1. Chapter 17
Acids and Bases

2. Arrhenius Definition
Acids produce hydrogen ions (H+1) in aqueous solution (HCl → H1+ + Cl1-)
Bases produce hydroxide ions (OH1-) when dissolved in water.
(NaOH → Na1+ + OH1-)
Limited to aqueous solutions.
Only one kind of base (hydroxides)
NH3 (ammonia) could not be an Arrhenius base: no OH1- produced.

3. Brønsted-Lowry A broader definition than Arrhenius
Acid is hydrogen-ion donor (H+ or proton); base is hydrogen-ion acceptor.
Acids and bases always come in pairs.
HCl is an acid.
When it dissolves in water, it gives it’s proton to water.
HCl(g) + H2O(l) ↔ H3O+(aq) + Cl-(aq)
Water is a base; makes hydronium ion.

4. Monoprotic Acids Compounds that donate one hydrogen atom
HNO3 (nitric acid)
HF (hydrofluoric acid)
HCl (hydrochloric acid)
CH3COOH (acetic acid)

5. Polyprotic Acids Compounds have more than one hydrogen to donate
H2SO4 (sulfuric acid) - diprotic - 2 H+
H3PO4 (phosphoric acid) - triprotic - 3 H+
H2CO3 (carbonic acid) - diprotic - 2 H+
Having more than one ionizable hydrogen does not mean stronger!

6. Amphoteric – a substance that can act as both an acid and base- as water shows
HCl + H2O H3O+ + Cl-
acid base
NH3 + H2O NH OH-
base acid

7. Amphoteric – a substance that can act as both an acid and base- as water shows
H2PO H2O HPO H3O+
acid base
H2PO4- + H2O H3PO OH-
base acid

9. Conjugate Acid-Base Pairs
A “conjugate base” is the remainder of the original acid, after it donates it’s hydrogen ion
A “conjugate acid” is the particle formed when the original base gains a hydrogen ion
Thus, a conjugate acid-base pair is related by the loss or gain of a single hydrogen ion.

12. Hydrogen Ions from Water
Water ionizes, or falls apart into ions:
2H2O H3O+ + OH-
Called the “self ionization” of water
Occurs to a very small extent:
[H+1] = [OH-1] = 1 x 10-7 M
Since they are equal, a neutral solution results from water
Kw = [H+1] x [OH-1] = 1 x 10-14
Kw is called the “ionization constant” for water

13. Based on writing equilibrium constants, we do not include water in the expression.
Kw is just an equilibrium constant used for water.
The low number for concentrations
(1 x 10-7) means that very few water molecules ionize.

14. 2H2O ↔ H3O+ + OH-
Kw is constant in every aqueous solution: [H+] x [OH-] = 1 x 10-14
If [H+] > 10-7 then [OH-] < 10-7
If [H+] < 10-7 then [OH-] > 10-7
If we know one, other can be determined
If [H+] > 10-7 , the solution is acidic
If [H+] < 10-7 , the solution is basic

15. pH

16. Calculating pH definition: pH = -log[H+]
in neutral pH = -log(1 x 10-7) = 7
in acidic solution [H+] > 1 x 10-7
pH < 7 (from 0 to 7 is the acid range)
pH > 7 (7 to 14 is base range)

17. Calculating pOH pOH = -log [OH-] [H+] x [OH-] = 1 x 10-14
pH + pOH = 14

18. Calculating [H+] from pH
[H+] = 10-pH
Example
If the pH of a solution is 3.12, what is
the hydronium ion concentration?
[H+] = = 7.6 x 10-4 M

19. Calculating [OH-] from pOH
[OH-] = 10-pOH
Example
If the pH of a solution is 3.12, what is
the hydroxide ion concentration?
pH + pOH = 14, so the pOH is 10.88
[OH-] = = 1.3 x M

20. Problems, page 763 What is the pH of a 0.0012 M NaOH solution.
The pH of diet soda is What is the hydronium ion concentration?
If pH of a solution containing the strong base NaOH is 10.46, what is the concentration of NaOH?

