1. Chapter 15TMHsiung©2015Slide 1 of 68 Chapter 15 Acids and Bases 熊同銘 tmhsiung@gmail.com

2. Chapter 15TMHsiung©2015Slide 2 of 68 1.Heartburn 2.The Nature of Acids and Bases 3.Definitions of Acids and Bases 4.Acid Strength and the Acid Ionization Constant (K a ) 5.Autoionization of Water and pH 6.Finding the [H 3 O + ] and pH of Strong and Weak Acid Solutions 7.Base Solutions 8.The Acid–Base Properties of Ions and Salts Contents

3. Chapter 15TMHsiung©2015Slide 3 of 68 9.Polyprotic Acids 10.Acid Strength and Molecular Structure 11.Lewis Acids and Bases 12.Acid Rain (omit) Continued

4. Chapter 15TMHsiung©2015Slide 4 of 68 1.Heartburn Stomach Acid: The cells that line your stomach produce hydrochloric acid. ̶ To kill unwanted bacteria ̶ To help break down food ̶ To activate enzymes that break down food Acid reflux: If the stomach acid backs up into your esophagus, it irritates those tissues, resulting in heartburn. GERD (gastroesophageal reflux disease): The chronic leaking of stomach acid into the esophagus. ̶ In people with GERD, the muscles separating the stomach from the esophagus do not close tightly, allowing stomach acid to leak into the esophagus.

5. Chapter 15TMHsiung©2015Slide 5 of 68 2.The Nature of Acids and Bases 1)The Nature of Acids Properties of Acids: ̶ Sour taste ̶ Ability to dissolve many metals ̶ Ability to neutralize bases ̶ Change blue litmus paper to red

6. Chapter 15TMHsiung©2015Slide 6 of 68 Common Acids

7. Chapter 15TMHsiung©2015Slide 7 of 68 Structures of Acids ̶ Binary acids: Acid hydrogens attached to a nonmetal atom. HCl, HF

8. Chapter 15TMHsiung©2015Slide 8 of 68 ̶ Oxyacids: Acid hydrogens attached to an oxygen atom. H 2 SO 4, HNO 3

9. Chapter 15TMHsiung©2015Slide 9 of 68 ̶ Carboxylic acids: the acids have -COOH group. HC 2 H 3 O 2, H 3 C 6 H 5 O 7 *Only the first H in the formula is acidic (the H is on the -COOH).

10. Chapter 15TMHsiung©2015Slide 10 of 68 2)The Nature of Bases Properties of Bases: ̶ Taste bitter ̶ Feel slippery ̶ Change red litmus paper to blue ̶ Ability to neutralize acids

11. Chapter 15TMHsiung©2015Slide 11 of 68 Common Bases

12. Chapter 15TMHsiung©2015Slide 12 of 68 3.Definitions of Acids and Bases Arrhenius definition –Based on H + and OH – Brønsted–Lowry definition –Based on reactions in which H + is transferred Lewis definition (15.11)

13. Chapter 15TMHsiung©2015Slide 13 of 68 1)The Arrhenius Definition Arrhenius Acid: A substance that produces H + ions in aqueous solution. Example: HCl (aq) → H + (aq) + Cl − (aq) *The H + (hydrogen ions) produced by the acid react with water molecules to produce complex ions, mainly hydronium ion, H 3 O + : H + + H 2 O  H 3 O +

14. Chapter 15TMHsiung©2015Slide 14 of 68 Arrhenius Base: A substance that produces OH – (hydroxide ions) in aqueous solution. Example: NaOH (aq) → Na + (aq) + OH – (aq) Arrhenius Acid–Base Reactions: The H + from the acid combines with the OH − from the base to make a molecule of H 2 O (neutralization reaction). acid + base → salt + water HCl (aq) + NaOH (aq) → NaCl (aq) + H 2 O (l)

15. Chapter 15TMHsiung©2015Slide 15 of 68 Problems with Arrhenius Theory -It does not explain why molecular substances, such as NH 3 and Na 2 CO 3 dissolve in water to form basic solutions, even though they do not contain OH – ions. -It does not explain why molecular substances, such as CO 2, dissolve in water to form acidic solutions, even though they do not contain H + ions. -It does not explain acid–base reactions that take place outside aqueous solution.

16. Chapter 15TMHsiung©2015Slide 16 of 68 2)The Brønsted–Lowry Definition Acid: the substance act as H + donor Base: the substance act as H + acceptor *Amphoteric substances: the substances can act as either an acid or a base, example H 2 O. *Any reaction involving H + (proton) that transfers from one molecule to another is an acid–base reaction, regardless of whether it occurs in aqueous solution or if there is OH − present. *All reactions that fit the Arrhenius definition also fit the Brønsted– Lowry definition.

