1. Acids and Bases

2. Acid/Base Definitions
Arrhenius Model
Acids produce hydrogen ions in aqueous solutions
Bases produce hydroxide ions in aqueous solutions
Bronsted-Lowry Model
Acids are proton donors
Bases are proton acceptors

3. Acid Dissociation Acid Proton Conjugate base HA  H+ + A-
Alternately, H+ may be written in its hydrated form, H3O+ (hydronium ion)

4. HNO2(aq) + H2O(l)
NO2-(aq) H3O+(aq)
Acid
Base
Conjugate
base
Conjugate
acid
NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)
Base
Acid
Conjugate
acid
Conjugate
base

5. Dissociation of Strong Acids
Strong acids are assumed to dissociate completely in solution.
Large Ka or small Ka?
Reactant favored or product favored?

6. Dissociation Constants: Strong Acids
Formula
Conjugate Base Ka
 Perchloric 
 HClO4 
 ClO4- 
 Very large 
 Hydriodic 
 HI 
 I- 
 Hydrobromic 
 HBr 
 Br- 
 Hydrochloric 
 HCl 
 Cl- 
 Nitric 
 HNO3 
 NO3- 
 Sulfuric 
 H2SO4 
 HSO4- 
 Hydronium ion 
 H3O+ 
 H2O 
 1.0 

7. Dissociation of Weak Acids
Weak acids are assumed to dissociate only slightly (less than 5%) in solution.
Large Ka or small Ka?
Reactant favored or product favored?

8. Dissociation Constants: Weak Acids
Formula
Conjugate Base Ka
 Iodic 
 HIO3 
 IO3- 
 1.7 x 10-1 
 Oxalic 
 H2C2O4 
 HC2O4- 
 5.9 x 10-2 
 Sulfurous 
 H2SO3 
 HSO3- 
 1.5 x 10-2 
 Phosphoric 
 H3PO4 
 H2PO4- 
 7.5 x 10-3 
 Citric 
 H3C6H5O7 
 H2C6H5O7- 
 7.1 x 10-4 
 Nitrous 
 HNO2 
 NO2- 
 4.6 x 10-4 
 Hydrofluoric 
HF 
 F- 
 3.5 x 10-4 
 Formic 
 HCOOH 
 HCOO- 
 1.8 x 10-4 
 Benzoic 
 C6H5COOH 
 C6H5COO- 
 6.5 x 10-5 
 Acetic 
 CH3COOH 
 CH3COO- 
 1.8 x 10-5 
 Carbonic 
 H2CO3 
 HCO3- 
 4.3 x 10-7 
 Hypochlorous 
 HClO 
 ClO- 
 3.0 x 10-8 
 Hydrocyanic 
 HCN 
 CN- 
 4.9 x 10-10 

9. PROBLEM:
Write the simple dissociation (ionization) reaction
(omitting water) for each of the following acids.
Hydrochloric acid (HCl)
Acetic acid (HC2H3O2)
The ammonium ion (NH4+)
The anilinium ion (C6H5NH3+)
The hydrated aluminum(III) ion [Al(H2O)6]3+

10. Self-Ionization of Water
H2O + H2O  H3O+ + OH-
At 25, [H3O+] = [OH-] = 1 x 10-7
Kw is a constant at 25 C:
Kw = [H3O+][OH-]
Kw = (1 x 10-7)(1 x 10-7) = 1 x 10-14

11. pH Scale

12. Relationship between pH and pOH
Calculating pH, pOH
pH = -log10(H3O+)
pOH = -log10(OH-)
Relationship between pH and pOH
pH + pOH = 14
Finding [H3O+], [OH-] from pH, pOH
[H3O+] = 10-pH
[OH-] = 10-pOH

13. pH and pOH Calculations

14. Problem:
Calculate [H+] or [OH-] as required for each of the
following 25oC, and state whether the
solution is neutral, acidic, or basic.
1.0 x 10-5M OH-
1.0 x 10-7M OH-
10.0M H+

15. A Weak Acid Equilibrium Problem
What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ?
Step #1: Write the dissociation equation
HC2H3O2  C2H3O2- + H+

16. A Weak Acid Equilibrium Problem
What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ?
Step #2: ICE it!
HC2H3O2  C2H3O2- + H+ I C E
0.50
- x +x +x x x x

