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Unformatted text preview: BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE WebsiteCHAPTER4Reactions in Aqueous SolutionINTRODUCTIONIN4.1 GENERAL PROPERTIES OF AQUEOUSSOLUTIONSTHIS CHAPTER WE WILL APPLY OUR KNOWLEDGE OF STOICHIOMETRYTO THE STUDY OF REACTIONS THAT OCCUR WHEN THE REACTANTS AREDISSOLVED IN WATER.MOST4.2 PRECIPITATION REACTIONSPEOPLE TAKE WATER FOR GRANTED BECAUSE IT IS READILYAVAILABLE IN SEEMINGLY UNLIMITED AMOUNTS EXCEPT IN ARID CLIMATES AND IN TIMES OF DROUGHT, WHEN RESTRICTIONS ON ITS USE4.3 ACID-BASE REACTIONSMAKE US ACUTELY AWARE OF ITS IMPORTANCE AND BECAUSE WE USEIT FOR SO MANY ORDINARY PURPOSES.WE DRINK IT, BATHE IN IT, COOKWE SWIM IN IT, SKATE ON IT, SKI OVERIT, AND SAIL THROUGH IT. WE HARNESS ITS ENERGY FOR HEAT ANDPOWER. WITHOUT WATER WE LITERALLY COULD NOT EXIST: WATERAMOUNTS TO ABOUT 60 PERCENT OF THE MASS OF THE HUMAN BODY.WATERS GREAT VERSATILITY STEMS, IN PART, FROM A TENDENCY4.4 OXIDATION-REDUCTION REACTIONSTO FORM AQUEOUS SOLUTIONS BY DISSOLVING A LARGE VARIETY OF SOLIDS4.7 ACID-BASE TITRATIONSWITH IT, AND CLEAN WITH IT.4.5 CONCENTRATION OF SOLUTIONS4.6 GRAVIMETRIC ANALYSISAND OTHER LIQUIDS; THE FACT THAT IT EXISTS AT NORMAL AIR TEMPERATURE AS A LIQUID IS DUE TO THE UNIQUE PROPERTY OF ITS MOLECULES.EVENEARTHSFACE, IT IS RARE TO FIND PURE WATER IN NATURE.SEAWATER4.8 REDOX TITRATIONSSURANDTHOUGH WATER COVERS 70 PERCENT OFFRESHWATER SOURCES ALIKE CONTAIN DISSOLVED MINERALS AND CONTAMINANTS SUCH AS FERTILIZERS AND INDUSTRIAL POLLUTANTS.ASFORTHE WATER THAT COMES FROM THE TAP, IT GENERALLY CONTAINS FLUORIDES (ADDED TO REDUCE TOOTH DECAY) IN ADDITION TO MINERALS(PRINCIPALLY CHLORIDES, SULFATES, BICARBONATES OF SODIUM, POTASSIUM, CALCIUM, AND MAGNESIUM), AND POSSIBLY ADDITIONAL CHLORINE(TO KILL BACTERIA) AND LEAD (IF THE PIPES CARRYING IT ARE MORETHAN 80 YEARS OLD).MANYCHEMICALREACTIONSPROCESSES TAKE PLACE IN WATER.ANDINVIRTUALLYALLBIOLOGICALTHIS CHAPTER WE WILL DISCUSSTHREE MAJOR CATEGORIES OF REACTIONS THAT OCCUR IN AQUEOUS SOLUTIONS: PRECIPITATION REACTIONS, ACID-BASE REACTIONS, AND REDOXREACTIONS. IN LATER CHAPTERS WE WILL STUDY THE STRUCTURAL CHARACTERISTICS AND PROPERTIES OF WATER, THE SO-CALLED UNIVERSAL SOLVENT, AND ITS SOLUTIONS.109BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website110REACTIONS IN AQUEOUS SOLUTION4.1GENERAL PROPERTIES OF AQUEOUS SOLUTIONSA solution is a homogeneous mixture of two or more substances. The solute is the substance present in a smaller amount, and the solvent is the substance present in a largeramount. A solution may be gaseous (such as air), solid (such as an alloy), or liquid(seawater, for example). In this section we will discuss only aqueous solutions, inwhich the solute initially is a liquid or a solid and the solvent is water.ELECTROLYTIC PROPERTIESAll solutes that dissolve in water fit into one of two categories: electrolytes and nonelectrolytes. An electrolyte is a substance that, when dissolved in water, results in asolution that can conduct electricity. A nonelectrolyte does not conduct electricity whendissolved in water. Figure 4.1 shows an easy and straightforward method of distinguishing between electrolytes and nonelectrolytes. A pair of inert electrodes (copper orplatinum) is immersed in a beaker of water. To light the bulb, electric current mustflow from one electrode to the other, thus completing the circuit. Pure water is a verypoor conductor of electricity. However, if we add a small amount of sodium chloride(NaCl), the bulb will glow as soon as the salt dissolves in the water. Solid NaCl, anionic compound, breaks up into Na and Cl ions when it dissolves in water. The Naions are attracted to the negative electrode and the Cl ions to the positive electrode.This movement sets up an electric current that is equivalent to the flow of electronsalong a metal wire. Because the NaCl solution conducts electricity, we say that NaClis an electrolyte. Pure water contains very few ions, so it cannot conduct electricity.Comparing the light bulbs brightness for the same molar amounts of dissolvedsubstances helps us distinguish between strong and weak electrolytes. A characteristicof strong electrolytes is that the solute is assumed to be 100 percent dissociated intoions in solution. (By dissociation we mean the breaking up of the compound into cationsand anions.) Thus we can represent sodium chloride dissolving in water asHO2NaCl(s) 88n Na (aq)Cl (aq)This equation says that all sodium chloride that enters the solution ends up as Na andCl ions; there are no undissociated NaCl units in solution.FIGURE 4.1 An arrangementfor distinguishing between electrolytes and nonelectrolytes. A solutions ability to conduct electricity depends on the number ofions it contains. (a) A nonelectrolyte solution does not containions, and the light bulb is not lit.(b) A weak electrolyte solutioncontains a small number of ions,and the light bulb is dimly lit. (c)A strong electrolyte solution contains a large number of ions, andthe light bulb is brightly lit. Themolar amounts of the dissolvedsolutes are equal in all threecases.(a)BackForwardMain MenuTOC(b)Study Guide TOC(c)Textbook WebsiteMHHE Website4.1GENERAL PROPERTIES OF AQUEOUS SOLUTIONS111TABLE 4.1 Classification of Solutes in Aqueous SolutionSTRONG ELECTROLYTEWEAK ELECTROLYTENONELECTROLYTEHClHNO3HClO4H2SO4*NaOHBa(OH)2Ionic compoundsCH3COOHHFHNO2NH3H2O(NH2)2CO (urea)CH3OH (methanol)C2H5OH (ethanol)C6H12O6 (glucose)C12H22O11 (sucrose)*H2SO4 has two ionizable H ions.Pure water is an extremely weak electrolyte.Table 4.1 lists examples of strong electrolytes, weak electrolytes, and nonelectrolytes. Ionic compounds, such as sodium chloride, potassium iodide (KI), and calcium nitrate [Ca(NO3)2], are strong electrolytes. It is interesting to note that humanbody fluids contain many strong and weak electrolytes.Water is a very effective solvent for ionic compounds. Although water is an electrically neutral molecule, it has a positive region (the H atoms) and a negative region(the O atom), or positive and negative poles; for this reason it is a polar solvent.When an ionic compound such as sodium chloride dissolves in water, the threedimensional network of ions in the solid is destroyed. The Na and Cl ions are separated from each other and undergo hydration, the process in which an ion is surrounded by water molecules arranged in a specific manner. Each Na ion is surroundedby a number of water molecules orienting their negative poles toward the cation.Similarly, each Cl ion is surrounded by water molecules with their positive poles oriented toward the anion (Figure 4.2). Hydration helps to stabilize ions in solution andprevents cations from combining with anions.Acids and bases are also electrolytes. Some acids, including hydrochloric acid(HCl) and nitric acid (HNO3), are strong electrolytes. These acids ionize completelyin water; for example, when hydrogen chloride gas dissolves in water, it forms hydrated H and Cl ions:HO2HCl(g) 88n H (aq)Cl (aq)In other words, all the dissolved HCl molecules separate into hydrated H and Clions. Thus when we write HCl(aq), it is understood that it is a solution of only H (aq)and Cl (aq) ions and that there are no hydrated HCl molecules present. On the otherhand, certain acids, such as acetic acid (CH3COOH), which gives vinegar its tart flavor, do not ionize completely and are weak electrolytes. We represent the ionizationof acetic acid asFIGURE 4.2 Hydration of Naand Cl ions.+BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website112REACTIONS IN AQUEOUS SOLUTIONCH3COOH(aq) 34 CH3COO (aq)There are different types of chemical equilibrium. We will return tothis very important topic inChapter 14.4.2H (aq)where CH3COO is called the acetate ion. We use the term ionization to describe theseparation of acids and bases into ions. By writing the formula of acetic acid asCH3COOH we indicate that the ionizable proton is in the COOH group.The ionization of acetic acid is written with a double arrow to show that it is areversible reaction; that is, the reaction can occur in both directions. Initially, a number of CH3COOH molecules break up into CH3COO and H ions. As time goes on,some of the CH3COO and H ions recombine into CH3COOH molecules. Eventually,a state is reached in which the acid molecules ionize as fast as the ions recombine.Such a chemical state, in which no net change can be observed (although activity iscontinuous on the molecular level), is called chemical equilibrium. Acetic acid, then,is a weak electrolyte because its ionization in water is incomplete. By contrast, in ahydrochloric acid solution the H and Cl ions have no tendency to recombine andform molecular HCl. We use a single arrow to represent complete ionizations.PRECIPITATION REACTIONSOne common type of reaction that occurs in aqueous solution is the precipitation reaction, which results in the formation of an insoluble product, or precipitate. A precipitate is an insoluble solid that separates from the solution. Precipitation reactionsusually involve ionic compounds. For example, when an aqueous solution of lead nitrate [Pb(NO3)2] is added to an aqueous solution of sodium iodide (NaI), a yellow precipitate of lead iodide (PbI2) is formed:Pb(NO3)2(aq)2NaI(aq) 88n PbI2(s)2NaNO3(aq)Sodium nitrate remains in solution. Figure 4.3 shows this reaction in progress.SOLUBILITYFIGURE 4.3 Formation of yellow PbI2 precipitate as a solutionof Pb(NO3)2 is added to a solution of NaI.How can we predict whether a precipitate will form when a compound is added to asolution or when two solutions are mixed? It depends on the solubility of the solute,which is defined as the maximum amount of solute that will dissolve in a given quantity of solvent at a specific temperature. Chemists refer to substances as soluble, slightlysoluble, or insoluble in a qualitative sense. A substance is said to be soluble if a fairamount of it visibly dissolves when added to water. If not, the substance is describedas slightly soluble or insoluble. All ionic compounds are strong electrolytes, but theyare not equally soluble.Table 4.2 classifies a number of common ionic compounds as soluble or insoluble. Keep in mind, however, that even insoluble compounds dissolve to a certain extent. Figure 4.4 shows several precipitates.The following example applies the solubility rules in Table 4.2.EXAMPLE 4.1Classify the following ionic compounds as soluble or insoluble: (a) silver sulfate(Ag2SO4), (b) calcium carbonate (CaCO3), (c) sodium phosphate (Na3PO4).AnswerBackForwardMain Menu(a) According to Table 4.2, Ag2SO4 is insoluble. (b) This is a carbonateTOCStudy Guide TOCTextbook WebsiteMHHE Website4.2TABLE 4.2PRECIPITATION REACTIONS113Solubility Rules for Common Ionic Compounds in Water at 25CSOLUBLE COMPOUNDSEXCEPTIONSCompounds containingalkali metal ions (Li , Na ,K , Rb , Cs ) and theammonium ion (NH4 )Nitrates (NO3 ), bicarbonates(HCO3 ), and chlorates(ClO3 )Halides (Cl , Br , I )Halides of Ag , Hg2 , and Pb22Sulfates (SO2 )4Sulfates of Ag , Ca2 , Sr2 , Ba2 , and Pb2INSOLUBLE COMPOUNDSEXCEPTIONS(CO23Carbonates), phosphates(PO3 ), chromates (CrO2 ),44sulfides (S2 )Hydroxides (OH )Similar problems: 4.15, 4.16.Compounds containing alkali metal ionsand the ammonium ionCompounds containing alkalimetal ions and the Ba2 ionand Ca is a Group 2A metal. Therefore, CaCO3 is insoluble. (c) Sodium is an alkali metal (Group 1A) so Na3PO4 is soluble.PRACTICE EXERCISEClassify the following ionic compounds as soluble or insoluble: (a) CuS,(b) Ca(OH)2, (c) Zn(NO3)2.MOLECULAR EQUATIONS AND IONIC EQUATIONSThe equation describing the precipitation of lead iodide on page 112 is called a molecular equation because the formulas of the compounds are written as though allFIGURE 4.4 Appearance ofseveral precipitates. From left toright: CdS, PbS, Ni(OH)2,Al(OH)3.BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website114REACTIONS IN AQUEOUS SOLUTIONspecies existed as molecules or whole units. A molecular equation is useful because itidentifies the reagents (that is, lead nitrate and sodium iodide). If we wanted to bringabout this reaction in the laboratory, the molecular equation would be the one to use.However, a molecular equation does not accurately describe what actually is happening at the microscopic level.As pointed out earlier, when ionic compounds dissolve in water, they break apartcompletely into their component cations and anions. To be more realistic, the equations should show the dissociation of dissolved ionic compounds into ions. Therefore,returning to the reaction between sodium iodide and lead nitrate, we would writePb2 (aq)2NO3 (aq)2Na (aq)2I (aq) 88n PbI2(s)2Na (aq)2NO3 (aq)The above equation is an example of an ionic equation, which shows dissolved speciesas free ions. An ionic equation includes spectator ions, or ions that are not involvedin the overall reaction, in this case the Na and NO3 ions. Spectator ions appear onboth sides of the equation and are unchanged in the chemical reaction, so they can becanceled. To focus on the change that occurs, we write the net ionic equation, whichshows only the species that actually take part in the reaction:Pb2 (aq)2I (aq) 88n PbI2(s)Looking at another example, we find that when an aqueous solution of bariumchloride (BaCl2) is added to an aqueous solution of sodium sulfate (Na2SO4), a whiteprecipitate of barium sulfate (BaSO4) is formed (Figure 4.5). The molecular equationfor this reaction isBaCl2(aq)FIGURE 4.5 Formation ofBaSO4 precipitate.Na2SO4(aq) 88n BaSO4(s)2NaCl(aq)The ionic equation for the reaction isBa2 (aq)2Cl (aq)2Na (aq)SO2 (aq) 88n BaSO4(s)42Na (aq)2Cl (aq)Canceling the spectator ions (Na and Cl ) on both sides of the equation gives us thenet ionic equationBa2 (aq)SO2 (aq) 88n BaSO4(s)4The following steps summarize the procedure for writing ionic and net ionic equations.Write a balanced molecular equation for the reaction.Rewrite the equation to show the dissociated ions that form in solution. Rememberthat all strong electrolytes, when dissolved in solution, are completely dissociatedinto cations and anions. This procedure gives us the ionic equation. Identify and cancel spectator ions on both sides of the equation to arrive at the netionic equation.These steps are applied in Example 4.2.EXAMPLE 4.2Predict the products of the following reaction and write a net ionic equation for thereactionK3PO4(aq)BackForwardMain MenuTOCCa(NO3)2(aq) 88n ?Study Guide TOCTextbook WebsiteMHHE