Accounting

Accounting

Title: Accounting

Author: Carl S. Warren, James M. Reeve, Jonathan E. Duchac

Rating:

Document Preview

Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website CHAPTER 4 Reactions in Aqueous Solution INTRODUCTION IN 4.1 GENERAL PROPERTIES OF AQUEOUS SOLUTIONS THIS CHAPTER WE WILL APPLY OUR KNOWLEDGE OF STOICHIOMETRY TO THE STUDY OF REACTIONS THAT OCCUR WHEN THE REACTANTS ARE DISSOLVED IN WATER. MOST 4.2 PRECIPITATION REACTIONS PEOPLE TAKE WATER FOR GRANTED BECAUSE IT IS READILY...

Unformatted Document Excerpt
Coursehero

Course Hero has millions of student submitted documents similar to the one
below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.

Course Hero has millions of student submitted documents similar to the one below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.

Menu TOC Study Back Forward Main Guide TOC Textbook Website MHHE Website CHAPTER 4 Reactions in Aqueous Solution INTRODUCTION IN 4.1 GENERAL PROPERTIES OF AQUEOUS SOLUTIONS THIS CHAPTER WE WILL APPLY OUR KNOWLEDGE OF STOICHIOMETRY TO THE STUDY OF REACTIONS THAT OCCUR WHEN THE REACTANTS ARE DISSOLVED IN WATER. MOST 4.2 PRECIPITATION REACTIONS PEOPLE TAKE WATER FOR GRANTED BECAUSE IT IS READILY AVAILABLE IN SEEMINGLY UNLIMITED AMOUNTS — EXCEPT IN ARID CLIMATES AND IN TIMES OF DROUGHT, WHEN RESTRICTIONS ON ITS USE 4.3 ACID-BASE REACTIONS MAKE US ACUTELY AWARE OF ITS IMPORTANCE — AND BECAUSE WE USE IT FOR SO MANY ORDINARY PURPOSES. WE DRINK IT, BATHE IN IT, COOK WE SWIM IN IT, SKATE ON IT, SKI OVER IT, AND SAIL THROUGH IT. WE HARNESS ITS ENERGY FOR HEAT AND POWER. WITHOUT WATER WE LITERALLY COULD NOT EXIST: WATER AMOUNTS TO ABOUT 60 PERCENT OF THE MASS OF THE HUMAN BODY. WATER’S GREAT VERSATILITY STEMS, IN PART, FROM A TENDENCY 4.4 OXIDATION-REDUCTION REACTIONS TO FORM AQUEOUS SOLUTIONS BY DISSOLVING A LARGE VARIETY OF SOLIDS 4.7 ACID-BASE TITRATIONS WITH IT, AND CLEAN WITH IT. 4.5 CONCENTRATION OF SOLUTIONS 4.6 GRAVIMETRIC ANALYSIS AND OTHER LIQUIDS; THE FACT THAT IT EXISTS AT NORMAL AIR TEMPERATURE AS A LIQUID IS DUE TO THE UNIQUE PROPERTY OF ITS MOLECULES. EVEN EARTH’S FACE, IT IS RARE TO FIND PURE WATER IN NATURE. SEAWATER 4.8 REDOX TITRATIONS SURAND THOUGH WATER COVERS 70 PERCENT OF FRESHWATER SOURCES ALIKE CONTAIN DISSOLVED MINERALS AND CONTAMINANTS SUCH AS FERTILIZERS AND INDUSTRIAL POLLUTANTS. AS FOR THE 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 MORE THAN 80 YEARS OLD). MANY CHEMICAL REACTIONS PROCESSES TAKE PLACE IN WATER. AND IN VIRTUALLY ALL BIOLOGICAL THIS CHAPTER WE WILL DISCUSS THREE MAJOR CATEGORIES OF REACTIONS THAT OCCUR IN AQUEOUS SOLUTIONS: PRECIPITATION REACTIONS, ACID-BASE REACTIONS, AND REDOX REACTIONS. IN LATER CHAPTERS WE WILL STUDY THE STRUCTURAL CHARACTERISTICS AND PROPERTIES OF WATER, THE SO-CALLED “UNIVERSAL SOLVENT,” AND ITS SOLUTIONS. 109 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 110 REACTIONS IN AQUEOUS SOLUTION 4.1 GENERAL PROPERTIES OF AQUEOUS SOLUTIONS A 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 larger amount. 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, in which the solute initially is a liquid or a solid and the solvent is water. ELECTROLYTIC PROPERTIES All 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 a solution that can conduct electricity. A nonelectrolyte does not conduct electricity when dissolved in water. Figure 4.1 shows an easy and straightforward method of distinguishing between electrolytes and nonelectrolytes. A pair of inert electrodes (copper or platinum) is immersed in a beaker of water. To light the bulb, electric current must flow from one electrode to the other, thus completing the circuit. Pure water is a very poor 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, an ionic compound, breaks up into Na and Cl ions when it dissolves in water. The Na ions 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 electrons along a metal wire. Because the NaCl solution conducts electricity, we say that NaCl is an electrolyte. Pure water contains very few ions, so it cannot conduct electricity. Comparing the light bulb’s brightness for the same molar amounts of dissolved substances helps us distinguish between strong and weak electrolytes. A characteristic of strong electrolytes is that the solute is assumed to be 100 percent dissociated into ions in solution. (By dissociation we mean the breaking up of the compound into cations and anions.) Thus we can represent sodium chloride dissolving in water as HO 2 NaCl(s) 88n Na (aq) Cl (aq) This equation says that all sodium chloride that enters the solution ends up as Na and Cl ions; there are no undissociated NaCl units in solution. FIGURE 4.1 An arrangement for distinguishing between electrolytes and nonelectrolytes. A solution’s ability to conduct electricity depends on the number of ions it contains. (a) A nonelectrolyte solution does not contain ions, and the light bulb is not lit. (b) A weak electrolyte solution contains a small number of ions, and the light bulb is dimly lit. (c) A strong electrolyte solution contains a large number of ions, and the light bulb is brightly lit. The molar amounts of the dissolved solutes are equal in all three cases. (a) Back Forward Main Menu TOC (b) Study Guide TOC (c) Textbook Website MHHE Website 4.1 GENERAL PROPERTIES OF AQUEOUS SOLUTIONS 111 TABLE 4.1 Classification of Solutes in Aqueous Solution STRONG ELECTROLYTE WEAK ELECTROLYTE NONELECTROLYTE HCl HNO3 HClO4 H2SO4* NaOH Ba(OH)2 Ionic compounds CH3COOH HF HNO2 NH3 H2O† (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 human body 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 surrounded by 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 and prevents 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 completely in water; for example, when hydrogen chloride gas dissolves in water, it forms hydrated H and Cl ions: HO 2 HCl(g) 88n H (aq) Cl (aq) In other words, all the dissolved HCl molecules separate into hydrated H and Cl ions. 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 other hand, certain acids, such as acetic acid (CH3COOH), which gives vinegar its tart flavor, do not ionize completely and are weak electrolytes. We represent the ionization of acetic acid as FIGURE 4.2 Hydration of Na and Cl ions. + Back Forward Main Menu TOC – Study Guide TOC Textbook Website MHHE Website 112 REACTIONS IN AQUEOUS SOLUTION CH3COOH(aq) 34 CH3COO (aq) There are different types of chemical equilibrium. We will return to this very important topic in Chapter 14. 4.2 H (aq) where CH3COO is called the acetate ion. We use the term ionization to describe the separation of acids and bases into ions. By writing the formula of acetic acid as CH3COOH 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 a reversible 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 is continuous 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 a hydrochloric acid solution the H and Cl ions have no tendency to recombine and form molecular HCl. We use a single arrow to represent complete ionizations. PRECIPITATION REACTIONS One 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 reactions usually 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. SOLUBILITY FIGURE 4.3 Formation of yellow PbI2 precipitate as a solution of 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 a solution 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, slightly soluble, or insoluble in a qualitative sense. A substance is said to be soluble if a fair amount of it visibly dissolves when added to water. If not, the substance is described as slightly soluble or insoluble. All ionic compounds are strong electrolytes, but they are 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 4.1 Classify the following ionic compounds as soluble or insoluble: (a) silver sulfate (Ag2SO4), (b) calcium carbonate (CaCO3), (c) sodium phosphate (Na3PO4). Answer Back Forward Main Menu (a) According to Table 4.2, Ag2SO4 is insoluble. (b) This is a carbonate TOC Study Guide TOC Textbook Website MHHE Website 4.2 TABLE 4.2 PRECIPITATION REACTIONS 113 Solubility Rules for Common Ionic Compounds in Water at 25°C SOLUBLE COMPOUNDS EXCEPTIONS Compounds containing alkali metal ions (Li , Na , K , Rb , Cs ) and the ammonium ion (NH4 ) Nitrates (NO3 ), bicarbonates (HCO3 ), and chlorates (ClO3 ) Halides (Cl , Br , I ) Halides of Ag , Hg2 , and Pb2 2 Sulfates (SO2 ) 4 Sulfates of Ag , Ca2 , Sr2 , Ba2 , and Pb2 INSOLUBLE COMPOUNDS EXCEPTIONS (CO2 3 Carbonates ), phosphates (PO3 ), chromates (CrO2 ), 4 4 sulfides (S2 ) Hydroxides (OH ) Similar problems: 4.15, 4.16. Compounds containing alkali metal ions and the ammonium ion Compounds containing alkali metal ions and the Ba2 ion and Ca is a Group 2A metal. Therefore, CaCO3 is insoluble. (c) Sodium is an alkali metal (Group 1A) so Na3PO4 is soluble. PRACTICE EXERCISE Classify the following ionic compounds as soluble or insoluble: (a) CuS, (b) Ca(OH)2, (c) Zn(NO3)2. MOLECULAR EQUATIONS AND IONIC EQUATIONS The 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 all FIGURE 4.4 Appearance of several precipitates. From left to right: CdS, PbS, Ni(OH)2, Al(OH)3. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 114 REACTIONS IN AQUEOUS SOLUTION species existed as molecules or whole units. A molecular equation is useful because it identifies the reagents (that is, lead nitrate and sodium iodide). If we wanted to bring about 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 apart completely 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 write Pb2 (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 species as free ions. An ionic equation includes spectator ions, or ions that are not involved in the overall reaction, in this case the Na and NO3 ions. Spectator ions appear on both sides of the equation and are unchanged in the chemical reaction, so they can be canceled. To focus on the change that occurs, we write the net ionic equation, which shows 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 barium chloride (BaCl2) is added to an aqueous solution of sodium sulfate (Na2SO4), a white precipitate of barium sulfate (BaSO4) is formed (Figure 4.5). The molecular equation for this reaction is BaCl2(aq) FIGURE 4.5 Formation of BaSO4 precipitate. Na2SO4(aq) 88n BaSO4(s) 2NaCl(aq) The ionic equation for the reaction is Ba2 (aq) 2Cl (aq) 2Na (aq) SO2 (aq) 88n BaSO4(s) 4 2Na (aq) 2Cl (aq) Canceling the spectator ions (Na and Cl ) on both sides of the equation gives us the net ionic equation Ba2 (aq) SO2 (aq) 88n BaSO4(s) 4 The 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. Remember that all strong electrolytes, when dissolved in solution, are completely dissociated into 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 net ionic equation. • • These steps are applied in Example 4.2. EXAMPLE 4.2 Predict the products of the following reaction and write a net ionic equation for the reaction K3PO4(aq) Back Forward Main Menu TOC Ca(NO3)2(aq) 88n ? Study Guide TOC Textbook Website MHHE Website 4.3 ACID-BASE REACTIONS 115 From 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, 4 calcium phosphate [Ca3(PO4)2]. Therefore this is a precipitation reaction. The other product, potassium nitrate (KNO3), is soluble and remains in solution. The molecular equation is Answer 2K3PO4(aq) 3Ca(NO3)2(aq) 88n 6KNO3(aq) Ca3(PO4)2(s) and the ionic equation is 6K (aq) Precipitate formed by the reaction between K3PO4(aq) and Ca(NO3)2(aq). 