REACTIONS IN AQUEOUS SOLUTION - Dr. Steve W. …drsticky.weebly.com/uploads/1/7/5/3/17533705/ch04_rxn_in...118 CHAPTER 4 Reactions in Aqueous Solution Water is a very effective solvent

4.5 CONCENTRATIONS OF SOLUTIONSWe learn how the amount of a compound dissolved in a givenvolume of a solution can be expressed as a concentration.Concentration can be defined in a number of ways, the mostcommonly used being moles of compound per liter of solution(molarity).

4.6 SOLUTION STOICHIOMETRY AND CHEMICAL ANALYSISWe see how the concepts of stoichiometry and concentration canbe used to calculate amounts or concentrations of substances insolution through a process called titration.

REACTIONSIN AQUEOUSSOLUTIONWATER COVERS NEARLY TWO-THIRDS of our planet, and thissimple substance has been the key to much of Earth’sevolutionary history. Life almost certainly originated inwater, and the need for water by all forms of life hashelped determine diverse biological structures.

115

Given the importance of water for life, it should come as no surprise that developmentof civilizations has been closely tied to reliable sources of fresh water. During the firstmillennium, Mayan civilization was one of the most advanced on Earth. Mayan city-states covered much of the Yucatan peninsula in what is now easternmost Mexico.The northern half of the peninsula is a flat shelf of land composed largely of carbonaterocks, such as limestone. Because there are no rivers and very few lakes, the Mayansdepended on sinkholes called cenotes, for their fresh water (! FIGURE 4.1).

Chemical reactions that occur in water are responsible for creation of cenotes.When carbon dioxide, CO2, dissolves in water, the resulting solution is slightly acidicand reacts with CaCO3 in the limestone:

[4.1]

A solution in which water is the dissolving medium is called an aqueous solution.In this chapter we examine chemical reactions that take place in aqueous solutions. Inaddition, we extend the concepts of stoichiometry learned in Chapter 3 by consideringhow solution concentrations are expressed and used.

CaCO31s2 + H2O1l2 + CO21aq2 ¡ Ca1HCO3221aq2

116 CHAPTER 4 Reactions in Aqueous Solution

4.1 | GENERAL PROPERTIES OFAQUEOUS SOLUTIONS

A solution is a homogeneous mixture of two or more substances. •(Section 1.2) Thesubstance present in the greatest quantity is usually called the solvent, and the othersubstances are called solutes; they are said to be dissolved in the solvent. When a smallamount of sodium chloride (NaCl) is dissolved in a large quantity of water, for example,water is the solvent and sodium chloride is the solute.

Electrolytic PropertiesAt a young age we learn not to bring electrical devices into the bathtub so as not to elec-trocute ourselves. That’s a useful lesson because most of the water you encounter indaily life is electrically conducting. Pure water, however, is a very poor conductor of elec-tricity. The conductivity of bathwater originates from the substances dissolved in thewater, not from the water itself.

Not all substances that dissolve in water make the resulting solution conducting.Imagine preparing two aqueous solutions—one by dissolving a teaspoon of table salt(sodium chloride) in a cup of water and the other by dissolving a teaspoon of tablesugar (sucrose) in a cup of water (! FIGURE 4.2). Both solutions are clear and color-less, but they possess very different electrical conductivities: the salt solution is a goodconductor of electricity, whereas the sugar solution is not.

In order for the bulb in the device of Figure 4.2 to light up, there must be a current(that is, a flow of electrically charged particles) between two electrodes immersed in thesolution. The conductivity of pure water is not sufficient to complete the electrical cir-cuit and light the bulb. The situation changes when ions are present in solution becausethe ions carry electrical charge from one electrode to the other, completing the circuit.Thus, the conductivity of NaCl solutions indicates the presence of ions. The lack of con-ductivity of sucrose solutions indicates the absence of ions. When NaCl dissolves inwater, the solution contains and ions, each surrounded by water molecules.When sucrose (C12H22O11) dissolves in water, the solution contains only neutral sucrosemolecules surrounded by water molecules.

Cl-Na+

CO2

!

!

!

!

"

"

"2

H2O

HCO3!

HCO3!

CaCO3 Ca2"

H"

H"

Acidic water reacts with CaCO3, dissolving limestone and forming a cenote

2Atmospheric CO2 dissolves in groundwater, forming H+

and HCO3–

1

" FIGURE 4.1 Cenote formation.

SECTION 4.1 General Properties of Aqueous Solutions 117

Pure water,H2O(l)

does not conduct electricity

Sucrose solution,C12H22O11(aq)Nonelectrolyte

does not conduct electricity

Sodium chloride solution,NaCl(aq)Electrolyte

conducts electricity

Not lit Not lit Lit ! FIGURE 4.2Electricalconductivities ofwater and twoaqueous solutions.

A substance (such as NaCl) whose aqueous solutions contain ions is called anelectrolyte. A substance (such as C12H22O11) that does not form ions in solution iscalled a nonelectrolyte. The different classifications of NaCl and C12H22O11 ariselargely because NaCl is ionic, whereas C12H22O11 is molecular.

Ionic Compounds in WaterRecall from Figure 2.21 that solid NaCl consists of an orderly arrangement of and

ions. When NaCl dissolves in water, each ion separates from the solid structure anddisperses throughout the solution [" FIGURE 4.3(a)]. The ionic solid dissociates into itscomponent ions as it dissolves.

Cl-Na+

(a) Ionic compounds like sodium chloride, NaCl, form ions when they dissolve

Ionic compounddissolves in water

Na!Cl"

Methanol

(b) Molecular substances like methanol, CH3OH, dissolve without forming ions

H2O molecules separate Na+ and Cl–

ions from solid NaCl

1

H2O molecules surround Na+ and Cl– ions

2

Na+ and Cl– ions dispersethroughout the solution

3

# FIGURE 4.3 Dissolution in water. (a) When an ionic compound, such as sodium chloride, NaCl, dissolves in water, H2O moleculesseparate, surround, and uniformly disperse the ions into the liquid. (b) Molecular substances that dissolve in water, such as methanol, CH3OH,usually do so without forming ions. We can think of this as a simple mixing of two molecular species. In both (a) and (b) the water molecules havebeen moved apart so that the solute particles can be seen clearly.

G O F I G U R EWhich solution, NaCl(aq) or CH3OH(aq), conducts electricity?


Water is a very effective solvent for ionic compounds. Although H2O is an electri-cally neutral molecule, the O atom is rich in electrons and has a partial negative charge,denoted by . Each H atom has a partial positive charge, denoted by . Cations areattracted by the negative end of H2O, and anions are attracted by the positive end.

As an ionic compound dissolves, the ions become surrounded by H2O molecules, asshown in Figure 4.3(a). The ions are said to be solvated. In chemical equations, we de-note solvated ions by writing them as and , where aq is an abbreviationfor “aqueous.” •(Section 3.1) Solvation helps stabilize the ions in solution and pre-vents cations and anions from recombining. Furthermore, because the ions and theirshells of surrounding water molecules are free to move about, the ions become dis-persed uniformly throughout the solution.

We can usually predict the nature of the ions in a solution of an ionic compoundfrom the chemical name of the substance. Sodium sulfate (Na2SO4), for example, disso-ciates into sodium ions and sulfate ions . You must remember theformulas and charges of common ions (Tables 2.4 and 2.5) to understand the forms inwhich ionic compounds exist in aqueous solution.

G I V E I T S O M E T H O U G H TWhat dissolved species are present in a solution of a. KCN,b. NaClO4?

Molecular Compounds in WaterWhen a molecular compound dissolves in water, the solution usually consists of intactmolecules dispersed throughout the solution. Consequently, most molecular com-pounds are nonelectrolytes. As we have seen, table sugar (sucrose) is a nonelectrolyte. Asanother example, a solution of methanol (CH3OH) in water consists entirely of CH3OHmolecules dispersed in the water [Figure 4.3(b)].

A few molecular substances have aqueous solutions that contain ions. Acids are themost important of these solutions. For example, when HCl(g) dissolves in water to formhydrochloric acid, HCl(aq), it ionizes; that is, it dissociates into and ions.

Strong and Weak ElectrolytesElectrolytes differ in the extent to which they conduct electricity. Strong electrolytesare those solutes that exist in solution completely or nearly completely as ions. Essen-tially all water-soluble ionic compounds (such as NaCl) and a few molecular com-pounds (such as HCl) are strong electrolytes. Weak electrolytes are those solutes thatexist in solution mostly in the form of neutral molecules with only a small fraction inthe form of ions. For example, in a solution of acetic acid (CH3COOH) most of thesolute is present as CH3COOH(aq) molecules. Only a small fraction (about 1%) of theCH3COOH has dissociated into and ions.*

We must be careful not to confuse the extent to which an electrolyte dissolves (itssolubility) with whether it is strong or weak. For example, CH3COOH is extremely sol-uble in water but is a weak electrolyte. Ca(OH)2, on the other hand, is not very solublein water, but the amount that does dissolve dissociates almost completely. Thus,Ca(OH)2 is a strong electrolyte.

When a weak electrolyte, such as acetic acid, ionizes in solution, we write the reac-tion in the form

[4.2]

The half-arrows pointing in opposite directions mean that the reaction is significantin both directions. At any given moment some CH3COOH molecules are ionizing to

CH3COOH(aq2 ! CH3COO-1aq2 + H+1aq2

CH3COO-1aq2H+(aq)

Cl-1aq2H+(aq)

1SO42-21Na+2

Cl-1aq2Na+(aq)

d+d-

*The chemical formula of acetic acid is sometimes written HC2H3O2 so that the formula looks like that ofother common acids such as HCl. The formula CH3COOH conforms to the molecular structure of acetic acid,with the acidic H on the O atom at the end of the formula.

d!

d!

d"

SECTION 4.2 Precipitation Reactions 119

form and ions but and ions are recombining to formCH3COOH. The balance between these opposing processes determines the relativenumbers of ions and neutral molecules. This balance produces a state of chemicalequilibrium in which the relative numbers of each type of ion or molecule in the reac-tion are constant over time. Chemists use half-arrows pointing in opposite directions torepresent the ionization of weak electrolytes and a single arrow to represent the ioniza-tion of strong electrolytes. Because HCl is a strong electrolyte, we write the equation forthe ionization of HCl as

[4.3]

The absence of a left-pointing arrow indicates that the and ions have notendency to recombine to form HCl molecules.

In the following sections we will look at how a compound’s composition lets us pre-dict whether it is a strong electrolyte, weak electrolyte, or nonelectrolyte. For the mo-ment, you need only to remember that water-soluble ionic compounds are strongelectrolytes. Ionic compounds can usually be identified by the presence of both metalsand nonmetals [for example, NaCl, FeSO4, and Al(NO3)3]. Ionic compounds containingthe ammonium ion, [for example, NH4Br and (NH4)2CO3], are exceptions to thisrule of thumb.

G I V E I T S O M E T H O U G H TWhich solute will cause the lightbulb in Figure 4.2 to glow most brightly, CH3OH,NaOH, or CH3COOH?

SAMPLE EXERCISE 4.1 Relating Relative Numbers of Anionsand Cations to Chemical Formulas

The accompanying diagram represents an aqueous solution of either MgCl2, KCl, or K2SO4.Which solution does the drawing best represent?

NH4+

Cl-H+

HCl1aq2 ¡ H+1aq2 + Cl-1aq2

CH3COO-H+CH3COO-H+

2!

2!

2!2!

"

"

"

"

"

" "

"

SOLUTIONAnalyze We are asked to associate the charged spheres in the diagram with ions present in asolution of an ionic substance.

Plan We examine each ionic substance given to determine the relative numbers and chargesof its ions. We then correlate these ionic species with the ones shown in the diagram.

Solve The diagram shows twice as many cations as anions, consistent with the formulationK2SO4.

Check Notice that the net charge in the diagram is zero, as it must be if it is to represent anionic substance.

PRACTICE EXERCISEIf you were to draw diagrams representing aqueous solutions of (a) NiSO4, (b) Ca(NO3)2,(c) Na3PO4, (d) Al2(SO4)3, how many anions would you show if each diagram contained sixcations?

Answers: (a) 6, (b) 12, (c) 2, (d) 9

4.2 | PRECIPITATION REACTIONS! FIGURE 4.4 shows two clear solutions being mixed. One solution contains potas-sium iodide, KI, dissolved in water and the other contains lead nitrate, Pb(NO3)2, dis-solved in water. The reaction between these two solutes produces a water-insolubleyellow solid. Reactions that result in the formation of an insoluble product are calledprecipitation reactions. A precipitate is an insoluble solid formed by a reaction insolution. In Figure 4.4 the precipitate is lead iodide (PbI2), a compound that has a verylow solubility in water:

[4.4]

The other product of this reaction, potassium nitrate (KNO3), remains in solution.

Pb1NO3221aq2 + 2 KI1aq2 ¡ PbI21s2 + 2 KNO31aq2


PbI2(s) ! 2 KI(aq)2 KI(aq) ! Pb(NO3)2(aq)ProductsReactants

Pb2!

Pb2!(aq) and I–(aq) combineto form precipitate

K!

NO3"

I"

! FIGURE 4.4 A precipitation reaction.

Precipitation reactions occur when pairs of oppositely charged ions attract eachother so strongly that they form an insoluble ionic solid. To predict whether certaincombinations of ions form insoluble compounds, we must consider some guidelinesconcerning the solubilities of common ionic compounds.

Solubility Guidelines for Ionic CompoundsThe solubility of a substance at a given temperature is the amount of the substance thatcan be dissolved in a given quantity of solvent at the given temperature. In our discus-sions, any substance with a solubility less than will be referred to as insoluble.In those cases the attraction between the oppositely charged ions in the solid is too greatfor the water molecules to separate the ions to any significant extent; the substance re-mains largely undissolved.

Unfortunately, there are no rules based on simple physical properties such as ioniccharge to guide us in predicting whether a particular ionic compound will be soluble.Experimental observations, however, have led to guidelines for predicting solubility forionic compounds. For example, experiments show that all common ionic compoundsthat contain the nitrate anion, , are soluble in water. " TABLE 4.1 summarizes thesolubility guidelines for common ionic compounds. The table is organized according tothe anion in the compound, but it also reveals many important facts about cations. Notethat all common ionic compounds of the alkali metal ions (group 1A of the periodic table)and of the ammonium ion are soluble in water.1NH4

+2NO3

-

0.01 mol/L

G O F I G U R EWhich ions remain in solution after PbI2 precipitation is complete?


SAMPLE EXERCISE 4.2 Using Solubility Rules

Classify these ionic compounds as soluble or insoluble in water: (a) sodium carbonate,Na2CO3, (b) lead sulfate, PbSO4.

SOLUTIONAnalyze We are given the names and formulas of two ionic compounds and asked to predictwhether 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 anion ineach compound because the table is organized by anions.

Solve(a) According to Table 4.1, most carbonates are insoluble. But carbonates of the alkali metalcations (such as sodium ion) are an exception to this rule and are soluble. Thus, Na2CO3 issoluble in water.

(b) Table 4.1 indicates that although most sulfates are water soluble, the sulfate of is anexception. Thus, PbSO4 is insoluble in water.

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

Answers: (a) insoluble, (b) soluble, (c) soluble

Pb2+

TABLE 4.1 • Solubility Guidelines for Common Ionic Compounds in Water

Soluble Ionic Compounds Important Exceptions

Compounds containing NO3- None

CH3COO- None

Cl- Compounds of , , and Pb2+Hg22+Ag+

Br- Compounds of , , and Pb2+Hg22+Ag+

I- Compounds of , , and Pb2+Hg22+Ag+

SO42- Compounds of , , , and Pb2+Hg2

2+Ba2+Sr2+

Insoluble Ionic Compounds Important Exceptions

Compounds containing S2- Compounds of , the alkali metal cations, , , and Ba2+Sr2+Ca2+NH4+

CO32- Compounds of and the alkali metal cationsNH4

+

PO43- Compounds of and the alkali metal cationsNH4

+

OH- Compounds of , the alkali metal cations, , , and Ba2+Sr2+Ca2+NH4+

To predict whether a precipitate forms when we mix aqueous solutions of twostrong electrolytes, we must (1) note the ions present in the reactants, (2) consider thepossible cation-anion combinations, and (3) use Table 4.1 to determine if any of thesecombinations is insoluble. For example, will a precipitate form when solutions ofMg(NO3)2 and NaOH are mixed? Both substances are soluble ionic compounds andstrong electrolytes. Mixing the solutions first produces a solution containing ,

, , and ions. Will either cation interact with either anion to form aninsoluble compound? Knowing from Table 4.1 that Mg(NO3)2 and NaOH are both sol-uble in water, our only possibilities are with and with . FromTable 4.1 we see that hydroxides are generally insoluble. Because is not an exception,Mg(OH)2 is insoluble and thus forms a precipitate. NaNO3, however, is soluble, so and remain in solution. The balanced equation for the precipitation reaction is

[4.5]

Exchange (Metathesis) ReactionsNotice in Equation 4.5 that the reactant cations exchange anions— ends up with

, and ends up with . The chemical formulas of the products are based onthe charges of the ions—two ions are needed to give a neutral compound with ,Mg2+OH-

NO3-Na+OH-

Mg2+

Mg1NO3221aq2 + 2 NaOH1aq2 ¡ Mg1OH221s2 + 2 NaNO31aq2NO3-

Na+Mg2+

NO3-Na+OH-Mg2+

OH-Na+NO3-

Mg2+

and one ion is needed to give a neutral compound with •(Section 2.7) Theequation can be balanced only after the chemical formulas of the products have beendetermined.

