predicting the out-come of reactions. The key to gaining this “chemical intuition” isunderstanding how to categorize reactions.

There are so many individual reactions in chemistry thatmemorizing them all is a futile task. It is far more fruitful to try touse pattern recognition to determine the general category of a reac-tion, such as metathesis or oxidation-reduction. Thus, when youare faced with the challenge of predicting the outcome of a chem-ical reaction, ask yourself the following pertinent questions:

• What are the reactants in the reaction?• Are they electrolytes or nonelectrolytes?• Are they acids and bases?• If the reactants are electrolytes, will metathesis produce a

precipitate? Water? A gas?

Molarity (M) is concentrationexpressed as moles of solute per literof solution, not per liter of solvent.Students frequently confuse thevolume on which molarity is based.

4.5 Concentrations of Solutions

The behavior of solutions often depends not only on the nature of the solutesbut also on their concentrations. Scientists use the term concentration to desig-nate the amount of solute dissolved in a given quantity of solvent or solution. Theconcept of concentration is intuitive: The greater the amount of solute dissolvedin a certain amount of solvent, the more concentrated the resulting solution. Inchemistry we often need to express the concentrations of solutions quantitatively.

Molarity

Molarity (symbol M) expresses the concentration of a solution as the number ofmoles of solute in a liter of solution (soln):

[4.33]

A 1.00 molar solution (written 1.00 M) contains 1.00 mol of solute in everyliter of solution. Figure 4.16 » shows the preparation of 250 mL of a 1.00 Msolution of by using a volumetric flask that is calibrated to hold exact-ly 250 mL. First, 0.250 mol of (39.9 g) is weighed out and placed in thevolumetric flask. Water is added to dissolve the salt, and the resultant solutionis diluted to a total volume of 250 mL. The molarity of the solution is

SAMPLE EXERCISE 4.11Calculate the molarity of a solution made by dissolving 23.4 g of sodium sulfate

in enough water to form 125 mL of solution.(Na2SO4)

(0.250 mol CuSO4)>(0.250 L soln) = 1.00 M.

CuSO4

CuSO4

Molarity =moles solute

volume of solution in liters

ANIMATIONSolution Formation from a Solid

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4.5 Concentrations of Solutions 135

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

SolutionAnalyze: We are given the number of grams of solute (23.4 g), its chemical formula

and the volume of the solution (125 mL), and we are asked to calculate themolarity of the solution.Plan: We can calculate molarity using Equation 4.33. To do so, we must convert the num-ber of grams of solute to moles and the volume of the solution from milliliters to liters.Solve: The number of moles of is obtained from its molar mass.

Converting the volume of the solution to liters:

Thus, the molarity is

Check: Because the numerator is only slightly larger than the denominator, it’s rea-sonable for the answer to be a little over 1 M. The units (mol L) are appropriate formolarity, and three significant figures are appropriate for the answer because each ofthe initial pieces of data had three significant figures.

PRACTICE EXERCISECalculate the molarity of a solution made by dissolving 5.00 g of glucose insufficient water to form exactly 100 mL of solution.Answer: 0.278 M

Expressing the Concentration of an Electrolyte

When an ionic compound dissolves, the relative concentrations of the ionsintroduced into the solution depend on the chemical formula of the

(C6H12O6)

>

Molarity =0.165 mol Na2SO4

0.125 L soln= 1.32

mol Na2SO4

L soln= 1.32 M

Liters soln = (125 mL )a 1 L1000 mL

b = 0.125 L

Moles Na2SO4 = (23.4 g Na2SO4 )¢ 1 mol Na2SO4

142 g Na2SO4 ≤ = 0.165 mol Na2SO4

Na2SO4

(Na2SO4),

Á Figure 4.16 Procedure for preparation of 0.250 L of 1.00 M solution of (a) Weighout 0.250 mol (39.9 g) of (formula amu). (b) Put the (solute)into a 250-mL volumetric flask, and add a small quantity of water. (c) Dissolve the solute byswirling the flask. (d) Add more water until the solution just reaches the calibration mark etchedon the neck of the flask. Shake the stoppered flask to ensure complete mixing.

CuSO4weight = 159.6CuSO4

CuSO4 .

ANIMATIONDissolution of KMnO4

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compound. For example, a 1.0 M solution of NaCl is 1.0 M in ions and1.0 M in ions. Similarly, a 1.0 M solution of is 2.0 M in ionsand 1.0 M in ions. Thus, the concentration of an electrolyte solution canbe specified either in terms of the compound used to make the solution

or in terms of the ions that the solution containsand .

SAMPLE EXERCISE 4.12What are the molar concentrations of each of the ions present in a 0.025 M aqueous solu-tion of calcium nitrate?

SolutionAnalyze: We are given the concentration of the ionic compound used to make the solu-tion and asked to determine the concentrations of the ions in the solution.Plan: We can use the subscripts in the chemical formula of the compound to deter-mine the relative concentrations of the ions.Solve: Calcium nitrate is composed of calcium ions and nitrate ions soits chemical formula is Because there are two ions for each ion inthe compound, each mole of that dissolves dissociates into 1 mol of and2 mol of Thus, a solution that is 0.025 M in is 0.025 M in and

in Check: The concentration of ions is twice that of ions, as the subscript 2 afterthe in the chemical formula suggests it should be.

PRACTICE EXERCISEWhat is the molar concentration of ions in a 0.015 M solution of potassiumcarbonate?Answer:

Interconverting Molarity, Moles, and Volume

The definition of molarity (Equation 4.33) contains three quantities—molarity,moles solute, and liters of solution. If we know any two of these, we can calcu-late the third. For example, if we know the molarity of a solution, we can calcu-late the number of moles of solute in a given volume. Molarity, therefore, is aconversion factor between volume of solution and moles of solute. Calculationof the number of moles of in 2.0 L of 0.200 M solution illustratesthe conversion of volume to moles:

Dimensional analysis can be used in this conversion if we express molarity asmoles/liter soln. To obtain moles, therefore, we multiply liters and molarity:

To illustrate the conversion of moles to volume, let’s calculate the volume of0.30 M solution required to supply 2.0 mol of

In this case we must use the reciprocal of molarity in the conversion:liters = moles * 1>M.

Liters soln = (2.0 mol HNO3 )¢ 1 L soln0.30 mol HNO3

≤ = 6.7 L soln

HNO3 :HNO3

moles = liters * molarity.

= 0.40 mol HNO3

Moles HNO3 = (2.0 L soln )¢0.200 mol HNO3

1 L soln ≤

HNO3HNO3

0.030 M K+

K+

Ca(NO3)2NO3

-Ca2+NO3

-NO3

-.2 * 0.025 M = 0.050 MCa2+Ca(NO3)2NO3

-.Ca2+Ca(NO3)2

Ca2+NO3

-Ca(NO3)2 .(NO3

-),(Ca2+)

1.0 M SO4

2-)(2.0 M Na+(1.0 M Na2SO4)

SO4

2-Na+Na2SO4Cl-

Na+Arthur M. Last, “A Cyclist’s Guide toIonic Concentration,” J. Chem. Educ.,Vol. 75, 1998, 1433.

BLB04-112-151-V4 2/21/02 1:34 PM Page 136

SAMPLE EXERCISE 4.13How many grams of are required to make 0.350 L of 0.500 M

SolutionAnalyze: We are given the volume of the solution (0.350 L), its concentration (0.500 M), and the identity of the solute andasked to calculate the number of grams of the solute in the solution.

Plan: We can use the definition ofmolarity (Equation 4.33) to determinethe number of moles of solute, andthen convert moles to grams using themolar mass of the solute:

Solve: Calculating the moles ofusing the molarity and vol-

ume of solution gives:

Because each mole of weighs142 g, the required number of gramsof is

Check: The magnitude of the answer, the units, and the number of significant figures are all appropriate.

PRACTICE EXERCISE(a) How many grams of are there in 15 mL of 0.50 M (b) How many milliliters of 0.50 M solution are need-ed to provide 0.038 mol of this salt?Answers: (a) 1.1 g; (b) 76 mL

Na2SO4Na2SO4 ?Na2SO4

Na2SO4

Na2SO4

Na2SO4

(Na2SO4)

Na2SO4 ?Na2SO4

4.5 Concentrations of Solutions 137

Grams Na2SO4 = (0.175 mol Na2SO4 )¢ 142 g Na2SO4

1 mol Na2SO4 ≤ = 24.9 g Na2SO4

= 0.175 mol Na2SO4

= (0.350 L soln)¢0.500 mol Na2SO4

1 L soln ≤

Moles Na2SO4 = liters soln * MNa2SO4

MNa2SO4=

moles Na2SO4

liters soln

* In diluting a concentrated acid or base, the acid or base should be added to water and then furtherdiluted by adding more water. Adding water directly to concentrated acid or base can cause spat-tering because of the intense heat generated.

