1. Acids and Bases Chapter 16 Acids and Bases

2. Acids and Bases Some Definitions Arrhenius  Acid:Substance that, when dissolved in water, increases the concentration of hydrogen ions.  Base:Substance that, when dissolved in water, increases the concentration of hydroxide ions.

3. Acids and Bases Some Definitions Brønsted–Lowry  Acid:Proton donor  Base:Proton acceptor

4. Acids and Bases A Brønsted–Lowry acid… …must have a removable (acidic) proton. A Brønsted–Lowry base… …must have a pair of nonbonding electrons.

5. Acids and Bases If it can be either…...it is amphiprotic. HCO 3 − HSO 4 − H2OH2O

6. Acids and Bases What Happens When an Acid Dissolves in Water? Water acts as a Brønsted–Lowry base and abstracts a proton (H + ) from the acid. As a result, the conjugate base of the acid and a hydronium ion are formed.

7. Acids and Bases Conjugate Acids and Bases: From the Latin word conjugare, meaning “to join together.” Reactions between acids and bases always yield their conjugate bases and acids.

8. Acids and Bases SAMPLE EXERCISE 16.1 Identifying Conjugate Acids and Bases (a) What is the conjugate base of each of the following acids: HClO 4, H 2 S, PH 4 +, HCO 3 – ? (b) What is the conjugate acid of each of the following bases: CN –, SO 4 2–, H 2 O, HCO 3 – ? PRACTICE EXERCISE Write the formula for the conjugate acid of each of the following: HSO 3 –, F –, PO 4 3–, CO. Answers: H 2 SO 3, HF, HPO 4 2–, HCO +

9. Acids and Bases Acid and Base Strength Strong acids are completely dissociated in water.  Their conjugate bases are quite weak. Weak acids only dissociate partially in water.  Their conjugate bases are weak bases.

10. Acids and Bases Acid and Base Strength Substances with negligible acidity do not dissociate in water.  Their conjugate bases are exceedingly strong.

11. Acids and Bases Acid and Base Strength In any acid-base reaction, the equilibrium will favor the reaction that moves the proton to the stronger base. HCl (aq) + H 2 O (l)  H 3 O + (aq) + Cl − (aq) H 2 O is a much stronger base than Cl −, so the equilibrium lies so far to the right K is not measured (K>>1).

12. Acids and Bases Acid and Base Strength Acetate is a stronger base than H 2 O, so the equilibrium favors the left side (K<1). C 2 H 3 O 2 (aq) + H 2 O (l) H 3 O + (aq) + C 2 H 3 O 2 − (aq)

13. Acids and Bases SAMPLE EXERCISE 16.3 Predicting the Position of a Proton-Transfer Equilibrium For the following proton-transfer reaction, use Figure 16.4 to predict whether the equilibrium lies predominantly to the left (that is, K c 1):

14. Acids and Bases Answers: (a) left, (b) right SAMPLE EXERCISE 16.3 continued PRACTICE EXERCISE For each of the following reactions, use Figure 16.4 to predict whether the equilibrium lies predominantly to the left or to the right:

15. Acids and Bases Autoionization of Water As we have seen, water is amphoteric. In pure water, a few molecules act as bases and a few act as acids. This is referred to as autoionization. H 2 O (l) + H 2 O (l) H 3 O + (aq) + OH − (aq)

16. Acids and Bases Ion-Product Constant The equilibrium expression for this process is K c = [H 3 O + ] [OH − ] This special equilibrium constant is referred to as the ion-product constant for water, K w. At 25°C, K w = 1.0  10 −14

17. Acids and Bases SAMPLE EXERCISE 16.4 Calculating [H + ] for Pure Water Calculate the values of [H + ] and [OH – ] in a neutral solution at 25°C. PRACTICE EXERCISE Indicate whether solutions with each of the following ion concentrations are neutral, acidic, or basic: (a) [H + ] = 4  10 –9 M; (b) [OH – ] = 1  10 –7 M; (c) [OH – ] = 7  10 –13 M. Answers: (a) basic, (b) neutral, (c) acidic

18. Acids and Bases SAMPLE EXERCISE 16.5 Calculating [H + ] from [OH – ] Calculate the concentration of H + (aq) in (a) a solution in which [OH – ] is 0.010 M, (b) a solution in which [OH – ] is 1.8  10 –9 M. Note: In this problem and all that follow, we assume, unless stated otherwise, that the temperature is 25°C.