2PO3 (aq) 4 3Ca2 (aq) 6NO3 (aq) 88n 6K (aq) 6NO3 (aq) Ca3(PO4)2(s) Canceling the spectator K and NO3 ions, we obtain the net ionic equation 3Ca2 (aq) 2PO3 (aq) 88n Ca3(PO4)2(s) 4 Note that because we balanced the molecular equation first, the net ionic equation is 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 EXERCISE Predict the precipitate produced by the following reaction, and write a net ionic equation for the reaction Al(NO3)3(aq) NaOH(aq) 88n ? The Chemistry in Action essay on p. 116 discusses some practical problems associated with precipitation reactions. 4.3 ACID-BASE REACTIONS Acids and bases are as familiar as aspirin and milk of magnesia although many people do not know their chemical names — acetylsalicylic acid (aspirin) and magnesium hydroxide (milk of magnesia). In addition to being the basis of many medicinal and household products, acid-base chemistry is important in industrial processes and essential in sustaining biological systems. Before we can discuss acid-base reactions, we need to know more about acids and bases themselves. GENERAL PROPERTIES OF ACIDS AND BASES In Section 2.7 we defined acids as substances that ionize in water to produce H ions and bases as substances that ionize in water to produce OH ions. These definitions were formulated in the late nineteenth century by the Swedish chemist Svante Arrhenius† to classify substances whose properties in aqueous solutions were well known. † 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 known as panspermia. Arrhenius was awarded the Nobel Prize in Chemistry in 1903. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 116 REACTIONS IN AQUEOUS SOLUTION Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry An Undesirable Precipitation Reaction Limestone (CaCO3) and dolomite (CaCO3 MgCO3), which are widespread on Earth’s surface, often enter the water supply. According to Table 4.2, calcium carbonate is insoluble in water. However, in the presence of dissolved carbon dioxide (from the atmosphere), calcium carbonate is converted to soluble calcium bicarbonate [Ca(HCO3)2]: CaCO3(s) CO2(aq) H2O(l ) 88n Ca2 (aq) 2HCO3 (aq) where HCO3 is the bicarbonate ion. Water containing Ca2 and/or Mg2 ions is called hard water, and water that is mostly free of these ions is called soft water. Hard water is unsuitable for some household and industrial uses. When water containing Ca2 and HCO3 ions is heated or boiled, the solution reaction is reversed to produce the CaCO3 precipitate Ca2 (aq) 2HCO3 (aq) 88n CaCO3(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 of scale reduces heat transfer and decreases the efficiency and durability of boilers, pipes, and appliances. In household hot-water pipes it can restrict or totally block the flow of water. A simple method used by plumbers to remove scale deposits is to introduce a 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 main component of the scale that accumulates in boilers, 2HCl(aq) 88n CaCl2(aq) H2O(l ) CO2(g) In this way, CaCO3 is converted to soluble CaCl2. Acids Acids have a sour taste; for example, vinegar owes its sourness to acetic acid, and lemons 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 mostly CaCO3, reacts with hydrochloric acid. Back Forward HCl(aq) Main Menu H2(g) Acids react with carbonates and bicarbonates, such as Na2CO3, CaCO3, and NaHCO3, 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. TOC Study Guide TOC Textbook Website MHHE Website 4.3 ACID-BASE REACTIONS 117 Bases Bases 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. • • • BRØNSTED ACIDS AND BASES Arrhenius’s definitions of acids and bases are limited in that they apply only to aqueous solutions. Broader definitions were proposed by the Danish chemist Johannes Brønsted† in 1932; a Brønsted acid is a proton donor, and a Brønsted base is a proton acceptor. Note that Brønsted’s definitions do not require acids and bases to be in aqueous solution. Hydrochloric acid is a Brønsted 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 a bare proton. The size of a proton is about 10 15 m, compared to a diameter of 10 10 m for an average atom or ion. Such an exceedingly small charged particle cannot exist as a separate entity in aqueous solution owing to its strong attraction for the negative pole (the O atom) in H2O. Consequently, the proton exists in the hydrated form as shown in Figure 4.7. Therefore, the ionization of hydrochloric acid should be written as HCl(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 Brønsted acid (HCl) donates a proton to a Brønsted base (H2O). Experiments show that the hydronium ion is further hydrated so that the proton may have several water molecules associated with it. Since the acidic properties of the proton are unaffected by the degree of hydration, in this text we will generally use H (aq) to represent the hydrated proton. This notation is for convenience, but H3O is 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), nitric acid (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 one hydrogen ion upon ionization: † Johannes Nicolaus Brønsted (1879 – 1947). Danish chemist. In addition to his theory of acids and bases, Brønsted worked on thermodynamics and the separation of mercury into its isotopes. FIGURE 4.7 Ionization of HCl in water to form the hydronium ion and the chloride ion. + HCl Back Forward Main Menu TOC + H2O H3O+ Study Guide TOC Textbook Website + + Cl– MHHE Website 118 REACTIONS IN AQUEOUS SOLUTION HCl(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 Table 4.1). 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 up two H ions, in two separate steps: H2SO4(aq) 88n H (aq) HSO4 (aq) HSO4 (aq) 34 H (aq) SO2 (aq) 4 H2SO4 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 best known triprotic acid is phosphoric acid, whose ionizations are H3PO4(aq) 34 H (aq) H2PO4 (aq) H2PO4 (aq) 34 H (aq) HPO2 (aq) 4 HPO2 (aq) 34 H (aq) 4 PO3 (aq) 4 All three species (H3PO4, H2PO4 , and HPO2 ) in this case are weak acids, and we use 4 the double arrows to represent each ionization step. Anions such as H2PO4 and HPO2 4 are 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: HO 2 NaOH(s) 88n Na (aq) HO 2 Ba(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 Brønsted base. Ammonia (NH3) is classified as a Brønsted 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 hydroxide ion. 34 + NH3 Back Forward Main Menu TOC + H2O 34 Study Guide TOC + + NH4 + Textbook Website OH– MHHE Website 4.3 ACID-BASE REACTIONS 119 Ammonia 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 is cheap and soluble. (In fact, all of the alkali metal hydroxides are soluble.) The most commonly 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, where M 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 of other metals, such as Al(OH)3 and Zn(OH)2 are insoluble and are less commonly used. The following example classifies substances as Brønsted acids or Brønsted bases. EXAMPLE 4.3 Classify each of the following species as a Brønsted acid or base: (a) HBr, (b) NO2 , (c) HCO3 . Register to View AnswerHBr dissolves in water to yield H and Br ions: HBr(aq) 88n H (aq) Br (aq) Therefore HBr is a Brønsted 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 Brønsted base. (c) The bicarbonate ion is a Brønsted acid because it ionizes in solution as follows: HCO3 (aq) 34 H (aq) CO2 (aq) 3 It is also a Brønsted 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 both acidic and basic properties. The double arrows show that both reactions are reversible. PRACTICE EXERCISE Classify each of the following species as a Brønsted acid or base: (a) SO2 , (b) HI. 4 ACID-BASE NEUTRALIZATION A 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 made up of a cation other than H and an anion other than OH or O2 : Acid-base reactions generally go to completion. acid base 88n salt water All salts are strong electrolytes. The substance we know as table salt, NaCl, is a familiar example. It is a product of the acid-base reaction HCl(aq) NaOH(aq) 88n NaCl(aq) H2O(l) However, since both the acid and the base are strong electrolytes, they are completely ionized in solution. The ionic equation is Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 120 REACTIONS IN AQUEOUS SOLUTION H (aq) Cl (aq) Na (aq) OH (aq) 88n Na (aq) Cl (aq) H2O(l) Therefore, the reaction can be represented by the net ionic equation H (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 the base, 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, represented by 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 becomes HNO3(aq) 4.4 NH4 (aq) OH (aq) 88n NH4NO3(aq) H2O(l) OXIDATION-REDUCTION REACTIONS Whereas acid-base reactions can be characterized as proton-transfer processes, the class of reactions called oxidation-reduction, or redox, reactions are considered electrontransfer reactions. Oxidation-reduction reactions are very much a part of the world around 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 by the process of oxidation or reduction. Many important redox reactions take place in water, but not all redox reactions occur 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). For convenience, we can think of this process as two separate steps, one involving the loss of four electrons by the two Ca atoms and the other being the gain of four electrons by an O2 molecule: 2Ca 88n 2Ca2 O2 4e 88n 2O 4e 2 Each 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: 2Ca O2 4e 88n 2Ca2 2O2 4e or, if we cancel the electrons that appear on both sides of the equation, 2Ca Back Forward Main Menu TOC O2 88n 2Ca2 Study Guide TOC 2O2 Textbook Website MHHE Website 4.4 OXIDATION-REDUCTION REACTIONS 121 Finally, the Ca2 and O2 ions combine to form CaO: 2Ca2 Oxidizing agents are always reduced, and reducing agents are always oxidized. This statement may be somewhat confusing, but it is simply a consequence of the definitions of the two processes. A useful mnemonic for redox is OILRIG: Oxidation Is Loss (of electrons) and Reduction Is Gain (of electrons). 