1. Acids and Bases © 2009, Prentice-Hall, Inc. Chapter 16 Acids and Bases Q: Do you guys have equation solving calculators? Use for quadratic formula? John D. Bookstaver St. Charles Community College Cottleville, MO Chemistry, The Central Science, 11th edition Theodore L. Brown, H. Eugene LeMay, Jr., and Bruce E. Bursten

2. Acids and Bases © 2009, Prentice-Hall, Inc. Some Definitions Arrhenius –An acid is a substance that, when dissolved in water, increases the concentration of hydrogen ions (H+). –A base is a substance that, when dissolved in water, increases the concentration of hydroxide ions (OH-). Ex: HCl, HC 2 H 3 O 2 NaOH, NH 3 NH 3 + H 2 O  NH 4 + + OH- HCl + H 2 O  H 3 O + + Cl-

3. Acids and Bases © 2009, Prentice-Hall, Inc. Some Definitions Brønsted-Lowry –An acid is a proton (H+) donor. –A base is a proton (H+) acceptor.

4. Acids and Bases © 2009, Prentice-Hall, Inc. A Brønsted-Lowry acid… …must have a removable (acidic) proton. A Brønsted-Lowry base… …must have a pair of nonbonding electrons. …where the H+, with no electrons for bonding, can attach itself.

5. Acids and Bases © 2009, Prentice-Hall, Inc. If it can be either… …it is amphiprotic. (or ≈ amphoteric) HCO 3 - HSO 4 - H2OH2O

6. Acids and Bases © 2009, Prentice-Hall, Inc. If it can be either… …it is amphiprotic. (or ≈ amphoteric) HCO 3 - HSO 4 - H2OH2O + H +  H 2 CO 3 CO 3 2-  - H +

7. Acids and Bases © 2009, Prentice-Hall, Inc. 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.

8. Acids and Bases © 2009, Prentice-Hall, Inc. Conjugate Acids and Bases The term conjugate comes from the Latin word “conjugare,” meaning “to join together.” Reactions between acids and bases always yield their conjugate bases and acids. A conjugate base has one less H+ than its conjugate acid.

9. 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 – ? Solution Plan: The conjugate base of a substance is simply the parent substance minus one proton, and the conjugate acid of a substance is the parent substance plus one proton. Solve: (a) HClO 4 less one proton (H + ) is ClO 4 –. The other conjugate bases are HS –, PH 3, and CO 3 2–. (b) CN – plus one proton (H + ) is HCN. The other conjugate acids are HSO 4 –, H 3 O +, and H 2 CO 3. Notice that the hydrogen carbonate ion (HCO 3 – ) is amphiprotic. It can act as either an acid or a base.

10. Acids and Bases 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 + © 2009, Prentice-Hall, Inc. Practice Exercise

11. Acids and Bases © 2009, Prentice-Hall, Inc. 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.

12. Acids and Bases © 2009, Prentice-Hall, Inc. Acid and Base Strength Substances with negligible acidity do not dissociate in water. Their conjugate bases are exceedingly strong.

13. Acids and Bases © 2009, Prentice-Hall, Inc. Acid and Base Strength In any acid-base reaction, the equilibrium will favor the reaction that moves the proton to the stronger base. *** (use table 16.4 from p672 or prior slide!) HCl (aq) + H 2 O (l)  H 3 O + (aq) + Cl - (aq) H 2 O is a much stronger base than Cl -, so H 2 O will get the proton, so equilibrium lies so far to the right that K is not measured (K>>1). base

14. Acids and Bases © 2009, Prentice-Hall, Inc. Acid and Base Strength In any acid-base reaction, the equilibrium will favor the reaction that moves the proton to the stronger base. Acetate is a stronger base than H 2 O, so acetate will get the proton, and the equilibrium favors the left side (K<1). (confirm that this is true!) HC 2 H 3 O 2 (aq) + H 2 O (l) H 3 O + (aq) + C 2 H 3 O 2 - (aq) base

15. Acids and Bases © 2009, Prentice-Hall, Inc. 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 © 2009, Prentice-Hall, Inc. Ion-Product Constant The equilibrium expression for this process is K c = [H 3 O + ] [OH - ] (ignore the H2O reactants b/c they are liquid) 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 © 2009, Prentice-Hall, Inc. pH pH is defined as the negative base-10 logarithm of the concentration of hydronium ion. pH = -log [H 3 O + ]

