1. Acids and Bases Chapter 15

2. Properties of Acids Sour to the taste. React with a variety of metals. Turn blue litmus pink. Denature proteins. Cancel the effect of a base (neutralization). Properties of Bases Bitter to the taste. Slippery to the touch. Turn pink litmus blue. Denature proteins. Cancel the effect of an acid (neutralization). Acid-Base Properties

3. 3 Binary acids have acid hydrogens attached to a nonmetal atom: HCl, HF Carboxylic acids have COOH group Structure of Acids Oxy acids have acid hydrogens attached to an oxygen atom: H 2 SO 4, HNO 3

4. 4 Common Acids

5. 5 Structure of Bases Most ionic bases contain OH ions NaOH, Ca(OH) 2 Some contain CO 3 2- ions CaCO 3, NaHCO 3 Molecular bases contain structures that react with H + mostly amine groups

6. 6 Common Bases

7. Acid-Base Concepts Concepts of acid-base theory including:  The Arrhenius concept  The Bronsted Lowry concept  The Lewis concept

8. 8 Arrhenius Theory HCl ionizes in water, producing H + and Cl – ions NaOH dissociates in water, producing Na + and OH – ions

9. Strong Acids and Bases In the Arrhenius concept, a strong acid is a substance that ionizes completely in aqueous solution to give H 3 O + (aq) and an anion. In the Arrhenius concept, a strong base is a substance that ionizes completely in aqueous solution to give OH - (aq) and a cation.

10. The STRONG ACIDS (all dissociate completely in water) Hydrochloric acid: HCl Hydrobromic acid: HBr Hydroiodic acid: HI Sulfuric acid: H 2 SO 4 Nitric acid: HNO 3 Perchloric acid: HClO 4 Some other acids that are sometimes considered strong are: chloric acid (HClO 3 ), bromic acid (HBrO 3 ), perbromic acid (HBrO 4 ), iodic acid (HIO 3 ), and per-iodic acid (HIO 4 ).

11. The STRONG BASES (all dissociate completely in water) Lithium hydroxide: LiOH Sodium hydroxide: NaOH Potassium hydroxide: KOH Rubidium hydroxide: RbOH Cesium hydroxide: CsOH Magnesium hydroxide: Mg(OH) 2 Calcium hydroxide: Ca(OH) 2 Strontium hydroxide: Sr(OH) 2 Barium hydroxide: Ba(OH) 2 Note: Alkali and some alkaline earth metals

12. Weak Acids and Bases Most other acids and bases that you encounter are weak. They are not completely ionized and exist in reversible reaction with the corresponding ions.  Ammonia, NH 3, is a weak base.  An example is acetic acid, HC 2 H 3 O 2.

13. 13 Arrhenius Acid-Base Reactions The H + from the acid combines with the OH - from the base to make a molecule of H 2 O it is often helpful to think of H 2 O as H-OH The cation from the base combines with the anion from the acid to make a salt acid + base → salt + water HCl(aq) + NaOH(aq) → NaCl(aq) + H 2 O(l)

14. 14 Brønsted-Lowry Concept of Acids and Bases A base is the species accepting the proton in a proton-transfer reaction. oIn any reversible acid-base reaction, both forward and reverse reactions involve proton transfer. According to the Brønsted-Lowry concept, an acid is the species donating the proton in a proton-transfer reaction.

15. 15 Brønsted-Lowry Concept of Acids and Bases Consider the reaction of NH 3 and H 2 O. oIn the forward reaction, NH 3 accepts a proton from H 2 O. Thus, NH 3 is a base and H 2 O is an acid. H+H+ baseacid

16. 16 Amphoteric Substances Amphoteric substances can act as either an acid or a base have both transferable H and atom with lone pair Water acts as base, accepting H + from HCl HCl(aq) + H 2 O(l) → Cl – (aq) + H 3 O + (aq) Water acts as acid, donating H + to NH 3 NH 3 (aq) + H 2 O(l)  NH 4 + (aq) + OH – (aq)

17. 17 Brønsted-Lowry Concept of Acids and Bases  In the Brønsted-Lowry concept: 2.Acids and bases can be ions as well as molecular substances. 3.Acid-base reactions are not restricted to aqueous solution. 4.Some species can act as either acids or bases depending on what the other reactant is. 1.A base is a species that accepts protons; OH - is only one example of a base.