Website4.3ACID-BASE REACTIONS115From the unbalanced equation we see that a solution of potassium phosphate is mixed or reacted with a solution of calcium nitrate. According to Table 4.2,calcium ions (Ca2 ) and phosphate ions (PO3 ) can form an insoluble compound,4calcium phosphate [Ca3(PO4)2]. Therefore this is a precipitation reaction. The otherproduct, potassium nitrate (KNO3), is soluble and remains in solution. The molecular equation isAnswer2K3PO4(aq)3Ca(NO3)2(aq) 88n 6KNO3(aq)Ca3(PO4)2(s)and the ionic equation is6K (aq)Precipitate formed by the reactionbetween K3PO4(aq) andCa(NO3)2(aq).2PO3 (aq)43Ca2 (aq)6NO3 (aq) 88n6K (aq)6NO3 (aq)Ca3(PO4)2(s)Canceling the spectator K and NO3 ions, we obtain the net ionic equation3Ca2 (aq)2PO3 (aq) 88n Ca3(PO4)2(s)4Note that because we balanced the molecular equation first, the net ionic equationis balanced in terms of the number of atoms on each side and the number of positive and negative charges on the left-hand side.Similar problems: 4.17, 4.18.PRACTICE EXERCISEPredict the precipitate produced by the following reaction, and write a net ionicequation for the reactionAl(NO3)3(aq)NaOH(aq) 88n ?The Chemistry in Action essay on p. 116 discusses some practical problems associated with precipitation reactions.4.3ACID-BASE REACTIONSAcids and bases are as familiar as aspirin and milk of magnesia although many people do not know their chemical names acetylsalicylic acid (aspirin) and magnesiumhydroxide (milk of magnesia). In addition to being the basis of many medicinal andhousehold products, acid-base chemistry is important in industrial processes and essential in sustaining biological systems. Before we can discuss acid-base reactions, weneed to know more about acids and bases themselves.GENERAL PROPERTIES OF ACIDS AND BASESIn Section 2.7 we defined acids as substances that ionize in water to produce H ionsand bases as substances that ionize in water to produce OH ions. These definitionswere formulated in the late nineteenth century by the Swedish chemist SvanteArrhenius to classify substances whose properties in aqueous solutions were wellknown.Svante August Arrhenius (1859 1927). Swedish chemist. Arrhenius made important contributions in the study of chemical kinetics and electrolyte solutions. He also speculated that life had come to Earth from other planets, a theory now knownas panspermia. Arrhenius was awarded the Nobel Prize in Chemistry in 1903.BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website116REACTIONS IN AQUEOUS SOLUTIONChemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry inChemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action ChemistryAn Undesirable Precipitation ReactionLimestone (CaCO3) and dolomite (CaCO3 MgCO3),which are widespread on Earths surface, often enterthe water supply. According to Table 4.2, calcium carbonate is insoluble in water. However, in the presenceof dissolved carbon dioxide (from the atmosphere),calcium carbonate is converted to soluble calcium bicarbonate [Ca(HCO3)2]:CaCO3(s)CO2(aq)H2O(l ) 88nCa2 (aq)2HCO3 (aq)where HCO3 is the bicarbonate ion.Water containing Ca2 and/or Mg2 ions iscalled hard water, and water that is mostly free ofthese ions is called soft water. Hard water is unsuitable for some household and industrial uses.When water containing Ca2 and HCO3 ions isheated or boiled, the solution reaction is reversed toproduce the CaCO3 precipitateCa2 (aq)2HCO3 (aq) 88nCaCO3(s)CO2(aq)H2O(l )and gaseous carbon dioxide is driven off:Boiler scale almost fills this hot-water pipe. The deposits consist mostly of CaCO3 with some MgCO3.water heaters, pipes, and teakettles. A thick layer ofscale reduces heat transfer and decreases the efficiency and durability of boilers, pipes, and appliances. In household hot-water pipes it can restrict ortotally block the flow of water. A simple method usedby plumbers to remove scale deposits is to introducea small amount of hydrochloric acid, which reacts with(and therefore dissolves) CaCO3:CaCO3(s)CO2(aq) 88n CO2(g)Solid calcium carbonate formed in this way is the maincomponent of the scale that accumulates in boilers,2HCl(aq) 88nCaCl2(aq)H2O(l )CO2(g)In this way, CaCO3 is converted to soluble CaCl2.AcidsAcids have a sour taste; for example, vinegar owes its sourness to acetic acid, andlemons and other citrus fruits contain citric acid. Acids cause color changes in plant dyes; for example, they change the color of litmus from blue to red. Acids react with certain metals, such as zinc, magnesium, and iron, to produce hydrogen gas. A typical reaction is that between hydrochloric acid and magnesium:2HCl(aq)FIGURE 4.6 A piece of blackboard chalk, which is mostlyCaCO3, reacts with hydrochloricacid.BackForwardHCl(aq)Main MenuH2(g)Acids react with carbonates and bicarbonates, such as Na2CO3, CaCO3, andNaHCO3, to produce carbon dioxide gas (Figure 4.6). For example,2HCl(aq)Mg(s) 88n MgCl2(aq)CaCO3(s) 88n CaCl2(aq)NaHCO3(s) 88n NaCl(aq)H2O(l)H2O(l)CO2(g)CO2(g)Aqueous acid solutions conduct electricity.TOCStudy Guide TOCTextbook WebsiteMHHE Website4.3ACID-BASE REACTIONS117BasesBases have a bitter taste.Bases feel slippery; for example, soaps, which contain bases, exhibit this property.Bases cause color changes in plant dyes; for example, they change the color of litmus from red to blue. Aqueous base solutions conduct electricity.BRNSTED ACIDS AND BASESArrheniuss definitions of acids and bases are limited in that they apply only to aqueous solutions. Broader definitions were proposed by the Danish chemist JohannesBrnsted in 1932; a Brnsted acid is a proton donor, and a Brnsted base is a proton acceptor. Note that Brnsteds definitions do not require acids and bases to be inaqueous solution.Hydrochloric acid is a Brnsted acid since it donates a proton in water:HCl(aq) 88n H (aq)Cl (aq)Note that the H ion is a hydrogen atom that has lost its electron; that is, it is just abare proton. The size of a proton is about 10 15 m, compared to a diameter of 10 10m for an average atom or ion. Such an exceedingly small charged particle cannotexist as a separate entity in aqueous solution owing to its strong attraction for thenegative pole (the O atom) in H2O. Consequently, the proton exists in the hydratedform as shown in Figure 4.7. Therefore, the ionization of hydrochloric acid shouldbe written asHCl(aq)H2O(l) 88n H3O (aq)Cl (aq)The hydrated proton, H3O , is called the hydronium ion. This equation shows a reaction in which a Brnsted acid (HCl) donates a proton to a Brnsted base (H2O).Experiments show that the hydronium ion is further hydrated so that the protonmay have several water molecules associated with it. Since the acidic properties of theproton are unaffected by the degree of hydration, in this text we will generally useH (aq) to represent the hydrated proton. This notation is for convenience, but H3Ois closer to reality. Keep in mind that both notations represent the same species in aqueous solution.Acids commonly used in the laboratory include hydrochloric acid (HCl), nitricacid (HNO3), acetic acid (CH3COOH), sulfuric acid (H2SO4), and phosphoric acid(H3PO4). The first three are monoprotic acids; that is, each unit of the acid yields onehydrogen ion upon ionization:Johannes Nicolaus Brnsted (1879 1947). Danish chemist. In addition to his theory of acids and bases, Brnsted workedon thermodynamics and the separation of mercury into its isotopes.FIGURE 4.7 Ionization of HClin water to form the hydroniumion and the chloride ion.+HClBackForwardMain MenuTOC+H2OH3O+Study Guide TOCTextbook Website++ClMHHE Website118REACTIONS IN AQUEOUS SOLUTIONHCl(aq) 88n H (aq)HNO3(aq) 88n H (aq)Cl (aq)NO3 (aq)CH3COOH(aq) 34 CH3COO (aq)H (aq)As mentioned earlier, because the ionization of acetic acid is incomplete (note the double arrows), it is a weak electrolyte. For this reason it is called a weak acid (see Table4.1). On the other hand, HCl and HNO3 are strong acids because they are strong electrolytes, so they are completely ionized in solution (note the use of single arrows).Sulfuric acid (H2SO4) is a diprotic acid because each unit of the acid gives uptwo H ions, in two separate steps:H2SO4(aq) 88n H (aq)HSO4 (aq)HSO4 (aq) 34 H (aq)SO2 (aq)4H2SO4 is a strong electrolyte or strong acid (the first step of ionization is complete),but HSO4 is a weak acid or weak electrolyte, and we need a double arrow to represent its incomplete ionization.Triprotic acids, which yield three H ions, are relatively few in number. The bestknown triprotic acid is phosphoric acid, whose ionizations areH3PO4(aq) 34 H (aq)H2PO4 (aq)H2PO4 (aq) 34 H (aq)HPO2 (aq)4HPO2 (aq) 34 H (aq)4PO3 (aq)4All three species (H3PO4, H2PO4 , and HPO2 ) in this case are weak acids, and we use4the double arrows to represent each ionization step. Anions such as H2PO4 and HPO24are found in aqueous solutions of phosphates such as NaH2PO4 and Na2HPO4.Table 4.1 shows that sodium hydroxide (NaOH) and barium hydroxide [Ba(OH)2]are strong electrolytes. This means that they are completely ionized in solution:HO2NaOH(s) 88n Na (aq)HO2Ba(OH)2(s) 88n Ba2 (aq)OH (aq)2OH (aq)The OH ion can accept a proton as follows:H (aq)OH (aq) 88n H2O(l)Thus OH is a Brnsted base.Ammonia (NH3) is classified as a Brnsted base because it can accept a H ion(Figure 4.8):NH3(aq)H2O(l) 34 NH4 (aq)OH (aq)FIGURE 4.8 Ionization of ammonia in water to form the ammonium ion and the hydroxideion.34+NH3BackForwardMain MenuTOC+H2O34Study Guide TOC++NH4+Textbook WebsiteOHMHHE Website4.3ACID-BASE REACTIONS119Ammonia is a weak electrolyte (and therefore a weak base) because only a small fraction of dissolved NH3 molecules react with water to form NH4 and OH ions.The most commonly used strong base in the laboratory is sodium hydroxide. It ischeap and soluble. (In fact, all of the alkali metal hydroxides are soluble.) The mostcommonly used weak base is aqueous ammonia solution, which is sometimes erroneously called ammonium hydroxide; there is no evidence that the species NH4OH actually exists. All of the Group 2A elements form hydroxides of the type M(OH)2, whereM denotes an alkaline earth metal. Of these hydroxides, only Ba(OH)2 is soluble.Magnesium and calcium hydroxides are used in medicine and industry. Hydroxides ofother metals, such as Al(OH)3 and Zn(OH)2 are insoluble and are less commonly used.The following example classifies substances as Brnsted acids or Brnsted bases.EXAMPLE 4.3Classify each of the following species as a Brnsted acid or base: (a) HBr, (b) NO2 ,(c) HCO3 .Answer(a) HBr dissolves in water to yield H and Br ions:HBr(aq) 88n H (aq)Br (aq)Therefore HBr is a Brnsted acid.(b) In solution the nitrite ion can accept a proton to form nitrous acid:NO2 (aq)H (aq) 88n HNO2(aq)This property makes NO2 a Brnsted base.(c) The bicarbonate ion is a Brnsted acid because it ionizes in solution as follows:HCO3 (aq) 34 H (aq)CO2 (aq)3It is also a Brnsted base because it can accept a proton to form carbonic acid:HCO3 (aq)Similar problems: 4.27, 4.28.H (aq) 34 H2CO3(aq)Comment The HCO3 species is said to be amphoteric because it possesses bothacidic and basic properties. The double arrows show that both reactions are reversible.PRACTICE EXERCISEClassify each of the following species as a Brnsted acid or base: (a) SO2 , (b) HI.4ACID-BASE NEUTRALIZATIONA neutralization reaction is a reaction between an acid and a base. Generally, aqueous acid-base reactions produce water and a salt, which is an ionic compound madeup of a cation other than H and an anion other than OH or O2 :Acid-base reactions generally goto completion.acidbase 88n saltwaterAll salts are strong electrolytes. The substance we know as table salt, NaCl, is a familiar example. It is a product of the acid-base reactionHCl(aq)NaOH(aq) 88n NaCl(aq)H2O(l)However, since both the acid and the base are strong electrolytes, they are completelyionized in solution. The ionic equation isBackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website120REACTIONS IN AQUEOUS SOLUTIONH (aq)Cl (aq)Na (aq)OH (aq) 88n Na (aq)Cl (aq)H2O(l)Therefore, the reaction can be represented by the net ionic equationH (aq)OH (aq) 88n H2O(l)Both Na and Cl are spectator ions.If we had started the above reaction with equal molar amounts of the acid and thebase, at the end of the reaction we would have only a salt and no leftover acid or base.This is a characteristic of acid-base neutralization reactions.The following are also examples of acid-base neutralization reactions, representedby molecular equations:HF(aq)H2SO4(aq)KOH(aq) 88n KF(aq)H2O(l)2NaOH(aq) 88n Na2SO4(aq)HNO3(aq)2H2O(l)NH3(aq) 88n NH4NO3(aq)The last equation looks different because it does not show water as a product. However,if we express NH3(aq) as NH4 (aq) and OH (aq), as discussed earlier, then the equation becomesHNO3(aq)4.4NH4 (aq)OH (aq) 88n NH4NO3(aq)H2O(l)OXIDATION-REDUCTION REACTIONSWhereas acid-base reactions can be characterized as proton-transfer processes, the classof reactions called oxidation-reduction, or redox, reactions are considered electrontransfer reactions. Oxidation-reduction reactions are very much a part of the worldaround us. They range from the burning of fossil fuels to the action of household bleach.Additionally, most metallic and nonmetallic elements are obtained from their ores bythe process of oxidation or reduction.Many important redox reactions take place in water, but not all redox reactionsoccur in aqueous solution. Consider the formation of calcium oxide (CaO) from calcium and oxygen:2Ca(s)O2(g) 88n 2CaO(s)Calcium oxide (CaO) is an ionic compound made up of Ca2 and O2 ions. In this reaction, two Ca atoms give up or transfer four electrons to two O atoms (in O2). Forconvenience, we can think of this process as two separate steps, one involving the lossof four electrons by the two Ca atoms and the other being the gain of four electronsby an O2 molecule:2Ca 88n 2Ca2O24e 88n 2O4e2Each of these steps is called a half-reaction, which explicitly shows the electrons involved in a redox reaction. The sum of the half-reactions gives the overall reaction:2CaO24e 88n 2Ca22O24eor, if we cancel the electrons that appear on both sides of the equation,2CaBackForwardMain MenuTOCO2 88n 2Ca2Study Guide TOC2O2Textbook WebsiteMHHE Website4.4OXIDATION-REDUCTION REACTIONS121Finally, the Ca2 and O2 ions combine to form CaO:2Ca2Oxidizing agents are always reduced, and reducing agents are always oxidized. This statementmay be somewhat confusing, butit is simply a consequence of thedefinitions of the two processes.A useful