18. Acids and Bases © 2009, Prentice-Hall, Inc. pH In pure water, K w = [H 3 O + ] [OH - ] = 1.0  10 -14 Since in pure water [H 3 O + ] = [OH - ], [H 3 O + ] = 1.0  10 -14 = 1.0  10 -7

19. Acids and Bases © 2009, Prentice-Hall, Inc. 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.

20. Acids and Bases Practice Exercise : (p 675) Indicate whether solutions with each of the following ion concentrations are neutral, acidic, or basic: (a) [H+] = 4 x 10 -9 M (b) [OH-] = 1 x 10 -7 M (c) [OH-] = 7 x 10 -13 M © 2009, Prentice-Hall, Inc.

21. Acids and Bases Sample 16.5: Calculating [H+] from [OH-] (p 675) © 2009, Prentice-Hall, Inc. 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 ºC. Analyze: We are asked to calculate the hydronium ion concentration in an aqueous solution where the hydroxide concentration is known. Plan: We can use the equilibrium-constant expression for the autoionization of water and the value of K w to solve for each unknown concentration.

22. Acids and Bases Sample 16.5: Solution, cont. © 2009, Prentice-Hall, Inc. Solve: a) Using Equation 16.16, we have: This solution is basic because (b) In this instance This solution is acidic because

23. Acids and Bases © 2009, Prentice-Hall, Inc. pH (p676) These are the pH values for several common substances.

24. Acids and Bases Sample Ex 16.6 Calculating pH from [H + ] (p 677) Calculate the pH values for the two solutions described in Sample Ex 16.5. Analyze: We are asked to determine the pH of aqueous solutions for which we have already calculated [H + ]. Plan: We can calculate pH using its defining equation, 16.17. Solve: (a) In the first instance we found [H+]. to be 1.0 ×10 –12 M. Because 1.0 ×10 –12 has two significant figures, the pH has two decimal places, 12.00. (b) For the second solution, [H + ] = 5.6 × 10 –6 M. Before performing the calculation, it is helpful to estimate the pH. To do so, we note that [H + ] lies between 1 × 10 –6 and 1 × 10 –5 Thus, we expect the pH to lie between 6.0 and 5.0. We use Equation 16.17 to calculate the pH.

25. Acids and Bases Check your work Check: After calculating a pH, it is useful to compare it to your prior estimate. In this case the pH, as we predicted, falls between 6 and 5. Had the calculated pH and the estimate not agreed, we should have reconsidered our calculation or estimate or both. © 2009, Prentice-Hall, Inc.

26. Acids and Bases © 2009, Prentice-Hall, Inc. Sample Exercise 16.7 Calculating [H + ] from pH A sample of freshly pressed apple juice has a pH of 3.76. Calculate [H + ]. Analyze: We need to calculate [H + ] from pH. Plan: We will use Equation 16.17, pH = –log[H + ], for the calculation. Solve: From Equation 16.17, we have Thus, To find [H + ], we need to determine the antilog of –3.76. Scientific calculators have an antilog function (sometimes labeled INV log or 10 x ) that allows us to perform the calculation:

27. Acids and Bases © 2009, Prentice-Hall, Inc. Other “p” Scales The “p” in pH tells us to take the negative base-10 logarithm of the quantity (in this case, hydronium ions). Some similar examples are –pOH: -log [OH - ] –pK w : -log K w

28. Acids and Bases © 2009, Prentice-Hall, Inc. 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

29. Acids and Bases © 2009, Prentice-Hall, Inc. How Do We Measure pH?(p679) For less accurate measurements, one can use –Litmus paper “Red” paper turns blue above ~pH = 8 “Blue” paper turns red below ~pH = 5 –Or an indicator.

30. Acids and Bases © 2009, Prentice-Hall, Inc. How Do We Measure pH? For more accurate measurements, one uses a pH meter, which measures the voltage in the solution.

31. Acids and Bases © 2009, Prentice-Hall, Inc. 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 monoprotic strong acids, [H 3 O + ] = [acid].