Dilution

Solutions that are used routinely in the laboratory are often purchased or pre-pared in concentrated form (called stock solutions). Hydrochloric acid, for exam-ple, is purchased as a 12 M solution (concentrated HCl). Solutions of lowerconcentrations can then be obtained by adding water, a process called dilution.*

To illustrate the preparation of a dilute solution from a concentrated one,suppose we wanted to prepare 250 mL (that is, 0.250 L) of 0.100 M solu-tion by diluting a stock solution containing 1.00 M When solvent isadded to dilute a solution, the number of moles of solute remains unchanged.

[4.34]

Because we know both the volume and concentration of the dilute solution, wecan calculate the number of moles of it contains.

Now we can calculate the volume of the concentrated solution needed to provide0.0250 mol

L of conc soln = (0.0250 mol CuSO4 )¢ 1 L soln1.00 mol CuSO4

≤ = 0.0250 L

CuSO4 :

Mol CuSO4 in dil soln = (0.250 L soln)¢0.100

mol CuSO4

L soln ≤ = 0.0250 mol CuSO4

CuSO4

Moles solute before dilution = moles solute after dilution

CuSO4.CuSO4

ANIMATIONSolution Formation by Dilution

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(a) (b) (c)

Á Figure 4.17 Procedure for preparing 250 mL of 0.100 M by dilution of0.100 M (a) Draw 25.0 mL of the 1.00 M solution into a pipet. (b) Add this to a250-mL volumetric flask. (c) Add water to dilute the solution to a total volume of 250 mL.

CuSO4 .CuSO4

It is important to emphasize theproper use of the equation

A commonerror is to use moles, instead ofmolarity, in this calculation.

MconcVconc = MdilVdil .

Students may already be familiarwith the notation where c refers to the concentratedsolution and d refers to the dilutesolution.

McVc = MdVd ,

Thus, this dilution is achieved by withdrawing 0.0250 L (that is, 25.0 mL) of the1.0 M solution using a pipet, adding it to a 250-mL volumetric flask, and thendiluting it to a final volume of 250 mL, as shown in Figure 4.17 Á. Notice that thediluted solution is less intensely colored than the concentrated one.

In laboratory situations, calculations of this sort are often made very quick-ly with a simple equation that can be derived by remembering that the numberof moles of solute is the same in both the concentrated and dilute solutions andthat

[4.35]

The molarity of the more concentrated stock solution is always largerthan the molarity of the dilute solution Because the volume of the solutionincreases upon dilution, is always larger than Although Equation 4.35is derived in terms of liters, any volume unit can be used so long as that sameunit is used on both sides of the equation. For example, in the calculation wedid for the solution, we have

Solving for gives mL as before.

SAMPLE EXERCISE 4.14How many milliliters of 3.0 M are needed to make 450 mL of 0.10 M

SolutionAnalyze: We need to dilute a concentrated solution. We are given the molarity of amore concentrated solution (3.0 M) and the volume and molarity of a more dilute onecontaining the same solute (450 mL of 0.10 M solution). We must calculate the volumeof the concentrated solution needed to prepare the dilute solution.Plan: We can calculate the number of moles of solute, in the dilute solution andthen calculate the volume of the concentrated solution needed to supply this amountof solute. Alternatively, we can directly apply Equation 4.35. Let’s compare the twomethods.Solve: Calculating the moles of in the dilute solution:

Moles H2SO4 in dilute solution = (0.450 L soln )¢0.10 mol H2SO4

1 L soln ≤ = 0.045 mol H2SO4

H2SO4

H2SO4,

H2SO4 ?H2SO4

Vconc = 25.0Vconc

(1.00 M)(Vconc) = (0.100 M)(250 mL)

CuSO4

Vconc.Vdil

(Mdil).(Mconc)

Mconc * Vconc = Mdil * Vdil

Moles solute in conc soln = moles solute in dil soln

moles = molarity * liters:

Lloyd J. McElroy, “TeachingDilutions,” J. Chem. Educ., Vol. 73,1996, 765–766.

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4.6 Solution Stoichiometry and Chemical Analysis 139

Irwin L. Shaprio, “On the Use ofIntravenous Solutions to Teach SomePrinciples of Solution Chemistry,” J.Chem. Educ., Vol. 59, 1982, 725.

Calculating the volume of concentrated solution that contains 0.045 mol

Converting liters to milliliters gives 15 mL.If we apply Equation 4.35, we get the same result:

Either way, we see that if we start with 15 mL of 3.0 M and dilute it to a totalvolume of 450 mL, the desired 0.10 M solution will be obtained.Check: The calculated volume seems reasonable because a small volume of concen-trated solution is used to prepare a large volume of dilute solution.

PRACTICE EXERCISE(a) What volume of 2.50 M lead nitrate solution contains 0.0500 mol of (b) Howmany milliliters of 5.0 M solution must be diluted to prepare 250 mL of 0.10 Msolution? (c) If 10.0 mL of a 10.0 M stock solution of NaOH is diluted to 250 mL, whatis the concentration of the resulting solution?Answers: (a) (b) 5.0 mL; (c) 0.40 M

4.6 Solution Stoichiometry and Chemical Analysis

Imagine that you have to determine the concentrations of several ions in a sam-ple of lake water. Although many instrumental methods have been developedfor such analyses, chemical reactions such as those discussed in this chapter con-tinue to be used. In Chapter 3 we learned that if you know the chemical equa-tion and the amount of one reactant consumed in the reaction, you can calculatethe quantities of other reactants and products. In this section we briefly exploresuch analyses of solutions.

Recall that the coefficients in a balanced equation give the relative numberof moles of reactants and products. • (Section 3.6) To use this information, wemust convert the quantities of substances involved in a reaction into moles. Whenwe are dealing with grams of substances, as we were in Chapter 3, we use themolar mass to achieve this conversion. When we are working with solutions ofknown molarity, however, we use molarity and volume to determine the num-ber of moles Figure 4.18 ¥ summarizes this approachto using stoichiometry.

(moles solute = M * L).

0.0200 L = 20.0 mL;

K2Cr2O7

Pb2+?

H2SO4

Vconc =(0.10 M )(450 mL)

3.0 M = 15 mL

(3.0 M)(Vconc) = (0.10 M)(450 mL)

L conc soln = (0.045 mol H2SO4 )¢ 1 L soln3.0 mol H2SO4

≤ = 0.015 L soln

H2SO4 :

Laboratory unitsLaboratory units Chemical units

Gramsof

substance A

Molesof

substance A

Molesof

substance B

Usemolar

mass of A

Use M= mol/L

Volume ormolarity ofsubstance A

Gramsof

substance B

Usemolar

mass of A

Use M =mol/L

Use stoichiometriccoeffiecients of A and B

Volume ormolarity ofsubstance B

Á Figure 4.18 Outline of the procedure used to solve stoichiometry problems thatinvolve measured (laboratory) units of mass, solution concentration (molarity), or volume.

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SAMPLE EXERCISE 4.15How many grams of are needed to neutralize 25.0 mL of 0.100 M

SolutionAnalyze: The reactants are an acid, and a base, The volume and molarity of are given, and we are askedhow many grams of are needed to neutralize this quantity of

Plan: We can use the molarity and vol-ume of the solution to calculatethe number of moles of Wethen use the balanced equation torelate the moles of to moles of

Finally, we can convertmoles of to grams. Thesesteps can be summarized as follows:

Solve: The product of the molar con-centration of a solution and its volumein liters gives the number of moles ofsolute:

Because this is an acid-base neutral-ization reaction, and react to form and the salt con-taining and

Thus, Therefore,

Check: The size of the answer is reasonable. A small volume of dilute acid will require only a small amount of base to neutralize it.

PRACTICE EXERCISE(a) How many grams of NaOH are needed to neutralize 20.0 mL of 0.150 M solution? (b) How many liters of 0.500 MHCl(aq) are needed to react completely with 0.100 mol of forming a precipitate of Answers: (a) 0.240 g; (b) 0.400 L

PbCl2(s)?Pb(NO3)2(aq),H2SO4

Ca(OH)2.2 mol HNO3 � 1 mol

NO3

-:Ca2+H2O

Ca(OH)2HNO3

Ca(OH)2

Ca(OH)2.HNO3

HNO3.HNO3

HNO3.Ca(OH)2

HNO3Ca(OH)2.HNO3,

HNO3 ?Ca(OH)2

= 0.0926 g Ca(OH)2

Grams Ca(OH)2 = (2.50 * 10-3 mol HNO3 )a1 mol Ca(OH)2

2 mol HNO3 b a 74.1 g Ca(OH)2

1 mol Ca(OH)2 b

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

= 2.50 * 10-3 mol HNO3

Moles HNO3 = LHNO3* MHNO3

= (0.0250 L )a0.100

mol HNO3

L b

LHNO3* MHNO3

Q mol HNO3 Q mol Ca(OH)2 Q g Ca(OH)2


Titrations

To determine the concentration of a particular solute in a solution, chemists oftencarry out a titration, which involves combining a sample of the solution with areagent solution of known concentration, called a standard solution. Titrationscan be conducted using acid-base, precipitation, or oxidation-reduction reac-tions. Suppose we have an HCl solution of unknown concentration and an NaOHsolution we know to be 0.100 M. To determine the concentration of the HCl solu-tion, we take a specific volume of that solution, say 20.00 mL. We then slowly addthe standard NaOH solution to it until the neutralization reaction between theHCl and NaOH is complete. The point at which stoichiometrically equivalentquantities are brought together is known as the equivalence point of the titration.

