630 CHAPTER 15 Chemical Equilibrium 15.89 When dissolved in water, glucose (corn sugar) and fructose (f ruit sugar) exist in equilibrium as follows: fructose •• ==' glucose A chemist prepared a 0.244 M fructose solution at 25°C. At equilibrium, it was found that its concentration had decreased to 0.113 M. (a) Calculate the equilibrium constant for the reaction. (b) At equilibrium, what percentage of fructose was converted to glucose? 15.90 At room temperature, solid iodine is in equilibrium with its vapor through sublimation and deposition. De scribe ho w you would use radioactive iodine, in either solid or vapor form, to show that there is a dynamic equilibrium between these two phases. 15.91 At 1024°C, the pressure of oxygen gas from the decomposition of copper(II) oxide (CuO) is 0.49 atm: 4CuO(s). ' 2CU20(S) + 0 2( g) (a) What is Kp for the reaction? (b) Calculate the fraction of CuO that will decompose if 0.16 mole of it is placed in a 2.0-L flask at 1024°C. (c) What would the fraction be if a l.O-mole sa mple of CuO were used? ( d) What is the smallest amount of CuO ( in moles) that would establish the equilibrium? 15.92 A mixture containing 3.9 moles of NO and 0.88 mole of CO 2 was allowed to react in a flask at a certain temperature according to the equation 15.93 15.94 NO(g ) + COi g) +. =~, N0 2 (g) + CO(g) At equilibrium, 0.11 mole of CO 2 was present. Calculate the equilibrium constant Kc of this reaction. The equilibrium constant Kc for the reaction is 54.3 at 430 °C. At the start of the reaction, there are 0.714 mole of H 2> 0.984 mole of 1 2> and 0.886 mole of HI in a 2.40-L reaction chamber. Calculate the concentrations of the gases at equilibrium. When heated, a gaseous compound A di ssociates as follows: A(g ) :;:::. =:::!:' B(g) + C(g) In an experiment, A was heated at a certain temperature until its equilibrium pressure reached 0.14p, where P is the total pressure. Calculate the equilibrium constant Kp of this reaction. 15.95 When a gas was heated under atmospheric conditions, its color deepened. Heating above 150°C caused the color to fade, and at 550°C the color was barely detectable. Howe ver, at 550°C, the color was partially restored by increasing the pressure of the system. Which of the following best fits the preceding description: . (a) a mixture of hydrogen and bromine, (b) pure bromine, (c) a mixture of nitrogen dioxide and dinitrogen tetroxide. (Hint: Bromine has a reddish color, and nitrogen diox ide is a brown gas. The other gases are colorless.) Justify your choice. 15.96 Both Mg 2+ and Ca 2 + are important biological ion s. One of their functions is to bind to the phosphate group of ATP molecules or amino acids of proteins. For Group 2A metals in general, the equilibrium constant for binding to the anions increases in the order Ba 2+ < Sr 2+ < Ca 2+ < Mg 2+ . What property of the Group 2A metal cations might account for this trend? • 15.97 The equilibrium constant Kc for the following reaction is l.2 at 375°C. 15.98 (a) What is the value of Kp for this reaction? (b) What is the value ofthe equilibrium constant Kc for 2NH 3 (g) • ' N 2 (g) + 3Hig)? (c) What is Kc for i N 2 (g) + ~Hig) • ' NH 3 (g)? (d) What are the values of Kpfor the reactions described in parts (b) and (c)? In this chapter we learned that a catalyst has no effect on the position of an equilibrium because it speeds up both the forward and reverse rates to the same extent. To test this statement, consider a situation in which an equilibrium of the type 2A(g) :;:::. =:::!:' B(g) is establis hed inside a cylinder fitted with a weightless piston. The piston is attached by a string to the cover of a box containing a catalyst. When the piston moves upward (expanding against atmospheric pressure), the cover is lifted and the catalyst is exposed to the gases. When the piston moves downward, the box is closed. Assume that the catalyst speeds up the forward reaction (2A ' B) but does not affect the reverse process (B ' 2A). Suppose the catalyst is suddenly exposed to the equilibrium system as shown here. Describe what would happen subsequently. How does this "thought experiment" convince you that no such catalyst can exist? 2A • 'B ·, String ~ Catalyst ~ _ . _ ;; 15.99 A sealed glass bulb contains a mixture of N0 2 and N 2 0 4 gases. Describe what happens to the following properties of the gases when the bulb is heated from 20°C to 40° C: (a) color, (b) pressure, (c) average molar mass, (d) degree of dissociation (from N 2 0 4 to N0 2 ), (e) density. Assume that volume remains constant. (Hint: N0 2 is a brown gas; N 2 0 4 is colorless.) 