2. Arrhenius Theory Acids release hydrogen ions (H + ) Acids release hydrogen ions (H + ) HCl → H + + Cl - HCl → H + + Cl - Bases release hydroxide ions (OH - ) Bases release hydroxide ions (OH - ) NaOH → Na + + OH -

4. 4 ACID-BASE THEORIES The most general theory for common aqueous acids and bases is the BRØNSTED - LOWRY theory ACIDS DONATE H + IONS BASES ACCEPT H + IONS

8. You Practice NH 3 + H 2 O → NH 4 + + OH - NH 3 + H 2 O → NH 4 + + OH - HNO 3 + NaOH → H 2 O + NaNO 3 HNO 3 + NaOH → H 2 O + NaNO 3 NaHCO 3 + HCl → NaCl + H 2 CO 3 NaHCO 3 + HCl → NaCl + H 2 CO 3

9. 9 Lewis acid = a substance that accepts an electron pair Lewis Acids & Bases Lewis base = a substance that donates an electron pair

10. 10 Formation of hydronium ion is also an excellent example. Lewis Acids & Bases Electron pair of the new O-H bond originates on the Lewis base.

11. 11 Lewis Acid/Base Reaction Lewis Acid/Base Reaction

12. Why is Water Neutral? When one water gains, another loses a H+ [H 3 O+ ] = [OH-] [H 3 O+ ] = [OH-]

13. Relative ion concentrations pH is a relative measure of the hydrogen ion concentration pH is a relative measure of the hydrogen ion concentration pH is a rating; pH is a rating; ranges from 0 – 14 0 = most, 7 equal, 14 = least

14. Any pX Scales In general pX = -log X pOH = - log [OH - ] pH = - log [H + ] pH = - log [H + ]

15. Determining pOH pH + pOH = 14 If know one can determine the other. If pH = 13, what is the pOH? 13 + pOH = 14 pOH = 14 – 13 = 1

17. Why at a pH = 7 ? Determined by concentration [ X ] of each ion Determined by concentration [ X ] of each ion [H+ ]= [OH-] = 10 -7 M [H+ ]= 10 -pH M [OH-] = 10 -pOH M

20. 20 Water H 2 O can function as both an ACID and a BASE. Equilibrium constant for water = K w K w = [H 3 O + ] [OH - ] = 1.00 x 10 -14 at 25 o C AUTOIONIZATION

21. 21 Water K w = [H 3 O + ] [OH - ] = 1.00 x 10 -14 at 25 o C In a neutral solution [H 3 O + ] = [OH - ] so [H 3 O + ] = [OH - ] = 1.00 x 10 -7 M Autoionization

22. 22 Acid-Base Reactions Chapter 17

23. 23 ACIDSACIDS Nonmetal oxides can be acids CO 2 (aq) + H 2 O(liq) ---> H 2 CO 3 (aq) SO 3 (aq) + H 2 O(liq) ---> H 2 SO 4 (aq) and can come from burning coal and oil.

24. 24 BASES Metal oxides are bases CaO in water. Indicator shows solution is basic. CaO(s) + H 2 O(liq) --> Ca(OH) 2 (aq)

25. 25 Hydrolysis

26. 26

27. 27 Acid-Base Reactions sometimes called NEUTRALIZATIONS because the solution is neither acidic nor basic at the end. The other product of the A-B reaction is a SALT, MX. HX + MOH ---> MX + H 2 O M n+ comes from base & X n- comes from acid This is one way to make compounds!

28. 28 Acid-Base Reactions The “driving force” is the formation of water. NaOH(aq) + HCl(aq) ---> NaCl(aq) + H 2 O(liq) Net ionic equation OH - (aq) + H + (aq) ---> H 2 O(liq) This applies to ALL reactions of STRONG acids and bases.

30. 30 Weak Acids and Bases Acid Conjugate Base acetic, CH 3 CO 2 H CH 3 CO 2 -, acetate ammonium, NH 4 + NH 3, ammonia bicarbonate, HCO 3 - CO 3 2-, carbonate A weak acid (or base) is one that ionizes to a VERY small extent (< 5%).