19. Acids and Bases pH pH is defined as the negative base-10 logarithm of the hydronium ion concentration. pH = −log [H 3 O + ]

20. Acids and Bases pH In pure water, K w = [H 3 O + ] [OH − ] = 1.0  10 −14 Because in pure water [H 3 O + ] = [OH − ], [H 3 O + ] = (1.0  10 −14 ) 1/2 = 1.0  10 −7

21. Acids and Bases pH Therefore, in pure water, pH = −log (1.0  10 −7 ) = 7.00 An acid has a higher [H 3 O + ] than pure water, so its pH is <7 A base has a lower [H 3 O + ] than pure water, so its pH is >7.

22. Acids and Bases pH These are the pH values for several common substances.

23. Acids and Bases Other “p” Scales The “p” in pH tells us to take the negative log of the quantity (in this case, hydrogen ions). Some similar examples are  pOH −log [OH − ]  pK w −log K w

24. Acids and Bases Watch This! Because [H 3 O + ] [OH − ] = K w = 1.0  10 −14, we know that −log [H 3 O + ] + −log [OH − ] = −log K w = 14.00 or, in other words, pH + pOH = pK w = 14.00

25. Acids and Bases SAMPLE EXERCISE 16.6 Calculating pH from [H + ] Calculate the pH values for the two solutions described in Sample Exercise 16.5.

26. Acids and Bases SAMPLE EXERCISE 16.6 continued PRACTICE EXERCISE (a) In a sample of lemon juice [H + ] is 3.8  10 –4 M. What is the pH? (b) A commonly available window- cleaning solution has a [H + ] of 5.3  10 –9 M. What is the pH? Answers: (a) 3.42, (b) 8.28

27. Acids and Bases SAMPLE EXERCISE 16.7 Calculating [H + ] from pH A sample of freshly pressed apple juice has a pH of 3.76. Calculate [H + ]. PRACTICE EXERCISE A solution formed by dissolving an antacid tablet has a pH of 9.18. Calculate [H + ]. Answer: [H + ] = 6.6  10 –10 M

28. Acids and Bases How Do We Measure pH? For less accurate measurements, one can use  Litmus paper “Red” paper turns blue above ~pH = 8 “Blue” paper turns red below ~pH = 5  An indicator

29. Acids and Bases How Do We Measure pH? For more accurate measurements, one uses a pH meter, which measures the voltage in the solution.

30. Acids and Bases Strong Acids You will recall that the seven strong acids are HCl, HBr, HI, HNO 3, H 2 SO 4, HClO 3, and HClO 4. These are, by definition, strong electrolytes and exist totally as ions in aqueous solution. For the monoprotic strong acids, [H 3 O + ] = [acid].

31. Acids and Bases Strong Bases Strong bases are the soluble hydroxides, which are the alkali metal and heavier alkaline earth metal hydroxides (Ca 2+, Sr 2+, and Ba 2+ ). Again, these substances dissociate completely in aqueous solution.

32. Acids and Bases SAMPLE EXERCISE 16.8 Calculating the pH of a Strong Acid What is the pH of a 0.040 M solution of HClO 4 ? PRACTICE EXERCISE An aqueous solution of HNO 3 has a pH of 2.34. What is the concentration of the acid? Answer: 0.0046 M

33. Acids and Bases SAMPLE EXERCISE 16.9 Calculating the pH of a Strong Base What is the pH of (a) a 0.028 M solution of NaOH, (b) a 0.0011 M solution of Ca(OH) 2 ?