15.100 At 20 °C, the vapor pressure of water is 0.0231 atm. Calculate Kp and Kc for the process 15.101 Industrially, sodium metal is obtained by electrolyzing molten sodium chloride. The reaction at the cathode is Na + + e - , Na. We might expect that potassium metal would also be prepared by electrolyzing molten potassium chloride. However, potassium metal is soluble in molten potassium chloride and therefore is hard to recover. Furthermore, potassium vaporizes readily at the operating temperature, creating hazardous conditions. Instead, potassium is prepared by the distillation of molten potassium chloride in the presence of sodium vapor at 892° C: Na(g) + KCI(l) :;:. ==' NaCl(l) + K(g) In view of the fact that potassium is a stronger reducing agent than sodium, explain why this approach works. (The boiling points of sodium and potassium are 892°C and 770°C, respectively .) 15.102 In the gas phase, nitrogen dioxide is actually a mixture of nitrogen dioxide (N0 2 ) and dinitrogen tetroxide (N 2 0 4 ). If the density of such a mixture is 2.3 gIL at 74 °C and 1.3 atm, calculate the partial pressures of the gases and Kp for the dissociation of N 2 0 4 . 15.103 A 2.50-mole sample of NOCI was initially in a 1.50-L reaction chamber at 400°C. After equilibrium was established, it was found that 28.0 percent of the NOCI had dissociated: 15.104 2NOCl(g) +:. :::=:=' 2NO(g) + Cl 2 (g) Calculate the equilibrium constant Kc for the reaction. About 75 percent of hydrogen for industrial use is produced by the steam-reforming process. This process is carried out in two stages called primary and secondary reforming. In the primary stage, a mixture of steam and methane at about 30 atm is heated over a nickel catalyst at 800°C to give hydrogen and carbon monoxide: LlH o = 206 kllmol The secondary stage is can'ied out at about 1000°C, in the presence of air, to convert the remaining methane to hydrogen: (a) What conditions of temperature and pressure would favor the formation of products in both the primary and secondary stages? (b) The equilibrium constant Kc for the primary stage is 18 at 800°C. (i) Calculate Kp for the reaction. (ii) If the partial pressures of methane and steam were both 15 atm at the start, what are the pressures of all the gases at equilibrium? 15.105 Photosynthesis can be represented by 6C0 2 (g) + 6H 2 0(l) +:. ==' C6 H, 20 6 (S ) + 60 2 (g) LlW = 2801 kJ/mol Explain how the equilibrium would be affected by the following changes: (a) partial pressure of CO 2 is increased, (b) O 2 is removed from the mixture, (c) C6 H' 20 6 (glucose) is removed from the mixture, (d) more water is added, (e) a catalyst is added, (f) temperature is decreased. 15.106 Consider the decomposition of ammonium chloride at a certain temperature: Calculate the equilibrium constant Kp if the total pressure is 2.2 atm at that temperature. 15.107 Water is a very weak electrolyte that undergoes the following ionization (called autoionization): kJ + H 2 0(l) • L J ' H (aq) + OH - (aq) (a) Ifk, = 2.4 X lO- s s- ' and L , = 1.3 X lO"IM' s,calculatethe equilibrium constant K where K = [H+ ][OH - ]/[H 2 0]. (b) Calculate the product [H+][OH-], [H+], and [OH- ]. (Hint: Calculate the concentration of liquid water using its density, 1.0 g/mL.) 15.108 Consider the following reaction, which takes place in a single elementary step: 2A + B • LI • A2B If the equilibrium constant Kc is 12.6 at a certain temperature and if k, = 5.1 X 10- 2 S-I, calculate the value of L,. QUESTIONS AND PROBLEMS 631 • 15.109 At 25°C, the equilibrium partial pressures of N0 2 and N 2 0 4 are 0.15 atm and 0.20 atm, respectively. If the volume is doubled at constant temperature, calculate the partial pressures of the gases when a new equilibrium is established. 15.110 In 1899 the German chemist Ludwig Mond developed a process for purifying nickel by converting it to the volatile nickel tetracarbonyl [Ni(CO)4] (b.p. = 42.2 °C): 15.111 Ni(s) + 4CO(g). ' Ni(COMg) (a) Describe how you can separate nickel and its solid impuritie s. (b) How would you recover nickel? [LlH f? for Ni(CO)4 is -602.9 kllmo!.] Consider the equilibrium reaction described in Problem 15.30. A quantity of 2. 50 g of PCIs is placed in an evacuated 0.500-L flask and heated to 250°C. (a) Calculate the pressure of PCIs, assuming it does not dissociate. (b) Calculate the partial pressure of PCls at equilibrium. (c) What is the total pressure at equilibrium? (d) What is the degree of dissociation of PCl s ? (The degree of dissociation is given by the fraction of PCls that has undergone dissociation.) 15.112 Consider the equilibrium system 3A • ' B. Sketch the changes in the concentrations of A and B over time for the following situations: (a) initially only A is present, (b) initially only B is present, (c) initially both A and B are present (with A in higher concentration). In each case, assume that the concentration of B is higher than that of A at equilibrium. 