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Raffi Hovasapian

Raffi Hovasapian

Buffers, Part III

Slide Duration:

Table of Contents

I. Review
Naming Compounds

41m 24s

Intro
0:00
Periodic Table of Elements
0:15
Naming Compounds
3:13
Definition and Examples of Ions
3:14
Ionic (Symbol to Name): NaCl
5:23
Ionic (Name to Symbol): Calcium Oxide
7:58
Ionic - Polyatoms Anions: Examples
12:45
Ionic - Polyatoms Anions (Symbol to Name): KClO
14:50
Ionic - Polyatoms Anions (Name to Symbol): Potassium Phosphate
15:49
Ionic Compounds Involving Transition Metals (Symbol to Name): Co₂(CO₃)₃
20:48
Ionic Compounds Involving Transition Metals (Name to Symbol): Palladium 2 Acetate
22:44
Naming Covalent Compounds (Symbol to Name): CO
26:21
Naming Covalent Compounds (Name to Symbol): Nitrogen Trifluoride
27:34
Naming Covalent Compounds (Name to Symbol): Dichlorine Monoxide
27:57
Naming Acids Introduction
28:11
Naming Acids (Name to Symbol): Chlorous Acid
35:08
% Composition by Mass Example
37:38
Stoichiometry

37m 19s

Intro
0:00
Stoichiometry
0:25
Introduction to Stoichiometry
0:26
Example 1
5:03
Example 2
10:17
Example 3
15:09
Example 4
24:02
Example 5: Questions
28:11
Example 5: Part A - Limiting Reactant
30:30
Example 5: Part B
32:27
Example 5: Part C
35:00
II. Aqueous Reactions & Stoichiometry
Precipitation Reactions

31m 14s

Intro
0:00
Precipitation Reactions
0:53
Dissociation of ionic Compounds
0:54
Solubility Guidelines for ionic Compounds: Soluble Ionic Compounds
8:15
Solubility Guidelines for ionic Compounds: Insoluble ionic Compounds
12:56
Precipitation Reactions
14:08
Example 1: Mixing a Solution of BaCl₂ & K₂SO₄
21:21
Example 2: Mixing a Solution of Mg(NO₃)₂ & KI
26:10
Acid-Base Reactions

43m 21s

Intro
0:00
Acid-Base Reactions
1:00
Introduction to Acid: Monoprotic Acid and Polyprotic Acid
1:01
Introduction to Base
8:28
Neutralization
11:45
Example 1
16:17
Example 2
21:55
Molarity
24:50
Example 3
26:50
Example 4
30:01
Example 4: Limiting Reactant
37:51
Example 4: Reaction Part
40:01
Oxidation Reduction Reactions

47m 58s

Intro
0:00
Oxidation Reduction Reactions
0:26
Oxidation and Reduction Overview
0:27
How Can One Tell Whether Oxidation-Reduction has Taken Place?
7:13
Rules for Assigning Oxidation State: Number 1
11:22
Rules for Assigning Oxidation State: Number 2
12:46
Rules for Assigning Oxidation State: Number 3
13:25
Rules for Assigning Oxidation State: Number 4
14:50
Rules for Assigning Oxidation State: Number 5
15:41
Rules for Assigning Oxidation State: Number 6
17:00
Example 1: Determine the Oxidation State of Sulfur in the Following Compounds
18:20
Activity Series and Reduction Properties
25:32
Activity Series and Reduction Properties
25:33
Example 2: Write the Balance Molecular, Total Ionic, and Net Ionic Equations for Al + HCl
31:37
Example 3
34:25
Example 4
37:55
Stoichiometry Examples

31m 50s

Intro
0:00
Stoichiometry Example 1
0:36
Example 1: Question and Answer
0:37
Stoichiometry Example 2
6:57
Example 2: Questions
6:58
Example 2: Part A Solution
12:16
Example 2: Part B Solution
13:05
Example 2: Part C Solution
14:00
Example 2: Part D Solution
14:38
Stoichiometry Example 3
17:56
Example 3: Questions
17:57
Example 3: Part A Solution
19:51
Example 3: Part B Solution
21:43
Example 3: Part C Solution
26:46
III. Gases
Pressure, Gas Laws, & The Ideal Gas Equation

49m 40s

Intro
0:00
Pressure
0:22
Pressure Overview
0:23
Torricelli: Barometer
4:35
Measuring Gas Pressure in a Container
7:49
Boyle's Law
12:40
Example 1
16:56
Gas Laws
21:18
Gas Laws
21:19
Avogadro's Law
26:16
Example 2
31:47
Ideal Gas Equation
38:20
Standard Temperature and Pressure (STP)
38:21
Example 3
40:43
Partial Pressure, Mol Fraction, & Vapor Pressure

32m

Intro
0:00
Gases
0:27
Gases
0:28
Mole Fractions
5:52
Vapor Pressure
8:22
Example 1
13:25
Example 2
22:45
Kinetic Molecular Theory and Real Gases

31m 58s

Intro
0:00
Kinetic Molecular Theory and Real Gases
0:45
Kinetic Molecular Theory 1
0:46
Kinetic Molecular Theory 2
4:23
Kinetic Molecular Theory 3
5:42
Kinetic Molecular Theory 4
6:27
Equations
7:52
Effusion
11:15
Diffusion
13:30
Example 1
19:54
Example 2
23:23
Example 3
26:45
AP Practice for Gases

25m 34s

Intro
0:00
Example 1
0:34
Example 1
0:35
Example 2
6:15
Example 2: Part A
6:16
Example 2: Part B
8:46
Example 2: Part C
10:30
Example 2: Part D
11:15
Example 2: Part E
12:20
Example 2: Part F
13:22
Example 3
14:45
Example 3
14:46
Example 4
18:16
Example 4
18:17
Example 5
21:04
Example 5
21:05
IV. Thermochemistry
Energy, Heat, and Work

37m 32s

Intro
0:00
Thermochemistry
0:25
Temperature and Heat
0:26
Work
3:07
System, Surroundings, Exothermic Process, and Endothermic Process
8:19
Work & Gas: Expansion and Compression
16:30
Example 1
24:41
Example 2
27:47
Example 3
31:58
Enthalpy & Hess's Law

32m 34s

Intro
0:00
Thermochemistry
1:43
Defining Enthalpy & Hess's Law
1:44
Example 1
6:48
State Function
13:11
Example 2
17:15
Example 3
24:09
Standard Enthalpies of Formation

23m 9s

Intro
0:00
Thermochemistry
1:04
Standard Enthalpy of Formation: Definition & Equation
1:05
∆H of Formation
10:00
Example 1
11:22
Example 2
19:00
Calorimetry

39m 28s

Intro
0:00
Thermochemistry
0:21
Heat Capacity
0:22
Molar Heat Capacity
4:44
Constant Pressure Calorimetry
5:50
Example 1
12:24
Constant Volume Calorimetry
21:54
Example 2
24:40
Example 3
31:03
V. Kinetics
Reaction Rates and Rate Laws

36m 24s

Intro
0:00
Kinetics
2:18
Rate: 2 NO₂ (g) → 2NO (g) + O₂ (g)
2:19
Reaction Rates Graph
7:25
Time Interval & Average Rate
13:13
Instantaneous Rate
15:13
Rate of Reaction is Proportional to Some Power of the Reactant Concentrations
23:49
Example 1
27:19
Method of Initial Rates