34. Acids and Bases SAMPLE EXERCISE 16.9 continued Answers: (a) 7.8  10 –3 M, (b) 2.4  10 –13 M PRACTICE EXERCISE What is the concentration of a solution of (a) KOH for which the pH is 11.89; (b) Ca(OH) 2 for which the pH is 11.68?

35. Acids and Bases Dissociation Constants For a generalized acid dissociation, the equilibrium expression would be This equilibrium constant is called the acid-dissociation constant, K a. [H 3 O + ] [A − ] [HA] K c = HA (aq) + H 2 O (l) A − (aq) + H 3 O + (aq)

36. Acids and Bases Dissociation Constants The greater the value of K a, the stronger the acid.

37. Acids and Bases Calculating K a from the pH The pH of a 0.10 M solution of formic acid, HCOOH, at 25°C is 2.38. Calculate K a for formic acid at this temperature. We know that [H 3 O + ] [COO − ] [HCOOH] K a =

38. Acids and Bases Calculating K a from the pH The pH of a 0.10 M solution of formic acid, HCOOH, at 25°C is 2.38. Calculate K a for formic acid at this temperature. To calculate K a, we need the equilibrium concentrations of all three things. We can find [H 3 O + ], which is the same as [HCOO − ], from the pH.

39. Acids and Bases Calculating K a from the pH pH = −log [H 3 O + ] 2.38 = −log [H 3 O + ] −2.38 = log [H 3 O + ] 10 −2.38 = 10 log [H 3 O + ] = [H 3 O + ] 4.2  10 −3 = [H 3 O + ] = [HCOO − ]

40. Acids and Bases Calculating K a from pH Now we can set up a table… [HCOOH], M[H 3 O + ], M[HCOO − ], M Initially0.1000 Change −4.2  10 -3 +4.2  10 -3 +4.2  10 −3 At Equilibrium 0.10 − 4.2  10 −3 = 0.0958 = 0.10 4.2  10 −3

41. Acids and Bases Calculating K a from pH [4.2  10 −3 ] [0.10] K a = = 1.8  10 −4

42. Acids and Bases Calculating Percent Ionization Percent Ionization =  100 In this example [H 3 O + ] eq = 4.2  10 −3 M [HCOOH] initial = 0.10 M [H 3 O + ] eq [HA] initial

43. Acids and Bases Calculating Percent Ionization Percent Ionization =  100 4.2  10 −3 0.10 = 4.2%

44. Acids and Bases Calculating pH from K a Calculate the pH of a 0.30 M solution of acetic acid, HC 2 H 3 O 2, at 25°C. HC 2 H 3 O 2 (aq) + H 2 O (l) H 3 O + (aq) + C 2 H 3 O 2 − (aq) K a for acetic acid at 25°C is 1.8  10 −5.

45. Acids and Bases Calculating pH from K a The equilibrium constant expression is [H 3 O + ] [C 2 H 3 O 2 − ] [HC 2 H 3 O 2 ] K a =

46. Acids and Bases Calculating pH from K a We next set up a table… [C 2 H 3 O 2 ], M[H 3 O + ], M[C 2 H 3 O 2 − ], M Initially0.3000 Change−x−x+x+x+x+x At Equilibrium 0.30 − x  0.30 xx We are assuming that x will be very small compared to 0.30 and can, therefore, be ignored.