In order to titrate an unknown with a standard solution, there must be someway to determine when the equivalence point of the titration has been reached.In acid-base titrations, dyes known as acid-base indicators are used for this pur-pose. For example, the dye known as phenolphthalein is colorless in acidic solu-tion but is pink in basic solution. If we add phenolphthalein to an unknownsolution of acid, the solution will be colorless, as seen in Figure 4.19(a) ». Wecan then add standard base from a buret until the solution barely turns from col-orless to pink, as seen in Figure 4.19(b). This color change indicates that the acidhas been neutralized and the drop of base that caused the solution to become

Bassam Z. Shakhashiri, “ColorfulAcid-Base Indicators,“ ChemicalDemonstrations: A Handbook forTeachers of Chemistry, Vol. 3 (TheUniversity of Wisconsin Press,Madison, 1989) pp. 33–40.

The equivalence point of a titration isthe point where the stoichiometricallycorrect number of moles of eachreactant is present. The end point of atitration is the point where theindicator changes. They are not thesame, although we choose anindicator that will change as close tothe equivalence point as possible.

ANIMATIONAcid-Base Titration

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(a) (b) (c)

Á Figure 4.19 Change in appearance of a solution containing phenolphthalein indicatoras base is added. Before the end point, the solution is colorless (a). As the end point isapproached, a pale pink color forms where the base is added (b). At the end point, this palepink color extends throughout the solution after the mixing. As even more base is added,the intensity of the pink color increases (c).

20.0 mL ofacidsolution

Buret

Initial volumereading

Pipet

20.0 mL ofacidsolution Standard

NaOHsolution

Final volumereading

Neutralizedsolution(indicatorhas changedcolor)

(a) (b) (c)

Á Figure 4.20 Procedure for titrating an acid against a standardized solution of NaOH.(a) A known quantity of acid is added to a flask. (b) An acid-base indicator is added, andstandardized NaOH is added from a buret. (c) Equivalence point is signaled by a colorchange in the indicator.

colored has no acid to react with. The solution therefore becomes basic, and thedye turns pink. The color change signals the end point of the titration, which usu-ally coincides very nearly with the equivalence point. Care must be taken tochoose indicators whose end points correspond to the equivalence point of thetitration. We will consider this matter in Chapter 17. The titration procedure issummarized in Figure 4.20 ¥.

Bassam Z. Shakhashiri, “RainbowColors with Mixed Acid-BaseIndicators,“ Chemical Demonstrations:A Handbook for Teachers of Chemistry,Vol. 3 (The University of WisconsinPress, Madison, DC, 1989)pp. 41–46.

ACTIVITYAcid-Base Titration

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Bassam Z. Shakhashiri, “Acid-BaseIndicators Extracted from Plants,“Chemical Demonstrations: AHandbook for Teachers of Chemistry,Vol. 3 (The University of WisconsinPress, Madison, 1989) pp. 50–57.

Ara S. Kooser, Judith L. Jenkins, andLawrence E. Welch, “Acid-BaseIndicators: a New Look at an OldTopic,” J. Chem. Educ., Vol. 78, 2001,1504–1506.

SAMPLE EXERCISE 4.16The quantity of in a water supply is determined by titrating the sample with

(a) How many grams of chloride ion are in a sample of the water if 20.2 mL of 0.100 Mis needed to react with all the chloride in the sample? (b) If the sample has a mass

of 10.0 g, what percent does it contain?

SolutionAnalyze: We are given the volume (20.2 mL) and molarity (0.100 M) of a solution of

and the chemical equation for reaction of this ion with We are asked first tocalculate the number of grams of in the sample and, second, to calculate the masspercent of in the sample.

(a) Plan: We begin by using the volume and molarity of to calculate thenumber of moles of used in the titration. We can then use the balanced equationto determine the moles of and from that the grams of

Solve:

From the balanced equation we see that 1 mol Using this informa-tion and the molar mass of Cl, we have

(b) Plan: To calculate the percentage of in the sample, we compare the numberof grams of in the sample, with the original mass of the sample, 10.0 g.

Solve:

Comment: Chloride ion is one of the most common ions in water and sewage. Oceanwater contains 1.92% Whether water containing tastes salty depends on theother ions present. If the only accompanying ions are a salty taste may be detect-ed with as little as 0.03%

PRACTICE EXERCISE

A sample of an iron ore is dissolved in acid, and the iron is converted to Thesample is then titrated with 47.20 mL of 0.02240 M solution. The oxidation-reduction reaction that occurs during titration is as follows:

(a) How many moles of wereadded to the solution? (b) How many moles of were in the sample? (c) How manygrams of iron were in the sample? (d) If the sample had a mass of 0.8890 g, what is thepercentage of iron in the sample?Answers: (a) (b) (c) 0.2952 g; (d) 33.21%5.286 * 10-3 mol Fe2+;1.057 * 10-3 mol MnO4

-;

Fe2+MnO4

-8H+(aq) ¡ Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)MnO4

-(aq) + 5Fe2+(aq) +

MnO4

-Fe2+.

Cl-.Na+,Cl-Cl-.

%Cl- =7.17 * 10-3 g

10.0 g* 100% = 0.717% Cl-

7.17 * 10-2 g,Cl-Cl-

Grams Cl- = (2.02 * 10-3 mol Ag+ )a 1 mol Cl-

1 mol Ag+

b a 35.5 g Cl-

1 mol Cl- b = 7.17 * 10-2 g Cl-

Ag+ � 1 mol Cl-.

= 2.02 * 10-3 mol Ag+

Mol Ag+ = (20.2 mL soln )a 1 L soln

1000 mL soln b a0.100

mol Ag+

L soln b

Cl-.Cl-Ag+

Ag+Cl-

Cl-Cl-.Ag+

Cl-Ag+

Ag+(aq) + Cl-(aq) ¡ AgCl(s)

Ag+.Cl-

Infusions from a series of herbal teasprovide a source of natural pHindicators in this simpledemonstration. Dianne N. Epp, “Teas as Natural Indicators,” J. Chem.Educ., Vol. 70, 1993, 326.

SAMPLE EXERCISE 4.17One commercial method used to peel potatoes is to soak them in a solution of NaOH for a short time, remove them from theNaOH, and spray off the peel. The concentration of NaOH is normally in the range of 3 to 6 M. The NaOH is analyzed periodi-cally. In one such analysis, 45.7 mL of 0.500 M is required to neutralize a 20.0-mL sample of NaOH solution. What is theconcentration of the NaOH solution?

SolutionAnalyze: We are given the volume (45.7 mL) and molarity (0.500 M) of an solution that reacts completely with a 20.0-mLsample of NaOH. We are asked to calculate the molarity of the NaOH solution.Plan: We can use the volume and molarity of the to calculate the number of moles of this substance. Then, we can use thisquantity and the balanced equation for the reaction to calculate the number of moles of NaOH. Finally, we can use the moles ofNaOH and the volume of this solution to calculate molarity.

H2SO4

H2SO4

H2SO4

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Solve: The number of moles of is given by the product of the volumeand molarity of this solution:

Acids react with metal hydroxides toform water and a salt. Thus, the bal-anced equation for the neutralizationreaction is

According to the balanced equation,1 mol mol NaOH. There-fore,

Knowing the number of moles ofNaOH present in 20.0 mL of solutionallows us to calculate the molarity ofthis solution:

PRACTICE EXERCISEWhat is the molarity of an NaOH solution if 48.0 mL is needed to neutralize 35.0 mL of 0.144 MAnswer: 0.210 M

H2SO4 ?

H2SO4 � 2

H2SO4


= 2.28 mol NaOH

L soln= 2.28 M

Molarity NaOH =mol NaOH

L soln= a4.56 * 10-2 mol NaOH

20.0 mL soln b a1000 mL soln

1 L solnb

= 4.56 * 10-2 mol NaOH

Moles NaOH = (2.28 * 10-2 mol H2SO4 )a 2 mol NaOH1 mol H2SO4

b

H2SO4(aq) + 2NaOH(aq) ¡ 2H2O(l) + Na2SO4(aq)

= 2.28 * 10-2 mol H2SO4

Moles H2SO4 = (45.7 mL soln )a 1 L soln

1000 mL soln b a0.500

mol H2SO4

L soln b

SAMPLE INTEGRATIVE EXERCISE 4: Putting Concepts TogetherNote: Integrative exercises require skills from earlier chapters as well as ones from thepresent chapter.

A sample of 70.5 mg of potassium phosphate is added to 15.0 mL of 0.050 M sil-ver nitrate, resulting in the formation of a precipitate. (a) Write the molecular equationfor the reaction. (b) What is the limiting reactant in the reaction? (c) Calculate the the-oretical yield, in grams, of the precipitate that forms.