2O2 88n 2CaO By convention, we do not show the charges in the formula of an ionic compound, so that 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 with oxygen. However, it now has a broader meaning that includes reactions not involving oxygen. A reduction reaction is a half-reaction that involves gain of electrons. In the formation of calcium oxide, calcium is oxidized. It is said to act as a reducing agent because it donates electrons to oxygen and causes oxygen to be reduced. Oxygen is reduced 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 must be equal to the extent of reduction; that is, the number of electrons lost by a reducing agent 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 than others. 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 are Zn 88n Zn2 2 Cu 2e 2e 88n Cu Similarly, metallic copper reduces silver ions in a solution of silver nitrate (AgNO3): Cu(s) 2AgNO3(aq) 88n Cu(NO3)2(aq) 2Ag(s) or Cu(s) 2Ag (aq) 88n Cu2 (aq) 2Ag(s) FIGURE 4.9 Left to right: When a piece of Zn metal is placed in an aqueous CuSO4 solution, Zn atoms enter the solution as Zn2 ions, and Cu2 ions are converted to solid Cu (first beaker). In time, the blue color of the CuSO4 solution disappears (second beaker). When a piece of Cu wire is placed in an aqueous AgNO3 solution, Cu atoms enter the solution as Cu2 ions, and Ag ions are converted to solid Ag (third beaker). Gradually, the solution acquires the blue color characteristic of hydrated Cu2 ions (fourth beaker). Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 122 REACTIONS IN AQUEOUS SOLUTION OXIDATION NUMBER The definitions of oxidation and reduction in terms of loss and gain of electrons apply to the formation of ionic compounds such as CaO and the reduction of Cu2 ions by Zn. 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 (from H 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 atom’s oxidation number, also called oxidation state, signifies the number of charges the atom would have in a molecule (or an ionic compound) if electrons were transferred completely. For example, we can rewrite the above equations for the formation of HCl and SO2 as follows: 0 0 H2(g) 0 S(s) 11 Cl2(g) 88n 2HCl(g) 0 4 2 O2(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 their oxidation 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. The oxidation numbers reflect the number of electrons “transferred.” Oxidation numbers allow us to identify elements that are oxidized and reduced at a 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 their oxidation numbers show a decrease from their initial values. Note that the sum of the oxidation numbers of H and Cl in HCl ( 1 and 1) is zero. Likewise, if we add the charges on S ( 4) and two atoms of O [2 ( 2)], the total is zero. The reason is that the 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 metals have an oxidation number of 1 and all alkaline earth metals have an oxidation number of 2 in their compounds. Aluminum has an oxidation number of 3 in all 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. 2 1. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 4.4 OXIDATION-REDUCTION REACTIONS 123 The oxidation number of hydrogen is 1, except when it is bonded to metals in binary compounds. In these cases (for example, LiH, NaH, CaH2), its oxidation number 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 their compounds. 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 be zero. In a polyatomic ion, the sum of oxidation numbers of all the elements in the ion 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 the oxidation numbers is 3 4( 1) 1, which is equal to the net charge of the ion. 7. Oxidation numbers do not have to be integers. For example, the oxidation number of O in the superoxide ion, O2 , is 1 . 2 4. We apply the above rules to assign oxidation numbers in Example 4.4 EXAMPLE 4.4 Assign oxidation numbers to all the elements in the following compounds and ion: (a) Li2O, (b) HNO3, (c) Cr2O2 7 Register to View AnswerBy rule 2 we see that lithium has an oxidation number of and oxygen’s 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 other group (the nitrate ion) must have a net oxidation number of 1. Since oxygen has an oxidation number of 2, the oxidation number of nitrogen (labeled x) is given by 1 or x x 3( 2) 5 (c) From rule 6 we see that the sum of the oxidation numbers in Cr2O2 must be 7 2. 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 is 2 y or Similar problems: 4.43, 4.45. 2(y) 7( 2) 6 PRACTICE EXERCISE Assign 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, arranged according to their positions in the periodic table. We can summarize the content of this figure as follows: • Back Forward Main Menu Metallic elements have only positive oxidation numbers, whereas nonmetallic ele- TOC Study Guide TOC Textbook Website MHHE Website 124 REACTIONS IN AQUEOUS SOLUTION 1 1A 18 8A 1 2 H He +1 –1 2 2A 13 3A 5 14 4A 6 15 5A 7 16 6A 8 17 7A 3 4 9 10 Li Be B C N O F Ne +1 +2 +3 +4 +2 –4 +5 +4 +3 +2 +1 –3 +2 –1 2 –1 –2 –1 15 11 12 13 14 16 17 18 Na Mg Al Si P S Cl Ar +1 +2 +3 +4 –4 +5 +3 –3 +6 +4 +2 –2 +7 +6 +5 +4 +3 +1 –1 3 3B 4 4B 5 5B 6 6B 7 7B 8 9 8B 10 11 1B 12 2B 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr +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 +4 –4 +5 +3 –3 +6 +4 –2 +5 +3 +1 –1 +4 +2 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe +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 +3 –3 +6 +4 –2 +7 +5 +1 –1 +6 +4 +2 55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn +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 +2 –1 FIGURE 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, 3B–8B) usually have several possible oxidation numbers. • TYPES OF REDOX REACTIONS There are four general types of redox reactions: combination reactions, decomposition reactions, displacement reactions, and disproportionation reactions. Displacement reactions have widespread applications in industry, so we will study them in some detail. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 4.4 (a) OXIDATION-REDUCTION REACTIONS 125 (b) (c) FIGURE 4.11 Some simple combination redox reactions. (a) Hydrogen gas burning in air to form water. (b) Sulfur burning in air to form sulfur dioxide. (c) Magnesium burning in air to form magnesium oxide and magnesium nitride. (d) Sodium burning in chlorine to form sodium chloride. (e) Aluminum reacting with bromine to form aluminum bromide. (d) (e) Combination Reactions A combination reaction may be represented by A B 88n C If 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, 0 S(s) 0 3Mg(s) 0 42 O2(g) 88n SO2(g) 0 2 3 N2(g) 88n Mg3N2(s) Decomposition Reactions Decomposition reactions are the opposite of combination reactions. Specifically, a decomposition reaction is the breakdown of a compound into two or more components: Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 126 REACTIONS IN AQUEOUS SOLUTION FIGURE 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 of the wood splint. (a) (b) C 88n A B If either A or B is an element, then the reaction is redox in nature (Figure 4.12). For example 2 2 0 0 2HgO(s) 88n 2Hg(l) 52 O2(g) 1 0 2KClO3(s) 88n 2KCl(s) 11 3O2(g) 0 0 2NaH(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 Reactions In a displacement reaction, an ion (or atom) in a compound is replaced by an ion (or atom) of another element.: A BC 88n AC B Most 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 Na than with Ca. Back Forward 2Na(s) 0 Ca(s) 1 1 1 2H2O(l) 88n 2NaOH(aq) 1 2 1 2H2O(l) 88n Ca(OH)2(s) 0 H2(g) 0 H2(g) Less reactive metals, such as aluminum and iron, react with steam to give hydrogen gas: Main Menu TOC Study Guide TOC Textbook Website MHHE Website 4.4 OXIDATION-REDUCTION REACTIONS 127 FIGURE 4.14 Left to right: Reactions of iron (Fe), zinc (Zn), and magnesium (Mg) with hydrochloric acid to form hydrogen gas and the metal chlorides (FeCl2, ZnCl2, MgCl2). The reactivity of these metals is reflected in the rate of hydrogen gas evolution, which is slowest for the least reactive metal, Fe, and fastest for the most reactive metal, Mg. (a) (b) 0 2Al(s) 1 3 3H2O(g) 88n Al2O3(s) 0 2Fe(s) (c) 1 3 3H2O(g) 88n Fe2O3(s) 0 3H2(g) 0 3H2(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: 0 1 Displace hydrogen from cold water Displace hydrogen from steam Li K Ba Ca Na Mg Al Zn Cr Fe Cd Co Ni Sn Pb H Cu Hg Ag Pt Au Displace hydrogen from acids Zn(s) Forward 1 0 H2(g) 2 2HCl(aq) 88n MgCl2(aq) 0 H2(g) The ionic equations are 0 Zn(s) 0 Mg(s) 1 2 2H (aq) 88n Zn2 (aq) 1 2 2H (aq) 88n Mg2 (aq) 0 H2(g) 0 H2(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 the laboratory. 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 are arranged according to their ability to displace hydrogen from an acid or water. Li (lithium) is the most reactive metal, and Au (gold) is the least reactive. Back 0 Mg(s) 2 2HCl(aq) 88n ZnCl2(aq) Main Menu 2. Metal Displacement. A metal in a compound can be displaced by another metal in the elemental state. We have already seen examples of zinc replacing copper ions and copper replacing silver ions (see p. 121). Reversing the roles of the metals would result 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 will actually 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 of the results of many possible displacement reactions similar to the ones already dis- TOC Study Guide TOC Textbook Website MHHE Website 128 REACTIONS IN AQUEOUS SOLUTION FIGURE 4.16 (a) An aqueous KBr solution. (b) After chlorine gas has been bubbled through the solution, most of the bromide ions are converted (oxidized) to liquid bromine, which is slightly soluble 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 an acid. 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 from copper sulfate. Metal displacement reactions find many applications in metallurgical processes, the goal of which is to separate pure metals from their ores. For example, vanadium is 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 reaction TiCl4(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 see more examples of this type of reaction in Chapter 20. 3. Halogen Displacement. Another activity series summarizes the halogens’ behavior in halogen displacement reactions: F2 Cl2 Br2 I2 The power of these elements as oxidizing agents decreases as we move down Group 7A from fluorine to iodine, so molecular fluorine can replace chloride, bromide, and iodide 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 Figure 4.16 shows. The displacement equations are Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 4.4 0 OXIDATION-REDUCTION REACTIONS 1 Cl2(g) 1 0 2KBr(aq) 88n 2KCl(aq) 0 1 Cl2(g) 129 Br2(l) 1 0 2NaI(aq) 88n 2NaCl(aq) I2(s) The ionic equations are 0 1 Cl2(g) 0 1 1 Cl2(g) 0 2Br(aq) 88n 2Cl (aq) 1 Br2(l) 0 2I (aq) 88n 2Cl (aq) I2(s) Molecular bromine, in turn, can displace iodide ion in solution: 0 Br2(l) 1 1 0 2I (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 strong oxidizing 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 far the most important industrial chemical. In 1995 the amount of chlorine produced was 25 billion pounds, making chlorine the tenth-ranking industrial chemical. The annual production of bromine is only one-hundredth that of chlorine, while the amounts of fluorine and iodine produced are even less. Recovering the halogens from their halides requires an oxidation process, which is represented by 2X 88n X2 Bromine is a fuming red liquid. 