32. Acids and Bases © 2009, Prentice-Hall, Inc. Sample Exercise 16.8 Calculating the pH of a Strong Acid What is the pH of a 0.040 M solution of HClO 4 ? Analyze and Plan: Because HClO 4 is a strong acid, it is completely ionized, giving [H + ] = [ClO 4 - ] = 0.040 M Solve: The pH of the solution is given by pH = –log(0.040) = 1.40. Check: Because [H + ] lies between 1 × 10 –2 and 1 × 10 –1, the pH will be between 2.0 and 1.0. Our calculated pH falls within the estimated range. Furthermore, because the concentration has two significant figures, the pH has two decimal places.

33. Acids and Bases © 2009, Prentice-Hall, Inc. 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.

34. Acids and Bases © 2009, Prentice-Hall, Inc. 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 ? Analyze: We are asked to calculate the pH of two solutions of strong bases. Plan: We can calculate each pH by either of two equivalent methods. First, we could use Equation 16.16 to calculate [H + ] and then use Equation 16.17 to calculate the pH. Alternatively, we could use [OH – ] to calculate pOH and then use Equation 16.20 to calculate the pH. (b) Ca(OH) 2 is a strong base that dissociates in water to give two OH – ions per formula unit. Thus, the concentration of OH – (aq) for the solution in part (b) is 2 × (0.0011 M) = 0.0022 M

35. Acids and Bases © 2009, Prentice-Hall, Inc. Sample Exercise 16.9 Calculating the pH of a Strong Base Solve: (a) NaOH dissociates in water to give one OH – ion per formula unit. Therefore, the OH – concentration for the solution in (a) equals the stated concentration of NaOH, namely 0.028 M.

36. Acids and Bases © 2009, Prentice-Hall, Inc. (b) Ca(OH) 2 is a strong base that dissociates in water to give two OH – ions per formula unit. Thus, the concentration of OH – (aq) for the solution in part (b) is 2 × (0.0011 M) = 0.0022 M Sample Exercise 16.9 Calculating the pH of a Strong Base

37. Acids and Bases © 2009, Prentice-Hall, Inc. 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) = K a

38. Acids and Bases © 2009, Prentice-Hall, Inc. Dissociation Constants (p682) The greater the value of K a, the stronger is the acid.

39. Acids and Bases © 2009, Prentice-Hall, Inc. 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] Ka=Ka=

40. Acids and Bases © 2009, Prentice-Hall, Inc. Calculating K a from the pH SOLUTION: 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.

41. Acids and Bases © 2009, Prentice-Hall, Inc. 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 - ]

42. Acids and Bases © 2009, Prentice-Hall, Inc. Calculating K a from pH Now we can set up an ICE table… [HCOOH], M[H 3 O + ], M[HCOO - ], M Initially0.1000 Change - 4.2  10 -3 + 4.2  10 -3 At Equilibrium 0.10 - 4.2  10 -3 = 0.0958 = 0.10 4.2  10 -3

43. Acids and Bases © 2009, Prentice-Hall, Inc. Calculating K a from pH [4.2  10 -3 ] [0.10] K a = = 1.8  10 -4

44. Acids and Bases © 2009, Prentice-Hall, Inc. 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 Percent Ionization =  100 4.2  10 -3 0.10 = 4.2%

45. Acids and Bases © 2009, Prentice-Hall, Inc. Sample Exercise 16.11 Calculating Percent Ionization A 0.10 M solution of formic acid (HCOOH) contains 4.2 × 10 –3 M H + (aq). Calculate the percentage of the acid that is ionized. Analyze: We are given the molar concentration of an aqueous solution of weak acid and the equilibrium concentration of H + (aq) and asked to determine the percent ionization of the acid. Plan: The percent ionization is given by Equation 16.27. Solve:

46. Acids and Bases © 2009, Prentice-Hall, Inc. 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.

47. Acids and Bases © 2009, Prentice-Hall, Inc. 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 =

48. Acids and Bases © 2009, Prentice-Hall, Inc. 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 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.