Solution (a) Potassium phosphate and silver nitrate are both ionic compounds. Potas-sium phosphate contains and ions, so its chemical formula is Silvernitrate contains and ions, so its chemical formula is Because bothreactants are strong electrolytes, the solution contains and ionsbefore the reaction occurs. According to the solubility guidelines in Table 4.1, and

form an insoluble compound, so will precipitate from the solution. Incontrast, and will remain in solution because is water soluble. Thus,the balanced molecular equation for the reaction is

(b) To determine the limiting reactant, we must examine the number of moles ofeach reactant. • (Section 3.7) The number of moles of is calculated from themass of the sample using the molar mass as a conversion factor. • (Section 3.4) Themolar mass of is Converting mil-ligrams to grams and then to moles, we have

We determine the number of moles of from the volume and molarity of the solu-tion. • (Section 4.5) Converting milliliters to liters and then to moles, we have

Comparing the amounts of the two reactants, we find that there aretimes as many moles of as there are moles

of According to the balanced equation, however, 1 mol requires 3 molof Thus, there is insufficient to consume the and isthe limiting reactant.

AgNO3K3PO4,AgNO3AgNO3.K3PO4K3PO4.

AgNO317.5 * 10-42>13.32 * 10-42 = 2.3

(15.0 mL )a10-3 L

1 mL b a0.050 mol AgNO3

L b = 7.5 * 10-4 mol AgNO3

AgNO3

(70.5 mg K3PO4 )a10-3 g K3PO4

1 mg K3PO4 b a 1 mol K3PO4

212.3 g K3PO4 b = 3.32 * 10-4 mol K3PO4

212.3 g>mol.=3139.12 + 31.0 + 4116.02K3PO4

K3PO4

K3PO4(aq) + 3AgNO3(aq) ¡ Ag3PO4(s) + 3KNO3(aq)

KNO3NO3

-K+Ag3PO4PO4

3-Ag+

NO3

-Ag+,K+, PO4

3-,AgNO3.NO3

-Ag+K3PO4.PO4

3-K+

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(c) The precipitate is whose molar mass is To calculate the number of grams of that could be produced in

this reaction (the theoretical yield), we use the number of moles of the limiting reac-tant, converting mol We use the coefficients inthe balanced equation to convert moles of to moles and we use themolar mass of to convert the number of moles of this substance to grams.

The answer has only two significant figures because the quantity of is givento only two significant figures.

AgNO3

(7.5 * 10-4 mol AgNO3 )a1 mol Ag3PO4

3 mol AgNO3 b a418.7 g Ag3PO4

1 mol Ag3PO4 b = 0.10 g Ag3PO4

Ag3PO4

Ag3PO4,AgNO3

Ag3PO4.Ag3PO4 Q gAgNO3 Q mol

Ag3PO4418.7 g>mol.31107.92 + 31.0 + 4116.02 = Ag3PO4,

Summary and Key Terms

Introduction and Section 4.1 Solutions in which wateris the dissolving medium are called aqueous solutions.The component of the solution that is in the greater quan-tity is the solvent. The other components are solutes.

Any substance whose aqueous solution contains ionsis called an electrolyte. Any substance that forms a solutioncontaining no ions is a nonelectrolyte. Those electrolytesthat are present in solution entirely as ions are strong elec-trolytes, whereas those that are present partly as ions andpartly as molecules are weak electrolytes. Ionic compoundsdissociate into ions when they dissolve, and they are strongelectrolytes. Most molecular compounds are nonelectrolytes,although some are weak electrolytes and a few are strongelectrolytes. When representing the ionization of a weakelectrolyte in solution, a double arrow is used, indicatingthat the forward and reverse reactions can achieve a chem-ical balance called a chemical equilibrium.

Section 4.2 Precipitation reactions are those in whichan insoluble product, called a precipitate, forms. Solubil-ity guidelines help determine whether or not an ionic com-pound will be soluble in water. (The solubility of asubstance is the amount that dissolves in a given quanti-ty of solvent.) Reactions such as precipitation reactions, inwhich cations and anions appear to exchange partners, arecalled exchange reactions, or metathesis reactions.

Chemical equations can be written to show whetherdissolved substances are present in solution predominant-ly as ions or molecules. When the complete chemical for-mulas of all reactants and products are used, the equationis called a molecular equation. A complete ionic equationshows all dissolved strong electrolytes as their componentions. In a net ionic equation, those ions that go throughthe reaction unchanged (spectator ions) are omitted.

Section 4.3 Acids and bases are important electrolytes.Acids are proton donors; they increase the concentrationof in aqueous solutions to which they are added.Bases are proton acceptors; they increase the concentra-tion of in aqueous solutions. Those acids andbases that are strong electrolytes are called strong acidsand strong bases, respectively. Those that are weak elec-

OH-(aq)

H+(aq)

trolytes are weak acids and weak bases. When solutionsof acids and bases are mixed, a neutralization reaction re-sults. The neutralization reaction between an acid and ametal hydroxide produces water and a salt. Gases can alsobe formed as a result of acid-base reactions. The reactionof a sulfide with an acid forms the reaction be-tween a carbonate and an acid forms

Section 4.4 Oxidation is the loss of electrons by a sub-stance, whereas reduction is the gain of electrons by a sub-stance. Oxidation numbers help us keep track of electronsduring chemical reactions and are assigned to atoms byusing specific rules. The oxidation of an element results inan increase in its oxidation number, whereas reduction isaccompanied by a decrease in oxidation number. Oxida-tion is always accompanied by reduction, givingoxidation-reduction, or redox, reactions.

Many metals are oxidized by acids, and salts. Theredox reactions between metals and acids and betweenmetals and salts are called displacement reactions. Theproducts of these displacement reactions are always anelement ( or a metal) and a salt. Comparing such reac-tions allows us to rank metals according to their ease ofoxidation. A list of metals arranged in order of decreasingease of oxidation is called an activity series. Any metal onthe list can be oxidized by ions of metals (or ) below itin the series.

Section 4.5 The composition of a solution expresses therelative quantities of solvent and solutes that it contains.One of the common ways to express the concentration ofa solute in a solution is in terms of molarity. The molarityof a solution is the number of moles of solute per liter ofsolution. Molarity makes it possible to interconvert solu-tion volume and number of moles of solute. Solutions ofknown molarity can be formed either by weighing out thesolute and diluting it to a known volume or by the dilutionof a more concentrated solution of known concentration(a stock solution). Adding solvent to the solution (theprocess of dilution) decreases the concentration of thesolute without changing the number of moles of solute inthe solution 1Mconc * Vconc = Mdil * Vdil2.

H+

H2

O2,

CO2(g).H2S(g);

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Exercises 145

CQ

CQ

CQ

CQ

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CQ

�

�

�

�

�

�

�

��

�

AX AY AZ

(a) (b) (c)

�

�

� �

�

�

�

�

�

��

Section 4.6 In the process called titration, we react a so-lution of known concentration (a standard solution) witha solution of unknown concentration in order to determinethe unknown concentration or the quantity of solute in theunknown. The point in the titration at which stoichiomet-

rically equivalent quantities of reactants are brought to-gether is called the equivalence point. An indicator canbe used to show the end point of the titration, which co-incides closely with the equivalence point.

Exercises

4.8 The two diagrams represent aqueous solutions of twodifferent substances, AX and BY. Are these substancesstrong electroytes, weak electrolytes, or nonelectrolytes?Which do you expect to be the better conductor of elec-tricity? Explain.

Electrolytes

4.1 Although pure water is a poor conductor of electricity,we are cautioned not to operate electrical appliancesaround water. Why?

4.2 When asked what causes electrolyte solutions to conductelectricity, a student responds that it is due to the move-ment of electrons through the solution. Is the student cor-rect? If not, what is the correct response?

4.3 When methanol, is dissolved in water, a non-conducting solution results. When acetic acid, dissolves in water, the solution is weakly conducting andacidic in nature. Describe what happens upon dissolutionin the two cases, and account for the different results.

4.4 We have learned in this chapter that many ionic solidsdissolve in water as strong electrolytes, that is, as sepa-rated ions in solution. What properties of water facilitatethis process?

4.5 Specify how each of the following strong electrolytes ion-izes or dissociates into ions upon dissolving in water:(a) (b) (c) (d)

4.6 Specify how each of the following strong electrolytes ion-izes or dissociates into ions upon dissolving in water:(a) (b) (c) (d)

4.7 Aqueous solutions of three different substances, AX, AY,and AZ are represented by the three diagrams below.Identify each substance as a strong electrolyte, weak elec-trolyte, or nonelectrolyte.

(NH4)2SO4.HClO4;Al(NO3)3 ;MgI2;

Ca1OH22 .K2SO4;HNO3;ZnCl2;

HC2H3O2,CH3OH,

4.9 Formic acid, is a weak electrolyte. What soluteparticles are present in an aqueous solution of this com-pound? Write the chemical equation for the ionization of

4.10 Acetone, is a nonelectrolyte; hypochlorousacid, HClO, is a weak electrolyte; and ammonium chlo-ride, is a strong electrolyte. (a) What are the soluteparticles present in aqueous solutions of each compound?(b) If 0.1 mol of each compound is dissolved in solution,which one contains 0.2 mol of solute particles, which con-tains 0.1 mol of solute particles, and which contains some-where between 0.1 and 0.2 mol of solute particles?

