1. CHAPTER 16 – ACIDS AND BASES ACID – A compound the produces hydrogen ions in a water solution HCl (g) → H + (aq) + Cl - (aq) BASE – A compound that produces hydroxide ions in a water solution NaOH (s) → Na + (aq) + OH - (aq) 8A-1 (of 41) 1884 SVANTE ARRHENIUS Proposed the first definitions of acids and bases

2. 1923 Expanded the definitions of acids and bases THOMAS LOWRY 8A-2

3. ACID – A hydrogen ion (or proton) donor BASE – A hydrogen ion (or proton) acceptor 1923 Expanded the definitions of acids and bases JOHANNES BRØNSTED 8A-3

4. HCl + H 2 O → acidbase HYDRONIUM ION – H 3 O +, formed when a hydrogen ion attaches to a water +- Cl - + H 3 O + 8A-4

5. NH 3 + H 2 O → baseacid AMPHOTERIC – A substance that can act as an acid or a base +- NH 4 + + OH - 8A-5

6. Acids turn into bases, and bases turn into acids HCl + H 2 O → Cl - + H 3 O + acidbase +- acidbase 8A-6

7. Acids turn into bases, and bases turn into acids HCl + H 2 O → Cl - + H 3 O + acidbaseconjugate base of HCl conjugate acid of H 2 O +- acidbase acid 8A-7

8. NH 3 + H 2 O → NH 4 + + OH - baseacid +- baseacid 8A-8

9. NH 3 + H 2 O → NH 4 + + OH - baseacidconjugate acid of NH 3 conjugate base of H 2 O +- baseacid base 8A-9

10. Conjugate base of HClO 4 When HClO 4 acts as an acid, it becomes: ClO 4 - Conjugate base of H 2 CO 3 When H 2 CO 3 acts as an acid, it becomes: HCO 3 - Conjugate acid of CF 3 NH 2 When CF 3 NH 2 acts as a base, it becomes: CF 3 NH 3 + 8A-10

11. POLYPROTIC ACID – An acid with more than one ionizable hydrogen ion H 2 CO 3 H 3 PO 4 Hydrogen ions become successively more difficult to ionize diprotic triprotic H 2 SO 4 (aq) → H + (aq) + HSO 4 - (aq) Strong Acid HSO 4 - (aq) → H + (aq) + SO 4 2- (aq) Weaker Acid 8A-11

12. WATER Water ionizes to a small extent 8A-12

13. WATER Water ionizes to a small extent 2H 2 O (l) → H 3 O + (aq) + OH - (aq) - + Makes solutions acidicMakes solutions basic Equal amounts of H 3 O + and OH - make a solution NEUTRAL  pure water is neutral 8A-13

14. Ionization: 2H 2 O (l) → H 3 O + (aq) + OH - (aq) Association: H 3 O + (aq) + OH - (aq)→ 2H 2 O (l) Equilibrium 2H 2 O (l) ↔ H 3 O + (aq) + OH - (aq) Because of this equilibrium, the product of the molarity of the hydronium ions and the molarity of the hydroxide ions is a constant K = [H 3 O + ][OH - ] 8A-14

15. Ionization: 2H 2 O (l) → H 3 O + (aq) + OH - (aq) Association: H 3 O + (aq) + OH - (aq)→ 2H 2 O (l) Equilibrium 2H 2 O (l) ↔ H 3 O + (aq) + OH - (aq) Because of this equilibrium, the product of the molarity of the hydronium ions and the molarity of the hydroxide ions is a constant K w = [H 3 O + ][OH - ] ION-PRODUCT CONSTANT FOR WATER (K w ) – The product of the hydronium ion and hydroxide ion molarities in any water solution 8A-15

16. The K w depends on temperature 10 25 40 0.29 x 10 -14 M 2 1.00 x 10 -14 M 2 2.92 x 10 -14 M 2 Temp (ºC)K w 8A-16

17. Calculate the hydronium ion and hydroxide ion molarities in pure water at 25ºC. (x)(x)1.00 x 10 -14 M 2 = 2H 2 O (l) → H 3 O + (aq) + OH - (aq) Molarities before Ionization: Molarities at Equilibrium: K w = [H 3 O + ][OH - ] = x 2 x1.00 x 10 -7 M = In pure water, [H + ] = [OH - ] = 1.00 x 10 -7 M 0 x 8A-17