2e where 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 F ions to F2 by chemical means. The only way to carry out the oxidation is by electrolytic means, the details of which will be discussed in Chapter 19. Industrially, chlorine, like fluorine, is produced electrolytically. Bromine is prepared industrially by oxidizing Br ions with chlorine, which is a strong enough oxidizing agent to oxidize Br ions but not water: 2Br (aq) 88n Br2(l) 2e One 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 (Figure 4.17). Iodine is also prepared from seawater and natural brine by the oxidation of I ions 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 from Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 130 REACTIONS IN AQUEOUS SOLUTION FIGURE 4.17 The industrial manufacture of liquid bromine by oxidizing an aqueous solution containing Br ions with chlorine gas. 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 Reaction Elements that are most likely to undergo 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 and reduced. One reactant in a disproportionation reaction always contains an element that can 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: 1 2 0 2H2O2(aq) 88n 2H2O(l) O2(g) Here the oxidation number of oxygen in the reactant ( 1) both increases to zero in O2 and decreases to 2 in H2O. Another example is the reaction between molecular chlorine and NaOH solution: 0 1 Cl2(g) 1 2OH (aq) 88n ClO (aq) Cl (aq) H2O(l) This reaction describes the formation of household bleaching agents, for it is the hypochlorite 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. They are analogous in that acid-base reactions involve the transfer of protons while redox reactions involve the transfer of electrons. However, while acid-base reactions are quite easy to recognize (since they always involve an acid and a base), there is no simple procedure for identifying a redox process. The only sure way is to compare the oxi- Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 4.5 CONCENTRATION OF SOLUTIONS 131 dation 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 4.5 Classify 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 converted to two different types of products. The oxidation number of N changes from 1 to 0 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 Pb2 ion. The oxidation number of Ni increases from 0 to 2 while that of Pb decreases from 2 to 0. (d) The oxidation number of N is 4 in NO2, 3 in HNO2, and 5 in HNO3. Since its oxidation number both increases and decreases, this is a disproportionation reaction. Answer Similar problems: 4.51, 4.52. PRACTICE EXERCISE Identify 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 makes use of a redox reaction to apprehend drunk drivers. 4.5 CONCENTRATION OF SOLUTIONS To study solution stoichiometry we must know how much of the reactants are present in a solution and also how to control the amounts of reactants used to bring about a reaction 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 or a solid and the solvent is a liquid.) The concentration of a solution can be expressed in many different ways, as we will see in Chapter 12. Here we will consider one of the 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 the equation Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 132 REACTIONS IN AQUEOUS SOLUTION Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry Breath Analyzer Every year in the United States about 25,000 people are killed and 500,000 more are injured as a result of drunk driving. In spite of efforts to educate the public about the dangers of driving while intoxicated and stiffer penalties for drunk driving offenses, law enforcement agencies still have to devote a great deal of work to removing drunk drivers from America’s roads. The police often use a device called a breath analyzer to test drivers suspected of being drunk. The chemical basis of this device is a redox reaction. A sample of the driver’s breath is drawn into the breath analyzer, where it is treated with an acidic solution of potassium dichromate. The alcohol (ethanol) in the breath is converted to acetic acid as shown in the following equation: 3CH3CH3OH 2K2Cr2O7 ethanol potassium dichromate (orange yellow) 2Cr2(SO4)3 2K2SO4 acetic acid chromium(III) sulfate (green) potassium sulfate Photocell detector K2Cr2O7 solution M 11H2O In this reaction the ethanol is oxidized to acetic acid and the chromium(VI) in the orange-yellow dichromate ion is reduced to the green chromium(III) ion (see Figure 4.22). The driver’s blood alcohol level can be determined readily by measuring the degree of this color 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. Meter Filter sulfuric acid 3CH3COOH Breath Light source 8H2SO4 88n Schematic diagram of a breath analyzer. The alcohol in the driver’s breath is reacted with a potassium dichromate solution. The change in the absorption of light is registered by the detector and shown on a meter, which directly displays the alcohol content in blood. The filter selects only one wavelength of light for measurement. moles of solute liters of soln molarity (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, contains 0.52 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 solution to liters. Thus, a 500-mL solution containing 0.730 mole of C6H12O6 also has a concentration of 1.46 M: M molarity 0.730 mol 0.500 L 1.46 mol/L Back Forward Main Menu TOC Study Guide TOC 1.46 M Textbook Website MHHE Website 4.5 FIGURE 4.18 Preparing a solution of known molarity. (a) A known amount of a solid solute is put into the volumetric flask; then water is added through a funnel. (b) The solid is slowly dissolved by gently shaking the flask. (c) After the solid has completely dissolved, more water is added to bring the level of solution to the mark. Knowing the volume of the solution and the amount of solute dissolved in it, we can calculate the molarity of the prepared solution. 133 Meniscus Marker showing known volume of 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 much of the solution is present. It is important to keep in mind that molarity refers only to the amount of solute originally 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 urea are nonelectrolytes, so a 1.00 M urea solution has 1.00 mole of the urea molecules in 1 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: HO 2 KCl(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 no KCl units are present. The concentrations of the ions can be expressed as [K ] 1 M and [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] solution HO 2 Ba(NO3)2(s) 88n Ba2 (aq) 2NO3 (aq) 2 In general, it is more convenient to measure the volume of a liquid than to determine its mass. Back Forward Main Menu we 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, the solute 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 dissolve the solid. After all the solid has dissolved, more water is added slowly to bring the level of solution exactly to the volume mark. Knowing the volume of the solution in the 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 does not require knowing the amount of water added, as long as the volume of the final solution is known. TOC Study Guide TOC Textbook Website MHHE Website 134 REACTIONS IN AQUEOUS SOLUTION Example 4.6 illustrates how to prepare a solution of known molarity. EXAMPLE 4.6 How 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 mL of a 2.16 M solution: Answer moles of K2Cr2O7 250 mL soln 2.16 mol K2Cr2O7 1000 mL soln 0.540 mol K2Cr2O7 A K2Cr2O7 solution. The molar mass of K2Cr2O7 is 294.2 g, so we write grams of K2Cr2O7 needed 0.540 mol K2Cr2O7 294.2 g K2Cr2O7 1 K2Cr2O7 159 mol g K2Cr2O7 Similar problems: 4.58, 4.59. PRACTICE EXERCISE What is the molarity of an 85.0-mL ethanol (C2H5OH) solution containing 1.77 g of ethanol? DILUTION OF SOLUTIONS Concentrated solutions are often stored in the laboratory stockroom for use as needed. Frequently we dilute these “stock” solutions before working with them. Dilution is the procedure 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 there is 1.00 mole of KMnO4 in 1 liter, or 1000 mL, of a 1.00 M KMnO4 solution, there is 0.400 mole of KMnO4 in 0.400 1000 mL, or 400 mL of the same solution: 1.00 mol 1000 mL soln Two KMnO4 solutions of different concentrations. 0.400 mol 400 mL soln Therefore, we must withdraw 400 mL from the 1.00 M KMnO4 solution and dilute it to 1000 mL by adding water (in a 1-liter volumetric flask). This method gives us 1 liter 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 of the solution without changing the number of moles of solute present in the solution (Figure 4.19). In other words, moles of solute before dilution moles of solute after dilution Since molarity is defined as moles of solute in one liter of solution, we see that the number of moles of solute is given by M Back Forward Main Menu TOC    volume of soln (in liters) moles of solute    moles of solute liters of soln V Study Guide TOC Textbook Website MHHE Website 4.5 CONCENTRATION OF SOLUTIONS 135 FIGURE 4.19 The dilution of a more concentrated solution (a) to a less concentrated solution (b) does not change the total number of moles of solute. (a) (b) or MV moles of solute Because all the solute comes from the original stock solution, we can conclude that MiVi MfVf moles of solute before dilution moles of solute after dilution (4.2) where Mi and Mf are the initial and final concentrations of the solution in molarity and Vi and Vf are the initial and final volumes of the solution, respectively. Of course, the units of Vi and Vf must be the same (mL or L) for the calculation to work. To check the reasonableness of your results, be sure that Mi Mf and Vf Vi. We apply Equation (4.2) in the following example. EXAMPLE 4.7 Describe 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 our data: Answer Mi 8.61 M Mf 1.75 M Vi ? Vf 5.00 102 mL (1.75 M)(5.00 102 mL) Substituting in Equation (4.2), (8.61 M)(Vi) (1.75 M)(5.00 102 mL) 8.61 M Vi 102 mL Similar problems: 4.67, 4.68. Thus we must dilute 102 mL of the 8.61 M H2SO4 solution with sufficient water to give a final volume of 5.00 102 mL in a 500-mL volumetric flask to obtain the desired concentration. PRACTICE EXERCISE How would you prepare 2.00 a 5.07 M stock solution? Back Forward Main Menu TOC 102 mL of a 0.866 M NaOH solution, starting with Study Guide TOC Textbook Website MHHE Website 136 REACTIONS IN AQUEOUS SOLUTION Now that we have discussed the concentration and dilution of solutions, we can examine the quantitative aspects of reactions in aqueous solution, or solution stoichiometry. Sections 4.6–4.8 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 a substance in a sample. 