NH4Cl,

CH3COCH3,HCHO2.

HCHO2,

reaction. (a) and (b) and(c) and

4.14 Identify the precipitate (if any) that forms when the fol-lowing solutions are mixed, and write a balanced equa-tion for each reaction. (a) and NaOH; (b) NaOHand (c) and

4.15 Write the balanced complete ionic equations and net ionicequations for the reactions that occur when each of thefollowing solutions is mixed.(a) and MgSO4(aq)Na2CO3(aq)

Cu(C2H3O2)2 .Na2SK2SO4;Sn(NO3)2

Pb(NO3)2 .FeSO4NiSO4;NaNO3AgNO3;Na2CO3Precipitation Reactions and Net Ionic Equations

4.11 Using solubility guidelines, predict whether each of thefollowing compounds is soluble or insoluble in water: (a) (b) (c) (d)(e)

4.12 Predict whether each of the following compounds is sol-uble in water: (a) (b) (c)(d) (e)

4.13 Will precipitation occur when the following solutions aremixed? If so, write a balanced chemical equation for the

AgC2H3O2.AlPO4;Ba(NO3)2 ;PbSO4;Ni(OH)2;

(NH4)2SO4.SrCO3;Cs3PO4;Ag2S;NiCl2 ;

CQ

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CQ

CQ

CQ

CQ

CQ

(b) and (c) and

4.16 Write balanced net ionic equations for the reactions thatoccur in each of the following cases. Identify the specta-tor ion or ions in each reaction.(a)(b)(c)

4.17 Separate samples of a solution of an unknown salt aretreated with dilute solutions of HBr, and NaOH.A precipitate forms only with Which of the fol-lowing cations could the solution contain:

4.18 Separate samples of a solution of an unknown ionic com-pound are treated with dilute andPb(NO3)2 ,AgNO3,

Ba2+ ?Pb2+;K+;

H2SO4.H2SO4,

Pb(NO3)2(aq) + KOH(aq) ¡AgNO3(aq) + K2SO4(aq) ¡Cr2(SO4)3(aq) + (NH4)2CO3(aq) ¡

CaCl2(aq)(NH4)3PO4(aq)Na2S(aq)Pb(NO3)2(aq) Precipitates form in all three cases. Which of the

following could be the anion of the unknown salt:

4.19 The labels have fallen off two bottles, one containingand the other containing You have

a bottle of dilute How could you use it to test aportion of each solution to identify which solution iswhich?

4.20 You know that an unlabeled bottle contains one of thefollowing: or A friend sug-gests that you test a portion of the bottle with and then with NaCl. What behavior would you expectwhen each of these compounds is added to the unlabeledbottle?

Ba(NO3)2

Al2(SO4)3 .CaCl2 ,AgNO3,

H2SO4.Pb(NO3)2 .Mg(NO3)2

NO3

- ?CO3

2-;Br-;

BaCl2 .

4.29 Complete and balance the following molecular equations,and then write the net ionic equation for each:(a)(b)(c)

4.30 Write the balanced molecular and net ionic equations foreach of the following neutralization reactions:(a) Aqueous acetic acid is neutralized by aqueous potas-

sium hydroxide.(b) Solid chromium(III) hydroxide reacts with nitric acid.(c) Aqueous hypochlorous acid and aqueous calcium

hydroxide react.4.31 Write balanced molecular and net ionic equations for the

following reactions, and identify the gas formed in each:(a) solid cadmium sulfide reacts with an aqueous solu-tion of sulfuric acid; (b) solid magnesium carbonate reactswith an aqueous solution of perchloric acid.

4.32 Write a balanced molecular equation and a net ionic equa-tion for the reaction that occurs when (a) solid reacts with an aqueous solution of nitric acid; (b) solidiron(II) sulfide reacts with an aqueous solution of hydro-bromic acid.

4.33 Because the oxide ion is basic, metal oxides react readilywith acids. (a) Write the net ionic equation for the followingreaction:

(b) Based on the example in part (a), write the netionic equation for the reaction that occurs between and an aqueous solution of nitric acid.

4.34 As dissolves in water, the oxide ion reacts with watermolecules to form hydroxide ions. Write the molecularand net ionic equations for this reaction. Based on thedefinitions of acid and base, what ion is the base in thisreaction? What is the acid? What is the spectator ion inthe reaction?

K2O

NiO(s)H2O(l).

FeO(s) + 2HClO4(aq) ¡ Fe(ClO4)2(aq) +

CaCO3

Al(OH)3(s) + HNO3(aq) ¡Cu(OH)2(s) + HClO4(aq) ¡HBr(aq) + Ca(OH)2(aq) ¡

Acid-Base Reactions

4.21 What is the difference between: (a) a monoprotic acid anda diprotic acid; (b) a weak acid and a strong acid; (c) anacid and a base?

4.22 Explain the following observations: (a) contains noions, and yet its aqueous solutions are basic; (b) HF

is called a weak acid, and yet it is very reactive;(c) although sulfuric acid is a strong electrolyte, an aque-ous solution of contains more ions than

ions. Explain.

4.23 Classify each of the following as a strong or weak acid orbase: (a) (b) (c) (d)

4.24 Classify each of the following as a strong or weak acid orbase: (a) CsOH (b) (c) (d)

4.25 Label each of the following substances as an acid, base,salt, or none of the above. Indicate whether the substanceexists in aqueous solution entirely in molecular form, en-tirely as ions, or as a mixture of molecules and ions.(a) HF; (b) acetonitrile, (c) (d)

4.26 An aqueous solution of an unknown solute is tested withlitmus paper and found to be acidic. The solution is weak-ly conducting compared with a solution of NaCl of thesame concentration. Which of the following substancescould the unknown be: KOH,

(acetone)?

4.27 Classify each of the following substances as a nonelec-trolyte, weak electrolyte, or strong electrolyte in water:(a) (b) (ethanol); (c) (d)(e)

4.28 Classify each of the following aqueous solutions as a non-electrolyte, weak electrolyte, or strong electrolyte:(a) HBrO; (b) (c) KOH; (d) (acetone);(e) (f) (sucrose).C12H22O11CoSO4;

CH3COCH3HNO3;

Cu(NO3)2 .KClO3;NH3;C2H5OHH2SO3;

CH3COCH3H3PO3,KClO2,HNO3,NH3,

Ba(OH)2.NaClO4;CH3CN;

H2SO4.HC7H5O2H3PO4;

Ba(OH)2.NH3;HClO2;HClO4;

SO4

2-HSO4

-H2SO4

OH-NH3

Oxidation-Reduction Reactions

4.35 Define oxidation and reduction in terms of (a) electrontransfer and (b) oxidation numbers.

4.36 Can oxidation occur without accompanying reduction?Explain.

4.37 Where, in general, do the most easily oxidized metalsoccur in the periodic table? Where do the least easily ox-idized metals occur in the periodic table?

4.38 Why are platinum and gold called noble metals? Whyare the alkali metals and alkaline earth metals calledactive metals?

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Exercises 147

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CQ

CQ

CQ

4.39 Determine the oxidation number for the indicated ele-ment in each of the following substances: (a) S in (b) C in (c) Mn in (d) Br in HBrO; (e) Asin (f) O in

4.40 Determine the oxidation number for the indicated ele-ment in each of the following compounds: (a) Ti in (b) Sn in (c) C in (d) N in (e) N in (f) Cr in

4.41 Which element is oxidized, and which is reduced in thefollowing reactions?(a)(b)(c)(d)

4.42 Which of the following are redox reactions? For those thatare, indicate which element is oxidized and which isreduced. For those that are not, indicate whether they areprecipitation or acid-base reactions.(a)

(b)(c)

(d)

4.43 Write balanced molecular and net ionic equations forthe reactions of (a) manganese with sulfuric acid;(b) chromium with hydrobromic acid; (c) tin with hy-drochloric acid; (d) aluminum with formic acid,

4.44 Write balanced molecular and net ionic equations for thereactions of (a) hydrochloric acid with nickel; (b) sulfuricacid with iron; (c) hydrobromic acid with magnesium;(d) acetic acid, with zinc.HC2H3O2,

HCHO2.