18. Calculate the hydroxide ion molarity in a water solution in at 25ºC in which the [H 3 O + ] = 1.00 x 10 -5. (1.00 x 10 -5 M)(x)1.00 x 10 -14 M 2 = K w = [H 3 O + ][OH - ] 1.00 x 10 -9 M = x 1.00 x 10 -14 M 2 _________________ 1.00 x 10 -5 M = x = [OH - ] 8A-18

19. Calculate the hydronium ion molarity in a water solution in at 25ºC in which the [OH - ] = 8.00 x 10 -3. 8A-19

20. [H 3 O + ] = [OH - ]: [H 3 O + ] > [OH - ]: [H 3 O + ] < [OH - ]: the solution is NEUTRAL the solution is ACIDIC the solution is BASIC All water solutions contain both hydronium and hydroxide ions For all water solutions: K w = [H 3 O + ][OH - ] 8A-20

21. THE pH SCALE pH – The negative logarithm of the hydronium ion molarity of a solution The common logarithm of a number is: the exponent to which 10 must be raised to equal the number 10010000.001 100 = 10 2 1000 = 10 3 0.001 = 10 -3 200 200 = 10 2.3 log 100 = 2log 1000 = 3log 0.001 = -3log 200 = 2.3 8A-21

22. [H 3 O + ]log [H 3 O + ]pH 0.1 M 0.01M 0.001 M 0.02M -2.0 -3.0 -1.7 1.0 2.0 3.0 1.7 For logarithmic numbers, only the digits after the decimal point (called the MANTISSA) are significant figures, not the digits before (called the CHARACTERISTIC) 8A-22

23. Calculate the pH of pure water at 25ºC. = -log(1.00 x 10 -7 M) pH= -log[H 3 O + ] For pure water: [H 3 O + ] = 1.00 x 10 -7 M = 7.000 8A-23

24. pH 7 = Neutral < 7 Acidic > 7 Basic 10. M 1.M 0. 1M 0.01M 0.001M 0.0001M 0.00001M 0.000001M 0.0000001M 0.00000001M 0.000000001M 0.0000000001M 0.00000000001M 0.000000000001M 0.0000000000001M 0.00000000000001M [H 3 O + ][OH - ] 0.000000000000001 M 0.00000000000001M 0.0000000000001M 0.000000000001M 0.00000000001M 0.0000000001M 0.000000001M 0.00000001M 0.0000001M 0.000001M 0.00001M 0.0001M 0.001M 0.01M 0.1M 1M1M 8A-24

25. pH 7 = Neutral < 7 Acidic > 7 Basic Battery Acid Gastric Fluid Soda Orange Juice Tap Water “Pure” Water Sea Water Baking Soda Ammonia Bleach Drain Cleaner 10. M 1.M 0. 1M 0.01M 0.001M 0.0001M 0.00001M 0.000001M 0.0000001M 0.00000001M 0.000000001M 0.0000000001M 0.00000000001M 0.000000000001M 0.0000000000001M 0.00000000000001M [H 3 O + ]Common Substances 8A-25

26. Calculate the pH of orange juice if it has a hydronium ion concentration of 1.6 x 10 -3 M. = -log(1.6 x 10 -3 M) pH= -log[H 3 O + ] [H 3 O + ] = 1.6 x 10 -3 M = 2.80 8A-26

27. Calculate the pH of toothpaste that has a hydroxide ion concentration of 5.6 x 10 -5 M. = -log(1.79 x 10 -10 M) pH= -log[H 3 O + ] = 9.74 (x)(5.6 x 10 -5 M) K w = [H 3 O + ][OH - ] = 1.79 x 10 -10 M1.00 x 10 -14 M 2 = x _________________ 5.6 x 10 -5 M 1.00 x 10 -14 M 2 = [OH - ] = 5.6 x 10 -5 M = [H 3 O + ] 8A-27