mnemonic for redox isOILRIG: Oxidation Is Loss (ofelectrons) and Reduction Is Gain(of electrons).2O2 88n 2CaOBy convention, we do not show the charges in the formula of an ionic compound, sothat calcium oxide is normally represented as CaO rather than Ca2 O2 .The term oxidation reaction refers to the half-reaction that involves loss of electrons. Chemists originally used oxidation to denote the combination of elements withoxygen. However, it now has a broader meaning that includes reactions not involvingoxygen. A reduction reaction is a half-reaction that involves gain of electrons. In theformation of calcium oxide, calcium is oxidized. It is said to act as a reducing agentbecause it donates electrons to oxygen and causes oxygen to be reduced. Oxygen isreduced and acts as an oxidizing agent because it accepts electrons from calcium, causing calcium to be oxidized. Note that the extent of oxidation in a redox reaction mustbe equal to the extent of reduction; that is, the number of electrons lost by a reducingagent must be equal to the number of electrons gained by an oxidizing agent.The occurrence of electron transfer is more apparent in some redox reactions thanothers. When metallic zinc is added to a solution containing copper(II) sulfate (CuSO4),zinc reduces Cu2 by donating two electrons to it:Zn(s)CuSO4(aq) 88n ZnSO4(aq)Cu(s)In the process the solution loses the blue color that characterizes the presence of hydrated Cu2 ions (Figure 4.9):Zn(s)Cu2 (aq) 88n Zn2 (aq)Cu(s)The oxidation and reduction half-reactions areZn 88n Zn22Cu2e2e 88n CuSimilarly, metallic copper reduces silver ions in a solution of silver nitrate (AgNO3):Cu(s)2AgNO3(aq) 88n Cu(NO3)2(aq)2Ag(s)orCu(s)2Ag (aq) 88n Cu2 (aq)2Ag(s)FIGURE 4.9 Left to right:When a piece of Zn metal isplaced in an aqueous CuSO4 solution, Zn atoms enter the solutionas Zn2 ions, and Cu2 ions areconverted to solid Cu (firstbeaker). In time, the blue color ofthe CuSO4 solution disappears(second beaker). When a pieceof Cu wire is placed in an aqueous AgNO3 solution, Cu atomsenter the solution as Cu2 ions,and Ag ions are converted tosolid Ag (third beaker).Gradually, the solution acquiresthe blue color characteristic of hydrated Cu2 ions (fourth beaker).BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website122REACTIONS IN AQUEOUS SOLUTIONOXIDATION NUMBERThe definitions of oxidation and reduction in terms of loss and gain of electrons applyto the formation of ionic compounds such as CaO and the reduction of Cu2 ions byZn. However, these definitions do not accurately characterize the formation of hydrogen chloride (HCl) and sulfur dioxide (SO2):H2(g)S(s)Cl2(g) 88n 2HCl(g)O2(g) 88n SO2(g)Because HCl and SO2 are not ionic but molecular compounds, no electrons are actually transferred in the formation of these compounds, as they are in the case of CaO.Nevertheless, chemists find it convenient to treat these reactions as redox reactions because experimental measurements show that there is a partial transfer of electrons (fromH to Cl in HCl and from S to O in SO2).To keep track of electrons in redox reactions, it is useful to assign oxidation numbers to the reactants and products. An atoms oxidation number, also called oxidationstate, signifies the number of charges the atom would have in a molecule (or an ioniccompound) if electrons were transferred completely. For example, we can rewrite theabove equations for the formation of HCl and SO2 as follows:00H2(g)0S(s)11Cl2(g) 88n 2HCl(g)042O2(g) 88n SO2(g)The numbers above the element symbols are the oxidation numbers. In both of the reactions shown, there is no charge on the atoms in the reactant molecules. Thus theiroxidation number is zero. For the product molecules, however, it is assumed that complete electron transfer has taken place and that atoms have gained or lost electrons. Theoxidation numbers reflect the number of electrons transferred.Oxidation numbers allow us to identify elements that are oxidized and reduced ata glance. The elements that show an increase in oxidation number hydrogen and sulfur in the above examples are oxidized. Chlorine and oxygen are reduced, so theiroxidation numbers show a decrease from their initial values. Note that the sum of theoxidation numbers of H and Cl in HCl ( 1 and 1) is zero. Likewise, if we add thecharges on S ( 4) and two atoms of O [2 ( 2)], the total is zero. The reason is thatthe HCl and SO2 molecules are neutral, so the charges must cancel.We use the following rules to assign oxidation numbers:In free elements (that is, in the uncombined state), each atom has an oxidation number of zero. Thus each atom in H2, Br2, Na, Be, K, O2, and P4 has the same oxidation number: zero.2. For ions composed of only one atom (that is, monatomic ions) the oxidation number is equal to the charge on the ion. Thus Li has an oxidation number of 1;Ba2 ion, 2; Fe3 , 3; I ion, 1; O2 ion, 2; and so on. All alkali metalshave an oxidation number of 1 and all alkaline earth metals have an oxidationnumber of 2 in their compounds. Aluminum has an oxidation number of 3 inall its compounds.3. The oxidation number of oxygen in most compounds (for example, MgO and H2O)is 2, but in hydrogen peroxide (H2O2) and peroxide ion (O2 ), it is 1.21.BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website4.4OXIDATION-REDUCTION REACTIONS123The oxidation number of hydrogen is 1, except when it is bonded to metals inbinary compounds. In these cases (for example, LiH, NaH, CaH2), its oxidationnumber is 1.5. Fluorine has an oxidation number of 1 in all its compounds. Other halogens (Cl,Br, and I) have negative oxidation numbers when they occur as halide ions in theircompounds. When combined with oxygen for example in oxoacids and oxoanions (see Section 2.7) they have positive oxidation numbers.6. In a neutral molecule, the sum of the oxidation numbers of all the atoms must bezero. In a polyatomic ion, the sum of oxidation numbers of all the elements in theion must be equal to the net charge of the ion. For example, in the ammonium ion,NH4 , the oxidation number of N is 3 and that of H is 1. Thus the sum of theoxidation numbers is 3 4( 1)1, which is equal to the net charge of theion.7. Oxidation numbers do not have to be integers. For example, the oxidation numberof O in the superoxide ion, O2 , is 1 .24.We apply the above rules to assign oxidation numbers in Example 4.4EXAMPLE 4.4Assign oxidation numbers to all the elements in the following compounds and ion:(a) Li2O, (b) HNO3, (c) Cr2O27Answer (a) By rule 2 we see that lithium has an oxidation number ofand oxygens oxidation number is 2 (O2 ).1 (Li )(b) This is the formula for nitric acid, which yields a H ion and a NO3 ion in solution. From rule 4 we see that H has an oxidation number of 1. Thus the othergroup (the nitrate ion) must have a net oxidation number of 1. Since oxygen hasan oxidation number of 2, the oxidation number of nitrogen (labeled x) is givenby1orxx3( 2)5(c) From rule 6 we see that the sum of the oxidation numbers in Cr2O2 must be72. We know that the oxidation number of O is 2, so all that remains is to determine the oxidation number of Cr, which we call y. The sum of the oxidation numbers for the ion is2yorSimilar problems: 4.43, 4.45.2(y)7( 2)6PRACTICE EXERCISEAssign oxidation numbers to all the elements in the following compound and ion:(a) PF3, (b) MnO4 .Figure 4.10 shows the known oxidation numbers of the familiar elements, arrangedaccording to their positions in the periodic table. We can summarize the content of thisfigure as follows:BackForwardMain MenuMetallic elements have only positive oxidation numbers, whereas nonmetallic ele-TOCStudy Guide TOCTextbook WebsiteMHHE Website124REACTIONS IN AQUEOUS SOLUTION11A188A12HHe+1122A133A5144A6155A7166A8177A34910LiBeBCNOFNe+1+2+3+4+24+5+4+3+2+13+2121211511121314161718NaMgAlSiPSClAr+1+2+3+44+5+33+6+4+22+7+6+5+4+3+1133B44B55B66B77B898B10111B122B192021222324252627282930313233343536KCaScTiVCrMnFeCoNiCuZnGaGeAsSeBrKr+1+2+3+4+3+2+5+4+3+2+6+5+4+3+2+7+6+4+3+2+3+2+3+2+2+2+1+2+3+44+5+33+6+42+5+3+11+4+2373839404142434445464748495051525354RbSrYZrNbMoTcRuRhPdAgCdInSnSbTeIXe+1+2+3+4+5+4+6+4+3+7+6+4+8+6+4+3+4+3+2+4+2+1+2+3+4+2+5+33+6+42+7+5+11+6+4+2555657727374757677787980818283848586CsBaLaHfTaWReOsIrPtAuHgTlPbBiPoAtRn+1+2+3+4+5+6+4+7+6+4+8+4+4+3+4+2+3+1+2+1+3+1+4+2+5+3+21FIGURE 4.10 The oxidation numbers of elements in their compounds. The more common oxidation numbers are in color.ments may have either positive or negative oxidation numbers.The highest oxidation number a representative element can have is its group number in the periodic table. For example, the halogens are in Group 7A, so their highest possible oxidation number is 7. The transition metals (Groups 1B, 3B8B) usually have several possible oxidationnumbers.TYPES OF REDOX REACTIONSThere are four general types of redox reactions: combination reactions, decompositionreactions, displacement reactions, and disproportionation reactions. Displacement reactions have widespread applications in industry, so we will study them in some detail.BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website4.4(a)OXIDATION-REDUCTION REACTIONS125(b)(c)FIGURE 4.11 Some simplecombination redox reactions. (a)Hydrogen gas burning in air toform water. (b) Sulfur burning inair to form sulfur dioxide. (c)Magnesium burning in air to formmagnesium oxide and magnesium nitride. (d) Sodium burningin chlorine to form sodium chloride. (e) Aluminum reacting withbromine to form aluminum bromide.(d)(e)Combination ReactionsA combination reaction may be represented byAB 88n CIf either A or B is an element, then the reaction is redox in nature. Combination reactions are reactions in which two or more substances combine to form a single product. Figure 4.11 shows some common combination reactions. For example,0S(s)03Mg(s)042O2(g) 88n SO2(g)023N2(g) 88n Mg3N2(s)Decomposition ReactionsDecomposition reactions are the opposite of combination reactions. Specifically, a decomposition reaction is the breakdown of a compound into two or more components:BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website126REACTIONS IN AQUEOUS SOLUTIONFIGURE 4.12 (a) On heating,mercury(II) oxide (HgO) decomposes to form mercury and oxygen. (b) Heating potassium chlorate (KClO3) produces oxygen,which supports the combustion ofthe wood splint.(a)(b)C 88n ABIf either A or B is an element, then the reaction is redox in nature (Figure 4.12). Forexample22002HgO(s) 88n 2Hg(l)52O2(g)102KClO3(s) 88n 2KCl(s)113O2(g)002NaH(s) 88n 2Na(s)H2(g)Note that we show oxidation numbers only for elements that are oxidized or reduced.Thus potassium is not given an oxidation number in the decomposition of KClO3.(a)Displacement ReactionsIn a displacement reaction, an ion (or atom) in a compound is replaced by an ion (oratom) of another element.:ABC 88n ACBMost displacement reactions fit into one of three subcategories: hydrogen displacement, metal displacement, or halogen displacement.1. Hydrogen Displacement. All alkali metals and some alkaline earth metals (Ca,Sr, and Ba), which are the most reactive of the metallic elements, will displace hydrogen from cold water (Figure 4.13):0(b)FIGURE 4.13 Reactions of (a)sodium (Na) and (b) calcium (Ca)with cold water. Note that the reaction is more vigorous with Nathan with Ca.BackForward2Na(s)0Ca(s)1112H2O(l) 88n 2NaOH(aq)1212H2O(l) 88n Ca(OH)2(s)0H2(g)0H2(g)Less reactive metals, such as aluminum and iron, react with steam to give hydrogengas:Main MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website4.4OXIDATION-REDUCTION REACTIONS127FIGURE 4.14 Left to right:Reactions of iron (Fe), zinc (Zn),and magnesium (Mg) with hydrochloric acid to form hydrogengas and the metal chlorides(FeCl2, ZnCl2, MgCl2). The reactivity of these metals is reflectedin the rate of hydrogen gas evolution, which is slowest for theleast reactive metal, Fe, andfastest for the most reactivemetal, Mg.(a)(b)02Al(s)133H2O(g) 88n Al2O3(s)02Fe(s)(c)133H2O(g) 88n Fe2O3(s)03H2(g)03H2(g)Many metals, including those that do not react with water, are capable of displacing hydrogen from acids. For example, zinc (Zn) and magnesium (Mg) do not react with water but do react with hydrochloric acid, as follows:01Displace hydrogenfrom cold waterDisplace hydrogenfrom steamLiKBaCaNaMgAlZnCrFeCdCoNiSnPbHCuHgAgPtAuDisplace hydrogen from acidsZn(s)Forward10H2(g)22HCl(aq) 88n MgCl2(aq)0H2(g)The ionic equations are0Zn(s)0Mg(s)122H (aq) 88n Zn2 (aq)122H (aq) 88n Mg2 (aq)0H2(g)0H2(g)Figure 4.14 shows the reactions between hydrochloric acid (HCl) and iron (Fe), zinc(Zn), and magnesium (Mg). These reactions are used to prepare hydrogen gas in thelaboratory. Some metals, such as copper (Cu), silver (Ag), and gold (Au), do not displace hydrogen when placed in hydrochloric acid.FIGURE 4.15 The activity series for metals. The metals arearranged according to their ability to displace hydrogen from anacid or water. Li (lithium) is themost reactive metal, and Au(gold) is the least reactive.Back0Mg(s)22HCl(aq) 88n ZnCl2(aq)Main Menu2. Metal Displacement. A metal in a compound can be displaced by another metalin the elemental state. We have already seen examples of zinc replacing copper ionsand copper replacing silver ions (see p. 121). Reversing the roles of the metals wouldresult in no reaction. Thus copper metal will not displace zinc ions from zinc sulfate,and silver metal will not displace copper ions from copper nitrate.An easy way to predict whether a metal or hydrogen displacement reaction willactually occur is to refer to an activity series (sometimes called the electrochemical series), shown in Figure 4.15. Basically, an activity series is a convenient summary ofthe results of many possible displacement reactions similar to the ones already dis-TOCStudy Guide TOCTextbook WebsiteMHHE Website128REACTIONS IN AQUEOUS SOLUTIONFIGURE 4.16 (a) An aqueousKBr solution. (b) After chlorinegas has been bubbled throughthe solution, most of the bromideions are converted (oxidized) toliquid bromine, which is slightlysoluble in water.