18. 18 Conjugated Acid-Base Pairs The Brønsted-Lowry concept introduced the idea of conjugate acid-base pairs and proton- transfer reactions. oIf an acid loses its H +, the resulting anion is now in a position to reaccept a proton, making it a Brønsted- Lowry base. oIt is logical to assume that if an acid is considered strong, its conjugate base (that is, its anion) would be weak, since it is unlikely to accept a hydrogen ion. conjugate acid-base pairs acid base

19. Copyright © Houghton Mifflin Company. All rights reserved. Label each species as an acid or base. Identify the conjugate acid-base pairs. a.HCO 3 - (aq) + HF(aq) H 2 CO 3 (aq) + F - (aq) b.HCO 3 - (aq) + OH - (aq) CO 3 2- (aq) + H 2 O(l) BaseAcid Conjugate base Conjugate acid AcidBase Conjugate acid Conjugate base

20. The conjugate acid of HSO 4 - is A. H 2 SO 4 B. HSO 3 + C. HSO 4 + D. H + E. SO 4 2-

21. The conjugate acid of HSO 4 - is A. H 2 SO 4 B. HSO 3 + C. HSO 4 + D. H + E. SO 4 2-

22. The conjugate base of HSO 4 - is A. H 2 SO 4 B. HSO 3 + C. HSO 4 + D. H + E. SO 4 2-

23. The conjugate base of HSO 4 - is A. H 2 SO 4 B. HSO 3 + C. HSO 4 + D. H + E. SO 4 2-

24. 24 Practice – Write the formula for the conjugate acid of the following H 2 O NH 3 CO 3 2− H 2 PO 4 1−

25. 25 Practice – Write the formula for the conjugate acid of the following H 2 OH 3 O + NH 3 NH 4 + CO 3 2− HCO 3 − H 2 PO 4 1− H 3 PO 4

26. 26 Practice – Write the formula for the conjugate base of the following H 2 O NH 3 CO 3 2− H 2 PO 4 1−

27. 27 Practice – Write the formula for the conjugate base of the following H 2 OHO − NH 3 NH 2 − CO 3 2− since CO 3 2− does not have an H, it cannot be an acid H 2 PO 4 1− HPO 4 2−

28. 28 Polyprotic Acids Acid molecules have more than one ionizable H 1 H = monoprotic, 2 H = diprotic, 3 H = triprotic  HCl,H 2 SO 4, H 3 PO 4 Polyprotic acids ionize in steps each ionizable H removed sequentially Removing of the first H automatically makes removal of the second H harder H 2 SO 4 is a stronger acid than HSO 4 

29. 29 Strengths of Acids & Bases Commonly, acid or base strength is measured by determining the equilibrium constant of a substance’s reaction with water HAcid + H 2 O  Acid -1 + H 3 O +1 Base: + H 2 O  HBase +1 + OH -1 The farther the equilibrium position lies to the products, the stronger the acid or base The position of equilibrium depends on the strength of attraction between the base form and the H + stronger attraction means stronger base or weaker acid

30. 30 Acid Ionization Constant, K a Acid strength measured by the size of the equilibrium constant when react with H 2 O HAcid + H 2 O  Acid -1 + H 3 O +1 The equilibrium constant is called the acid ionization constant, K a larger K a = stronger acid

31. 31

32. 32 Autoionization of Water H 2 O  H + + OH – H 2 O(l) + H 2 O(l)  H 3 O + (aq) + OH – (aq) All aqueous solutions contain both H 3 O + and OH – the concentration of H 3 O + and OH – are equal in water [H 3 O + ] = [OH – ] = 10 -7 M @ 25°C The ion product of water and has the symbol K w [H 3 O + ] x [OH – ] = K w = 1 x 10 -14 @ 25°C Self-ionization of Water