17. Chapter 15TMHsiung©2015Slide 17 of 68 Brønsted–Lowry Acid–Base Reactions  One of the advantages of Brønsted–Lowry theory is that it illustrates reversible reactions to be as follows: H–A + :B  :A – + H–B +  The original base has an extra H + after the reaction, so it will act as an acid in the reverse process. And the original acid has a lone pair of electrons after the reaction, so it will act as a base in the reverse process: :A– + H–B+  H–A + :B

18. Chapter 15TMHsiung©2015Slide 18 of 68 Conjugate acid–base pair -A base accepts a proton and becomes a conjugate acid. -An acid donates a proton and becomes a conjugate base.

19. Chapter 15TMHsiung©2015Slide 19 of 68 Solution a. b. In each reaction, identify the Brønsted–Lowry acid, the Brønsted–Lowry base, the conjugate acid, and the conjugate base. a. b. Example 15.1Identifying Brønsted–Lowry Acids and Bases and Their Conjugates

20. Chapter 15TMHsiung©2015Slide 20 of 68 1)Strong acids 4.Acid Strength and the Acid Ionization Constant (K a ) -A strong electrolyte, essentially completely ionized. -H 3 O + is a weaker acid than HCl and Cl − is a weaker base than H 2 O. The stronger the acid, the weaker the conjugate base. The stronger the base, the weaker the conjugate acid. An acid-base reaction is favored in the direction from the stronger member to the weaker member of each conjugate acid-base pair.

21. Chapter 15TMHsiung©2015Slide 21 of 68 -Important strong acids

22. Chapter 15TMHsiung©2015Slide 22 of 68 2)Weak acids -A weak electrolyte, only a small percentage of the molecules ionized. -H 3 O + is a stronger acid than HF and F − is a stronger base than H 2 O.

23. Chapter 15TMHsiung©2015Slide 23 of 68 -Important weak acids

24. Chapter 15TMHsiung©2015Slide 24 of 68 3)Ionic Attraction and Acid Strength

25. Chapter 15TMHsiung©2015Slide 25 of 68 Acid Ionization Constant, K a -Acid strength is measured by the size of the equilibrium constant when it reacts with H 2 O. -The equilibrium constant for this reaction is called the acid ionization constant, K a. * larger K a = stronger acid ***

26. Chapter 15TMHsiung©2015Slide 26 of 68 Table 15.5

27. Chapter 15TMHsiung©2015Slide 27 of 68 1)Autoionization of Water 5.Autoionization of Water and pH Water is amphoteric; it can act either as an acid or a base. About 2 out of every 1 billion water molecules form ions through a process called autoionization. Ion-product constant of water, K W : K W = [H 3 O + ][OH – ] At 25 o C, K W = 1x10 –14 *K W applies to all aqueous solutions—acids, bases, salts, and nonelectrolytes—not just to pure water.

28. Chapter 15TMHsiung©2015Slide 28 of 68 2)p function: –log pH = –log [H 3 O + ][H 3 O + ] = 10 –pH pOH = –log [OH – ][OH – ] = 10 –pOH pK W = –log K W pK a = –log K a pK b = –log K b Acidic solution:[H 3 O + ] > [OH – ]pH < pOH Basic solution:[H 3 O + ] pOH Aqueous solution at 25 o C: pK W = pH + pOH = 14.00 Aqueous solution at 25 o C: acidic solution:[H 3 O + ] > 1.0×10 –7 pH < 7.00 basic solution:[H 3 O + ] 7.00 neutral solution:[H 3 O + ] = 1.0×10 –7 pH = 7.00

29. Chapter 15TMHsiung©2015Slide 29 of 68 -When you take the log of a number written in scientific notation, the digits before the decimal point come from the exponent on 10, and the digits after the decimal point come from the decimal part of the number. log(2.0 x 10 6 ) = log(10 6 ) + log(2.0) = 6 + 0.30303… = 6.30303... -Because the part of the scientific notation number that determines the significant figures is the decimal part, the sig. figs. are the digits after the decimal point in the log. log(2.0 × 10 6 ) = 6.30 3)Sig., Figs., and Logs

30. Chapter 15TMHsiung©2015Slide 30 of 68 Solution a. b. Calculate the pH of each solution at 25 °C and indicate whether the solution is acidic or basic. a. [H 3 O + ] = 1.8 × 10 −4 M b. [OH − ] = 1.3 × 10 −2 M Example 15.3Calculating pH from [H 3 O + ] or [OH − ]