Reactions in which cations and anions appear to exchange partners conform to thegeneral equation

[4.6]

Example:

Such reactions are called either exchange reactions or metathesis reactions (meh-TATH-eh-sis, Greek for “to transpose”). Precipitation reactions conform to this pattern,as do many neutralization reactions between acids and bases, as we will see inSection 4.3.

To complete and balance the equation for a metathesis reaction, we followthese steps:

1. Use the chemical formulas of the reactants to determine which ions are present.2. Write the chemical formulas of the products by combining the cation from one

reactant with the anion of the other, using the ionic charges to determine thesubscripts in the chemical formulas.

3. Check the water solubilities of the products. For a precipitation reaction to occur, atleast one product must be insoluble in water.

4. Balance the equation.

AgNO31aq2 + KCl1aq2 ¡ AgCl1s2 + KNO31aq2AX + BY ¡ AY + BX

Na+.NO3-


SAMPLE EXERCISE 4.3 Predicting a Metathesis Reaction

(a) Predict the identity of the precipitate that forms when aqueous solutions of BaCl2 andK2SO4 are mixed. (b) Write the balanced chemical equation for the reaction.

SOLUTIONAnalyze We are given two ionic reactants and asked to predict the insoluble product thatthey form.

Plan We need to write the ions present in the reactants and exchange the anions betweenthe two cations. Once we have written the chemical formulas for these products, we can useTable 4.1 to determine which is insoluble in water. Knowing the products also allows us towrite the equation for the reaction.

Solve(a) The reactants contain , , , and ions. Exchanging the anions gives usBaSO4 and KCl. According to Table 4.1, most compounds of are soluble but those of

are not. Thus, BaSO4 is insoluble and will precipitate from solution. KCl is soluble.(b) From part (a) we know the chemical formulas of the products, BaSO4 and KCl. The bal-anced equation is

PRACTICE EXERCISE(a) What compound precipitates when aqueous solutions of Fe2(SO4)3 and LiOH are mixed?(b) Write a balanced equation for the reaction. (c) Will a precipitate form when solutions ofBa(NO3)2 and KOH are mixed?

Answers: (a) Fe(OH)3, (b), (c) no (both possible products, Ba(OH)2 and KNO3, are water soluble)3 Li2SO41aq2 Fe21SO4231aq2 + 6 LiOH1aq2 ¡ 2 Fe1OH231s2 +

BaCl21aq2 + K2SO41aq2 ¡ BaSO41s2 + 2 KCl1aq2Ba2+

SO42-

SO42 -K+Cl-Ba2+

Ionic EquationsIn writing equations for reactions in aqueous solution, it is often useful to indicatewhether the dissolved substances are present predominantly as ions or as molecules.Let’s reconsider the precipitation reaction between Pb(NO3)2 and 2 KI:

An equation written in this fashion, showing the complete chemical formulas of reac-tants and products, is called a molecular equation because it shows chemical formulas

Pb1NO3221aq2 + 2 KI1aq2 ¡ PbI21s2 + 2 KNO31aq2


without indicating ionic character. Because Pb(NO3)2, KI, and KNO3 are all water-soluble ionic compounds and therefore strong electrolytes, we can write the equation ina form that indicates which species exist as ions in the solution:

[4.7]

An equation written in this form, with all soluble strong electrolytes shown as ions, iscalled a complete ionic equation.

Notice that and appear on both sides of Equation 4.7. Ions thatappear in identical forms on both sides of a complete ionic equation, called spectatorions, play no direct role in the reaction. When spectator ions are omitted from theequation (they cancel out like algebraic quantities), we are left with the net ionicequation, which is one that includes only the ions and molecules directly involved inthe reaction:

[4.8]

Because charge is conserved in reactions, the sum of the ionic charges must bethe same on both sides of a balanced net ionic equation. In this case the charge ofthe cation and the two charges of the anions add to zero, the charge of the electri-cally neutral product. If every ion in a complete ionic equation is a spectator, no reac-tion occurs.

G I V E I T S O M E T H O U G H T

Which ions, if any, are spectator ions in the reaction?

Net ionic equations illustrate the similarities between various reactions involvingelectrolytes. For example, Equation 4.8 expresses the essential feature of the precipita-tion reaction between any strong electrolyte containing (aq) and any strong elec-trolyte containing (aq): The ions combine to form a precipitate of PbI2. Thus, a netionic equation demonstrates that more than one set of reactants can lead to the samenet reaction. For example, aqueous solutions of KI and MgI2 share many chemical sim-ilarities because both contain ions. Either solution when mixed with a Pb(NO3)2solution produces PbI2(s). The complete ionic equation, on the other hand, identifiesthe actual reactants that participate in a reaction.

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 soluble

strong electrolyte dissociates into its ions. Only strong electrolytes dissolved in aque-ous solution are written in ionic form.

3. Identify and cancel spectator ions.

I-

I-Pb2+

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

1-2+

Pb2+1aq2 + 2 I-1aq2 ¡ PbI21s2

NO3-1aq2K+(aq)

PbI21s2 + 2 K+1aq2 + 2 NO3-1aq2Pb2+1aq2 + 2 NO3

-1aq2 + 2 K+1aq2 + 2 I-1aq2 ¡

SAMPLE EXERCISE 4.4 Writing a Net Ionic Equation

Write the net ionic equation for the precipitation reaction that occurs when aqueous 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 write the chemical formulas of the reactants and products and then determinewhich product is insoluble. We then write and balance the molecular equation. Next, we writeeach soluble strong electrolyte as separated ions to obtain the complete ionic equation. Finally,we eliminate the spectator ions to obtain the net ionic equation.

Solve Calcium chloride is composed of calcium ions, , and chloride ions, ; hence, anaqueous solution of the substance is CaCl2(aq). Sodium carbonate is composed of ions and

ions; hence, an aqueous solution of the compound is Na2CO3(aq). In the molecularCO32 -

Na+Cl-Ca2+


! FIGURE 4.5 Vinegar and lemon juiceare common household acids. Ammoniaand baking soda (sodium bicarbonate)are common household bases.

Acetic acid, CH3COOH

Nitric acid, HNO3

Hydrochloric acid, HCl

H

Cl

N

O

C

! FIGURE 4.6 Molecular models ofthree common acids.

equations for precipitation reactions, the anions and cations appear to exchange partners. Thus,we put and together to give CaCO3 and and together to give NaCl. Accord-ing to the solubility guidelines in Table 4.1, CaCO3 is insoluble and NaCl is soluble. The balancedmolecular equation is

In a complete ionic equation, only dissolved strong electrolytes (such as soluble ionic com-pounds) are written as separate ions. As the (aq) designations remind us, CaCl2, Na2CO3, andNaCl are all dissolved in the solution. Furthermore, they are all strong electrolytes. CaCO3 is anionic compound, but it is not soluble. We do not write the formula of any insoluble compoundas its component ions. Thus, the complete ionic 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 electric chargeare balanced. Each side has one Ca, one C, and three O, and the net charge on each side equals 0.

Comment If none of the ions in an ionic equation is removed from solution or changed insome way, all ions are spectator ions and a reaction does not occur.

PRACTICE EXERCISEWrite the net ionic equation for the precipitation reaction that occurs when aqueous solutionsof silver nitrate and potassium phosphate are mixed.

Answer:

4.3 | ACIDS, BASES, AND NEUTRALIZATIONREACTIONS

Many acids and bases are industrial and household substances (" FIGURE 4.5), andsome are important components of biological fluids. Hydrochloric acid, for example, isan important industrial chemical and the main constituent of gastric juice in yourstomach. Acids and bases are also common electrolytes.

AcidsAs noted in Section 2.8, acids are substances that ionize in aqueous solution to form hy-drogen ions . Because a hydrogen atom consists of a proton and an electron,is simply a proton. Thus, acids are often called proton donors. Molecular models of threecommon acids are shown in " FIGURE 4.6.

Protons in aqueous solution are solvated by water molecules, just as other cationsare [Figure 4.3(a)]. In writing chemical equations involving protons in water, therefore,we write .

Molecules of different acids ionize to form different numbers of ions. Both HCland HNO3 are monoprotic acids, yielding one per molecule of acid. Sulfuric acid,H2SO4, is a diprotic acid, one that yields two per molecule of acid. The ionization ofH2SO4 and other diprotic acids occurs in two steps:

[4.9]

[4.10]

Although H2SO4 is a strong electrolyte, only the first ionization (Equation 4.9) iscomplete. Thus, aqueous solutions of sulfuric acid contain a mixture of ,

and .The molecule CH3COOH (acetic acid) that we have mentioned frequently is the

primary component in vinegar. Acetic acid has four hydrogens, as Figure 4.6 shows, butonly one of them, the H in the COOH group, is ionized in water. The three other hydro-gens are bound to carbon and do not break their bonds in water.C ¬ H

SO42-1aq2HSO4

-1aq2, H+1aq2HSO4

-1aq2 ! H+1aq2 + SO42-1aq2H2SO41aq2 ¡ H+1aq2 + HSO4

-1aq2H+

H+H+

H+1aq2H+H+1aq2

3 Ag+1aq2 + PO43-1aq2 ¡ Ag3PO41s2

Ca2+1aq2 + CO32-1aq2 ¡ CaCO31s2Na+Cl-

CaCO31s2 + 2 Na+1aq2 + 2 Cl-1aq2Ca2+1aq2 + 2 Cl-1aq2 + 2 Na+1aq2 + CO32-1aq2 ¡

CaCl21aq2 + Na2CO31aq2 ¡ CaCO31s2 + 2 NaCl1aq2Cl-Na+CO3

2-Ca2+

SECTION 4.3 Acids, Bases, and Neutralization Reactions 125

G I V E I T S O M E T H O U G H TThe structural formula of citric acid, a main component of citrusfruits, is

How many can be generated by each citric acid molecule dissolved in water?

BasesBases are substances that accept (react with) ions. Bases produce hydroxide ions

when they dissolve in water. Ionic hydroxide compounds, such as NaOH, KOH,and Ca(OH)2, are among the most common bases. When dissolved in water, they disso-ciate into ions, introducing ions into the solution.

Compounds that do not contain ions can also be bases. For example, ammo-nia (NH3) is a common base. When added to water, it accepts an ion from a watermolecule and thereby produces an ion (! FIGURE 4.7):

[4.11]

Ammonia is a weak electrolyte because only about 1% of the NH3 forms andions.

Strong and Weak Acids and BasesAcids and bases that are strong electrolytes (completely ionized in solution) are strongacids and strong bases. Those that are weak electrolytes (partly ionized) are weakacids and weak bases. When reactivity depends only on concentration, strongacids are more reactive than weak acids. The reactivity of an acid, however, can dependon the anion as well as on concentration. For example, hydrofluoric acid (HF) isa weak acid (only partly ionized in aqueous solution), but it is very reactive and vigor-ously attacks many substances, including glass. This reactivity is due to the combinedaction of and .

" TABLE 4.2 lists the strong acids and bases we are most likely to encounter. Youneed to commit this information to memory in order to correctly identify strong elec-trolytes and write net ionic equations. The brevity of this list tells us that most acids areweak. (For H2SO4, as we noted earlier, only the first proton completely ionizes.) Theonly common strong bases are the common soluble metal hydroxides. Most other metalhydroxides are insoluble in water. The most common weak base is NH3, which reactswith water to form ions (Equation 4.11).OH-

F-1aq2H+1aq2H+1aq2 H+1aq2

OH-NH4

+NH31aq2 + H2O1l2 ! NH4

+1aq2 + OH-1aq2OH-H+

OH-OH-

1OH-2 H+

H+(aq)

C COOH

COOH

COOH

H

HO

H

C

C

H

H H2O OH!NH3

" "

NH4"

# FIGURE 4.7 Hydrogen ion transfer.An H2O molecule acts as a proton donor(acid), and NH3 acts as a proton acceptor(base). Only a fraction of the NH3 moleculesreact with H2O. Consequently, NH3 is aweak electrolyte.

TABLE 4.2 • Common Strong Acids and Bases

Strong Acids Strong Bases

Hydrochloric, HCl Group 1A metal hydroxides [LiOH, NaOH, KOH, RbOH, CsOH]Hydrobromic, HBr Heavy group 2A metal hydroxides [Ca(OH)2, Sr(OH)2, Ba(OH)2]Hydroiodic, HIChloric, HClO3

Perchloric, HClO4

Nitric, HNO3

Sulfuric, H2SO4


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

Strong Electrolyte Weak Electrolyte Nonelectrolyte

Ionic All None NoneMolecular Strong acids (see Table 4.2) Weak acids, weak bases All other compounds

G I V E I T S O M E T H O U G H TWhy isn’t Al(OH)3 classified as a strong base?

SAMPLE EXERCISE 4.5 Comparing Acid Strengths

The following diagrams represent aqueous solutions of acids HX, HY, and HZ, with watermolecules omitted for clarity. Rank the acids from strongest to weakest.

SOLUTIONAnalyze We are asked to rank three acids from strongest to weakest, based on schematicdrawings of their solutions.

Plan We can determine the relative numbers of uncharged molecular species in the diagrams.The strongest acid is the one with the most ions and fewest undissociated molecules insolution. The weakest acid is the one with the largest number of undissociated molecules.

Solve The order is . HY is a strong acid because it is totally ionized (no HYmolecules in solution), whereas both HX and HZ are weak acids, whose solutions consist of amixture of molecules and ions. Because HZ contains more ions and fewer molecules thanHX, it is a stronger acid.

PRACTICE EXERCISEImagine a diagram showing 10 ions and 10 ions. If this solution were mixed with theone pictured above for HY, what species would be present in a diagram that represents thecombined solutions after any possible reaction?

Answer: The diagram would show 10 ions, 2 ions, 8 ions, and 8 H2O molecules.

Identifying Strong and Weak ElectrolytesIf we remember the common strong acids and bases (Table 4.2) and also remember thatNH3 is a weak base, we can make reasonable predictions about the electrolytic strength ofa great number of water-soluble substances. ! TABLE 4.3 summarizes our observationsabout electrolytes. To classify a soluble substance as strong electrolyte, weak electrolyte,or nonelectrolyte, we work our way down and across this table. We first ask whether thesubstance is ionic or molecular. If it is ionic, it is a strong electrolyte. The second columnof Table 4.3 tells us that all ionic compounds are strong electrolytes. If the substance ismolecular, we ask whether it is an acid or a base. (It is an acid if it either has H first in thechemical formula or contains a COOH group.) If it is an acid, we use Table 4.2 to deter-mine whether it is a strong or weak electrolyte: All strong acids are strong electrolytes,and all weak acids are weak electrolytes. If an acid is not listed in Table 4.2, it is probablya weak acid and therefore a weak electrolyte.

Y-OH-Na+

OH-Na+

H+

HY 7 HZ 7 HX

H+

!

!"

!

!

!!

!!

! !

"

!

!

!

"

"

"

"

""

"

""

"

"

HX HY HZ


If our substance is a base, we use Table 4.2 to determine whether it is a strong base.NH3 is the only molecular base that we consider in this chapter, and Table 4.3 tells us itis a weak electrolyte. Finally, any molecular substance that we encounter in this chapterthat is not an acid or NH3 is probably a nonelectrolyte.

SAMPLE EXERCISE 4.6 Identifying Strong, Weak, and Nonelectrolytes

Classify these dissolved substances as strong, weak, or nonelectrolyte: CaCl2, HNO3, C2H5OH(ethanol), HCOOH (formic acid), KOH.

SOLUTIONAnalyze We are given several chemical formulas and asked to classify each substance as astrong electrolyte, weak electrolyte, or nonelectrolyte.

Plan The approach we take is outlined in Table 4.3. We can predict whether a substance isionic or molecular based on its composition. As we saw in Section 2.7, most ionic compoundswe encounter in this text are composed of a metal and a nonmetal, whereas most molecularcompounds are composed only of nonmetals.

Solve Two compounds fit the criteria for ionic compounds: CaCl2 and KOH. Because Table4.3 tells us that all ionic compounds are strong electrolytes, that is how we classify these twosubstances. The three remaining compounds are molecular. Two, HNO3 and HCOOH, areacids. Nitric acid, HNO3, is a common strong acid, as shown in Table 4.2, and therefore is astrong electrolyte. Because most acids are weak acids, our best guess would be that HCOOH isa weak acid (weak electrolyte). This is correct. The remaining molecular compound, C2H5OH,is neither an acid nor a base, so it is a nonelectrolyte.

Comment Although C2H5OH has an OH group, it is not a metal hydroxide and so not abase. Rather, it is a member of a class of organic compounds that have C—OH bonds, whichare known as alcohols. •(Section 2.9) Organic compounds containing the COOH groupare called carboxylic acids (Chapter 16). Molecules that have this group are weak acids.

PRACTICE EXERCISEConsider solutions in which 0.1 mol of each of the following compounds is dissolved in 1 L ofwater: Ca(NO3)2 (calcium nitrate), C6H12O6 (glucose), NaCH3COO (sodium acetate), andCH3COOH (acetic acid). Rank the solutions in order of increasing electrical conductivity,based on the fact that the greater the number of ions in solution, the greater the conductivity.