28. Calculate the pH of milk if it has a hydroxide ion concentration of 3.2 x 10 -9 M. 8A-28

29. Calculate the hydrogen ion concentration in blood, which has a pH of 7.4. = 0.0000000398 M pH= -log[H 3 O + ] -pH= log[H 3 O + ] antilog (-pH)= [H 3 O + ] antilog (-7.4)= [H 3 O + ] For logarithmic numbers, only the digits after the decimal point (called the MANTISSA) are significant figures, not the digits before (called the CHARACTERISTIC) = 3.98 x 10 -8 M = 4 x 10 -8 M 8A-29

30. Calculate the hydroxide ion concentration in egg yolks, which have a pH of 5.65. 8A-30

31. pH OF STRONG ACID SOLUTIONS Strong acids completely ionize HCl (g) HCl (aq) H + (aq) + Cl - (aq) 8A-31

32. Calculate the pH of a 0.015 M hydrochloric acid solution. HCl (aq) + H 2 O (l) → H 3 O + (aq) + Cl - (aq) x 1 M 0.015 1 M = -log(0.015 M) pH= -log[H 3 O + ] = 1.82 x 1 M H 3 O + ____________ 1 M HCl = 0.015 M H 3 O + 0.015 M HCl 8A-32

33. Calculate the pH of a 0.0400 M sulfuric acid solution. H 2 SO 4 (aq) + 2H 2 O (l) → 2H 3 O + (aq) + SO 4 2- (aq) x 2 M 0.0400 1 M = -log(0.0800 M) pH= -log[H 3 O + ] = 1.097 x 2 M H 3 O + ______________ 1 M H 2 SO 4 = 0.0800 M H 3 O + 0.0400 M H 2 SO 4 8A-33

34. pH OF STRONG BASE SOLUTIONS Strong bases completely dissociate NaOH (s) NaOH (aq) Na + (aq) + OH - (aq) 8A-34

35. Calculate the pH of a 0.40 M sodium hydroxide solution. NaOH (aq) → Na + (aq) + OH - (aq) x 1 M 0.40 1 M = -log(2.50 x 10 -14 M)pH= -log[H 3 O + ]= 13.60 (x)(0.40 M) K w = [H 3 O + ][OH - ] = 2.50 x 10 -14 M1.00 x 10 -14 M 2 = x _________________ 0.40 M 1.00 x 10 -14 M 2 = = [H 3 O + ] x 1 M OH - ______________ 1 M NaOH = 0.40 M OH - 0.40 M NaOHnot [H 3 O + ] 8A-35

36. BUFFERS BUFFER – A solution that resists a change in its pH even when a strong acid or base is added to it 7.00 pH 1.00 L H 2 O 0.10 moles HCl 8A-36

37. BUFFERS BUFFER – A solution that resists a change in its pH even when a strong acid or base is added to it 1.00 pH 1.00 L H 2 O 0.10 moles HCl 3.45 pH 1.00 L 1.0 M HF, 1.0 M F - 0.10 moles HCl 8A-37

38. BUFFERS BUFFER – A solution that resists a change in its pH even when a strong acid or base is added to it 1.00 pH 1.00 L H 2 O 0.10 moles HCl 3.37 pH 1.00 L 1.0 M HF, 1.0 M F - 0.10 moles HCl 8A-38

39. A solution is a buffer if is contains a weak acid and its conjugate base Weak acid: Conjugate base: HF F-F- 1.00 L 1.0 M HF, 1.0 M F - 0.10 moles HCl If a strong acid is added to the buffer: 8A-39

40. H + + F - → HF The strong acid (H + ) is reacted away by the conjugate base (F - ) A solution is a buffer if is contains a weak acid and its conjugate base Weak acid: Conjugate base: HF F - 1.00 L 1.0 M HF, 1.0 M F - 0.10 moles HCl  no more strong acid in the solution If a strong acid is added to the buffer: 8A-40

41. A solution is a buffer if it contains a weak acid and its conjugate base Weak acid: Conjugate base: HF F - 1.00 L 1.0 M HF, 1.0 M F - If a strong base is added to the buffer: OH - + HF → H 2 O + F - The strong base (OH - ) is reacted away by the weak acid (HF)  no more strong base in the solution 0.10 moles OH - 8A-41