15.113 The vapor pressure of mercury is 0.0020 mrnHg at 26°C. (a) Calculate Kc and Kp for the process Hg(l). ' Hg(g). (b) A chemist breaks a thermometer and spills mercury onto the floor of a laboratory measuring 6.1 m long, 5.3 m wide, and 3.1 m high. Calculate the mass of mercury (in grams) vaporized at equilibrium and the concentration of mercury vapor (in mg/m\ Does this concentration exceed the safety limit of 0.05 mg/m 3 ? (Ignore the volume of furniture and other objects in the laboratory.) 15.114 At 25°C, a mixture of N0 2 and N 2 0 4 gases are in equilibrium in a cylinder fitted with a movable piston. The concentrations are [N0 2 ] = 0.0475 M and [N 2 0 4 ] = 0.487 M. The volume of the gas mixture is halved by pushing down on the piston at constant temperature. Calculate the concentrations of the gases when equilibrium is reestablished. Will the color become darker or lighter after the change? [Hint: Kc for the dissociation of N 2 0 4 is 4.63 X 10- 3 . N 2 0i g) is colorless, and N0 2 (g) has a brown color.] 15.115 A student placed a few ice cubes in a drinking glass with water. A few minutes later she noticed that some of the ice cubes were fused together. Explain what happened. 15.116 Consider the potential energy diagrams for two types of reactions A • • B. In each ca se, answer the following questions for the system at equilibrium. (a) How would a catalyst affect the forward and reverse rates of the reaction? (b) How would a catalyst affect the energies of the reactant and product? (c) How would an increase in temperature affect the equilibrium constant? ( d) If the only effect of a catalyst is to lower the activation energies for the forward and rev er se reactions, show that the equilibrium constant remains unchanged if a catalyst is added to the reacting mixture. > > bJ) OJ) \ '- '" A '" B c: c: '" '" A ~ ~ co co ~ B ~ J:: c: '" '" ~ ~ 0 0 0., 0., Reaction progress Reaction progress 632 CHAPTER 15 Chemical Equilibrium 15.117 The equilibrium constant Ke for the reaction is 0.83 at 375° C. A 14.6-g sample of ammonia is placed in a 4.00-L fla sk and heated to 375° C. Calculate the concentrations of all the gases when equilibrium is reached. 15.118 The dependence of the equilibrium constant of a reaction on temperature is given by the van't Hoff equation: -tll/ 0 In K = RT + C where C is a constant. The following table gives the equilibrium constant (Kp) for the reaction at various temperatures. 2NO(g) + 0 2(g) +=. ====' 2N0 2 (g) 138 5.12 0.436 0.0626 0.0130 T(K) 600 700 800 900 1000 Determine graphically the 6H o for the reaction. 15.119 Cons id er the reaction between N0 2 and N 2 0 4 in a closed container: Initially, 1 mole of N 2 0 4 is present. At equilibrium, x mole of N 2 0 4 has dissociated to form N0 2 . (a) Derive an expression for Kp in terms of x and P, the total pressure. (b) How does the expression in part (a) help you predict the shift in equilibrium due to an increase in P? Doe s yo ur prediction agree with Le Chiitelier's principle? 15.120 (a) Use the van't Hoff equation in Problem 15.] 18 to derive the following expression, which relates the equilibrium constants at two different temperatures: In KI = t: H o (J _ J ) K2 R T2 TI How does this equation support the prediction ba sed on Le Chiitelier's principle about the shift in equilibrium with temperature? (b) The vapor pressures of water are 31.82 mmHg at 30 °C and 92.51 mmHg at 50° C. Calculate the molar heat of vaporization of water. 15.121 The Kp for the reaction S0 2C I2(g) :;:. ==' S0 2(g) + CI 2 (g) is 2 .05 at 648 K. A sample of S0 2Cl2 is placed in a container and heated to 648 K, while the total pressure is kept constant at 9.00 atm. Calculate the partial pressures of the gases at equilibrium. 15.122 The "boat" form and the "chair" form of cyclohexane (C 6 H I2 ) interconvert as shown here: > • Boat Chair In this representation, the H atoms are omitted and a C atom is assumed to be at each intersection of two lines (bonds). The convers i on is fi rst order in each direction. The activation energy for the chair > boat conversion is 41 kJ/mol. If the frequency factor is 1.0 X 10 12 S- I, what is kl at 298 K? The equilibrium constant Ke for the reaction is 9.83 X 10 3 at 298 K. 15.123 Consider the following reaction at a certain temperature 15 .124 A2 + B2 :;:. ==' 2AB The mixing of 1 mole of A2 with 3 moles of B2 gives rise to x mole of AB at equilibrium. The addition of 2 more moles of A2 produces another x mole of AB. What is the equilibrium constant for the reaction? Iodine is sparingly soluble in water but much more