Answers: C6H12O6 (weak electrolyte, existing mainly inthe form of molecules with few ions) (strong electrolyte that provides twoions, and (strong electrolyte that provides three ions,and )

Neutralization Reactions and SaltsThe properties of acidic solutions are quite different from those of basic solutions. Acidshave a sour taste, whereas bases have a bitter taste.* Acids change the colors of certaindyes in a way that differs from the way bases affect the same dyes. This is the principlebehind the indicator known as litmus paper (! FIGURE 4.8). In addition, acidic andbasic solutions differ in chemical properties in several other important ways that weexplore in this chapter and in later chapters.

When a solution of an acid and a solution of a base are mixed, a neutralizationreaction occurs. The products of the reaction have none of the characteristic propertiesof either the acidic solution or the basic solution. For example, when hydrochloric acidis mixed with a solution of sodium hydroxide, the reaction is

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

HCl1aq2 + NaOH1aq2 ¡ H2O1l2 + NaCl (aq)

2 NO3-

Ca2+CH3COO-2 6 Ca1NO322Na+6 NaCH3COO

1nonelectrolyte2 6 CH3COOH

Base turns litmus paper blue

Acid turns litmus paper red

" FIGURE 4.8 Litmus paper. Litmuspaper is coated with dyes that change colorin response to exposure to either acids orbases.

*Tasting chemical solutions is not a good practice. However, we have all had acids such as ascorbic acid (vita-min C), acetylsalicylic acid (aspirin), and citric acid (in citrus fruits) in our mouths, and we are familiar withtheir characteristic sour taste. Soaps, which are basic, have the characteristic bitter taste of bases.

Water and table salt, NaCl, are the products of the reaction. By analogy to this reaction,the term salt has come to mean any ionic compound whose cation comes from a base(for example, from NaOH) and whose anion comes from an acid (for example,from HCl). In general, a neutralization reaction between an acid and a metal hydroxideproduces water and a salt.

Because HCl, NaOH, and NaCl are all water-soluble strong electrolytes, thecomplete ionic equation associated with Equation 4.12 is

[4.13]Therefore, the net ionic equation is

[4.14]

Equation 4.14 summarizes the main feature of the neutralization reaction between anystrong acid and any strong base: and ions combine to form H2O.

! FIGURE 4.9 shows the neutralization reaction between hydrochloric acid andthe water-insoluble base Mg(OH)2:

Molecular equation:[4.15]

Net ionic equation:[4.16]

Notice that the ions (this time in a solid reactant) and ions combine to formH2O. Because the ions exchange partners, neutralization reactions between acids andmetal hydroxides are metathesis reactions.

H+OH-Mg1OH221s2 + 2 H+1aq2 ¡ Mg2+1aq2 + 2 H2O1l2

Mg1OH221s2 + 2 HCl1aq2 ¡ MgCl21aq2 + 2 H2O1l2OH-1aq2H+(aq)

H+1aq2 + OH-1aq2 ¡ H2O1l2H2O1l2 + Na+1aq2 + Cl-1aq2H+1aq2 + Cl-1aq2 + Na+1aq2 + OH-1aq2 ¡

Cl-Na+


MgCl2(aq) ! 2 H2O(l)Mg(OH)2(s) ! 2 HCl(aq)

+

ProductsReactants

H!(aq) combines withhydroxide ions in Mg(OH)2(s),

forming H2O(l)

Mg(OH)2

H!

Mg2!

Cl– Cl–

H2O

" FIGURE 4.9 Neutralization reaction between Mg(OH)2(s) and hydrochloric acid. Milk of magnesia is a suspension of water-insolublemagnesium hydroxide, Mg(OH)2(s), in water. When sufficient hydrochloric acid, HCl(aq), is added a reaction ensues that leads to an aqueoussolution containing Mg2+(aq) and Cl–(aq) ions.

G O F I G U R EAdding just a few drops of hydrochloric acid would not be sufficient to dissolve all the Mg(OH)2(s). Why not?


SAMPLE EXERCISE 4.7 Writing Chemical Equations for a Neutralization Reaction

For the reaction between aqueous solutions of acetic acid (CH3COOH) and barium hydroxide, Ba(OH)2,write (a) the balanced molecular equation, (b) the complete ionic equation, (c) the net ionic equation.

(a) The salt contains the cation of the base and the anionof the acid . Thus, the salt formula is Ba(CH3COO)2.According to Table 4.1, this compound is soluble in water. Theunbalanced molecular equation for the neutralization reaction is

1CH3COO-2 1Ba2+2CH3COOH1aq2 + Ba1OH221aq2 ¡ H2O1l2 + Ba1CH3COO221aq2

To balance this equation, we must provide two molecules ofCH3COOH to furnish the two ions and to supplythe two ions needed to combine with the two ions ofthe base. The balanced molecular equation is

OH-H+CH3COO-

2 H2O1l2 + Ba1CH3COO221aq22 CH3COOH1aq2 + Ba1OH221aq2 ¡

(b) To write the complete ionic equation, we identify the strongelectrolytes and break them into ions. In this case Ba(OH)2 andBa(CH3COO)2 are both water-soluble ionic compounds andhence strong electrolytes. Thus, the complete ionic equation is 2 H2O1l2 + Ba2+1aq2 + 2 CH3COO-1aq22 CH3COOH1aq2 + Ba2+1aq2 + 2 OH-1aq2 ¡

(c) Eliminating the spectator ion, Ba2+, and simplifying coeffi-cients gives the net ionic equation:

CH3COOH 1aq2 + OH-1aq2 ¡ H2O1l2 + CH3COO-1aq22 CH3COOH1aq2 + 2 OH-1aq2 ¡ 2 H2O1l2 + 2 CH3COO-1aq2

PRACTICE EXERCISEFor the reaction of phosphorous acid (H3PO3) and potassium hydroxide (KOH), write (a) the balancedmolecular equation and (b) the net ionic equation.

Answers: (a) ,(b) . (H3PO3 is a weak acid and therefore a weakelectrolyte, whereas KOH, a strong base, and K3PO3, an ionic compound, are strong electrolytes.)

H3PO31aq2 + 3 OH-1aq2 ¡ 3 H2O1l2 + PO33-1aq2H3PO31aq2 + 3 KOH1aq2 ¡ 3 H2O1l2 + K3PO31aq2

Neutralization Reactions with Gas FormationMany bases besides react with to form molecular compounds. Two of thesethat you might encounter in the laboratory are the sulfide ion and the carbonate ion.Both of these anions react with acids to form gases that have low solubilities in water.Hydrogen sulfide (H2S), the substance that gives rotten eggs their foul odor, forms whenan acid such as HCl(aq) reacts with a metal sulfide such as Na2S:

Molecular equation:

[4.17]

Net ionic equation:

[4.18]

Carbonates and bicarbonates react with acids to form CO2(g). Reaction ofor with an acid first gives carbonic acid (H2CO3). For example, when hy-drochloric acid is added to sodium bicarbonate, the reaction is

[4.19]HCl1aq2 + NaHCO31aq2 ¡ NaCl1aq2 + H2CO31aq2HCO3

-CO3

2-

2 H+1aq2 + S2-1aq2 ¡ H2S1g22 HCl1aq2 + Na2S1aq2 ¡ H2S1g2 + 2 NaCl1aq2

H+OH-

SOLUTIONAnalyze We are given the chemical formulas for an acid and abase and asked to write a balanced molecular equation, a completeionic equation, and a net ionic equation for their neutralizationreaction.

Plan As Equation 4.12 and the italicized statement that follows itindicate, neutralization reactions form two products, H2O and asalt. We examine the cation of the base and the anion of the acid todetermine the composition of the salt.

Solve

Check We can determine whether the molecular equation is bal-anced by counting the number of atoms of each kind on both sidesof the arrow (10 H, 6 O, 4 C, and 1 Ba on each side). However, itis often easier to check equations by counting groups: There are

2 CH3COO groups, as well as 1 Ba, and 4 additional H atoms and 2additional O atoms on each side of the equation. The net ionicequation checks out because the numbers of each kind of elementand the net charge are the same on both sides of the equation.


Antacids

Your stomach secretes acids to help digest foods. Theseacids, which include hydrochloric acid, contain about

0.1 mol of per liter of solution. The stomach anddigestive tract are normally protected from the

corrosive effects of stomach acid by a mucosallining. Holes can develop in this lining, however, allowing the acid toattack the underlying tissue, causing painful damage. These holes,known as ulcers, can be caused by the secretion of excess acids or bya weakness in the digestive lining. Studies indicate, however, thatmany ulcers are caused by bacterial infection. Between 10 and 20%of Americans suffer from ulcers at some point in their lives. Many

H+

! FIGURE 4.10 Antacids. These products all serve as acid-neutralizing agents in the stomach.

TABLE 4.4 • Some Common Antacids

Commercial Name Acid-Neutralizing Agents

Alka-Seltzer® NaHCO3

Amphojel® Al(OH)3

Di-Gel® Mg(OH)2 and CaCO3

Milk of Magnesia Mg(OH)2

Maalox® Mg(OH)2 and Al(OH)3

Mylanta® Mg(OH)2 and Al(OH)3

Rolaids® NaAl(OH)2CO3

Tums® CaCO3

CHEMISTRY PUT TO WORK

Carbonic acid is unstable. If present in solution in sufficient concentrations, itdecomposes to H2O and CO2, which escapes from the solution as a gas:

[4.20]

The overall reaction is summarized by the equations

Molecular equation:

[4.21]

Net ionic equation:

[4.22]

Both NaHCO3(s) and Na2CO3(s) are used as neutralizers in acid spills, either salt isadded until the fizzing caused by CO2(g) formation stops. Sometimes sodium bicarbon-ate is used as an antacid to soothe an upset stomach. In that case the reacts withstomach acid to form CO2(g).

G I V E I T S O M E T H O U G H TBy analogy to examples given in the text, predict what gas forms whenNa2SO3(s) reacts with HCl(aq).

HCO3-

H2O1l2 + CO21g2H+1aq2 + HCO3-1aq2 ¡

NaCl1aq2 + H2O1l2 + CO21g2HCl1aq2 + NaHCO31aq2 ¡

H2CO31aq2 ¡ H2O1l2 + CO21g2

others experience occasional indigestion or heartburn due to diges-tive acids entering the esophagus.

We can address the problem of excess stomach acid in two ways:(1) removing the excess acid or (2) decreasing the production of acid.Substances that remove excess acid are called antacids, whereasthose that decrease acid production are called acid inhibitors." FIGURE 4.10 shows several common over-the-counter antacids,which usually contain hydroxide, carbonate, or bicarbonate ions(# TABLE 4.4). Antiulcer drugs, such as Tagamet® and Zantac®, areacid inhibitors. They act on acid-producing cells in the lining of thestomach. Formulations that control acid in this way are now availableas over-the-counter drugs.

RELATED EXERCISE: 4.95

SECTION 4.4 Oxidation-Reduction Reactions 131

4.4 | OXIDATION-REDUCTION REACTIONSIn precipitation reactions, cations and anions come together to form an insoluble ioniccompound. In neutralization reactions, ions and ions come together to formH2O molecules. Now let’s consider a third kind of reaction, one in which electrons aretransferred from one reactant to another. Such reactions are called either oxidation-reduction reactions or redox reactions. In this chapter we concentrate on redox reac-tions where one of the reactants is a metal in its elemental form.

Oxidation and ReductionOne of the most familiar redox reactions is corrosion of a metal (! FIGURE 4.11). Insome instances corrosion is limited to the surface of the metal, with the green coatingthat forms on copper roofs and statues being one such case. In other instances the cor-rosion goes deeper, eventually compromising the structural integrity of the metal. Ironrusting is an important example.

Corrosion is the conversion of a metal into a metal compound by a reaction betweenthe metal and some substance in its environment. When a metal corrodes, each metalatom loses electrons and so forms a cation, which can combine with an anion to form anionic compound. The green coating on the Statue of Liberty contains Cu2+ combinedwith carbonate and hydroxide anions, rust contains Fe3+ combined with oxide andhydroxide anions, and silver tarnish contains Ag+ combined with sulfide anions.

When an atom, ion, or molecule becomes more positively charged (that is, when itloses electrons), we say that it has been oxidized. Loss of electrons by a substance is calledoxidation. The term oxidation is used because the first reactions of this sort to be stud-ied were reactions with oxygen. Many metals react directly with O2 in air to form metaloxides. In these reactions the metal loses electrons to oxygen, forming an ionic com-pound of the metal ion and oxide ion. The familiar example of rusting involves the reac-tion between iron metal and oxygen in the presence of water. In this process Fe isoxidized (loses electrons) to form .

The reaction between iron and oxygen tends to be relatively slow, but other metals,such as the alkali and alkaline earth metals, react quickly upon exposure to air." FIGURE 4.12 shows how the bright metallic surface of calcium tarnishes as CaOforms in the reaction

[4.23]

In this reaction Ca is oxidized to Ca2+ and neutral O2 is transformed to ions. Whenan atom, ion, or molecule becomes more negatively charged (gains electrons), we saythat it is reduced. The gain of electrons by a substance is called reduction. When one reac-tant loses electrons (that is, when it is oxidized), another reactant must gain them. Inother words, oxidation of one substance must be accompanied by reduction of someother substance.

O2-2 Ca1s2 + O21g2 ¡ 2 CaO1s2

Fe3+

OH-H+

(a) (b) (c)

# FIGURE 4.11 Familiar corrosion products. (a) A green coating forms when copper is oxidized. (b) Rust forms when iron corrodes.(c) A black tarnish forms as silver corrodes.


Oxidation NumbersBefore we can identify an oxidation-reduction reaction, we must have a bookkeepingsystem—a way of keeping track of electrons gained by the substance being reduced andelectrons lost by the substance being oxidized. The concept of oxidation numbers (alsocalled oxidation states) was devised as a way of doing this. Each atom in a neutral sub-stance or ion is assigned an oxidation number. For monatomic ions the oxidation num-ber is the same as the charge. For neutral molecules and polyatomic ions, the oxidationnumber of a given atom is a hypothetical charge. This charge is assigned by artificiallydividing up the electrons among the atoms in the molecule or ion. We use the followingrules for assigning oxidation numbers:

1. For an atom in its elemental form, the oxidation number is always zero. Thus, eachH atom in the H2 molecule has an oxidation number of 0 and each P atom in theP4 molecule has an oxidation number of 0.

2. For any monatomic ion the oxidation number equals the ionic charge. Thus, hasan oxidation number of , has an oxidation number of , and so forth. Inionic compounds the alkali metal ions (group 1A) always have a charge andtherefore an oxidation number of . The alkaline earth metals (group 2A) are al-ways , and aluminum (group 3A) is always in ionic compounds. (In writingoxidation 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 can sometimesbe positive:

(a) The oxidation number of oxygen is usually in both ionic and molecular com-pounds. The major exception is in compounds called peroxides, which containthe ion, giving each oxygen an oxidation number of .

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

(c) The oxidation number of fluorine is in all compounds. The other halogenshave an oxidation number of in most binary compounds. When combinedwith oxygen, as in oxyanions, however, they have positive 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 hydrogen is

and that of oxygen is . Thus, the sum of the oxidation numbers is, which equals the net charge of the ion. This rule is useful in

obtaining the oxidation number of one atom in a compound or ion if you know theoxidation numbers of the other atoms, as illustrated in Sample Exercise 4.8.

3(+1) + 1-2) = +1-2+1

H3O+

-1-1

-1+1

-1O22-

-2

+3+2+1

1+-2S2-+1

K+

O2(g) is reduced(gains electrons)

2 CaO(s)2 Ca(s) ! O2(g)ProductsReactants

Ca(s) is oxidized(loses electrons)

Ca2! and O2" ionscombine to form CaO(s)

e"

e"

e"

! FIGURE 4.12 Oxidation of calcium metal by molecular oxygen. The oxidation involves transfer of electrons from thecalcium metal to the O2, leading to formation of CaO.


It’s important to remember that in every oxidation-reduction reaction, the oxida-tion numbers of at least two atoms must change. The oxidation number increases forany atom that is oxidized and decreases for any atom that is reduced.

G I V E I T S O M E T H O U G H TWhat is the oxidation number of nitrogen (a) in aluminum nitride, AlN, and (b) innitric acid, HNO3?

SAMPLE EXERCISE 4.8 Determining Oxidation Numbers

Determine the oxidation number of sulfur in (a) H2S, (b) S8, (c) SCl2, (d) Na2SO3, (e) .

SOLUTIONAnalyze We are asked to determine the oxidation number of sulfur in two molecular species,in the elemental form, and in two substances containing ions.

Plan In each species the sum of oxidation numbers of all the atoms must equal the charge onthe species. We will use the rules outlined previously to assign oxidation numbers.

Solve(a) When bonded to a nonmetal, hydrogen has an oxidation number of (rule 3b). Becausethe H2S molecule is neutral, the sum of the oxidation numbers must equal zero (rule 4). Let-ting x equal the oxidation number of S, we have . Thus, S has an oxidationnumber 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 oxidation number of(rule 3c). The sum of the oxidation numbers must equal zero (rule 4). Letting x equal theoxidation number of S, we have . Consequently, the oxidation number of Smust be .

