1. Acids and Bases
Chapter 14

2. Some Properties of Acids
Produce H+ (as H3O+) ions in water (the hydronium ion is a hydrogen ion attached to a water molecule)
Taste sour
Corrode metals
Electrolytes
React with bases to form a salt and water
pH is less than 7
Turns blue litmus paper to red “Blue to Red A-CID”

3. Some Properties of Bases
Produce OH- ions in water
Taste bitter, chalky
Are electrolytes
Feel soapy, slippery
React with acids to form salts and water
pH greater than 7
Turns red litmus paper to blue “Basic Blue”

4. Acid Nomenclature Review
No Oxygen
w/Oxygen
An easy way to remember which goes with which…
“In the cafeteria, you ATE something ICky”

5. Acid/Base definitions Definition 1: Arrhenius
Arrhenius acid is a substance that donates H+ (H3O+) in water
Arrhenius base is a substance that donates OH- in water
NaOH  Na+ + OH-
4.3

6. base is a proton acceptor
Definition #2: Brønsted – Lowry
acid is a proton donor
base is a proton acceptor
conjugate acid
conjugate base
base
acid

7. Conjugate Pairs

8. Learning Check!
Label the acid, base, conjugate acid, and conjugate base in each reaction:
HCl + OH-    Cl- + H2O
Acid
Base
Conj.Base
Conj.Acid
H2O + H2SO4    HSO4- + H3O+
Conj.Base
Conj.Acid
Base
Acid

9. Definition #3 – Lewis
Lewis acid - a substance that accepts an electron pair
Lewis base - a substance that donates an electron pair

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

11. Pairs HA(aq) + H2O(l) H3O+(aq) + A-(aq)
Acid Base Conjugate + Conjugate
acid base
This is an equilibrium.
Competition for H+ between the bases, H2O and A-
If H2O is a stronger base: Equilibrium moves right.
If A- is a stronger base: Equilibrium moves to left.

12. Conjugate acid-base pairs:
Strong acid : weak conjugate base. (weaker than H2O)
HCl + H2O H3O+ + Cl-
Weak acid : strong conjugate base.
HF + H2O H3O+ + F-
H3O+ is the strongest acid that can exist in aqueous solution.
The OH- ion is the strongest base that can exist in aqueous solution.
Equilibrium?
Equilibrium?
15.4

13. Acid dissociation constant Ka
HA(aq) + H2O(l) H3O+(aq) + A-(aq)
Ka = [H3O+][A-] [H+][A-] [HA] [HA]
H3O+ is often written H+ ignoring the water in equation (it is implied). =

14. Acid Dissociation (Ionization) Reactions
Write the dissociation reaction for the following:
Hydrochloric acid
Acetic acid
Anilinium ion C6H5NH3+
Hydrated aluminum (III) ion [Al(H2O)6]3+
HCl(aq) H+(aq) + Cl-(aq)
HC2H3O2 ↔ H+(aq) + C2H3O2-(aq)
↔ H+(aq) + C6H5NH2(aq)
[Al(H2O)6]3+(aq) H+(aq) + [Al(H2O)5OH]2+(aq)

15. Back to Pairs Strong acids Weak acids Ka is large Ka is small
[H+] is equal to [HA]
Weak conjugate base: A- is a weaker base than water
Weak acids
Ka is small
[H+] <<< [HA]
Strong conjugate base: A- is a stronger base than water

16. The use of conjugate acid-base pairs allows us to make a very simple statement about relative strengths of acids and bases. The stronger an acid, the weaker its conjugate base, and, conversely, the stronger a base, the weaker its conjugate acid.
Table 1 gives a list of some of the more important conjugate acid-base pairs in order of increasing strength of the base. This table enables us to see how readily a given acid will react with a given base. The reactions with most tendency to occur are between the strong acids in the top left-hand comer of the table and the strong bases in the bottom right-hand comer. If a line is drawn from acid to base for such a reaction, it will have adownhill slope. By contrast, reactions with little or no tendency to occur (between the weak acids at the bottom left and the weak bases at the top right) correspond to a line from acid to base with an uphill slope. When the slope of the line is not far from horizontal, the conjugate pairs are not very different in strength, and the reaction goes only part way to completion. Thus, for example, if the acid HF is compared with the base CH3COO–, we expect the reaction to go part way to completion since the line is barely downhill.

17. Strong and Weak Acids/Bases
The strength of an acid (or base) is determined by the amount of IONIZATION.
HNO3, HCl, HBr, HI, H2SO4 and HClO4 HClO3 are the strong acids.