(a)(b)cussed. According to this series, any metal above hydrogen will displace it from water or from an acid, but metals below hydrogen will not react with either water or anacid. In fact, any species listed in the series will react with any species (in a compound)below it. For example, Zn is above Cu, so zinc metal will displace copper ions fromcopper sulfate.Metal displacement reactions find many applications in metallurgical processes,the goal of which is to separate pure metals from their ores. For example, vanadiumis obtained by treating vanadium(V) oxide with metallic calcium:V2O5(s)5Ca(l ) 88n 2V(l )5CaO(s)Similarly, titanium is obtained from titanium(IV) chloride according to the reactionTiCl4(g)2Mg(l ) 88n Ti(s)2MgCl2(l)In each case the metal that acts as the reducing agent lies above the metal that is reduced (that is, Ca is above V and Mg is above Ti) in the activity series. We will seemore examples of this type of reaction in Chapter 20.3. Halogen Displacement. Another activity series summarizes the halogens behavior in halogen displacement reactions:F2Cl2Br2I2The power of these elements as oxidizing agents decreases as we move down Group7A from fluorine to iodine, so molecular fluorine can replace chloride, bromide, andiodide ions in solution. In fact, molecular fluorine is so reactive that it also attacks water; thus these reactions cannot be carried out in aqueous solutions. On the other hand,molecular chlorine can displace bromide and iodide ions in aqueous solution as Figure4.16 shows. The displacement equations areBackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website4.40OXIDATION-REDUCTION REACTIONS1Cl2(g)102KBr(aq) 88n 2KCl(aq)01Cl2(g)129Br2(l)102NaI(aq) 88n 2NaCl(aq)I2(s)The ionic equations are01Cl2(g)011Cl2(g)02Br(aq) 88n 2Cl (aq)1Br2(l)02I (aq) 88n 2Cl (aq)I2(s)Molecular bromine, in turn, can displace iodide ion in solution:0Br2(l)1102I (aq) 88n 2Br (aq)I2(s)Reversing the roles of the halogens produces no reaction. Thus bromine cannot displace chloride ions, and iodine cannot displace bromide and chloride ions.The halogen displacement reactions have a direct industrial application. The halogens as a group are the most reactive of the nonmetallic elements. They are all strongoxidizing agents. As a result, they are found in nature in the combined state (with metals) as halides and never as free elements. Of these four elements, chlorine is by farthe most important industrial chemical. In 1995 the amount of chlorine produced was25 billion pounds, making chlorine the tenth-ranking industrial chemical. The annualproduction of bromine is only one-hundredth that of chlorine, while the amounts offluorine and iodine produced are even less.Recovering the halogens from their halides requires an oxidation process, whichis represented by2X 88n X2Bromine is a fuming red liquid.2ewhere X denotes a halogen element. Seawater and natural brine (for example, underground water in contact with salt deposits) are rich sources of Cl , Br , and I ions.Minerals such as fluorite (CaF2) and cryolite (Na3A1F6) are used to prepare fluorine.Because fluorine is the strongest oxidizing agent known, there is no way to convert Fions to F2 by chemical means. The only way to carry out the oxidation is by electrolyticmeans, the details of which will be discussed in Chapter 19. Industrially, chlorine, likefluorine, is produced electrolytically.Bromine is prepared industrially by oxidizing Br ions with chlorine, which is astrong enough oxidizing agent to oxidize Br ions but not water:2Br (aq) 88n Br2(l)2eOne of the richest sources of Br ions is the Dead Sea about 4000 parts per million(ppm) by mass of all dissolved substances in the Dead Sea is Br. Following the oxidation of Br ions, bromine is removed from the solution by blowing air over the solution, and the air-bromine mixture is then cooled to condense the bromine (Figure4.17).Iodine is also prepared from seawater and natural brine by the oxidation of Iions with chlorine. Because Br and I ions are invariably present in the same source,they are both oxidized by chlorine. However, it is relatively easy to separate Br2 fromBackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website130REACTIONS IN AQUEOUS SOLUTIONFIGURE 4.17 The industrialmanufacture of liquid bromine byoxidizing an aqueous solutioncontaining Br ions with chlorinegas.I2 because iodine is a solid that is sparingly soluble in water. The air-blowing procedure will remove most of the bromine formed but will not affect the iodine present.Disproportionation ReactionElements that are most likely toundergo disproportionation include N, P, O, S, Cl, Br, I, Mn,Cu, Au, and Hg.A special type of redox reaction is the disproportionation reaction. In a disproportionation reaction, an element in one oxidation state is simultaneously oxidized andreduced. One reactant in a disproportionation reaction always contains an element thatcan have at least three oxidation states. The reactant itself is in an intermediate oxidation state; that is, both higher and lower oxidation states exist for that element. The decomposition of hydrogen peroxide is an example of a disproportionation reaction:1202H2O2(aq) 88n 2H2O(l)O2(g)Here the oxidation number of oxygen in the reactant ( 1) both increases to zero in O2and decreases to 2 in H2O. Another example is the reaction between molecular chlorine and NaOH solution:01Cl2(g)12OH (aq) 88n ClO (aq)Cl (aq)H2O(l)This reaction describes the formation of household bleaching agents, for it is thehypochlorite ion (ClO ) that oxidizes the color-bearing substances in stains, converting them to colorless compounds.Finally, it is interesting to compare redox reactions and acid-base reactions. Theyare analogous in that acid-base reactions involve the transfer of protons while redoxreactions involve the transfer of electrons. However, while acid-base reactions are quiteeasy to recognize (since they always involve an acid and a base), there is no simpleprocedure for identifying a redox process. The only sure way is to compare the oxi-BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website4.5CONCENTRATION OF SOLUTIONS131dation numbers of all the elements in the reactants and products. Any change in oxidation number guarantees that the reaction is redox in nature.The classification of different types of redox reactions is illustrated in the following example.EXAMPLE 4.5Classify the following redox reactions and indicate changes in the oxidation numbers of the elements:(a) 2N2O(g) 88n 2N2(g) O2(g)(b) 6Li(s) N2(g) 88n 2Li3N(s)(c) Ni(s) Pb(NO3)2(aq) 88n Pb(s) Ni(NO3)2(aq)(d) 2NO2(g) H2O(l) 88n HNO2(aq) HNO3(aq)(a) This is a decomposition reaction since one type of reactant is convertedto two different types of products. The oxidation number of N changes from 1 to0 while that of O changes from 2 to 0.(b) This is a combination reaction (two reactants form a single product). The oxidation number of Li changes from 0 to 1 while that of N changes from 0 to 3.(c) This is a metal displacement reaction. The Ni metal replaces (reduces) the Pb2ion. The oxidation number of Ni increases from 0 to 2 while that of Pb decreasesfrom 2 to 0.(d) The oxidation number of N is 4 in NO2, 3 in HNO2, and 5 in HNO3. Sinceits oxidation number both increases and decreases, this is a disproportionation reaction.AnswerSimilar problems: 4.51, 4.52.PRACTICE EXERCISEIdentify the following redox reactions by type:(a) Fe H2SO4 88n FeSO4 H2(b) S 3F2 88n SF6(c) 2CuCl 88n Cu CuCl2(d) 2Ag PtCl2 88n 2AgCl Pt.The Chemistry in Action essay on p. 132 describes how law enforcement makesuse of a redox reaction to apprehend drunk drivers.4.5CONCENTRATION OF SOLUTIONSTo study solution stoichiometry we must know how much of the reactants are presentin a solution and also how to control the amounts of reactants used to bring about areaction in aqueous solution.The concentration of a solution is the amount of solute present in a given quantity of solvent or solution. (For this discussion we will assume the solute is a liquid ora solid and the solvent is a liquid.) The concentration of a solution can be expressedin many different ways, as we will see in Chapter 12. Here we will consider one ofthe most commonly used units in chemistry, molarity (M), or molar concentration,which is the number of moles of solute in 1 liter of solution. Molarity is defined by theequationBackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website132REACTIONS IN AQUEOUS SOLUTIONChemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry inChemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action ChemistryBreath AnalyzerEvery year in the United States about 25,000 peopleare killed and 500,000 more are injured as a resultof drunk driving. In spite of efforts to educate the public about the dangers of driving while intoxicated andstiffer penalties for drunk driving offenses, law enforcement agencies still have to devote a great dealof work to removing drunk drivers from Americasroads.The police often use a device called a breath analyzer to test drivers suspected of being drunk. Thechemical basis of this device is a redox reaction. Asample of the drivers breath is drawn into the breathanalyzer, where it is treated with an acidic solution ofpotassium dichromate. The alcohol (ethanol) in thebreath is converted to acetic acid as shown in the following equation:3CH3CH3OH2K2Cr2O7ethanolpotassiumdichromate(orange yellow)2Cr2(SO4)32K2SO4acetic acidchromium(III)sulfate(green)potassiumsulfatePhotocelldetectorK2Cr2O7solutionM11H2OIn this reaction the ethanol is oxidized to acetic acidand the chromium(VI) in the orange-yellow dichromateion is reduced to the green chromium(III) ion (seeFigure 4.22). The drivers blood alcohol level can bedetermined readily by measuring the degree of thiscolor change (read from a calibrated meter on the instrument). The current legal limit of blood alcohol content in the United States is 0.1 percent by mass.Anything higher constitutes intoxication.MeterFiltersulfuricacid3CH3COOHBreathLightsource8H2SO4 88nSchematic diagram ofa breath analyzer. Thealcohol in the driversbreath is reacted witha potassium dichromatesolution. The change inthe absorption of lightis registered by the detector and shown on ameter, which directlydisplays the alcoholcontent in blood. Thefilter selects only onewavelength of light formeasurement.moles of soluteliters of solnmolarity(4.1)where soln denotes solution. Thus, a 1.46 molar glucose (C6H12O6) solution, written 1.46 M C6H12O6, contains 1.46 moles of the solute (C6H12O6) in 1 liter of the solution; a 0.52 molar urea [(NH2)2CO] solution, written 0.52 M (NH2)2CO, contains0.52 mole of (NH2)2CO (the solute) in 1 liter of solution; and so on.Of course, we do not always work with solution volumes of exactly 1 liter (L).This is not a problem as long as we remember to convert the volume of the solutionto liters. Thus, a 500-mL solution containing 0.730 mole of C6H12O6 also has a concentration of 1.46 M:Mmolarity0.730 mol0.500 L1.46 mol/LBackForwardMain MenuTOCStudy Guide TOC1.46 MTextbook WebsiteMHHE Website4.5FIGURE 4.18 Preparing a solution of known molarity. (a) Aknown amount of a solid solute isput into the volumetric flask; thenwater is added through a funnel.(b) The solid is slowly dissolvedby gently shaking the flask. (c)After the solid has completely dissolved, more water is added tobring the level of solution to themark. Knowing the volume of thesolution and the amount of solutedissolved in it, we can calculatethe molarity of the prepared solution.133MeniscusMarker showingknown volumeof solution(a)Number of moles is given by volume (L) molarity.CONCENTRATION OF SOLUTIONS(b)(c)As you can see, the unit of molarity is moles per liter, so a 500-mL solution containing 0.730 mole of C6H12O6 is equivalent to 1.46 mol/L or 1.46 M. Note that concentration, like density, is an intensive property, so its value does not depend on how muchof the solution is present.It is important to keep in mind that molarity refers only to the amount of soluteoriginally dissolved in water and does not take into account any subsequent processes,such as the dissociation of a salt or the ionization of an acid. Both glucose and ureaare nonelectrolytes, so a 1.00 M urea solution has 1.00 mole of the urea molecules in1 liter of solution. But consider what happens when a sample of potassium chloride(KCl), a strong electrolyte, is dissolved in enough water to make a 1 M solution:HO2KCl(s) 88n K (aq)Cl (aq)Because KCl is a strong electrolyte, it undergoes complete dissociation in solution.Thus, a 1 M KCl solution contains 1 mole of K ions and 1 mole of Cl ions, and noKCl units are present. The concentrations of the ions can be expressed as [K ] 1 Mand [Cl ] 1 M, where the square brackets [ ] indicate that the concentration is expressed in molarity. Similarly, in a 1 M barium nitrate [Ba(NO3)2] solutionHO2Ba(NO3)2(s) 88n Ba2 (aq)2NO3 (aq)2In general, it is more convenientto measure the volume of a liquidthan to determine its mass.BackForwardMain Menuwe have [Ba ] 1 M and [NO3 ] 2 M and no Ba(NO3)2 units at all.The procedure for preparing a solution of known molarity is as follows. First, thesolute is accurately weighed and transferred to a volumetric flask through a funnel(Figure 4.18). Next, water is added to the flask, which is carefully swirled to dissolvethe solid. After all the solid has dissolved, more water is added slowly to bring thelevel of solution exactly to the volume mark. Knowing the volume of the solution inthe flask and the quantity of compound (the number of moles) dissolved, we can calculate the molarity of the solution using Equation (4.1). Note that this procedure doesnot require knowing the amount of water added, as long as the volume of the final solution is known.TOCStudy Guide TOCTextbook WebsiteMHHE Website134REACTIONS IN AQUEOUS SOLUTIONExample 4.6 illustrates how to prepare a solution of known molarity.EXAMPLE 4.6How many grams of potassium dichromate (K2Cr2O7) are required to prepare a 250mL solution whose concentration is 2.16 M?The first step is to determine the number of moles of K2Cr2O7 in 250 mLof a 2.16 M solution:Answermoles of K2Cr2O7250 mL soln2.16 mol K2Cr2O71000 mL soln0.540 mol K2Cr2O7A K2Cr2O7 solution.The molar mass of K2Cr2O7 is 294.2 g, so we writegrams of K2Cr2O7 needed0.540 mol K2Cr2O7294.2 g K2Cr2O71 mol K2Cr2O7159 g K2Cr2O7Similar problems: 4.58, 4.59.PRACTICE EXERCISEWhat is the molarity of an 85.0-mL ethanol (C2H5OH) solution containing 1.77 gof ethanol?DILUTION OF SOLUTIONSConcentrated solutions are often stored in the laboratory stockroom for use as needed.Frequently we dilute these stock solutions before working with them. Dilution is theprocedure for preparing a less concentrated solution from a more concentrated one.Suppose that we want to prepare 1 liter of a 0.400 M KMnO4 solution from a solution of 1.00 M KMnO4. For this purpose we need 0.400 mole of KMnO4. Since thereis 1.00 mole of KMnO4 in 1 liter, or 1000 mL, of a 1.00 M KMnO4 solution, there is0.400 mole of KMnO4 in 0.400 1000 mL, or 400 mL of the same solution:1.00 mol1000 mL solnTwo KMnO4 solutions of differentconcentrations.0.400 mol400 mL solnTherefore, we must withdraw 400 mL from the 1.00 M KMnO4 solution and dilute itto 1000 mL by adding water (in a 1-liter volumetric flask). This method gives us 1liter of the desired solution of 0.400 M KMnO4.In carrying out a dilution process, it is useful to remember that adding more solvent to a given amount of the stock solution changes (decreases) the concentration ofthe solution without changing the number of moles of solute present in the solution(Figure 4.19). In other words,moles of solute before dilutionmoles of solute after dilutionSince molarity is defined as moles of solute in one liter of solution, we see that thenumber of moles of solute is given byMBackForwardMain MenuTOCvolume of soln (in liters)moles of solutemoles of soluteliters of solnVStudy Guide TOCTextbook WebsiteMHHE Website4.5CONCENTRATION OF SOLUTIONS135FIGURE 4.19 The dilution of amore concentrated solution (a) toa less concentrated solution (b)does not change the total numberof moles of solute.