1. AP CHEMISTRY UNIT 14 WHAT ARE ACIDS AND BASES?

2. UNIT 14: ACIDS AND BASES The Nature of Acids and Bases Acid Strength The pH Scale Calculating the pH of Strong Acid Solutions Calculating the pH of Weak Acid Solutions Bases Polyprotic Acids Acid-Base Properties of Salts The Effect of Structure on Acid-Base Properties Acid-Base Properties of Oxides The Lewis Acid-Base Model Strategy for Solving Acid-Base Problems

3. THE NATURE OF ACIDS AND BASES 14.1

4. ARRHENIUS CONCEPT The first person to recognize the essential nature of acids and bases was Svante Arrhenius. He stated that acids produce hydrogen ions in aqueous solutions, while bases produce hydroxide ion. Major step in quantifying acid-base chemistry. Concept is very limited because it applies to aqueous solutions only and it only allows for only one kind of base – the hydroxide ion.

5. BRONSTED-LOWRY MODEL A more general definition of acids and bases was suggested by the Danish chemist, Bronsted, and the English chemist Lowry. An acid is defined as a proton (H+) donor, and a base is a proton acceptor. includes reactions of gases and not just aqueous reactions.

6. WATER AS A BASE Water can act as a base because the water molecule has two unshared electrons pairs that allow a covalent bond with a proton (H + ). When water accepts a proton, H 3 O +, hydronium is formed. HA is the acid (proton donor) and H 2 O is the base (proton acceptor).

7. CONJUGATE ACID-BASE PAIR A conjugate acid-base pair consists of two substances that are related to each other by the donation and acceptance of a proton. H 3 O + is the conjugate acid. The result of the accepted proton. A - is the conjugate base. The result of the donated proton. These two components are considered conjugate acid base pairs.

8. CONJUGATE ACID-BASE PAIR H 2 O and A -, the base and conjugate base are both competing for a proton. If H 2 O is a stronger base, the reaction will lie to the right because water will accept the proton before A - will.

9. CONJUGATE ACID-BASE PAIR H 2 O and A -, the base and conjugate base are both competing for a proton. If A- is a stronger base, the reaction will lie to the left because A - will accept the proton before H 2 O will.

10. ACID DISSOCIATION CONSTANT The equilibrium expression for an acid-base reaction gives K a, the acid dissociation constant: **Note that H 2 O is not in the expression and H + is commonly substituted for hydronium.

11. ACID DISSOCIATION CONSTANT Strong acids have large dissociation constants. Weak acids have small dissociation constants.

12. PRACTICE PROBLEM #1 Write the simple dissociation (ionization) reaction (omitting water) for each of the following acids. a.hydrochloric acid b.acetic acid c.the ammonium ion d.the anilinium ion (C 6 H 5 NH 3 + ) e.the hydrated aluminum (III) ion [Al(H 2 O) 6 ] 3+

13. ACID STRENGTH 14.2

14. ACID STRENGTH The strength of an acid is defined by the equilibrium position of its dissociation reaction. A strong acid has an equilibrium that lies to the right. Most HA has dissociated. A strong acid produces a weak conjugate base. Water is the stronger base

15. ACID STRENGTH The strength of an acid is defined by the equilibrium position of its dissociation reaction. A weak acid has an equilibrium that lies to the left. Most HA is still present. A weak acid produces a strong conjugate base. Water is the weaker base

16. ACID STRENGTH SUMMARY Acid strength and conjugate base strength are indirectly related.

17. COMMON STRONG ACIDS 1.HCl 2.HNO 3 3.HClO 4 4.H 2 SO 4  Sulfuric acid is a diprotic acid, an acid with two acidic protons. H 2 SO 4 is a strong acid and dissociates completely, but HSO 4 - is a weak acid.

18. COMMON WEAK OXYACIDS 1.H 3 PO 4 2.HNO 2 3.HOCl Most acids are oxyacids, in which the acidic proton is attached to an oxygen atom.