so in carbon tetrachloride (CCI 4 ) . The equilibrium constant, also called the partition coefficient, for the distribution of 12 between these two phases is 83 at 20°C. (a) A student adds 0.030 L of CC1 4 to 0.200 L of an aqueous solution containing 0.032 g of 1 2 , The mixture at 20 °C is shaken, and the two phases are then allowed to separate. Calculate the fraction of 12 remaining in the aqueous phase. (b) The student now repeats the extraction OfI2 with another 0.0 30 L of CCI 4 . Calculate the fraction of the 12 from the original solution that remains in the aqueous phase. (c) Compare the result in part (b) with a single extraction using 0.060 L of CCI 4 . Comment on the difference. PRE-PROFESSIONAL PRACTICE EXAM PROBLEMS: PHYSICAL AND BIOLOGICAL SCIENCES Lime (CaO) is used to prevent S0 2 from escaping from the s moke stacks of coal-burning power plants via the formation of solid CaS04 . 2H 2 0 (gyp- sum). One of the important reactions in the overall process is the decom- position of CaC0 3 : CaC0 3 (s) :;:. ==> CaO(s) + CO 2 (g) which has an equilibrium constant (K e) of 3.0 X 10- 6 at ns o c. 1. Calculate the value of Kp for the decomposition of CaC0 3 at 725° C. a) 3.0 X 10- 6 c) 2.0 X 10- 2 b) 2.5 X 10 - 4 d) 3.7 X 10- 8 2. If a 12.0-g sample of solid CaC0 3 is placed in an evacuated vessel at ns o c, what will the pressure of CO 2 be when the system reaches equilibrium? 3. 4. it) 3.0 X 10- 6 atm b) 3.3 X 10 5 atm c) 1.7 X 10- 3 atm d) 1.0 atrn Which of the following actions will cause an increase in the pressure of CO 2 in the vessel described in Question 2? a) Addition of He gas b) Addition of S02 gas c) Addition of more CaC0 3 solid d) Increasing the volume of the vessel If a 12.0-g sample of solid CaC0 3 is placed in a vessel at nsoc in which the pressure of CO 2 is 2.5 atm, wha t mass of CaO will form? a) 6.n g c) 6.00 g b) 12.0 g d) None ANSWERS TO IN-CHAPTER MATERIALS 63 3 ANSWERS TO IN-CHAPTER MATERIALS lS.2B (a) Hz + Cl z • ' 2HCl, (b) H+ + F- :;::. ==:' HF , (c) Cr H + 40H - . ' Cr(OH ); , (d) HC10 • ' H+ + ClO- , (e) H Z S0 3 • ' H+ + HS0 3 , (f) 2NO + Br z . ' 2NOBr. [HClt I lS.3A (a) Kc = , (b) Kc = , [SiCl 4 ] [H2f [Hg2+] [C1 - f [ N'(CO) ] [Z 2+] [NH+ ][OW] (c)K c= 1 4,(d)K c = n .lS.3B (a) Kc = 4 , [Cot [Fe 2 +] [ NH 3] [ Aa (N H 3 ) ~ ][Cl-] (b) Kc = [Ag+][Cl- ], (c) Kc = [Ba 2 + ][F - f , (d) Kc = b -2 [NH 3 ] • lS.4A (a) 2.3 X 10 24 , (b) 6.6 X 10- 13 , (c) 2.8 X 10- 15 lS.4B (a) 1.5 X 34 29 38 (P eal 10 , (b) 1.3 X 10 , (c) 8.5 X 10 . lS.SA (a) Kp = , (Peoi (P o ) 2 2 (P NH ) (b) Kp = P eo , (c ) Kp = 3' lS.SB (a) 2NO z + O 2 • ' 2N 0 3 , 2 (P N )(P H ) 2 2 (b) CH 4 + 2H z O. ' COZ + 4H z , (c) I z + Hz • ' 2HI. lS.6A 8.1 X 10- 6 lS.6B 0.104. 15.7 A Left. lS.7B Right. lS.8A [H z] = [I z ] = 0.056 M; [HI] = 0.413 M. lS.8B [Hz] = [I z ] = 0.043 M; [HI] = 0.314 M. 15.9 [Br] = 8.8 X 10- 3 , [Brz] = 6.5 X 10- 2 lS.10A [PH] = 2 [PI] = 0.37 atm, PHI = 2.75 atm. lS.10B PH = PI = 0.294 atm, P HI = 222 2.16 atm.1S.11 A (a) Right, (b) left, (c) right, (d) left.1S.12A (a) Right, (b) right, (c) right. lS.12B Right shift: remove HF; left shift: remove Hz; no shift: reduce volume of container. Answers to Checkpoints 15.2.1 e. 15.2.2 a. 15.3.1 d. 15.3.2 a. 15.3.3 c. 15.3.4 c. 15.4.1 d. 15.4.2 b. 15.5.1 a, c. 15.5.2 b, d, e. 15.5.3 a, d, e. 15.5.4 b. Answers to Applying What You've learned a) [OD17X] . b) 3.05 X 10- 5 . c) [ODl7X-A77] 30.5. [ODlA] [OF2A] [ODlA][OF 2 A][A77IAj' d) Qc = 3 .6 X 10- 5 ; therefore, the mixture is not at equilibrium. It will have to proceed to the left (rev er se reaction) in order to achieve equilibrium. e) At equilibrium, [ODIA] = 0.05 M, [OF2A] = 0.05 M, [ODl 7X] = 0.95 M. s an ases 16.1 Bf0nsted Acids and Bases 16.2 The Acid-Base Properties of Water 16.3 The pH Scale 16.4 Strong Acids and Bases • Strong Acids • Strong Bases 16.5 Weak Acids and Acid Ionization Constants • The Ionization Constant, Ka • Calculating pH from Ka • Using pH to Determine Ka 16.6 Weak Bases and Base Ionization Constants · • The Ionization Constant, Kb • Calculating pH from Kb • Using pH to Determine Kb 16.7 Conjugate Acid-Base Pairs • The Strength of a Conjugate Acid or Base • The Relationship Between Ka and Kb of a Conjugate Acid-Base Pair 16.8 Diprotic and Polyprotic Acids 16.9 Molecular Structure and Acid Strength • Hydrohalic Acids • Oxoacids • Carboxylic Acids 16.10 Acid-Base Properties of Salt Solutions • Basic Salt Solutions • Acidic Salt Solutions • Neutral Salt Solutions • Salts in Which Both the Cation and the Anion Hydrolyze 16.11 Acid-Base Properties of Oxides and Hydroxides • Oxides of Metals and of Nonmetals • Basic and Amphoteric Hydroxides 16.12 Lewis Acids and Bases An Acid Essential to Good