17. A Weak Acid Equilibrium Problem
What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ?
Step #3: Set up the law of mass action
HC2H3O2  C2H3O2- + H+ E x x x

18. A Weak Acid Equilibrium Problem
What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ?
Step #4: Solve for x, which is also [H+]
HC2H3O2  C2H3O2- + H+ E x x x
[H+] = 3.0 x 10-3 M

19. A Weak Acid Equilibrium Problem
What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ?
Step #5: Convert [H+] to pH
HC2H3O2  C2H3O2- + H+ E x x x
pH = -log(3.0 x 10-3) = 2.52

20. Percent Dissociation/Ionization
Amount dissociated (mol/L) =
X 100%
Initial concentration (mol/L)
So….Calculate the percent dissociation for
the previous problem.
60%?
JK…How about 0.6%? Better?
Tip of the day!
For a given weak acid….the percent dissociation
INCREASES as the acid becomes more dilute.
Larger Percent dissociation for which one?
1.0 M HC2H3O2 or M HC2H3O2?

21. HC9H7O4  C9H7O4- + H+ HANDOUT!!! DO ALL WORK ON THIS!!!
Aspirin, is a weak organic acid whose molecular formula is HC9H7O4. An aqueous solution of aspirin has a total volume of mL and contains 1.26 g of aspirin. The pH of the solution is found to be Calculate Ka for aspirin.culate the Ka for aspirin
Step #1: Write the dissociation equation
HC9H7O4  C9H7O4- + H+

22. I C E HC9H7O4  C9H7O4- + H+ Step #2: Determine concentration of acid!
1.26 g x 1 mol
.350L g
= 0.02M HC9H7O4
Step #3: ICE it!
HC9H7O4  C9H7O4- + H+ I C E
.02
- x +x +x
.02 - x x x

23. HC9H7O4  C9H7O4- + H+ E Step #4: Determine the [H+] pH = 2.60
[H+] = 2.5 x 10-3 M
Step #5: Set up the law of mass action
HC9H7O4  C9H7O4- + H+ E
.02 - x x x 2 x2
.0025
Ka =
Ka =
Ka = 3.6 x 10-4
.02 - x

24. Dissociation of Strong Bases
MOH(s)  M+(aq) + OH-(aq)
Strong bases are metallic hydroxides
Group I hydroxides (NaOH, KOH) are very soluble
Group II hydroxides (Ca, Ba, Mg, Sr) are less soluble
pH of strong bases is calculated directly from the concentration of the base in solution

25. Reaction of Weak Bases with Water
The generic reaction for a base reacting with water, producing its conjugate acid and hydroxide ion:
B + H2O  BH+ + OH-
(Yes, all weak bases do this – DO NOT
endeavor to make this complicated!)

26. Reaction of Weak Bases with Water
The base reacts with water, producing its conjugate acid and hydroxide ion:
CH3NH2 + H2O  CH3NH3+ + OH- Kb = 4.38 x 10-4

27. Kb for Some Common Weak Bases
Many students struggle with identifying weak bases and their conjugate acids.What patterns do you see that may help you?
Base
Formula
Conjugate Acid Kb
Ammonia 
 NH3
 NH4+
 1.8 x 10-5 
 Methylamine
 CH3NH2
 CH3NH3+
 4.38 x 10-4 
 Ethylamine
 C2H5NH2
 C2H5NH3+
 5.6 x 10-4 
 Diethylamine
 (C2H5)2NH
 (C2H5)2NH2+
 1.3 x 10-3 
 Triethylamine
  (C2H5)3N
  (C2H5)3NH+
 4.0 x 10-4 
 Hydroxylamine
 HONH2 
 HONH3+  
 1.1 x 10-8 
 Hydrazine
H2NNH2 
H2NNH3+  
 3.0 x 10-6 
 Aniline
 C6H5NH2 
 C6H5NH3+  
 3.8 x 10-10 
 Pyridine
 C5H5N 
 C5H5NH+  
 1.7 x 10-9 

28. A Weak Base Equilibrium Problem
What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ?
Step #1: Write the equation for the reaction
NH3 + H2O  NH4+ + OH-

29. A Weak Base Equilibrium Problem
What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ?
Step #2: ICE it!
NH3 + H2O  NH4+ + OH- I C E
0.50
- x +x +x x x x

30. A Weak Base Equilibrium Problem
What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ?
Step #3: Set up the law of mass action
NH3 + H2O  NH4+ + OH- E x x x