31. Chapter 15TMHsiung©2015Slide 31 of 68 Solution Calculate the [H 3 O + ] for a solution with a pH of 4.80. Example 15.4Calculating [H 3 O + ] from pH

32. Chapter 15TMHsiung©2015Slide 32 of 68 There are two potential sources of H 3 O + in an aqueous solution of a acid—the acid and the water. 6.Finding the [H 3 O + ] and pH of Strong and Weak Acid Solutions Except in extremely dilute acid solutions (generally < 1 × 10 −4 M), the autoionization of water contributes a negligibly small amount of H 3 O + compared to the ionization of the acid.

33. Chapter 15TMHsiung©2015Slide 33 of 68 1)Strong Acids Unless the solution is extremely dilute, the strong acids dissociated completely. Therefore, the concentration of H 3 O + in a strong acid solution is equal to the concentration of the strong acid. [H 3 O + ]  C HCl C HCl : initial concentration of HCl

34. Chapter 15TMHsiung©2015Slide 34 of 68

35. Chapter 15TMHsiung©2015Slide 35 of 68 For extremely dilute solution of a strong acid and strong base, 1.0 x 10 –8 M HCl for example:

36. Chapter 15TMHsiung©2015Slide 36 of 68 Equilibrium of monoprotic acid, HA HA + H 2 O  H 3 O + + A – C HA – x x x Therefore Assume C HA – x  C HA *C HA – x  C HA, using: *5% rule: 2)Weak Acids

37. Chapter 15TMHsiung©2015Slide 37 of 68 Solution Find the [H 3 O + ] of a 0.100 M HCN solution. Example 15.5Finding the [H 3 O + ] of a Weak Acid Solution the approximation is valid. [H 3 O + ] = 7.0 × 10 −6 M

38. Chapter 15TMHsiung©2015Slide 38 of 68 Solution Find the pH of a 0.100 M HClO 2 solution. Example 15.7Finding the pH of a Weak Acid Solution in Cases Where the x is small Approximation Does Not Work the approximation is not valid.

39. Chapter 15TMHsiung©2015Slide 39 of 68

40. Chapter 15TMHsiung©2015Slide 40 of 68 Solution Use the given pH to find the equilibrium concentration of [H 3 O + ]. Then write the balanced equation for the ionization of the acid and use it as a guide to prepare an ICE table showing all known concentrations. A 0.100 M weak acid (HA) solution has a pH of 4.25. Find K a for the acid. Example 15.8Finding the Equilibrium Constant from pH *****

41. Chapter 15TMHsiung©2015Slide 41 of 68 3)Percent Ionization Another way to measure the strength of an acid is to determine the percentage of acid molecules that ionize when dissolved in water; this is called the percent ionization.

42. Chapter 15TMHsiung©2015Slide 42 of 68 Solution. Therefore, [H 3 O + ] = 0.034 M. Find the percent ionization of a 2.5 M HNO 2 solution. Example 15.9Finding the Percent Ionization of a Weak Acid *****

43. Chapter 15TMHsiung©2015Slide 43 of 68 *****

44. Chapter 15TMHsiung©2015Slide 44 of 68 4) Finding the pH of Mixtures of Acids Generally, you can ignore the contribution of the weaker acid to the [H 3 O + ] equil. For a mixture of a strong acid with a weak acid, the complete ionization of the strong acid provides more than enough [H 3 O + ] to shift the weak acid equilibrium to the left so far that the weak acid’s added [H 3 O + ] is negligible. For mixtures of weak acids, you generally only need to consider the stronger for the same reasons, as long as one is significantly stronger than the other and their concentrations are similar.

45. Chapter 15TMHsiung©2015Slide 45 of 68 1)Strong Bases Unless the solution is extremely dilute, the strong acids dissociated completely. Therefore, [OH − ] = (number OH − ) × [Base]. 7.Base Solutions

46. Chapter 15TMHsiung©2015Slide 46 of 68 Solution a. b. What is the OH − concentration and pH in each solution? a. 0.225 M KOH b. 0.0015 M Sr(OH) 2 Example 15.11Finding the [OH − ] and pH of a Strong Base Solution *****

47. Chapter 15TMHsiung©2015Slide 47 of 68 Equilibrium of monobasic base, B Assume C B – x  C B *C B – x  C B, using: *5% rule: B + H 2 O  HB + + OH – C B – x x x Therefore 2)Weak Bases