Health Scurvy is a devastating disease caused by a deficiency of vitamin C. It ha s a host of terrible symptoms including extreme fatigue and weakness, hemorrhaging, and lo ss of teeth due to inflamed gums. Untreated, it results in death, usually from pneumonia or other acute infection-or from heart failure. Most mammals make their own vitamin C, but humans and other primates, as well as fruit bats and guinea pigs, must obtain it from their diet. Since ancient times, people were at risk for contracting scurvy when their diets did not include an adequate supply of fresh fruits and vegetables. One segment of the population historically deprived of an appropriate diet was sailors. During the fourteenth and fifteenth centuries, as the development of ships and sails made voyages on the open ocean possible, and returns to shore became less frequent, scurvy became . . . . . . . . . - - . commonplace among sailors. Although reports of cures for scurvy date back to the sixteenth century, it was not until 1747 that the effectiveness of citrus was proven with the experiments of Scottish naval surgeon James Lind. It was another 40 years, though, before the British navy began supplying citrus juice to its crews. It was the compulsory iss ue of the juice of lemons or limes that gave rise to the term lime y, referring to a British sailor. Vitamin C, or ascorbic acid (H 2 C 6 H 6 0 6 ), acts as an antioxidant, reacting with oxidizing species such as the ·OH radical, which can damage an organism's DNA. It is one of many acids that are important to human health. • Ascorbic acid • One hundred of Portuguese explorer Vasco de Gama 's 160 - member crew died from scurvy by the time they had sailed around the sou thern tip of Africa in 1497. In This Chapter, You Will Learn more about the properties of acids and bases and how those properties are related to molecular structure. Before you begin, you should review • The list of strong acids and the list of strong bases [~ . Section 4.3, Table 4.4] • How to solve equilibrium problems [ ~ . Section 15.4] Acids such as citric acid and ascorbic acid (v itamin C) are responsible for the sour tast e of citrus fruits. Media Player/ MPEG Content Chapter in Review 635 636 CHAPTER 16 Acids and Bases Conjugate Species Base CH 3 COOH CH 3 COO - H 2 O OH - NH3 NH2" H 2 SO 4 HS0 4 Conjugate Species Acid NH 3 NH t H 2 O H3 0 + OH- H 2 O H 2 NCONH 2 H 2 NCON H! (urea) Think About It A species does not need to be what we think of as an acid in order for it to have a conjugate bas e. For example, we would not ref er to the hydroxide ion (OH - ) as an acid- but it does have a conjugate base, the oxide ion (0 2 - ) . Furthermore, a species that can either lose or gain a proton, such as HC0 .1 , has b ot h a conjugate base ( CO ~-) and a conjugate acid (H 2 C0 3 ). Br¢nsted Acids and Bases In Chapter 4 we learned that a Br ¢nsted acid is a substance that can donate a proton and a Br¢nsted ba se is a substance that can accept a proton [ ~~ Section 4.3]. In this chapter we extend our di s- cussion of Br¢nsted acid-base theory to include conjugate acids and conjugate bases. When a Br¢nsted acid donates a proton, what remains of the acid is known as a conjugate base. For example, in the ionization of HCI in water, HCI(aq) + H 2 0(l) +=. ~. H30 +(aq) + Cl - (aq) acid con jug ate base HCI donates a proton to water, producing the hydronium ion CH30 +) and the chloride ion CCI-), which is the conjugate base of HCl. The two species, HCI and Cl- , are known as a conjugate acid-base pair or simply a conjugate pair. Table 16.1 li s ts the conjugate bases of several familiar species. Conversely, when a Br¢n sted base accepts a proton, the newly formed protonated species is known as a conjugate acid. When ammonia (NH3) ionizes in water, NH 3(aq) + H 2 0(l) +=. ~. NH t Caq) + OH - (aq) base con jug ate acid NH 3 accepts a proton from water to become the ammonium ion (NH)t . The ammonium ion is the conjugate acid of ammonia. Table 16.2 lists the conjugate acids of several common species. Any reaction that we describe using Br¢nsted acid-base theory involves an acid and a base. The acid donates the proton, and the base accepts it. Furthermore, the products of such a reaction are always a conjugate base and a conjugate acid. It is useful to identify and label each species in a Br¢nsted acid-base reaction. For the ionization of HCl in water, the species are labeled as follows: loses a proton 1 - , gains a proton HCI(aq) + H2 0 (l) • H3 0 +(aq) + CI-(aq) • acid base r. ;.:., ~d ~ conjugate • 1 ell! base And for the ionization of NH 3 in water, gains a proton loses a proton NH3(aq) + H 2 °(l) NHt(aq) • + OH-(aq) • base acid • • ~ . r , conjugate Ck base