4Zn2+(aq) + N2O(g) + 5H2O(l)4Zn(s) + 10H+(aq) + 2NO3

-(aq) ¡SrSO4(s) + 2HNO3(aq)

Sr(NO3)2(aq) + H2SO4(aq) ¡Fe2O3(s) + 3CO(g) ¡ 2Fe(s) + 3CO2(g)

Cu(NO3)2(aq) + 2H2O(l)Cu(OH)2(s) + 2HNO3(aq) ¡

PbS(s) + 4H2O2(aq) ¡ PbSO4(s) + 4H2O(l)Cl2(aq) + 2NaI(aq) ¡ I2(aq) + 2NaCl(aq)3Fe(NO3)2(aq) + 2Al(s) ¡ 3Fe(s) + 2Al(NO3)3(aq)Ni(s) + Cl2(g) ¡ NiCl2(s)

Cr2O7

2-.HNO2;(NH4)2SO4;C2O4

2-;SnCl4 ;TiO2;

K2O2.As4;MnO4

-;COCl2;SO3;

4.45 Based on the activity series (Table 4.5), what is the out-come of each of the following reactions?(a)(b)(c)(d)(e)

4.46 Using the activity series (Table 4.5), write balanced chem-ical equations for the following reactions. If no reactionoccurs, simply write NR. (a) Iron metal is added to a solu-tion of copper(II) nitrate; (b) zinc metal is added to a solu-tion of magnesium sulfate; (c) hydrobromic acid is addedto tin metal; (d) hydrogen gas is bubbled through anaqueous solution of nickel(II) chloride; (e) aluminummetal is added to a solution of cobalt(II) sulfate.

4.47 The metal cadmium tends to form ions. The fol-lowing observations are made: (i) When a strip of zincmetal is placed in cadmium metal is deposit-ed on the strip. (ii) When a strip of cadmium metal isplaced in nickel metal is deposited on thestrip. (a) Write net ionic equations to explain each of theobservations made above. (b) What can you concludeabout the position of cadmium in the activity series?(c) What experiments would you need to perform to lo-cate more precisely the position of cadmium in the ac-tivity series?

4.48 (a) Use the following reactions to prepare an activ-ity series for the halogens:

(b) Relate the positions of the halogens in theperiodic table with their locations in this activity series.(c) Predict whether a reaction occurs when the followingreagents are mixed: and KI(aq); andLiCl(aq).

Br2(aq)Cl2(aq)

Br2(aq).+2NaCl(aq)Cl2(aq) + 2NaBr(aq) ¡2NaBr(aq) + I2(aq);

Br2(aq) + 2NaI(aq) ¡

Ni(NO3)2(aq),

CdCl2(aq),

Cd2+

H2(g) + CuCl2(aq) ¡Mn(s) + HBr(aq) ¡Cr(s) + NiSO4(aq) ¡Ag(s) + Pb(NO3)2(aq) ¡Al(s) + NiCl2(aq) ¡

Solution Composition; Molarity

4.49 (a) Is the concentration of a solution an intensive or anextensive property? (b) What is the difference between0.50 mol HCl and 0.50 M HCl?

4.50 (a) Suppose you prepare 500 mL of a 0.10 M solution ofsome salt and then spill some of it. What happens to theconcentration of the solution left in the container? (b) Acertain volume of a 0.50 M solution contains 4.5 g of asalt. What mass of the salt is present in the same volumeof a 2.50 M solution?

4.51 (a) Calculate the molarity of a solution that contains0.0345 mol in exactly 400 mL of solution. (b) Howmany moles of are present in 35.0 mL of a 2.20 Msolution of nitric acid? (c) How many milliliters of 1.50 MKOH solution are needed to provide 0.125 mol of KOH?

4.52 (a) Calculate the molarity of a solution made by dissolv-ing 0.145 mol in enough water to form exactly750 mL of solution. (b) How many moles of arepresent in 125 mL of a 0.0850 M solution? (c) How manymilliliters of 11.6 M HCl solution are needed to obtain0.255 mol of HCl?

4.53 Calculate (a) the number of grams of solute in 0.250 L of0.150 M KBr; (b) the molar concentration of a solutioncontaining 4.75 g of in 0.200 L; (c) the volumeCa(NO3)2

KMnO4

Na2SO4

HNO3

NH4Cl

of 1.50 M in milliliters that contains 5.00 g ofsolute.

4.54 (a) How many grams of solute are present in 50.0 mL of0.850 M (b) If 2.50 g of is dissolvedin enough water to form 250 mL of solution, what is themolarity of the solution? (c) How many milliliters of0.387 M contain 1.00 g of solute?

4.55 (a) Which will have the highest concentration of potassi-um ion: 0.20 M KCl, 0.15 M or 0.080 M(b) Which will contain the greater number of moles ofpotassium ion: 30.0 mL of 0.15 M or 25.0 mL of0.080 M

4.56 (a) Without doing detailed calculations, rank the follow-ing solutions in order of increasing concentration of ions: 0.10 M 0.15 M KCl, a solution formed by dis-solving 0.10 mol of NaCl in enough water to form 250 mLof solution. (b) Which will contain the greater number ofmoles of chloride ion: 40.0 mL of 0.35 M NaCl or 25.0 mLof 0.25 M

4.57 Indicate the concentration of each ion or molecule pres-ent in the following solutions: (a) 0.14 M NaOH;(b) 0.25 M (c) 0.25 M (d) a mixture of50.0 mL of 0.10 M and 25.0 mL of 0.20 MAssume the volumes are additive.

Na2SO4.KClO3

CH3OH;CaBr2;

CaCl2 ?

CaCl2 ,Cl-

K3PO4 ?K2CrO4

K3PO4 ?K2CrO4,

CuSO4

(NH4)2SO4K2Cr2O7 ?

Na3PO4

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4.58 Indicate the concentration of each ion present in the solu-tion formed by mixing: (a) 20 mL of 0.100 M HCl and10.0 mL of 0.500 M HCl; (b) 15.0 mL of 0.300 Mand 10.0 mL of 0.200 M KCl; (c) 3.50 g of NaCl in 50.0 mLof 0.500 M solution. (Assume that the volumes areadditive.)

4.59 (a) You have a stock solution of 14.8 M How manymilliliters of this solution should you dilute to make100.0 mL of 0.250 M (b) If you take a 10.0-mL por-tion of the stock solution and dilute it to a total volumeof 0.250 L, what will be the concentration of the final so-lution?

4.60 (a) How many milliliters of a stock solution of 12.0 Mwould you have to use to prepare 0.500 L of

0.500 M ? (b) If you dilute 25.0 mL of the stock solu-tion to a final volume of 0.500 L, what will be the con-centration of the diluted solution?

4.61 (a) Starting with solid sucrose, describe howyou would prepare 125 mL of 0.150 M sucrose solution.

C12H22O11,

HNO3

HNO3

NH3 ?

NH3.

CaCl2

Na2SO4

(b) Describe how you would prepare 400.0 mL of 0.100 Mstarting with 2.00 L of 1.50 M

4.62 (a) How would you prepare 100.0 mL of 0.200 Msolution starting with pure solute? (b) An experimentcalls for you to use 250 mL of 1.0 M solution. Allyou have available is a bottle of 6.0 M How wouldyou prepare the desired solution?

[4.63] Pure acetic acid, known as glacial acetic acid, is a liquidwith a density of 1.049 g mL at Calculate the mo-larity of a solution of acetic acid made by dissolving20.00 mL of glacial acetic acid at in enough water tomake 250.0 mL of solution.

[4.64] Glycerol, is a substance used extensively in themanufacture of cosmetics, foodstuffs, antifreeze, and plas-tics. Glycerol is a water-soluble liquid with a density of1.2656 at Calculate the molarity of a solutionof glycerol made by dissolving 50.000 mL glycerol at in enough water to make 250.00 mL of solution.

15°C15°C.g>mL

C3H8O3,

25°C

25°C.>

HNO3.HNO3

AgNO3

C12H22O11.C12H22O11

Solution Stoichiometry; Titrations

4.65 What mass of NaCl is needed to precipitate all the silverions from 20.0 mL of 0.100 M solution?

4.66 What mass of NaOH is needed to precipitate all the ions from 25.0 mL of 0.500 M solution?

4.67 (a) What volume of 0.115 M solution is needed toneutralize 50.00 mL of 0.0875 M NaOH? (b) What vol-ume of 0.128 M HCl is needed to neutralize 2.87 g of

(c) If 25.8 mL of is needed to precip-itate all the ions in a 785-mg sample of KCl (formingAgCl), what is the molarity of the solution? (d) If45.3 mL of 0.108 M HCl solution is needed to neutralizea solution of KOH, how many grams of KOH must bepresent in the solution?

4.68 (a) How many milliliters of 0.120 M HCl are needed tocompletely neutralize 50.0 mL of 0.101 M solu-tion? (b) How many milliliters of 0.125 M areneeded to neutralize 0.200 g of NaOH? (c) If 55.8 mL of

solution is needed to precipitate all the sulfate ionin a 752-mg sample of what is the molarity ofthe solution? (d) If 42.7 mL of 0.208 M HCl solution isneeded to neutralize a solution of how manygrams of must be in the solution?

4.69 Some sulfuric acid is spilled on a lab bench. It can be neu-tralized by sprinkling sodium bicarbonate on it and thenmopping up the resultant solution. The sodium bicar-bonate reacts with sulfuric acid as follows:

Sodium bicarbonate is added until the fizzing due to theformation of stops. If 27 mL of 6.0 M wasspilled, what is the minimum mass of that mustbe added to the spill to neutralize the acid?

4.70 The distinctive odor of vinegar is due to acetic acid,Acetic acid reacts with sodium hydroxide in

the following fashion:HC2H3O2.