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

(e) The oxidation state of O is (rule 3a). The sum of the oxidation numbers equals , thenet charge of the ion (rule 4). Thus, we have . From this relation weconclude that the oxidation number of S in this ion is .

Comment 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 these exam-ples, range from to .

PRACTICE EXERCISEWhat is the oxidation state of the boldfaced element in (a) P2O5, (b) NaH, (c)(d) SnBr4, (e) BaO2?

Answers: (a) , (b) , (c) , (d) , (e)

Oxidation of Metals by Acids and SaltsThe reaction between a metal and either an acid or a metal salt conforms to the generalpattern

[4.24]

Examples:

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

Mn1s2 + Pb1NO3221aq2 ¡ Mn1NO3221aq2 + Pb1s2Zn1s2 + 2 HBr1aq2 ¡ ZnBr21aq2 + H21g2A + BX ¡ AX + B

-1+4+6-1+5

Cr2O72-,

+6-2

+6x + 41-22 = -2SO4

2--2-2

+421+12 + x + 31-22 = 0

-2+1

+2x + 21-12 = 0

-1

-221+12 + x = 0

+1

SO42-


Many metals undergo displacement reactions with acids, producing salts andhydrogen gas. For example, magnesium metal reacts with hydrochloric acid to formmagnesium chloride and hydrogen gas (! FIGURE 4.13):

The oxidation number of Mg changes from 0 to , an increase that indicates the atomhas lost electrons and has therefore been oxidized. The oxidation number of in theacid decreases from to 0, indicating that this ion has gained electrons and has there-fore been reduced. Chlorine has an oxidation number of both before and after thereaction, indicating that it is neither oxidized nor reduced. In fact the Cl– ions are spec-tator ions, dropping out of the net ionic equation:

[4.26]

Metals can also be oxidized by aqueous solutions of various salts. Iron metal, forexample, is oxidized to by aqueous solutions of such as Ni(NO3)2(aq):

Molecular equation: [4.27]

Net ionic equation: [4.28]

The oxidation of Fe to in this reaction is accompanied by the reduction ofto Ni. Remember: Whenever one substance is oxidized, another substance must be reduced.

Ni2+Fe2+

Fe1s2 + Ni2+1aq2 ¡ Fe2+1aq2 + Ni1s2Fe1s2 + Ni1NO3221aq2 ¡ Fe1NO3221aq2 + Ni1s2Ni2+Fe2+

Mg1s2 + 2 H+1aq2 ¡ Mg2+1aq2 + H21g2-1

+1H+

+2

Mg(s) ! 2 HCl(aq) MgCl2(aq) ! H2(g)

!1 "1 00Oxidation number !2 "1

[4.25]

e!

e!

e!

H!(aq) is reduced(gains electrons)

H!

Mg2!

H2

MgCl"

"

"

Mg(s) is oxidized(loses electrons)

H2(g) ! MgCl2(aq)2 HCl(aq) ! Mg(s)Oxidationnumber

!1 !2 "10"1 0

ProductsReactants

! FIGURE 4.13 Reaction of magnesium metal with hydrochloric acid. The metal is readily oxidized by the acid, producing hydrogengas, H2(g), and MgCl2(aq).


SAMPLE EXERCISE 4.9 Writing Equations for Oxidation-ReductionReactions

Write the balanced molecular and net ionic equations for the reaction of aluminum withhydrobromic acid.

SOLUTIONAnalyze We must write two equations—molecular and net ionic—for the redox reactionbetween a metal and an acid.

Plan Metals react with acids to form salts and H2 gas. To write the balanced equations, wemust write the chemical formulas for the two reactants and then determine the formula of thesalt, which is composed of the cation formed by the metal and the anion of the acid.

Solve The reactants are Al and HBr. The cation formed by Al is , and the anion fromhydrobromic acid is . Thus, the salt formed in the reaction is AlBr3. Writing the reactantsand products and then balancing the equation gives the molecular equation:

Both HBr and AlBr3 are soluble strong electrolytes. Thus, the complete ionic equation is

Because is a spectator ion, the net ionic equation is

Comment The substance oxidized is the aluminum metal because its oxidation state changesfrom 0 in the metal to in the cation, thereby increasing in oxidation number. The isreduced because its oxidation state changes from in the acid to 0 in H2.

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

Answers: (a) ;, (b) Mg is oxidized and is reduced.

The Activity SeriesCan we predict whether a certain metal will be oxidized either by an acid or by a partic-ular salt? This question is of practical importance as well as chemical interest. Accordingto Equation 4.27, for example, it would be unwise to store a solution of nickel nitrate inan iron container because the solution would dissolve the container. When a metal is ox-idized, it forms various compounds. Extensive oxidation can lead to the failure of metalmachinery parts or the deterioration of metal structures.

Different metals vary in the ease with which they are oxidized. Zn is oxidized byaqueous solutions of , for example, but Ag is not. Zn, therefore, loses electronsmore readily than Ag; that is, Zn is easier to oxidize than Ag.

A list of metals arranged in order of decreasing ease of oxidation, such as! TABLE 4.5, is called an activity series. The metals at the top of the table, such as thealkali metals and the alkaline earth metals, are most easily oxidized; that is, they reactmost readily to form compounds. They are called the active metals. The metals at thebottom of the activity series, such as the transition elements from groups 8B and 1B, arevery stable and form compounds less readily. These metals, which are used to makecoins and jewelry, are called noble metals because of their low reactivity.

The activity series can be used to predict the outcome of reactions between metalsand either metal salts or acids. Any metal on the list can be oxidized by the ions of elementsbelow it. For example, copper is above silver in the series. Thus, copper metal is oxidizedby silver ions:

[4.29]Cu1s2 + 2 Ag+1aq2 ¡ Cu2+1aq2 + 2 Ag1s2

Cu2+

Co2+Mg2+1aq2 + Co1s2Mg1s2 + Co2+1aq2 ¡Mg1s2 + CoSO41aq2 ¡ MgSO41aq2 + Co1s2

+1H++3

2 Al1s2 + 6 H+1aq2 ¡ 2 Al3+1aq2 + 3 H21g2Br-

2 Al1s2 + 6 H+1aq2 + 6 Br-1aq2 ¡ 2 Al3+1aq2 + 6 Br-1aq2 + 3 H21g22 Al1s2 + 6 HBr1aq2 ¡ 2 AlBr31aq2 + 3 H21g2Br-

Al3+


The oxidation of copper to copper ions is accompanied by the reduction of silver ions tosilver metal. The silver metal is evident on the surface of the copper wire in ! FIGURE 4.14.The copper(II) nitrate produces a blue color in the solution, as can be seen most clearly inthe photograph on the right of Figure 4.14.

G I V E I T S O M E T H O U G H TDoes a reaction occur (a) when an aqueous solution of NiCl2(aq) is added to atest tube containing strips of metallic zinc, and (b) when NiCl2(aq) is added to atest tube containing Zn(NO3)2(aq)?

Only metals above hydrogen in the activity series are able to react with acids toform H2. For example, Ni reacts with HCl(aq) to form H2:

[4.30]

Because elements below hydrogen in the activity series are not oxidized by , Cudoes not react with HCl(aq). Interestingly, copper does react with nitric acid, as shownin Figure 1.11, but the reaction is not oxidation of Cu by ions. Instead, the metal isoxidized to by the nitrate ion, accompanied by the formation of brown nitrogendioxide, NO2(g):

[4.31]

As the copper is oxidized in this reaction, , where the oxidation number of nitro-gen is , is reduced to NO2, where the oxidation number of nitrogen is . We willexamine reactions of this type in Chapter 20.

+4+5NO3

-

Cu1s2 + 4 HNO31aq2 ¡ Cu1NO3221aq2 + 2 H2O1l2 + 2 NO21g2Cu2+

H+

H+

Ni1s2 + 2 HCl1aq2 ¡ NiCl21aq2 + H21g2

TABLE 4.5 • Activity Series of Metals in Aqueous Solution

Metal Oxidation Reaction

Ease

of o

xida

tion

incr

ease

s

Lithium Li1s2 ¡ Li+1aq2 + e-

Potassium K1s2 ¡ K+1aq2 + e-

Barium Ba1s2 ¡ Ba2+1aq2 + 2e-

Calcium Ca1s2 ¡ Ca2+1aq2 + 2e-

Sodium Na1s2 ¡ Na+1aq2 + e-

Magnesium Mg1s2 ¡ Mg2+1aq2 + 2e-

Aluminum Al1s2 ¡ Al3+1aq2 + 3e-

Manganese Mn1s2 ¡ Mn2+1aq2 + 2e-

Zinc Zn1s2 ¡ Zn2+1aq2 + 2e-

Chromium Cr1s2 ¡ Cr3+1aq2 + 3e-

Iron Fe1s2 ¡ Fe2+1aq2 + 2e-

Cobalt Co1s2 ¡ Co2+1aq2 + 2e-

Nickel Ni1s2 ¡ Ni2+1aq2 + 2e-

Tin Sn1s2 ¡ Sn2+1aq2 + 2e-

Lead Pb1s2 ¡ Pb2+1aq2 + 2e-

Hydrogen H21g2 ¡ 2 H+1aq2 + 2e-

Copper Cu1s2 ¡ Cu2+1aq2 + 2e-

Silver Ag1s2 ¡ Ag+1aq2 + e-

Mercury Hg1l2 ¡ Hg2+1aq2 + 2e-

Platinum Pt1s2 ¡ Pt2+1aq2 + 2e-

Gold Au1s2 ¡ Au3+1aq2 + 3e-


e!

e!

Cu(s) is oxidized(loses electrons)

Ag!(aq) is reduced(gains electrons)

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

NO3"Ag! Ag

NO3" Cu2!

Cu

++

! FIGURE 4.14 Reaction of coppermetal with silver ion. When copper metal isplaced in a solution of silver nitrate, a redoxreaction forms silver metal and a bluesolution of copper(II) nitrate.

SAMPLE EXERCISE 4.10 Determining When an Oxidation-ReductionReaction Can Occur

Will an aqueous solution of iron(II) chloride oxidize magnesium metal? If so, write thebalanced molecular and net ionic equations for the reaction.

SOLUTIONAnalyze We are given two substances—an aqueous salt, FeCl2, and a metal, Mg—and askedif they react with each other.

Plan A reaction occurs if the reactant that is a metal in its elemental form (Mg) is locatedabove the reactant that is a metal in its oxidized form (Fe2+) in Table 4.5. If the reaction occurs,the ion in FeCl2 is reduced to Fe, and the Mg is oxidized to .

Solve Because Mg is above Fe in the table, the reaction occurs. To write the formula for thesalt produced in the reaction, we must remember the charges on common ions. Magnesium isalways present in compounds as ; the chloride ion is . The magnesium salt formed inthe reaction is MgCl2, meaning the balanced molecular equation is

Both FeCl2 and MgCl2 are soluble strong electrolytes and can be written in ionic form, whichshows us that 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.

Check Note that the net ionic equation is balanced with respect to both charge and mass.

PRACTICE EXERCISEWhich of the following metals will be oxidized by Pb(NO3)2: Zn, Cu, Fe?

Answer: Zn and Fe

Fe2+

Mg1s2 + Fe2+1aq2 ¡ Mg2+1aq2 + Fe1s2Cl-

Mg1s2 + FeCl21aq2 ¡ MgCl21aq2 + Fe1s2Cl-Mg2+

Mg2+Fe2+


• Are they acids and bases?

• If the reactants are electrolytes, will metathesis produce aprecipitate? Water? A gas?

• If metathesis cannot occur, can the reactants engage in anoxidation-reduction reaction? This requires that there be both areactant that can be oxidized and a reactant that can be reduced.

By asking questions such as these, you should be able to predictwhat happens during the reaction. You might not always be entirelycorrect, but if you keep your wits about you, you will not be far off.As you gain experience, you will begin to look for reactants thatmight not be immediately obvious, such as water from the solutionor oxygen from the atmosphere.

One of the greatest tools available to chemists is experimenta-tion. If you perform an experiment in which two solutions are mixed,you can make observations that help you understand what is hap-pening. For example, using Table 4.1 to predict whether a precipitatewill form is not nearly as exciting as seeing the precipitate form, as inFigure 4.4. Careful observations in the laboratory portion of thecourse will make your lecture material easier to master.

ANALYZING CHEMICAL REACTIONS

In this chapter you have been introduced to a greatnumber of chemical reactions. A major difficulty stu-

dents face in trying to master material of this sort isgaining a “feel” for what happens when chemicals

react. In fact, you might marvel at the ease withwhich your professor or teaching assistant can figure out the resultsof a chemical reaction. One of our goals in this textbook is to helpyou become more adept at predicting the outcomes of reactions.The key to gaining this “chemical intuition” is understanding how tocategorize reactions.

Attempting to memorize individual reactions would be a futiletask. It is far more fruitful to recognize patterns to determine the gen-eral category of a reaction, such as metathesis or oxidation-reduction.Thus, when you are faced with the challenge of predicting the out-come of a chemical reaction, ask yourself the following questions:

• What are the reactants?

• Are they electrolytes or nonelectrolytes?

STRATEGIES IN CHEMISTRY

Many of the early studies on gold arose from alchemy, in whichpeople attempted to turn cheap metals, such as lead, into gold. Al-chemists discovered that gold can be dissolved in a 3:1 mixture ofconcentrated hydrochloric and nitric acids, known as aqua regia(“royal water”). The action of the nitric acid on gold is similar to thaton copper (Equation 4.31) in that the nitrate ion, rather than , ox-idizes the metal to . The ions interact with to formhighly stable ions. The net ionic equation is

All the gold ever mined would fit in a cube 21 m on a side andweighing about 1.6 ! 108 kg. More than 90% of this amount has beenproduced since the 1848 California gold rush. Annual worldwideproduction of gold is about 2.4 ! 106 kg. By contrast 16,000 timesmore aluminum, over 3.97 ! 1010 kg, are produced annually.

Roughly three-quarters of gold production goes to make jew-elry, where it is often alloyed with other metals. Approximately 12%of gold production is used to meet a variety of industrial applica-tions, most notably in electronic devices where its excellent conduc-tivity and corrosion resistance make it a valuable component.A typical touch-tone telephone contains 33 gold-plated contacts.Gold is also used in computers and other microelectronic deviceswhere fine gold wire is used to link components.

Because of its resistance to corrosion, gold is an ideal metal fordental crowns and caps, which accounts for about of the annualuse of the element. The pure metal is too soft to use in dentistry, so itis combined with other metals to form alloys.

RELATED EXERCISE: 4.91

3%

AuCl4-1aq2 + 2 H2O1l2 + NO1g2Au1s2 + NO3

-1aq2 + 4 H+1aq2 + 4 Cl-1aq2 ¡

AuCl4-

Au3+Cl-Au3+H+

THE AURA OF GOLD

Throughout history, people have cherished gold, foughtfor it, and even died for it.

The physical and chemical properties of goldmake it special. First, its intrinsic beauty and rar-

ity make it precious. Second, it is soft and can beeasily formed into jewelry, coins, and other objects. Third, it is one ofthe least active metals (Table 4.5). It is not oxidized in air and doesnot react with water (! FIGURE 4.15), with basic solutions, or withmost acidic solutions.

A CLOSER LOOK

" FIGURE 4.15 The chemical inertnessof gold. Two, stamped, weighed, andpriced (in the 1800's) gold bars recentlyrecovered from a steamship that sank offthe coast of Central America in 1857.

SECTION 4.5 Concentrations of Solutions 139

4.5 | CONCENTRATIONS OF SOLUTIONSScientists use the term concentration to designate the amount of solute dissolved in agiven quantity of solvent or quantity of solution. The greater the amount of solutedissolved in a certain amount of solvent, the more concentrated the resulting solution.In chemistry we often need to express the concentrations of solutions quantitatively.

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

[4.32]

A 1.00 molar solution (written 1.00 M) contains 1.00 mol of solute in every liter of solu-tion. ! FIGURE 4.16 shows the preparation of 250.0 mL of a 1.00 M solution of CuSO4.The molarity of the solution is .

G I V E I T S O M E T H O U G H TWhich is more concentrated, a solution prepared by dissolving 21.0 g of NaF(0.500 mol) in enough water to make 500 mL of solution or a solution preparedby dissolving 10.5 g (0.250 mol) of NaF in enough water to make 100 mL ofsolution?

10.250 mol CuSO42>10.250 L soln2 = 1.00 M

Molarity = moles solutevolume of solution in liters

Weigh out 39.9 g (0.250 mol) CuSO4

1 Put CuSO4 (solute) into 250-mL volumetric flask; add water and swirl to dissolve solute

2 Add water until solution just reaches calibration mark on neck of flask

3 " FIGURE 4.16 Preparing0.250 L of a 1.00 M solution ofCuSO4.

SAMPLE EXERCISE 4.11 Calculating Molarity

Calculate the molarity of a solution made by dissolving 23.4 g of sodium sulfate (Na2SO4) in enough water to form 125 mL of solution.

SOLUTIONAnalyze We are given the number of grams of solute (23.4 g), itschemical formula (Na2SO4), and the volume of the solution (125mL) and asked to calculate the molarity of the solution.

Plan We can calculate molarity using Equation 4.32. To do so, wemust convert the number of grams of solute to moles and the vol-ume of the solution from milliliters to liters.


