1. 1 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 9 Lecture Outline Prepared by Andrea D. Leonard University of Louisiana at Lafayette

2. 9.1Introduction to Acids and Bases 2 The earliest definition was given by Arrhenius: An acid contains a hydrogen atom and dissolves in water to form a hydrogen ion, H +. A base contains hydroxide and dissolves in water to form OH −. HCl(g) acid H + (aq) + Cl − (aq) NaOH(s) base Na + (aq) + OH − (aq)

3. 9.1Introduction to Acids and Bases 3 The Arrhenius definition correctly predicts the behavior of many acids and bases. However, this definition is limited and sometimes inaccurate. For example, H + does not exist in water. Instead, it reacts with water to form the hydronium ion, H 3 O +. H 3 O + (aq)H + (aq) + H 2 O(l) hydrogen ion: does not really exist in solution hydronium ion: actually present in aqueous solution

4. 9.1Introduction to Acids and Bases 4 The Brønsted–Lowry definition: A Brønsted–Lowry acid is a proton (H + ) donor. A Brønsted–Lowry base is a proton (H + ) acceptor. H 3 O + (aq) + Cl − (aq)HCl(g) + H 2 O(l) This proton is donated. HCl is a Brønsted–Lowry acid because it donates a proton to the solvent water. H 2 O is a Brønsted–Lowry base because it accepts a proton from HCl.

5. 9.1Introduction to Acids and Bases A. Brønsted–Lowry Acids 5 A Brønsted–Lowry acid must contain a hydrogen atom. Common Brønsted–Lowry acids (HA): HCl hydrochloric acid HBr hydrobromic acid C H H H C O O H acidic H atom acetic acid H 2 SO 4 sulfuric acid HNO 3 nitric acid

6. 9.1Introduction to Acids and Bases A. Brønsted–Lowry Acids 6 A monoprotic acid contains one acidic proton. HCl A diprotic acid contains two acidic protons. H 2 SO 4 A triprotic acid contains three acidic protons. H 3 PO 4 A Brønsted–Lowry acid may be neutral or it may carry a net positive or negative charge. HCl, H 3 O +, HSO 4 −

7. 9.1Introduction to Acids and Bases B. Brønsted–Lowry Bases 7 A Brønsted–Lowry base is a proton acceptor, so it must be able to form a bond to a proton. A base must contain a lone pair of electrons that can be used to form a new bond to the proton. N H H H + H 2 O(l) N H H H H + + OH − (aq) Brønsted–Lowry base This e − pair forms a new bond to a H from H 2 O.

8. 9.1Introduction to Acids and Bases B. Brønsted–Lowry Bases 8 Common Brønsted–Lowry Bases (B ): NaOH sodium hydroxide KOH potassium hydroxide Mg(OH) 2 magnesium hydroxide Ca(OH) 2 calcium hydroxide H 2 O water NH 3 ammonia Lone pairs make these neutral compounds bases. The OH − is the base in each metal salt.

9. 9.2The Reaction of a Brønsted – Lowry Acid with a Brønsted – Lowry Base 9 HA+ B A − HB+B+ + gain of H + acidbase loss of H + This e − pair stays on A. This e − pair forms a new bond to H +.

10. 9.2The Reaction of a Brønsted – Lowry Acid with a Brønsted – Lowry Base 10 The product formed by loss of a proton from an acid is called its conjugate base. HA+ B A − HB+B+ + gain of H + acidbaseconjugate base conjugate acid loss of H + The product formed by gain of a proton by a base is called its conjugate acid.

11. 9.2The Reaction of a Brønsted – Lowry Acid with a Brønsted – Lowry Base 11 HBr+ + gain of H + acidbaseconjugate base conjugate acid loss of H + H2OH2O Br − H3O+H3O+ HBr and Br − are a conjugate acid–base pair. H 2 O and H 3 O + are a conjugate acid–base pair. The net charge must be the same on both sides of the equation.