19. ORGANIC ACIDS Organic acids are those with a carbon atom “backbone”, commonly contain the carboxyl group: Acids of this type are usually weak. 1. acetic acid (HC 2 H 3 O 2 or CH 3 COOH) 2. benzoic acid (C 6 H 5 COOH) *note: the rest of the hydrogens are not acidic

20. COMMON MONOPROTIC ACIDS

21. PRACTICE PROBLEM #2 Using Table 14.2, arrange the following species according to their strengths as bases: H 2 O, F -, Cl -, NO 2 -, and CN -. Cl - < H 2 O < F - < NO 2 -, < CN -

22. WATER AS AN ACID AND A BASE A substance that can behave either as an acid or as a base is called amphoteric. Water is the most common amphoteric substance by autoionization.

23. WATER AS AN ACID AND A BASE The equilibrium expression for this reaction is the dissociation constant of water, also called the ion- product constant. At 25°C, the ion-product constant, K w, = 1.0 x 10 -14 [H + ] = [OH - ]= 1.0 x 10 -7 M

24. USING THE ION-PRODUCT CONSTANT It is important to recognize the meaning of K s. In any aqueous solution at 25°C, no matter what it contains, the product of [H + ] and [OH - ] must always equal 1.0 x 10 -14. [H + ] = [OH - ] : neutral solution [H + ] > [OH - ] : acidic solution [H + ] < [OH - ] : basic solution

25. PRACTICE PROBLEM #3 Calculate [H + ] or [OH - ] as required for each of the following solutions at 25°C, and state whether the solution is neutral, acidic, or basic. a.1.0 x 10 -5 M OH - b.1.0 x 10 -7 M OH - c.10.0 M H + a. basic b. neutral c. acidic

26. PRACTICE PROBLEM #4 At 60°C, the value of K w is 1 x 10 -13. a.Using Le Chatelier’s principle, predict whether the below is exothermic or endothermic. b.Calculate [H + ] and [OH - ] in a neutral solution at 60°C. K w increases with temperature – endothermic [H + ] =[OH - ] = 3 x 10 -7 M

27. THE PH SCALE 14.3

28. PH The pH scale represents solution acidity with a log based scale. [H + ] = 1.0 x 10 -7 pH = 7.00 *The number of decimal places in the log is equal to the number of sig figs in the original number

29. PH Because the pH scale is based on a scale of 10, the pH changes by 1 for every power of 10 change in [H + ] example: a solution of pH = 3 has 10 times the concentration of that a solution of pH 4 and 100 times the concentration of a solution of pH 5.

30. POH The pOH scale is similar to the pH scale [OH - ] = 1.0 x 10 -7 pOH = 7.00 *Because the ion product constant of water, pOH + pH always = 14.

31. PRACTICE PROBLEM #5 Calculate pH and POH for each of the following solutions at 25°C. a.1.0 x 10 -3 M OH - b.1.0 M OH - a. pH=11.00, pOH=3.00 b. pH=0.00, pOH=14.00

32. PRACTICE PROBLEM #6 The pH of a sample of human blood was measured to be 7.41 at 25°C. Calculate pOH, [H + ], and [OH - ] for the sample. pOH=6.59, [H + ]= 3.9x10 -8, [OH - ]= 2.6x10 -7 M

33. CALCULATING THE PH OF STRONG ACID SOLUTIONS 14.4

34. SOLVING ACID-BASE PROBLEMS Strategies: 1. Think chemistry: Focus on the solution components and their reactions. Choose the reaction that is most important. 2. Be systematic: Acid-base problems require a step by step approach. 3. Be flexible: Treat each problem as a separate entity. Look for both similarities and the differences to other problems.

35. SOLVING ACID-BASE PROBLEMS Strategies: 4. Be patient: The complete solution to a complicated problem cannot be seen immediately in all its detail. Pick the problem apart into its workable steps. 5. Be confident: Do not rely on memorizing solutions to problems. Understand and think, do not memorize.

36. CALCULATING THE PH OF STRONG ACID SOLUTIONS When calculating acid-base equilibria, focus on the main solution components: example: 1.0 M HCl contains virtually no HCl molecules. Because HCl is a strong acid, it is completely dissociated. The main solution components are H +, Cl -, and H 2 O. The first step of solving acid-base problems is the writing of the major species in the solution.