(a)(b)orMVmoles of soluteBecause all the solute comes from the original stock solution, we can conclude thatMiViMfVfmoles of solutebefore dilutionmoles of soluteafter dilution(4.2)where Mi and Mf are the initial and final concentrations of the solution in molarity andVi and Vf are the initial and final volumes of the solution, respectively. Of course, theunits of Vi and Vf must be the same (mL or L) for the calculation to work. To checkthe reasonableness of your results, be sure that Mi Mf and Vf Vi.We apply Equation (4.2) in the following example.EXAMPLE 4.7Describe how you would prepare 5.00 102 mL of 1.75 M H2SO4 solution, starting with an 8.61 M stock solution of H2SO4.Since the concentration of the final solution is less than that of the original one, this is a dilution process. We prepare for the calculation by tabulating ourdata:AnswerMi8.61 MMf1.75 MVi?Vf5.00102 mL(1.75 M)(5.00102 mL)Substituting in Equation (4.2),(8.61 M)(Vi)(1.75 M)(5.00 102 mL)8.61 MVi102 mLSimilar problems: 4.67, 4.68.Thus we must dilute 102 mL of the 8.61 M H2SO4 solution with sufficient water togive a final volume of 5.00 102 mL in a 500-mL volumetric flask to obtain thedesired concentration.PRACTICE EXERCISEHow would you prepare 2.00a 5.07 M stock solution?BackForwardMain MenuTOC102 mL of a 0.866 M NaOH solution, starting withStudy Guide TOCTextbook WebsiteMHHE Website136REACTIONS IN AQUEOUS SOLUTIONNow that we have discussed the concentration and dilution of solutions, we canexamine the quantitative aspects of reactions in aqueous solution, or solution stoichiometry. Sections 4.64.8 focus on two techniques for studying solution stoichiometry: gravimetric analysis and titration. These techniques are important tools of quantitative analysis, which is the determination of the amount or concentration of asubstance in a sample.4.6GRAVIMETRIC ANALYSISGravimetric analysis is an analytical technique based on the measurement of mass.One type of gravimetric analysis experiment involves the formation, isolation, and massdetermination of a precipitate. Generally this procedure is applied to ionic compounds.First, a sample substance of unknown composition is dissolved in water and allowedto react with another substance to form a precipitate. Then the precipitate is filteredoff, dried, and weighed. Knowing the mass and chemical formula of the precipitateformed, we can calculate the mass of a particular chemical component (that is, the anion or cation) of the original sample. Finally, from the mass of the component and themass of the original sample, we can determine the percent composition by mass of thecomponent in the original compound.A reaction that is often studied in gravimetric analysis, because the reactants canbe obtained in pure form, isAgNO3(aq)NaCl(aq) 88n NaNO3(aq)AgCl(s)The net ionic equation isAg (aq)Cl (aq) 88n AgCl(s)The precipitate is silver chloride (see Table 4.2). As an example, let us say that wewanted to determine experimentally the percent by mass of Cl in NaCl. First we wouldaccurately weigh out a sample of NaCl and dissolve it in water. Next, we would addenough AgNO3 solution to the NaCl solution to cause the precipitation of all the Clions present in solution as AgCl. In this procedure NaCl is the limiting reagent andAgNO3 the excess reagent. The AgCl precipitate is separated from the solution by filtration, dried, and weighed. From the measured mass of AgCl, we can calculate themass of Cl using the percent by mass of Cl in AgCl. Since this same amount of Cl waspresent in the original NaCl sample, we can calculate the percent by mass of Cl inNaCl. Figure 4.20 shows how this procedure is performed.Gravimetric analysis is a highly accurate technique, since the mass of a samplecan be measured accurately. However, this procedure is applicable only to reactionsthat go to completion, or have nearly 100 percent yield. Thus, if AgCl were slightlysoluble instead of being insoluble, it would not be possible to remove all the Cl ionsfrom the NaCl solution and the subsequent calculation would be in error.The following example shows the calculations involved in a gravimetric experiment.EXAMPLE 4.8A 0.5662-g sample of an ionic compound containing chloride ions and an unknownmetal is dissolved in water and treated with an excess of AgNO3. If 1.0882 g ofAgCl precipitate forms, what is the percent by mass of Cl in the original compound?BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website4.6(a)GRAVIMETRIC ANALYSIS(b)137(c)FIGURE 4.20 Basic steps for gravimetric analysis. (a) A solution containing a known amount ofNaCl in a beaker. (b) The precipitation of AgCl upon the addition of AgNO3 solution from ameasuring cylinder. In this reaction, AgNO3 is the excess reagent and NaCl is the limitingreagent. (c) The solution containing the AgCl precipitate is filtered through a preweighed sintereddisk crucible, which allows the liquid (but not the precipitate) to pass through. The crucible is thenremoved from the apparatus, dried in an oven, and weighed again. The difference between thismass and that of the empty crucible gives the mass of the AgCl precipitate.The mass of Cl in AgCl (and hence in the original compound) is determined by multiplying the mass of AgCl by the percent by mass of Cl in AgCl:Answermass of CI1.0882 g AgCl1 mol AgCl143.4 g AgCl1 mol Cl1 mol AgCl35.45 g Cl1 mol Cl0.2690 g ClNext, we calculate the percent by mass of Cl in the unknown sample asmass of Clmass of sample%Cl by mass0.2690 g0.5662 g100%100%47.51%Similar problem: 4.74.CommentAs a comparison, calculate the percent by mass of Cl in KCl.PRACTICE EXERCISEA sample of 0.3220 g of an ionic compound containing the bromide ion (Br ) isdissolved in water and treated with an excess of AgNO3. If the mass of the AgBrprecipitate that forms is 0.6964 g, what is the percent by mass of Br in the originalcompound?Note that gravimetric analysis does not establish the whole identity of the unknown. Thus in Example 4.8 we still do not know what the cation is. However, know-BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website138REACTIONS IN AQUEOUS SOLUTIONing the percent by mass of Cl greatly helps us in narrowing the possibilities. Becauseno two compounds containing the same anion (or cation) have the same percent composition by mass, comparison of the percent by mass obtained from gravimetric analysis with that calculated from a series of known compounds would reveal the identityof the unknown.4.7ACID-BASE TITRATIONSQuantitative studies of acid-base neutralization reactions are most conveniently carriedout using a technique known as titration. In titration, a solution of accurately knownconcentration, called a standard solution, is added gradually to another solution ofunknown concentration, until the chemical reaction between the two solutions is complete. If we know the volumes of the standard and unknown solutions used in the titration, along with the concentration of the standard solution, we can calculate the concentration of the unknown solution.Sodium hydroxide is one of the bases commonly used in the laboratory. However,it is difficult to obtain solid sodium hydroxide in a pure form because it has a tendencyto absorb water from air, and its solution reacts with carbon dioxide. For these reasons,a solution of sodium hydroxide must be standardized before it can be used in accurateanalytical work. We can standardize the sodium hydroxide solution by titrating it againstan acid solution of accurately known concentration. The acid often chosen for this taskis a monoprotic acid called potassium hydrogen phthalate (KHP), for which the molecular formula is KHC8H4O4. KHP is a white, soluble solid that is commercially available in highly pure form. The reaction between KHP and sodium hydroxide isKHC8H4O4(aq)Potassium hydrogen phthalate.NaOH(aq) 88n KNaC8H4O4(aq)H2O(l)and the net ionic equation isHC8H4O4 (aq)OH (aq) 88n C8H4O2 (aq)4H2O(l)The procedure for the titration is shown in Figure 4.21. First, a known amount of KHPis transferred to an Erlenmeyer flask and some distilled water is added to make up asolution. Next, NaOH solution is carefully added to the KHP solution from a buret until we reach the equivalence point, that is, the point at which the acid has completelyreacted with or been neutralized by the base. The equivalence point is usually signaledby a sharp change in the color of an indicator in the acid solution. In acid-base titrations, indicators are substances that have distinctly different colors in acidic and basic media. One commonly used indicator is phenolphthalein, which is colorless in acidicand neutral solutions but reddish pink in basic solutions. At the equivalence point, allthe KHP present has been neutralized by the added NaOH and the solution is still colorless. However, if we add just one more drop of NaOH solution from the buret, thesolution will immediately turn pink because the solution is now basic. Example 4.9 illustrates such a titration.EXAMPLE 4.9In a titration experiment, a student finds that 0.5468 g of KHP is needed to completely neutralize 23.48 mL of a NaOH solution. What is the concentration (in molarity) of the NaOH solution?BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website4.7ACID-BASE TITRATIONS139FIGURE 4.21 (a) Apparatusfor acid-base titration. A NaOHsolution is added from the buretto a KHP solution in anErlenmeyer flask. (b) A reddishpink color appears when theequivalence point is reached. Thecolor here has been intensifiedfor visual display.(a)(b)The balanced equation for this reaction is shown above. First we calculate the number of moles of KHP used in the titration:Answermoles of KHP0.5468 g KHP2.6781031 mol KHP204.2 g KHPmol KHPSince 1 mol KHP 1 mol NaOH, there must be 2.678 10 3 mole of NaOH in23.48 mL of NaOH solution. Finally, we calculate the molarity of the NaOH as follows:molarity of NaOH soln2.678 10 3 mol NaOH23.48 mL soln1000 mL soln1 L soln0.1141 MSimilar problems: 4.79, 4.80.PRACTICE EXERCISEHow many grams of KHP are needed to neutralize 18.64 mL of a 0.1004 M NaOHsolution?The neutralization reaction between NaOH and KHP is one of the simplest typesof acid-base neutralization known. Suppose, though, that instead of KHP, we wantedto use a diprotic acid such as H2SO4 for the titration. The reaction is represented by2NaOH(aq)H2SO4(aq) 88n Na2SO4(aq)2H2O(l)Since 2 mol NaOH 1 mol H2SO4, we need twice as much NaOH to react completelywith a H2SO4 solution of the same molar concentration and volume as a KHP solu-BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website140REACTIONS IN AQUEOUS SOLUTIONtion. On the other hand, we would need twice the amount of HCl to neutralize a Ba(OH)2solution compared to a NaOH solution having the same concentration and volume because 1 mole of Ba(OH)2 yields two moles of OH ions:2HCl(aq)Ba(OH)2(aq) 88n BaCl2(aq)2H2O(l)In calculations involving acid-base titrations, regardless of the acid or base that takespart in the reaction, keep in mind that the total number of moles of H ions that havereacted at the equivalence point must be equal to the total number of moles of OHions that have reacted. The number of moles of an acid in a certain volume is givenbymoles of acidmolarity (mol/L)volume (L)MVwhere M is the molarity and V the volume in liters. A similar expression can be written for a base.The following example shows the titration of a NaOH solution with a diproticacid.EXAMPLE 4.10How many milliliters (mL) of a 0.610 M NaOH solution are needed to completelyneutralize 20.0 mL of a 0.245 M H2SO4 solution?The equation for the neutralization reaction is shown on p. 139. First wecalculate the number of moles of H2SO4 in a 20.0-mL solution:Answermoles H2SO40.245 mol H2SO41 L soln4.901031 L soln1000 mL soln20.0 mL solnmol H2SO4From the stoichiometry we see that 1 mol H2SO42 mol NaOH. Therefore, thenumber of moles of NaOH reacted must be 2 4.90 10 3 mole, or 9.80 10 3mole. From the definition of molarity [see Equation (4.1)] we havemoles of solutemolarityliters of solnorvolume of NaOH9.8010 3 mol NaOH0.610 mol/L0.0161 L or 16.1 mLSimilar problems: 4.79, 4.80.PRACTICE EXERCISEHow many milliliters of a 1.28 M H2SO4 solution are needed to neutralize 60.2 mLof a 0.427 M KOH solution?4.8REDOX TITRATIONSAs mentioned earlier, redox reactions involve the transfer of electrons, and acid-basereactions involve the transfer of protons. Just as an acid can be titrated against a base,we can titrate an oxidizing agent against a reducing agent, using a similar procedure.BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website4.8FIGURE 4.22There are not as many redox indicators as there are acid-base indicators.REDOX TITRATIONSLeft to right: Solutions containing the MnO4 , Mn2 , Cr2O2 , and Cr37141ions.We can, for example, carefully add a solution containing an oxidizing agent to a solution containing a reducing agent. The equivalence point is reached when the reducingagent is completely oxidized by the oxidizing agent.Like acid-base titrations, redox titrations normally require an indicator that clearlychanges color. In the presence of large amounts of reducing agent, the color of the indicator is characteristic of its reduced form. The indicator assumes the color of its oxidized form when it is present