Expressing the Concentration of an ElectrolyteWhen an ionic compound dissolves, the relative concentrations of the ions in the solu-tion depend on the chemical formula of the compound. For example, a 1.0 M solutionof NaCl is 1.0 M in ions and 1.0 M in ions, and a 1.0 M solution of Na2SO4 is2.0 M in ions and 1.0 M in ions. Thus, the concentration of an electrolytesolution can be specified either in terms of the compound used to make the solution(1.0 M Na2SO4) or in terms of the ions in the solution (2.0 M and 1.0 M ).SO4

2-Na+

SO42-Na+

Cl-Na+

Solve The number of moles of Na2SO4 is

obtained by using its molar mass: Moles Na2SO4 = 123.4 g Na2SO42¢ 1 mol Na2SO4

142 g Na2SO4! = 0.165 mol Na2SO4

Converting the volume of the solution to liters: Liters soln = 1125 mL2¢ 1 L1000 mL

! = 0.125 L

Thus, the molarity is Molarity =0.165 mol Na2SO4

0.125 L soln= 1.32

mol Na2SO4

L soln= 1.32 M

PRACTICE EXERCISECalculate the molarity of a solution made by dissolving 5.00 g of glucose (C6H12O6) in sufficient water to form exactly 100 mL of solution.

Answer: 0.278 M

SAMPLE EXERCISE 4.12 Calculating Molar Concentrations of Ions

What is the molar concentration of each ion present in a 0.025 M aqueous solution of calciumnitrate?

SOLUTION:Analyze We are given the concentration of the ionic compound used to make the solutionand 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 determine therelative ion concentrations.

Solve Calcium nitrate is composed of calcium ions and nitrate ions , so itschemical formula is Ca(NO3)2. Because there are two ions for each ion, each moleof Ca(NO3)2 that dissolves dissociates into 1 mol of and 2 mol of . Thus, a solution that is 0.025 M in Ca(NO3)2 is 0.025 M in and in :

Check The concentration of ions is twice that of ions, as the subscript 2 after thein the chemical formula Ca(NO3)2 suggests it should be.

PRACTICE EXERCISEWhat is the molar concentration of ions in a 0.015 M solution of potassium carbonate?

Answer:

Interconverting Molarity, Moles, and VolumeIf we know any two of the three quantities in Equation 4.32, we can calculate the third.For example, if we know the molarity of an HNO3 solution to be 0.200 M, which means0.200 mol of HNO3 per liter of solution, we can calculate the number of moles of solutein a given volume, say 2.0 L. Molarity therefore is a conversion factor between volume ofsolution and moles of solute:

Moles HNO3 = 12.0 L soln2¢ 0.200 mol HNO3

1 L soln! = 0.40 mol HNO3

0.030 M

K+

NO3-

Ca2+NO3-

mol NO3-

L= ¢ 0.025 mol Ca1NO322

L! ¢ 2 mol NO3

-

1 mol Ca1NO322 ! = 0.050 M

NO3-2 * 0.025 M = 0.050 MCa2+

NO3-Ca2+

Ca2+NO3-

1NO3-21Ca2+2

Check Because the numerator is only slightly larger than thedenominator, it is reasonable for the answer to be a little over 1 M.The units are appropriate for molarity, and three signifi-

cant figures are appropriate for the answer because each of theinitial pieces of data had three significant figures.1mol>L2


To illustrate the conversion of moles to volume, let’s calculate the volume of 0.30 MHNO3 solution required to supply 2.0 mol of HNO3:

In this case we must use the reciprocal of molarity in the conversion:

SAMPLE EXERCISE 4.13 Using Molarity to Calculate Grams of Solute

How many grams of Na2SO4 are required to make 0.350 L of 0.500 M Na2SO4?

SOLUTIONAnalyze We are given the volume of the solution (0.350 L), its concentration (0.500 M), andthe identity of the solute Na2SO4 and asked to calculate the number of grams of the solute inthe solution.

Plan We can use the definition of molarity (Equation 4.32) to determine the number ofmoles of solute, and then convert moles to grams using the molar mass of the solute.

Solve Calculating the moles of Na2SO4 using the molarity and volume of solution gives

Because each mole of Na2SO4 has a mass of 142 g, the required number of grams of Na2SO4 is

Check The magnitude of the answer, the units, and the number of significant figures are allappropriate.

PRACTICE EXERCISE(a) How many grams of Na2SO4 are there in 15 mL of 0.50 M Na2SO4? (b) How many milli-liters of 0.50 M Na2SO4 solution are needed to provide 0.038 mol of this salt?Answers: (a) 1.1 g, (b) 76 mL

DilutionSolutions used routinely in the laboratory are often purchased or prepared in concen-trated form (called stock solutions). Solutions of lower concentrations can then beobtained by adding water, a process called dilution.*

Let’s see how we can prepare a dilute solution from a concentrated one. Suppose wewant to prepare 250.0 mL (that is, 0.2500 L) of 0.100 M CuSO4 solution by diluting a1.00 M CuSO4 stock solution. The main point to remember is that when solvent isadded to a solution, the number of moles of solute remains unchanged:

[4.33]Moles solute before dilution = moles solute after dilution

Grams Na2SO4 = 10.175 mol Na2SO42¢ 142 g Na2SO4

1 mol Na2SO4! = 24.9 g Na2SO4

= 0.175 mol Na2SO4

= 10.350 L soln2¢ 0.500 mol Na2SO4

1 L soln! Moles Na2SO4 = liters soln * MNa2SO4

MNa2SO4=

moles Na2SO4

liters soln

MNa2SO4=

moles Na2SO4

liters soln

Liters = moles * 1>M = moles * liters>mole.

Liters soln = 12.0 mol HNO32¢ 1 L soln0.30 mol HNO3

! = 6.7 L soln

*In diluting a concentrated acid or base, the acid or base should be added to water and then further diluted byadding more water. Adding water directly to concentrated acid or base can cause spattering because of theintense heat generated.


Because we know both the volume (250 mL) and the concentration (0.100 mol>L) of thedilute solution, we can calculate the number of moles of CuSO4 it contains:

The volume of stock solution needed to provide 0.0250 mol CuSO4 is therefore:

! FIGURE 4.17 shows the dilution carried out in the laboratory. Notice that thediluted solution is less intensely colored than the concentrated one.

G I V E I T S O M E T H O U G H THow is the molarity of a 0.50 M KBr solution changed when water is added todouble its volume?

In laboratory situations, calculations of this sort are often made with an equationderived by remembering that the number of moles of solute is the same in both the con-centrated and dilute solutions and that :

[4.34]

Although we derived Equation 4.34 in terms of liters, any volume unit can be usedas long as it is used on both sides of the equation. For example, in the calculation we didfor the CuSO4 solution, we have

Solving for Vconc gives as before.Vconc = 25.0 mL

11.00 M21Vconc2 = 10.100 M21250. mL2Mconc * Vconc = Mdil * Vdil

Moles solute in conc soln = moles solute in dilute soln

moles = molarity * liters

Liters of conc soln = 10.0250 mol CuSO42¢ 1 L soln1.00 mol CuSO4

! = 0.0250 L

= 0.0250 mol CuSO4

Moles CuSO4 in dilute soln = 10.2500 L soln2¢ 0.100 mol CuSO4

L soln!

Draw 25.0 mL of 1.00 M stock solution into pipet

1 Add concentrated solution in pipet to 250–mL volumetric flask

2 Dilute with water until solution reaches calibration mark on neck of flask and mix to create 0.100 Msolution

3

" FIGURE 4.17 Preparing 250 mL of 0.100 M CuSO4 by dilution of 1.00 M CuSO4.


marathoner named Hillary Bellamy, running in the Marine Corpsmarathon in 2003, collapsed near mile 22 and died the next day. Onephysician who treated her said that she died from hyponatremia-induced brain swelling, the result of drinking too much water beforeand during the race.

The normal blood sodium level is 135 to 145 mM (millimolar).When that level drops to 125 mM, dizziness and confusion set in.A concentration below 120 mM can be critical. Dangerously low lev-els can occur in any active athlete who is sweating out salt (NaCl) atthe same time that excessive amounts of NaCl-free water are beingdrunk to compensate for water loss. The condition affects womenmore than men because of differences in body composition and pat-terns of metabolism. Drinking a sport drink that contains some elec-trolytes helps to prevent hyponatremia.

RELATED EXERCISES: 4.63, 4.64

DRINKING TOO MUCH WATERCAN KILL YOU

For a long time dehydration was considered a poten-tial danger for people engaged in extended vigorous

activity. Thus, athletes were encouraged to drinklots of water while engaged in active sport. The

trend toward extensive hydration has spread throughout society, sothat today many people carry water bottles everywhere and dutifullykeep well hydrated.

In some circumstances, however, drinking too much water is agreater danger than not drinking enough. Excess water consumptioncan lead to hyponatremia, a condition in which the concentration ofsodium ion in the blood is too low. In the past decade at least fourmarathon runners have died from hyponatremia-related trauma,and dozens more have become seriously ill. For example, a first-time

CHEMISTRY AND LIFE

SAMPLE EXERCISE 4.14 Preparing a Solution by Dilution

How many milliliters of 3.0 M H2SO4 are needed to make 450 mL of 0.10 M H2SO4?

SOLUTIONAnalyze We need to dilute a concentrated solution. We are given the molarity of a moreconcentrated solution (3.0 M) and the volume and molarity of a more dilute one containingthe same solute (450 mL of 0.10 M solution). We must calculate the volume of the concen-trated solution needed to prepare the dilute solution.

Plan We can calculate the number of moles of solute, H2SO4, in the dilute solution and thencalculate the volume of the concentrated solution needed to supply this amount of solute.Alternatively, we can directly apply Equation 4.34. Let’s compare the two methods.

Solve Calculating the moles of H2SO4 in the dilute solution:

Calculating the volume of the concentrated solution that contains 0.045 mol H2SO4:

Converting liters to milliliters gives 15 mL.If we apply Equation 4.34, we get the same result:

Either way, we see that if we start with 15 mL of 3.0 M H2SO4 and dilute it to a total volume of450 mL, the desired 0.10 M solution will be obtained.

Check The calculated volume seems reasonable because a small volume of concentratedsolution is used to prepare a large volume of dilute solution.

Comment The first approach can also be used to find the final concentration when twosolutions of different concentrations are mixed, whereas the second approach, using Equation4.34, can be used only for diluting a concentrated solution with pure solvent.

PRACTICE EXERCISE(a) What volume of 2.50 M lead(II) nitrate solution contains 0.0500 mol of ? (b) Howmany milliliters of 5.0 M K2Cr2O7 solution must be diluted to prepare 250 mL of 0.10 M solu-tion? (c) If 10.0 mL of a 10.0 M stock solution of NaOH is diluted to 250 mL, what is the con-centration of the resulting stock solution?

Answers: (a) , (b) 5.0 mL, (c) 0.40 M0.0200 L = 20.0 mL

Pb2+

1Vconc2 =10.10 M21450 mL2

3.0 M= 15 mL

13.0 M21Vconc2 = 10.10 M21450 mL2L conc soln = 10.045 mol H2SO42¢ 1 L soln

3.0 mol H2SO4! = 0.015 L soln

= 0.045 mol H2SO4

Moles H2SO4 in dilute solution = 10.450 L soln2¢ 0.10 mol H2SO4

1 L soln!


4.6 | SOLUTION STOICHIOMETRY ANDCHEMICAL ANALYSIS

In Chapter 3 we learned that given the chemical equation for a reaction and the amount ofone reactant consumed in the reaction, you can calculate the quantities of other reactantsand products. In this section we extend this concept to reactions involving solutions.

Recall that the coefficients in a balanced equation give the relative number of molesof reactants and products. •(Section 3.6) To use this information, we must convertthe masses of substances involved in a reaction into moles. When dealing with pure sub-stances, as we were in Chapter 3, we use molar mass to convert between grams andmoles of the substances. This conversion is not valid when working with a solution be-cause both solute and solvent contribute to its mass. However, if we know the soluteconcentration, we can use molarity and volume to determine the number of moles(moles ). ! FIGURE 4.18 summarizes this approach to using stoi-chiometry for the reaction between a pure substance and a solution.

SAMPLE EXERCISE 4.15 Using Mass Relations in a Neutralization Reaction

How many grams of Ca(OH)2 are needed to neutralize 25.0 mL of 0.100 M HNO3?

SOLUTIONAnalyze The reactants are an acid, HNO3, and a base, Ca(OH)2. The volume and molarity ofHNO3 are given, and we are asked how many grams of Ca(OH)2 are needed to neutralize thisquantity of HNO3.

Plan Following the steps outlined by the green arrows in Figure 4.18, we use the molarity andvolume of the HNO3 solution (substance B in Figure 4.18) to calculate the number of moles ofHNO3. We then use the balanced equation to relate moles of HNO3 to moles of Ca(OH)2(substance A). Finally, we use the molar mass to convert moles to grams of Ca(OH)2:

Solve The product of the molar concentration of a solution and its volume in liters gives thenumber of moles of solute:

= 2.50 * 10-3 mol HNO3

Moles HNO3 = VHNO3* MHNO3

= 10.0250 L2¢ 0.100 mol HNO3

L!VHNO3

* MHNO3Q mol HNO3Q mol Ca1OH22Q g Ca1OH22

solute = M * V

Molarity ofsolution containing

substance B

Volume ofsolution containing

substance B

Use molarmass of A

Use coefficients frombalanced equation

Use volumeof solution

containing B

Use molarityof solution

containing B

Moles ofsubstance A

Moles ofsubstance B

Grams ofsubstance A" FIGURE 4.18 Procedure

for solving stoichiometryproblems involving reactionsbetween a pure substanceA and a solution containing aknown concentration ofsubstance B. Starting from aknown mass of substance A,we follow the red arrows todetermine either the volume ofthe solution containing B (if themolarity of B is known) or themolarity of the solutioncontaining B (if the volume of Bis known). Starting from eithera known volume or knownmolarity of the solutioncontaining B, we follow thegreen arrows to determinethe mass of substance A.

SECTION 4.6 Solution Stoichiometry and Chemical Analysis 145

Because this is a neutralization reaction, HNO3 and Ca(OH)2 react to form H2O and the saltcontaining and :

Thus, . Therefore,

Check The answer is reasonable because a small volume of dilute acid requires only a smallamount of base to neutralize it.

PRACTICE EXERCISE(a) How many grams of NaOH are needed to neutralize 20.0 mL of 0.150 M H2SO4 solution?(b) How many liters of 0.500 M HCl(aq) are needed to react completely with 0.100 mol ofPb(NO3)2(aq), forming a precipitate of PbCl2(s)?

Answers: (a) 0.240 g, (b) 0.400 L

TitrationsTo determine the concentration of a particular solute in a solution, chemists often carryout a titration, which involves combining a solution where the solute concentration isnot known with a reagent solution of known concentration, called a standard solution.Just enough standard solution is added to completely react with the solute in thesolution of unknown concentration. The point at which stoichiometrically equivalentquantities are brought together is known as the equivalence point.

Titrations can be conducted using neutralization, precipitation, or oxidation-reduction reactions. ! FIGURE 4.19 illustrates a typical neutralization titration, one be-tween a HCl solution of unknown concentration and a NaOH solution we know to have aconcentration of 0.100 M (our standard solution). To determine the HCl concentration,we first add a specific volume of the HCl solution, 20.0 mL in this example, to a flask. Next

= 0.0926 g Ca1OH22 Grams Ca(OH)2 = 12.50 * 10-3 mol HNO32¢ 1 mol Ca(OH)2

2 mol HNO3! ¢ 74.1 g Ca(OH)2

1 mol Ca(OH)2!2 mol HNO3 ! mol Ca1OH222 HNO31aq2 + Ca1OH221s2 ¡ 2 H2O1l2 + Ca1NO3221aq2NO3

-Ca2+

Final volumereading

Solution becomes basic on passing equivalence point, triggering indicator color changeBuretInitial volume

reading

20.0 mL of acid solution added to flask

1 A few drops of acid–base indicator added

2 Standard NaOH solution added from buret

3 4

" FIGURE 4.19 Procedure for titrating an acid against a standard solution of NaOH.The acid–base indicator, phenolphthalein, is colorless in acidic solution but takes on a pink color in basic solution.

G O F I G U R EHow would the volume of standard solution added change if that solution were Ba(OH)2(aq) instead of NaOH(aq)?


a few drops of an acid–base indicator are added. The acid–base indicator is a dye thatchanges color on passing the equivalence point.* For example, the dye phenolphthalein iscolorless in acidic solution but pink in basic solution. The standard solution is then slowlyadded until the solution turns pink, telling us that the neutralization reaction betweenHCl and NaOH is complete. The standard solution is added from a buret so that we canaccurately determine the added volume of NaOH solution. Knowing the volumes of bothsolutions and the concentration of the standard solution we can calculate the concentra-tion of the unknown solution as diagrammed in ! FIGURE 4.20.

Volumestandard solutionneeded to reach

equivalence point

Concentration (molarity)of unknown solution

Use molarity ofstandard solution

Use coefficientsfrom balanced

equation

Use volume ofunknown solution

Moles solutein unknown

solution

Moles solutein standard

solution

" FIGURE 4.20 Procedure fordetermining the concentration of asolution from titration with a standardsolution.