37. PRACTICE PROBLEM #7, PART 1 Calculate the pH of 0.10 M HNO 3. major species: H +, NO - 3, and H 2 O concentration of species contributing: H + = 0.10 M H + from autoionization of water is negligible compared to H + from HNO 3. pH pH = 1.00

38. PRACTICE PROBLEM #7 PART 2 Calculate the pH of 1.0 x 10 -10 M HCl. major species: H +, Cl -, and H 2 O concentration of species contributing: H + from HCl is negligible. H + from autoionization of water is most important. pH pH = 7.00

39. CALCULATING THE PH OF WEAK ACID SOLUTIONS 14.5

40. CALCULATING THE PH OF A WEAK ACID We will walk through a systematic approach to solving pH for a weak acid: Look for major species. Narrow down contributing species and its concentration. Write balanced equation Write equilibrium expression. ICE Use approximations when able Check validity

41. CALCULATING THE PH OF A WEAK ACID What is the pH of a 1.00 M solution of HF, K a = 7.2x10 -4 Major Species: HF and H 2 O because both have small dissociation constants. Concentration of species contributing: H + from HF is far more significant than H + from H 2 O K a = 7.2x10 -4 vs K w = 1.0x10 -14

42. CALCULATING THE PH OF A WEAK ACID What is the pH of a 1.00 M solution of HF, K a = 7.2x10 -4 strategy: pH  [H + ]  K a initial [HF] = 1.0 M Equilibrium [H + ] = ? ICE!

43. CALCULATING THE PH OF A WEAK ACID What is the pH of a 1.00 M solution of HF, K a = 7.2x10 -4 ICE: InitialChangeEquilibrium H+H+ F-F- HF

44. CALCULATING THE PH OF A WEAK ACID What is the pH of a 1.00 M solution of HF, K a = 7.2x10 -4 ICE: InitialChangeEquilibrium H+H+ 0xx F-F- 0xx HF1.0-x1.0-x

45. CALCULATING THE PH OF A WEAK ACID What is the pH of a 1.00 M solution of HF, K a = 7.2x10 -4 Since K a for HF is so small, HF will dissociate only slightly, and x is expected to be small. If x is very small compared to concentration (2-3+ exponents), the denominator can be simplified.

46. CALCULATING THE PH OF A WEAK ACID What is the pH of a 1.00 M solution of HF, K a = 7.2x10 -4 The validity of this answer needs to be checked due to the approximation made in the calculations. Typically K a values are known to an accuracy of ±5%. We will use this same accuracy for answers to be considered valid.

47. CALCULATING THE PH OF A WEAK ACID What is the pH of a 1.00 M solution of HF, K a = 7.2x10 -4 Checking validity: x=2.7x10 -2, [HF] = 1.00 M x is valid. [H + ] = 2.7x10 -2, pH=1.57

48. PRACTICE PROBLEM #8 The hypochlorite ion (OCl - ) is a strong oxidizing agent often found in household bleaches and disinfectants. It is also the active ingredient that forms when swimming pool water is treated with chlorine. In addition to its oxidizing abilities, the hypochlorite ion has a relatively high affinity for protons (it is much stronger base than Cl -, for example) and forms the weakly acidic hypochlorous acid (HOCl, K a =3.5x10 -8 ). Calculate the pH of a 0.100M aqueous solution of hypochlorous acid.

49. PRACTICE PROBLEM #8 Major species: HOCl and H 2 O Narrow down contributing species and its concentration. Write balanced equation HOCl K a =3.5x10 -8 significant ≅ 0.100M H 2 O K w =1.0x10 -14 negligible

50. PRACTICE PROBLEM #8 Write equilibrium expression. ICE InitialChangeEquilibrium H+H+ OCl - HOCl

51. PRACTICE PROBLEM #8 Use approximations when able: x ≅ 5.9x10 -5

52. PRACTICE PROBLEM #8 Check validity pH: [H + ] = 5.7x10 -8, pH=4.23

53. PRACTICE PROBLEM #9 Calculate the pH of a solution that contains 1.00M HCN (K a =6.2 x 10 -10 ) and 5.00 M HNO 2 (K a =4.0 x 10 -4 ). Also calculate the concentration of cyanide ion (CN - ) in this solution at equilibrium.