4.6 GRAVIMETRIC ANALYSIS Gravimetric analysis is an analytical technique based on the measurement of mass. One type of gravimetric analysis experiment involves the formation, isolation, and mass determination of a precipitate. Generally this procedure is applied to ionic compounds. First, a sample substance of unknown composition is dissolved in water and allowed to react with another substance to form a precipitate. Then the precipitate is filtered off, dried, and weighed. Knowing the mass and chemical formula of the precipitate formed, 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 the mass of the original sample, we can determine the percent composition by mass of the component in the original compound. A reaction that is often studied in gravimetric analysis, because the reactants can be obtained in pure form, is AgNO3(aq) NaCl(aq) 88n NaNO3(aq) AgCl(s) The net ionic equation is Ag (aq) Cl (aq) 88n AgCl(s) The precipitate is silver chloride (see Table 4.2). As an example, let us say that we wanted to determine experimentally the percent by mass of Cl in NaCl. First we would accurately weigh out a sample of NaCl and dissolve it in water. Next, we would add enough AgNO3 solution to the NaCl solution to cause the precipitation of all the Cl ions present in solution as AgCl. In this procedure NaCl is the limiting reagent and AgNO3 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 the mass of Cl using the percent by mass of Cl in AgCl. Since this same amount of Cl was present in the original NaCl sample, we can calculate the percent by mass of Cl in NaCl. Figure 4.20 shows how this procedure is performed. Gravimetric analysis is a highly accurate technique, since the mass of a sample can be measured accurately. However, this procedure is applicable only to reactions that go to completion, or have nearly 100 percent yield. Thus, if AgCl were slightly soluble instead of being insoluble, it would not be possible to remove all the Cl ions from the NaCl solution and the subsequent calculation would be in error. The following example shows the calculations involved in a gravimetric experiment. EXAMPLE 4.8 A 0.5662-g sample of an ionic compound containing chloride ions and an unknown metal is dissolved in water and treated with an excess of AgNO3. If 1.0882 g of AgCl precipitate forms, what is the percent by mass of Cl in the original compound? Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 4.6 (a) GRAVIMETRIC ANALYSIS (b) 137 (c) FIGURE 4.20 Basic steps for gravimetric analysis. (a) A solution containing a known amount of NaCl in a beaker. (b) The precipitation of AgCl upon the addition of AgNO3 solution from a measuring cylinder. In this reaction, AgNO3 is the excess reagent and NaCl is the limiting reagent. (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 then removed from the apparatus, dried in an oven, and weighed again. The difference between this mass 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: Answer mass of CI 1.0882 g AgCl 1 mol AgCl 143.4 g AgCl 1 mol Cl 1 mol AgCl 35.45 g Cl 1 mol Cl 0.2690 g Cl Next, we calculate the percent by mass of Cl in the unknown sample as mass of Cl mass of sample %Cl by mass 0.2690 g 0.5662 g 100% 100% 47.51% Similar problem: 4.74. Comment As a comparison, calculate the percent by mass of Cl in KCl. PRACTICE EXERCISE A sample of 0.3220 g of an ionic compound containing the bromide ion (Br ) is dissolved in water and treated with an excess of AgNO3. If the mass of the AgBr precipitate that forms is 0.6964 g, what is the percent by mass of Br in the original compound? 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- Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 138 REACTIONS IN AQUEOUS SOLUTION ing the percent by mass of Cl greatly helps us in narrowing the possibilities. Because no 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 identity of the unknown. 4.7 ACID-BASE TITRATIONS Quantitative studies of acid-base neutralization reactions are most conveniently carried out using a technique known as titration. In titration, a solution of accurately known concentration, called a standard solution, is added gradually to another solution of unknown 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 tendency to 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 accurate analytical work. We can standardize the sodium hydroxide solution by titrating it against an acid solution of accurately known concentration. The acid often chosen for this task is 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 is KHC8H4O4(aq) Potassium hydrogen phthalate. NaOH(aq) 88n KNaC8H4O4(aq) H2O(l) and the net ionic equation is HC8H4O4 (aq) OH (aq) 88n C8H4O2 (aq) 4 H2O(l) The procedure for the titration is shown in Figure 4.21. First, a known amount of KHP is transferred to an Erlenmeyer flask and some distilled water is added to make up a solution. 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 completely reacted with or been neutralized by the base. The equivalence point is usually signaled by 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 acidic and neutral solutions but reddish pink in basic solutions. At the equivalence point, all the 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, the solution will immediately turn pink because the solution is now basic. Example 4.9 illustrates such a titration. EXAMPLE 4.9 In 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? Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 4.7 ACID-BASE TITRATIONS 139 FIGURE 4.21 (a) Apparatus for acid-base titration. A NaOH solution is added from the buret to a KHP solution in an Erlenmeyer flask. (b) A reddishpink color appears when the equivalence point is reached. The color here has been intensified for 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: Answer moles of KHP 0.5468 g KHP 2.678 10 3 1 mol KHP 204.2 g KHP mol KHP Since 1 mol KHP 1 mol NaOH, there must be 2.678 10 3 mole of NaOH in 23.48 mL of NaOH solution. Finally, we calculate the molarity of the NaOH as follows: molarity of NaOH soln 2.678 10 3 mol NaOH 23.48 mL soln 1000 mL soln 1 L soln 0.1141 M Similar problems: 4.79, 4.80. PRACTICE EXERCISE How many grams of KHP are needed to neutralize 18.64 mL of a 0.1004 M NaOH solution? The neutralization reaction between NaOH and KHP is one of the simplest types of acid-base neutralization known. Suppose, though, that instead of KHP, we wanted to use a diprotic acid such as H2SO4 for the titration. The reaction is represented by 2NaOH(aq) H2SO4(aq) 88n Na2SO4(aq) 2H2O(l) Since 2 mol NaOH 1 mol H2SO4, we need twice as much NaOH to react completely with a H2SO4 solution of the same molar concentration and volume as a KHP solu- Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 140 REACTIONS IN AQUEOUS SOLUTION tion. On the other hand, we would need twice the amount of HCl to neutralize a Ba(OH)2 solution 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 takes part in the reaction, keep in mind that the total number of moles of H ions that have reacted at the equivalence point must be equal to the total number of moles of OH ions that have reacted. The number of moles of an acid in a certain volume is given by moles of acid molarity (mol/L) volume (L) MV where 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 diprotic acid. EXAMPLE 4.10 How many milliliters (mL) of a 0.610 M NaOH solution are needed to completely neutralize 20.0 mL of a 0.245 M H2SO4 solution? The equation for the neutralization reaction is shown on p. 139. First we calculate the number of moles of H2SO4 in a 20.0-mL solution: Answer moles H2SO4 0.245 mol H2SO4 1 L soln 4.90 10 3 1 L soln 1000 mL soln 20.0 mL soln mol H2SO4 From the stoichiometry we see that 1 mol H2SO4 2 mol NaOH. Therefore, the number of moles of NaOH reacted must be 2 4.90 10 3 mole, or 9.80 10 3 mole. From the definition of molarity [see Equation (4.1)] we have moles of solute molarity liters of soln or volume of NaOH 9.80 10 3 mol NaOH 0.610 mol/L 0.0161 L or 16.1 mL Similar problems: 4.79, 4.80. PRACTICE EXERCISE How many milliliters of a 1.28 M H2SO4 solution are needed to neutralize 60.2 mL of a 0.427 M KOH solution? 4.8 REDOX TITRATIONS As mentioned earlier, redox reactions involve the transfer of electrons, and acid-base reactions 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. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 4.8 FIGURE 4.22 There are not as many redox indicators as there are acid-base indicators. REDOX TITRATIONS Left to right: Solutions containing the MnO4 , Mn2 , Cr2O2 , and Cr3 7 141 ions. 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 reducing agent is completely oxidized by the oxidizing agent. Like acid-base titrations, redox titrations normally require an indicator that clearly changes 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 equivalence point, a sharp change in the indicator ’s color will occur as it changes from one form to 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 and permanganate anions are distinctly different from those of the reduced species: Cr2O2 88n Cr3 7 orange yellow green MnO4 88n Mn2 purple light pink Thus these oxidizing agents can themselves be used as an internal indicator in a redox titration 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 stoichiometry tend to be more complex for redox reactions. The following is an example of a redox titration. EXAMPLE 4.11 A 16.42-mL volume of 0.1327 M KMnO4 solution is needed to oxidize 20.00 mL of a FeSO4 solution in an acidic medium. What is the concentration of the FeSO4 solution? The net ionic equation is Addition of a KMnO4 solution from a buret to a FeSO4 solution. Back Forward Main Menu 5Fe2 TOC MnO4 8H 88n Mn2 Study Guide TOC 5Fe3 4H2O Textbook Website MHHE Website 142 REACTIONS IN AQUEOUS SOLUTION Answer The number of moles of KMnO4 in 16.42 mL of the solution is moles of KMnO4 0.1327 mol KMnO4 1 L soln 16.42 mL 2.179 10 3 mol KMnO4 From the equation we see that 5 mol Fe2 of moles of FeSO4 oxidized is 1 mol MnO4 . Therefore, the number 2.179 10 3 10 2 5 mol FeSO4 1 mol KMnO4 mol KMnO4 1.090 moles FeSO4 1 L soln 1000 mL soln mol FeSO4 The concentration of the FeSO4 solution in moles of FeSO4 per liter of solution is [FeSO4] 1.090 10 2 mol FeSO4 20.00 mL soln 1000 mL soln 1 L soln 0.5450 M Similar problems: 4.83, 4.84. PRACTICE EXERCISE How many milliliters of a 0.206 M HI solution are needed to reduce 22.5 mL of a 0.374 M KMnO4 solution according to the following equation: 10HI 2KMnO4 3H2SO4 88n 5I2 2MnSO4 K2SO4 8H2O The Chemistry in Action essay on p. 143 describes an industrial process that involves the types of reactions discussed in this chapter. SUMMARY OF KEY EQUATIONS • molarity (M) • MiVi SUMMARY OF FACTS AND CONCEPTS Back Forward moles of solute liters of soln MfVf (4.2) (4.1) Calculating molarity. Dilution of solution. 