12. 9.2The Reaction of a Brønsted – Lowry Acid with a Brønsted – Lowry Base 12 When a species gains a proton (H + ), it gains a +1 charge. H 2 O base zero charge add H + H 3 O + +1 charge When a species loses a proton (H + ), it effectively gains a −1 charge. HBr acid zero charge lose H + Br − −1 charge

13. 9.2The Reaction of a Brønsted – Lowry Acid with a Brønsted – Lowry Base 13 Amphoteric compound: A compound that contains both a hydrogen atom and a lone pair of e − ; it can be either an acid or a base. HOH H 2 O as a base add H + HOH H + conjugate acid HOH H 2 O as an acid remove H + HO − conjugate base

14. 9.3Acid and Base Strength A. Relating Acid and Base Strength 14 When a covalent acid dissolves in water, the proton transfer that forms H 3 O + is called dissociation. When a strong acid dissolves in water, 100% of the acid dissociates into ions. Common strong acids are HI, HBr, HCl, H 2 SO 4, and HNO 3. A single reaction arrow can be used, because the product is greatly favored at equilibrium. H 3 O + (aq) + Cl − (aq)HCl(g) + H 2 O(l)

15. 9.3Acid and Base Strength A. Relating Acid and Base Strength 15 When a weak acid dissolves in water, only a small fraction of the acid dissociates into ions. Unequal reaction arrows may be used, because the reactants are usually favored at equilibrium. H 3 O + (aq) + CH 3 COO − (aq)CH 3 COOH(l) + H 2 O(l) Common weak acids are H 3 PO 4, HF, H 2 CO 3, and HCN.

16. 9.3Acid and Base Strength A. Relating Acid and Base Strength 16 A strong acid, HCl, is completely dissociated into H 3 O + (aq) and Cl − (aq). A weak acid, CH 3 COOH, contains mostly undissociated acid.

17. 9.3Acid and Base Strength A. Relating Acid and Base Strength 17 When a strong base dissolves in water, 100% of the base dissociates into ions. Na + (aq) + − OH(aq)NaOH(s) + H 2 O(l) Common strong bases are NaOH and KOH. When a weak base dissolves in water, only a small fraction of the base dissociates into ions. NH 4 + (aq) + − OH(aq)NH 3 (g) + H 2 O(l)

18. 9.3Acid and Base Strength A. Relating Acid and Base Strength 18 A strong base, NaOH, is completely dissociated into Na + (aq) and − OH(aq). A weak base contains mostly undissociated base, NH 3.

19. 9.3Acid and Base Strength A. Relating Acid and Base Strength 19 A strong acid readily donates a proton, forming a weak conjugate base. HCl strong acid Cl − weak conjugate base A strong base readily accepts a proton, forming a weak conjugate acid. OH − strong base H 2 O weak conjugate acid

20. 9.3Acid and Base Strength A. Relating Acid and Base Strength 20 9.2 -----

21. 9.3Acid and Base Strength B. Using Acid Strength to Predict the Direction of Equilibrium 21 The position of the equilibrium depends upon the strengths of the acids and bases. The stronger acid reacts with the stronger base to form the weaker acid and the weaker base. A Brønsted–Lowry acid–base reaction represents an equilibrium. HA + B A − HB+B+ + acidbaseconjugate base conjugate acid

22. 9.3Acid and Base Strength B. Using Acid Strength to Predict the Direction of Equilibrium 22 When the stronger acid and base are the reactants on the left side, the reaction readily occurs and the reaction proceeds to the right. HA+ B A − HB+B+ + stronger acid stronger base weaker base weaker acid A larger forward arrow means that products are favored.

23. 9.3Acid and Base Strength B. Using Acid Strength to Predict the Direction of Equilibrium 23 If an acid–base reaction would form the stronger acid and base, equilibrium favors the reactants and little product forms. HA+ B A − HB+B+ + weaker acid weaker base stronger base stronger acid A larger reverse arrow means that reactants are favored.

24. 9.3Acid and Base Strength 24 HOW TO Predict the Direction of Equilibrium in an Acid–Base Reaction Example Are the reactants or products favored in the following acid–base reaction? Step [1] Identify the acid in the reactants and the conjugate acid in the products. acidbase conjugate base conjugate acid + + HCN(g) − OH(aq) − CN(aq) H 2 O(l) gain of H + loss of H +

25. 9.3Acid and Base Strength 25 HOW TO Predict the Direction of Equilibrium in an Acid–Base Reaction Step [2] Determine the relative strength of the acid and the conjugate acid. From Table 9.2, HCN is a stronger acid than H 2 O. Step [3] Equilibrium favors the formation of the weaker acid. + + HCN(g) − OH(aq) − CN(aq) H 2 O(l) stronger acid weaker acid Products are favored.