30m 48s

Intro
0:00
Kinetics
0:33
Rate
0:34
Idea
2:24
Example 1: NH₄⁺ + NO₂⁻ → NO₂ (g) + 2 H₂O
5:36
Example 2: BrO₃⁻ + 5 Br⁻ + 6 H⁺ → 3 Br₂ + 3 H₂O
19:29
Integrated Rate Law & Reaction Half-Life

32m 17s

Intro
0:00
Kinetics
0:52
Integrated Rate Law
0:53
Example 1
6:26
Example 2
15:19
Half-life of a Reaction
20:40
Example 3: Part A
25:41
Example 3: Part B
28:01
Second Order & Zero-Order Rate Laws

26m 40s

Intro
0:00
Kinetics
0:22
Second Order
0:23
Example 1
6:08
Zero-Order
16:36
Summary for the Kinetics Associated with the Reaction
21:27
Activation Energy & Arrhenius Equation

40m 59s

Intro
0:00
Kinetics
0:53
Rate Constant
0:54
Collision Model
2:45
Activation Energy
5:11
Arrhenius Proposed
9:54
2 Requirements for a Successful Reaction
15:39
Rate Constant
17:53
Arrhenius Equation
19:51
Example 1
25:00
Activation Energy & the Values of K
32:12
Example 2
36:46
AP Practice for Kinetics

29m 8s

Intro
0:00
Kinetics
0:43
Example 1
0:44
Example 2
6:53
Example 3
8:58
Example 4
11:36
Example 5
16:36
Example 6: Part A
21:00
Example 6: Part B
25:09
VI. Equilibrium
Equilibrium, Part 1

46m

Intro
0:00
Equilibrium
1:32
Introduction to Equilibrium
1:33
Equilibrium Rules
14:00
Example 1: Part A
16:46
Example 1: Part B
18:48
Example 1: Part C
22:13
Example 1: Part D
24:55
Example 2: Part A
27:46
Example 2: Part B
31:22
Example 2: Part C
33:00
Reverse a Reaction
36:04
Example 3
37:24
Equilibrium, Part 2

40m 53s

Intro
0:00
Equilibrium
1:31
Equilibriums Involving Gases
1:32
General Equation
10:11
Example 1: Question
11:55
Example 1: Answer
13:43
Example 2: Question
19:08
Example 2: Answer
21:37
Example 3: Question
33:40
Example 3: Answer
35:24
Equilibrium: Reaction Quotient

45m 53s

Intro
0:00
Equilibrium
0:57
Reaction Quotient
0:58
If Q > K
5:37
If Q < K
6:52
If Q = K
7:45
Example 1: Part A
8:24
Example 1: Part B
13:11
Example 2: Question
20:04
Example 2: Answer
22:15
Example 3: Question
30:54
Example 3: Answer
32:52
Steps in Solving Equilibrium Problems
42:40
Equilibrium: Examples

31m 51s

Intro
0:00
Equilibrium
1:09
Example 1: Question
1:10
Example 1: Answer
4:15
Example 2: Question
13:04
Example 2: Answer
15:20
Example 3: Question
25:03
Example 3: Answer
26:32
Le Chatelier's principle & Equilibrium

40m 52s

Intro
0:00
Le Chatelier
1:05
Le Chatelier Principle
1:06
Concentration: Add 'x'
5:25
Concentration: Subtract 'x'
7:50
Example 1
9:44
Change in Pressure
12:53
Example 2
20:40
Temperature: Exothermic and Endothermic
24:33
Example 3
29:55
Example 4
35:30
VII. Acids & Bases
Acids and Bases

50m 11s

Intro
0:00
Acids and Bases
1:14
Bronsted-Lowry Acid-Base Model
1:28
Reaction of an Acid with Water
4:36
Acid Dissociation
10:51
Acid Strength
13:48
Example 1
21:22
Water as an Acid & a Base
25:25
Example 2: Part A
32:30
Example 2: Part B
34:47
Example 3: Part A
35:58
Example 3: Part B
39:33
pH Scale
41:12
Example 4
43:56
pH of Weak Acid Solutions

43m 52s

Intro
0:00
pH of Weak Acid Solutions
1:12
pH of Weak Acid Solutions
1:13
Example 1
6:26
Example 2
14:25
Example 3
24:23
Example 4
30:38
Percent Dissociation: Strong & Weak Bases

43m 4s

Intro
0:00
Bases
0:33
Percent Dissociation: Strong & Weak Bases
0:45
Example 1
6:23
Strong Base Dissociation
11:24
Example 2
13:02
Weak Acid and General Reaction
17:38
Example: NaOH → Na⁺ + OH⁻
20:30
Strong Base and Weak Base
23:49
Example 4
24:54
Example 5
33:51
Polyprotic Acids

35m 34s

Intro
0:00
Polyprotic Acids
1:04
Acids Dissociation
1:05
Example 1
4:51
Example 2
17:30
Example 3
31:11
Salts and Their Acid-Base Properties

41m 14s

Intro
0:00
Salts and Their Acid-Base Properties
0:11
Salts and Their Acid-Base Properties
0:15
Example 1
7:58
Example 2
14:00
Metal Ion and Acidic Solution
22:00
Example 3
28:35
NH₄F → NH₄⁺ + F⁻
34:05
Example 4
38:03
Common Ion Effect & Buffers

41m 58s

Intro
0:00
Common Ion Effect & Buffers
1:16
Covalent Oxides Produce Acidic Solutions in Water
1:36
Ionic Oxides Produce Basic Solutions in Water
4:15
Practice Example 1
6:10
Practice Example 2
9:00
Definition
12:27
Example 1: Part A
16:49
Example 1: Part B
19:54
Buffer Solution
25:10
Example of Some Buffers: HF and NaF
30:02
Example of Some Buffers: Acetic Acid & Potassium Acetate
31:34
Example of Some Buffers: CH₃NH₂ & CH₃NH₃Cl
33:54
Example 2: Buffer Solution
36:36
Buffer

32m 24s

Intro
0:00
Buffers
1:20
Buffer Solution
1:21
Adding Base
5:03
Adding Acid
7:14
Example 1: Question
9:48
Example 1: Recall
12:08
Example 1: Major Species Upon Addition of NaOH
16:10
Example 1: Equilibrium, ICE Chart, and Final Calculation
24:33
Example 1: Comparison
29:19
Buffers, Part II

40m 6s

Intro
0:00
Buffers
1:27
Example 1: Question
1:32
Example 1: ICE Chart
3:15
Example 1: Major Species Upon Addition of OH⁻, But Before Rxn
7:23
Example 1: Equilibrium, ICE Chart, and Final Calculation
12:51
Summary
17:21
Another Look at Buffering & the Henderson-Hasselbalch equation
19:00
Example 2
27:08
Example 3
32:01
Buffers, Part III

38m 43s

Intro
0:00
Buffers
0:25
Buffer Capacity Part 1
0:26
Example 1
4:10
Buffer Capacity Part 2
19:29
Example 2
25:12
Example 3
32:02
Titrations: Strong Acid and Strong Base