1. Aqueous solutions are electrically conducting if the solutes are electrolytes. If the solutes are 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 to give OH ions. Brønsted acids donate protons, and Brønsted 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 is characterized 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 atoms in a compound or ion according to specific rules. Oxidation can be defined as an increase in 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 Menu TOC Study Guide TOC Textbook Website MHHE Website SUMMARY OF FACTS AND CONCEPTS 143 Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry Metal from the Sea Magnesium is a valuable, lightweight metal used as a structural material as well as in alloys, in batteries, and in chemical synthesis. Although magnesium is plentiful in Earth’s crust, it is cheaper to “mine” the metal from seawater. Magnesium forms the second most abundant cation in the sea (after sodium); there are about 1.3 g of magnesium in a kilogram of seawater. The process for obtaining magnesium from seawater employs all three types of reactions discussed in this chapter: precipitation, acid-base, and redox reactions. In the first stage in the recovery of magnesium, limestone (CaCO3) is heated at high temperatures to produce quicklime, or calcium oxide (CaO): CaCO3(s) 88n CaO(s) CO2(g) When calcium oxide is treated with seawater, it forms calcium 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 settling ponds at the Dow Chemical Company in Freeport, Texas. After the water is evaporated, the solid magnesium chloride is melted in a steel cell. The molten magnesium chloride contains both Mg2 and Cl ions. In a process called electrolysis, an electric current is passed through the cell to reduce the Mg2 ions and oxidize the Cl ions. The half-reactions are The 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 reacted with hydrochloric acid to form magnesium chloride (MgCl2): Mg(OH)2(s) 2HCl(aq) 88n MgCl2(aq) 2H2O(l ) Mg2 2e 88n Mg 2Cl 88n Cl2 2e The overall reaction is MgCl2(l ) 88n Mg(l ) Cl2(g) This is how magnesium metal is produced. The chlorine gas generated can be converted to hydrochloric acid 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 in the solution. 11. Gravimetric analysis is a technique for determining the identity of a compound and/or the concentration of a solution by measuring mass. Gravimetric experiments often involve precipitation reactions. 12. In acid-base titration, a solution of known concentration (say, a base) is added gradually to 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 called the 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. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 144 REACTIONS IN AQUEOUS SOLUTION KEY WORDS Activity series, p. 127 Aqueous solution, p. 110 Brønsted acid, p. 117 Brønsted base, p. 117 Combustion reaction, p. 125 Concentration of a solution, p. 131 Decomposition reaction, p. 125 Dilution, p. 134 Diprotic acid, p. 118 Displacement reaction, p. 126 Disproportionation reaction, p. 130 Electrolyte, p. 110 Equivalence point, p. 138 Gravimetric analysis, p. 136 Half-reaction, p. 120 Hydration, p. 111 Hydronium ion, p. 117 Indicator, p. 138 Ionic equation, p. 114 Molar concentration, p. 131 Molarity (M), p. 131 Molecular equation, p. 113 Monoprotic acid, p. 117 Net ionic equation, p. 114 Neutralization reaction, p. 119 Nonelectrolyte, p. 110 Oxidation number, p. 122 Oxidation state, p. 122 Oxidation reaction, p. 121 Oxidation-reduction reaction, p. 120 Oxidizing agent, p. 121 Precipitate, p. 112 Precipitation reaction, p. 112 Quantitative analysis, p. 136 Redox reaction, p. 120 Reducing agent, p. 121 Reduction reaction, p. 121 Reversible reaction, p. 112 Salt, p. 119 Solubility, p. 112 Solute, p. 110 Solution, p. 110 Solvent, p. 110 Spectator ion, p. 114 Standard solution, p. 138 Titration, p. 138 Triprotic acid, p. 118 QUESTIONS AND PROBLEMS PROPERTIES OF AQUEOUS SOLUTIONS Review Questions 4.1 Define solute, solvent, and solution by describing the process of dissolving a solid in a liquid. 4.2 What is the difference between a nonelectrolyte and an electrolyte? Between a weak electrolyte and a strong electrolyte? 4.3 Describe hydration. What properties of water enable its 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 therefore cannot conduct electricity. Why are we often cautioned not to operate electrical appliances when our hands are wet. 4.6 Lithium fluoride (LiF) is a strong electrolyte. What species are present in LiF(aq)? Problems 4.7 Identify each of the following substances as a strong electrolyte, weak electrolyte, or nonelectrolyte: (a) H2O, (b) KCl, (c) HNO3, (d) CH3COOH, (e) C12H22O11. 4.8 Identify each of the following substances as a strong electrolyte, 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) electrons only, (b) cations only, (c) anions only, (d) both cations and anions. 4.10 Predict and explain which of the following systems are electrically conducting: (a) solid NaCl, (b) molten NaCl, (c) an aqueous solution of NaCl. 4.11 You are given a water-soluble compound X. Describe Back Forward Main Menu TOC how you would determine whether it is an electrolyte or a nonelectrolyte. If it is an electrolyte, how would you determine whether it is strong or weak? 4.12 Explain why a solution of HCl in benzene does not conduct electricity but in water it does. PRECIPITATION REACTIONS Review Questions 4.13 What is the difference between an ionic equation and a molecular equation? 4.14 What is the advantage of writing net ionic equations? Problems 4.15 Characterize the following compounds as soluble or insoluble in water: (a) Ca3(PO4)2, (b) Mn(OH)2, (c) AgClO3, (d) K2S. 4.16 Characterize the following compounds as soluble or 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 following reactions: (a) 2AgNO3(aq) Na2SO4(aq) 88n (b) BaCl2(aq) ZnSO4(aq) 88n (c) (NH4)2CO3(aq) CaCl2(aq) 88n 4.18 Write ionic and net ionic equations for the following reactions: (a) Na2S(aq) ZnCl2(aq) 88n (b) 2K3PO4(aq) 3Sr(NO3)2(aq) 88n (c) Mg(NO3)2(aq) 2NaOH(aq) 88n 4.19 Which of the following processes will likely result in a 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 TOC Textbook Website MHHE Website QUESTIONS AND PROBLEMS 4.20 With reference to Table 4.2, suggest one method by which you might separate (a) K from Ag , (b) Ag from 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 REACTIONS Review Questions 4.21 List the general properties of acids and bases. 4.22 Give Arrhenius’s and Brønsted’s definitions of an acid and a base. Why are Brønsted’s definitions more useful 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? Specify which 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 or base: (a) NH3, (b) H3PO4, (c) LiOH, (d) HCOOH (formic acid), (e) H2SO4, (f) HF, (g) Ba(OH)2. Problems 4.27 Identify each of the following species as a Brønsted acid, base, or both: (a) HI, (b) CH3COO , (c) H2PO4 , (d) HSO4 . 4.28 Identify each of the following species as a Brønsted acid, 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) 88n 4.30 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) 88n OXIDATION-REDUCTION REACTIONS Review Questions 4.31 Give an example of a combination redox reaction, a decomposition redox reaction, and a displacement redox reaction. 4.32 All combustion reactions are redox reactions. True or false? Explain. 4.33 What is an oxidation number? How is it used to identify redox reactions? Explain why, except for ionic Back Forward Main Menu TOC 145 compounds, oxidation number does not have any physical significance. 4.34 (a) Without referring to Figure 4.10, give the oxidation numbers of the alkali and alkaline earth metals in their compounds. (b) Give the highest oxidation numbers that the Groups 3A–7A elements can have. 4.35 How is the activity series organized? How is it used in 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 oxidation occurs and reduction does not? Explain. 4.38 What is the requirement for an element to undergo disproportionation reactions? Name five common elements that are likely to take part in such reactions. Problems 4.39 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 Mg3N2 4.40 For the complete redox reactions given below, write the 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 2HCl 4.41 Arrange the following species in order of increasing oxidation 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 atoms in 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 in the following molecules and ions: (a) Cs2O, (b) CaI2, (c) Al2O3, (d) H3AsO3, (e) TiO2, (f) MoO2 , (g) 4 PtCl2 , (h) PtCl2 , (i) SnF2, (j) ClF3, (k) SbF6 . 4 6 4.46 Give the oxidation numbers of the underlined atoms Study Guide TOC Textbook Website MHHE Website 146 REACTIONS IN AQUEOUS SOLUTION 4.47 4.48 4.49 4.50 4.51 4.52 in the following molecules and ions: (a) Mg3N2, (b) CsO2, (c) CaC2, (d) CO2 , (e) C2O2 , (f) ZnO2 , (g) 3 4 2 NaBH4, (h) WO2 . 4 Nitric acid is a strong oxidizing agent. State which of the following species is least likely to be produced when nitric acid reacts with a strong reducing agent such 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, one of the following oxides would not react with molecular oxygen: NO, N2O, SO2, SO3, P4O6. Which one is it? Why? Predict the outcome of the reactions represented by the 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) 88n Classify 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 2HBr Classify 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 2NO CONCENTRATION OF SOLUTIONS Review Questions 4.53 Write the equation for calculating molarity. Why is molarity a convenient concentration unit in chemistry? 4.54 Describe the steps involved in preparing a solution of known molar concentration using a volumetric flask. Problems 4.55 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 of solution. What is the molarity of the solution? 4.57 How many moles of MgCl2 are present in 60.0 mL of 0.100 M MgCl2 solution? 4.58 How many grams of KOH are present in 35.0 mL of a 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 of solution, (b) 15.4 g of sucrose (C12H22O11) in 74.0 Back Forward Main Menu TOC mL 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 1.50 102 mL 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 to provide the following: (a) 2.14 g of sodium chloride from a 0.270 M solution, (b) 4.30 g of ethanol from a 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 following solutes would be needed to make 2.50 102 mL of a 0.100 M solution: (a) cesium iodide (CsI), (b) sulfuric acid (H2SO4), (c) sodium carbonate (Na2CO3), (d) potassium dichromate (K2Cr2O7), (e) potassium permanganate (KMnO4). DILUTION OF SOLUTIONS Review Questions 4.63 Describe the basic steps involved in diluting a solution of known concentration. 