21. Problems, page 763
What is the pH of a M NaOH solution. Answer is 11.08
The pH of diet soda is What is the hydronium ion concentration? 4.8 x 10-5
If pH of a solution containing the strong base NaOH is 10.46, what is the concentration of NaOH? 2.9 x 10-4

23. Ionization of Acids and Bases
Strong acid:
HCl H+ + Cl-
Weak acid:
CH3COOH CH3COO- + H+

24. Ionization of Acids and Bases
Strong acid:
[HA] = [H+]
Weak acid:
[HA] >>> [H+]

25. Ionization of Acids and Bases
Strong base:
NaOH Na+ + OH-
Weak base:
NH NH4+ + OH-
H2O

26. Ionization of Acids and Bases
Strong base:
[B] = [OH-] in H2O solution
Weak base:
[B] >>> [OH-] in H2O solution

28. Increasing Acid Strength
HCO3- HClO HF
Ka = 4.8 x x x 10-4
Increasing acid strength

29. pKa and Ka
pKa = - log Ka
The higher the Ka value, the lower the pKa value.
Lower pKa value – the stronger the acid
Common in Organic and Biochemistry
This is similar to pH = -log[H+]

30. Increasing Acid Strength
CH3CH2COOH CH3COOH HCOOH
Ka = 1.3 x x x 10-4
pKa =
Increasing acid strength

31. Direction of Acid/Base Reactions
All acid/base reactions proceed from the stronger acid and base, to the weaker acid and base

32. Direction of Acid/Base Reactions
All acid/base reactions proceed from the stronger acid and base, to the weaker acid and base

33. Using a table of acid strength (Ka values), you can predict the direction of an acid/base reaction.
Is the reaction product favored or reactant favored.

34. Example 17.3, page 773
Write a balanced, net ionic equation for the reaction that occurs between acetic acid (CH3COOH) and sodium bicarbonate (NaHCO3).
Decide which direction does the equilibrium lie.

35. CH3COOH + NaHCO3 CH3COO-Na+ + H2CO3
CH3COOH + HCO CH3COO- + H2CO3
Ka =
1.8 x x 10-7
All acid/base reactions proceed from the stronger acid and base, to the weaker acid and base

36. Types of Acid-Base Reactions

37. Calculations with Equilibrium Constants
Values of Ka and Kb are determined through experiments.
Ka and Kb values are determined in a similar fashion to other K values.
To determine Ka and Kb values, you use ICE tables.

38. What is the value of Ka for lactic acid?
Example 17.4, page 776
A 0.10 M aqueous solution of lactic acid, CH3CHOHCOOH, has a pH of 2.43.
What is the value of Ka for lactic acid?
CH3CHOHCOOH + H2O CH3CHOHCOO H3O+

39. To determine Ka, you need to know the equilibrium concentrations of each species. The pH allows you to calculate the [H3O+].
pH = 2.43
[H3O+] = 10-pH
[H3O+] =
[H3O+] = 3.7 x 10-3 M

40. CH3CHOHCOOH + H2O CH3CHOHCOO- + H3O+
(I)
(C) x x x
(E) –x x x 10-3
x = 3.7 x 10-3

41. [CH3CHOHCOO-] [H3O+]
Ka =
[CH3CHOHCOOH]
(3.7 x 10-3) (3.7 x 10-3)
Ka =
( x 10-3)
(3.7 x 10-3) (3.7 x 10-3)
Ka =
(0.096)
Ka = 1.4 x 10-4

42. Knowing the values of equilibrium constants for weak acids and bases enables you to calculate the pH of a solution.

43. Example 17.5, page 778
Calculate the pH of a M aqueous solution of benzoic acid (C6H5CO2H) given that the value of Ka = 6.3 x 10-5.
C6H5COOH + H2O C6H5COO H3O+

44. C6H5COOH + H2O C6H5COO H3O+
(I)
(C) x x x
(E) –x x x

45. [C6H5COO-] [H3O+]
Ka =
[C6H5COOH]
(x) (x)
6.3 x =
( x)
(x)2
6.3 x 10-5 =
(0.02)
x =

46. Using the value of x calculated, return to the ICE table and substitute x into the equilibrium line.
X = [H3O+]
[H3O+] = 0.001M
pH = -log(0.001)
pH = 2.96

47. Example 17.7, page 781
Calculate the pH of a M aqueous solution of sodium acetate (Na+CH3CO2-) given that the value of Kb = 5.6 x
CH3COO- Na+ + H2O CH3COOH + OH-