Sample Problems 16.1 and 16.2 let you practice identifying conjugate pairs and the species in a Br¢nsted acid-base reaction. " Sample Problem 16.1 ".""" What is (a) the conju ga te base of H N0 3 , (b) the conjugate acid of 0 2 - , (c) the conjugate base of HS0 4 , and (d) the conjugate acid of HC0 .1 ? Strategy To find the conjugate base of a specie s, remove a proton from the formula. To find the conjugate acid of a species, add a proton to the formula. Setup The word proton, in this context, refers to H+. Thus, the formula and the charge will both be affected by the addition or removal of H+. Solution (a) NO .1 (b) OW (c) SO ~  (d) H 2 C0 3 SECTION 16.2 The Acid-Base Properties of Water 637 Practice Problem A What is (a) the conjugate acid of Cl0 4 , (b) the conjugate acid of S2 - , (c) the conjugate base of H 2 S, and (d) the conjugate base of H 2 C 2 0 4 ? Practice Problem B HS0 3 is the conjugate acid of what species? HS0 3 is the conjugate base of what species? Label each of the species in the following equations as an acid, base , conjugate base, or conjugate acid: (a) HF(aq) + NH 3 (aq). • F- (aq) + NH r (aq) (b) CH 3 COO- (aq) + H 2 0(l). • CH 3 COOH(aq) + OW(aq) Strategy In each equation, the reactant that loses a proton is the acid and the reactant that gains a proton is the base. Each product is the conjugate of one of the reactants. Two species that differ only by a proton constitute a conjugate pair . . Setup (a) HF loses a proton and becomes F- ; NH 3 gains a proton and becomes NH r . (b) CH 3 COO - gains a proton to become CH 3 COOH; H 2 0 loses a proton to become OH Solution (a) HF(aq) + NH 3 (aq) ::0:. =~. F-(aq) + Nm(aq) acid base conjugate conjugate base acid (b) CH 3 COO- (aq) + H 2 0(l) ::o:.=~' CH 3 COOH(aq) + OH-(aq) base acid conjugate acid conjugate ba se Practice Problem A Identify and label the species in each reaction. (a) NHr(aq) + H 2 0(l). • NH 3(aq) + H30 +(aq) (b) CN - (aq) + H 2 0(l). • HCN(aq) + OW(aq) Practice Problem B (a) Write an equation in which HS0 4 reacts (with water) to form its conjugate base. (b) Write an equation in which HS0 4 reacts (with water) to form its conjugate acid. Checkpoint 16.1 Bnmsted Acids and Bases 16 .1.1 Which of the following pairs of species are conjugate pairs? (Select all that apply.) a) H 2 S and S2- b) NH2 and NH 3 c) O 2 and H 2 0 2 d) HBr and Br - e) HCI and OH- 16.1.2 Which of the following species does not have a conjugate base? (Select all that apply.) a) HC 2 0 4 b) OW c) 0 2 - d) CO ~- e) HCIO The Acid-Base Properties of Water Water is often referred to as the "universal solvent," because it is so common and so important to life on Earth. In addition, most of the acid-base chemistry that you will encounter takes place in aqueous solution. In this section, we take a closer look at water's ability to act as either a Br¢nsted acid (as in the ionization of NH 3 ) or a Brlilnsted base (as in the ionization of Hel). A species that can behave either as a Br¢nsted acid or a Br¢nsted base is called amphoteric. Water is a very weak electrolyte, but it does undergo ionization to a small extent: Thi nk About It In a Br¢nsted acid-base reaction, there is always an acid and a base, and whether a substance behaves as an acid or a base depends on what it is combined with. Water, for example, behaves as a base when combined with HCI but behaves as an acid when combined with NH 3 . 638 CHAPTER 16 Acids and Bases ." t. ~ ~. Mu ltimedia Chemical Equilibrium equilibrium (interactive). Recall that in a heterogeneous equilibr iu m such as this, liquids and solids do not appear in the equilibrium express i on [ ~~ Section 15.3] . The co nst an t Kw is so metimes referr ed to as the ion-product constant. Re call that we disregard the units when we s ub stitute conc entrations into an equilib riu m expression [ ~~ Section 15.5]. Bec ause the product of H30+ and OW concentrations is a constant, we cannot alter the conc en trations independently. Any change in o ne a lso affects the oth er . Think About It Remember that equilibri urn constants are temperature dependent. The value of Kw is 1.0 X 10 - 14 only at 25 ° C. This reaction is known as the autoionization o/water. Because we can represent the aqueous pro- ton as either H+ or H30 + [ ~~ Section 4.3] , we can also write the auto ionization of water as H30 +(aq) + OH-(aq) + • acid base . • conjugate acid + + conjugate base where one water molecule acts as an acid and the other acts as a base . As ·lildkatecn:' y" ih·e ·do ubie ·arrov,i"iu· the· e· ci.uatlOii : the· i-e ·actlon IS a il equiiibii"um: The equilib- rium expression for the autoionization of water is or Because the autoionization of water is an important equilibrium that you will encounter frequently in the study of acids and base s, we use the subscript w to indicate that the equilibrium constant is · . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . , that specifically for the autoionization of water. It is important to realize, though, that Kw is simply a Kc for a specific reaction. We will frequently replace the c in Kc expressions with a letter or a series of letters to indicate the specific type of reaction to which the Kc refers. For example, Kc for the ionization of a weak acid is called Ka, and Kc for the ionization of a weak base is called K b . In Chapter 17 we will encounter Ks p, where "s p" stands for "solubility product." Each specially subscripted K is simply a Kc for a specific type of reaction. In pure water, autoionization is the only source ofH 30 + and OH-, and the stoichiometry of the reaction tells us that their concentrations are equal. At 25 °C, the concentrations of hydronium and hydroxide ion s in pure water are [H30 +] = [OH - ] = 1.0 X 10 - 7 M. Using the equilibrium · . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . express ion , we can calculate the value of Kw at 25 °C as follows: Kw = [H 3 0 + ][OH-] = (1.0 X 10 - 7 )(1.0 X 10- 7 ) = 1.0 X 10 - 14 FUltherrnore, in any aqueous solution at 25 ° C, the product of H30 + and OH - concentrations is equal to 1.0 X 10 - 14 . Equation 16.1 Although their product is a constant , the individual concentrations of hydronium and hydroxide · . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . + can be influenced by the addition of an acid or a base. The relative amounts of H30 and OH - determine whether a solution is neutral, acidic, or basic. • When [H30 +] = [OH - ], the solution is neutral. • When [H30 +] > [OH - ], the solution is acidic. • When [H30 +] < [OH - ], the solution is basic. Sample Problem 16.3 shows how to use Equation 16.1. Sample Problem 16.3 • The concentration of hydronium ions in stomach acid is 0. 10 M. Calculate the concentration of hydroxide ions in stomach acid at 25 ° C. Strategy Use the value of Kw to determine [OH - ] when [H 30 +] = 0.10 M. Setup Kw = [H 3 0 + ][OH - ] = 1.0 X 10 - 14 at 25 ° C. Rearranging Equation 16.1 to solve for [OH - ], [OH - ] = 1.0 X 10 - 14 [H 3 0 +] Solution [OH - ] = 1.0 X 10 - 14 = 1.0 X 10 - 13 M 0.10 Practice Problem A The concentration of hydroxide ions in the antacid milk of magnesia is 5.0 X 10 - 4 M. Calculate the concentration of hydronium ions at 25 ° C. Practice Problem B The value of Kw at normal body temperature (37 ° C) is 2.8 X 10 - 14 . Calculate th e concentration of hydroxide ions in stomach acid at body temperature. ( [H 30 +] = 0.10 M) Checkpoint 16.2 The Acid-Base Properties of Water 16.2.1 Calculate [OH-] in a solution in which [H30 +] = 0.0012 M at 25°C. a) l.2 X 10- 3 M b) 8.3 X 10- 17 M c) l.0 X 10- 14 M d) 8.3 X 10- 12 M e) 1.2XlO II M The pH Scale 16.2.2 Calculate [H30 +] in a solution in which [OH-] = 0.25 M at 25°C. a) 4.0 X 10- 14 M b) 1.0 X 10- 14 M c) 2.5 X 10 13 M d) 1.0 X 10- 7 M e) 4.0 X 10- 7 M The acidity of an aqueous solution depends on the concentration of hydronium ions, [H30 + ]. This concentration can range over many orders of magnitude, which can make reporting the numbers cumbersome. To describe the acidity of a solution, rather than report the molar concentration of hydronium ions, we typically use the more convenient pH scale. The pH of a solution is defined as SECTION 16.3 The pH Scale 639 the negative base-l 0 logarithm ' of the' hydi-oni' um ' {on 'coricenii-iti'on ' (In ' motA.)". ·········· . Equation 16 .2 converts numbers that can span an enormous range (-10 ' to 10- 14 ) to numbers generally ranging from -1 to 14. or Equation 16.2 The pH of a solution is a dimensionless quantity, so the units of concentration must be removed from [H30 +] before taking the logarithm. Because [H30 +] = [OH- ] = 1.0 X 10 - 7 M in pure water at 25°C, the pH of pure water at 25°C is -log (1.0 X 10 - 7 ) = 7.00 " . A word about significant figu res: When we take the log of a number with two sign ificant figures, we report the result to two places past the decimal point. Thus, pH 7.00 has two signifi cant fi gures, not three. Remember, too, that a solution in which [H30 +] = [OH- ] is neutral. At 25°C, therefore, a neutral solution has pH 7.00. An acidic solution, one in which [H30 +] > [OH-] , ha s pH < 7.00, whereas a basic solution, in which [H30 +] < [OH-], ha s pH > 7.00. Table 16.3 shows the calculation of pH for solutions ranging from 0.10 M to 1.0 X 10- 14 M. In the laboratory, pH is measured with a pH meter (Figure 16.1). Table 16.4 lists the pH val- ues of a number of common fluids. Note that the pH of body fluids