33. Example– Calculate the [OH  ] at 25°C when the [H 3 O + ] = 1.5 x 10 -9 M, and determine if the solution is acidic, basic, or neutral The units are correct. The fact that the [H 3 O + ] < [OH  ] means the solution is basic [H 3 O + ] = 1.5 x 10 -9 M [OH  ] Check: Solution: Concept Plan: Relationships: Given: Find: [H 3 O + ][OH  ]

34. 34 pH pH = -log[H 3 O + ], [H 3 O + ] = 10 -pH exponent on 10 with a positive sign pH water = -log[10 -7 ] = 7 pH 7 is basic, pH = 7 is neutral Another way of expressing the acidity/basicity of a solution is pOH pOH = -log[OH  ], [OH  ] = 10 -pOH pH + pOH = 14.00

35. Example– Calculate the pH at 25°C when the [OH  ] = 1.3 x 10 -2 M, and determine if the solution is acidic, basic, or neutral pH is unitless. The fact that the pH > 7 means the solution is basic [OH  ] = 1.3 x 10 -2 M pH Check: Solution: Concept Plan: Relationships: Given: Find: [H 3 O + ][OH  ]pH

36. 36 pKpK A way of expressing the strength of an acid or base is pK pK a = -log(K a ), K a = 10 -pKa pK b = -log(K b ), K b = 10 -pKb The stronger the acid, the smaller the pK a larger K a = smaller pK a

37. 37 The pH of a Strong Acid There are two sources of H 3 O + in an aqueous solution of a strong acid – the acid and the water The contribution of the water to the total [H 3 O + ] is negligible shifts the K w equilibrium to the left so far that [H 3 O + ] water is too small to be significant For a monoprotic strong acid [H 3 O + ] = [HAcid] for polyprotic acids, the other ionizations can generally be ignored 0.10 M HCl has [H 3 O + ] = 0.10 M and pH = 1.00

38. 38 Calculate the concentration of H + ion in 0.010 M NaOH. Because you started with 0.010 M NaOH (a strong base) the reaction will produce 0.010 M OH - (aq). – Substituting [OH - ]=0.010 into the ion-product expression, we get: The pH of a Strong Base

39. 39 Acid-Ionization Equilibria For a weak acid, the equilibrium concentrations of ions in solution are determined by the acid- ionization constant (also called the acid- dissociation constant).

40. 40 Acid-Ionization Equilibria Since the concentration of water remains relatively constant, we rearrange the equation to get: Thus, K a, the acid-ionization constant, equals the constant [H 2 O]K c.

41. 41 [HNO 2 ][NO 2 - ][H 3 O + ] initial change equilibrium Example- Find the pH of 0.200 M HNO 2 (aq) solution @ 25°C Write the reaction for the acid with water Construct an ICE table for the reaction Enter the initial concentrations – assuming the [H 3 O + ] from water is ≈ 0 HNO 2 + H 2 O  NO 2  + H 3 O + [HNO 2 ][NO 2 - ][H 3 O + ] initial 0.200 0 ≈ 0 change equilibrium K a for HNO 2 = 4.6 x 10 -4

42. 42 [HNO 2 ][NO 2 - ][H 3 O + ] initial 0.200 00 change equilibrium represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression +x+x +x+x xx 0.200  x xx Example- Find the pH of 0.200 M HNO 2 (aq) solution @ 25°C K a for HNO 2 = 4.6 x 10 -4

43. 43 determine the value of K a from Table 15.5 since K a is very small, approximate the [HNO 2 ] eq = [HNO 2 ] init and solve for x K a for HNO 2 = 4.6 x 10 -4 [HNO 2 ][NO 2 - ][H 3 O + ] initial 0.200 0 ≈ 0 change -x-x+x+x+x+x equilibrium 0.200xx 0.200  x Example- Find the pH of 0.200 M HNO 2 (aq) solution @ 25°C