18. Strong and Weak Acids/Bases
Weak acids are much less than 100% ionized in water.
*One of the best known is acetic acid = CH3CO2H

19. Types of Acids
Polyprotic Acids- more than 1 acidic hydrogen (diprotic, triprotic).
Oxyacids - Proton is attached to the oxygen of an ion.
Organic acids contain the Carboxyl group -COOH with the H attached to O
Generally very weak.

21. Strong and Weak Acids/Bases
Strong Base: 100% dissociated in water.
NaOH (aq)  Na+ (aq) + OH- (aq)
Other common strong bases include KOH and Ca(OH)2.
CaO (lime) + H2O -->
Ca(OH)2 (slaked lime)
CaO
Strong bases are the group I hydroxides
Calcium, strontium, and barium hydroxides are strong, but only soluble in water to 0.01 M

22. Lime-soda process water softener
Add CaO (lime) and Na2CO3 (soda ash) CaO (lime) + H2O  Ca(OH)2 (slaked lime) Ca(OH)2 + Ca2+ + 2HCO3- 2CaCO3 + 2H2O

23. Strong and Weak Acids/Bases
Weak base: less than 100% ionized in water
One of the best known weak bases is ammonia
NH3 (aq) + H2O (l) ↔ NH4+ (aq) + OH- (aq)

24. Relative Base Strength
Order of increasing base strength: H2O, F-, Cl-, NO2-, CN-
Cl-, H2O, F-, NO2-, CN-
(H2O is a stronger base than
the conjugate base of a strong
acid, but weaker than the
conjugate base of a weak acid.)
*Common Exam Question!*

25. H2O(l) + H2O(l)   H3O+(aq)+ OH (aq)
Amphoteric : H2O can function as both an ACID and a BASE.
In pure water there can be AUTOIONIZATION
H2O(l) + H2O(l)   H3O+(aq)+ OH (aq)
acid base conj conj
acid base 1
Equilibrium (ion product) constant for water = Kw
Kw = [H3O+] [OH-] = x at 25 oC

26. In any aqueous solution, the product of [H+] and [OH-] must always equal 1.0 x 10-14
neutral [H+] = [OH-] = 1.00 x 10-7 M
Acidic [H+] > [OH-]
Basic [OH-] > [H+]

27. At 60°C, the value of Kw is 1 x 10-13
At 60°C, the value of Kw is 1 x Using Le Chatelier’s principle, predict whether the reaction 2H2O(l) ↔ H3O+(aq) + OH-(aq) is exothermic or endothermic. b. calculate [H+] and [OH-] in a neutral solution at 60°C. √
[H+][OH-] = 1x10-13
= 3x10-7M

28. Subset of Amphoteric:
Amphiprotic substance: can donate or accept a proton, H+
For a substance to be amphiprotic it must
contain a hydrogen atom which is able to be donated in the presence of a base.
be able to accept a hydrogen ion in the presence of an acid.
 H2O, HCO3- , HSO4- , HPO42- and H2PO4-
HSO4- + OH-   H2O + SO4-
HSO H+   H2SO4

29. The pH scale is a way of expressing the strength of acids and bases
The pH scale is a way of expressing the strength of acids and bases. Instead of using very small numbers, we just use the NEGATIVE power of 10 on the Molarity of the H+ (or OH-) ion. Under 7 = acid 7 = neutral Over 7 = base

30. pH = - log [H+] Calculating the pH
(Remember that the [ ] mean Molarity) **pH changes 1 for every power of 10**
Example: If [H+] = 1 X pH = - log 1 X 10-10
pH = - (- 10)
pH = 10.0
Example: If [H+] = 1.8 X 10-5 pH = - log 1.8 X 10-5
pH = - (- 4.74)
pH = 4.74
(Sig figs: the number
of decimal places in the log
is equal to the number of sig
figs in the original number)

31. pH calculations – Solving for H+
If the pH of Coke is 3.12, [H+] = ???
Because pH = - log [H+] then
- pH = log [H+]
Take antilog (10x) of both sides and get
10-pH = [H+]
[H+] = = 7.6 x 10-4 M
*** to find antilog on your calculator, look for “Shift” or “2nd function” and then the log button

32. pOH Since acids and bases are opposites, pH and pOH are opposites!
pOH does not really exist, but it is useful for changing bases to pH.
pOH looks at the perspective of a base
pOH = - log [OH-]
Since pH and pOH are on opposite ends,
pH + pOH = 14