47. Acids and Bases Calculating pH from K a Now, (x) 2 (0.30) 1.8  10 −5 = (1.8  10 −5 ) (0.30) = x 2 5.4  10 −6 = x 2 2.3  10 −3 = x

48. Acids and Bases Calculating pH from K a pH = −log [H 3 O + ] pH = −log (2.3  10 −3 ) pH = 2.64

49. Acids and Bases SAMPLE EXERCISE 16.10 Calculating K a and Percent Ionization from Measured pH A student prepared a 0.10 M solution of formic acid (HCHO 2 ) and measured its pH using a pH meter of the type illustrated in Figure 16.6. The pH at 25°C was found to be 2.38. (a) Calculate K a for formic acid at this temperature. (b) What percentage of the acid is ionized in this 0.10 M solution?

50. Acids and Bases A 0.020 M solution of niacin has a pH of 3.26. (a) What percentage of the acid is ionized in this solution? (b) What is the acid-dissociation constant, K a, for niacin? Answers: (a) 2.7%, (b) 1.5  10 –5 SAMPLE EXERCISE 16.10 continued PRACTICE EXERCISE Niacin, one of the B vitamins, has the following molecular structure:

51. Acids and Bases SAMPLE EXERCISE 16.11 Using K a to Calculate pH Calculate the pH of a 0.20 M solution of HCN. (Refer to Table 16.2 or Appendix D for the value of K a.)

52. Acids and Bases PRACTICE EXERCISE The K a for niacin (Practice Exercise 16.10) is 1.5  10 –5. What is the pH of a 0.010 M solution of niacin? Answer: 3.42 SAMPLE EXERCISE 16.11 continued

53. Acids and Bases SAMPLE EXERCISE 16.12 Using K a to Calculate Percent Ionization Calculate the percentage of HF molecules ionized in (a) a 0.10 M HF solution, (b) a 0.010 M HF solution.

54. Acids and Bases SAMPLE EXERCISE 16.12 continued PRACTICE EXERCISE In Practice Exercise 16.10, we found that the percent ionization of niacin (K a = 1.5  10 –5 ) in a 0.020 M solution is 2.7%. Calculate the percentage of niacin molecules ionized in a solution that is (a) 0.010 M, (b) 1.0  10 –3 M. Answers: (a) 3.8%, (b) 12%

55. Acids and Bases Polyprotic Acids Have more than one acidic proton. If the difference between the K a for the first dissociation and subsequent K a values is 10 3 or more, the pH generally depends only on the first dissociation.

56. Acids and Bases SAMPLE EXERCISE 16.13 Calculating the pH of a Polyprotic Acid Solution The solubility of CO 2 in pure water at 25°C and 0.1 atm pressure is 0.0037 M. The common practice is to assume that all of the dissolved CO 2 is in the form of carbonic acid (H 2 CO 3 ), which is produced by reaction between the CO 2 and H 2 O: What is the pH of a 0.0037 M solution of H 2 CO 3 ?

57. Acids and Bases SAMPLE EXERCISE 16.13 continued PRACTICE EXERCISE (a) Calculate the pH of a 0.020 M solution of oxalic acid (H 2 C 2 O 4 ). (See Table 16.3 for K a1 and K a2.) (b) Calculate the concentration of oxalate ion, [C 2 O 4 2– ], in this solution. Answers: (a) pH = 1.80, (b) [C 2 O 4 2– ] = 6.4  10 –5 M

58. Acids and Bases Weak Bases Bases react with water to produce hydroxide ion.

59. Acids and Bases Weak Bases The equilibrium constant expression for this reaction is [HB] [OH − ] [B − ] K b = where K b is the base-dissociation constant.