varies greatly, depending on the location and function of the fluid. The low pH (high acidity) of gastric juices is vital for digestion of food, whereas the higher pH of blood is required to facilitate the transport of oxygen. [H30+](M) -log [H30+] 0.10 - log (1.0 X 10- 1 ) 0.010 -log (1.0 X 10- 2 ) 1.0 X 10 3 -log (1.0 X 10- 3 ) 1.0 X 10 4 -log (1.0 X 10- 4 ) 1.0 X 10 - 5 ;;;''''I====~ l = o = g = (l = .0 = X = I = 0 ~ 5 ) 1.0 X 10- 6 -log (1.0 X 10- 6 ) 1.0 X 10- 7 - log (1.0 X 10- 7 ) 1.0 X 10- 8 -log (1.0 X 10 - 8 ) 1.0 X 10- 9 -log (1.0 X 10- 9 ) 1.0 X 10- 10 -log (1.0 X 10- 1 °) 1.0 X 10- 11 -log (1.0 X 10- 11 ) 1.0 X 10- 12 -log (1.0 X 10- 12 ) 1.0 X 10- 13 - log (1.0 X 10- 13 ) 1.0 X 10 - 14 -log (1.0 X 10- 14 ) pH 1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.00 11.00 12.00 13.00 14.00 Acidic Neutral Basic Figure 16.1 ApHmeteris commonly used in the laboratory to determine the pH of a solution. Although many pH meters have a ran ge of 1 to 14, pH values can actually be less than 1 and greater than 14. [...]... a solution at 25 °C in which the pOH is Ca) 13.02, (b) 5.14, and (c) 6.98 Practice Problem B Calculate the hydroxide ion concentration in a solution at 2YC in which the pOH is C 11 .26, (b) 3.69, and (c) 1.60 a) Bringing Chemistry to Life Antacids and the pH Balance in Your Stomach An average adult produces between 2 and 3 L of gastric juice daily Gastric juice is an acidic digestive fluid secreted... [H30 +] = 10- 1195 (c) [H30 +] = 10- Think About It If you use the calculated hydronium ion concentrations to recalculate pH, you will get numbers slightly different from those given in the problem In part (a), for example, -log (1.7 X 10- 5) = 4.77 The small difference between this and 4.76 (the pH given in the problem) is due to a rounding error Remember that a concentration derived from a pH with... On the pOH scale, numbers below 7 indicate a basic solution, whereas numbers above 7 indicate an acidic solution The pOH benchmarks (abbreviated in Table 16.5) work the same way the pH benchmarks do In part (a), for example, a hydroxide ion concentration between 1 X 10- 4 M and 1 X 10- 5 M conesponds to a pOH between 4 and 5: [OH- ] (M) pOH 4 4.00 4.43 t 5.00 1.0 X 103.7 X 10- 5 * 1.0 X 10- 5 Determine... which the pOH is (a) 4.91, C 9.03, and (c) 10.55 b) Strategy Given pOH, use Equation 16.5 to calculate [OH- ] Setup (a) [OH-] benchmark pOH values to determine whether these solutions are reasonable In part (a), for example, the pOH between 4 and 5 corresponds to [OH- ] between 1 X 10- 4 M and 1 X 10- 5 M 10- 491 (b) [OH- ] Think About It Use the = = 10- 903 (c) [OH-] = 10- 1055 Solution (a) [OW] = 1.2... -log (8 8 X 10- 11 ) (c) pH = 10.06 Think About It When a hydronium ion concentration falls between two "benchmark" concentrations in Table 16.3, the pH falls between the two corresponding pH values In part (c), for example, the hydronium ion concentration (8.8 X 10- 11 M) is greater than 1.0 X 10- 11 Mbut less than 1.0 X 10- 10 M Therefore, we expect the pH to be between 11.00 and 10.00 pH 1.0 X 10-... aqueous solution Think About It Again, note that when a hydronium ion concentration falls between two of the benchmark concentrations in Table 16.3, the pH falls between the two corresponding pH values In part (b), for example, the hydronium ion concentration of 1.2 X 10- 4 M is greater than 1.0 X 10- 4 M and less than 1.0 X 10- 3 M Therefore, we expect the pH to be between 4.00 and 3.00 Calculate the pH... twice that of the base: Ba(OHhCaq) - _ Ba2+(aq) Think About It These are basic pOH values, which is what we should expect for the solutions described in the problem Note that while the sol utions in parts (a) and (b) have the same base concentration, they do not have the same hydroxide concentration and therefore do not have the same pOH + 20H - (aq) Therefore, [OW] = 2 X [Ba(OH)2] = 2(0.013 M) =... base such as the acetate ion is introduced into a solution by dissolving a solu ble salt containing acetate Sodium acetate, for example, can be used to supply the acetate ion The sodium ion does not take part in the reaction-it is a spectator ion [ ~~ Section 4.2] . 25°C, the equilibrium partial pressures of N0 2 and N 2 0 4 are 0.15 atm and 0.20 atm, respectively. If the volume is doubled at constant temperature, calculate the partial pressures of. hydroxide ion concentration in a solution at 2YC in which the pOH is Ca) 11 .26, (b) 3.69, and (c) 1.60. Bringing Chemistry to Life Antacids and the pH Balance in Your Stomach An average. (N 2 0 4 ). If the density of such a mixture is 2.3 gIL at 74 °C and 1.3 atm, calculate the partial pressures of the gases and Kp for the dissociation of N 2 0 4 . 15.103 A 2.50-mole