42m 42s

Intro
0:00
Titrations: Strong Acid and Strong Base
1:11
Definition of Titration
1:12
Sample Problem
3:33
Definition of Titration Curve or pH Curve
9:46
Scenario 1: Strong Acid- Strong Base Titration
11:00
Question
11:01
Part 1: No NaOH is Added
14:00
Part 2: 10.0 mL of NaOH is Added
15:50
Part 3: Another 10.0 mL of NaOH & 20.0 mL of NaOH are Added
22:19
Part 4: 50.0 mL of NaOH is Added
26:46
Part 5: 100.0 mL (Total) of NaOH is Added
27:26
Part 6: 150.0 mL (Total) of NaOH is Added
32:06
Part 7: 200.0 mL of NaOH is Added
35:07
Titrations Curve for Strong Acid and Strong Base
35:43
Titrations: Weak Acid and Strong Base

42m 3s

Intro
0:00
Titrations: Weak Acid and Strong Base
0:43
Question
0:44
Part 1: No NaOH is Added
1:54
Part 2: 10.0 mL of NaOH is Added
5:17
Part 3: 25.0 mL of NaOH is Added
14:01
Part 4: 40.0 mL of NaOH is Added
21:55
Part 5: 50.0 mL (Total) of NaOH is Added
22:25
Part 6: 60.0 mL (Total) of NaOH is Added
31:36
Part 7: 75.0 mL (Total) of NaOH is Added
35:44
Titration Curve
36:09
Titration Examples & Acid-Base Indicators

52m 3s

Intro
0:00
Examples and Indicators
0:25
Example 1: Question
0:26
Example 1: Solution
2:03
Example 2: Question
12:33
Example 2: Solution
14:52
Example 3: Question
23:45
Example 3: Solution
25:09
Acid/Base Indicator Overview
34:45
Acid/Base Indicator Example
37:40
Acid/Base Indicator General Result
47:11
Choosing Acid/Base Indicator
49:12
VIII. Solubility
Solubility Equilibria

36m 25s

Intro
0:00
Solubility Equilibria
0:48
Solubility Equilibria Overview
0:49
Solubility Product Constant
4:24
Definition of Solubility
9:10
Definition of Solubility Product
11:28
Example 1
14:09
Example 2
20:19
Example 3
27:30
Relative Solubilities
31:04
Solubility Equilibria, Part II

42m 6s

Intro
0:00
Solubility Equilibria
0:46
Common Ion Effect
0:47
Example 1
3:14
pH & Solubility
13:00
Example of pH & Solubility
15:25
Example 2
23:06
Precipitation & Definition of the Ion Product
26:48
If Q > Ksp
29:31
If Q < Ksp
30:27
Example 3
32:58
Solubility Equilibria, Part III

43m 9s

Intro
0:00
Solubility Equilibria
0:55
Example 1: Question
0:56
Example 1: Step 1 - Check to See if Anything Precipitates
2:52
Example 1: Step 2 - Stoichiometry
10:47
Example 1: Step 3 - Equilibrium
16:34
Example 2: Selective Precipitation (Question)
21:02
Example 2: Solution
23:41
Classical Qualitative Analysis
29:44
Groups: 1-5
38:44
IX. Complex Ions
Complex Ion Equilibria

43m 38s

Intro
0:00
Complex Ion Equilibria
0:32
Complex Ion
0:34
Ligan Examples
1:51
Ligand Definition
3:12
Coordination
6:28
Example 1
8:08
Example 2
19:13
Complex Ions & Solubility

31m 30s

Intro
0:00
Complex Ions and Solubility
0:23
Recall: Classical Qualitative Analysis
0:24
Example 1
6:10
Example 2
16:16
Dissolving a Water-Insoluble Ionic Compound: Method 1
23:38
Dissolving a Water-Insoluble Ionic Compound: Method 2
28:13
X. Chemical Thermodynamics
Spontaneity, Entropy, & Free Energy, Part I

56m 28s

Intro
0:00
Spontaneity, Entropy, Free Energy
2:25
Energy Overview
2:26
Equation: ∆E = q + w
4:30
State Function/ State Property
8:35
Equation: w = -P∆V
12:00
Enthalpy: H = E + PV
14:50
Enthalpy is a State Property
17:33
Exothermic and Endothermic Reactions
19:20
First Law of Thermodynamic
22:28
Entropy
25:48
Spontaneous Process
33:53
Second Law of Thermodynamic
36:51
More on Entropy
42:23
Example
43:55
Spontaneity, Entropy, & Free Energy, Part II

39m 55s

Intro
0:00
Spontaneity, Entropy, Free Energy
1:30
∆S of Universe = ∆S of System + ∆S of Surrounding
1:31
Convention
3:32
Examining a System
5:36
Thermodynamic Property: Sign of ∆S
16:52
Thermodynamic Property: Magnitude of ∆S
18:45
Deriving Equation: ∆S of Surrounding = -∆H / T
20:25
Example 1
25:51
Free Energy Equations
29:22
Spontaneity, Entropy, & Free Energy, Part III

30m 10s

Intro
0:00
Spontaneity, Entropy, Free Energy
0:11
Example 1
2:38
Key Concept of Example 1
14:06
Example 2
15:56
Units for ∆H, ∆G, and S
20:56
∆S of Surrounding & ∆S of System
22:00
Reaction Example
24:17
Example 3
26:52
Spontaneity, Entropy, & Free Energy, Part IV

30m 7s

Intro
0:00
Spontaneity, Entropy, Free Energy
0:29
Standard Free Energy of Formation
0:58
Example 1
4:34
Reaction Under Non-standard Conditions
13:23
Example 2
16:26
∆G = Negative
22:12
∆G = 0
24:38
Diagram Example of ∆G
26:43
Spontaneity, Entropy, & Free Energy, Part V

44m 56s

Intro
0:00
Spontaneity, Entropy, Free Energy
0:56
Equations: ∆G of Reaction, ∆G°, and K
0:57
Example 1: Question
6:50
Example 1: Part A
9:49
Example 1: Part B
15:28
Example 2
17:33
Example 3
23:31
lnK = (- ∆H° ÷ R) ( 1 ÷ T) + ( ∆S° ÷ R)
31:36
Maximum Work
35:57
XI. Electrochemistry
Oxidation-Reduction & Balancing

39m 23s

Intro
0:00
Oxidation-Reduction and Balancing
2:06
Definition of Electrochemistry
2:07
Oxidation and Reduction Review
3:05
Example 1: Assigning Oxidation State
10:15
Example 2: Is the Following a Redox Reaction?
18:06
Example 3: Step 1 - Write the Oxidation & Reduction Half Reactions
22:46
Example 3: Step 2 - Balance the Reaction
26:44
Example 3: Step 3 - Multiply
30:11
Example 3: Step 4 - Add
32:07
Example 3: Step 5 - Check
33:29
Galvanic Cells

43m 9s

Intro
0:00
Galvanic Cells
0:39
Example 1: Balance the Following Under Basic Conditions
0:40
Example 1: Steps to Balance Reaction Under Basic Conditions
3:25
Example 1: Solution
5:23
Example 2: Balance the Following Reaction
13:56
Galvanic Cells
18:15
Example 3: Galvanic Cells
28:19
Example 4: Galvanic Cells
35:12
Cell Potential