SAMPLE EXERCISE 4.16 Determining Solution Concentration by anAcid–Base Titration

One commercial method used to peel potatoes is to soak them in a NaOH solution for a shorttime, then remove them and spray off the peel. The NaOH concentration is normally 3 to 6 M,and the solution must be analyzed periodically. In one such analysis, 45.7 mL of 0.500 MH2SO4 is required to neutralize 20.0 mL of NaOH solution. What is the concentration of theNaOH solution?

SOLUTIONAnalyze We are given the volume (45.7 mL) and molarity (0.500 M) of an H2SO4 solution(the standard solution) that reacts completely with 20.0 mL of NaOH solution. We are asked tocalculate the molarity of the NaOH solution.

Plan Following the steps of Figure 4.20, we use the H2SO4 volume and molarity to calculatethe number of moles of H2SO4. Then we can use this quantity and the balanced equation forthe reaction to calculate moles of NaOH. Finally, we can use moles of NaOH and the NaOHvolume to calculate NaOH molarity.

Solve The number of moles of H2SO4 is the product of the volume and molarity of thissolution:

Acids react with metal hydroxides to form water and a salt. Thus, the balanced equation for theneutralization reaction is

According to the balanced equation, . Therefore,

= 4.56 * 10-2 mol NaOH

Moles NaOH = 12.28 * 10-2 mol H2SO42¢ 2 mol NaOH1 mol H2SO4

!1 mol H2SO4 ! 2 mol NaOH

H2SO41aq2 + 2 NaOH1aq2 ¡ 2 H2O1l2 + Na2SO41aq2= 2.28 * 10-2 mol H2SO4

Moles H2SO4 = 145.7 mL soln2¢ 1 L soln1000 mL soln

! ¢ 0.500 mol H2SO4

L soln!

*More precisely, the color change of an indicator signals the end point of the titration, which if the proper indicatoris chosen lies very near the equivalence point. Acid–base titrations are discussed in more detail in Section 17.3.

SECTION 4.6 Solution Stoichiometry and Chemical Analysis 147

Knowing the number of moles of NaOH in 20.0 mL of solution allows us to calculate themolarity of this solution:

PRACTICE EXERCISEWhat is the molarity of an NaOH solution if 48.0 mL neutralizes 35.0 mL of 0.144 M H2SO4?

Answer: 0.210 M

= 2.28mol NaOH

L soln= 2.28 M

Molarity NaOH = mol NaOHL soln

= ¢ 4.56 * 10-2 mol NaOH20.0 mL soln

! ¢ 1000 mL soln1 L soln

!

SAMPLE EXERCISE 4.17 Determining the Quantity of Solute by Titration

The quantity of in a municipal water supply is determined by titrating the sample with. The precipitation reaction taking place during the titration is

The end point in this type of titration is marked by a change in color of a special type of indi-cator. (a) How many grams of chloride ion are in a sample of the water if 20.2 mL of 0.100 M

is needed to react with all the chloride in the sample? (b) If the sample has a mass of10.0 g, what percent does it contain?

SOLUTIONAnalyze We are given the volume (20.2 mL) and molarity (0.100 M) of a solution of andthe chemical equation for reaction of this ion with . We are asked to calculate the numberof grams of in the sample and the mass percent of in the sample.

(a) Plan We can use the procedure outlined by the green arrows in Figure 4.18. We begin byusing the volume and molarity of to calculate the number of moles of used in thetitration. We then use the balanced equation to determine the moles of in the sample andfrom that the grams of .

Solve

From the balanced equation we see that . Using this information andthe molar mass of Cl, we have

(b) Plan To calculate the percentage of in the sample, we compare the number of gramsof 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. Ocean watercontains Whether water containing tastes salty depends on the other ionspresent. If the only accompanying ions are , a salty taste may be detected with as little as

.Cl-0.03%Na+

Cl-Cl-.1.92%

Percent Cl- =7.17 * 10-2 g

10.0 g* 100% = 0.717% Cl-

7.17 * 10-2 g,Cl-Cl-

= 7.17 * 10-2 g Cl-

Grams Cl- = 12.02 * 10-3 mol Ag+ 2¢ 1 mol Cl-

1 mol Ag+ ! ¢ 35.5 g Cl-

mol Cl-!1 mol Ag+ ! 1 mol Cl-

= 2.02 * 10-3 mol Ag+

Moles Ag+ = 120.2 mL soln 2¢ 1 L soln 1000 mL soln

! ¢ 0.100 mol Ag+

L soln !Cl-

Cl-Ag+Ag+

Cl-Cl-Cl-

Ag+

Cl-Ag+

Ag+1aq2 + Cl-1aq2 ¡ AgCl1s2Ag+Cl-


PRACTICE EXERCISEA sample of an iron ore is dissolved in acid, and the iron is converted to . The sample isthen titrated with 47.20 mL of 0.02240 M solution. The oxidation-reduction reactionthat occurs during titration is

(a) How many moles of were added to the solution? (b) How many moles of werein the sample? (c) How many grams of iron were in the sample? (d) If the sample had a mass of0.8890 g, what is the percentage of iron in the sample?

Answers: (a) , (b) , (c) 0.2952 g, (d)

SAMPLE INTEGRATIVE EXERCISE Putting Concepts Together

Note: Integrative exercises require skills from earlier chapters as well as ones from the presentchapter.

A sample of 70.5 mg of potassium phosphate is added to 15.0 mL of 0.050 M silvernitrate, resulting in the formation of a precipitate. (a) Write the molecular equation for thereaction. (b) What is the limiting reactant in the reaction? (c) Calculate the theoretical yield, ingrams, of the precipitate that forms.

SOLUTION(a) Potassium phosphate and silver nitrate are both ionic compounds. Potassium phosphatecontains and ions, so its chemical formula is K3PO4. Silver nitrate contains and

ions, so its chemical formula is AgNO3. Because both reactants are strong electrolytes,the solution contains , , , and ions before the reaction occurs. According tothe solubility guidelines in Table 4.1, and form an insoluble compound, so Ag3PO4will precipitate from the solution. In contrast, and will remain in solution becauseKNO3 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 of each reac-tant. •(Section 3.7) The number of moles of K3PO4 is calculated from the mass of the sam-ple using the molar mass as a conversion factor. •(Section 3.4) The molar mass of K3PO4 is

. Converting milligrams to grams and then tomoles, we have

We determine the number of moles of AgNO3 from the volume and molarity of the solution.(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 AgNO3 as there are moles of

K3PO4. According to the balanced equation, however, 1 mol K3PO4 requires 3 mol AgNO3.Thus, there is insufficient AgNO3 to consume the K3PO4, and AgNO3 is the limiting reactant.

(c) The precipitate is Ag3PO4, whose molar mass is To calculate the number of grams of Ag3PO4 that could be produced in this reaction (thetheoretical yield), we use the number of moles of the limiting reactant, converting mol

. We use the coefficients in the balanced equation toconvert moles of AgNO3 to moles Ag3PO4, and we use the molar mass of Ag3PO4 to convert thenumber of moles of this substance to grams.

The answer has only two significant figures because the quantity of AgNO3 is given to only twosignificant figures.

17.5 * 10-4 mol AgNO32¢ 1 mol Ag3PO4

3 mol AgNO3! ¢ 418.7 g Ag3PO4

1 mol Ag3PO4! = 0.10 g Ag3PO4

AgNO3Q mol Ag3PO4Q g Ag3PO4

31107.92 + 31.0 + 4116.02 = 418.7 g>mol.

17.5 * 10-42>13.32 * 10-42 = 2.3

(15.0 mL2¢ 10-3 L1 mL

! ¢ 0.050 mol AgNO3

L! = 7.5 * 10-4 mol AgNO3

(70.5 mg K3PO42¢ 10-3 g K3PO4

1 mg K3PO4! ¢ 1 mol K3PO4

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

3139.12 + 31.0 + 4116.02 = 212.3 g/mol

K3PO41aq2 + 3 AgNO31aq2 ¡ Ag3PO41s2 + 3 KNO31aq2NO3

-K+PO4

3-Ag+NO3

-Ag+PO43-K+

NO3-

Ag+PO43-K+

33.21%5.286 * 10-3 mol Fe2+1.057 * 10-3 mol MnO4-

Fe2+MnO4-

MnO4-1aq2 + 5 Fe2+1aq2 + 8 H+1aq2 ¡ Mn2+1aq2 + 5 Fe3+1aq2 + 4 H2O1l2

MnO4-

Fe2+

Key Equations 149

CHAPTER SUMMARY AND KEY TERMS

INTRODUCTION AND SECTION 4.1 Solutions in which water isthe dissolving medium are called aqueous solutions. The componentof the solution that is present in the greatest quantity is the solvent.The other components are solutes.

Any substance whose aqueous solution contains ions is called anelectrolyte. Any substance that forms a solution containing no ions isa nonelectrolyte. Electrolytes that are present in solution entirely asions are strong electrolytes, whereas those that are present partly asions and partly as molecules are weak electrolytes. Ionic compoundsdissociate into ions when they dissolve, and they are strong elec-trolytes. The solubility of ionic substances is made possible bysolvation, the interaction of ions with polar solvent molecules. Mostmolecular compounds are nonelectrolytes, although some are weakelectrolytes, and a few are strong electrolytes. When representing theionization of a weak electrolyte in solution, half-arrows in both direc-tions are used, indicating that the forward and reverse reactions canachieve a chemical balance called a chemical equilibrium.

SECTION 4.2 Precipitation reactions are those in which an insolu-ble product, called a precipitate, forms. Solubility guidelines helpdetermine whether or not an ionic compound will be soluble in water.(The solubility of a substance is the amount that dissolves in a givenquantity of solvent.) Reactions such as precipitation reactions, inwhich cations and anions appear to exchange partners, are calledexchange reactions, or metathesis reactions.

Chemical equations can be written to show whether dissolvedsubstances are present in solution predominantly as ions or molecules.When the complete chemical formulas of all reactants and productsare used, the equation is called a molecular equation. A completeionic equation shows all dissolved strong electrolytes as their compo-nent ions. In a net ionic equation, those ions that go through the reac-tion unchanged (spectator ions) are omitted.

SECTION 4.3 Acids and bases are important electrolytes. Acids areproton donors; they increase the concentration of in aqueoussolutions to which they are added. Bases are proton acceptors; theyincrease the concentration of in aqueous solutions. Thoseacids and bases that are strong electrolytes are called strong acids andstrong bases, respectively. Those that are weak electrolytes are weakacids and weak bases. When solutions of acids and bases are mixed,a neutralization reaction occurs. The neutralization reaction betweenan acid and a metal hydroxide produces water and a salt. Gases can

OH-1aq2 H+1aq2

also be formed as a result of neutralization reactions. The reaction of asulfide with an acid forms H2S(g); the reaction between a carbonateand an acid forms CO2(g).

SECTION 4.4 Oxidation is the loss of electrons by a substance,whereas reduction is the gain of electrons by a substance. Oxidationnumbers keep track of electrons during chemical reactions and areassigned to atoms using specific rules. The oxidation of an elementresults in an increase in its oxidation number, whereas reduction isaccompanied by a decrease in oxidation number. Oxidation is alwaysaccompanied by reduction, giving oxidation-reduction, or redox,reactions.

Many metals are oxidized by O2, acids, and salts. The redox reac-tions between metals and acids as well as those between metals andsalts are called displacement reactions. The products of these dis-placement reactions are always an element (H2 or a metal) and a salt.Comparing such reactions allows us to rank metals according to theirease of oxidation. A list of metals arranged in order of decreasing easeof oxidation is called an activity series. Any metal on the list can beoxidized by ions of metals (or ) below it in the series.

SECTION 4.5 The concentration of a solution expresses theamount of a solute dissolved in the solution. One of the common waysto express the concentration of a solute is in terms of molarity. Themolarity of a solution is the number of moles of solute per liter ofsolution. Molarity makes it possible to interconvert solution volumeand number of moles of solute. Solutions of known molarity can beformed either by weighing out the solute and diluting it to a knownvolume or by the dilution of a more concentrated solution of knownconcentration (a stock solution). Adding solvent to the solution(the process of dilution) decreases the concentration of the solutewithout changing the number of moles of solute in the solution

.

SECTION 4.6 In the process called titration, we combine a solutionof known concentration (a standard solution) with a solution ofunknown concentration to determine the unknown concentration orthe quantity of solute in the unknown. The point in the titration atwhich stoichiometrically equivalent quantities of reactants are broughttogether is called the equivalence point. An indicator can be used toshow the end point of the titration, which coincides closely with theequivalence point.

1Mconc * Vconc = Mdil * Vdil2

H+

KEY SKILLS• Recognize compounds as acids or bases, and as strong, weak, or nonelectrolytes. (Sections 4.1 and 4.3)• Recognize reactions by type and be able to predict the products of simple acid–base, precipitation, and redox reactions. (Sections 4.2–4.4)• Be able to calculate molarity and use it to convert between moles of a substance in solution and volume of the solution. (Section 4.5)• Understand how to carry out a dilution to achieve a desired solution concentration. (Section 4.5)• Understand how to perform and interpret the results of a titration. (Section 4.6)

KEY EQUATIONS

• [4.32] Molarity is the most commonly used unit ofconcentration in chemistry.

• [4.34] When adding solvent to a concentrated solution tomake a dilute solution, molarities and volumes ofboth concentrated and dilute solutions can becalculated if three of the quantities are known.

Mconc * Vconc = Mdil * Vdil

Molarity = moles solutevolume of solution in liters


GENERAL PROPERTIES OF AQUEOUS SOLUTIONS (section 4.1)

4.11 When asked what causes electrolyte solutions to conduct elec-tricity, a student responds that it is due to the movement ofelectrons through the solution. Is the student correct? If not,what is the correct response?

4.12 When methanol, CH3OH, is dissolved in water, a nonconduct-ing solution results. When acetic acid, CH3COOH, dissolves inwater, the solution is weakly conducting and acidic in nature.Describe what happens upon dissolution in the two cases, andaccount for the different results.

4.13 We have learned in this chapter that many ionic solids dissolvein water as strong electrolytes, that is, as separated ions in so-lution. What properties of water facilitate this process? Wouldyou expect ionic compounds to be soluble in elemental liquidslike bromine or mercury, just as they are in water? Explain.

4.14 What does it mean to say that ions are solvated when an ionicsubstance dissolves in water?

4.15 Specify what ions are present in solution upon dissolving eachof the following substances in water: (a) ZnCl2, (b) HNO3,(c) (NH4)2SO4, (d) Ca(OH)2.

4.16 Specify what ions are present upon dissolving each of the fol-lowing substances in water: (a) MgI2, (b) Al(NO3)3,(c) HClO4, (d) NaCH3COO.

4.17 Formic acid, HCOOH, is a weak electrolyte. What solute par-ticles are present in an aqueous solution of this compound?Write the chemical equation for the ionization of HCOOH.

4.18 Acetone, CH3COCH3, is a nonelectrolyte; hypochlorous acid,HClO, is a weak electrolyte; and ammonium chloride, NH4Cl,is a strong electrolyte. (a) What are the solute particles presentin aqueous solutions of each compound? (b) If 0.1 mol of eachcompound is dissolved in solution, which one contains0.2 mol of solute particles, which contains 0.1 mol of soluteparticles, and which contains somewhere between 0.1 and0.2 mol of solute particles?

4.1 Which of the following schematic drawings best describes asolution of Li2SO4 in water (water molecules not shown forsimplicity)? [Section 4.1]

4.2 Aqueous solutions of three different substances, AX, AY, and AZ,are represented by the three accompanying diagrams. Identifyeach substance as a strong electrolyte, weak electrolyte, or non-electrolyte. [Section 4.1]

4.3 Use the molecular representations shown here to classify eachcompound as either a nonelectrolyte, a weak electrolyte, or astrong electrolyte (see inside back cover for element colorscheme). [Sections 4.1 and 4.3]

(a) (b) (c)

!

!

!

!

!

"

"

""

"

AX AY AZ

(a) (b) (c)

(a) (b) (c)

2"2"

2"2"2"

2"

2"

2"

2"

2"

2"2"

!

!!

!

!!

!

!!

!

!!

!

!

!

4.4 A 0.1 M solution of acetic acid, CH3COOH, causes the light-bulb in the apparatus of Figure 4.2 to glow about as brightly asa 0.001 M solution of HBr. How do you account for this fact?[Section 4.1]

4.5 You are presented with a white solid and told that due to care-less labeling it is not clear if the substance is barium chloride,lead chloride, or zinc chloride. When you transfer the solid toa beaker and add water, the solid dissolves to give a clear solu-tion. Next a Na2SO4(aq) solution is added and a white precip-itate forms. What is the identity of the unknown white solid?[Section 4.2]

4.6 We have seen that ions in aqueous solution are stabilizedby the attractions between the ions and the water molecules.Why then do some pairs of ions in solution form precipitates?[Section 4.2]

4.7 Which of the following ions will always be a spectator ion in aprecipitation reaction? (a) , (b) , (c) , (d) ,(e) Explain briefly. [Section 4.2]

4.8 The labels have fallen off three bottles containing powderedsamples of metals; one contains zinc, one lead, and the otherplatinum. You have three solutions at your disposal: 1 Msodium nitrate, 1 M nitric acid, and 1 M nickel nitrate. Howcould you use these solutions to determine the identities ofeach metal powder? [Section 4.4]

4.9 Explain how a redox reaction involves electrons in the sameway that a neutralization reaction involves protons. [Sections4.3 and 4.4]

4.10 If you want to double the concentration of a solution, howcould you do it? [Section 4.5]

SO42-.