32. 32 Equilibrium Constants for Weak Acids Weak acid has K a < 1 Leads to small [H 3 O + ] and a pH of 2 - 7

33. 33 Equilibrium Constants for Weak Bases Weak base has K b < 1 Leads to small [OH - ] and a pH of 12 - 7

34. 34 AcidsAcids ConjugateBasesConjugateBases K w = K a * K b

36. 36 Weak Acids and Bases acetic acid, CH 3 CO 2 H (HOAc) HOAc + H 2 O  H 3 O + + OAc - Acid Conj. base (K is designated K a for ACID) [H 3 O + ] and [OAc - ] are SMALL, K a << 1.

37. 37 Equilibria Involving A Weak Acid You have 1.00 M HOAc. Calc. the pH. And the equilibrium concs. of EACH Step 1. ICE table. [HOAc][H 3 O + ][OAc - ] I C E

38. 38 Equilibria Involving A Weak Acid [HOAc] [H 3 O + ] [OAc - ] I 1.00 0 0 C -x +x +x E 1.00-x x x Note that we neglect [H 3 O + ] from H 2 O.

39. 39 Equilibria Involving A Weak Acid Assume x is very small because K a is so small. Now we can more easily solve this approximate expression.

40. 40 Equilibria Involving A Weak Acid Step 3. Solve K a approximate expression x = [ H 3 O + ] = [ OAc - ] = [K a 1.00] 1/2 x = [ H 3 O + ] = [ OAc - ] = 4.2 x 10 -3 M pH = -log (4.2 x 10 -3 ) = 2.37

41. © 2011 Pearson Education, Inc. Chapter 17 – Acids and Bases What is the [H + ] of a 1.0 M acetic acid if K a = 1.8 x 10 -5 for the solution: A. 1.0 M B. 0.3 M C. 0.0042 M D. 1.8 x 10 -5 M

42. © 2011 Pearson Education, Inc. Chapter 17 – Acids and Bases A 0.100 M solution of HX has a pH of is 4.5 Determine the K a for the acid, HX. A. 1.0 x 10 -8 B. 3.0 x 10 -7 C. 2.5 x 10 -8 D. 4.5 x 10 -9

43. Weak Bases

44. 44 Equilibria Involving A Weak Base You have 0.010 M NH 3. Calc. the pH. NH 3 + H 2 O  NH 4 + + OH - K b = 1.8 x 10 -5 Step 1; ICE table [NH 3 ][NH 4 + ][OH - ] I C E

45. 45 Weak Base Step 1. ICE table [NH 3 ][NH 4 + ][OH - ] I 0.010 0 0 C -x +x +x E 0.010 - x x x

46. 46 Weak Base Step 2. Solve the equilibrium expression Assume x is small (100K b < C o ), so x = [OH - ] = [NH 4 + ] = 4.2 x 10 -4 M [NH 3 ] = 0.010 - 4.2 x 10 -4 ≈ 0.010 M The approximation is valid!

47. Neutralization Reactions NaOH + HCl → NaCl + H 2 0 NaOH + HCl → NaCl + H 2 0 REACT a base with an acid. REACT a base with an acid. forms salt and water forms salt and water

48. 19.4 Neutralization Reactions > 48 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. A reaction between an acid and a base will go to completion when the solutions contain equal moles of hydrogen ions and hydroxide ions. Acid-Base Reactions The balanced equation provides the correct ratio of acid to base.

49. 19.4 Neutralization Reactions > 49 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. EQUAL Neutralization occurs when the number of moles of hydrogen ions is EQUAL to the number of moles of hydroxide ions.Titration

50. 19.4 Neutralization Reactions > 50 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Acid-Base Reactions For hydrochloric acid and sodium hydroxide, the mole ratio is 1:1. HCl(aq) + NaOH(aq) → NaCl(aq) + H 2 O(l) 1 mol

51. 19.4 Neutralization Reactions > 51 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. For sulfuric acid and sodium hydroxide, the ratio is 1:2. Acid-Base Reactions 1 mol2 mol1 mol2 mol Two moles of the base are required to neutralize one mole of the acid. H 2 SO 4 (aq) + 2NaOH(aq) → Na 2 SO 4 (aq) + 2H 2 O(l)

52. 19.4 Neutralization Reactions > 52 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Acid-Base Reactions Similarly, hydrochloric acid and calcium hydroxide react in a 2:1 ratio. 2HCl(aq) + Ca(OH) 2 (aq) → CaCl 2 (aq) + 2H 2 O(l) 2 mol1 mol 2 mol

53. 19.4 Neutralization Reactions > 53 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Sample Problem 19.7 Finding the Moles Needed for Neutralization How many moles of sulfuric acid are required to neutralize 0.50 mol of sodium hydroxide? The equation for the reaction is H 2 SO 4 (aq) + 2NaOH(aq) → Na 2 SO 4 (aq) + 2H 2 O.