44. 44 K a for HNO 2 = 4.6 x 10 -4 [HNO 2 ][NO 2 - ][H 3 O + ] initial 0.200 0 ≈ 0 change -x-x+x+x+x+x equilibrium 0.200 xx check if the approximation is valid by seeing if x < 5% of [HNO 2 ] init the approximation is valid x = 9.6 x 10 -3 Example- Find the pH of 0.200 M HNO 2 (aq) solution @ 25°C

45. 45 K a for HNO 2 = 4.6 x 10 -4 [HNO 2 ][NO 2 - ][H 3 O + ] initial 0.200 0 ≈ 0 change -x-x+x+x+x+x equilibrium 0.200-x xx x = 9.6 x 10 -3 substitute x into the equilibrium concentration definitions and solve [HNO 2 ][NO 2 - ][H 3 O + ] initial 0.200 0 ≈ 0 change -x-x+x+x+x+x equilibrium 0.1900.0096 Example- Find the pH of 0.200 M HNO 2 (aq) solution @ 25°C

46. 46 K a for HNO 2 = 4.6 x 10 -4 substitute [H 3 O + ] into the formula for pH and solve [HNO 2 ][NO 2 - ][H 3 O + ] initial 0.200 0 ≈ 0 change -x-x+x+x+x+x equilibrium 0.1900.0096 Example- Find the pH of 0.200 M HNO 2 (aq) solution @ 25°C

47. A solution has a hydroxide ion concentration of 0.001 M. What is the pOH of the solution? A. 3 B. 7 C. 11 D. 13 47

48. The pH of a solution is 5.50. What is its hydronium concentration? A. 5.50 M B. 3.2 x 10 -5 M C. 3.2 x 10 -6 M D. 3.2 x 10 -7 M 48

49. 49 Base-Ionization Equilibria Equilibria involving weak bases are treated similarly to those for weak acids. –In general, a weak base B with the base ionization has a base ionization constant equal to

50. 50 [NH 3 ][NH 4 + ][OH  ] initial change equilibrium Example - Find the pH of 0.100 M NH 3 (aq) solution Write the reaction for the base with water Construct an ICE table for the reaction Enter the initial concentrations – assuming the [OH  ] from water is ≈ 0 NH 3 + H 2 O  NH 4 + + OH  [NH 3 ][NH 4 + ][OH  ] initial 0.100 0 ≈ 0 change equilibrium K b for NH 3 = 1.76 x 10 -5

51. 51 [NH 3 ][NH 4 + ][OH  ] initial 0.100 00 change equilibrium Example - Find the pH of 0.100 M NH 3 (aq) solution represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression +x+x +x+x xx 0.100  x xx K b for NH 3 = 1.76 x 10 -5

52. 52 determine the value of K b from Table 15.8 since K b is very small, approximate the [NH 3 ] eq = [NH 3 ] init and solve for x K b for NH 3 = 1.76 x 10 -5 [NH 3 ][NH 4 + ][OH  ] initial 0.100 0 ≈ 0 change -x-x+x+x+x+x equilibrium 0.100xx 0.100  x Example - Find the pH of 0.100 M NH 3 (aq) solution

53. 53 K b for NH 3 = 1.76 x 10 -5 [NH 3 ][NH 4 + ][OH  ] initial 0.100 0 ≈ 0 change -x-x+x+x+x+x equilibrium 0.100 xx check if the approximation is valid by seeing if x < 5% of [NH 3 ] init the approximation is valid x = 1.33 x 10 -3 Example - Find the pH of 0.100 M NH 3 (aq) solution

54. 54 substitute x into the equilibrium concentration definitions and solve K b for NH 3 = 1.76 x 10 -5 [NH 3 ][NH 4 + ][OH  ] initial 0.100 0 ≈ 0 change -x-x+x+x+x+x equilibrium 0.100  x xx x = 1.33 x 10 -3 [NH 3 ][NH 4 + ][OH  ] initial 0.100 0 ≈ 0 change -x-x+x+x+x+x equilibrium 0.0991.33E-3 Example - Find the pH of 0.100 M NH 3 (aq) solution