33. [H+], [OH-] and pH What is the pH of the 0.0010 M NaOH solution?
[OH-] = (or 1.0 X 10-3 M)
pOH = - log
pOH = 3
pH = 14 – 3 = 11
OR Kw = [H+] [OH-]
1 x 10-14/ 1.0 x 10 -3
[H+] = 1.0 x M
pH = - log (1.0 x 10-11) = 11.00

34. Acid/Base Equilibria Calculate the pH of 1.0M HCl
List the major species: HCl and H2O
H+, Cl-, H2O
Concentrate on the species that furnish H+. (pH)
H+, H2O Which gives more H2O or H+ (??)
H2O  H+ + OH- (negligable)
autoionization at 25°C [H+] is 10-7M.
In 1.0M HCl the water produces even less H+. (Le Chatelier’s)
HCl  H+ + Cl-
Therefore [H+] is 1.0M
pH = -log[H+] = -log(1.0) = 0

35. What is the pH of a 2 x 10-3 M HNO3 solution?
HNO3 is a strong acid – 100% dissociation.
Start
0.002 M
0.0 M
0.0 M
HNO3 (aq) + H2O (l) H3O+ (aq) + NO3- (aq)
End
0.0 M
0.002 M
0.002 M
pH = -log [H+] = -log [H3O+] = -log(0.002) = 2.7
What is the pH of a 1.8 x 10-2 M Ba(OH)2 solution?
Ba(OH)2 is a strong base – 100% dissociation.
Start
0.018 M
0.0 M
0.0 M
Ba(OH)2 (s) Ba2+ (aq) + 2OH- (aq)
End
0.0 M
0.018 M
0.036 M
pH = – pOH = log(0.036) = 12.56
15.4

36. Equilibria Involving A Weak Acid
You have 1.00 M HOAc (CH3COOH) Calculate the equilibrium concentrations of HOAc, H+, OAc- (CH3COO-), and the pH.
Step 1. List major species
HOAc, H2O
Step 2. Choose the species that produces the H+ . Write the balanced equation.
HOAc  H+ + OAc- Ka = 1.8 x 10-5
H2O  H+ + OH Kw = 1.0 x 10-14
Step 3. Which
one
dominates?

37. Step 4. Write Ka expression Ka = 1.8 x 10-5 = [H+][OAc-] [HOAc]
HOAc  H+ + OAc- Ka = 1.8 x 10-5
Step 4. Write Ka expression
Ka = 1.8 x 10-5 = [H+][OAc-]
[HOAc]
Step 5-7. Define [equilibrium] in ICE table.
[HOAc] [H+] [OAc-]
initial
change
equilib
-x +x +x
1.00-x x x

38. Step 9. Solve for x the easy way!
Step 8. Substitute into the expression
Step 9. Solve for x the easy way!
x = [H+] = [OAc-] = 4.2 x 10-3 M
Make an approximation if x is very small! (Rule of thumb: ±5% or smaller is ok)

39. Step 10. Validate the approximation
x = [H+] = [OAc-] = 4.2 x 10-3 M
pH = - log [H+] = -log (4.2 x 10-3) = 2.37
4.2 x 10-3 M
1.00M
% dissociation
X 100 = .42%
Step 11. Calculate [H+] and pH.

40. Equilibria Involving A Weak Acid
Calculate the pH of a M solution of formic acid, HCO2H. Ka = 1.8 x 10-4
HCO2H H2O ↔ HCO H3O+
Approximate solution
[H3O+] = 4.2 x 10-4 M, pH = 3.37
Exact Solution
[H3O+] = [HCO2-] = 3.4 x 10-4 M
[HCO2H] = x 10-4 = M
pH = 3.47

41. A mixture of Weak Acids The process is the same.
Determine the major species.
The stronger acid will predominate.
Bigger Ka if concentrations are comparable
Calculate the pH of a mixture 1.20 M HF (Ka = 7.2 x 10-4) and 3.4 M HOC6H5 (Ka = 1.6 x 10-10)

42. Percent dissociation = amount dissociated x 100 initial concentration
For a weak acid percent dissociation increases as acid becomes more dilute.
Calculate the % dissociation of M and M Acetic acid (Ka = 1.8 x 10-5) x2
.100 m
1.3 x 10-3M
As [HA]0 decreases [H+] decreases but % dissociation increases.
Le Châtelier HOAc  H+ + OAc-
4.2 x 10-3 M
1.00M
X 100 = .42%
X = [H+] = 1.3 x 10-3M 100 = .42%
X 100 = 1.3%

43. The other way
What is the Ka of a weak acid that is 8.1 % dissociated as M solution?
(assume you can estimate)
.081M2 / .100M = 6.5 x 10-4