26. 9.4Equilibrium and Acid Dissociation Constants 26 For the reaction where an acid (HA) dissolves in water, H 3 O + (aq) + (aq)HA(g) + H 2 O(l)A − the following equilibrium constant can be written: K = [H 3 O + ][ ] A − [HA][H 2 O]

27. 9.4Equilibrium and Acid Dissociation Constants 27 Multiplying both sides by [H 2 O] forms a new (more useful) constant, called the acid dissociation constant, K a. K a = K[H 2 O] = [H 3 O + ][ ] A − [HA] acid dissociation constant The stronger the acid, the larger the K a value. Equilibrium favors formation of the weaker acid, the acid with the smaller K a value.

28. 9.4Equilibrium and Acid Dissociation Constants 28

29. 9.5Dissociation of Water 29 H OH base HOH H + conjugate acid HOH conjugate base HO − ++ loss of H + gain of H + Water can behave as both a Brønsted–Lowry acid and a Brønsted–Lowry base. Thus, two water molecules can react together in an acid–base reaction:

30. 9.5Dissociation of Water 30 [H 3 O + ][OH − ] [H 2 O] 2 From the reaction of two water molecules, the following equilibrium constant expression can be written: Multiplying both sides by [H 2 O] 2 yields K w, the ion-product constant for water. K w = [H 3 O + ][OH − ] ion-product constant K =

31. 9.5Dissociation of Water 31 Experimentally it can be shown that: [H 3 O + ] = [OH − ] = 1.0 x 10 −7 M at 25 o C K w = [H 3 O + ] [OH − ] K w = (1.0 x 10 −7 ) x (1.0 x 10 −7 ) K w = 1.0 x 10 −14 K w is a constant, 1.0 x 10 −14, for all aqueous solutions at 25 o C.

32. 9.5Dissociation of Water 32 To calculate [OH − ] when [H 3 O + ] is known: To calculate [H 3 O + ] when [OH − ] is known: K w = [H 3 O + ][OH − ]

33. 9.5Dissociation of Water 33 If the [H 3 O + ] in a cup of coffee is 1.0 x 10 −5 M, then the [OH − ] can be calculated as follows: [OH − ] = KwKw [H 3 O + ] = 1.0 x 10 −14 1.0 x 10 −5 =1.0 x 10 −9 M In this cup of coffee, therefore, [H 3 O + ] > [OH – ], and the solution is acidic overall.

34. 9.5Dissociation of Water 34

35. 9.6The pH Scale A. Calculating pH 35 The lower the pH, the higher the concentration of H 3 O + : Acidic solution: [H 3 O + ] > 1 x 10 −7  pH < 7 Basic solution: [H 3 O + ] 7 Neutral solution: [H 3 O + ] = 1 x 10 −7  pH = 7

36. 9.6The pH Scale A. Calculating pH 36

37. 9.6The pH Scale B. Calculating pH Using a Calculator 37 If [H 3 O + ] = 1.2 x 10 –5 M for a solution, what is its pH? pH = –log [H 3 O + ] pH = –(–4.92) The solution is acidic because the pH < 7. A logarithm has the same number of places after the decimal as there are digits in the original number. Example: pH = –log (1.2 x 10 –5 )2 digits pH = 4.92 2 decimal places

38. 9.6The pH Scale B. Calculating pH Using a Calculator 38 If the pH of a solution is 8.50, what is the [H 3 O + ]? pH = −log [H 3 O + ] 8.50 = −log [H 3 O + ] −8.50 = log [H 3 O + ] antilog (−8.50 ) = [H 3 O + ] [H 3 O + ] = 3.2 x 10 −9 M The solution is basic because [H 3 O + ] > 1 x 10 –7 M. 2 decimal places 2 digits

39. 9.6The pH Scale 39

40. 9.7Common Acid–Base Reactions A. Reaction of Acids with Hydroxide Bases 40 Neutralization reaction: An acid-base reaction that produces a salt and water as products. HA(aq) + MOH(aq) acid base OH(l) + MA(aq) H watersalt The acid HA donates a proton (H + ) to the OH − base to form H 2 O. The anion A − from the acid combines with the cation M + from the base to form the salt MA.