4.64 Write the equation that enables us to calculate the concentration of a diluted solution. Give units for all the terms. Problems 4.65 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 500 mL. What is the concentration of the final solution? 4.67 How would you prepare 60.0 mL of 0.200 M HNO3 from a stock solution of 4.00 M HNO3? 4.68 You have 505 mL of a 0.125 M HCl solution and you want 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 with 16.7 mL of 0.892 M KMnO4 solution. Calculate the concentration 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 ANALYSIS Review Questions 4.71 Describe the basic steps involved in gravimetric analysis. How does this procedure help us determine the identity of a compound or the purity of a compound if its formula is known? Study Guide TOC Textbook Website MHHE Website 147 QUESTIONS AND PROBLEMS 4.72 Distilled water must be used in the gravimetric analysis of chlorides. Why? Problems Problems 4.73 If 30.0 mL of 0.150 M CaCl2 is added to 15.0 mL of 0.100 M AgNO3, what is the mass in grams of AgCl precipitate? 4.74 A sample of 0.6760 g of an unknown compound containing barium ions (Ba2 ) is dissolved in water and treated with an excess of Na2SO4. If the mass of the BaSO4 precipitate formed is 0.4105 g, what is the percent by mass of Ba in the original unknown compound? 4.75 How many grams of NaCl are required to precipitate most of the Ag ions from 2.50 102 mL of 0.0113 M AgNO3 solution? Write the net ionic equation for the reaction. 4.76 The concentration of Cu2 ions in the water (which also contains sulfate ions) discharged from a certain industrial plant is determined by adding excess sodium sulfide (Na2S) solution to 0.800 liter of the water. The molecular equation is Na2S(aq) CuSO4(aq) 88n Na2SO4(aq) CuS(s) Write the net ionic equation and calculate the molar concentration of Cu2 in the water sample if 0.0177 g of solid CuS is formed. ACID-BASE TITRATIONS Review Questions 4.77 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? Problems 4.79 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 solution 4.80 What volume of a 0.50 M HCl solution is needed to neutralize 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 solution REDOX TITRATIONS Review Questions 4.81 What are the similarities and differences between acid-base titrations and redox titrations? 4.82 Explain why potassium permanganate (KMnO4) and Back Forward potassium dichromate (K2Cr2O7) can serve as internal indicators in redox titrations. Main Menu TOC 4.83 Iron(II) can be oxidized by an acidic K2Cr2O7 solution according to the net ionic equation: Cr2O2 7 6Fe2 14H 88n 2Cr3 6Fe3 7H2O If it takes 26.0 mL of 0.0250 M K2Cr2O7 to titrate 25.0 mL of a solution containing Fe2 , what is the molar concentration of Fe2 ? 4.84 The SO2 present in air is mainly responsible for the acid rain phenomenon. Its concentration can be determined by titrating against a standard permanganate solution as follows: 5SO2 2MnO4 2H2O 88n 5SO2 4 2Mn2 4H Calculate the number of grams of SO2 in a sample of air if 7.37 mL of 0.00800 M KMnO4 solution are required for the titration. 4.85 A sample of iron ore weighing 0.2792 g was dissolved in dilute acid solution, and all the Fe(II) was converted to Fe(III) ions. The solution required 23.30 mL of 0.0194 M K2Cr2O7 for titration. Calculate the percent by mass of iron in the ore. (Hint: See Problem 4.83 for the balanced equation.) 4.86 The concentration of a hydrogen peroxide solution can be conveniently determined by titration against a standardized potassium permanganate solution in an acidic medium according to the following equation: 2MnO4 5H2O2 6H 88n 5O2 2Mn2 8H2O If 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. 3 The half-reaction for the oxidation is SO2 3 H2O 88n SO2 4 2H 2e A 100.0-mL sample of solution containing 1.390 g of KIO3 reacts with 32.5 mL of 0.500 M Na2SO3. What is the final oxidation state of the iodine after the reaction has occurred? 4.88 Oxalic acid (H2C2O4) is present in many plants and vegetables. If 24.0 mL of 0.0100 M KMnO4 solution is needed to titrate 1.00 g of H2C2O4 to the equivalence point, what is the percent by mass of H2C2O4 in the sample? The net ionic equation is 2MnO4 Study Guide TOC 16H 5C2O2 88n 4 2Mn2 Textbook Website 10CO2 8H2O MHHE Website 148 REACTIONS IN AQUEOUS SOLUTION 4.89 A quantity of 25.0 mL of a solution containing both Fe2 and Fe3 ions is titrated with 23.0 mL of 0.0200 M KMnO4 (in dilute sulfuric acid). As a result, all of the Fe2 ions are oxidized to Fe3 ions. Next, the solution is treated with Zn metal to convert all of the Fe3 ions to Fe2 ions. Finally, the solution containing only the Fe2 ions requires 40.0 mL of the same KMnO4 solution for oxidation to Fe3 . Calculate the molar concentrations of Fe2 and Fe3 in the original solution. The net ionic equation is MnO4 5Fe2 8H 88n Mn2 5Fe3 4H2O 4.90 Calcium oxalate (CaC2O4) is insoluble in water. For this reason it can be used to determine the amount of Ca2 ions in fluids such as blood. The calcium oxalate isolated from blood is dissolved in acid and titrated against a standardized KMnO4 solution as shown in Problem 4.88. In one test it is found that the calcium oxalate isolated from a 10.0-mL sample of blood requires 24.2 mL of 9.56 10 4 M KMnO4 for titration. Calculate the number of milligrams of calcium per milliliter of blood. ADDITIONAL PROBLEMS 4.91 Classify the following reactions according to the types discussed in the chapter: ClO H2O (a) Cl2 2OH 88n Cl CO2 88n CaCO3 (b) Ca2 3 (c) NH3 H 88n NH4 (d) 2CCl4 CrO2 88n 2COCl2 4 CrO2Cl2 2Cl (e) Ca F2 88n CaF2 (f ) 2Li H2 88n 2LiH Na2SO4 88n 2NaNO3 BaSO4 (g) Ba(NO3)2 (h) CuO H2 88n Cu H2O (i) Zn 2HCl 88n ZnCl2 H2 (j) 2FeCl2 Cl2 88n 2FeCl3 4.92 Oxygen (O2) and carbon dioxide (CO2) are colorless and odorless gases. Suggest two chemical tests that would allow you to distinguish between these two gases. 4.93 Which of the following aqueous solutions would you expect 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)2 4.94 A 5.00 102-mL sample of 2.00 M HCl solution is treated with 4.47 g of magnesium. Calculate the concentration of the acid solution after all the metal has reacted. Assume that the volume remains unchanged. Back Forward Main Menu TOC 4.95 Calculate the volume of a 0.156 M CuSO4 solution that would react with 7.89 g of zinc. 4.96 Sodium carbonate (Na2CO3) is available in very pure form and can be used to standardize acid solutions. What is the molarity of a HCl solution if 28.3 mL of the solution are required to react with 0.256 g of Na2CO3? 4.97 A 3.664-g sample of a monoprotic acid was dissolved in water. It took 20.27 mL of a 0.1578 M NaOH solution to neutralize the acid. Calculate the molar mass of the acid. 4.98 Acetic acid (CH3COOH) is an important ingredient of vinegar. A sample of 50.0 mL of a commercial vinegar 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 needed for 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 AgNO3 solution, silver metal formed on the surface of the strip. After some time had passed, the strip was removed from the solution, dried, and weighed. If the mass of the strip was 3.37 g, calculate the mass of Ag and Zn metals present. 4.101 Calculate the mass of the precipitate formed when 2.27 L of 0.0820 M Ba(OH)2 are mixed with 3.06 L of 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 HNO3 are 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 upset stomach) needed to react with two spoonfuls (approximately 10 mL) of milk of magnesia [at 0.080 g Mg(OH)2/mL]. 4.104 A 1.00-g sample of a metal X (that is known to form X2 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 and identify the element. 4.105 A quantitative definition of solubility is the number of grams of a solute that will dissolve in a given volume of water at a particular temperature. Describe an experiment that would enable you to determine the solubility of a soluble compound. 4.106 A 60.0-mL 0.513 M glucose (C6H12O6) solution is mixed with 120.0 mL of 2.33 M glucose solution. Study Guide TOC Textbook Website MHHE Website QUESTIONS AND PROBLEMS 4.107 4.108 4.109 4.110 4.111 4.112 4.113 4.114 4.115 4.116 4.117 4.118 Back What 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 the compound does indeed dissolve in water to a certain extent? 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 show that it is water. Using the apparatus shown in Figure 4.1, a student found that a sulfuric acid solution caused the light bulb to glow brightly. However, after the addition of a certain amount of a barium hydroxide [Ba(OH)2] solution, the light began to dim even though Ba(OH)2 is also a strong electrolyte. Explain. You are given a soluble compound of unknown molecular formula. (a) Describe three tests that would show that the compound is an acid. (b) Once you have established that the compound is an acid, describe how you would determine its molar mass using a NaOH solution of known concentration. (Assume the acid is monoprotic.) (c) How would you find out whether the acid is weak or strong? You are provided with a sample of NaCl and an apparatus like that shown in Figure 4.1 for comparison. You are given two colorless solutions, one containing NaCl and the other sucrose (C12H22O11). Suggest a chemical and a physical test that would allow you to distinguish between these two solutions. The concentration of lead ions (Pb2 ) in a sample of polluted water that also contains nitrate ions (NO3 ) is determined by adding solid sodium sulfate (Na2SO4) to exactly 500 mL of the water. (a) Write the molecular and net ionic equations for the reaction. (b) Calculate the molar concentration of Pb2 if 0.00450 g of Na2SO4 was needed for the complete precipitation of Pb2 ions as PbSO4. Hydrochloric acid is not an oxidizing agent in the sense that sulfuric acid and nitric acid are. Explain why the chloride ion is not a strong oxidizing agent like SO2 and NO3 . 