in an oxidizing medium. At or near the equivalencepoint, a sharp change in the indicator s color will occur as it changes from one formto the other, so the equivalence point can be readily identified.Two common oxidizing agents are potassium dichromate (K2Cr2O7) and potassium permanganate (KMnO4). As Figure 4.22 shows, the colors of the dichromate andpermanganate anions are distinctly different from those of the reduced species:Cr2O2 88n Cr37orangeyellowgreenMnO4 88n Mn2purplelightpinkThus these oxidizing agents can themselves be used as an internal indicator in a redoxtitration because they have distinctly different colors in the oxidized and reduced forms.Redox titrations require the same type of calculations (based on the mole method)as acid-base neutralizations. The difference is that the equations and the stoichiometrytend to be more complex for redox reactions. The following is an example of a redoxtitration.EXAMPLE 4.11A 16.42-mL volume of 0.1327 M KMnO4 solution is needed to oxidize 20.00 mLof a FeSO4 solution in an acidic medium. What is the concentration of the FeSO4solution? The net ionic equation isAddition of a KMnO4 solutionfrom a buret to a FeSO4 solution.BackForwardMain Menu5Fe2TOCMnO48H 88n Mn2Study Guide TOC5Fe34H2OTextbook WebsiteMHHE Website142REACTIONS IN AQUEOUS SOLUTIONAnswerThe number of moles of KMnO4 in 16.42 mL of the solution ismoles of KMnO40.1327 mol KMnO41 L soln16.42 mL2.179103mol KMnO4From the equation we see that 5 mol Fe2of moles of FeSO4 oxidized is1 mol MnO4 . Therefore, the number2.1791031025 mol FeSO41 mol KMnO4mol KMnO41.090moles FeSO41 L soln1000 mL solnmol FeSO4The concentration of the FeSO4 solution in moles of FeSO4 per liter of solution is[FeSO4]1.09010 2 mol FeSO420.00 mL soln1000 mL soln1 L soln0.5450 MSimilar problems: 4.83, 4.84.PRACTICE EXERCISEHow many milliliters of a 0.206 M HI solution are needed to reduce 22.5 mL of a0.374 M KMnO4 solution according to the following equation:10HI2KMnO43H2SO4 88n 5I22MnSO4K2SO48H2OThe Chemistry in Action essay on p. 143 describes an industrial process that involves the types of reactions discussed in this chapter.SUMMARY OFKEY EQUATIONS molarity (M) MiViSUMMARY OF FACTSAND CONCEPTSBackForwardmoles of soluteliters of solnMfVf (4.2)(4.1)Calculating molarity.Dilution of solution.1. Aqueous solutions are electrically conducting if the solutes are electrolytes. If the solutesare nonelectrolytes, the solutions do not conduct electricity.2. Three major categories of chemical reactions that take place in aqueous solution are precipitation reactions, acid-base reactions, and oxidation-reduction reactions.3. From general rules about solubilities of ionic compounds, we can predict whether a precipitate will form in a reaction.4. Arrhenius acids ionize in water to give H ions, and Arrhenius bases ionize in water togive OH ions. Brnsted acids donate protons, and Brnsted bases accept protons.5. The reaction of an acid and a base is called neutralization.6. In redox reactions, oxidation and reduction always occur simultaneously. Oxidation ischaracterized by the loss of electrons, reduction by the gain of electrons.7. Oxidation numbers help us keep track of charge distribution and are assigned to all atomsin a compound or ion according to specific rules. Oxidation can be defined as an increasein oxidation number; reduction can be defined as a decrease in oxidation number.8. Many redox reactions can be subclassified as combination, decomposition, displacement,or disproportionation reactions.9. The concentration of a solution is the amount of solute present in a given amount of solution. Molarity expresses concentration as the number of moles of solute in 1 liter of solution.Main MenuTOCStudy Guide TOCTextbook WebsiteMHHE WebsiteSUMMARY OF FACTS AND CONCEPTS143Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in ActionChemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action ChemistryMetal from the SeaMagnesium is a valuable, lightweight metal used asa structural material as well as in alloys, in batteries,and in chemical synthesis. Although magnesium isplentiful in Earths crust, it is cheaper to mine themetal from seawater. Magnesium forms the secondmost abundant cation in the sea (after sodium); thereare about 1.3 g of magnesium in a kilogram of seawater. The process for obtaining magnesium from seawater employs all three types of reactions discussedin this chapter: precipitation, acid-base, and redox reactions.In the first stage in the recovery of magnesium,limestone (CaCO3) is heated at high temperatures toproduce quicklime, or calcium oxide (CaO):CaCO3(s) 88n CaO(s)CO2(g)When calcium oxide is treated with seawater, it formscalcium hydroxide [Ca(OH)2], which is slightly soluble and ionizes to give Ca2 and OH ions:CaO(s)H2O(l ) 88n Ca2 (aq)2OH (aq)Magnesium hydroxide is obtained from seawater in settlingponds at the Dow Chemical Company in Freeport, Texas.After the water is evaporated, the solid magnesiumchloride is melted in a steel cell. The molten magnesium chloride contains both Mg2 and Cl ions. In aprocess called electrolysis, an electric current is passedthrough the cell to reduce the Mg2 ions and oxidizethe Cl ions. The half-reactions areThe surplus hydroxide ions cause the much less soluble magnesium hydroxide to precipitate:Mg2 (aq)2OH (aq) 88n Mg(OH)2(s)The solid magnesium hydroxide is filtered and reactedwith hydrochloric acid to form magnesium chloride(MgCl2):Mg(OH)2(s)2HCl(aq) 88n MgCl2(aq)2H2O(l )Mg22e 88n Mg2Cl 88n Cl22eThe overall reaction isMgCl2(l ) 88n Mg(l )Cl2(g)This is how magnesium metal is produced. The chlorine gas generated can be converted to hydrochloricacid and recycled through the process.10. Adding a solvent to a solution, a process known as dilution, decreases the concentration(molarity) of the solution without changing the total number of moles of solute present inthe solution.11. Gravimetric analysis is a technique for determining the identity of a compound and/or theconcentration of a solution by measuring mass. Gravimetric experiments often involveprecipitation reactions.12. In acid-base titration, a solution of known concentration (say, a base) is added graduallyto a solution of unknown concentration (say, an acid) with the goal of determining the unknown concentration. The point at which the reaction in the titration is complete is calledthe equivalence point.13. Redox titrations are similar to acid-base titrations. The point at which the oxidationreduction reaction is complete is called the equivalence point.BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website144REACTIONS IN AQUEOUS SOLUTIONKEY WORDSActivity series, p. 127Aqueous solution, p. 110Brnsted acid, p. 117Brnsted base, p. 117Combustion reaction, p. 125Concentration of a solution,p. 131Decomposition reaction,p. 125Dilution, p. 134Diprotic acid, p. 118Displacement reaction, p. 126Disproportionation reaction,p. 130Electrolyte, p. 110Equivalence point, p. 138Gravimetric analysis, p. 136Half-reaction, p. 120Hydration, p. 111Hydronium ion, p. 117Indicator, p. 138Ionic equation, p. 114Molar concentration, p. 131Molarity (M), p. 131Molecular equation, p. 113Monoprotic acid, p. 117Net ionic equation, p. 114Neutralization reaction, p. 119Nonelectrolyte, p. 110Oxidation number, p. 122Oxidation state, p. 122Oxidation reaction, p. 121Oxidation-reduction reaction,p. 120Oxidizing agent, p. 121Precipitate, p. 112Precipitation reaction, p. 112Quantitative analysis, p. 136Redox reaction, p. 120Reducing agent, p. 121Reduction reaction, p. 121Reversible reaction, p. 112Salt, p. 119Solubility, p. 112Solute, p. 110Solution, p. 110Solvent, p. 110Spectator ion, p. 114Standard solution, p. 138Titration, p. 138Triprotic acid, p. 118QUESTIONS AND PROBLEMSPROPERTIES OF AQUEOUS SOLUTIONSReview Questions4.1 Define solute, solvent, and solution by describing theprocess of dissolving a solid in a liquid.4.2 What is the difference between a nonelectrolyte andan electrolyte? Between a weak electrolyte and astrong electrolyte?4.3 Describe hydration. What properties of water enableits molecules to interact with ions in solution?4.4 What is the difference between the following symbols in chemical equations: 88n and 34 ?4.5 Water is an extremely weak electrolyte and thereforecannot conduct electricity. Why are we often cautioned not to operate electrical appliances when ourhands are wet.4.6 Lithium fluoride (LiF) is a strong electrolyte. Whatspecies are present in LiF(aq)?Problems4.7 Identify each of the following substances as a strongelectrolyte, weak electrolyte, or nonelectrolyte: (a)H2O, (b) KCl, (c) HNO3, (d) CH3COOH, (e)C12H22O11.4.8 Identify each of the following substances as a strongelectrolyte, weak electrolyte, or nonelectrolyte: (a)Ba(NO3)2, (b) Ne, (c) NH3, (d) NaOH.4.9 The passage of electricity through an electrolyte solution is caused by the movement of (a) electronsonly, (b) cations only, (c) anions only, (d) both cationsand anions.4.10 Predict and explain which of the following systemsare electrically conducting: (a) solid NaCl, (b) moltenNaCl, (c) an aqueous solution of NaCl.4.11 You are given a water-soluble compound X. DescribeBackForwardMain MenuTOChow you would determine whether it is an electrolyteor a nonelectrolyte. If it is an electrolyte, how wouldyou determine whether it is strong or weak?4.12 Explain why a solution of HCl in benzene does notconduct electricity but in water it does.PRECIPITATION REACTIONSReview Questions4.13 What is the difference between an ionic equation anda molecular equation?4.14 What is the advantage of writing net ionic equations?Problems4.15 Characterize the following compounds as soluble orinsoluble in water: (a) Ca3(PO4)2, (b) Mn(OH)2, (c)AgClO3, (d) K2S.4.16 Characterize the following compounds as solubleor insoluble in water: (a) CaCO3, (b) ZnSO4,(c) Hg(NO3)2, (d) HgSO4, (e) NH4ClO4.4.17 Write ionic and net ionic equations for the followingreactions:(a) 2AgNO3(aq) Na2SO4(aq) 88n(b) BaCl2(aq) ZnSO4(aq) 88n(c) (NH4)2CO3(aq) CaCl2(aq) 88n4.18 Write ionic and net ionic equations for the followingreactions:(a) Na2S(aq) ZnCl2(aq) 88n(b) 2K3PO4(aq) 3Sr(NO3)2(aq) 88n(c) Mg(NO3)2(aq) 2NaOH(aq) 88n4.19 Which of the following processes will likely result ina precipitation reaction? (a) Mixing a NaNO3 solution with a CuSO4 solution. (b) Mixing a BaCl2 solution with a K2SO4 solution. Write a net ionic equation for the precipitation reaction.Study Guide TOCTextbook WebsiteMHHE WebsiteQUESTIONS AND PROBLEMS4.20 With reference to Table 4.2, suggest one method bywhich you might separate (a) K from Ag , (b) Agfrom Pb2 , (c) NH4 from Ca2 , (d) Ba2 from Cu2 .All cations are assumed to be in aqueous solution,and the common anion is the nitrate ion.ACID-BASE REACTIONSReview Questions4.21 List the general properties of acids and bases.4.22 Give Arrheniuss and Brnsteds definitions of anacid and a base. Why are Brnsteds definitions moreuseful in describing acid-base properties?4.23 Give an example of a monoprotic acid, a diprotic acid,and a triprotic acid.4.24 What are the characteristics of an acid-base neutralization reaction?4.25 What factors qualify a compound as a salt? Specifywhich of the following compounds are salts: CH4,NaF, NaOH, CaO, BaSO4, HNO3, NH3, KBr?4.26 Identify the following as a weak or strong acid orbase: (a) NH3, (b) H3PO4, (c) LiOH, (d) HCOOH(formic acid), (e) H2SO4, (f) HF, (g) Ba(OH)2.Problems4.27 Identify each of the following species as a Brnstedacid, base, or both: (a) HI, (b) CH3COO , (c) H2PO4 ,(d) HSO4 .4.28 Identify each of the following species as a Brnstedacid, base, or both: (a) PO3 , (b) ClO2 , (c) NH4 ,4(d) HCO3 .4.29 Balance the following equations and write the corresponding ionic and net ionic equations (if appropriate):(a) HBr(aq) NH3(aq) 88n(HBr is a strong acid)(b) Ba(OH)2(aq) H3PO4(aq) 88n(c) HClO4(aq) Mg(OH)2(s) 88n4.30 Balance the following equations and write the corresponding ionic and net ionic equations (if appropriate):(a) CH3COOH(aq) KOH(aq) 88n(b) H2CO3(aq) NaOH(aq) 88n(c) HNO3(aq) Ba(OH)2(aq) 88nOXIDATION-REDUCTION REACTIONSReview Questions4.31 Give an example of a combination redox reaction, adecomposition redox reaction, and a displacement redox reaction.4.32 All combustion reactions are redox reactions. True orfalse? Explain.4.33 What is an oxidation number? How is it used to identify redox reactions? Explain why, except for ionicBackForwardMain MenuTOC145compounds, oxidation number does not have anyphysical significance.4.34 (a) Without referring to Figure 4.10, give the oxidation numbers of the alkali and alkaline earth metalsin their compounds. (b) Give the highest oxidationnumbers that the Groups 3A7A elements can have.4.35 How is the activity series organized? How is it usedin the study of redox reactions?4.36 Use the following reaction to define redox reaction,half-reaction, oxidizing agent, reducing agent:4Na(s)O2(g) 88n 2Na2O(s)4.37 Is it possible to have a reaction in which oxidationoccurs and reduction does not? Explain.4.38 What is the requirement for an element to undergodisproportionation reactions? Name five common elements that are likely to take part in such reactions.Problems4.39 For the complete redox reactions given below, (i)break down each reaction into its half-reactions; (ii)identify the oxidizing agent; (iii) identify the reducing agent.(a) 2Sr O2 88n 2SrO(b) 2Li H2 88n 2LiH(c) 2Cs Br2 88n 2CsBr(d) 3Mg N2 88n Mg3N24.40 For the complete redox reactions given below, writethe half-reactions and identify the oxidizing and reducing agents:(a) 4Fe 3O2 88n 2Fe2O3(b) Cl2 2NaBr 88n 2NaCl Br2(c) Si 2F2 88n SiF4(d) H2 Cl2 88n 2HCl4.41 Arrange the following species in order of increasingoxidation number of the sulfur atom: (a) H2S, (b) S8,(c) H2SO4, (d) S2 , (e) HS , (f) SO2, (g) SO3.4.42 Phosphorus forms many oxoacids. Indicate the oxidation number of phosphorus in each of the following acids: (a) HPO3, (b) H3PO2, (c) H3PO3, (d)H3PO4, (e) H4P2O7, (f) H5P3O10.4.43 Give the oxidation number of the underlined atomsin the following molecules and ions: (a) ClF, (b) IF7,(c) CH4, (d) C2H2, (e) C2H4, (f) K2CrO4, (g)K2Cr2O7, (h) KMnO4, (i) NaHCO3, (j) Li2, (k)NaIO3, (l) KO2, (m) PF6 , (n) KAuCl4.4.44 Give the oxidation number for the following species:H2, Se8, P4, O, U, As4, B12.4.45 Give oxidation numbers for the underlined atoms inthe following molecules and ions: (a) Cs2O, (b) CaI2,(c) Al2O3, (d) H3AsO3, (e) TiO2, (f) MoO2 , (g)4PtCl2 , (h) PtCl2 , (i) SnF2, (j) ClF3, (k) SbF6 .464.46 Give the oxidation numbers of the underlined atomsStudy Guide TOCTextbook WebsiteMHHE Website146REACTIONS IN AQUEOUS SOLUTION4.474.484.494.504.514.52in the following molecules and ions: (a) Mg3N2, (b)CsO2, (c) CaC2, (d) CO2 , (e) C2O2 , (f) ZnO2 , (g)342NaBH4, (h) WO2 .4Nitric acid is a strong oxidizing agent. State whichof the following species is least likely to be producedwhen nitric acid reacts with a strong reducing agentsuch as zinc metal, and explain why: N2O, NO, NO2,N2O4, N2O5, NH4 .Which of the following metals can react with water?