60. Acids and Bases Weak Bases K b can be used to find [OH − ] and, through it, pH.

61. Acids and Bases pH of Basic Solutions What is the pH of a 0.15 M solution of NH 3 ? [NH 4 + ] [OH − ] [NH 3 ] K b = = 1.8  10 −5 NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH − (aq)

62. Acids and Bases pH of Basic Solutions Tabulate the data. [NH 3 ], M[NH 4 + ], M[OH − ], M Initially0.1500 At Equilibrium 0.15 - x  0.15 xx

63. Acids and Bases pH of Basic Solutions (1.8  10 −5 ) (0.15) = x 2 2.7  10 −6 = x 2 1.6  10 −3 = x 2 (x) 2 (0.15) 1.8  10 −5 =

64. Acids and Bases pH of Basic Solutions Therefore, [OH − ] = 1.6  10 −3 M pOH = −log (1.6  10 −3 ) pOH = 2.80 pH = 14.00 − 2.80 pH = 11.20

65. Acids and Bases SAMPLE EXERCISE 16.14 Using K b to Calculate [OH – ] Calculate the concentration of OH – in a 0.15 M solution of NH 3.

66. Acids and Bases PRACTICE EXERCISE Which of the following compounds should produce the highest pH as a 0.05 M solution: pyridine, methylamine, or nitrous acid? Answer: methylamine (because it has the largest K b value) SAMPLE EXERCISE 16.14 continued

67. Acids and Bases SAMPLE EXERCISE 16.15 Using pH to Determine the Concentration of a Salt A solution made by adding solid sodium hypochlorite (NaClO) to enough water to make 2.00 L of solution has a pH of 10.50. Using the information in Equation 16.37, calculate the number of moles of NaClO that were added to the water.

68. Acids and Bases Answer: 0.12 M PRACTICE EXERCISE A solution of NH 3 in water has a pH of 11.17. What is the molarity of the solution? SAMPLE EXERCISE 16.15 continued

69. Acids and Bases K a and K b K a and K b are related in this way: K a  K b = K w Therefore, if you know one of them, you can calculate the other.

70. Acids and Bases SAMPLE EXERCISE 16.16 Calculating K a or K b for a Conjugate Acid-Base Pair Calculate (a) the base-dissociation constant, K b, for the fluoride ion (F – ); (b) the acid-dissociation constant, K a, for the ammonium ion (NH 4 + ).

71. Acids and Bases Its conjugate acid is listed in handbooks as having a pK a of 4.90. What is the base-dissociation constant for quinoline? Answers: (a) PO 4 3– (K b = 2.4  10 –2 ), (b) 7.9  10 –10 SAMPLE EXERCISE 16.16 continued PRACTICE EXERCISE (a) Which of the following anions has the largest base-dissociation constant: NO 2 –, PO 4 3–, or N 3 – ? (b) The base quinoline has the following structure:

72. Acids and Bases Reactions of Anions with Water Anions are bases. As such, they can react with water in a hydrolysis reaction to form OH − and the conjugate acid: X − (aq) + H 2 O (l) HX (aq) + OH − (aq)

73. Acids and Bases Reactions of Cations with Water Cations with acidic protons (like NH 4 + ) will lower the pH of a solution. Most metal cations that are hydrated in solution also lower the pH of the solution.

74. Acids and Bases Reactions of Cations with Water Attraction between nonbonding electrons on oxygen and the metal causes a shift of the electron density in water. This makes the O-H bond more polar and the water more acidic. Greater charge and smaller size make a cation more acidic.

75. Acids and Bases Effect of Cations and Anions 1.An anion that is the conjugate base of a strong acid will not affect the pH. 2.An anion that is the conjugate base of a weak acid will increase the pH. 3.A cation that is the conjugate acid of a weak base will decrease the pH.

76. Acids and Bases Effect of Cations and Anions 4.Cations of the strong Arrhenius bases will not affect the pH. 5.Other metal ions will cause a decrease in pH. 6.When a solution contains both the conjugate base of a weak acid and the conjugate acid of a weak base, the affect on pH depends on the K a and K b values.

