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CHAPTER 16 – ACIDS AND BASES

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

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CHAPTER 16 – ACIDS AND BASES

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  1. CHAPTER 16 – ACIDS AND BASES 1884 SVANTE ARRHENIUS Proposed the first definitions of acids and bases ACID – A compound that 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 51)

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

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

  4. HCl + H2O → Cl- + H3O+ acid base HYDRONIUM ION – H3O+, formed when a hydrogen ion attaches to a water - + 8A-4

  5. NH3 + H2O → NH4+ + OH- base acid AMPHOTERIC – A substance that can act as an acid or a base + - 8A-5

  6. Acids turn into bases, and bases turn into acids HCl + H2O → Cl- + H3O+ acid base - + acid base 8A-6

  7. Acids turn into bases, and bases turn into acids HCl + H2O → Cl- + H3O+ acid base conjugate base of HCl conjugate acid of H2O - + acid base base acid Strong acids (or bases) have non-conjugates 8A-7

  8. NH3+ H2O → NH4+ + OH- base acid + - base acid 8A-8

  9. NH3 + H2O → NH4++ OH- base acid conjugate acid of NH3 conjugate base of H2O + - base acid acid base Weak bases (or acids) have weak conjugates 8A-9

  10. Conjugate base of HClO4 When HClO4 acts as an acid, it becomes: ClO4- Strong Acid Non Base Conjugate acid of CH3NH2 When CH3NH2acts as a base, it becomes: CH3NH3+ Weak Base Non Acid 8A-10

  11. POLYPROTIC ACID – An acid with more than one ionizable hydrogen ion H2CO3 H3PO4 diprotic triprotic Hydrogen ions become successively more difficult to ionize H2SO4(aq) + H2O(l) → H3O+(aq) + HSO4-(aq) Strong Acid HSO4-(aq) + H2O(l) → H3O+(aq) + SO42-(aq) Weaker Acid 8A-11

  12. Conjugate base of H2CO3 When H2CO3 acts as an acid, it becomes: HCO3- Conjugate base of HCO3- When HCO3- acts as an acid, it becomes: CO32- Conjugate acid of HCO3- When HCO3- acts as an base, it becomes: H2CO3 8A-12

  13. WATER Water ionizes to a small extent 8A-13

  14. WATER Water ionizes to a small extent - + 2H2O (l)→ H3O+(aq) + OH-(aq) Makes solutions acidic Makes solutions basic Equal amounts of H3O+ and OH- make a solution NEUTRAL  pure water is neutral 8A-14

  15. Ionization: 2H2O (l)→ H3O+(aq) + OH-(aq) Association: H3O+(aq) + OH-(aq) → 2H2O (l) Equilibrium 2H2O (l) ⇆ H3O+(aq) + OH-(aq) Because of this dynamic equilibrium, the product of the molarity of the hydronium ions and the molarity of the hydroxide ions is a constant If we use [ ] to represent the molarity of a substance K= [H3O+][OH-] Where K is a constant 8A-15

  16. Ionization: 2H2O (l)→ H3O+(aq) + OH-(aq) Association: H3O+(aq) + OH-(aq) → 2H2O (l) Equilibrium 2H2O (l) ⇆ H3O+(aq) + OH-(aq) Because of this dynamic equilibrium, the product of the molarity of the hydronium ions and the molarity of the hydroxide ions is a constant If we use [ ] to represent the molarity of a substance Kw= [H3O+][OH-] ION-PRODUCT CONSTANT FOR WATER (Kw) – The product of the hydronium ion and hydroxide ion molarities in any water solution 8A-16

  17. The Kw depends on temperature Temp (ºC)Kw 10 25 40 0.29 x 10-14 M2 1.00 x 10-14 M2 2.92 x 10-14 M2 8A-17

  18. Calculate the hydronium ion and hydroxide ion molarities in pure water at 25ºC. 2H2O (l) ⇆H3O+(aq) + OH-(aq) Molarities before any Ionization: Molarity Change: Molarities at Equilibrium: 0 0 +x +x x x Kw= [H3O+][OH-] 1.00 x 10-14 M2 = (x)(x) = x2 1.00 x 10-7 M = x In pure water, [H3O+] = [OH-] = 1.00 x 10-7 M 8A-18

