1. Chapter 12 Acids and Bases
12.1 The Nature of Acids and Bases
12.2 Acid Strength
12.3 The pH Scale
12.4 Calculating the pH of Strong Acid Solutions
12.5 Calculating the pH of Weak Acid Solutions
12.6 Bases
12.7 Polyprotic Acids
12.8 Acid-Base Properties of Salts
12.9 Acid Solutions in Which Water Contributes to the H+ Concentration (skip)
12.10 Strong Acid Solutions in Which Water Contributes to the H+ Concentration (skip)
12.11 Strategy for Solving Acid-Base Problems: A Summary

2. Acids and Bases and Their Reactions
Definitions
Arrhenius Acids and Bases Acids are
H+ donors
Bases are OH- donors
Arrhenius Broadened Definition
Acids increase H+ concentration or [H+] increases
Bases increase OH- concentration or [OH-] increases
Brønsted-Lowry Acids and Bases (1923)
Acids donate H+
Bases accept H+
Arrhenius 1903
Nobel Prize

3. Brønsted-Lowry Acids and Bases
Acids-Bases
Brønsted-Lowry Acids and Bases
A Brønsted-Lowry acid is a substance that can donate a hydrogen ion (aka H+, proton).
A Brønsted-Lowry base is a substance that can accept a hydrogen ion.
Acids and bases occur as conjugate acid - base pairs.

4. Conjugate Base - subtract an H+ from the acid
Conjugate Acid add H+ to the base
Examples
OH– is the conjugate base of
H2O is the conjugated base of
H2O is the conjugated acid of
H3O+ (or often shown as H+) is the conjugate acid of

5. CH3CO2H + H2O H3O+ + CH3CO2- Pairs 1 2 2 1
Acetic Acid
Acetate Ion
Pairs
acid
base
acid
base
Point of View #1
acid
base
Conjugate acid of H2O
Conjugate base of CH3CO2H
CH3CO2H H2O H3O CH3CO2-
Point of View #2
acid
base
Conjugate base of H3O+
Conjugate acid of CH3CO2-
CH3CO2H H2O H3O CH3CO2-

6. Nomenclature
When H+ is hydrated, it is H3O+ and is called a hydronium ion. A hydronium ion has the same molecular geometry as NH3. +
111.7°

7. HA + H2O H3O+ + A – HA = generic acid
There is Competition for the proton between two bases H2O and A–
If H2O is a much stronger base than A– the equilibrium lies far to the right.
If A– is a much stronger base than H2O the equilibrium lies far to the left.

8. HA + H2O H3O+ + A – HA + H2O H3O+ + A –
Acid Strength: graphical representation of the behavior of acids of different strengths in aqueous solution.
A strong acid: equilibrium lies far to the right
HA + H2O H3O+ + A –
A weak acid: equilibrium lies far to the left
HA + H2O H3O+ + A –
A weak acid yields a relatively strong conjugate base

9. Relationship of acid strength and conjugate base strength
HA + H2O H3O+ + A-

10. HA + H2O H3O+ + A–
SAME AS
HA H+ + A–

11. Ka is the Acidity constant

12. Autoionization of H2O H2O + H2O H3O+ + OH- Pairs 1 2 2 1
acid
base
acid
base
Point of View #1
acid
base
Conjugate acid of H2O #2
Conjugate base of H2O #1
H2O H2O H3O OH-
# 1
# 2
Point of View #2
acid
base
Conjugate base of H3O+
Conjugate acid of OH-
H2O + H2O H3O OH-
Amphoterism - an ion or molecule can act as an acid or base depending upon the reaction conditions

13. 1.) Water in NH3 serves as an acid
Amphoterism - an ion or molecule can act as an acid or base depending upon the reaction conditions
1.) Water in NH3 serves as an acid
H2O NH NH OH-
acid
base
acid
base
2.) Water in acetic acid serves as a base
H2O CH3CO2H H3O CH3CO2-
base
acid
acid
base
Conjugate acid
of H2O
Conjugate base
of acetic acid

14. Kw = [H3O+][OH-] Water as an Acid and a Base
Autoionization of water:
2 H2O (l) H3O+ (aq) + OH- (aq)
[H3O+][OH-]
[H2O(l)]2
Kw = [H3O+][OH-] = [H+][OH-]
Kw = 1.0 x (at 25oC)
In pure water [H+] = [OH-]
Kw = [H3O+][OH-]