77. Acids and Bases SAMPLE EXERCISE 16.17 Predicting the Relative Acidity of Salt Solutions List the following solutions in order of increasing pH: (i) 0.1 M Ba(C 2 H 3 O 2 ) 2, (ii) 0.1 M NH 4 Cl, (iii) 0.1 M NH 3 CH 3 Br, (iv) 0.1 M KNO 3. Answers: (a) Fe(NO 3 ) 3, (b) KBr, (c) CH 3 NH 3 Cl, (d) NH 4 NO 3 PRACTICE EXERCISE In each of the following, indicate which salt will form the more acidic (or less basic) 0.010 M solution: (a) NaNO 3, Fe(NO 3 ) 3 ; (b) KBr, KBrO; (c) CH 3 NH 3 Cl, BaCl 2, (d) NH 4 NO 2, NH 4 NO 3.

78. Acids and Bases SAMPLE EXERCISE 16.18 Predicting Whether the Solution of an Amphiprotic Anion is Acidic or Basic Predict whether the salt Na 2 HPO 4 will form an acidic solution or a basic solution on dissolving in water.

79. Acids and Bases SAMPLE EXERCISE 16.18 continued PRACTICE EXERCISE Predict whether the dipotassium salt of citric acid (K 2 HC 6 H 5 O 7 ) will form an acidic or basic solution in water (see Table 16.3 for data). Answer: acidic

80. Acids and Bases Factors Affecting Acid Strength The more polar the H-X bond and/or the weaker the H-X bond, the more acidic the compound. Acidity increases from left to right across a row and from top to bottom down a group.

81. Acids and Bases Factors Affecting Acid Strength In oxyacids, in which an OH is bonded to another atom, Y, the more electronegative Y is, the more acidic the acid.

82. Acids and Bases Factors Affecting Acid Strength For a series of oxyacids, acidity increases with the number of oxygens.

83. Acids and Bases Factors Affecting Acid Strength Resonance in the conjugate bases of carboxylic acids stabilizes the base and makes the conjugate acid more acidic.

84. Acids and Bases Lewis Acids Lewis acids are defined as electron-pair acceptors. Atoms with an empty valence orbital can be Lewis acids.

85. Acids and Bases Lewis Bases Lewis bases are defined as electron-pair donors. Anything that could be a Brønsted–Lowry base is a Lewis base. Lewis bases can interact with things other than protons, however.

86. Acids and Bases SAMPLE INTEGRATIVE EXERCISE Putting Concepts Together (a) Explain why H 3 PO 3 is diprotic and not triprotic. (b) A 25.0-mL sample of a solution of H 3 PO 3 is titrated with 0.102 M NaOH. It requires 23.3 mL of NaOH to neutralize both acidic protons. What is the molarity of the H 3 PO 3 solution? (c) This solution has a pH of 1.59. Calculate the percent ionization and K a1 for H 3 PO 3, assuming that K a1 >> K a2. (d) How does the osmotic pressure of a 0.050 M solution of HCl compare with that of a 0.050 M solution of H 3 PO 3 ? Explain. Phosphorous acid (H 3 PO 3 ) has the following Lewis structure.

87. Acids and Bases Buffers: Solutions of a weak conjugate acid-base pair. They are particularly resistant to pH changes, even when strong acid or base is added.

88. Acids and Bases Buffers If a small amount of hydroxide is added to an equimolar solution of HF in NaF, for example, the HF reacts with the OH − to make F − and water.

89. Acids and Bases Buffers If acid is added, the F − reacts to form HF and water.

90. Acids and Bases Buffer Calculations Consider the equilibrium constant expression for the dissociation of a generic acid, HA: [H 3 O + ] [A − ] [HA] K a = HA + H 2 OH 3 O + + A −

91. Acids and Bases Buffer Calculations Rearranging slightly, this becomes [A − ] [HA] K a = [H 3 O + ] Taking the negative log of both side, we get [A − ] [HA] −log K a = −log [H 3 O + ] + − log pKapKa pH acid base

92. Acids and Bases Henderson–Hasselbalch Equation What is the pH of a buffer that is 0.12 M in lactic acid, HC 3 H 5 O 3, and 0.10 M in sodium lactate? K a for lactic acid is 1.4  10 −4.