54. 19.4 Neutralization Reactions > 54 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. KNOWNS mol NaOH = 0.50 mol 1 mol H 2 SO 4 /2 mol NaOH (from balanced equation) Analyze List the knowns and the unknown. 1 UNKNOWN mol H 2 SO 4 = ? mol To determine the number of moles of acid, you need to know the number of moles of base and the mole ratio of acid to base. Sample Problem 19.7

55. 19.4 Neutralization Reactions > 55 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Calculate Solve for the unknown. 2 Use the mole ratio of acid to base to determine the number of moles of acid. Sample Problem 19.7 0.50 mol NaOH × 1 mol H 2 SO 4 2 mol NaOH = 0.25 mol H 2 SO 4

56. 19.4 Neutralization Reactions > 56 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Evaluate Does the result make sense? 3 Because the mole ratio of H 2 SO 4 to NaOH is 1:2, the number of moles of H 2 SO 4 should be half the number of the moles of NaOH. Sample Problem 19.7

57. 19.4 Neutralization Reactions > 57 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. You can use a neutralization reaction to determine the concentration of an acid or base. The process of adding a measured amount of a solution of known concentration to a solution of unknown concentration is called a titration.Titration

58. 19.4 Neutralization Reactions > 58 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.Titration A flask with a known volume of acids (and an indicator) is placed beneath a buret that is filled with a base of known concentration. The base is slowly added from the buret to the acid. A change in the color of the solution is the signal that neutralization has occurred.

59. 19.4 Neutralization Reactions > 59

60. 19.4 Neutralization Reactions > 60 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. The indicator that is chosen for a titration must change color at or near the pH of the equivalence point.Titration The point at which the indicator changes color is the end point of the titration.

61. 19.2 Hydrogen Ions and Acidity > 61 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. An indicator (HIn) is a weak acid or base that dissociates in a known pH range. Indicators work because their acid form and base form have different colors in solution. Acid-Base Indicators The acid form of the indicator (HIn) is dominant at low pH and high [H + ]. The base form (In − ) is dominant at high pH and high [OH − ].

62. 19.4 Neutralization Reactions > 62 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. The pH change of a solution during the titration of a strong acid (HCl) with a strong base (NaOH). The equivalence point for this reaction occurs at a pH of 7. As the titration nears the equivalence point, the pH rises dramatically because hydrogen ions are being used up.

63. 19.4 Neutralization Reactions > 63 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Sample Problem 19.8 Determining Concentration by Titration A 25-mL solution of H 2 SO 4 is neutralized by 18 mL of 1.0M NaOH. What is the concentration of the H 2 SO 4 solution? H 2 SO 4 (aq) + 2NaOH(aq) → Na 2 SO 4 (aq) + 2H 2 O(l).

64. 19.4 Neutralization Reactions > 64 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. KNOWNS [NaOH] = 1.0M V NaOH = 18 mL = 0.018 L V H 2 SO 4 = 18 mL = 0.018 L Analyze List the knowns and the unknown. 1 UNKNOWN [H 2 SO 4 ] = ?M The conversion steps are as follows: L NaOH → mol NaOH → mol H 2 SO 4 → M H 2 SO 4. Sample Problem 19.8 Convert volume to liters because molarity is in moles per liter.

65. 19.4 Neutralization Reactions > 65 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Calculate Solve for the unknown. 2 Use the molarity to convert the volume of base to moles of base. Sample Problem 19.8 0.018 L NaOH × 1.0 mol NaOH 1 L NaOH = 0.018 mol NaOH

66. 19.4 Neutralization Reactions > 66 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Calculate Solve for the unknown. 2 Use the mole ratio to find the moles of acid. Sample Problem 19.8 0.018 mol NaOH × 1.0 mol H 2 SO 4 2 mol NaOH = 0.0090 mol H 2 SO 4

67. 19.4 Neutralization Reactions > 67 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Calculate Solve for the unknown. 2 Calculate the molarity by dividing moles of acid by liters of solution. molarity = mol of solute L of solution 0.0090 mol 0.025 L = = 0.36M H 2 SO 4 Sample Problem 19.8