55. 55 use the [OH - ] to find the [H 3 O + ] using K w substitute [H 3 O + ] into the formula for pH and solve K b for NH 3 = 1.76 x 10 -5 [NH 3 ][NH 4 + ][OH  ] initial 0.100 0 ≈ 0 change -x-x+x+x+x+x equilibrium 0.0991.33E-3 Example - Find the pH of 0.100 M NH 3 (aq) solution

56. 56 use the [OH - ] to find the pOH use pOH to find pH K b for NH 3 = 1.76 x 10 -5 [NH 3 ][NH 4 + ][OH  ] initial 0.100 0 ≈ 0 change -x-x+x+x+x+x equilibrium 0.0991.33E-3 Example - Find the pH of 0.100 M NH 3 (aq) solution

57. 57 To predict the acidity or basicity of a salt, you must examine the acidity or basicity of the ions composing the salt. 1.A salt of a strong base and a strong acid. The salt gives a neutral aqueous solution. NaCl. 2.A salt of a strong base and a weak acid. The salt gives a basic solution. NaCN 3.A salt of a weak base and a strong acid. The salt gives an acidic solution. NH 4 Cl 4.A salt of a weak base and a weak acid. K a > K b  the solution is acidic. K b > K a  the solution is basic. Classifying Salt Solutions as Acidic, Basic, or Neutral

58. 58 Example- Determine whether a solution of the following salts is acidic, basic, or neutral a) SrCl 2 Sr 2+ is the counterion of a strong base, pH neutral Cl − is the conjugate base of a strong acid, pH neutral solution will be pH neutral b) CH 3 NH 3 NO 3 CH 3 NH 3 + is the conjugate acid of a weak base, acidic NO 3 − is the conjugate base of a strong acid, pH neutral solution will be acidic

59. 59 Example- Determine whether a solution of the following salts is acidic, basic, or neutral c) NaCHO 2 Na + is the counterion of a strong base, pH neutral CHO 2 − is the conjugate base of a weak acid, basic solution will be basic d) NH 4 F NH 4 + is the conjugate acid of a weak base, acidic F − is the conjugate base of a weak acid, basic K a (NH 4 + ) > K b (F − ); solution will be acidic

60. 60 If K a > K b, the solution is acidic. If K a < K b, the solution is basic. If K a = K b, the solution is neutral.

61. An aqueous solution of __________ will produce a basic solution. A. LiCl B. NH 4 ClO 4 C. CaSO 4 D. KBr E. Na 2 CO 3 61

62. 62 K a and K b Relationship Reference tables only provide K a values because K b values can be found from them when you add equations, you multiply the K’s

63. 63 Example- Find the pH of 0.100 M NaCHO 2 (aq) solution Na + is the cation of a strong base – pH neutral. The CHO 2 − is the anion of a weak acid – pH basic Write the reaction for the anion with water Construct an ICE table for the reaction Enter the initial concentrations – assuming the [OH  ] from water is ≈ 0 CHO 2 − + H 2 O  HCHO 2 + OH  [CHO 2 − ][HCHO 2 ][OH  ] initial 0.1000≈ 0 change equilibrium

64. 64 0.100  x Example- Find the pH of 0.100 M NaCHO 2 (aq) solution represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x Calculate the value of K b from the value of K a of the weak acid from Table 15.5 substitute into the equilibrium constant expression +x+x+x+x xx xx [CHO 2 − ][HCHO 2 ][OH  ] initial 0.1000≈ 0 change equilibrium

65. 65 Example- Find the pH of 0.100 M NaCHO 2 (aq) solution since K b is very small, approximate the [CHO 2 − ] eq = [CHO 2 − ] init and solve for x K b for CHO 2 − = 5.6 x 10 -11 [CHO 2 − ][HCHO 2 ][OH  ] initial 0.1000≈ 0 change -x-x+x+x+x+x equilibrium 0.100xx 0.100  x