93. Acids and Bases Henderson–Hasselbalch Equation pH = pK a + log [base] [acid] pH = −log (1.4  10 −4 ) + log (0.10) (0.12) pH = 3.85 + (−0.08) pH = 3.77

94. Acids and Bases pH Range The pH range is the range of pH values over which a buffer system works effectively. It is best to choose an acid with a pK a close to the desired pH.

95. Acids and Bases When Strong Acids or Bases Are Added to a Buffer… …it is safe to assume that all of the strong acid or base is consumed in the reaction.

96. Acids and Bases Addition of Strong Acid or Base to a Buffer 1.Determine how the neutralization reaction affects the amounts of the weak acid and its conjugate base in solution. 2.Use the Henderson–Hasselbalch equation to determine the new pH of the solution.

97. Acids and Bases Calculating pH Changes in Buffers A buffer is made by adding 0.300 mol HC 2 H 3 O 2 and 0.300 mol NaC 2 H 3 O 2 to enough water to make 1.00 L of solution. The pH of the buffer is 4.74. Calculate the pH of this solution after 0.020 mol of NaOH is added.

98. Acids and Bases Calculating pH Changes in Buffers Before the reaction, since mol HC 2 H 3 O 2 = mol C 2 H 3 O 2 − pH = pK a = −log (1.8  10 −5 ) = 4.74

99. Acids and Bases Calculating pH Changes in Buffers The 0.020 mol NaOH will react with 0.020 mol of the acetic acid: HC 2 H 3 O 2 (aq) + OH − (aq)  C 2 H 3 O 2 − (aq) + H 2 O (l) HC 2 H 3 O 2 C2H3O2−C2H3O2− OH − Before reaction0.300 mol 0.020 mol After reaction0.280 mol0.320 mol0.000 mol

100. Acids and Bases Calculating pH Changes in Buffers Now use the Henderson–Hasselbalch equation to calculate the new pH: pH = 4.74 + log (0.320) (0. 200) pH = 4.74 + 0.06 pH = 4.80

101. Acids and Bases SAMPLE EXERCISE 17.3 Calculating the pH of a Buffer What is the pH of a buffer that is 0.12 M in lactic acid (HC 3 H 5 O 3 ) and 0.10 M in sodium lactate? For lactic acid, K a = 1.4  10 –4.

102. Acids and Bases SAMPLE EXERCISE 17.3 continued PRACTICE EXERCISE Calculate the pH of a buffer composed of 0.12 M benzoic acid and 0.20 M sodium benzoate. (Refer to Appendix D.) Answer: 4.42

103. Acids and Bases SAMPLE EXERCISE 17.4 Preparing a Buffer How many moles of NH 4 Cl must be added to 2.0 L of 0.10 M NH 3 to form a buffer whose pH is 9.00? (Assume that the addition of NH 4 Cl does not change the volume of the solution.)

104. Acids and Bases SAMPLE EXERCISE 17.4 continued Answer: 0.13 M PRACTICE EXERCISE Calculate the concentration of sodium benzoate that must be present in a 0.20 M solution of benzoic acid (HC 7 H 5 O 2 ) to produce a pH of 4.00.

105. Acids and Bases SAMPLE EXERCISE 17.5 Calculating pH Changes in Buffers A buffer is made by adding 0.300 mol HC 2 H 3 O 2 and 0.300 mol NaC 2 H 3 O 2 to enough water to make 1.00 L of solution. The pH of the buffer is 4.74 (Sample Exercise 17.1). (a) Calculate the pH of this solution after 0.020 mol of NaOH is added. (b) For comparison, calculate the pH that would result if 0.020 mol of NaOH was added to 1.00 L of pure water (neglect any volume changes).