15. NH H+ + NH3
Ka/Kb/Kw
NH3 + H2O OH- + NH4+

16. NH H+ + NH3
Ka/Kb/Kw
NH3 + H2O OH- + NH4+
H2O OH- + H+

17. The pH Scale pH = -log10[H3O+] pH = -log10[H+]
SAME AS
pH = -log10[H+]
pH < 7 acidic solution
[H3O+] > [OH-]
pH = 7 neutral solution
[H3O+] = [OH-]
pH > 7 basic solution
[H3O+] < [OH-]
Sig figs: for logs: the number of decimal places in the log is equal to the number of sig figs in the original number

18. Calculate the pH of Strong Acid-Base Solutions
pH = -log10[H3O+]
EXAMPLE
Calculate the pH (at 25oC) of an aqueous solution that has an
OH-(aq) concentration of 1.2 x 10-6 M (i.e., mol/liter)
Solution
The concentration of H+ (aq) is
[H+][OH-] = KW
[H+] = KW/[OH-] = 10-14/1.2x10-6 = 8.3 x 10-9
pH = -log[8.3 x 10-9] = 8.1

19. [H+][OH-] = Kw Strong Acids
A strong acid is one that dissociates completely in water to produce H+(aq).
E.g., Hydrochloric acid (HCl) is a strong acid:
HCl (aq) → H+ (aq) + Cl- (aq) (reaction essentially complete)
Dissolving 0.10 mol of HCl in enough water to make 1.0 L of solution gives a final concentration of 0.10 M for H+(aq).
H2O (l) H+ (aq) + OH- (aq)
[H+][OH-] = Kw
[OH-] = Kw / [H+] = 10-14/10-1 = 10-13

20. Write the major species in solution and Ka values. Which is dominant?
Weak Acids
Write the major species in solution and Ka values. Which is dominant?
#1: HA H+ + A –
#2: H2O H+ + OH –
Compare the value for Ka and Kw. Which is larger? This is the dominate source of H+

21. Problem: (a) Calculate pH and (b) the fraction of CH3CO2H ionized at equilibrium. The concentration of CH3CO2H is 1 M (initial, or total). The Ka for acetic acid is 1.8 x 10-5
Estimate major species in solution CH3CO2H (a weak acid) and H2O.
CH3CO2H H+ + CH3CO2─ Ka = 1.8 x 10-5
H2O H+ + OH – Kw = 1.0 x 10-14

22. Problem: (a) Calculate pH and (b) the fraction of CH3CO2H ionized at equilibrium.
CH3CO2H H CH3CO2─
Initial M ~ 0 0
Change - y + y y
Equilibrium – y y y

23. Problem: (a) Calculate pH and (b) the fraction of CH3CO2H ionized at equilibrium.
CH3CO2H H CH3CO2─
Initial M ~ 0 0
Change - y + y y
Equilibrium – y y y

24. Problem: (a) Calculate pH and (b) the fraction of CH3CO2H ionized at equilibrium.
CH3CO2H H CH3CO2─
Initial M ~ 0 0
Change - y + y y
Equilibrium – y y y

25. Effect of dilution on the % dissociation and [H+] and pH
HA H+ + A –
Le Chatelier’s Priniciple:
If a chemical system at equilibrium experiences a change in concentration (or temperature, volume, or partial pressure) then the system shifts to counteract the imposed change.
The more dilute the weak acid solution, the greater the percent dissociation

26. NaOH (s) → Na+ (aq) + OH- (aq)
Strong Bases A strong base reacts completely with water to produce OH-(aq) ions.
Strong Acids and Bases
Sodium hydroxide (NaOH) is a strong base:
Others are KOH–, NH2– (amide ion) and H– (Hydride ion)
NaOH (s) → Na+ (aq) + OH- (aq)
(reaction essentially complete)
Dissolving 0.10 mol of NaOH in enough water to make 1.0 L of solution gives a final concentration of 0.10 M for OH- (aq). From this you can calculate pH and pOH.
[H+][OH-] = Kw
[H+] = Kw/[OH-] =10-14/10-1 = 10-13
[OH-] = [H+] = pOH = pH = 13

27. B (aq) + H2O (l) BH+ (aq) + OH– (aq)
Weak Bases
B (aq) + H2O (l) BH+ (aq) + OH– (aq)
B = Base
Kb is the basicity constant
Kb = [BH+][OH–]/[B]