S2-NH4+NO3

-Cl-

AddNa2SO4(aq)

AddH2O

EXERCISESVISUALIZING CONCEPTS

Exercises 151

PRECIPITATION REACTIONS (section 4.2)

4.19 Using solubility guidelines, predict whether each of the fol-lowing compounds is soluble or insoluble in water: (a) MgBr2,(b) PbI2, (c) (NH4)2CO3, (d) Sr(OH)2, (e) ZnSO4.

4.20 Predict whether each of the following compounds is solublein water: (a) AgI, (b) Na2CO3, (c) BaCl2, (d) Al(OH)3,(e) Zn(CH3COO)2.

4.21 Will precipitation occur when the following solutions aremixed? If so, write a balanced chemical equation for the reac-tion. (a) Na2CO3 and AgNO3, (b) NaNO3 and NiSO4,(c) FeSO4 and Pb(NO3)2.

4.22 Identify the precipitate (if any) that forms when the followingsolutions are mixed, and write a balanced equation for eachreaction. (a) NaCH3COO and HCl, (b) KOH and Cu(NO3)2,(c) Na2S and CdSO4.

4.23 Name the spectator ions in any reactions that may be involvedwhen each of the following pairs of solutions are mixed.(a) Na2CO3(aq) and MgSO4(aq)(b) Pb(NO3)2(aq) and Na2S(aq)(c) (NH4)3PO4(aq) and CaCl2(aq)

4.24 Write balanced net ionic equations for the reactions that occurin each of the following cases. Identify the spectator ion orions in each reaction.

(a)(b)(c)

4.25 Separate samples of a solution of an unknown salt are treatedwith dilute solutions of HBr, H2SO4, and NaOH. A precipitateforms in all three cases. Which of the following cations couldthe solution contain: , , ?

4.26 Separate samples of a solution of an unknown ionic com-pound are treated with dilute AgNO3, Pb(NO3)2, and BaCl2.Precipitates form in all three cases. Which of the followingcould be the anion of the unknown salt: , , ?

4.27 You know that an unlabeled bottle contains a solution of oneof the following: AgNO3, CaCl2, or Al2(SO4)3. A friend sug-gests that you test a portion of the solution with Ba(NO3)2and then with NaCl solutions. Explain how these two teststogether would be sufficient to determine which salt is presentin the solution.

4.28 Three solutions are mixed together to form a single solution.One contains 0.2 mol Pb(CH3COO)2, the second contains0.1 mol Na2S, and the third contains 0.1 mol CaCl2. (a) Writethe net ionic equations for the precipitation reaction orreactions that occur. (b) What are the spectator ions in thesolution?

NO3-CO3

2-Br-

Ba2+Pb2+K+

Fe1NO3221aq2 + KOH1aq2 ¡Ba1NO3221aq2 + K2SO41aq2 ¡Cr21SO4231aq2 + 1NH422CO31aq2 ¡

ACIDS, BASES, AND NEUTRALIZATION REACTIONS (section 4.3)

4.29 Which of the following solutions has the largest concentrationof solvated protons: (a) 0.2 M LiOH, (b) 0.2 M HI, (c) 1.0 Mmethyl alcohol (CH3OH)? Explain

4.30 Which of the following solutions is the most basic? (a) 0.6 MNH3, (b) 0.150 M KOH, (c) 0.100 M Ba(OH)2. Explain.

4.31 What is the difference between (a) a monoprotic acid and adiprotic acid, (b) a weak acid and a strong acid, (c) an acid anda base?

4.32 Explain the following observations: (a) NH3 contains no ions, and yet its aqueous solutions are basic; (b) HF is called aweak acid, and yet it is very reactive; (c) although sulfuric acidis a strong electrolyte, an aqueous solution of H2SO4 containsmore ions than ions.

4.33 Is there any correlation between the anions that form wheneach of the strong acids in Table 4.2 dissociates and the anionsthat normally form soluble ionic compounds (Table 4.1)?Which anions if any are exceptions to the general trend?

4.34 What is the relationship between the solubility rules in Table4.1 and the list of strong bases in Table 4.2? Another way ofasking this question is, why is Cd(OH)2, for example, notlisted as a strong base in Table 4.2?

4.35 Label each of the following substances as an acid, base, salt, ornone of the above. Indicate whether the substance exists inaqueous solution entirely in molecular form, entirely as ions,or as a mixture of molecules and ions. (a) HF, (b) acetonitrile,CH3CN, (c) NaClO4, (d) Ba(OH)2.

4.36 An aqueous solution of an unknown solute is tested with litmuspaper and found to be acidic. The solution is weakly conductingcompared with a solution of NaCl of the same concentration.

SO42-HSO4

-

OH-

Which of the following substances could the unknown be:KOH, NH3, HNO3, KClO2, H3PO3, CH3COCH3 (acetone)?

4.37 Classify each of the following substances as a nonelectrolyte,weak electrolyte, or strong electrolyte in water: (a) H2SO3,(b) C2H5OH (ethanol), (c) NH3, (d) KClO3, (e) Cu(NO3)2.

4.38 Classify each of the following aqueous solutions as a nonelec-trolyte, weak electrolyte, or strong electrolyte: (a) LiClO4,(b) HClO, (c) CH3CH2CH2OH (propanol), (d) HClO3,(e) CuSO4, (f) C12H22O11 (sucrose).

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

4.40 Write the balanced molecular and net ionic equations for eachof the following neutralization reactions:(a) Aqueous acetic acid is neutralized by aqueous barium

hydroxide.(b) Solid chromium(III) hydroxide reacts with nitrous acid.(c) Aqueous nitric acid and aqueous ammonia react.

4.41 Write balanced molecular and net ionic equations for the fol-lowing reactions, and identify the gas formed in each: (a) solidcadmium sulfide reacts with an aqueous solution of sulfuricacid; (b) solid magnesium carbonate reacts with an aqueoussolution of perchloric acid.

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

FeO1s2 + 2 HClO41aq2 ¡ Fe1ClO4221aq2 + H2O1l2

Al1OH231s2 + HNO31aq2 ¡Cu1OH221s2 + HClO41aq2 ¡HBr1aq2 + Ca1OH221aq2 ¡


(b) Based on the equation in part (a), write the net ionicequation for the reaction that occurs between NiO(s) and anaqueous solution of nitric acid.

4.43 Magnesium carbonate, magnesium oxide, and magnesiumhydroxide are all white solids that react with acidic solutions.(a) Write a balanced molecular equation and a net ionicequation for the reaction that occurs when each substance re-acts with a hydrochloric acid solution. (b) By observing thereactions in part (a) could you distinguish any of the three

magnesium substances from the other two? If so how? (c) Ifexcess HCl(aq) is added, would the clear solutions left behindafter the reaction is complete contain the same or differentions in each case?

4.44 As K2O dissolves in water, the oxide ion reacts with water mol-ecules to form hydroxide ions. Write the molecular and netionic equations for this reaction. Based on the definitions ofacid and base, what ion is the base in this reaction? What is theacid? What is the spectator ion in the reaction?

OXIDATION-REDUCTION REACTIONS (section 4.4)

4.45 Define oxidation and reduction in terms of (a) electron trans-fer and (b) oxidation numbers.

4.46 Can oxidation occur without oxygen? Can oxidation occurwithout reduction?

4.47 Which region of the periodic table shown here contains themost readily oxidized elements? Which region contains theleast readily oxidized?

4.48 Determine the oxidation number of sulfur in each of the fol-lowing substances: (a) barium sulfate, BaSO4, (b) sulfurousacid, H2SO3, (c) strontium sulfide, SrS, (d) hydrogen sulfide,H2S. (e) Based on these compounds what is the range of oxi-dation numbers seen for sulfur? Is there any relationship be-tween the range of accessible oxidation states and sulfur’sposition on the periodic table?

4.49 Determine the oxidation number for the indicated element ineach of the following substances: (a) S in SO2, (b) C in COCl2,(c) Mn in KMnO4, (d) Br in HBrO, (e) As in As4, (f) O inK2O2.

4.50 Determine the oxidation number for the indicated elementin each of the following compounds: (a) Co in LiCoO2, (b) Alin NaAlH4, (c) C in CH3OH (methanol), (d) N in GaN, (e) Cl inHClO2, (f) Cr in BaCrO4.

4.51 Which element is oxidized and which is reduced in the follow-ing reactions?(a)(b)

(c)(d)

4.52 Which of the following are redox reactions? For those that are,indicate which element is oxidized and which is reduced. Forthose that are not, indicate whether they are precipitation orneutralization reactions.(a) P4(s) 10 HClO(aq) 6 H2O(l)

4 H3PO4(aq) 10 HCl(aq)(b) Br2(l) 2 K(s) 2 KBr(s)¡+

+¡++

PbS1s2 + 4 H2O21aq2 ¡ PbSO41s2 + 4 H2O1l2Cl21aq2 + 2 NaI1aq2 ¡ I21aq2 + 2 NaCl1aq23 Fe1s2 + 2 Al1NO3231aq23 Fe1NO3221aq2 + 2 Al1s2 ¡N21g2 + 3 H21g2 ¡ 2 NH31g2

A

B C

D

(c) CH3CH2OH(l) 3 O2(g) 3 H2O(l) 2 CO2(g)(d) ZnCl2(aq) 2 NaOH(aq) Zn(OH)2(s) 2 NaCl(aq)

4.53 Write balanced molecular and net ionic equations for the reac-tions of (a) manganese with dilute sulfuric acid, (b) chromiumwith hydrobromic acid, (c) tin with hydrochloric acid, (d) alu-minum with formic acid, HCOOH.

4.54 Write balanced molecular and net ionic equations for the re-actions of (a) hydrochloric acid with nickel, (b) dilute sulfuricacid with iron, (c) hydrobromic acid with magnesium,(d) acetic acid, CH3COOH, with zinc.

4.55 Using the activity series (Table 4.5), write balanced chemicalequations for the following reactions. If no reaction occurs,simply write NR. (a) Iron metal is added to a solution of cop-per(II) nitrate; (b) zinc metal is added to a solution of magne-sium sulfate; (c) hydrobromic acid is added to tin metal;(d) hydrogen gas is bubbled through an aqueous solution ofnickel(II) chloride; (e) aluminum metal is added to a solutionof cobalt(II) sulfate.

4.56 Using the activity series (Table 4.5), write balanced chemicalequations for the following reactions. If no reaction occurs,simply write NR. (a) Nickel metal is added to a solution ofcopper(II) nitrate; (b) a solution of zinc nitrate is added to asolution of magnesium sulfate; (c) hydrochloric acid is addedto gold metal; (d) chromium metal is immersed in an aqueoussolution of cobalt(II) chloride; (e) hydrogen gas is bubbledthrough a solution of silver nitrate.

4.57 The metal cadmium tends to form ions. The followingobservations are made: (i) When a strip of zinc metal is placedin CdCl2(aq), cadmium metal is deposited on the strip. (ii)When a strip of cadmium metal is placed in Ni(NO3)2(aq),nickel metal is deposited on the strip. (a) Write net ionic equa-tions to explain each of the preceding observations. (b) Whatcan you conclude about the position of cadmium in the activ-ity series? (c) What experiments would you need to perform tolocate more precisely the position of cadmium in the activityseries?

4.58 (a) Use the following reactions to prepare an activity series forthe halogens:

(b) Relate the positions of the halogens in the periodic tablewith their locations in this activity series. (c) Predict whether areaction occurs when the following reagents are mixed:Cl2(aq) and KI(aq); Br2(aq) and LiCl(aq).

Cl21aq2 + 2 NaBr1aq2 ¡ 2 NaCl1aq2 + Br21aq2Br21aq2 + 2 NaI1aq2 ¡ 2 NaBr1aq2 + I21aq2

Cd2+

+¡++¡+

Exercises 153

CONCENTRATIONS OF SOLUTIONS (section 4.5)

4.59 (a) Is the concentration of a solution an intensive or an exten-sive property? (b) What is the difference between 0.50 molHCl and 0.50 M HCl?

4.60 (a) Suppose you prepare 500 mL of a 0.10 M solution of somesalt and then spill some of it. What happens to the concentra-tion of the solution left in the container? (b) Suppose you pre-pare 500 mL of a 0.10 M aqueous solution of some salt and letit sit out, uncovered, for a long time, and some water evapo-rates. What happens to the concentration of the solution leftin the container? (c) A certain volume of a 0.50 M solutioncontains 4.5 g of a salt. What mass of the salt is present in thesame volume of a 2.50 M solution?

4.61 (a) Calculate the molarity of a solution that contains 0.175 molZnCl2 in exactly 150 mL of solution. (b) How many moles ofHCl are present in 35.0 mL of a 4.50 M solution of nitric acid?(c) How many milliliters of 6.00 M NaOH solution are neededto provide 0.325 mol of NaOH?

4.62 (a) Calculate the molarity of a solution made by dissolving12.5 grams of Na2CrO4 in enough water to form exactly550 mL of solution. (b) How many moles of KBr are present in150 mL of a 0.275 M solution? (c) How many milliliters of6.1 M HCl solution are needed to obtain 0.100 mol of HCl?

4.63 The average adult human male has a total blood volume of5.0 L. If the concentration of sodium ion in this average indi-vidual is 0.135 M, what is the mass of sodium ion circulatingin the blood?

4.64 A person suffering from hyponatremia has a sodium ion con-centration in the blood of 0.118 M and a total blood volume of4.6 L. What mass of sodium chloride would need to be addedto the blood to bring the sodium ion concentration up to0.138 M, assuming no change in blood volume?

4.65 The concentration of alcohol (CH3CH2OH) in blood, calledthe “blood alcohol concentration” or BAC, is given in units ofgrams of alcohol per 100 mL of blood. The legal definition ofintoxication, in many states of the United States, is that theBAC is 0.08 or higher. What is the concentration of alcohol, interms of molarity, in blood if the BAC is 0.08?

4.66 The average adult male has a total blood volume of 5.0 L. Afterdrinking a few beers, he has a BAC of 0.10 (see Exercise 4.65).What mass of alcohol is circulating in his blood?

4.67 Calculate (a) the number of grams of solute in 0.250 L of0.175 M KBr, (b) the molar concentration of a solution con-taining 14.75 g of Ca(NO3)2 in 1.375 L, (c) the volume of1.50 M Na3PO4 in milliliters that contains 2.50 g of solute.

4.68 (a) How many grams of solute are present in 15.0 mL of0.736 M K2Cr2O7? (b) If 14.00 g of (NH4)2SO4 is dissolved inenough water to form 250 mL of solution, what is the molarityof the solution? (c) How many milliliters of 0.0455 M CuSO4contain 3.65 g of solute?

4.69 (a) Which will have the highest concentration of potassiumion: 0.20 M KCl, 0.15 M K2CrO4, or 0.080 M K3PO4? (b)Which will contain the greater number of moles of potassiumion: 30.0 mL of 0.15 M K2CrO4 or 25.0 mL of 0.080 M K3PO4?

4.70 In each of the following pairs, indicate which has the higherconcentration of ion: (a) 0.10 M BaI2 or 0.25 M KI solution,(b) 100 mL of 0.10 M KI solution or 200 mL of 0.040 M ZnI2solution, (c) 3.2 M HI solution or a solution made by dissolv-ing 145 g of NaI in water to make 150 mL of solution.

4.71 Indicate the concentration of each ion or molecule present inthe following solutions: (a) 0.25 M NaNO3, (b)MgSO4, (c) 0.0150 M C6H12O6, (d) a mixture of 45.0 mL of0.272 M NaCl and 65.0 mL of 0.0247 M (NH4)2CO3. Assumethat the volumes are additive.

4.72 Indicate the concentration of each ion present in the solutionformed by mixing (a) 42.0 mL of 0.170 M NaOH and 37.6 mLof 0.400 M NaOH, (b) 44.0 mL of 0.100 M and Na2SO4 and25.0 mL of 0.150 M KCl, (c) 3.60 g KCl in 75.0 mL of 0.250 MCaCl2 solution. Assume that the volumes are additive.

4.73 (a) You have a stock solution of 14.8 M NH3. How many milli-liters of this solution should you dilute to make 1000.0 mL of0.250 M NH3? (b) If you take a 10.0-mL portion of the stocksolution and dilute it to a total volume of 0.500 L, what will bethe concentration of the final solution?

4.74 (a) How many milliliters of a stock solution of 6.0 M HNO3would you have to use to prepare 110 mL of 0.500 M HNO3?(b) If you dilute 10.0 mL of the stock solution to a final vol-ume of 0.250 L, what will be the concentration of the dilutedsolution?

4.75 (a) Starting with solid sucrose, C12H22O11, describe howyou would prepare 250 mL of a 0.250 M sucrose solution.(b) Describe how you would prepare 350.0 mL of 0.100 MC12H22O11 starting with 3.00 L of 1.50 M C12H22O11.

4.76 (a) How would you prepare 175.0 mL of 0.150 M AgNO3 solu-tion starting with pure AgNO3? (b) An experiment calls foryou to use 100 mL of 0.50 M HNO3 solution. All you haveavailable is a bottle of 3.6 M HNO3. How would you preparethe desired solution?