48. Chapter 15TMHsiung©2015Slide 48 of 68 Table 15.8 page 721 Common Weak Bases

49. Chapter 15TMHsiung©2015Slide 49 of 68 Structure of Amines

50. Chapter 15TMHsiung©2015Slide 50 of 68 1)Ions as Acids and Bases (1)NaCl (aq)  Na + (aq) + Cl – (aq) Neutral Na + (aq) + H 2 O  Cl – (aq) + H 2 O  (2)NH 4 Cl (aq)  NH 4 + (aq) + Cl – (aq) Acidic NH 4 + (aq) + H 2 O  NH 3(aq) + H 3 O + (aq) Cl – (aq) + H 2 O  (3)CH 3 COONa (aq)  Na + (aq) + CH 3 COO – (aq) Basic Na + (aq) + H 2 O  CH 3 COO – (aq) + H 2 O  CH 3 COOH (aq) + OH – (aq) (4)CH 3 COONH 4(aq)  NH 4 + (aq) + CH 3 COO – (aq) ????? NH 4 + (aq) + H 2 O  NH 3(aq) + H 3 O + (aq) CH 3 COO – (aq) + H 2 O  CH 3 COOH (aq) + OH – (aq) X X X X 8.Acid–Base Properties of Ions and Salts *****

51. Chapter 15TMHsiung©2015Slide 51 of 68 Type of SaltExample Ion That Undergo Hydrolysis pH of Solution Cation from strong base/ Anion from strong acid NaCl, KI, KNO 3, RbBr, BaCl 2 None 77 Cation from strong base/ Anion from weak acid CH 3 COONa, KNO 2, NaOCl Anion>7 Cation from weak base/ Anion from strong acid NH 4 Cl, NH 4 NO 3 Cation<7 Cation from weak base/ Anion from weak acid NH 4 NO 2, NH 4 CN, CH 3 COONH 4 Anion and Cation <7 if K b <K a  7 if K b  K a >7 if K b >K a Cation is small, highly charged/ Anion from strong acid AlCl 3, Fe(NO 3 ) 3 Hydrated Cation <7 2)The pH of Salt Solutions *****

52. Chapter 15TMHsiung©2015Slide 52 of 68 *Cations of small, highly charged metals are weakly acidic. Al(H 2 O) 6 3+ (aq) + H 2 O (l)  Al(H 2 O) 5 (OH) 2+ (aq) + H 3 O + (aq) * Hydrolysis of hydrated metal ion

53. Chapter 15TMHsiung©2015Slide 53 of 68 2)Relationship between K a of an Acid and K b of Its Conjugate Base Many reference books only give tables of K a values because K b values can be found from them. When you add equations, you multiply the K’s.

54. Chapter 15TMHsiung©2015Slide 54 of 68 Solution Find the pH of a 0.100 M NaCHO 2 solution. The salt completely dissociates into Na + (aq) and CHO 2 − (aq), and the Na + ion has no acid or base properties. K a of HCHO 2 is 1.8x10 -4 Example 15.14Determining the pH of a Solution Containing an Anion Acting as a Base *****

55. Chapter 15TMHsiung©2015Slide 55 of 68 9. Polyprotic Acids

56. Chapter 15TMHsiung©2015Slide 56 of 68  Diprotic acid H 2 A + H 2 O  HA – + H 3 O + HA – + H 2 O  A 2– + H 3 O + For conjugate base: A 2– + H 2 O  HA – + OH – HA – + H 2 O  H 2 A + OH – K a1 x K b2 = K w K a2 x K b1 = K w

57. Chapter 15TMHsiung©2015Slide 57 of 68  Triprotic acid K a1 x K b3 = K w K a2 x K b2 = K w K a3 x K b1 = K w Phosphoric acid for example: H 3 PO 4 + H 2 O  H 2 PO 4 – + H 3 O + H 2 PO 4 – + H 2 O  HPO 4 2– + H 3 O + HPO 4 2– + H 2 O  PO 4 3– + H 3 O + Ionization constants for polyprotic acid progressively decrease: K a1 > K a2 > K a3 > ….. Except in very dilute solutions, essentially all of the H 3 O + ions come from the first ionization step alone.