48. CHCOO- + H2O CH3COOH OH-
(I)
(C) x x x
(E) –x x x

49. [CH3COOH] [OH-]
Kb =
[CH3COO-]
(x) (x)
5.6 x =
( x)
(x)2
5.6 x =
(0.015)
x = 2.9 x 10-6

50. Using the value of x calculated, return to the ICE table and substitute x into the equilibrium line to solve for [OH].
[OH-] = 2.9 x 10-6
pOH = -log(2.9 x 10-6)
pOH = 5.54
pH + pOH = 14
pH = 14 – 5.54 = 8.46

51. Example 17.8, page 782 Skip
What is the pH of a solution that results from mixing 25 mL of M aqueous NH3 with 25 mL of M aqueous HCl?
NH3 + HCl + H2O NH H2O + Cl-
Since you are solving for pH, you need the [H3O+].
Rewrite the equation ignoring Cl:
NH3 + H3O NH H2O

52. +

53. To solve this type of problem between a strong acid and weak base, reverse the reaction.
NH H2O NH H3O+
(I)
(C) x x x
(E) –x x x

54. [NH3] [H3O+]
Ka =
[NH4+]
(x) (x)
5.6 x =
( x)
(x)2
5.6 x =
(0.008)
x = 2.1 x 10-6

55. [H3O+] = 2.1 x 10-6 M
pH = -log(2.1 x 10-6 )
pH = 5.67
The final solution is acidic, which is predicted when mixing a strong acid with a weak base.

56. Ionization of Strong Acids and Bases

57. Strong acids and bases are completely ionized in aqueous solutions:
HCl H Cl-
NaOH Na OH-

58. Therefore:
NaOH Na OH-
[1 x 10-2 M] [1 x 10-2 M]
of Na+ and OH-

59. Problem 17.1, page 762
What are the hydroxide and hydronium ion concentrations in a M aqueous solution of NaOH?
NaOH is a strong base and is completely ionized:
NaOH Na+ + OH-

60. NaOH Na+ + OH-
[OH-] = M
Kw = [H+1] x [OH-1] = 1 x 10-14
[H+1] x [0.0012] = 1 x 10-14
[H+1] = 1 x M2 /
= 8.3 x M

61. Calculate the hydroxide ion concentration for a 2.5 M Ca(OH)2 solution.
Ca(OH)2 is a strong base and is completely ionized:
Ca(OH) Ca OH-
[Ca(OH)2] = [Ca+2] [OH-]
2.5 M M M

62. Oxyacids Oxyacids are acids containing an oxygen
atom, hydrogen atom and another non-metal atom (N, S, P).
Examples
HNO2, HNO3, H2SO4, H3PO4, HOCl

63. HClO4 > HClO3 > HClO2 > HClO
In a series of related compounds, acid strength increases as the number of oxygen atoms bonded to the central atom increases.
HNO3 > HNO2
Stronger weaker
HClO4 > HClO3 > HClO2 > HClO
Stronger weaker

65. After the H+ is removed from the molecule, you
are left with a negative charge. The more electronegative atoms in the molecule (O atoms), the better able to accommodate the negative charge.

66. Carboxylic Acids Organic acids represented by the structure
Acetic Acid

67. Acetic Acid

68. Resonance Structures

69. Electronegative Groups in the Molecule

70. Hydrated Metal Cations
Hydrated metal cations can act as acids.
This occurs with transition elements with a +2 or +3 charge.
[Fe(H2O)6]+3
[Ni(H2O)6]+2
[Fe(H2O)6]+2
[Cu(H2O)6]+3

71. Examples [Fe(H2O)6]+3 + H2O [Fe(H2O)5OH]+2 + H3O+
[Cu(H2O)6] H2O [Cu(H2O)5OH] H3O+

72. Lewis Acids and Bases Lewis Acid – electron pair acceptor
Lewis Base – electron pair donor
A B: B-A
acid base

73. Typical Lewis Acids
1) Molecules with a + charge
2) Molecules that do not meet the octet rule
(BH3)
Typical Lewis Bases
Molecules with a – charge
Molecules with lone pairs of electrons
(H2O, NH3)

74. Fe+2 + 6 H2O [Fe(H2O)6]+2 Cu+2 + 4 NH3 [Cu(NH3)4]+2 acid base

75. Lewis Acid – Base Examples

76. Lewis Acid – Base Examples
BH3
AlCl3