28. B (aq) + H2O (l) BH+ (aq) + OH– (aq)
Calculations for solutions of weak bases are similar to those for weak acids. Bases (B) compete with OH–, a very strong base, for H+ ions.
B (aq) + H2O (l) BH+ (aq) + OH– (aq)

29. pKa + pKb = pKw Base Strength
Acid-Base Equilibria
Base Strength
strong acids have weak conjugate bases
weak acids have strong conjugate bases
The strength of a base is inversely related to the strength of its conjugate acid; the weaker the acid, the stronger its conjugate base, and vice versa
pKa + pKb = pKw
This equation applies to an acid and its conjugate base. 3

30. NH3 + H20 NH4+ + OH- acid1 base2 acid2 base1
Weak Bases
NH3 is a base Kb = 1.8 x 10-5
NH3 + H NH4+ + OH- acid1 base2 acid2 base1
NH4+ is an acid Ka = 5.6 x 10-10
NH NH3 + H+ acid base1

31. Weak Bases
NH3 is a base Kb = 1.8 x 10-5
NH3 + H NH4+ + OH-
NH4+ is an acid Ka = 5.6 x 10-10
NH NH3 + H+
Ka Kb = Kw = (5.6 x 10-10) (1.8 x 10-5) = 10-14
pKa + pKb = pKw = (-9.25) + (-4.75) = -14

32. Weak Bases with Weak Acids
NH4+ is an acid Ka(am) = 5.6 x 10-10
NH NH3 + H+
CH3COOH is an acid Ka(aa) = 1.8 x 10-5
CH3COOH CH3COO- + H+
change the direction of the aa reaction and add them together
NH4+ + CH3COO NH3 + CH3COOH
K = Ka(am)/Ka(aa) = 5.6 x 10-10/1.8 x 10-5
=3.1 x 10-5

33. Assuming 0.1M NH3 initial, calculate the pH of the resulting solution
H20 + NH NH OH-

34. Assuming 0.1M NH3 initial, calculate the pH of the resulting solution
H20 + NH NH OH-
Init. conc. 0.1M 0 ~0
Change - y + y y
Equil. conc. 0.1– y y y

35. Assuming 0.1M NH3 initial, calculate the pH of the resulting solution
H20 + NH NH OH-
Init. conc. 0.1M 0 ~0
Change - y + y y
Equil. conc. 0.1– y y y
[H+] =
Kw/[OH-] =
10-14/y
pH=-log[H+]

36. H2SO4 (aq) → H+ (aq) + HSO4– (aq) Ka1 ≈ 1.0x102
Polyprotic Acids
It is always easier to remove the first proton in a     polyprotic acid than the second. That is, Ka1 > Ka2 > Ka3
Note: Sulfuric acid is a strong acid in its first dissociation step and a weak acid in its second step.
H2SO4 (aq) → H+ (aq) + HSO4– (aq) Ka1 ≈ 1.0x102
HSO4– (aq) H+ (aq) + SO42– (aq) Ka2 ≈ 1.2x10-2

37. Polyprotic Acids H2SO3 (aq) + H2O (l) → H3O+ (aq) + HSO3- (aq)
Polyprotic acids have more than one ionizable proton.
The protons are removed in steps, not all at once.
The extent of the steps can be calculated sequentially.
From the successive acidity constants, the equilibrium concentrations of all species present can be calculated at any value of the pH.
     H2SO3 (aq) + H2O (l) → H3O+ (aq) + HSO3- (aq)
Ka1 = 103
     HSO3- (aq) + H2O (l) H3O+ (aq) + SO32- (aq)                                                                       
Ka2 = 1.2 x 10-2

38. Ka1 >> Ka2 >> Ka3
Polyprotic Acids
pH calculations for solutions of Polyprotic acids appear complicated, most common cases (with weak acids) are surprising straightforward.
Setup equations, K values, set up ICE type table, assume dissociation is small (check assumption) and solve.
For typical weak Polyprotic acids.
Ka1 >> Ka2 >> Ka3
The first dissociation step dominates the H+ concentration and thus the pH

39. A plot of the fractions of H2CO3, HCO-3 and CO32-
At pH = 9.00 H2CO3 ≈ 0%, HCO-3 = 95% and CO32- = 5%
At pH = H2CO3 ≈ 0%, HCO-3 = 68% and CO32- = 32%