106. Acids and Bases SAMPLE EXERCISE 17.5 continued

107. Acids and Bases PRACTICE EXERCISE Determine (a) the pH of the original buffer described in Sample Exercise 17.5 after the addition of 0.020 mol HCl, and (b) the pH of the solution that would result from the addition of 0.020 mol HCl to 1.00 L of pure water. Answers: (a) 4.68, (b) 1.70 SAMPLE EXERCISE 17.5 continued

108. Acids and Bases Titration A known concentration of base (or acid) is slowly added to a solution of acid (or base).

109. Acids and Bases Titration A pH meter or indicators are used to determine when the solution has reached the equivalence point, at which the stoichiometric amount of acid equals that of base.

110. Acids and Bases Titration of a Strong Acid with a Strong Base From the start of the titration to near the equivalence point, the pH goes up slowly.

111. Acids and Bases Titration of a Strong Acid with a Strong Base Just before and after the equivalence point, the pH increases rapidly.

112. Acids and Bases Titration of a Strong Acid with a Strong Base At the equivalence point, moles acid = moles base, and the solution contains only water and the salt from the cation of the base and the anion of the acid.

113. Acids and Bases Titration of a Strong Acid with a Strong Base As more base is added, the increase in pH again levels off.

114. Acids and Bases SAMPLE EXERCISE 17.6 Calculating pH for a Strong Acid–Strong Base Titration Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution: (a) 49.0 mL, (b) 51.0 mL.

115. Acids and Bases PRACTICE EXERCISE Calculate the pH when the following quantities of 0.100 M HNO 3 have been added to 25.0 mL of 0.100 M KOH solution: (a) 24.9 mL, (b) 25.1 mL. Answers: (a) 10.30, (b) 3.70 SAMPLE EXERCISE 17.6 continued

116. Acids and Bases Titration of a Weak Acid with a Strong Base Unlike in the previous case, the conjugate base of the acid affects the pH when it is formed. The pH at the equivalence point will be >7. Phenolphthalein is commonly used as an indicator in these titrations.

117. Acids and Bases Titration of a Weak Acid with a Strong Base At each point below the equivalence point, the pH of the solution during titration is determined from the amounts of the acid and its conjugate base present at that particular time.

118. Acids and Bases Titration of a Weak Acid with a Strong Base With weaker acids, the initial pH is higher and pH changes near the equivalence point are more subtle.

119. Acids and Bases SAMPLE EXERCISE 17.7 Calculating pH for a Weak Acid–Strong Base Titration Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M HC 2 H 3 O 2 (K a = 1.8  10 –5 ).

120. Acids and Bases SAMPLE EXERCISE 17.7 continued PRACTICE EXERCISE (a) Calculate the pH in the solution formed by adding 10.0 mL of 0.050 M NaOH to 40.0 mL of 0.0250 M benzoic acid (HC 7 H 5 O 2, K a = 6.3  10 –5 ). (b) Calculate the pH in the solution formed by adding 10.0 mL of 0.100 M HCl to 20.0 mL of 0.100 M NH 3. Answers: (a) 4.20, (b) 9.26

121. Acids and Bases Titration of a Weak Base with a Strong Acid The pH at the equivalence point in these titrations is < 7. Methyl red is the indicator of choice.

122. Acids and Bases Titrations of Polyprotic Acids In these cases there is an equivalence point for each dissociation.

123. Acids and Bases SAMPLE EXERCISE 17.8 Calculating the pH at the Equivalence Point Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M HC 2 H 3 O 2 with 0.100 M NaOH.

124. Acids and Bases Answers: (a) 8.21, (b) 5.28 PRACTICE EXERCISE Calculate the pH at the equivalence point when (a) 40.0 mL of 0.025 M benzoic acid (HC 7 H 5 O 2, K a = 6.3  10 –5 ) is titrated with 0.050 M NaOH; (b) 40.0 mL of 0.100 M NH 3 is titrated with 0.100 M HCl. SAMPLE EXERCISE 17.8 continued