(a) Au, (b) Li, (c) Hg, (d) Ca, (e) Pt.On the basis of oxidation number considerations, oneof the following oxides would not react with molecular oxygen: NO, N2O, SO2, SO3, P4O6. Which oneis it? Why?Predict the outcome of the reactions represented bythe following equations by using the activity series,and balance the equations.(a) Cu(s) HCl(aq) 88n(b) I2(s) NaBr(aq) 88n(c) Mg(s) CuSO4(aq) 88n(d) Cl2(g) KBr(aq) 88nClassify the following redox reactions:(a) 2H2O2 88n 2H2O O2(b) Mg 2AgNO3 88n Mg(NO3)2 2Ag(c) NH4NO2 88n N2 2H2O(d) H2 Br2 88n 2HBrClassify the following redox reactions:(a) P4 10Cl2 88n 4PCl5(b) 2NO 88n N2 O2(c) Cl2 2KI 88n 2KCl I2(d) 3HNO2 88n HNO3 H2O 2NOCONCENTRATION OF SOLUTIONSReview Questions4.53 Write the equation for calculating molarity. Why ismolarity a convenient concentration unit in chemistry?4.54 Describe the steps involved in preparing a solution ofknown molar concentration using a volumetric flask.Problems4.55 Calculate the mass of KI in grams required to prepare 5.00 102 mL of a 2.80 M solution.4.56 A quantity of 5.25 g of NaNO3 is dissolved in a sufficient amount of water to make up exactly 1 liter ofsolution. What is the molarity of the solution?4.57 How many moles of MgCl2 are present in 60.0 mLof 0.100 M MgCl2 solution?4.58 How many grams of KOH are present in 35.0 mL ofa 5.50 M solution?4.59 Calculate the molarity of each of the following solutions: (a) 29.0 g of ethanol (C2H5OH) in 545 mL ofsolution, (b) 15.4 g of sucrose (C12H22O11) in 74.0BackForwardMain MenuTOCmL of solution, (c) 9.00 g of sodium chloride (NaCl)in 86.4 mL of solution.4.60 Calculate the molarity of each of the following solutions: (a) 6.57 g of methanol (CH3OH) in 1.50 102mL of solution, (b) 10.4 g of calcium chloride (CaCl2)in 2.20 102 mL of solution, (c) 7.82 g of naphthalene (C10H8) in 85.2 mL of benzene solution.4.61 Calculate the volume in mL of a solution required toprovide the following: (a) 2.14 g of sodium chloridefrom a 0.270 M solution, (b) 4.30 g of ethanol froma 1.50 M solution, (c) 0.85 g of acetic acid(CH3COOH) from a 0.30 M solution.4.62 Determine how many grams of each of the followingsolutes would be needed to make 2.50 102 mL ofa 0.100 M solution: (a) cesium iodide (CsI), (b) sulfuric acid (H2SO4), (c) sodium carbonate (Na2CO3),(d) potassium dichromate (K2Cr2O7), (e) potassiumpermanganate (KMnO4).DILUTION OF SOLUTIONSReview Questions4.63 Describe the basic steps involved in diluting a solution of known concentration.4.64 Write the equation that enables us to calculate theconcentration of a diluted solution. Give units for allthe terms.Problems4.65 Describe how to prepare 1.00 L of 0.646 M HCl solution, starting with a 2.00 M HCl solution.4.66 Water is added to 25.0 mL of a 0.866 M KNO3 solution until the volume of the solution is exactly 500mL. What is the concentration of the final solution?4.67 How would you prepare 60.0 mL of 0.200 M HNO3from a stock solution of 4.00 M HNO3?4.68 You have 505 mL of a 0.125 M HCl solution and youwant to dilute it to exactly 0.100 M. How much water should you add?4.69 A 35.2-mL, 1.66 M KMnO4 solution is mixed with16.7 mL of 0.892 M KMnO4 solution. Calculate theconcentration of the final solution.4.70 A 46.2-mL, 0.568 M calcium nitrate [Ca(NO3)2] solution is mixed with 80.5 mL of 1.396 M calcium nitrate solution. Calculate the concentration of the final solution.GRAVIMETRIC ANALYSISReview Questions4.71 Describe the basic steps involved in gravimetricanalysis. How does this procedure help us determinethe identity of a compound or the purity of a compound if its formula is known?Study Guide TOCTextbook WebsiteMHHE Website147QUESTIONS AND PROBLEMS4.72 Distilled water must be used in the gravimetric analysis of chlorides. Why?ProblemsProblems4.73 If 30.0 mL of 0.150 M CaCl2 is added to 15.0 mL of0.100 M AgNO3, what is the mass in grams of AgClprecipitate?4.74 A sample of 0.6760 g of an unknown compound containing barium ions (Ba2 ) is dissolved in water andtreated with an excess of Na2SO4. If the mass of theBaSO4 precipitate formed is 0.4105 g, what is thepercent by mass of Ba in the original unknown compound?4.75 How many grams of NaCl are required to precipitatemost of the Ag ions from 2.50 102 mL of 0.0113M AgNO3 solution? Write the net ionic equation forthe reaction.4.76 The concentration of Cu2 ions in the water (whichalso contains sulfate ions) discharged from a certainindustrial plant is determined by adding excesssodium sulfide (Na2S) solution to 0.800 liter of thewater. The molecular equation isNa2S(aq)CuSO4(aq) 88n Na2SO4(aq)CuS(s)Write the net ionic equation and calculate the molarconcentration of Cu2 in the water sample if 0.0177g of solid CuS is formed.ACID-BASE TITRATIONSReview Questions4.77 Describe the basic steps involved in an acid-base titration. Why is this technique of great practical value?4.78 How does an acid-base indicator work?Problems4.79 Calculate the volume in mL of a 1.420 M NaOH solution required to titrate the following solutions:(a) 25.00 mL of a 2.430 M HCl solution(b) 25.00 mL of a 4.500 M H2SO4 solution(c) 25.00 mL of a 1.500 M H3PO4 solution4.80 What volume of a 0.50 M HCl solution is needed toneutralize completely each of the following:(a) 10.0 mL of a 0.30 M NaOH solution(b) 10.0 mL of a 0.20 M Ba(OH)2 solutionREDOX TITRATIONSReview Questions4.81 What are the similarities and differences betweenacid-base titrations and redox titrations?4.82 Explain why potassium permanganate (KMnO4) andBackForwardpotassium dichromate (K2Cr2O7) can serve as internal indicators in redox titrations.Main MenuTOC4.83 Iron(II) can be oxidized by an acidic K2Cr2O7 solution according to the net ionic equation:Cr2O276Fe214H 88n2Cr36Fe37H2OIf it takes 26.0 mL of 0.0250 M K2Cr2O7 to titrate25.0 mL of a solution containing Fe2 , what is themolar concentration of Fe2 ?4.84 The SO2 present in air is mainly responsible for theacid rain phenomenon. Its concentration can be determined by titrating against a standard permanganatesolution as follows:5SO22MnO42H2O 88n5SO242Mn24HCalculate the number of grams of SO2 in a sampleof air if 7.37 mL of 0.00800 M KMnO4 solution arerequired for the titration.4.85 A sample of iron ore weighing 0.2792 g was dissolvedin dilute acid solution, and all the Fe(II) was converted to Fe(III) ions. The solution required 23.30 mLof 0.0194 M K2Cr2O7 for titration. Calculate the percent by mass of iron in the ore. (Hint: See Problem4.83 for the balanced equation.)4.86 The concentration of a hydrogen peroxide solutioncan be conveniently determined by titration against astandardized potassium permanganate solution in anacidic medium according to the following equation:2MnO45H2O26H 88n 5O22Mn28H2OIf 36.44 mL of a 0.01652 M KMnO4 solution are required to completely oxidize 25.00 mL of a H2O2 solution, calculate the molarity of the H2O2 solution.4.87 Iodate ion, IO3 , oxidizes SO2 in acidic solution.3The half-reaction for the oxidation isSO23H2O 88n SO242H2eA 100.0-mL sample of solution containing 1.390 g ofKIO3 reacts with 32.5 mL of 0.500 M Na2SO3. Whatis the final oxidation state of the iodine after the reaction has occurred?4.88 Oxalic acid (H2C2O4) is present in many plants andvegetables. If 24.0 mL of 0.0100 M KMnO4 solutionis needed to titrate 1.00 g of H2C2O4 to the equivalence point, what is the percent by mass of H2C2O4in the sample? The net ionic equation is2MnO4Study Guide TOC16H5C2O2 88n42Mn2Textbook Website10CO28H2OMHHE Website148REACTIONS IN AQUEOUS SOLUTION4.89 A quantity of 25.0 mL of a solution containing bothFe2 and Fe3 ions is titrated with 23.0 mL of 0.0200M KMnO4 (in dilute sulfuric acid). As a result, all ofthe Fe2 ions are oxidized to Fe3 ions. Next, the solution is treated with Zn metal to convert all of theFe3 ions to Fe2 ions. Finally, the solution containing only the Fe2 ions requires 40.0 mL of thesame KMnO4 solution for oxidation to Fe3 .Calculate the molar concentrations of Fe2 and Fe3in the original solution. The net ionic equation isMnO45Fe28H 88nMn25Fe34H2O4.90 Calcium oxalate (CaC2O4) is insoluble in water. Forthis reason it can be used to determine the amount ofCa2 ions in fluids such as blood. The calcium oxalate isolated from blood is dissolved in acid andtitrated against a standardized KMnO4 solution asshown in Problem 4.88. In one test it is found thatthe calcium oxalate isolated from a 10.0-mL sampleof blood requires 24.2 mL of 9.56 10 4 M KMnO4for titration. Calculate the number of milligrams ofcalcium per milliliter of blood.ADDITIONAL PROBLEMS4.91 Classify the following reactions according to thetypes discussed in the chapter:ClOH2O(a) Cl2 2OH 88n ClCO2 88n CaCO3(b) Ca23(c) NH3 H 88n NH4(d) 2CCl4 CrO2 88n 2COCl24CrO2Cl2 2Cl(e) Ca F2 88n CaF2(f ) 2Li H2 88n 2LiHNa2SO4 88n 2NaNO3BaSO4(g) Ba(NO3)2(h) CuO H2 88n Cu H2O(i) Zn 2HCl 88n ZnCl2 H2(j) 2FeCl2 Cl2 88n 2FeCl34.92 Oxygen (O2) and carbon dioxide (CO2) are colorlessand odorless gases. Suggest two chemical tests thatwould allow you to distinguish between these twogases.4.93 Which of the following aqueous solutions would youexpect to be the best conductor of electricity at 25 C?Explain your answer.(a) 0.20 M NaCl(b) 0.60 M CH3COOH(c) 0.25 M HCl(d) 0.20 M Mg(NO3)24.94 A 5.00 102-mL sample of 2.00 M HCl solution istreated with 4.47 g of magnesium. Calculate the concentration of the acid solution after all the metal hasreacted. Assume that the volume remains unchanged.BackForwardMain MenuTOC4.95 Calculate the volume of a 0.156 M CuSO4 solutionthat would react with 7.89 g of zinc.4.96 Sodium carbonate (Na2CO3) is available in very pureform and can be used to standardize acid solutions.What is the molarity of a HCl solution if 28.3 mL ofthe solution are required to react with 0.256 g ofNa2CO3?4.97 A 3.664-g sample of a monoprotic acid was dissolvedin water. It took 20.27 mL of a 0.1578 M NaOH solution to neutralize the acid. Calculate the molar massof the acid.4.98 Acetic acid (CH3COOH) is an important ingredientof vinegar. A sample of 50.0 mL of a commercialvinegar is titrated against a 1.00 M NaOH solution.What is the concentration (in M) of acetic acid present in the vinegar if 5.75 mL of the base are neededfor the titration?4.99 A 15.00-mL solution of potassium nitrate (KNO3)was diluted to 125.0 mL, and 25.00 mL of this solution were then diluted to 1.000 103 mL. The concentration of the final solution is 0.00383 M.Calculate the concentration of the original solution.4.100 When 2.50 g of a zinc strip were placed in a AgNO3solution, silver metal formed on the surface of thestrip. After some time had passed, the strip was removed from the solution, dried, and weighed. If themass of the strip was 3.37 g, calculate the mass ofAg and Zn metals present.4.101 Calculate the mass of the precipitate formed when2.27 L of 0.0820 M Ba(OH)2 are mixed with 3.06 Lof 0.0664 M Na2SO4.4.102 Calculate the concentration of the acid (or base) remaining in solution when 10.7 mL of 0.211 M HNO3are added to 16.3 mL of 0.258 M NaOH.4.103 Milk of magnesia is an aqueous suspension of magnesium hydroxide [Mg(OH)2] used to treat acid indigestion. Calculate the volume of a 0.035 M HCl solution (a typical acid concentration in an upsetstomach) needed to react with two spoonfuls (approximately 10 mL) of milk of magnesia [at 0.080 gMg(OH)2/mL].4.104 A 1.00-g sample of a metal X (that is known to formX2 ions) was added to 0.100 liter of 0.500 M H2SO4.After all the metal had reacted, the remaining acid required 0.0334 L of 0.500 M NaOH solution for neutralization. Calculate the molar mass of the metal andidentify the element.4.105 A quantitative definition of solubility is the numberof grams of a solute that will dissolve in a given volume of water at a particular temperature. Describe anexperiment that would enable you to determine thesolubility of a soluble compound.4.106 A 60.0-mL 0.513 M glucose (C6H12O6) solution ismixed with 120.0 mL of 2.33 M glucose solution.Study Guide TOCTextbook WebsiteMHHE WebsiteQUESTIONS AND PROBLEMS4.1074.1084.1094.1104.1114.1124.1134.1144.1154.1164.1174.118BackWhat is the concentration of the final solution?Assume the volumes are additive.An ionic compound X is only slightly soluble in water. What test would you employ to show that thecompound does indeed dissolve in water to a certainextent?A student is given an unknown that is either iron(II)sulfate or iron(III) sulfate. Suggest a chemical procedure for determining its identity. (Both iron compounds are water soluble.)You are given a colorless liquid. Describe three chemical tests you would perform on the liquid to showthat it is water.Using the apparatus shown in Figure 4.1, a studentfound that a sulfuric acid solution caused the lightbulb to glow brightly. However, after the addition ofa certain amount of a barium hydroxide [Ba(OH)2]solution, the light began to dim even though Ba(OH)2is also a strong electrolyte. Explain.You are given a soluble compound of unknown molecular formula. (a) Describe three tests that wouldshow that the compound is an acid. (b) Once you haveestablished that the compound is an acid, describehow you would determine its molar mass using aNaOH solution of known concentration. (Assume theacid is monoprotic.) (c) How would you find outwhether the acid is weak or strong? You are providedwith a sample of NaCl and an apparatus like thatshown in Figure 4.1 for comparison.You are given two colorless solutions, one containing NaCl and the other sucrose (C12H22O11). Suggesta chemical and a physical test that would allow youto distinguish between these two solutions.The concentration of lead ions (Pb2 ) in a sample ofpolluted water that also contains nitrate ions (NO3 )is determined by adding solid sodium sulfate(Na2SO4) to exactly 500 mL of the water. (a) Writethe molecular and net ionic equations for the reaction. (b) Calculate the molar concentration of Pb2 if0.00450 g of Na2SO4 was needed for the completeprecipitation of Pb2 ions as PbSO4.Hydrochloric acid is not an oxidizing