48m 41s

Intro
0:00
Cell Potential
2:08
Definition of Cell Potential
2:17
Symbol and Unit
5:50
Standard Reduction Potential
10:16
Example Figure 1
13:08
Example Figure 2
19:00
All Reduction Potentials are Written as Reduction
23:10
Cell Potential: Important Fact 1
26:49
Cell Potential: Important Fact 2
27:32
Cell Potential: Important Fact 3
28:54
Cell Potential: Important Fact 4
30:05
Example Problem 1
32:29
Example Problem 2
38:38
Potential, Work, & Free Energy

41m 23s

Intro
0:00
Potential, Work, Free Energy
0:42
Descriptions of Galvanic Cell
0:43
Line Notation
5:33
Example 1
6:26
Example 2
11:15
Example 3
15:18
Equation: Volt
22:20
Equations: Cell Potential, Work, and Charge
28:30
Maximum Cell Potential is Related to the Free Energy of the Cell Reaction
35:09
Example 4
37:42
Cell Potential & Concentration

34m 19s

Intro
0:00
Cell Potential & Concentration
0:29
Example 1: Question
0:30
Example 1: Nernst Equation
4:43
Example 1: Solution
7:01
Cell Potential & Concentration
11:27
Example 2
16:38
Manipulating the Nernst Equation
25:15
Example 3
28:43
Electrolysis

33m 21s

Intro
0:00
Electrolysis
3:16
Electrolysis: Part 1
3:17
Electrolysis: Part 2
5:25
Galvanic Cell Example
7:13
Nickel Cadmium Battery
12:18
Ampere
16:00
Example 1
20:47
Example 2
25:47
XII. Light
Light

44m 45s

Intro
0:00
Light
2:14
Introduction to Light
2:15
Frequency, Speed, and Wavelength of Waves
3:58
Units and Equations
7:37
Electromagnetic Spectrum
12:13
Example 1: Calculate the Frequency
17:41
E = hν
21:30
Example 2: Increment of Energy
25:12
Photon Energy of Light
28:56
Wave and Particle
31:46
Example 3: Wavelength of an Electron
34:46
XIII. Quantum Mechanics
Quantum Mechanics & Electron Orbitals

54m

Intro
0:00
Quantum Mechanics & Electron Orbitals
0:51
Quantum Mechanics & Electron Orbitals Overview
0:52
Electron Orbital and Energy Levels for the Hydrogen Atom
8:47
Example 1
13:41
Quantum Mechanics: Schrodinger Equation
19:19
Quantum Numbers Overview
31:10
Principal Quantum Numbers
33:28
Angular Momentum Numbers
34:55
Magnetic Quantum Numbers
36:35
Spin Quantum Numbers
37:46
Primary Level, Sublevels, and Sub-Sub-Levels
39:42
Example
42:17
Orbital & Quantum Numbers
49:32
Electron Configurations & Diagrams

34m 4s

Intro
0:00
Electron Configurations & Diagrams
1:08
Electronic Structure of Ground State Atom
1:09
Order of Electron Filling
3:50
Electron Configurations & Diagrams: H
8:41
Electron Configurations & Diagrams: He
9:12
Electron Configurations & Diagrams: Li
9:47
Electron Configurations & Diagrams: Be
11:17
Electron Configurations & Diagrams: B
12:05
Electron Configurations & Diagrams: C
13:03
Electron Configurations & Diagrams: N
14:55
Electron Configurations & Diagrams: O
15:24
Electron Configurations & Diagrams: F
16:25
Electron Configurations & Diagrams: Ne
17:00
Electron Configurations & Diagrams: S
18:08
Electron Configurations & Diagrams: Fe
20:08
Introduction to Valence Electrons
23:04
Valence Electrons of Oxygen
23:44
Valence Electrons of Iron
24:02
Valence Electrons of Arsenic
24:30
Valence Electrons: Exceptions
25:36
The Periodic Table
27:52
XIV. Intermolecular Forces
Vapor Pressure & Changes of State

52m 43s

Intro
0:00
Vapor Pressure and Changes of State
2:26
Intermolecular Forces Overview
2:27
Hydrogen Bonding
5:23
Heat of Vaporization
9:58
Vapor Pressure: Definition and Example
11:04
Vapor Pressures is Mostly a Function of Intermolecular Forces
17:41
Vapor Pressure Increases with Temperature
20:52
Vapor Pressure vs. Temperature: Graph and Equation
22:55
Clausius-Clapeyron Equation
31:55
Example 1
32:13
Heating Curve
35:40
Heat of Fusion
41:31
Example 2
43:45
Phase Diagrams & Solutions

31m 17s

Intro
0:00
Phase Diagrams and Solutions
0:22
Definition of a Phase Diagram
0:50
Phase Diagram Part 1: H₂O
1:54
Phase Diagram Part 2: CO₂
9:59
Solutions: Solute & Solvent
16:12
Ways of Discussing Solution Composition: Mass Percent or Weight Percent
18:46
Ways of Discussing Solution Composition: Molarity
20:07
Ways of Discussing Solution Composition: Mole Fraction
20:48
Ways of Discussing Solution Composition: Molality
21:41
Example 1: Question
22:06
Example 1: Mass Percent
24:32
Example 1: Molarity
25:53
Example 1: Mole Fraction
28:09
Example 1: Molality
29:36
Vapor Pressure of Solutions

37m 23s

Intro
0:00
Vapor Pressure of Solutions
2:07
Vapor Pressure & Raoult's Law
2:08
Example 1
5:21
When Ionic Compounds Dissolve
10:51
Example 2
12:38
Non-Ideal Solutions
17:42
Negative Deviation
24:23
Positive Deviation
29:19
Example 3
31:40
Colligatives Properties

34m 11s

Intro
0:00
Colligative Properties
1:07
Boiling Point Elevation
1:08
Example 1: Question
5:19
Example 1: Solution
6:52
Freezing Point Depression
12:01
Example 2: Question
14:46
Example 2: Solution
16:34
Osmotic Pressure
20:20
Example 3: Question
28:00
Example 3: Solution
30:16
XV. Bonding
Bonding & Lewis Structure

48m 39s

Intro
0:00
Bonding & Lewis Structure
2:23
Covalent Bond
2:24
Single Bond, Double Bond, and Triple Bond
4:11
Bond Length & Intermolecular Distance
5:51
Definition of Electronegativity
8:42
Bond Polarity
11:48
Bond Energy
20:04
Example 1
24:31
Definition of Lewis Structure
31:54
Steps in Forming a Lewis Structure
33:26
Lewis Structure Example: H₂
36:53
Lewis Structure Example: CH₄
37:33
Lewis Structure Example: NO⁺
38:43
Lewis Structure Example: PCl₅
41:12
Lewis Structure Example: ICl₄⁻
43:05
Lewis Structure Example: BeCl₂
45:07
Resonance & Formal Charge

36m 59s

Intro
0:00
Resonance and Formal Charge
0:09
Resonance Structures of NO₃⁻
0:25
Resonance Structures of NO₂⁻
12:28
Resonance Structures of HCO₂⁻
16:28
Formal Charge
19:40
Formal Charge Example: SO₄²⁻
21:32
Formal Charge Example: CO₂
31:33
Formal Charge Example: HCN
32:44
Formal Charge Example: CN⁻
33:34
Formal Charge Example: 0₃
34:43
Shapes of Molecules