44. Equilibrium Constants for Weak Bases
Weak base has Kb < 1
Leads to small [OH-] and a pH of

45. Relation of Ka, Kb, [H3O+] and pH

46. Ionization Constants of Conjugate Acid-Base Pairs
HA (aq) H+ (aq) + A- (aq) Ka
A- (aq) + H2O (l) OH- (aq) + HA (aq) Kb
H2O (l) H+ (aq) + OH- (aq) Kw
KaKb = Kw
Weak Acid and Its Conjugate Base
Ka = Kw Kb
Kb = Kw Ka
15.7

47. Equilibria Involving A Weak Base
You have M NH3. Calc. the pH.
NH3 + H2O ↔ NH OH-
Kb = 1.8 x 10-5
Step 1. Define equilibrium concs. in ICE table
[NH3] [NH4+] [OH-]
initial
change
equilib
-x +x +x
x x x

48. Equilibria Involving A Weak Base
You have M NH3. Calc. the pH.
NH3 + H2O  NH OH-
Kb = 1.8 x 10-5
Step 2. Solve the equilibrium expression
Assume x is small, so
x = [OH-] = [NH4+] = 4.2 x 10-4 M
and [NH3] = x 10-4 ≈ M
The approximation is valid !

49. Equilibria Involving A Weak Base
You have M NH3. Calc. the pH.
NH3 + H2O  NH OH-
Kb = 1.8 x 10-5
Step 3. Calculate pH
[OH-] = 4.2 x 10-4 M
so pOH = - log [OH-] = 3.37
Because pH + pOH = 14,
pH = 10.63

50. Polyprotic acids Always dissociate stepwise.
The first H+ comes of much easier than the second.
Ka for the first step is much bigger than Ka for the second.
Denoted Ka1, Ka2, Ka3

51. Polyprotic acid H2CO3 H+ + HCO3- Ka1= 4.3 x 10-7
HCO3- H+ + CO Ka2= 4.3 x 10-10
Base in first step is acid in second.
In calculations we can normally ignore the second dissociation.

52. Calculate the Concentration
Of all the ions in a solution of 1.00 M Arsenic acid H3AsO4
Ka1 = 5.0 x 10-3
Ka2 = 8.0 x 10-8
Ka3 = 6.0 x 10-10

53. Sulfuric acid is special
In first step it is a strong acid.
In the second step Ka2 = 1.2 x 10-2
Calculate the concentrations in a 1.0 M solution of H2SO4
Calculate the concentrations in a 1.0x 10-2 M solution of H2SO4

54. Equilibria Involving A Polyprotic Acid
You have 1.00 x 10-2 M H2SO4. Calc. the pH.
H2SO4 ↔ H HSO4- Ka = very large
HSO4- ↔ H+ + SO Ka 1.2 x 10-2
Step 1. Define equilibrium concs. in ICE table
[HSO4-] [H+] [SO42-]
initial
change
equilib
-x +x +x
x x x
X = 4.5 x 10-3
= pH = 1.84

56. Salts as acids and bases
Salts are ionic compounds.
Salts of the cation of strong bases and the anion of strong acids are neutral.
for example NaCl, KNO3
There is no equilibrium for strong acids and bases.
We ignore the reverse reaction.

57. Acid-Base Properties of Salts
Neutral Solutions:
Salts containing an alkali metal or alkaline earth metal ion (except Be2+) and the conjugate base of a strong acid (e.g. Cl-, Br-, and NO3-).
NaCl (s) Na+ (aq) + Cl- (aq)
H2O
Basic Solutions:
Salts derived from a strong base and a weak acid.
NaCH3COO (s) Na+ (aq) + CH3COO- (aq)
H2O
CH3COO- (aq) + H2O (l) CH3COOH (aq) + OH- (aq)
Salt Hydrolysis
15.10

58. Basic Salts
If the anion of a salt is the conjugate base of a weak acid – solution is basic.
In an aqueous solution of NaF
The major species are Na+, F-, and H2O
F- + H2O HF + OH-
Kb =[HF][OH-] [F- ]
For HF- the acid form- Ka = [H+][F-] [HF]

59. Basic Salts The Ka for HF is known: 7.2 x 10-4 How do we find Kb?
Kw= Ka x Kb
Ka x Kb = [HF][OH-] [H+][F-] [F- ] [HF] X