31. A Weak Base Equilibrium Problem
What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ?
Step #4: Solve for x, which is also [OH-]
NH3 + H2O  NH4+ + OH- E x x x
[OH-] = 3.0 x 10-3 M

32. A Weak Base Equilibrium Problem
What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ?
Step #5: Convert [OH-] to pH
NH3 + H2O  NH4+ + OH- E x x x

33. ANOTHER TIP OF THE DAY
Ka x Kb = Kw

34. The calculation of concentration and pH for
weak acids is more complex than for strong
acids due to:
The incomplete ionization of weak acids.
The low Ka value for strong acids.
The more complex atomic structures of
strong acids.
D. The low percent ionization of strong acids.
E. The inconsistent Kb value for strong acids.
Answer: A Strong acids ionize 100%, therefore, the [H+]
is equal to the initial concentration of the strong acid.
This is not the case for weak acids. Since not all of the
acid ionizes, it is necessary to determine both how much
of the acid forms H+ and how much is left in molecular form.

35. The general reaction of an acid dissolving
in water may be shown as
HA (aq) + HOH (l) H3O (aq) + A-(aq)
A conjugate acid base pair for this reaction is:
HA and HOH
HA and A-
HOH and A-
H3O+ and A-
HA and H3O+
Answer: B The conjugate acid differs from its conjugate base by
a proton, (H+). The acid form has the proton (HA in this example)
and the base form has lost the proton (A- in this case).

36. Strong acids are those which
Have an equilibrium lying far to the left.
Yield a weak conjugate base when reacting with water
Have a conjugate base which is a stronger base than
water
D. Readily remove the H+ ions from water.
E. Are only slightly dissociated (ionized) at equilibrium.
Answer: B Strong acids give up their protons (H+ ions) easily.
If their conjugate base form were strong, then the acid formed
would hold strongly to the H+, which would not be a characteristic
of a strong acid.

37. When calculating the pOH of a hydrofluoric acid solution
(Ka = 7.2 x 10-4) from its concentration, the contribution
of water ionizing (Kw = 1.0 x 10-14) is usually ignored because
Hydrofluoric acid is such a weak acid.
Hydrofluoric acid can dissolve glass
The ionization of water provides relatively few H+ ions.
The [OH-] for pure water is unknown.
The conjugate base of HF is such a strong base.
Answer: C By comparing the Ka for HF with Kw (for water),
you can see that even though hydrofluoric acid is a very weak
acid it is a much stronger acid than is water, so much so that
the H+ ion contribution of water may be ignored.

38. The percent dissociation (percent ionization) for weak acids
Is always the same for a given acid, no matter what the
concentration.
B. Usually increases as the acid becomes more concentrated.
C. Compares the amount of acid that has dissociated at
equilibrium with the initial concentration of the acid.
D. May only be used to express the dissociation of weak
acids.
E. Has no meaning for polyprotic acids.
Answer: C the percent dissociation does change with the
concentration of the acid.

39. The [OH-] of a certain aqueous solution is 1.0 x 10-5M.
The pH of this same solution must be
1.0 x 10-14
5.00
7.00
9.00
12.00
Answer: D Referring to pH + pOH = at 25oC, if
[OH-] = 1 x 10-5M, pOH = 5, pH = =9.0

40. In many calculations for the pH of a weak acid from the
concentration of the acid, an assumption is made that
often takes the form [HA]o – x = [HA]o
This
Is valid because x is very small compared to the initial
concentration of the weak acid.
B. Is valid because the concentration of the acid changes
by such large amounts.
C. Is valid because actual value of x cannot be known.
D. Is valid because pH is not dependent upon the concentration
of the weak acid.
E. Approximation is always shown to be valid and so need not
be checked.
Answer: A. We can expect x to be very small (as indicated
by the low value for Ka for weak acids), so that the concentration
of the weak acid does not significantly change from its initial
value. As the final step in such problems, you must determine
that the change in the initial [HA] is less than 5% for the
approximation to be considered valid.