41. 9.7Common Acid–Base Reactions 41 HOW TO Draw a Balanced Equation for a Neutralization Reaction Between HA and MOH Example Write a balanced equation for the reaction of Mg(OH) 2 with HCl. Step [1] Identify the acid and base in the reactants and draw H 2 O as one product. HCl(aq) + Mg(OH) 2 (aq) acid base H 2 O(l) + water salt

42. 9.7Common Acid–Base Reactions 42 HOW TO Draw a Balanced Equation for a Neutralization Reaction between HA and MOH Step [2] Determine the structure of the salt. The salt is formed from the parts of the acid and base that are not used to form H 2 O. HCl H + reacts to form H 2 O Cl − used to form salt Mg(OH) 2 Mg 2+ used to form salt 2 OH − react to form water Mg 2+ and Cl − combine to form MgCl 2.

43. 9.7Common Acid–Base Reactions 43 HOW TO Draw a Balanced Equation for a Neutralization Reaction between HA and MOH Step [3] Balance the equation. HCl(aq) + Mg(OH) 2 (aq) acid base H 2 O(l) + water salt MgCl 2 2 Place a 2 to balance Cl. Place a 2 to balance O and H. 2

44. 9.7Common Acid–Base Reactions A. Reaction of Acids with Hydroxide Bases 44 A net ionic equation contains only the species involved in a reaction. HCl(aq) + NaOH(aq) H—OH(l) + NaCl(aq) Written as individual ions: H + (aq) + Cl − (aq) + Na + (aq) + OH − (aq) H—OH(l) + Na + (aq) + Cl − (aq) Omit the spectator ions, Na + and Cl –. H + (aq) + OH − (aq)H—OH(l) What remains is the net ionic equation:

45. 9.7Common Acid–Base Reactions B. Reaction of Acids with Bicarbonate Bases 45 A bicarbonate base, HCO 3 −, reacts with one H + to form carbonic acid, H 2 CO 3. H + (aq) + HCO 3 − (aq) Carbonic acid then decomposes into H 2 O and CO 2. H 2 O(l) + CO 2 (g) H 2 CO 3 (aq) HCl(aq) + NaHCO 3 (aq) H 2 O(l) + CO 2 (g) NaCl(aq) + H 2 CO 3 (aq) For example:

46. 9.7Common Acid–Base Reactions B. Reaction of Acids with Bicarbonate Bases 46 A carbonate base, CO 3 2–, reacts with two H + to form carbonic acid, H 2 CO 3. 2 H + (aq) + CO 3 2– (aq) H 2 O(l) + CO 2 (g) H 2 CO 3 (aq) 2 HCl(aq) + Na 2 CO 3 (aq) H 2 O(l) + CO 2 (g) 2 NaCl(aq) + H 2 CO 3 (aq) For example:

47. 9.8The Acidity and Basicity of Salt Solutions 47 A salt can form an acidic, basic, or neutral solution depending on whether its cation and anion are derived from a strong or weak acid and base. For the salt M + A − : The cation M + comes from the base. The anion A − comes from the acid HA.

48. 9.8The Acidity and Basicity of Salt Solutions 48 NaCl Na + from NaOH strong base Cl − from HCl strong acid A salt derived from a strong acid and strong base forms a neutral solution (pH = 7).

49. 9.8The Acidity and Basicity of Salt Solutions 49 NaHCO 3 Na + from NaOH strong base HCO 3 − from H 2 CO 3 weak acid A salt derived from a strong base and a weak acid forms a basic solution (pH > 7).

50. 9.8The Acidity and Basicity of Salt Solutions 50 NH 4 Cl NH 4 + from NH 3 weak base Cl − from HCl strong acid A salt derived from a weak base and a strong acid forms an acidic solution (pH < 7).

51. 9.8The Acidity and Basicity of Salt Solutions 51 Thus, the ion derived from the stronger acid or base determines whether the solution is acidic or basic.