66. 66 substitute x into the equilibrium concentration definitions and solve x = 2.4 x 10 -6 K b for CHO 2 − = 5.6 x 10 -11 [CHO 2 − ][HCHO 2 ][OH  ] initial 0.1000≈ 0 change -x-x+x+x+x+x equilibrium 0.100 −x xx [CHO 2 − ][HCHO 2 ][OH  ] initial 0.1000≈ 0 change -x-x+x+x+x+x equilibrium 0.1002.4E-6 Example- Find the pH of 0.100 M NaCHO 2 (aq) solution

67. 67 use the [OH - ] to find the [H 3 O + ] using K w substitute [H 3 O + ] into the formula for pH and solve K b for CHO 2 − = 5.6 x 10 -11 [CHO 2 − ][HCHO 2 ][OH  ] initial 0.1000≈ 0 change -x-x+x+x+x+x equilibrium 0.1002.4E-6 Example- Find the pH of 0.100 M NaCHO 2 (aq) solution

68. 68 use the [OH - ] to find the pOH use pOH to find pH K b for CHO 2 − = 5.6 x 10 -11 [CHO 2 − ][HCHO 2 ][OH  ] initial 0.1000≈ 0 change -x-x+x+x+x+x equilibrium 0.1002.4E-6 Example- Find the pH of 0.100 M NaCHO 2 (aq) solution

69. 69 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K b to the given K b though not exact, the answer is reasonably close K b for CHO 2 − = 5.6 x 10 -11 [CHO 2 − ][HCHO 2 ][OH  ] initial 0.1000≈ 0 change -x-x+x+x+x+x equilibrium 0.1002.4E-6 Example- Find the pH of 0.100 M NaCHO 2 (aq) solution

70. 70 Polyprotic Acids Some acids have two or more protons (hydrogen ions) to donate in aqueous solution. These are referred to as polyprotic acids. –The first proton is lost completely followed by a weak ionization of the hydrogen sulfate ion, HSO 4 -. –Sulfuric acid, for example, can lose two protons in aqueous solution.

71. 71 Polyprotic Acids For a weak diprotic acid like carbonic acid, H 2 CO 3, two simultaneous equilibria must be considered.

72. 72

73. 73 Strengths of Binary Acids The more  + H-X  - polarized the bond, the more acidic the bond The stronger the H-X bond, the weaker the acid Binary acid strength increases to the right across a period H-C < H-N < H-O < H-F Binary acid strength increases down the column H-F < H-Cl < H-Br < H-I ++ --

74. 74 Strengths of Oxyacids, H-O-Y The more electronegative the Y atom, the stronger the acid helps weakens the H-O bond The more oxygens attached to Y, the stronger the acid further weakens and polarizes the H-O bond

75. 75 Acid Strengths

76. 76 Acid Strengths

77. 77 Lewis Acid - Base Theory Electron donor = Lewis Base = nucleophile must have a lone pair of electrons Electron acceptor = Lewis Acid = electrophile electron deficient Covalent bond forms between the molecules Nucleophile : + Electrophile  Nucleophile : Electrophile Product called an adduct Other acid-base reactions also Lewis

78. 78 Example - Complete the Following Lewis Acid-Base Reactions Label the Nucleophile and Electrophile OH HCH  + OH -1  OH HCH OH HCH  + OH -1  Electrophile Nucleophile

79. Which of the following is a Lewis base? A. SiF 4 B. AlF 3 C. H 2 O D. C 5 H 12 79

80. 80 Practice - Complete the Following Lewis Acid-Base Reactions Label the Nucleophile and Electrophile CaO + SO 3  KI + I 2 

81. 81 Practice - Complete the Following Lewis Acid-Base Reactions Label the Nucleophile and Electrophile CaO + SO 3  Ca +2 SO 4 -2 KI + I 2  KI 3 O S O O Ca +2 O -2 + O S O O O -2 Ca +2 ElecNuc I K +1 I -1 + ElecNuc K +1 I I I -1