NaHCO3

H2SO4CO2(g)

Na2SO4(aq) + 2H2O(l) + 2CO2(g)2NaHCO3(s) + H2SO4(aq) ¡

Ca(OH)2

Ca(OH)2,

Na2SO4,BaCl2

H2SO4

Ba(OH)2

AgNO3

Cl-AgNO3Mg1OH22 ?

HClO4

Fe(NO3)2

Fe2+AgNO3

If 2.50 mL of vinegar needs 35.5 mL of 0.102 M NaOH toreach the equivalence point in a titration, how manygrams of acetic acid are in a 1.00-qt sample of thisvinegar?

4.71 A sample of solid is stirred in water at untilthe solution contains as much dissolved as itcan hold. A 100-mL sample of this solution is withdrawnand titrated with HBr. It requires 48.8 mLof the acid solution for neutralization. What is the mo-larity of the solution? What is the solubility of

in water, at in grams of per100 mL of solution?

4.72 In the laboratory 7.52 g of is dissolved in enoughwater to form 0.750 L. A 0.100-L sample is withdrawn fromthis stock solution and titrated with a 0.0425 M solution of

What volume of solution is needed toprecipitate all the as

4.73 A solution of 100.0 mL of 0.200 M KOH is mixed with asolution of 200.0 mL of 0.150 M (a) Write the bal-anced chemical equation for the reaction that occurs.(b) What precipitate forms? (c) What is the limiting reac-tant? (d) How many grams of this precipitate form?(e) What is the concentration of each ion that remains insolution?

4.74 A solution is made by mixing 12.0 g of NaOH and75.0 mL of 0.200 M (a) Write a balanced equationfor the reaction that occurs between the solutes.(b) Calculate the concentration of each ion remaining insolution. (c) Is the resultant solution acidic or basic?

[4.75] A 0.5895-g sample of impure magnesium hydroxide isdissolved in 100.0 mL of 0.2050 M HCl solution. The ex-cess acid then needed 19.85 mL of 0.1020 M NaOH forneutralization. Calculate the percent by mass of magne-sium hydroxide in the sample, assuming that it is the onlysubstance reacting with the HCl solution.

HNO3.

NiSO4.

SrCrO4 ?Sr2+(aq)Na2CrO4Na2CrO4.

Sr(NO3)2

Ca(OH)230°C,Ca(OH)2

Ca(OH)2

5.00 * 10-2 M

Ca(OH)2

30°CCa(OH)2

HC2H3O2(aq) + NaOH(aq) ¡ H2O(l) + NaC2H3O2(aq)

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Exercises 149

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CQ

CQ

Solution Solute Color of Solution

A YellowB ColorlessC ColorlessD ColorlessCaCl2

AgNO3

(NH4)2C2O4

Na2CrO4

[4.76] A 1.452-g sample of limestone rock is pulverized and thentreated with 25.00 mL of 1.035 M HCl solution. The excessacid then required 15.25 mL of 0.1010 M NaOH for neu-

tralization. Calculate the percent by mass of calcium car-bonate in the rock, assuming that it is the only substancereacting with the HCl solution.

When these solutions are mixed, the following observa-tions are made:

(a) Write a net ionic equation for the reaction that occursin each of the experiments. (b) Identify the precipitateformed, if any, in each of the experiments. (c) Based onthese limited observations, which ion tends to form themore soluble salts, chromate or oxalate?

Additional Exercises

4.77 The accompanying photo shows the reaction between asolution of and one of What is the iden-tity of the precipitate? What ions remain in solution?Write the net ionic equation for the reaction.

4.78 Suppose you have a solution that might contain any or allof the following cations: and Addi-tion of HCl solution causes a precipitate to form. After fil-tering off the precipitate, solution is added to theresultant solution and another precipitate forms. This isfiltered off, and a solution of NaOH is added to the result-ing solution. No precipitate is observed. Which ions arepresent in each of the precipitates? Which of the four ionslisted above must be absent from the original solution?

4.79 You choose to investigate some of the solubility guide-lines for two ions not listed in Table 4.1, the chromate ion

and the oxalate ion You are givensolutions (A, B, C, D) of four water-soluble salts:

(C2O4

2-).(CrO4

2-)

H2SO4

Mn2+.Sr2+,Ag+,Ni2+,

Na2S.Cd(NO3)2

4.80 Antacids are often used to relieve pain and promote heal-ing in the treatment of mild ulcers. Write balanced netionic equations for the reactions between the HCl(aq) inthe stomach and each of the following substances used invarious antacids: (a) (b)(c) (d) (e)

[4.81] Salts of the sulfite ion, react with acids in a way sim-ilar to that of carbonates. (a) Predict the chemical formula,and name the weak acid that forms when the sulfite ionreacts with acids. (b) The acid formed in part (a) decom-poses to form water and an insoluble gas. Predict the molec-ular formula, and name the gas formed. (c) Use a sourcebook such as the CRC Handbook of Chemistry and Physics toconfirm that the substance in part (b) is a gas under normalroom-temperature conditions. (d) Write balanced net ionicequations of the reaction of HCl(aq) with (i)(ii) (iii) and (iv)

4.82 The commercial production of nitric acid involves the fol-lowing chemical reactions:

(a) Which of these reactions are redox reactions? (b) Ineach redox reaction, identify the element undergoing oxi-dation and the element undergoing reduction.

4.83 Use Table 4.5 to predict which of the following ions canbe reduced to their metal forms by reacting with zinc:(a) (b) (c) (d)(e) (f) Write the balanced net ionicequation for each reaction that occurs.

4.84 Titanium(IV) ion, can be reduced to by the care-ful addition of zinc metal. (a) Write the net ionic equa-tion for this process. (b) Would it be appropriate to usethis reaction as a means of including titanium in the activ-ity series of Table 4.5? Why or why not?

[4.85] Lanthanum metal forms cations with a charge of Con-sider the following observations about the chemistry oflanthanum: When lanthanum metal is exposed to air, awhite solid (compound A) is formed that contains lan-thanum and one other element. When lanthanum metalis added to water, gas bubbles are observed and a differ-ent white solid (compound B) is formed. Both A and Bdissolve in hydrochloric acid to give a clear solution.When the solution from either A or B is evaporated, a sol-uble white solid (compound C) remains. If compound Cis dissolved in water and sulfuric acid is added, a whiteprecipitate (compound D) forms. (a) Propose identitiesfor the substances A, B, C, and D. (b) Write net ionic equa-tions for all the reactions described. (c) Based on the pre-ceding observations, what can be said about the positionof lanthanum in the activity series (Table 4.5)?

4.86 A 25.0-mL sample of 1.00 M KBr and a 75.0-mL sample of0.800 M KBr are mixed. The solution is then heated toevaporate water until the total volume is 50.0 mL. Whatis the molarity of the KBr in the final solution?

3+ .

Ti3+Ti4+,

Al3+(aq).Cu2+(aq);Fe2+(aq);Mg2+(aq);Pb2+(aq);Na+(aq);

3NO2(g) + H2O(l) ¡ 2HNO3(aq) + NO(g) 2NO(g) + O2(g) ¡ 2NO2(g)

4NH3(g) + 5O2(g) ¡ 4NO(g) + 6H2O(g)

ZnSO3(aq).KHSO3(s),Ag2SO3(s),Na2SO3(aq),

SO3

2-,CaCO3(s).NaAl(CO3)(OH)2(s);MgCO3(s);

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

Expt. SolutionsNumber Mixed Result

1 No precipitate, yellow solution2 Red precipitate forms3 No precipitate, yellow solution4 White precipitate forms5 White precipitate forms6 White precipitate formsC + D

B + DB + CA + DA + CA + B

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4.87 Calculate the molarity of the solution produced by mix-ing (a) 50.0 mL of 0.200 M NaCl and 75.0 mL of 0.100 MNaCl; (b) 24.5 mL of 1.50 M NaOH and 25.5 mL of0.750 M NaOH. (Assume that the volumes are additive.)

4.88 Using modern analytical techniques, it is possible todetect sodium ions in concentrations as low as 50What is this detection limit expressed in: (a) molarity of

(b) ions per cubic centimeter?4.89 Hard water contains and which inter-

fere with the action of soap and leave an insoluble coatingon the insides of containers and pipes when heated. Watersofteners replace these ions with If L of hardwater contains 0.010 M and 0.0050 M howmany moles of are needed to replace these ions?

4.90 Tartaric acid, has two acidic hydrogens. Theacid is often present in wines and precipitates from solutionas the wine ages. A solution containing an unknown con-centration of the acid is titrated with NaOH. It requires

H2C4H4O6,Na+

Mg2+,Ca2+1.0 * 103Na+.

Fe2+,Mg2+,Ca2+,Na+Na+;

pg>mL.

22.62 mL of 0.2000 M NaOH solution to titrate both acidicprotons in 40.00 mL of the tartaric acid solution. Write abalanced net ionic equation for the neutralization reaction,and calculate the molarity of the tartaric acid solution.