41m 21s

Intro
0:00
Shapes of Molecules
0:35
VSEPR
0:36
Steps in Determining Shapes of Molecules
6:18
Linear
11:38
Trigonal Planar
11:55
Tetrahedral
12:45
Trigonal Bipyramidal
13:23
Octahedral
14:29
Table: Shapes of Molecules
15:40
Example: CO₂
21:11
Example: NO₃⁻
24:01
Example: H₂O
27:00
Example: NH₃
29:48
Example: PCl₃⁻
32:18
Example: IF₄⁺
34:38
Example: KrF₄
37:57
Hybrid Orbitals

40m 17s

Intro
0:00
Hybrid Orbitals
0:13
Introduction to Hybrid Orbitals
0:14
Electron Orbitals for CH₄
5:02
sp³ Hybridization
10:52
Example: sp³ Hybridization
12:06
sp² Hybridization
14:21
Example: sp² Hybridization
16:11
σ Bond
19:10
π Bond
20:07
sp Hybridization & Example
22:00
dsp³ Hybridization & Example
27:36
d²sp³ Hybridization & Example
30:36
Example: Predict the Hybridization and Describe the Molecular Geometry of CO
32:31
Example: Predict the Hybridization and Describe the Molecular Geometry of BF₄⁻
35:17
Example: Predict the Hybridization and Describe the Molecular Geometry of XeF₂
37:09
XVI. AP Practice Exam
AP Practice Exam: Multiple Choice, Part I

52m 34s

Intro
0:00
Multiple Choice
1:21
Multiple Choice 1
1:22
Multiple Choice 2
2:23
Multiple Choice 3
3:38
Multiple Choice 4
4:34
Multiple Choice 5
5:16
Multiple Choice 6
5:41
Multiple Choice 7
6:20
Multiple Choice 8
7:03
Multiple Choice 9
7:31
Multiple Choice 10
9:03
Multiple Choice 11
11:52
Multiple Choice 12
13:16
Multiple Choice 13
13:56
Multiple Choice 14
14:52
Multiple Choice 15
15:43
Multiple Choice 16
16:20
Multiple Choice 17
16:55
Multiple Choice 18
17:22
Multiple Choice 19
18:59
Multiple Choice 20
20:24
Multiple Choice 21
22:20
Multiple Choice 22
23:29
Multiple Choice 23
24:30
Multiple Choice 24
25:24
Multiple Choice 25
26:21
Multiple Choice 26
29:06
Multiple Choice 27
30:42
Multiple Choice 28
33:28
Multiple Choice 29
34:38
Multiple Choice 30
35:37
Multiple Choice 31
37:31
Multiple Choice 32
38:28
Multiple Choice 33
39:50
Multiple Choice 34
42:57
Multiple Choice 35
44:18
Multiple Choice 36
45:52
Multiple Choice 37
48:02
Multiple Choice 38
49:25
Multiple Choice 39
49:43
Multiple Choice 40
50:16
Multiple Choice 41
50:49
AP Practice Exam: Multiple Choice, Part II

32m 15s

Intro
0:00
Multiple Choice
0:12
Multiple Choice 42
0:13
Multiple Choice 43
0:33
Multiple Choice 44
1:16
Multiple Choice 45
2:36
Multiple Choice 46
5:22
Multiple Choice 47
6:35
Multiple Choice 48
8:02
Multiple Choice 49
10:05
Multiple Choice 50
10:26
Multiple Choice 51
11:07
Multiple Choice 52
12:01
Multiple Choice 53
12:55
Multiple Choice 54
16:12
Multiple Choice 55
18:11
Multiple Choice 56
19:45
Multiple Choice 57
20:15
Multiple Choice 58
23:28
Multiple Choice 59
24:27
Multiple Choice 60
26:45
Multiple Choice 61
29:15
AP Practice Exam: Multiple Choice, Part III

32m 50s

Intro
0:00
Multiple Choice
0:16
Multiple Choice 62
0:17
Multiple Choice 63
1:57
Multiple Choice 64
6:16
Multiple Choice 65
8:05
Multiple Choice 66
9:18
Multiple Choice 67
10:38
Multiple Choice 68
12:51
Multiple Choice 69
14:32
Multiple Choice 70
17:35
Multiple Choice 71
22:44
Multiple Choice 72
24:27
Multiple Choice 73
27:46
Multiple Choice 74
29:39
Multiple Choice 75
30:23
AP Practice Exam: Free response Part I

47m 22s

Intro
0:00
Free Response
0:15
Free Response 1: Part A
0:16
Free Response 1: Part B
4:15
Free Response 1: Part C
5:47
Free Response 1: Part D
9:20
Free Response 1: Part E. i
10:58
Free Response 1: Part E. ii
16:45
Free Response 1: Part E. iii
26:03
Free Response 2: Part A. i
31:01
Free Response 2: Part A. ii
33:38
Free Response 2: Part A. iii
35:20
Free Response 2: Part B. i
37:38
Free Response 2: Part B. ii
39:30
Free Response 2: Part B. iii
44:44
AP Practice Exam: Free Response Part II

43m 5s

Intro
0:00
Free Response
0:12
Free Response 3: Part A
0:13
Free Response 3: Part B
6:25
Free Response 3: Part C. i
11:33
Free Response 3: Part C. ii
12:02
Free Response 3: Part D
14:30
Free Response 4: Part A
21:03
Free Response 4: Part B
22:59
Free Response 4: Part C
24:33
Free Response 4: Part D
27:22
Free Response 4: Part E
28:43
Free Response 4: Part F
29:35
Free Response 4: Part G
30:15
Free Response 4: Part H
30:48
Free Response 5: Diagram
32:00
Free Response 5: Part A
34:14
Free Response 5: Part B
36:07
Free Response 5: Part C
37:45
Free Response 5: Part D
39:00
Free Response 5: Part E
40:26
AP Practice Exam: Free Response Part III

28m 36s

Intro
0:00
Free Response
0:43
Free Response 6: Part A. i
0:44
Free Response 6: Part A. ii
3:08
Free Response 6: Part A. iii
5:02
Free Response 6: Part B. i
7:11
Free Response 6: Part B. ii
9:40
Free Response 7: Part A
11:14
Free Response 7: Part B
13:45
Free Response 7: Part C
15:43
Free Response 7: Part D
16:54
Free Response 8: Part A. i
19:15
Free Response 8: Part A. ii
21:16
Free Response 8: Part B. i
23:51
Free Response 8: Part B. ii
25:07
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Lecture Comments (8)

0 answers

Post by Bhupen Khanolkar on May 16, 2015

Dear Prof. Hovasapian,

Thank you very much for your help in this concept. I was a little confused on how you did Example 3 of this lecture. When calculating the molarity of NaPr, wouldn't you have to find the moles of Pr- that come from the dissociation of HPr, and subtract that from the total Pr- concentration to give the moles of Pr- that come from the dissociation of NaPr, and then find the molarity? When I solved it this way, I found the concentration of Pr- (from HPr)= 6.43*10^(-4). The concentration of Pr- (from NaPr) was 0.09691 mol, and dividing that by 0.250L gave me a molarity of 0.387. I was wondering if this method was correct.

Please let me know soon.