  19. Calculate the hydroxide ion molarity in a water solution at 25ºC in which the [H3O+] = 1.00 x 10-5 M. Kw= [H3O+][OH-] 1.00 x 10-14 M2 = (1.00 x 10-5 M)(x) 1.00 x 10-14 M2 _________________ 1.00 x 10-5 M = x 1.00 x 10-9 M = x = [OH-] 8A-19

  20. Calculate the hydronium ion molarity in a water solution at 25ºC in which the [OH-] = 8.00 x 10-3 M. Kw= [H3O+][OH-] 1.00 x 10-14 M2 = (x)(8.00 x 10-3 M) 1.00 x 10-14 M2 _________________ 8.00 x 10-3 M = x 1.25 x 10-12 M = x = [H3O+] 8A-20

  21. All water solutions contain both hydronium and hydroxide ions [H3O+] = [OH-] : [H3O+] > [OH-] : [H3O+] < [OH-] : the solution is NEUTRAL the solution is ACIDIC the solution is BASIC For all water solutions: Kw = [H3O+][OH-] 8A-21

  22. 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 100 1000 0.001 200 100 = 102 1000 = 103 0.001 = 10-3 200 = 102.3 log 100 = 2 log 1000 = 3 log 0.001 = -3 log 200 = 2.3 8A-22

  23. [H3O+] log [H3O+] pH 0.1 M 0.01 M 0.001 M 0.02 M -1.0 -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-23

  24. Calculate the pH of pure water at 25ºC. For pure water: [H3O+] = 1.00 x 10-7 M pH = -log[H3O+] = -log(1.00 x 10-7 M) = 7.000 8A-24

  25. 10. M 1. M 0. 1 M 0.01 M 0.001 M 0.0001 M 0.00001 M 0.000001 M 0.0000001 M 0.00000001 M 0.000000001 M 0.0000000001 M 0.00000000001 M 0.000000000001 M 0.0000000000001 M 0.00000000000001 M 0.000000000000001 M 0.00000000000001 M 0.0000000000001 M 0.000000000001 M 0.00000000001 M 0.0000000001 M 0.000000001 M 0.00000001 M 0.0000001 M 0.000001 M 0.00001 M 0.0001 M 0.001 M 0.01 M 0.1 M 1 M < 7 Acidic pH 7 = Neutral > 7 Basic [H3O+] [OH-] 8A-25

  26. 10. M 1. M 0. 1 M 0.01 M 0.001 M 0.0001 M 0.00001 M 0.000001 M 0.0000001 M 0.00000001 M 0.000000001 M 0.0000000001 M 0.00000000001 M 0.000000000001 M 0.0000000000001 M 0.00000000000001 M Battery Acid Gastric Fluid < 7 Acidic Soda Orange Juice Tap Water pH 7 = Neutral “Pure” Water Sea Water Baking Soda > 7 Basic Ammonia Bleach Drain Cleaner [H3O+] Common Substances 8A-26

  27. Calculate the pH of orange juice if it has a hydronium ion concentration of 1.6 x 10-3 M. [H3O+] = 1.6 x 10-3 M pH = -log[H3O+] = -log(1.6 x 10-3 M) = 2.80 8A-27

  28. Calculate the pH of toothpaste that has a hydroxide ion concentration of 5.6 x 10-5 M. [OH-] = 5.6 x 10-5 M Kw= [H3O+][OH-] 1.00 x 10-14 M2 = (x)(5.6 x 10-5 M) 1.00 x 10-14 M2 = x _________________ 5.6 x 10-5 M = 1.79 x 10-10 M = [H3O+] pH = -log[H3O+] = -log(1.79 x 10-10 M) = 9.75 8A-28