49. Acids and Bases © 2009, Prentice-Hall, Inc. 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

50. Acids and Bases © 2009, Prentice-Hall, Inc. Calculating pH from K a pH = -log [H 3 O + ] pH = -log (2.3  10 -3 ) pH = 2.64

51. Acids and Bases © 2009, Prentice-Hall, Inc. Sample Exercise 16.12 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.) Analyze: We are given the molarity of a weak acid and are asked for the pH. From Table 16.2, Ka for HCN is 4.9 × 10 –10. Plan: We proceed as in the example just worked in the text, writing the chemical equation and constructing a table of initial and equilibrium concentrations in which the equilibrium concentration of H + is our unknown.

52. Acids and Bases Solve: Writing both the chemical equation for the ionization reaction that forms H + (aq) and the equilibrium-constant (K a ) expression for the reaction: Next, we tabulate the concentration of the species involved in the equilibrium reaction, letting x = [H + ] at equilibrium: Substituting the equilibrium concentrations from the table into the equilibrium- constant expression yields Sample Exercise 16.12 Using K a to Calculate pH

53. Acids and Bases © 2009, Prentice-Hall, Inc. 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. p689

54. Acids and Bases © 2009, Prentice-Hall, Inc. Sample Exercise 16.13 Using K a to Calculate % Ionization Calculate the percentage of HF molecules ionized in (a) a 0.10 M HF solution, (b) a 0.010 M HF solution. Analyze: We are asked to calculate the percent ionization of two HF solutions of different concentration. From Appendix D, we find K a = 6.8 × 10 -4. Plan: We approach this problem as we would previous equilibrium problems. We begin by writing the chemical equation for the equilibrium and tabulating the known and unknown concentrations of all species. We then substitute the equilibrium concentrations into the equilibrium-constant expression and solve for the unknown concentration, that of H +.

55. Acids and Bases © 2009, Prentice-Hall, Inc. Solve: (a) The equilibrium reaction and equilibrium concentrations are as follows: The equilibrium-constant expression is When we try solving this equation using the approximation 0.10 – x = 0.10 (that is, by neglecting the concentration of acid that ionizes in comparison with the initial concentration), we obtain

56. Acids and Bases © 2009, Prentice-Hall, Inc. Sample Exercise 16.13 Using K a to Calculate % Ionization Because this value is greater than 5% of 0.10 M, we should work the problem without the approximation, using an equation-solving calculator or the quadratic formula. Rearranging our equation and writing it in standard quadratic form, we have This equation can be solved using the standard quadratic formula. Substituting the appropriate numbers gives Of the two solutions, only the one that gives a positive value for x is chemically reasonable. Thus, From our result, we can calculate the percent of molecules ionized: (b) Proceeding similarly for the 0.010 M solution, we have Solving the resultant quadratic expression, we obtain The percentage of molecules ionized is

57. Acids and Bases © 2009, Prentice-Hall, Inc. This equation can be solved using the standard quadratic formula. Substituting the appropriate numbers gives Of the two solutions, only the one that gives a positive value for x is chemically reasonable. Thus, From our result, we can calculate the percent of molecules ionized: (b) Proceeding similarly for the 0.010 M solution, we have Solving the resultant quadratic expression, we obtain The percentage of molecules ionized is

58. Acids and Bases © 2009, Prentice-Hall, Inc. Weak Bases Bases react with water to produce hydroxide ion.

59. Acids and Bases © 2009, Prentice-Hall, Inc. 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 © 2009, Prentice-Hall, Inc. Weak Bases (p691) K b can be used to find [OH - ] and, through it, pH.

61. Acids and Bases © 2009, Prentice-Hall, Inc. 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 © 2009, Prentice-Hall, Inc. 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 © 2009, Prentice-Hall, Inc. 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 © 2009, Prentice-Hall, Inc. 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 © 2009, Prentice-Hall, Inc. K a and K b (p 694 ) 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.

66. Acids and Bases © 2009, Prentice-Hall, Inc. 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)

67. Acids and Bases © 2009, Prentice-Hall, Inc. 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.

68. Acids and Bases © 2009, Prentice-Hall, Inc. 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.

69. Acids and Bases © 2009, Prentice-Hall, Inc. 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.

70. Acids and Bases © 2009, Prentice-Hall, Inc. 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.

71. Acids and Bases © 2009, Prentice-Hall, Inc. Factors Affecting Acid Strength The more polar the H-X bond and/or the weaker the H-X bond, the more acidic the compound. So acidity increases from left to right across a row and from top to bottom down a group.