58. Chapter 15TMHsiung©2015Slide 58 of 68 For a 3.0 M H 3 PO 4, calculate (a) [H 3 O + ]; (b) [H 2 PO 4 – ]; (c) [HPO 4 2– ]; (d) [PO 4 3– ]. Solve (a) assume that all H 3 O + the forms in the first ionization step H 3 PO 4 + H 2 O  H 2 PO 4 – + H 3 O + Initial conc. 3.0 - - Change –x ＋ x ＋ x Equilibrium (3.0 – x) x x x 2 = 0.021 x = [H 3 O + ] = 0.14 M Check: (x/C H 3 PO 4 ) x 100% = 4.7% < 5%, OK!! EXAMPLE *****

59. Chapter 15TMHsiung©2015Slide 59 of 68 (b) H 3 PO 4 + H 2 O  H 2 PO 4 – + H 3 O + Equilibrium (3.0 – x) x x [H 2 PO 4 – ]  [H 3 O + ] = 0.14 M [HPO 4 2– ] = 6.3 x 10 –8 M (c) H 2 PO 4 – + H 2 O  HPO 4 2– + H 3 O + Since [H 2 PO 4 – ]  [H 3 O + ], therefore *****

60. Chapter 15TMHsiung©2015Slide 60 of 68 Since [H 3 O + ] = 0.14 M and [HPO 4 2– ] = 6.3 x 10 –8 M (d) HPO 4 2– + H 2 O  PO 4 3– + H 3 O + [PO 4 3– ] = 1.9x 10 –19 M

61. Chapter 15TMHsiung©2015Slide 61 of 68 (1)H 2 SO 4 + H 2 O → HSO 4 – + H 3 O + K a1 large C H 2 SO 4 C H2SO4 C H2SO4 (2)HSO 4 – + H 2 O  SO 4 2– + H 3 O + C H2SO4 － x x x Concentrated solutions (>0.5 M H 2 SO 4 ): H 3 O + is predominated by first ionization step. e.g., 1.00 M H 2 SO 4, [H 3 O + ]  1.00 M. Very dilute solutions (< 0.001 M H 2 SO 4 ): both ionization steps are nearly completely dissociated, e.g., 0.001 M H 2 SO 4, [H 3 O + ]  0.002 M, [SO 4 2– ]  0.001 M. Intermediate concentrations (0.001 M< C H2SO4 <0.5 M), first ionization step is completely dissociated, the second ionization step is partially dissociated.  A Somewhat Different Case: H 2 SO 4 *****

62. Chapter 15TMHsiung©2015Slide 62 of 68 1)Binary Acids HY 10.Acid Strength and Molecular Structure Comparing binary acids of X in the same row* MoleculeCH 4 NH 3 H 2 OHF ΔEN0.40.91.41.9 Acidity:CH 4 <NH 3 <H 2 O <HF The higher polarity of the bond (the larger ΔEN (electronegativity difference)), the stronger acid. Small ΔEN has more covalent character Large ΔEN has more ionic character *****

63. Chapter 15TMHsiung©2015Slide 63 of 68 Comparing binary acids of X the same group The smaller bond energy (the weaker H—X bond) the stronger acid. MoleculeHFHClHBrHI BE (kJ/mol)565431364297 Anion radius (pm)136181195216 K a 6.6x10 –4 ~10 6 ~10 8 ~10 9 Acidity:HF < HCl <HBr <HI Other example: Acidity:H 2 O <H 2 S <H 2 Se <H 2 Te *** HF (aq) is a weak acid *****

64. Chapter 15TMHsiung©2015Slide 64 of 68 2)Oxoacids More examples: Acid strength:H 2 SeO 3 < H 2 SO 3 HBrO 4 < HClO 4 H–O–Y, Comparing the EN of Y The larger EN (electronegativity) of Y, the weaker H– O bond, the stronger acid. *****

65. Chapter 15TMHsiung©2015Slide 65 of 68 H–O–YO n, Comparing the n of O n The more number (n) of terminal O, the weaker protonated H–O bond, the stronger acid. *O is the element has second higher electronegativity. More examples: K a1 (H 2 SO 3 ) < K a1 (H 2 SO 4 ) *****

66. Chapter 15TMHsiung©2015Slide 66 of 68  Glossary and Definition Lewis acid: The species act as electron-pair acceptor. Lewis base: The species act as electron-pair donor. *In organic chemistry: Lewis acids called electrophiles (electron-loving) Lewis bases called nucleophiles (nucleus-loving) Example: Lewis base Lewis acid Complex (adduct) 11.Lewis Acids and Bases *****

67. Chapter 15TMHsiung©2015Slide 67 of 68 1)Molecules That Act as Lewis Acids 2)Cations That Act as Lewis Acids

68. Chapter 15TMHsiung©2015Slide 68 of 68 End of Chapter 15