4.77 Pure acetic acid, known as glacial acetic acid, is a liquid witha density of at . Calculate the molarity of asolution of acetic acid made by dissolving 20.00 mL of glacialacetic acid at in enough water to make 250.0 mL ofsolution.

4.78 Glycerol, C3H8O3, is a substance used extensively in the manu-facture of cosmetics, foodstuffs, antifreeze, and plastics. Glyc-erol is a water-soluble liquid with a density of at

. Calculate the molarity of a solution of glycerol made bydissolving 50.000 mL glycerol at in enough water tomake 250.00 mL of solution.

15 °C15 °C

1.2656 g>mL

25 °C

25 °C1.049 g>mL

1.3 * 10-2 M

I-

SOLUTION STOICHIOMETRY AND CHEMICAL ANALYSIS (section 4.6)

4.79 What mass of KCl is needed to precipitate the silver ions from15.0 mL of 0.200 M AgNO3 solution?

4.80 What mass of NaOH is needed to precipitate the ionsfrom 35.0 mL of 0.500 M Cd(NO3)2 solution?

Cd2+

4.81 (a) What volume of 0.115 M HClO4 solution is needed to neu-tralize 50.00 mL of 0.0875 M NaOH? (b) What volume of0.128 M HCl is needed to neutralize 2.87 g of Mg(OH)2? (c) If25.8 mL of AgNO3 is needed to precipitate all the ions in aCl-


785-mg sample of KCl (forming AgCl), what is the molarity ofthe AgNO3 solution? (d) If 45.3 mL of 0.108 M HCl solution isneeded to neutralize a solution of KOH, how many grams ofKOH must be present in the solution?

4.82 (a) How many milliliters of 0.120 M HCl are needed to com-pletely neutralize 50.0 mL of 0.101 M Ba(OH)2 solution?(b) How many milliliters of 0.125 M H2SO4 are needed toneutralize 0.200 g of NaOH? (c) If 55.8 mL of BaCl2 solution isneeded to precipitate all the sulfate ion in a 752-mg sampleof Na2SO4, what is the molarity of the solution? (d) If 42.7 mLof 0.208 M HCl solution is needed to neutralize a solutionof Ca(OH)2, how many grams of Ca(OH)2 must be in thesolution?

4.83 Some sulfuric acid is spilled on a lab bench. You can neutralizethe acid by sprinkling sodium bicarbonate on it and thenmopping up the resultant solution. The sodium bicarbonatereacts with sulfuric acid as follows:

Sodium bicarbonate is added until the fizzing due to the for-mation of CO2(g) stops. If 27 mL of 6.0 M H2SO4 was spilled,what is the minimum mass of NaHCO3 that must be added tothe spill to neutralize the acid?

4.84 The distinctive odor of vinegar is due to acetic acid,CH3COOH, which reacts with sodium hydroxide in thefollowing fashion:

If 3.45 mL of vinegar needs 42.5 mL of 0.115 M NaOH toreach the equivalence point in a titration, how many grams ofacetic acid are in a 1.00-qt sample of this vinegar?

4.85 A 4.36-g sample of an unknown alkali metal hydroxide is dis-solved in 100.0 mL of water. An acid–base indicator is addedand the resulting solution is titrated with 2.50 M HCl(aq)

H2O1l2 + NaC2H3O21aq2CH3COOH1aq2 + NaOH1aq2 ¡

Na2SO41aq2 + 2 H2O1l2 + 2 CO21g22 NaHCO31s2 + H2SO41aq2 ¡

solution. The indicator changes color signaling that the equiv-alence point has been reached after 17.0 mL of the hydrochlo-ric acid solution has been added. (a) What is the molar mass ofthe metal hydroxide? (b) What is the identity of the alkalimetal cation: Li+, Na+, K+, Rb+, or Cs+?

4.86 An 8.65-g sample of an unknown group 2A metal hydroxide isdissolved in 85.0 mL of water. An acid–base indicator is addedand the resulting solution is titrated with 2.50 M HCl(aq) so-lution. The indicator changes color signaling that the equiva-lence point has been reached after 56.9 mL of the hydrochloricacid solution has been added. (a) What is the molar mass ofthe metal hydroxide? (b) What is the identity of the metalcation: Ca2+, Sr2+, Ba2+?

4.87 A solution of 100.0 mL of 0.200 M KOH is mixed with a solu-tion of 200.0 mL of 0.150 M NiSO4. (a) Write the balancedchemical equation for the reaction that occurs. (b) What pre-cipitate forms? (c) What is the limiting reactant? (d) Howmany grams of this precipitate form? (e) What is the concen-tration of each ion that remains in solution?

4.88 A solution is made by mixing 15.0 g of Sr(OH)2 and 55.0 mLof 0.200 M HNO3. (a) Write a balanced equation for the reac-tion that occurs between the solutes. (b) Calculate the concen-tration of each ion remaining in solution. (c) Is the resultantsolution acidic or basic?

4.89 A 0.5895-g sample of impure magnesium hydroxide is dis-solved in 100.0 mL of 0.2050 M HCl solution. The excess acidthen needs 19.85 mL of 0.1020 M NaOH for neutralization.Calculate the percent by mass of magnesium hydroxide in thesample, assuming that it is the only substance reacting withthe HCl solution.

4.90 A 1.248-g sample of limestone rock is pulverized and thentreated with 30.00 mL of 1.035 M HCl solution. The excessacid then requires 11.56 mL of 1.010 M NaOH for neutraliza-tion. Calculate the percent by mass of calcium carbonate inthe rock, assuming that it is the only substance reacting withthe HCl solution.

ADDITIONAL EXERCISES

4.91 Gold is one of the few metals that can be obtained by panning,where a simple pan is used to separate gold from other de-posits found in or near a stream bed. What two properties ofgold make it possible to find gold, but not metals like copper,silver, lead, and aluminum, by panning?

4.92 The accompanying photo shows the reaction between a solu-tion of Cd(NO3)2 and one of Na2S. What is the identity of theprecipitate? What ions remain in solution? Write the net ionicequation for the reaction.

Additional Exercises 155

Solution Solute Color of Solution

A Na2CrO4 YellowB (NH4)2C2O4 ColorlessC AgNO3 ColorlessD CaCl2 Colorless

4.93 Suppose you have a solution that might contain any or all ofthe following cations: , , , and . Addition ofHCl solution causes a precipitate to form. After filtering offthe precipitate, H2SO4 solution is added to the resulting solu-tion and another precipitate forms. This is filtered off, and asolution of NaOH is added to the resulting solution. No pre-cipitate is observed. Which ions are present in each of theprecipitates? Which of the four ions listed above must beabsent from the original solution?

4.94 You choose to investigate some of the solubility guidelines fortwo ions not listed in Table 4.1, the chromate ion ( )and the oxalate ion ( ). You are given 0.01 M solutions(A, B, C, D) of four water-soluble salts:

C2O42-

CrO42-

Mn2+Sr2+Ag+Ni2+4.97 Consider the following reagents: zinc, copper, mercury (den-

sity 13.6 g>mL), silver nitrate solution, nitric acid solution.(a) Given a 500-mL Erlenmeyer flask and a balloon can youcombine two or more of the foregoing reagents to initiate achemical reaction that will inflate the balloon? Write a bal-anced chemical equation to represent this process. What is theidentity of the substance that inflates the balloon? (b) What isthe theoretical yield of the substance that fills the balloon? (c) Can you combine two or more of the foregoing reagents toinitiate a chemical reaction that will produce metallic silver?Write a balanced chemical equation to represent this process.What ions are left behind in solution? (d) What is the theoret-ical yield of silver?

[4.98] Lanthanum metal forms cations with a charge of . Con-sider the following observations about the chemistry of lan-thanum: When lanthanum metal is exposed to air, a whitesolid (compound A) is formed that contains lanthanum andone other element. When lanthanum metal is added to water,gas bubbles are observed and a different white solid (com-pound B) is formed. Both A and B dissolve in hydrochloricacid to give a clear solution. When either of these solutions isevaporated, a soluble white solid (compound C) remains. Ifcompound C is dissolved in water and sulfuric acid is added, awhite precipitate (compound D) forms. (a) Propose identitiesfor the substances A, B, C, and D. (b) Write net ionic equationsfor all the reactions described. (c) Based on the precedingobservations, what can be said about the position of lan-thanum in the activity series (Table 4.5)?

4.99 A 35.0-mL sample of 1.00 M KBr and a 60.0-mL sample of0.600 M KBr are mixed. The solution is then heated to evapo-rate water until the total volume is 50.0 mL. What is themolarity of the KBr in the final solution?

4.100 Using modern analytical techniques, it is possible to detectsodium ions in concentrations as low as . What isthis detection limit expressed in (a) molarity of , (b)ions per cubic centimeter?

Na+Na+50 pg>mL

3+

35.0 g Zn 42.0 g Cu 6.55 mL Hg

150 mL of 0.750 M AgNO3(aq) 150 mL of 3.00 M HNO3(aq)

ExptNumber

SolutionsMixed Result

1 A + B No precipitate, yellow solution2 A + C Red precipitate forms3 A + D Yellow precipitate forms4 B + C White precipitate forms5 B + D White precipitate forms6 C + D White precipitate forms

When these solutions are mixed, the following observationsare made:

(a) Write a net ionic equation for the reaction that occurs ineach of the experiments. (b) Identify the precipitate formed, ifany, in each of the experiments.

4.95 Antacids are often used to relieve pain and promote healing inthe treatment of mild ulcers. Write balanced net ionic equa-tions for the reactions between the HCl(aq) in the stomachand each of the following substances used in various antacids:(a) Al(OH)3(s), (b) Mg(OH)2(s), (c) MgCO3(s), (d)NaAl(CO3)(OH)2(s), (e) CaCO3(s).

[4.96] The commercial production of nitric acid involves the follow-ing chemical reactions:

(a) Which of these reactions are redox reactions? (b) In eachredox reaction identify the element undergoing oxidation andthe element undergoing reduction.

3 NO21g2 + H2O1l2 ¡ 2 HNO31aq2 + NO1g2 2 NO1g2 + O21g2 ¡ 2 NO21g2 4 NH31g2 + 5 O21g2 ¡ 4 NO1g2 + 6 H2O1g2


4.101 Hard water contains , , and , which interferewith the action of soap and leave an insoluble coating on theinsides of containers and pipes when heated. Water softenersreplace these ions with . (a) If 1500 L of hard water con-tains 0.020 M and 0.0040 M , how many moles of

are needed to replace these ions? (b) If the sodium isadded to the water softener in the form of NaCl, how manygrams of sodium chloride are needed?

4.102 Tartaric acid, H2C4H4O6, has two acidic hydrogens. The acid isoften present in wines and precipitates from solution as thewine ages. A solution containing an unknown concentrationof the acid is titrated with NaOH. It requires 24.65 mL of0.2500 M NaOH solution to titrate both acidic protons in50.00 mL of the tartaric acid solution. Write a balanced netionic equation for the neutralization reaction, and calculatethe molarity of the tartaric acid solution.

Na+Mg2+Ca2+

Na+

Fe2+Mg2+Ca2+ 4.103 (a) A strontium hydroxide solution is prepared by dissolving10.45 g of Sr(OH)2 in water to make 50.00 mL of solution.What is the molarity of this solution? (b) Next the strontiumhydroxide solution prepared in part (a) is used to titrate a ni-tric acid solution of unknown concentration. Write a balancedchemical equation to represent the reaction between stron-tium hydroxide and nitric acid solutions. (c) If 23.9 mL ofthe strontium hydroxide solution was needed to neutralize a31.5 mL aliquot of the nitric acid solution, what is the concen-tration (molarity) of the acid?

[4.104] A solid sample of Zn(OH)2 is added to 0.350 L of 0.500 Maqueous HBr. The solution that remains is still acidic. It is thentitrated with 0.500 M NaOH solution, and it takes 88.5 mL ofthe NaOH solution to reach the equivalence point. What massof Zn(OH)2 was added to the HBr solution?

INTEGRATIVE EXERCISES

4.105 Suppose you have 5.00 g of powdered magnesium metal,1.00 L of 2.00 M potassium nitrate solution, and 1.00 L of 2.00 Msilver nitrate solution. (a) Which one of the solutions will reactwith the magnesium powder? (b) What is the net ionic equationthat describes this reaction? (c) What volume of solution isneeded to completely react with the magnesium? (d) What is themolarity of the Mg2+ ions in the resulting solution?

4.106 (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) Anelemental analysis of the acid indicates that it is composed of

, , and by mass. What is its molecu-lar formula?

4.107 A 3.455-g sample of a mixture was analyzed for barium ionby adding a small excess of sulfuric acid to an aqueous solu-tion of the sample. The resultant reaction produced a precip-itate of barium sulfate, which was collected by filtration,washed, dried, and weighed. If 0.2815 g of barium sulfatewas obtained, what was the mass percentage of barium inthe sample?

4.108 A tanker truck carrying of concentrated sulfuricacid solution tips over and spills its load. If the sulfuric acid is

H2SO4 by mass and has a density of , howmany kilograms of sodium carbonate must be added to neu-tralize the acid?

4.109 A sample of 5.53 g of Mg(OH)2 is added to 25.0 mL of 0.200M HNO3. (a) Write the chemical equation for the reactionthat occurs. (b) Which is the limiting reactant in the reaction?(c) How many moles of Mg(OH)2, HNO3, and Mg(NO3)2 arepresent after the reaction is complete?

1.84 g>mL95.0%

5.0 * 103 kg

23.5% O70.6% C5.89% H

4.110 A sample of 1.50 g of lead(II) nitrate is mixed with 125 mL of0.100 M sodium sulfate solution. (a) Write the chemical equa-tion for the reaction that occurs. (b) Which is the limitingreactant in the reaction? (c) What are the concentrations of allions that remain in solution after the reaction is complete?

[4.111] The average concentration of bromide ion in seawater is65 mg of bromide ion per kg of seawater. What is the molarityof the bromide ion if the density of the seawater is

[4.112] The mass percentage of chloride ion in a 25.00-mL sample ofseawater was determined by titrating the sample with silvernitrate, precipitating silver chloride. It took 42.58 mL of0.2997 M silver nitrate solution to reach the equivalence pointin the titration. What is the mass percentage of chloride ion inthe seawater if its density is ?

4.113 The arsenic in a 1.22-g sample of a pesticide was converted toby suitable chemical treatment. It was then titrated

using to form Ag3AsO4 as a precipitate. (a) What is theoxidation state of As in ? (b) Name Ag3AsO4 by anal-ogy to the corresponding compound containing phosphorusin place of arsenic. (c) If it took 25.0 mL of 0.102 M toreach the equivalence point in this titration, what is the masspercentage of arsenic in the pesticide?

[4.114] The newest US standard for arsenate in drinking water, man-dated by the Safe Drinking Water Act, required that by January2006, public water supplies must contain no greater than10 parts per billion (ppb) arsenic. If this arsenic is present asarsenate, , what mass of sodium arsenate would bepresent in a 1.00-L sample of drinking water that just meetsthe standard? Parts per billion is defined on a mass basis as

ppb =g solute

g solution* 109.

AsO43-

Ag+

AsO43-

Ag+AsO4

3-

1.025 g>mL

1.025 g>mL?

Integrative Exercises 157

[4.115] Federal regulations set an upper limit of 50 parts per million(ppm) of NH3 in the air in a work environment [that is, 50molecules of NH3(g) for every million molecules in the air].Air from a manufacturing operation was drawn through a so-lution containing of 0.0105 M HCl. The NH3reacts with HCl as follows:

NH31aq2 + HCl1aq2 ¡ NH4Cl1aq21.00 * 102 mL

After drawing air through the acid solution for 10.0 min at arate of , the acid was titrated. The remaining acidneeded 13.1 mL of 0.0588 M NaOH to reach the equivalencepoint. (a) How many grams of NH3 were drawn into the acidsolution? (b) How many ppm of NH3 were in the air? (Air hasa density of and an average molar mass ofunder the conditions of the experiment.) (c) Is this manufac-turer in compliance with regulations?

29.0 g>mol1.20 g>L10.0 L/min

another or transferred between systems and surroundings. Theenergy possessed by a system is called its internal energy.Internal energy is a state function, a quantity whose valuedepends only on the current state of a system, not on how thesystem came to be in that state.

5.3 ENTHALPYNext, we encounter a state function called enthalpy that is usefulbecause the change in enthalpy measures the quantity of heatenergy gained or lost by a system in a process occurring underconstant pressure.

WHAT’S AHEAD5.1 THE NATURE OF ENERGYWe begin by considering the nature of energy and the forms ittakes, notably kinetic energy and potential energy. We discuss theunits used in measuring energy and the fact that energy can beused to do work or to transfer heat. To study energy changes, wefocus on a particular part of the universe, which we call thesystem. Everything else is called the surroundings.

5.2 THE FIRST LAW OF THERMODYNAMICSWe then explore the first law of thermodynamics: Energy cannotbe created or destroyed but can be transformed from one form to

5BIOENERGY. The sugars in sugarcane,produced from CO2, H2O, and sunshine viaphotosynthesis, can be converted intoethanol, which is used as an alternative togasoline. In certain climates, such as that inBrazil, the sugarcane crop replenishes itselfrapidly, making cane-based ethanol asustainable fuel source.

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