54. PRACTICE PROBLEM #9 Major species: Narrow down contributing species and its concentration. Write balanced equation

55. PRACTICE PROBLEM #9 Write equilibrium expression. ICE InitialChangeEquilibrium H+H+ NO 2 - HNO 2

56. PRACTICE PROBLEM #9 Use approximations when able: Check validity. pH= 1.35But wait….what about CN - ?

57. PRACTICE PROBLEM #9 Calculate the pH of a solution that contains 1.00M HCN (K a =6.2 x 10 -10 ) and 5.00 M HNO 2 (K a =4.0 x 10 -4 ). Also calculate the concentration of cyanide ion (CN - ) in this solution at equilibrium. We also want to calculate the CN - concentration from the dissociation of HCN HCN (a) H + (aq) + CN - (aq) K a =6.2x10 -10

58. PRACTICE PROBLEM #9 The molarity of the HCN is known and the [H + ] is equal to the prominent contribution of HNO 2. The source of the H + ions is not important in equilibrium. [CN - ]=1.4x10 -8 M

59. PERCENT DISSOCIATION The percent dissociation is: For a given weak acid, the percent dissociation increases as the acid becomes more dilute.

60. PRACTICE PROBLEM #10 Calculate the percent dissociation of acetic acid (K a = 1.8 x 10 -5 ) in each of the following solutions. a.1.00 M HC 2 H 3 O 2 Major species: HC 2 H 3 O 2 and H 2 O Narrow down contributing species and its concentration. Write balanced equation HC 2 H 3 O 2 K a = 1.8 x 10 -5, H 2 O K w = 1.0 x 10 -14 HC 2 H 3 O 2(aq) H + (aq) + C 2 H 3 O 2 - (aq)

61. PRACTICE PROBLEM #10 Calculate the percent dissociation of acetic acid (K a = 1.8 x 10 -5 ) in each of the following solutions. a.1.00 M HC 2 H 3 O 2 ICE InitialChangeEquilibrium H+H+ C2H3O2-C2H3O2- HC 2 H 3 O 2

62. PRACTICE PROBLEM #10 Calculate the percent dissociation of acetic acid (K a = 1.8 x 10 -5 ) in each of the following solutions. a.1.00 M HC 2 H 3 O 2 Use approximations when able: Apply Equation

63. PRACTICE PROBLEM #10 Calculate the percent dissociation of acetic acid (K a = 1.8 x 10 -5 ) in each of the following solutions. b. 0.100 M HC 2 H 3 O 2 Major species: HC 2 H 3 O 2 and H 2 O Narrow down contributing species and its concentration. Write balanced equation HC 2 H 3 O 2 K a = 1.8 x 10 -5, H 2 O K w = 1.0 x 10 -14 HC 2 H 3 O 2(aq) H + (aq) + C 2 H 3 O 2 - (aq)

64. PRACTICE PROBLEM #10 Calculate the percent dissociation of acetic acid (K a = 1.8 x 10 -5 ) in each of the following solutions. b. 0.100 M HC 2 H 3 O 2 ICE InitialChangeEquilibrium H+H+ C2H3O2-C2H3O2- HC 2 H 3 O 2

65. PRACTICE PROBLEM #10 Calculate the percent dissociation of acetic acid (K a = 1.8 x 10 -5 ) in each of the following solutions. b. 0.100 M HC 2 H 3 O 2 Use approximations when able: Apply Equation

66. PERCENT DISSOCIATION The previous example is a perfect example of the percent dissociation relationship. For a given weak acid, the percent dissociation increases as the acid becomes more dilute.

67. PRACTICE PROBLEM #11 Lactic acid (HC 3 H 5 O 3 ) is a waste product that accumulates in muscle tissue during exertion, leading to pain and a feeling of fatigue. In a 0.100 M aqueous solution, lactic acid is 3.7% dissociated. Calculate the value of K a for this acid.

68. PRACTICE PROBLEM #11 Major species: HC 3 H 5 O 3 and H 2 O Narrow down contributing species and its concentration. Write balanced equation HC 3 H 5 O 3(aq) H + (aq) + C 3 H 5 O 3 - (aq)

69. PRACTICE PROBLEM #11 Percent Dissociation Substitute in the expression K a =1.4x10 -4

70. UNIT 14 CONTINUED IN PART 2