4 Explain how you would prepare potassium iodide (KI) by means of (a) an acid-base reaction and (b) a reaction between an acid and a carbonate compound. Sodium reacts with water to yield hydrogen gas. Why is this reaction not used in the laboratory preparation of 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 the Forward Main Menu TOC 4.119 4.120 4.121 4.122 4.123 4.124 149 floor of a chemistry laboratory. To neutralize the acid, would it be preferrable to pour concentrated sodium hydroxide solution or spray solid sodium bicarbonate over the acid? Explain your choice and the chemical basis for the action. Describe in each case how you would separate the cations or anions in an aqueous solution: (a) NaNO3 and Ba(NO3)2, (b) Mg(NO3)2 and KNO3, (c) KBr and KNO3, (d) K3PO4 and KNO3, (e) Na2CO3 and NaNO3. The following are common household compounds: table salt (NaCl), table sugar (sucrose), vinegar (contains acetic acid), baking soda (NaHCO3), washing soda (Na2CO3 10H2O), boric acid (H3BO3, used in eyewash), epsom salt (MgSO4 7H2O), sodium hydroxide (used in drain openers), ammonia, milk of magnesia [Mg(OH)2], and calcium carbonate. Based on what you have learned in this chapter, describe test(s) that would allow you to identify each of these compounds. Sulfites (compounds containing the SO2 ions) are 3 used as preservatives in dried fruit and vegetables and in wine making. In an experiment to test the presence of sulfite in fruit, a student first soaked several dried apricots in water overnight and then filtered the solution to remove all solid particles. She then treated the solution with hydrogen peroxide (H2O2) to oxidize the sulfite ions to sulfate ions. Finally, the sulfate ions were precipitated by treating the solution with a few drops of a barium chloride (BaCl2) solution. Write a balanced equation for each of the above steps. A 0.8870-g sample of a mixture of NaCl and KCl is dissolved in water, and the solution is then treated with an excess of AgNO3 to yield 1.913 g of AgCl. Calculate the percent by mass of each compound in the 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 of rust (Fe2O3) from, say, bathtub rings according to the reaction Fe2O3(s) 6H2C2O4(aq) 88n 2Fe(C2O4)3 (aq) 3 3H2O 6H (aq) Calculate the number of grams of rust that can be removed by 5.00 102 mL of a 0.100 M solution of oxalic acid. 4.125 Acetylsalicylic acid (C9H8O4) is a monoprotic acid commonly known as “aspirin.” A typical aspirin tablet, however, contains only a small amount of the acid. In an experiment to determine its composition, an aspirin tablet was crushed and dissolved in water. Study Guide TOC Textbook Website MHHE Website 150 REACTIONS IN AQUEOUS SOLUTION 4.126 4.127 4.128 4.129 4.130 Back It took 12.25 mL of 0.1466 M NaOH to neutralize the 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 dissolved in water, and AgNO3 is added to the solution to form AgBr precipitate. If the mass of the precipitate is 1.6930 g, what is the percent by mass of NaBr in the original mixture? The following “cycle of copper” experiment is performed in some general chemistry laboratories. The series of reactions starts with copper and ends with metallic copper. The steps are as follows: (1) A piece of copper wire of known mass is allowed to react with concentrated 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 to yield 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 with an excess of zinc metal to form metallic copper. (6) The remaining zinc metal is removed by treatment with 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 why it is possible to recover most of the copper used at the start. A 325-mL sample of solution contains 25.3 g of CaCl2. (a) Calculate the molar concentration of Cl in this solution. (b) How many grams of Cl are in 0.100 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 can be generated by reacting CaF2 and NaCl with concentrated sulfuric acid. Write appropriate equations for 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 with concentrated sulfuric acid? (Hint: H2SO4 is a stronger oxidizing agent than both Br2 and I2.) (c) HBr can be prepared by reacting phosphorus tribromide (PBr3) with water. Write an equation for this reaction. Referring to the Chemistry in Action essay on page 143, answer the following questions: (a) Identify the precipitation, acid-base, and redox processes. (b) Instead of calcium oxide, why don’t we simply add sodium hydroxide to seawater to precipitate magnesium hydroxide? (c) Sometimes a mineral called Forward Main Menu TOC dolomite (a mixture of CaCO3 and MgCO3) is substituted for limestone to bring about the precipitation of magnesium hydroxide. What is the advantage of using dolomite? 4.131 Phosphoric acid (H3PO4) is an important industrial chemical used in fertilizers, in detergents, and in the food industry. It is produced by two different methods. In the electric furnace method elemental phosphorus (P4) is burned in air to form P4O10, which is then reacted with water to give H3PO4. In the wet process the mineral phosphate rock [Ca5(PO4)3F] is reacted with sulfuric acid to give H3PO4 (and HF and CaSO4). 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 can be analyzed by titrating a solution of NH4NO3 with a standard NaOH solution. In one experiment a 0.2041-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 and 4.8 to represent the following acid-base reactions: (a) OH H3O 88n 2H2O (b) NH4 NH2 88n 2NH3 Identify the Brønsted acid and base in each case. 4.136 The alcohol content in a 10.0-g sample of blood from a driver required 4.23 mL of 0.07654 M K2Cr2O7 for titration. Should the police prosecute the individual for drunken driving? (Hint: See Chemistry in Action essay on p. 132.) 4.137 On standing, a concentrated nitric acid gradually turns yellow in color. Explain. (Hint: Nitric acid slowly decomposes. Nitrogen dioxide is a colored gas.) 4.138 Describe the laboratory preparation for the following gases: (a) hydrogen, (b) oxygen, (c) carbon dioxide, and (d) nitrogen. Indicate the physical states of the reactants and products in each case. [Hint: Nitrogen can 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 stops even though none of the reactants is used up. Explain. (b) In the activity series, aluminum is above hydro- Study Guide TOC Textbook Website MHHE Website QUESTIONS AND PROBLEMS gen, yet the metal appears to be unreactive toward steam and hydrochloric acid. Why? (c) Sodium and potassium lie above copper in the activity series. Explain why Cu2 ions in a CuSO4 solution are not converted to metallic copper upon the addition of these metals. (d) A metal M reacts slowly with steam. There is no visible change when it is placed in a pale green iron(II) sulfate solution. Where should we place M in the activity series? (e) Before aluminum metal was obtained by electrolysis, it was produced by reducing its chloride (AlCl3) with an active metal. What metals would you use to produce aluminum in that way? 4.140 Metals and activity series crossword. 1. Displaces silver but not lead. 2. Honorary metal in many versions of the activity series. 3. Unreactive metal, a salt of which is used in the chloride test. 4. Most abundant transition metal in Earth’s crust. 5. This metal forms a nitrate which is hard to decompose. 6. Metal used in sacrificial protection of iron from corrosion. 7. First member of Group 1A element. 8. This metal does not react with water, but reacts with acid. (Taken from Metals and the reactivity series, InfoChem, issue no. 23, September 1993. Reprinted with permission of Education in Chemistry.) Back Forward Main Menu TOC 151 1 2 3 4 5 6 7 8 Answers to Practice Exercises: 4.1 (a) Insoluble, (b) in- soluble, (c) soluble. 4.2 Al3 (aq) 3OH (aq) 88n Al(OH)3(s). 4.3 (a) Brønsted base, (b) Brønsted acid. 4.4 (a) P: 3, F: 1; (b) Mn: 7, O: 2. 4.5 (a) Hydrogen displacement reaction, (b) combination reaction, (c) disproportionation reaction, (d) metal displacement reaction. 4.6 0.452 M. 4.7 Dilute 34.2 mL of the stock solution to 200 mL. 4.8 92.02%. 4.9 0.3822 g. 4.10 10.0 mL. 4.11 204 mL. Study Guide TOC Textbook Website MHHE Website C HEMICAL M YSTERY Who Killed Napoleon? A fter 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 of his life. In the 1960s, samples of his hair were analyzed and found to contain a high level of arsenic, suggesting that he might have been poisoned. The prime suspects are the governor of St. Helena, with whom Napoleon did not get along, and the French royal family, who wanted to prevent his return 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 if administered 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 demise so 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 Marsh’s name, combines hydrogen formed by the reaction between zinc and sulfuric acid with a sample of the suspected poison. If As2O3 is present, it reacts with hydrogen to form a toxic gas, arsine (AsH3). When arsine gas is 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 too late to do Napoleon any good, if, in fact, he was a victim of deliberate arsenic poisoning. Doubts about the conspiracy theory of Napoleon’s death developed in the early 1990s, when a sample of the wallpaper from his drawing room was found to contain copper arsenate (CuHAsO4), a green pigment that was commonly used at the time Napoleon lived. It has been suggested that the damp climate on St. Helena promoted the growth of molds on the wallpaper. To rid themselves of arsenic, the molds could have converted it to trimethyl arsine [(CH3)3As], which is a volatile and highly poisonous compound. Prolonged exposure to these vapors would have ruined Napoleon’s health and would also account for the presence of arsenic in his body, though it may not have been the primary cause of his death. This provocative theory is supported by the fact that Napoleon’s regular guests suffered from gastrointestinal disturbances and other 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 fascinating example 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 research to practical applications, such as quality control of commercial products and medical diagnosis. 152 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website CHEMICAL CLUES The arsenic in Napoleon’s hair was detected using a technique called neutron activation. When arsenic-75 is bombarded with high-energy neutrons, it is converted to 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 how much 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 symbols for the two isotopes of As, showing mass number and atomic number. (b) Name two advantages of analyzing the arsenic content by neutron activation instead of a chemical analysis. 2. Arsenic is not an essential element for the human body. (a) Based on its position in the periodic table, suggest a reason for its toxicity. (b) In addition to hair, where else 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 with As(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 in each step. 1. H2SO4 A lock of Napoleon’s hair. Hydrogen flame Shiny metallic ring As2O3 solution Zinc granules Apparatus for the Marsh test. Sulfuric acid is added to zinc metal and a solution containing arsenic(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. 153 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website