Thank you very much

1 answer

Last reply by: Professor Hovasapian
Wed Feb 4, 2015 10:08 PM

Post by Ravi Bala on February 4, 2015

Hello Prof. Raffi,

I really feel like I understand the concept behind buffers very well. I think it's for this reason why I don't really like using the henderson equation. However, I just want to make sure that I can always avoid using that equation when calculating for pH, or are there certain situations in where the problem can only be solved using that equation?

Also I am just wondering, considering we won't get a calculator for the AP chem multiple choice, what will the acid base questions look like on there?

1 answer

Last reply by: Professor Hovasapian
Thu Oct 30, 2014 1:26 AM

Post by Long Tran on October 28, 2014

Hi Prof
on the example 1 calculating STOICH part a, the changing of H+ is 0. could you explain ?

0 answers

Post by Chemutai Shiow on March 23, 2014

In Example 3, you use 150 L instead of 150 mL(what was in the question) which threw of the math but the concept was the same. No biggie

1 answer

Last reply by: Professor Hovasapian
Wed Nov 7, 2012 1:38 PM

Post by noha nasser on November 7, 2012

Hello prof Raffi,
im a little confused here about the ka of the HAc in example 1 at the begining you gave us = 1.8x10(-5) and then in the solution you used 4.74 where did the 4.74 come from? :))

Buffers, Part III

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Buffers 0:25
    • Buffer Capacity Part 1
    • Example 1
    • Buffer Capacity Part 2
    • Example 2
    • Example 3

Transcription: Buffers, Part III

Hello, and welcome back to Educator.com; welcome back to AP Chemistry.0000

Today, we are going to close out our discussion of buffering.0003

I know that we spent a fair amount of time on this, but it is important; it is an important application of acid-base chemistry, and so, today, we are going to round it out.0007

We are going to discuss something called buffer capacity and do some applications of buffer capacity.0016

Let's just start with a definition, and go forward from there.0023

Buffer capacity: buffer capacity is the amount of H+ or OH- a buffer solution can absorb without significant change in pH.0029

So again, you can create a buffer solution in any way you like, but the idea is: the more of the components in that buffer that you have that can actually swallow up any excess hydroxide or hydrogen ion, the greater your buffer capacity.0074

That means it has a greater capacity to actually buffer; it gives you a broader range of lack of change; that is the whole idea--that is the whole idea behind what a buffer is.0090

A buffer gives you a range where things can happen without anything changing in the overall system.0101

Recall: a buffer solution is HA, A-.0110

So, added base (which is OH-; anytime you add a base, you are essentially adding OH-) reacts with...these are the two things that are in the buffer.0129

Added base reacts with the HA; so HA essentially absorbs the OH- so it is not floating around freely in solution, changing the pH.0145

Here is how it does it: the HA reacts with the OH- to create A- + H2O.0179

It takes the OH-, and it pulls it out of solution; it pulls it out of solution as water, so that this floats around instead of the OH---that is what a buffer does.0190

Now, added H+ reacts with A-; so A- absorbs the H+, and it does so as follows.0199

All of these minuses and plusses...A- + H+ produces HA; any added acid is taken up by the A-, and it is locked away as the weak acid--it is locked away as the acid itself, the undissociated form, so you don't have any of this free hydrogen ion floating around, changing the pH.0224

Let's go ahead and do an example; I think it will be the best.0251

So, Example 1: Calculate the change in pH when 0.010 mol of gaseous hydrogen chloride is added to 1.0 liter of solution A, which is 4.0 Molar acetic acid, and a 4.0 Molar sodium acetate, and solution B is 0.040 Molar acetic acid and 0.040 Molar sodium acetate.0256

OK, so we have two buffer solutions: the first buffer solution is made with 4 Molar of the acetic acid and 4 Molar sodium acetate; another buffer solution is made with .04 Molar acetic acid and .04 Molar sodium acetate.0321

To both of these solutions (separately), we bubble in .010 moles of gaseous hydrogen chloride.0338

Now, notice: I said hydrogen chloride, not hydrochloric acid; as a gas, hydrogen chloride is a molecule; and when you bubble it into water, it dissolves because it's a strong acid in aqueous solution, and then you free H+ and Cl.0348

So, gaseous hydrogen chloride is added to 1 liter of each of these solutions; what is the change in pH?0362

OK, so what we are investigating here is this thing called buffer capacity, in case you wanted to know what it is that we are actually doing.0368

Well, so let's go ahead and use the Henderson-Hasselbalch equation to actually calculate the pH of both of these solutions before the hydrogen chloride is added.0376

So, let's write the Henderson-Hasselbalch equation up here: pH is equal to pKa + the log of A- over HA, the concentration of the base part over the acid.0388

And now, we want to say the Ka of acetic acid, of HAc, is equal to 1.8x10-5 (I didn't write that in the previous page, but there it is).0407

OK, so, well, if we take solution A...so, solution A: the pH is equal to the pKa (so the negative log of 1.8x10-5, which is 4.74), plus the logarithm of...well, the concentration of the acetate is 4 Molar; the concentration of the acetic acid is 4 Molar; well, this is...4/4 is 1; what is the logarithm of 1?--it's 0.0419

It goes to 0; so the pH equals the pKa, which is 4.74; interesting.0453

How about solution B?0459

Well, pH equals pKa (the same thing: 4.74), plus the log (this time, it is the log of 0.040, over .040; .040 Molar sodium acetate, .040 Molar acetic acid).0465

Well, this is still 1; the logarithm of 1 is still 0; it hasn't changed; so again, notice, we have different molarities of these solutions, but the ratio is 1; the logarithm of 1 is 0; so both of these have a pH of 4.74.0482

It's kind of interesting, isn't it?0497

Now, we want to see which one of these has a greater capacity to absorb anything that we add, whether it's acid or base (in this case, we are adding acid, not base) and remain close to 4.74, or as close as possible.0499

OK, so now, we do our major species upon addition of H+, but before any reaction.0513

We are doing accounting here; we are accounting for what is inside the solution, so we can decide what is happening.0534

Well, we have the acetic acid; it's a weak acid, so it's mostly in this form; we have dropped in the acetate, so we also have this floating around.0539

We have H2O floating around, but that is the solvent; we have sodium--that is the sodium acetate; and we have the H+, and we have Cl-, because that is the hydrochloric acid--the hydrogen chloride--that we have bubbled into it; now it's dissociated, because it is a strong acid in aqueous solution, so now we can call it hydrochloric acid.0547

OK, now, let's go ahead and do the stoichiometry.0569

We always have to do the stoichiometry, right?--when we add an acid or a base to a buffer solution, we do the stoichiometry before we do the equilibrium part.0575

The Henderson-Hasselbalch equation is for equilibrium part; it's not for the stoichiometry--there is no shortcut for the stoichiometry; we simply have to do the stoichiometry.0592

OK, the stoichiometry is the following: you have this floating around in solution, and this is going to react with the best source of base here (the best source of base is either this or this).0600

It is not going to react with this--it is not going to combine with this; it just separated from that.0614

It is going to combine with this; these two are going to dominate the stoichiometry.0619

H is going to react completely with this according to the following: it is going to be: Ac- + H+ is going to form HAc.0623

We have a Before; we have the Change; and we have the After--stoichiometry, not equilibrium.0637

Well, before anything happens, the concentration of acetate is 4.0 molarity; the H+ concentration is 0.010, because that is how many moles were bubbled in, and we have 4.0 moles of the acetic acid.0647