72. Acids and Bases © 2009, Prentice-Hall, Inc. 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.

73. Acids and Bases © 2009, Prentice-Hall, Inc. Factors Affecting Acid Strength For a series of oxyacids, acidity increases with the number of oxygens.

74. Acids and Bases © 2009, Prentice-Hall, Inc. Factors Affecting Acid Strength Resonance in the conjugate bases of carboxylic acids stabilizes the base and makes the conjugate acid more acidic.

75. Acids and Bases © 2009, Prentice-Hall, Inc. Lewis Acids Lewis acids are defined as electron-pair acceptors. Atoms with an empty valence orbital can be Lewis acids.

76. Acids and Bases © 2009, Prentice-Hall, Inc. 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.

77. Acids and Bases © 2009, Prentice-Hall, Inc. Buffers Buffers are solutions of a weak conjugate acid-base pair. They are particularly resistant to pH changes, even when strong acid or base is added.

78. Acids and Bases © 2009, Prentice-Hall, Inc. 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.

79. Acids and Bases © 2009, Prentice-Hall, Inc. Buffers Similarly, if acid is added, the F − reacts with it to form HF and water.

80. Acids and Bases © 2009, Prentice-Hall, Inc. 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 −

81. Acids and Bases © 2009, Prentice-Hall, Inc. 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

82. Acids and Bases © 2009, Prentice-Hall, Inc. Buffer Calculations So pK a = pH − log [base] [acid] Rearranging, this becomes pH = pK a + log [base] [acid] This is the Henderson–Hasselbalch equation.

83. Acids and Bases © 2009, Prentice-Hall, Inc. Henderson–Hasselbalch Equation What is the pH of a buffer that is 0.12 M in lactic acid, CH 3 CH(OH)COOH, and 0.10 M in sodium lactate? K a for lactic acid is 1.4  10 −4.

84. Acids and Bases © 2009, Prentice-Hall, Inc. Henderson–Hasselbalch Equation pH = pK a + log [base] [acid] pH = −log (1.4  10 −4 ) + log (0.10) (0.12) pH pH = 3.77 pH = 3.85 + (−0.08) pH

85. Acids and Bases © 2009, Prentice-Hall, Inc. 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.

86. Acids and Bases © 2009, Prentice-Hall, Inc. 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.

87. Acids and Bases © 2009, Prentice-Hall, Inc. 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.

88. Acids and Bases © 2009, Prentice-Hall, Inc. 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.

89. Acids and Bases © 2009, Prentice-Hall, Inc. 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

90. Acids and Bases © 2009, Prentice-Hall, Inc. 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

91. Acids and Bases © 2009, Prentice-Hall, Inc. 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 + 0.06pH pH = 4.80

92. Acids and Bases © 2009, Prentice-Hall, Inc. Titration In this technique a known concentration of base (or acid) is slowly added to a solution of acid (or base).

93. Acids and Bases © 2009, Prentice-Hall, Inc. 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.

94. Acids and Bases © 2009, Prentice-Hall, Inc. Titration of a Strong Acid with a Strong Base Just before (and after) the equivalence point, the pH increases rapidly.

95. Acids and Bases © 2009, Prentice-Hall, Inc. 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.

96. Acids and Bases © 2009, Prentice-Hall, Inc. 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.

97. Acids and Bases © 2009, Prentice-Hall, Inc. Titration of a Strong Acid with a Strong Base As more base is added, the increase in pH again levels off.

98. Acids and Bases © 2009, Prentice-Hall, Inc. 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. At the equivalence point the pH is >7. Phenolphthalein is commonly used as an indicator in these titrations.

99. Acids and Bases © 2009, Prentice-Hall, Inc. 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.

100. Acids and Bases © 2009, Prentice-Hall, Inc. 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.

101. Acids and Bases © 2009, Prentice-Hall, Inc. 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.

102. Acids and Bases © 2009, Prentice-Hall, Inc. Titrations of Polyprotic Acids When one titrates a polyprotic acid with a base there is an equivalence point for each dissociation.

103. Acids and Bases Objectives Met? You should be able to: © 2009, Prentice-Hall, Inc.