41. HA is a weak acid which is 4.0% dissociated at 0.100M.
Determine the Ka for this acid
0.0040
0.040
1.6
16.5
Answer: B
[HA]o = 0.100M
[H+] = [A-] = 0.004M
Ka = [H+][A-]/[HA] = (0.004 x 0.004) / =
(4 x 10-3)(4 x 10-3) / (10-1) = 1.6 x 10-4

42. SALTS

43. Acid-Base Properties of Salts
Type of Salt
Examples
Comment
pH of solution
Cation is from a strong base, anion from a strong acid
KCl, KNO3
NaCl
NaNO3
Both ions are neutral
Neutral
These salts simply dissociate in water:
KCl(s)  K+(aq) + Cl-(aq)

44. Acid-Base Properties of Salts
Type of Salt
Examples
Comment
pH of solution
Cation is from a strong base, anion from a weak acid
NaC2H3O2
KCN, NaF
Cation is neutral,
Anion is basic
Basic
The basic anion can accept a proton from water:
C2H3O H2O  HC2H3O2 + OH base acid acid base

45. Acid-Base Properties of Salts
Type of Salt
Examples
Comment
pH of solution
Cation is the conjugate acid of a weak base, anion is from a strong acid
NH4Cl,
NH4NO3
Cation is acidic,
Anion is neutral
Acidic
The acidic cation can act as a proton donor:
NH4+(aq)  NH3(aq) + H+(aq)
Acid Conjugate Proton
base

46. Acid-Base Properties of Salts
Type of Salt
Examples
Comment
pH of solution
Cation is the conjugate acid of a weak base, anion is conjugate base of a weak acid
NH4C2H3O2
NH4CN
Cation is acidic,
Anion is basic
See below
IF Ka for the acidic ion is greater than Kb for the basic ion, the solution is acidic
IF Kb for the basic ion is greater than Ka for the acidic ion, the solution is basic
IF Kb for the basic ion is equal to Ka for the acidic ion, the solution is neutral

47. Acid-Base Properties of Salts
Type of Salt
Examples
Comment
pH of solution
Cation is a highly charged metal ion; Anion is from strong acid
Al(NO3)2
FeCl3
Hydrated cation acts as an acid;
Anion is neutral
Acidic
Step #1:
AlCl3(s) + 6H2O  Al(H2O)63+(aq) + Cl-(aq)
Salt water Complex ion anion
Step #2:
Al(H2O)63+(aq)  Al(OH)(H2O)52+(aq) + H+(aq)
Acid Conjugate base Proton

48. Summary of Acid/Base Properties of Salts:
Cation From Anion From Solution
Strong Base
Weak Base
Strong Acid
Weak Acid
Neutral
Basic
Acidic
Must compare K
values!

49. Effect of Structure on Acid-Base Properties

50. Question: Do I really know this stuff?
Answer: Let’s find out!
Good luck

51. Let’s try this:
Determine the pH of a 0.10M solution of
NaC2H3O2. Ka for HC2H3O2 is 1.8 x 10-5
Good luck

52. PRACTICE TIME! Please complete problems: #44,50,52, and 64a-d
on a separate sheet of paper
(Will be collected at end of block!)
You may work with other students AT YOUR WORKSTATION!

53. PRACTICE TIME! pp. 673-75: #28,30,32,36,38,40,44,50 52 and 55.
PLEASE BEGIN TO SOLVE THE FOLLOWING PROBLEMS
RELATING TO: NATURE OF ACIDS-n-BASES, AUTOIONIZATION
OF WATER, pH SCALE AND THE SOLUTIONS OF ACIDS.
pp : #28,30,32,36,38,40,44,50
52 and 55.

54. Please complete problems: on a separate sheet of paper
PRACTICE TIME!
Please complete problems: Pp
64a-d and # 78,82,84,86,92 and 94
on a separate sheet of paper

55. PRACTICE TIME! Please complete problems: #44,50,52,58,59 and 64a-d
on a separate sheet of paper
(Will be collected at end of block!)

56. PRACTICE TIME!
PLEASE BEGIN TO SOLVE THE FOLLOWING PROBLEMS
RELATING TO: NATURE OF ACIDS-n-BASES, AUTOIONIZATION
OF WATER, pH SCALE AND THE SOLUTIONS OF ACIDS.
pp : #28-30,32,33-36,38,40,44,50
52,55,58,59, and 64a-d
Please complete problems
#28-30,32,33-36,38 and 40
on a separate sheet of paper
(Will be collected at end of block!)

57. PRACTICE TIME!
pp : #78,82,84,86,92 and 94

58. PRACTICE TIME!
pp : #
43-46,50,52,55,58,
59,62 and 64(a-d)