agent in thesense that sulfuric acid and nitric acid are. Explainwhy the chloride ion is not a strong oxidizing agentlike SO2 and NO3 .4Explain how you would prepare potassium iodide(KI) by means of (a) an acid-base reaction and (b) areaction between an acid and a carbonate compound.Sodium reacts with water to yield hydrogen gas. Whyis this reaction not used in the laboratory preparationof hydrogen?Describe how you would prepare the following compounds: (a) Mg(OH)2, (b) AgI, (c) Ba3(PO4)2.Someone spilled concentrated sulfuric acid on theForwardMain MenuTOC4.1194.1204.1214.1224.1234.124149floor of a chemistry laboratory. To neutralize the acid,would it be preferrable to pour concentrated sodiumhydroxide solution or spray solid sodium bicarbonateover the acid? Explain your choice and the chemicalbasis for the action.Describe in each case how you would separate thecations or anions in an aqueous solution: (a) NaNO3and Ba(NO3)2, (b) Mg(NO3)2 and KNO3, (c) KBr andKNO3, (d) K3PO4 and KNO3, (e) Na2CO3 andNaNO3.The following are common household compounds:table salt (NaCl), table sugar (sucrose), vinegar (contains acetic acid), baking soda (NaHCO3), washingsoda (Na2CO3 10H2O), boric acid (H3BO3, used ineyewash), epsom salt (MgSO4 7H2O), sodium hydroxide (used in drain openers), ammonia, milk ofmagnesia [Mg(OH)2], and calcium carbonate. Basedon what you have learned in this chapter, describetest(s) that would allow you to identify each of thesecompounds.Sulfites (compounds containing the SO2 ions) are3used as preservatives in dried fruit and vegetables andin wine making. In an experiment to test the presenceof sulfite in fruit, a student first soaked several driedapricots in water overnight and then filtered the solution to remove all solid particles. She then treatedthe solution with hydrogen peroxide (H2O2) to oxidize the sulfite ions to sulfate ions. Finally, the sulfate ions were precipitated by treating the solutionwith a few drops of a barium chloride (BaCl2) solution. Write a balanced equation for each of the abovesteps.A 0.8870-g sample of a mixture of NaCl and KCl isdissolved in water, and the solution is then treatedwith an excess of AgNO3 to yield 1.913 g of AgCl.Calculate the percent by mass of each compound inthe mixture.Chlorine forms a number of oxides with the following oxidation numbers: 1, 3, 4, 6, and 7.Write a formula for each of these compounds.A useful application of oxalic acid is the removal ofrust (Fe2O3) from, say, bathtub rings according to thereactionFe2O3(s)6H2C2O4(aq) 88n2Fe(C2O4)3 (aq)33H2O6H (aq)Calculate the number of grams of rust that can be removed by 5.00 102 mL of a 0.100 M solution ofoxalic acid.4.125 Acetylsalicylic acid (C9H8O4) is a monoprotic acidcommonly known as aspirin. A typical aspirintablet, however, contains only a small amount of theacid. In an experiment to determine its composition,an aspirin tablet was crushed and dissolved in water.Study Guide TOCTextbook WebsiteMHHE Website150REACTIONS IN AQUEOUS SOLUTION4.1264.1274.1284.1294.130BackIt took 12.25 mL of 0.1466 M NaOH to neutralizethe solution. Calculate the number of grains of aspirin in the tablet. (One grain 0.0648 g.)A 0.9157-g mixture of CaBr2 and NaBr is dissolvedin water, and AgNO3 is added to the solution to formAgBr precipitate. If the mass of the precipitate is1.6930 g, what is the percent by mass of NaBr in theoriginal mixture?The following cycle of copper experiment is performed in some general chemistry laboratories. Theseries of reactions starts with copper and ends withmetallic copper. The steps are as follows: (1) A pieceof copper wire of known mass is allowed to react withconcentrated nitric acid [the products are copper(II)nitrate, nitrogen dioxide, and water]. (2) The copper(II) nitrate is treated with a sodium hydroxide solution to form copper(II) hydroxide precipitate. (3)On heating, copper(II) hydroxide decomposes toyield copper(II) oxide. (4) The copper(II) oxide is reacted with concentrated sulfuric acid to yield copper(II) sulfate. (5) Copper(II) sulfate is treated withan excess of zinc metal to form metallic copper. (6)The remaining zinc metal is removed by treatmentwith hydrochloric acid, and metallic copper is filtered, dried, and weighed. (a) Write a balanced equation for each step and classify the reactions. (b)Assuming that a student started with 65.6 g of copper, calculate the theoretical yield at each step. (c)Considering the nature of the steps, comment on whyit is possible to recover most of the copper used atthe start.A 325-mL sample of solution contains 25.3 g ofCaCl2. (a) Calculate the molar concentration of Clin this solution. (b) How many grams of Cl are in0.100 liter of this solution?Hydrogen halides (HF, HCl, HBr, HI) are highly reactive compounds that have many industrial and laboratory uses. (a) In the laboratory, HF and HCl canbe generated by reacting CaF2 and NaCl with concentrated sulfuric acid. Write appropriate equationsfor the reactions. (Hint: These are not redox reactions.) (b) Why is it that HBr and HI cannot be prepared similarly, that is, by reacting NaBr and NaI withconcentrated sulfuric acid? (Hint: H2SO4 is a strongeroxidizing agent than both Br2 and I2.) (c) HBr can beprepared by reacting phosphorus tribromide (PBr3)with water. Write an equation for this reaction.Referring to the Chemistry in Action essay on page143, answer the following questions: (a) Identify theprecipitation, acid-base, and redox processes. (b)Instead of calcium oxide, why dont we simply addsodium hydroxide to seawater to precipitate magnesium hydroxide? (c) Sometimes a mineral calledForwardMain MenuTOCdolomite (a mixture of CaCO3 and MgCO3) is substituted for limestone to bring about the precipitationof magnesium hydroxide. What is the advantage ofusing dolomite?4.131 Phosphoric acid (H3PO4) is an important industrialchemical used in fertilizers, in detergents, and in thefood industry. It is produced by two different methods. In the electric furnace method elemental phosphorus (P4) is burned in air to form P4O10, which isthen reacted with water to give H3PO4. In the wetprocess the mineral phosphate rock [Ca5(PO4)3F] isreacted with sulfuric acid to give H3PO4 (and HF andCaSO4). Write equations for these processes and classify each step as precipitation, acid-base, or redox reaction.4.132 Ammonium nitrate (NH4NO3) is one of the most important nitrogen-containing fertilizers. Its purity canbe analyzed by titrating a solution of NH4NO3 witha standard NaOH solution. In one experiment a0.2041-g sample of industrially prepared NH4NO3 required 24.42 mL of 0.1023 M NaOH for neutralization. (a) Write a net ionic equation for the reaction.(b) What is the percent purity of the sample?4.133 Is the following reaction a redox reaction? Explain.3O2(g) 88n 2O3(g)4.134 What is the oxidation number of O in HFO?4.135 Use molecular models like those in Figures 4.7 and4.8 to represent the following acid-base reactions:(a) OHH3O 88n 2H2O(b) NH4NH2 88n 2NH3Identify the Brnsted acid and base in each case.4.136 The alcohol content in a 10.0-g sample of blood froma driver required 4.23 mL of 0.07654 M K2Cr2O7 fortitration. Should the police prosecute the individualfor drunken driving? (Hint: See Chemistry in Actionessay on p. 132.)4.137 On standing, a concentrated nitric acid graduallyturns yellow in color. Explain. (Hint: Nitric acidslowly decomposes. Nitrogen dioxide is a coloredgas.)4.138 Describe the laboratory preparation for the followinggases: (a) hydrogen, (b) oxygen, (c) carbon dioxide,and (d) nitrogen. Indicate the physical states of thereactants and products in each case. [Hint: Nitrogencan be obtained by heating ammonium nitrite(NH4NO2).]4.139 Give a chemical explanation for each of the following: (a) Calcium metal is added to a sulfuric acid solution. Hydrogen gas is generated. After a few minutes, the reaction slows down and eventually stopseven though none of the reactants is used up. Explain.(b) In the activity series, aluminum is above hydro-Study Guide TOCTextbook WebsiteMHHE WebsiteQUESTIONS AND PROBLEMSgen, yet the metal appears to be unreactive towardsteam and hydrochloric acid. Why? (c) Sodium andpotassium lie above copper in the activity series.Explain why Cu2 ions in a CuSO4 solution are notconverted to metallic copper upon the addition ofthese metals. (d) A metal M reacts slowly with steam.There is no visible change when it is placed in a palegreen iron(II) sulfate solution. Where should we placeM in the activity series? (e) Before aluminum metalwas obtained by electrolysis, it was produced by reducing its chloride (AlCl3) with an active metal. Whatmetals would you use to produce aluminum in thatway?4.140 Metals and activity series crossword. 1. Displaces silver but not lead. 2. Honorary metal in many versionsof the activity series. 3. Unreactive metal, a salt ofwhich is used in the chloride test. 4. Most abundanttransition metal in Earths crust. 5. This metal formsa nitrate which is hard to decompose. 6. Metal usedin sacrificial protection of iron from corrosion. 7.First member of Group 1A element. 8. This metaldoes not react with water, but reacts with acid. (Takenfrom Metals and the reactivity series, InfoChem, issue no. 23, September 1993. Reprinted with permission of Education in Chemistry.)BackForwardMain MenuTOC15112345678Answers to Practice Exercises: 4.1 (a) Insoluble, (b) in-soluble, (c) soluble. 4.2 Al3 (aq)3OH (aq) 88nAl(OH)3(s). 4.3 (a) Brnsted base, (b) Brnsted acid. 4.4 (a) P:3, F: 1; (b) Mn: 7, O: 2. 4.5 (a) Hydrogen displacementreaction, (b) combination reaction, (c) disproportionation reaction, (d) metal displacement reaction. 4.6 0.452 M. 4.7 Dilute34.2 mL of the stock solution to 200 mL. 4.8 92.02%. 4.9 0.3822g. 4.10 10.0 mL. 4.11 204 mL.Study Guide TOCTextbook WebsiteMHHE WebsiteCHEMICALMYSTERYWho Killed Napoleon?After his defeat at Waterloo in 1815, Napoleon was exiled to St. Helena,a small island in the Atlantic Ocean, where he spent the last six years ofhis life. In the 1960s, samples of his hair were analyzed and found tocontain a high level of arsenic, suggesting that he might have been poisoned. The prime suspects are the governor of St. Helena, with whomNapoleon did not get along, and the French royal family, who wanted to prevent hisreturn to France.Elemental arsenic is not that harmful. The commonly used poison is actually arsenic(III) oxide, As2O3, a white compound that dissolves in water, is tasteless, and ifadministered over a period of time, is hard to detect. It was once known as the inheritance powder because it could be added to grandfather s wine to hasten his demiseso that his grandson could inherit the estate!In 1832 the English chemist James Marsh devised a procedure for detecting arsenic. This test, which now bears Marshs name, combines hydrogen formed by the reaction between zinc and sulfuric acid with a sample of the suspected poison. If As2O3is present, it reacts with hydrogen to form a toxic gas, arsine (AsH3). When arsine gasis heated, it decomposes to form arsenic, which is recognized by its metallic luster.The Marsh test is an effective deterrent to murder by As2O3, but it was invented toolate to do Napoleon any good, if, in fact, he was a victim of deliberate arsenic poisoning.Doubts about the conspiracy theory of Napoleons death developed in the early1990s, when a sample of the wallpaper from his drawing room was found to containcopper arsenate (CuHAsO4), a green pigment that was commonly used at the timeNapoleon lived. It has been suggested that the damp climate on St. Helena promotedthe growth of molds on the wallpaper. To rid themselves of arsenic, the molds couldhave converted it to trimethyl arsine [(CH3)3As], which is a volatile and highly poisonous compound. Prolonged exposure to these vapors would have ruined Napoleonshealth and would also account for the presence of arsenic in his body, though it maynot have been the primary cause of his death. This provocative theory is supported bythe fact that Napoleons regular guests suffered from gastrointestinal disturbances andother symptoms of arsenic poisoning and that their health all seemed to improve whenever they spent hours working outdoors in the garden, their main hobby on the island.We will probably never know whether Napoleon died from arsenic poisoning, intentional or accidental, but this exercise in historical sleuthing provides a fascinatingexample of the use of chemical analysis. Not only is chemical analysis used in forensic science, but it also plays an essential part of endeavors ranging from pure researchto practical applications, such as quality control of commercial products and medicaldiagnosis.152BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE WebsiteCHEMICAL CLUESThe arsenic in Napoleons hair was detected using a technique called neutron activation. When arsenic-75 is bombarded with high-energy neutrons, it is convertedto the radioactive As-76 isotope. The energy of the rays emitted by the radioactive isotope is characteristic of arsenic, and the intensity of the rays establishes howmuch arsenic is present in a sample. With this technique, as little as 5 nanograms(5 10 9 g) of arsenic can be detected in 1 gram of material. (a) Write symbolsfor the two isotopes of As, showing mass number and atomic number. (b) Nametwo advantages of analyzing the arsenic content by neutron activation instead of achemical analysis.2. Arsenic is not an essential element for the human body. (a) Based on its positionin the periodic table, suggest a reason for its toxicity. (b) In addition to hair, whereelse might one look for the accumulation of the element if arsenic poisoning is suspected?3. The Marsh test for arsenic involves the following steps: (a) The generation of hydrogen gas when sulfuric acid is added to zinc. (b) The reaction of hydrogen withAs(III) oxide to produce arsine. (c) Conversion of arsine to arsenic by heating.Write equations representing these steps and identify the type of the reaction ineach step.1.H2SO4A lock of Napoleons hair.Hydrogen flameShiny metallic ringAs2O3 solutionZinc granulesApparatus for the Marsh test. Sulfuric acid is added to zinc metal and a solution containingarsenic(III) oxide. The hydrogen produced reacts with As2O3 to yield arsine (AsH3). On heating,arsine decomposes to elemental arsenic, which has a metallic appearance, and hydrogen gas.153BackForwardMain MenuTOCStudy Guide TOCTextbook WebsiteMHHE Website...
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