68. 19.5 Salts in Solution > 68 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Interpret Graphs One curve is for the addition of sodium hydroxide, a strong base, to ethanoic acid, a weak acid. An aqueous solution of sodium ethanoate exists at the equivalence point. CH 3 COOH(aq) + NaOH(aq) → CH 3 COONa(aq) + H 2 O(l)

69. 69 Acetic acid titrated with NaOH Figure 18.5 Weak acid titrated with a strong base

70. 70 Acid-Base Titration Section 18.3 You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final solution? HBz + NaOH ---> Na + + Bz - + H 2 O You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final solution? HBz + NaOH ---> Na + + Bz - + H 2 O C 6 H 5 CO 2 H = HBz Benzoate ion = Bz - K b = 1.6 x 10 -10

71. 71 QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. pH at equivalence point? pH of solution of benzoic acid, a weak acid Benzoic acid + NaOH pH at half-way point?

72. 72 Acid-Base Reactions Strategy — find the conc. of the conjugate base Bz - in the solution AFTER the titration, then calculate pH. This is a two-step problem 1. stoichiometry of acid-base reaction 2. equilibrium calculation Strategy — find the conc. of the conjugate base Bz - in the solution AFTER the titration, then calculate pH. This is a two-step problem 1. stoichiometry of acid-base reaction 2. equilibrium calculation QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final solution?

73. 73 STOICHIOMETRY PORTION M * V = mol 1.Calc. moles of NaOH req’d (0.100 L)(0.025 M) = 0.0025 mol HBz This requires 0.0025 mol NaOH 2.Calc. volume of NaOH req’d 0.0025 mol (1 L / 0.100 mol) = 0.025 L 25 mL of NaOH req’d M * V = mol 1.Calc. moles of NaOH req’d (0.100 L)(0.025 M) = 0.0025 mol HBz This requires 0.0025 mol NaOH 2.Calc. volume of NaOH req’d 0.0025 mol (1 L / 0.100 mol) = 0.025 L 25 mL of NaOH req’d

74. 74 STOICHIOMETRY PORTION 3. Moles Bz - produced = moles HBz = 0.0025 mol 4. Calc. conc. of Bz - There are 0.0025 mol of Bz - in a TOTAL SOLUTION VOLUME of 3. Moles Bz - produced = moles HBz = 0.0025 mol 4. Calc. conc. of Bz - There are 0.0025 mol of Bz - in a TOTAL SOLUTION VOLUME of 125 mL [Bz - ] = 0.0025 mol / 0.125 L = 0.020 M = 0.020 M

75. 75 Equivalence Point Most important species in solution is benzoate ion, Bz -. It will react to form the weak conjugate base, benzoic acid, HBz. Bz - + H 2 O  HBz + OH - K b = 1.6 x 10 -10 [Bz - ] [HBz] [OH - ] I 0.020 0 0 C - x +x+x E 0.020 - x x x Most important species in solution is benzoate ion, Bz -. It will react to form the weak conjugate base, benzoic acid, HBz. Bz - + H 2 O  HBz + OH - K b = 1.6 x 10 -10 [Bz - ] [HBz] [OH - ] I 0.020 0 0 C - x +x+x E 0.020 - x x x

76. 76 Acid-Base Reactions x = [OH - ] = 1.8 x 10 -6 pOH = 5.75 -----> pH = 8.25

77. 77 QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH at half-way point? pH at half-way point? Equivalence point pH = 8.25 Equivalence point pH = 8.25

78. 78 Acid-Base Reactions You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH. What is the pH at the half-way point? At the half-way point, [HBz] = [Bz - ] Therefore, [H 3 O + ] = K a = 6.3 x 10 -5 pH = 4.20 = pK a of the acid Both HBz and Bz - are present. This is a BUFFER! Both HBz and Bz - are present. This is a BUFFER!

82. 82 Figure 18.7 Weak base (NH 3 ) titrated with a strong acid (HCl)

83. 83 Acid-Base Reactions Strong acid + strong base HCl + NaOH ----> SALT WATER Strong acid + weak base HCl + NH 3 ---> ACID Weak acid + strong base HOAc + NaOH ---> BASIC Weak acid + weak base HOAc + NH 3 ---> Ka / Kb

84. 84

85. 85