40. Acid-Base Properties of Salts
“Hydrolysis”

41. Cation: M+2 + H2O ↔ H3O+ + M(OH)+
Hydrolysis is a Brønsted-Lowry Acid and Base Reaction
Anion: A─ + H2O ↔ HA + OH─
Cation: M+2 + H2O ↔ H3O+ + M(OH)+
E.g., Cations Ni+2 and Fe+2

42. Other salts behave similarly, NH4Cl and AlCl3 give acid solutions.
Hydrolysis is a term applied to reactions of aquated ions that change the pH from 7
When NaCl is placed in water, the resulting solution is observed to be neutral (pH = 7)
However when sodium acetate (NaC2H3O2) is dissolved in water the resulting solution is basic
Other salts behave similarly, NH4Cl and AlCl3 give acid solutions.
These interactions between salts and water are called hydrolysis 3

43. CH3CO2– + H2O ↔ CH3CO2H + OH– Example problem:
Suppose a 0.1 mole solution sodium acetate is dissolved in 1 liter of water. What is the pH of the solution?
CH3CO2– + H2O ↔ CH3CO2H + OH–
base
acid
acid
base
Init. conc. 0.1M 0 ~0
∆ conc. - y + y y
Equil. conc. 0.1– y y y
Find Kb
Find [OH-]
Find [H+]
Find pH

44. Example problem: Ka x Kb = Kw CH3CO2– + H2O ↔ CH3CO2H + OH–
Find Kb
Find [OH-]
Find [H+]
Find pH
Example problem:
What is the pH of the solution?
CH3CO2– + H2O ↔ CH3CO2H + OH–
Init. conc. 0.1M ~0
∆ conc. - y y y
Equil. conc. 0.1– y y y
Ka x Kb = Kw

45. a non-hydrolyzed cation (Na+) a hydrolyzed anion (acetate ion)
Hydrolysis is a term applied to reactions of aquated ions that change the pH from 7
When NaCl is placed in water, the resulting solution is observed to be neutral (pH = 7)
However when sodium acetate (NaC2H3O2) is dissolved in water the resulting solution is basic (Problem: found at 0.1M NaAc, pH =8.89)
a non-hydrolyzed cation (Na+)
a hydrolyzed anion (acetate ion)
Other salts behave similarly, NH4Cl and AlCl3 give acid solutions.
a non-hydrolyzed anion (Cl-)
a hydrolyzed cation (NH4+ or Al+3)
These interactions between salts and water are called hydrolysis 3

46. Anions Cations Raise pH Lower pH Non-Hyrolyzed Ions (a few)
Hydrolysis of
Result
Anions
Cations
Raise pH
Lower pH
Non-Hyrolyzed Ions (a few)
7 Anions, not hydrolyzed
Cl –, Br –, I –, HSO4–, NO3–, ClO3–, ClO4–
10 Cations, not hydrolyzed
Li+, Na+, K+, Rb+, Sc+, Mg++, Ca++, Sr++, Ba++, Ag+

47. Anions Cations Raise pH Lower pH Hydrolysis of Result
Predict pH of salts in water (relative pH)
Na3PO4 is basic (raised pH)
(a non hydrolyzed cation and a hydrolyzed anion)
Na3PO4?
FeCl3 is acidic (lowers pH)
(a hydrolyzed cation and a non hydrolyzed anion)
FeCl3?

48. Acid-Base Properties of Salts
Summary
Acid-Base Properties of Salts
“Hydrolysis”

49. Acid-Base Equilibrium Problems
Summary
Acid-Base Equilibrium Problems
Which major species are present
Does a reaction occur that can be assumed to go to completion?
Which equilibrium dominates the solution?
Set up ICE type table
Solve for equilibrium concentrations using know K values
Check any simplifying assumptions
Typically solve for pH or % species in solution

50. Chapter 7 Acids and Bases
7.1 The Nature of Acids and Bases
7.2 Acid Strength
7.3 The pH Scale
7.4 Calculating the pH of Strong Acid Solutions
7.5 Calculating the pH of Weak Acid Solutions
7.6 Bases
7.7 Polyprotic Acids
7.8 Acid-Base Properties of Salts
7.9 Acid Solutions in Which Water Contributes to the H+ Concentration (skip)
7.10 Strong Acid Solutions in Which Water Contributes to the H+ Concentration (skip)
7.11 Strategy for Solving Acid-Base Problems: A Summary