60. Ka tells us Kb The anion of a weak acid is a weak base.
Calculate the pH of a solution of 1.00 M NaF. Ka of HF is 7.2 x 10-4
H+ + F HF
The F- ion competes with OH- for the H+
(Weak acid strong conjugate base, stronger than that of water)

61. Acidic salts
A salt with the cation of a weak base and the anion of a strong acid will be acidic.
The same development as bases leads to
Ka x Kb = KW
Calculate the pH of a solution of 0.40 M NH4Cl (the Kb of NH3 1.8 x 10-5).
NH4Cl (s) NH4+ (aq) + Cl- (aq)
H2O
NH4+ (aq) NH3 (aq) + H+ (aq)

62. Hydrated metals make solution acidic
Highly charged metal ions pull the electrons of surrounding water molecules toward them.
Make it easier for H+ to come off.
Greater the charge: more acidic the solution. H
Al+3 O H
Al(H2O)6 (aq) Al(OH)(H2O)5 (aq) + H+ (aq) 3+ 2+

63. Acid-Base Properties of Salts
Solutions in which both the cation and the anion hydrolyze:
Kb for the anion > Ka for the cation, solution will be basic
Kb for the anion < Ka for the cation, solution will be acidic
Kb for the anion  Ka for the cation, solution will be neutral
15.10

64. Acid-Base Properties of Salts

65. Types of Acid/Base Reactions: Summary

66. Structure and Acid-Base Properties What causes it to be an acid or a base?
Any molecule with an H in it is a potential acid.
The stronger the X-H bond the less acidic (compare bond dissociation energies).
The more polar the X-H bond the stronger the acid (use electronegativities).

67. Molecular Structure and Acid Strength
H X
H+ + X-
Bond strength ↓ down .WHY?
Polarity ↓ down
The stronger the bond
The weaker the acid
HF << HCl < HBr < HI
Most polar least polar
Highest bond energy
15.9

68. Strength of oxyacids The more polar H-O-X bond -stronger acid.
The more oxygen hooked to the central atom, the more acidic the hydrogen.
HClO4 > HClO3 > HClO2 > HClO
Remember that the H is attached to an oxygen atom.
The oxygens are electronegative
Pull electrons away from hydrogen

69. Molecular Structure and Acid Strength
1. Oxyacids having different central atoms (Z) that are from the same group and that have the same oxidation number.
Acid strength increases with increasing electronegativity of Z
H O Cl O O •
H O Br O O •
Cl is more electronegative than Br
HClO3 > HBrO3
15.9

70. Molecular Structure and Acid Strength
2. Oxyacids having the same central atom (Z) but different numbers of attached groups.
Acid strength increases as the oxidation number of Z increases.
HClO4 > HClO3 > HClO2 > HClO
15.9

71. Strength of oxyacids
Electron Density Cl O H

72. Strength of oxyacids
Electron Density O Cl O H

73. Strength of oxyacids
Electron Density O Cl O H O

74. Strength of oxyacids
Electron Density O O Cl O H O

75. Acid-Base Properties of Oxides
Non-metal (covalent) oxides dissolved in water can make acids.
SO3 (g) + H2O(l) H2SO4(aq)
Ionic oxides dissolve in water to produce bases. (metal oxides)
CaO(s) + H2O(l) Ca(OH)2(aq)
Hydroxides

76. Amphoteric Metallic Oxides
Be(OH)2, Al(OH)3, Ga(OH)3, Sn(OH)2, Pb(OH)2, Cr(OH)3, Cu(OH)2, Zn(OH)2, Cd(OH)2
Be(OH)2(s) +2H+(aq)  Be2+(aq) +2H2O(l)
Be(OH)2(s) + 2OH-(aq)  Be(OH)42-(aq)
All amphoteric
hydroxides are
insoluble!!

77. Reactions CO2(g) + NaOH(aq)  Na2CO3(aq) + H2O(l) acid base salt water
anhydride
BaO(s) + 2HNO3(aq)  Ba(NO3)2(aq) + H2O(l)
basic acid salt water

78. Lewis Acids and Bases :N F H B F H F H Most general definition.
Acids are electron pair acceptors.
Bases are electron pair donors. F H B F :N H F H

79. Lewis Acids and Bases :N F H B F H F H
Boron triflouride wants more electrons. F H B F :N H F H

80. Lewis Acids and Bases F H F B N H F H
Boron triflouride wants more electrons.
BF3 is Lewis base NH3 is a Lewis Acid. F H F B N H F H

81. ( ) ( ) Lewis Acids and Bases Al+3 Al H O H +3 H O H 6 + 6 Lewis acid
Lewis base +3 ( ) 6 H Al O H