52. 9.9Titration 52 To determine the concentration of an acid or base in a solution, we carry out a titration. If we want to know the concentration of an acid solution, a base of known concentration is added slowly until the acid is neutralized. When the acid is neutralized: # of moles of acid = # of moles of base This is called the end point of the titration.

53. 9.9Titration 53

54. 9.9Titration 54 Determining an unknown molarity from titration data requires three operations: Moles of base Moles of base Volume of acid Volume of acid mole–mole conversion factor mole–mole conversion factor M (mol/L) conversion factor M (mol/L) conversion factor Moles of acid Moles of acid Volume of base M (mol/L) conversion factor [1] [2] [3]

55. 9.9Titration 55 HOW TO Determine the Molarity of an Acid Solution from Titration Example What is the molarity of an HCl solution if 22.5 mL of a 0.100 M NaOH solution are needed to titrate a 25.0 mL sample of the acid? volume of base (NaOH) 22.5 mL conc. of base (NaOH) 0.100 M volume of acid (HCl) 25.0 mL conc. of acid (HCl) ?

56. 9.9Titration 56 Step [1] Determine the number of moles of base used to neutralize the acid. 22.5 mL NaOH x 1 L 1000 mL x 0.100 mol NaOH 1 L = 0.00225 mol NaOH HOW TO Determine the Molarity of an Acid Solution from Titration Moles of base Moles of base Volume of base M (mol/L) conversion factor

57. 9.9Titration 57 HOW TO Determine the Molarity of an Acid Solution from Titration Step [2] Determine the number of moles of acid that react from the balanced chemical equation. HCl(aq) + NaOH(aq) H 2 O(l) + NaCl(aq) 0.00225 mol NaOH x 1 mol HCl 1 mol NaOH = 0.00225 mol HCl Moles of base Moles of base mole–mole conversion factor mole–mole conversion factor Moles of acid Moles of acid

58. 9.9Titration 58 HOW TO Determine the Molarity of an Acid Solution from Titration Step [3] Determine the molarity of the acid from the number of moles and the known volume. M= mol L = 0.00225 mol HCl 25.0 mL solution x 1000 mL 1 L =0.0900 M HCl 3 sig. fig. Answer mL—L conversion factor mL—L conversion factor

59. 9.10Buffers 59 A buffer is a solution whose pH changes very little when acid or base is added. Most buffers are solutions composed of roughly equal amounts of: A weak acid The salt of its conjugate base The buffer resists change in pH because Added base, OH −, reacts with the weak acid Added acid, H 3 O +, reacts with the conjugate base

60. 9.10Buffers A. General Characteristics of a Buffer 60 If an acid is added to the following buffer equilibrium, then the excess acid reacts with the conjugate base, so the overall pH does not change much.

61. 9.10Buffers A. General Characteristics of a Buffer 61 If a base is added to the following buffer equilibrium, then the excess base reacts with the conjugate acid, so the overall pH does not change much.

62. 9.10Buffers 62

63. 9.10Buffers B. Calculating the pH of a Buffer 63 The effective pH range of a buffer depends on its K a. [H 3 O + ][ ] A − [HA] K a = Rearranging this expression to solve for [H 3 O + ]: [ ] A − [HA] =[H 3 O + ]xKaKa determines the buffer pH H 3 O + (aq) + (aq)HA(aq) + H 2 O(l)A −

64. 9.11Focus on the Human Body Buffers in the Blood 64 Normal blood pH is between 7.35 and 7.45. The principal buffer in the blood is carbonic acid/ bicarbonate (H 2 CO 3 /HCO 3 − ). CO 2 (g) + H 2 O(l) H 2 CO 3 (aq) H2OH2O H 3 O + (aq) + HCO 3 − (aq) CO 2 is constantly produced by metabolic processes in the body. The amount of CO 2 is related to the pH of the blood.

65. 9.11Focus on the Human Body Buffers in the Blood 65 Respiratory acidosis results when the body fails to eliminate enough CO 2, due to lung disease or failure.

66. 9.11Focus on the Human Body Buffers in the Blood 66 Respiratory alkalosis is caused by hyperventilating; very little CO 2 is produced by the body.