Well, what change is going to take place?--well, all of this H is going to react with this completely; this reaction goes to completion--that is what the stoichiometry part of these buffer problems is.0677

You are assuming that this reaction of H + Ac- (or, if you add base, OH- plus the HAc) goes to completion, which means that all of this is going to be 0, which means that all of the H+ is going to be used up, and since this is 1:1, that means this is going to diminish by .010.0689

This is going to show up; so this is going to be +0.010.0714

Our final concentration, before the system comes to equilibrium, is going to be 3.99 moles; 0 moles; 4.01 moles.0719

These are moles, OK?--this is one liter of solution; that is why we had 4 Molar; but we are actually...because it's in one liter of solution, it's in moles.0737

OK, now let's see if we can...all right, now we are ready to do the equilibrium part, but we don't have to do the equilibrium part; we can use the Henderson-Hasselbalch equation--that is the equilibrium part.0747

Let us do that: we get: pH=pKa, plus the log of acetate concentration, over acetic acid concentration; pH = 4.74 plus the log of...now, we said it's moles, 3.99 moles, but we are in one liter, so the molarity is 3.99/1, so it's the same.0761

That is the nice thing about working with one liter of solution.0795

3.99: that is the acetate concentration; 4.01: that is the HAc concentration; it equals 4.74, plus log of (oh, you know what, I didn't even do the numbers; but actually, it doesn't matter; if you were to do this number, you would find that it's very, very small, so I'll just put a little box here for whichever number you get when you take the logarithm of) 3.99/4.01: you are going to find, to two significant figures, the answer is 4.74: no change in pH.0799

Very interesting: the reason there is no change in pH is because we had so much of the acetate and this, and we had it in a 1:1...the ratio here is 1, so that this changed very little down; this changed very little up; this ratio remained .996, .997, virtually 1.0842

That is why, as far as to two significant figures, there is virtually no change.0864

The pH didn't change here when we added .010 moles of acid to this solution.0868

.010 moles of acid is not...it sounds like a little, but it's actually a fair amount of acid; and yet, the pH didn't change at all--this has a high buffer capacity.0875

Now, let's do the same for the solution B.0885

Solution B: OK, well, we want to do the stoichiometry part.0891

Stoichiometry: again, we have acetate, plus any of the acid that we want to absorb, forming HAc; we have a Before; we have a Change; and we have an After.0900

This time, the Before concentration...number of moles; I'm sorry...the concentration is .040 moles per liter; we have one liter of the solution, so we have .040 moles.0920

We have 0.010 moles of the H+ floating around; and we have 0.040 moles, minus 0.010 (because this reaction is going to go to completion, so all of the H+ is going to be used up, which means that .010 moles is going to disappear from here); this is going to be -0.010, and this is going to be +0.010.0933

Here, we get 0.03; here, we get 0; here, we get 0.05; here we go.0965

These are how many moles of the acetate and the acetic acid that we have; well, again, since we are dealing with one liter of solution, it is just these things, divided by 1 liter, so the moles and the molarity is the same.0976

Now, let's go ahead and do pH=pKa, plus the log...actually, you know what, I'm just not going to rewrite the equation; you already know it by now, so I'm just going to write out pH is equal to 4.74, plus the logarithm of 0.03, divided by 0.05, and we end up with a pH of 4.52.0989

OK, now, in solution A, we went from a pH of 4.74 to a pH of 4.74.1022

In the second solution we had, we went from a pH of 4.74 to a pH of 4.52.1032

That is a reasonable change in pH under the circumstances, and the reason that this particular buffer solution did not regulate the change in pH was because it did not have as much buffer capacity as the other solution.1040

Buffer capacity means the amount of OH or H+ that can be absorbed without a significant change in pH.1058

This is significant: .52 to .74--that is .22 units.1065

Well, that means the buffering components, which are your base, your acetate, and your acetic acid--your acetate and your acetic acid--the amounts of these are so high, compared to what you add, that any change in these concentrations leaves this value close to 1.1069

But, the change here, even though it can act as a buffer--it is buffering here; it's not a problem--I mean, you are not getting a huge change--but now, the ratio has gone from 1 to .03 to .05; so let's take a look at what these ratios are.1088

This ratio here, for solution A...the ratio of A to HA is equal to 3.99 over 4.01, is 0.996; the ratio hasn't changed all that much from the 1.1105

Here, you have a ratio of 0.03 over 0.05, and the ratio is 0.6; it has gone from 1 (.04 over .04) to .03/.05; it has gone down to .6.1127

That is a 40% drop in the ratio; therefore, even though it is buffering, its capacity to buffer is not as good as this one, because you don't have as much of the acetate and the acetic acid--you have less of it.1143

Because you have less of it, the more of it that is used, it changes the ratio that much more of an amount.1159

That is what is going on here.1167

So, let me write this down (wow, it would be really, really nice if I could actually spell; OK): the greater concentration of buffering components A- and HA in solution gives a greater buffering capacity.1169

So, pH=pKa plus the logarithm of concentration of A-, over the concentration of HA; these are the components of the buffer, the weak acid and its conjugate base.1226

When there are enough of those--when there are enough of those to where they can actually absorb fair amounts of added acid or base, this value, this ratio, changes little.1244

When this ratio changes little, the pH changes little.1257

That is the whole idea behind buffer capacity.1261

OK, so let's close this out with a nice practical application of buffers.1264

In the lab, when preparing a buffer solution, you are choosing the pH you want beforehand; that is the whole idea.1270

So again, a buffer solution is a solution of a specific pH, prescribed, predetermined pH, that you want before you run your experiment--before you add acid or base to it.1306

Let's say you are running an experiment for a particular cell culture, and you need a pH of, I don't know, 4.6.1317

Who knows what it is?--4.6, 7.2, 3 point...you can choose any pH you want.1326

Well, the question is: OK, so let's say you have chosen a particular pH; how do you actually choose an acid and its conjugate base, or a base and a conjugate acid, that actually gives you the pH that you want?1332

Well, we are going to answer that question right now.1344

OK, so the question is: how do you choose which weak acid or base to use?1347

And the answer is: You choose the acid with a pKa as close as possible to your predetermined pH, to your desired pH.1375

And here is why--it's actually very, very obvious: pH equals pKa, plus the log of the base over the acid component; these are our concentrations.1406

We said that we want this number to change as little as possible--this ratio; so, when you add acid or base, these are going to...one is going to go up; one is going to go down.1418

Well, we want it to change as little as possible at our given pH; well, if we want this to change as little as possible, we want the ratio of this to be as close to 1 as possible, because we want as much of this and as much of this as possible.1433

We don't want something like 10 of this and 1 of this...you know, for every one of these, ten of these; or for every 1 of these, 10 of these, because then, any change that is made is going to get even bigger in those directions.1450

So, when this value--when this ratio--is equal to 1; well, when this ratio is equal to 1, it has the greatest buffer capacity and the greatest resistance to changes in pH, but a logarithm of 1 is 0.1461

Because the logarithm of 1 is 0, pH=pKa.1472

Therefore, if you can find an acid that has a pKa whose pH is close to that pKa, you can pretty much arrange it in such a way that these will always be close to 1.1477

That is your answer: if you want...the pH of the buffer solution that you want...you want to choose an acid that has a pKa as close to that pH as possible.1490