4.91 The concentration of hydrogen peroxide in a solution isdetermined by titrating a 10.0-mL sample of the solutionwith permanganate ion.

If it takes 13.5 mL of 0.109 M solution to reach theequivalence point, what is the molarity of the hydrogenperoxide solution?

[4.92] A solid sample of is added to 0.400 L of 0.500 Maqueous HBr. The solution that remains is still acidic. It isthen titrated with 0.500 M NaOH solution, and it takes98.5 mLof the NaOH solution to reach the equivalence point.What mass of was added to the HBr solution?Zn(OH)2

Zn(OH)2

MnO4

-

2Mn2+(aq) + 5O2(g) + 8H2O(l)

2MnO4

-(aq) + 5H2O2(aq) + 6H+(aq) ¡

Integrative Exercises

4.93 Calculate the number of sodium ions in 1.00 mL of a0.0100 M solution of sodium phosphate.

4.94 (a) By titration, 15.0 mL of 0.1008 M sodium hydroxide isneeded to neutralize a 0.2053-g sample of an organic acid.What is the molar mass of the acid if it is monoprotic?(b) An elemental analysis of the acid indicates that it iscomposed of 5.89% H, 70.6% C, and 23.5% O by mass.What is its molecular formula?

4.95 A 6.977-g sample of a mixture was analyzed for bariumion by adding a small excess of sulfuric acid to an aque-ous solution of the sample. The resultant reaction pro-duced a precipitate of barium sulfate, which was collectedby filtration, washed, dried, and weighed. If 0.4123 g ofbarium sulfate was obtained, what was the mass per-centage of barium in the sample?

[4.96] A tanker truck carrying kg of concentrated sul-furic acid solution tips over and spills its load. If the sul-furic acid is 95.0% by mass and has a density of1.84 how many kilograms of sodium carbonatemust be added to neutralize the acid?

4.97 A sample of 5.53 g of is added to 25.0 mL of0.200 M (a) Write the chemical equation for thereaction that occurs. (b) Which is the limiting reactant inthe reaction? (c) How many moles of and are present after the reaction is complete?

4.98 A sample of 1.50 g of lead(II) nitrate is mixed with 125 mLof 0.100 M sodium sulfate solution. (a) Write the chemi-cal equation for the reaction that occurs. (b) Which is thelimiting reactant in the reaction? (c) What are the con-centrations of all ions that remain in solution after thereaction is complete?

4.99 A mixture contains NaCl, 1.5% and 8.5%by mass. What is the molarity of ions in a

solution formed by dissolving 7.50 g of the mixture inenough water to form 500.0 mL of solution?

[4.100] The average concentration of bromide ion in seawater is65 mg of bromide ion per kg of seawater. What is themolarity of the bromide ion if the density of the seawa-ter is 1.025 g>mL?

Cl-Na2SO4

MgCl2,89.0%

Mg(NO3)2

HNO3,Mg(OH)2,

HNO3.Mg(OH)2

g>mL,H2SO4

5.0 * 103

[4.101] The mass percentage of chloride ion in a 25.00-mL sampleof seawater was determined by titrating the sample withsilver nitrate, precipitating silver chloride. It took 42.58 mLof 0.2997 M silver nitrate solution to reach the equivalencepoint in the titration. What is the mass percentage of chlo-ride ion in the seawater if its density is 1.025

4.102 The arsenic in a 1.22-g sample of a pesticide was con-verted to by suitable chemical treatment. It wasthen titrated using to form as a precipitate.(a) What is the oxidation state of As in (b) Name

by analogy to the corresponding compoundcontaining phosphorus in place of arsenic. (c) If it took25.0 mL of 0.102 M to reach the equivalence point inthis titration, what is the mass percentage of arsenic inthe pesticide?

[4.103] A 500-mg tablet of an antacid containing and an inert “binder” was dissolved in 50.0 mL

of 0.500 M HCl. The resulting solution, which was acidic,needed 30.9 mL of 0.255 M NaOH for neutralization.(a) Calculate the number of moles of ions in thetablet. (b) If the tablet contains 5.0% binder, how manymilligrams of and how many of doesthe tablet contain?

[4.104] Federal regulations set an upper limit of 50 parts per mil-lion (ppm) of in the air in a work environment (thatis, 50 molecules of for every million molecules inthe air). Air from a manufacturing operation was drawnthrough a solution containing mL of 0.0105 MHCl. The reacts with HCl as follows:

After drawing air through the acid solution for 10.0 minat a rate of 10.0 L min, the acid was titrated. The remain-ing acid needed 13.1 mL of 0.0588 M NaOH to reach theequivalence point. (a) How many grams of weredrawn into the acid solution? (b) How many ppm of were in the air? (Air has a density of 1.20 g L and an aver-age molar mass of 29.0 g mol under the conditions of theexperiment.) (c) Is this manufacturer in compliance withregulations?

>>

NH3

NH3

>

NH3(aq) + HCl(aq) ¡ NH4Cl(aq)

NH3

1.00 * 102

NH3(g)NH3

Al(OH)3Mg(OH)2

OH-

Al(OH)3,Mg(OH)2,

Ag+

Ag3AsO4

AsO4

3-?Ag3AsO4Ag+

AsO4

3-

g>mL?

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eMedia Exercises 151

eMedia Exercises

4.105 The Electrolytes and Non-Electrolytes movie (eChapter4.1) and the Aqueous Acids and Aqueous Bases movies(eChapter 4.3) illustrate the behavior of various substancesin aqueous solution. For each of the seven substancesmentioned in the movies, write the chemical equationthat corresponds to dissolution in water. (The chemicalformula of sugar is ) Where appropriate, usethe double arrow notation.

4.106 In the Strong and Weak Electrolytes movie (eChapter 4.1),the lightbulb glows brightly when the beaker containsaqueous hydrochloric acid, but relatively dimly when thebeaker contains aqueous acetic acid. (a) For each of thecompounds in Exercise 4.3, would you expect an aqueoussolution to cause the bulb to light? If so, how brightly?(b) Consider the use of aqueous solutions of each of thefollowing compounds in the apparatus shown in thedemonstration. For each compound, tell whether youwould expect the lightbulb to glow brightly, dimly, or notat all: NH4Cl; and HF.

4.107 (a) Use the solubility rules to predict what precipitate, ifany, will form as the result of each combination.(i) Na2CO3(aq) and Fe(NO3)2(aq); (ii) NH4NO3(aq) andK2SO4(aq); (iii) AlBr3(aq) and Fe2(SO4)3(aq); (iv) H2SO4(aq)and Pb(NO3)2(aq); (v) Na2S(aq) and (NH4)2SO4(aq). Usethe Ionic Compounds activity (eChapter 2.8) to check youranswers. (b) For each combination that produces a pre-cipitate, write a balanced net ionic equation. (c) WhenNH4Cl(aq) and Pb(NO3)2(aq) are combined, a precipitateforms.What ions are still present in the solution in sig-nificant concentration after the precipitation? Explain.

4.108 In the Redox Chemistry of Tin and Zinc movie (eChapter4.4), zinc is oxidized by a solution containing tin ions.(a) Write the equation corresponding to this redox reac-tion. (b) In addition to the reaction between zinc metaland tin ions, there is another process occurring. Write thenet ionic equation corresponding to this process. (Refer toExercise 4.44.)

CaF2;C2H5OH;H2CO3;

C12H22O11.

4.109 After watching the Solution Formation from a Solidmovie (eChapter 4.5), answer the following questions: (a) Ifwe neglect to account for the mass of the weighing paper,how would our calculated concentration differ from theactual concentration of the solution? (b) Describe theprocess of preparing an aqueous solution of known con-centration, starting with a solid. (c) Why is it necessary tomake the solution as described in the movie, rather thansimply filling the flask up to the mark with water andthen adding the solute? (d) Describe how you would pre-pare the solution in part (a) starting with the concentrat-ed stock solution in the Solution Formation by Dilutionmovie (eChapter 4.5).

4.110 Use the Titration simulation (eChapter 4.6) to determinethe concentration of an unknown acid by adding 0.40 MNaOH in increments of 1.0 mL. Repeat the titrationadding increments of 0.10 mL of base near the end point.Once more, repeat the titration, adding increments of0.05 mL of base near the end point. If your acid is diluteenough, repeat the titration three more times using 0.10 MNaOH in 1.0-mL, 0.50-mL, and 0.05-mL increments.(a) Tabulate the acid concentrations that you calculatefrom your titration data. Are the values all the same? Ifnot, why not? (b) Which value do you consider to be mostprecise and why?

4.112 (a) What is the maximum concentration of monoproticacid that could be titrated in the Titration simulation(eChapter 4.6) using 0.05 M NaOH? (b) What is the max-imum concentration of a diprotic acid that could be titrat-ed in this simulation using 0.10 M base? (c) All of theacid-base indicators available in the simulation changecolor within a pH range of to 10.5. What effect wouldthe use of an indicator such as metacresol purple haveon the experimentally determined value of an unknownacid’s concentration? (Metacresol purple changes colorsin the pH range of to 2.8.)'1.2

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