  29. Calculate the pH of milk if it has a hydroxide ion concentration of 3.2 x 10-9 M. [OH-] = 3.2 x 10-9 M Kw= [H3O+][OH-] 1.00 x 10-14 M2 = (x)(3.2 x 10-9 M) 1.00 x 10-14 M2 = x _________________ 3.2 x 10-9 M = 3.13 x 10-6 M = [H3O+] pH = -log[H3O+] = -log(3.13 x 10-6 M) = 5.51 8A-29

  30. Calculate the hydronium ion concentration in blood, which has a pH of 7.4. pH = -log [H3O+] -pH = log [H3O+] antilog (-pH) = [H3O+] = 4 x 10-8 M antilog (-7.4) = [H3O+] = 0.0000000398 M = 3.98 x 10-8 M For logarithmic numbers, only the digits after the decimal point are significant figures, not the digits before 8A-30

  31. Calculate the hydroxide ion concentration in egg yolks, which have a pH of 5.65. pH = -log [H3O+] -pH = log [H3O+] antilog (-pH) = [H3O+] antilog (-5.65) = [H3O+] = 0.00000224 M = 2.24 x 10-6 M Kw= [H3O+][OH-] 1.00 x 10-14 M2 = (2.24 x 10-6 M)(x) 1.00 x 10-14M2 = __________________ 2.24 x 10-6 M x = 4.5 x 10-9 M = [OH-] 8A-31

  32. pH OF STRONG ACID SOLUTIONS Strong acids completely ionize HCl(g) HCl(aq) H+(aq) + Cl-(aq) 8A-32

  33. Calculate the pH of a 0.015 M hydrochloric acid solution. HCl completely ionizes,  0.015 M H+ (H3O+) 0.015 M Cl- pH = -log[H3O+] = -log(0.015 M) = 1.82 8A-33

  34. Calculate the pH of a 0.0400 M sulfuric acid solution. H2SO4 completely ionizes,  0.0800 M H+ (H3O+) 0.0400 M SO42- pH = -log[H3O+] = -log(0.0800 M) = 1.097 8A-34

  35. pH OF STRONG BASE SOLUTIONS Strong bases completely dissociate NaOH(s) NaOH(aq) Na+(aq) + OH-(aq) 8A-35

  36. Calculate the pH of a 0.40 M sodium hydroxide solution. NaOH completely dissociates,  0.40 M Na+ 0.40 M OH- not [H3O+] Kw= [H3O+][OH-] 1.00 x 10-14 M2 = (x)(0.40 M) 1.00 x 10-14 M2 = x __________________ 0.40 M = 2.50 x 10-14 M = [H3O+] = 13.60 pH = -log[H3O+] = -log(2.50 x 10-14 M) 8A-36

  37. BUFFERS BUFFER – A solution that resists a change in its pH even when a strong acid or strong base is added to it 7.00 pH 1.00 L H2O 0.10 moles HCl 8A-37

  38. BUFFERS BUFFER – A solution that resists a change in its pH even when a strong acid or strong base is added to it 1.00 3.45 pH pH 1.00 L H2O 1.00 L 1.0 M HF, 1.0 M F- 0.10 moles HCl 0.10 moles HCl 8A-38

  39. BUFFERS BUFFER – A solution that resists a change in its pH even when a strong acid or strong base is added to it 1.00 3.37 pH pH 1.00 L H2O 1.00 L 1.0 M HF, 1.0 M F- 0.10 moles HCl 0.10 moles HCl 8A-39

  40. A solution is a buffer if is contains a weak acid and its conjugate base Weak acid: Conjugate base: HF F- If a strong acid is added to the buffer: 1.00 L 1.0 M HF, 1.0 M F- 0.10 moles HCl 8A-40

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

  42. A solution is a buffer if is contains a weak acid and its conjugate base Weak acid: Conjugate base: HF F- If a strong base is added to the buffer: The strong base (OH-) is reacted away by the weak acid (HF) OH- + HF → H2O + F- 1.00 L 1.0 M HF, 1.0 M F-  no more strong base in the solution 0.10 moles OH- 8A-42

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