If you want a buffer solution of a pH of 3.6, you need to find an acid that has a pKa as close to 3.6 as possible.1500

So, let's go ahead and do an example, and I think it will all make sense.1512

All right, let's see: Example...I think this is going to be Example 2: OK, a chemist needs a solution buffered at 4.80, and can choose from the following.1520

OK, our first choice is monochloroacetic acid; it has a Ka of 1.35x10-3.1561

Our second choice is propanoic acid, and our Ka for that is 1.3x10-5; we saw it in a previous example.1582

Our third choice is benzoic acid...very common, actually, benzoic acid...most of you probably use something called...when you put sunscreen on, that is actually paraamina benzoic acid; that is benzoic acid that has had some amino group attached to it.1595

That is what you are actually putting on; that is what sunscreen is--it absorbs ultraviolet light.1621

But, the standard acid, benzoic acid, has a Ka of 6.4x10-5.1626

And finally, we have something called hypochlorous acid, and it has a Ka of 3.5x10-8.1633

OK, well, which acid works best?1648

Which acid works best?...that is only part; and: What is the ratio that puts the pH at 4.80?1653

What is the ratio of concentration of base to acid that puts the pH at 4.80?1669

Now again, chances are very, very slim that you are going to find an acid that has exactly a pKa of the pH that you are looking for.1688

There is going to be some difference in the ratio to adjust, based on the Henderson-Hasselbalch equation.1695

So here, we want to know which acid works best; and once you have chosen the acid, what is the ratio of base to acid (the A-, the HA) that actually puts the pH at the 4.80 that we want?1701

OK, well, let's start with our Henderson-Hasselbalch equation: pKa + log(A-/HA).1717

I hope you will forgive me, but I am so tired of putting these parentheses on there; I'm just going to leave them like that--the parentheses and those brackets--they drive me crazy; we have done these enough to know that we are dealing with concentrations, so it shouldn't be a problem.1730

OK, we want the pH to be 4.80, and the pKa (oh, wait; we have to...what am I doing?--we haven't even chosen an acid yet!--we don't know what a pKa is!--OK).1741

So, let's go back; so, for acid A, which is monochloroacetic acid, we have a pKa...we just take the negative log of these values that we have right here (OK, let me circle them in red); we just take the negative log of these--that is the p.1754

That pKa is going to be 2.87; B: our pKa is going to be 4.89--that looks pretty good; C--benzoic acid: the pKa is going to be 4.19; and D: our pKa is equal to 7.46.1776

We want 4.8, is our target, right?--target pH; so it looks like we are going to go with propanoic acid.1798

That is the one we want to use: propanoic acid.1805

OK, so now that we have chosen HPr, now we can go ahead and find out what our ratio is.1809

So, let's move on: let's do pH is equal to pKa, plus the log of base over acid, Pr- over HPr concentrations.1817

You know what, I had better just "bite the bullet" and do it.1831

OK, we want 4.80 equals...the pKa is 4.89; plus the log of Pr- over HPr; go ahead and do the math here, and we end up with...well, you know what, let me just go ahead and do it.1836

-0.09=log(Pr- concentration/HPr concentration), and anytime you have a log, the inverse function of a logarithm function is the exponentiation with a base 10, so 10x.1854

So we raise both sides to the power of 10; that gets rid of the log function; we get a ratio of 10-0.09=0.813.1872

So, when we pick propanoic acid, we want to have a solution that has a pH of 4.8.1893

We pick propanoic acid because the pKa is 4.89; it is close to it--that means that the adjustment we have to make, to bring it down from the 4.89 to the 4.8--the ratio is going to be .813; that is nice--that is still pretty close to 1.1900

It's close enough to where we can actually get some good buffering out of it; that is the whole idea.1917

So now, we will continue this example; now we will do Example 3.1923

I'm just going to basically continue our discussion of this; this is a typical problem that you might see in a free response question that has multiple parts to it.1931

Example 3: Now, I want to prepare a 250-milliliter buffer solution like the previous problem.1942

In other words, I want to prepare the buffer solution at 4.8, using my propanoic acid.1969

OK, if I use 150 milliliters of a 0.20 Molar HPr (propanoic acid), what is the molarity of the sodium propanoate solution I must use to achieve the desired 4.80 pH?1975

OK, so we said we have a propanoic acid solution, and we know that the ratio has to be .813; that was our previous example.2025

Now, I want to actually prepare the buffer solution; I want to prepare a buffer solution, 250 milliliters of that buffer solution.2032

Well, I use 150 milliliters of a .20 propanoic acid; I know that I have to add another 100 milliliters of the sodium propanoate; but what is the molarity of the sodium propanoate solution that I must use in order to achieve the desired pH?2039

See...please, it is very, very important that you understand what this question is asking: I want the pH of 4.8; I have already calculated the fact that I have a ratio of .813; I am preparing 250 milliliters of that buffer solution.2060

I am using 150 milliliters of a .2 Molar HPr; well, 250 minus 150 leaves me 100 milliliters of solution of sodium propanoate.2075

However, what is the molarity of that sodium propanoate I need in order to get to a pH of 4.80?2083

This is actually just a stoichiometry problem, believe it or not.2090

Let's go ahead and do this: well, we know that our (actually, let me use the actual identities of species) ratio of propanoate to HPr is equal to 0.813.2093

OK, well, I have 0.150 liters (that is 150 milliliters), times 0.20 moles per liter of the propanoic acid.2112

That means there are .03 moles in that 150 milliliter: .030 moles of (wait a minute, I'm using HPr...yes, that is right) the HPr.2128

OK, now this 0.03 mol of HPr is in 250 milliliters, OK?--because again, I am adding another 100 liters of solution, so the Henderson-Hasselbalch equation talks about...I am talking about the molarity in the whole solution.2151

I know that I have used .030 moles, because I have used 150 milliliters of a .2 Molar; but now, it's swimming around in 250 milliliters.2179

0.03 moles, divided by 0.250 liters, gives me 0.12 Molar HPr.2187

Well, I know that my Pr- concentration, divided by my HPr concentration (which I just found is 0.12 Molar) is equal to 0.813.2203

Therefore, my propanoate concentration has to be 0.813 times 0.12, equals 0.09756 (I'm not worried about significant figures here) molarity.2217

I also know that I am going to have 100 milliliters of this, because I had 150 milliliters of the propanoic acid solution; I want a total volume of 250; so I need 100 milliliters.2237

Now, I can go, and I can actually make...I can take some solid sodium propanoate; I can drop it into solution, depending on...I weigh it out in terms of grams, because now I know what concentration I need.2249

I hope that this made sense.2263

I needed a buffer solution at a specific pH; I knew that, if I can find an acid that has a pKa close to that pH, and I knew that the ratio of the base to the acid is going to be .813, and based on the rest of the problem (given the volumes that I had and the molarity that I was dealing with), I was able to calculate the molarity of the sodium propanoate solution that I would need in order to achieve that 4.80 pH.2265

I would definitely recommend going through this problem a couple of times, just to make sure you understand what it is that we did.2297

This is sort of a whole bunch of problems tied together.2303

OK, well, this pretty much covers our discussion of buffer solutions.2307

Next time that we see each other, we are going to start our discussion of acid-base titrations and titration curves.2312

Thank you for joining us here at Educator.com.2319

We will see you next time; goodbye.2321