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Raffi Hovasapian

Raffi Hovasapian

Titrations: Weak Acid and Strong Base

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 (22)

2 answers

Last reply by: Magic Fu
Fri Feb 24, 2017 1:33 AM

Post by Magic Fu on February 21, 2017

Hi, Professor Hovasapian, can you go over want is the ending point of the Titration Curve and where could we find our endpoint?

0 answers

Post by Sun HaoHui on February 9, 2017

Hi, Professor Hovasapian, I have two quick questions. So when we deal with titration problems, do we always start off by writing out the stoichiometry reactions chart, then Equilibrium? And why?
Is there a big difference between solving buffer and titration problems?

2 answers

Last reply by: Professor Hovasapian
Sat May 2, 2015 1:24 AM

Post by Maya Balaji on May 1, 2015

Hello Mr. Hovasapian! Amazing lecture as always! quick question before my AP test this Monday.. How do I decide which species are dominating/ which species really deals with the chemistry? Thank you professor, your videos have helped me tremendously!

2 answers

Last reply by: sadia sarwar
Wed Feb 4, 2015 12:27 AM

Post by sadia sarwar on January 30, 2015

hello sir
is there any lecture on back titration??

1 answer

Last reply by: Professor Hovasapian
Tue Jun 3, 2014 7:17 PM

Post by husian alturki on June 1, 2014

Hello Raffi
thanks for the lovely videos they are helping me a lot. but i had one thing that keep confusing me
when you list the major species and compare it to water to see which one is dominating do you compare the ka of the acid to 1x10^-14 or to only the H^+ concentration which is 1x10^-7
so on example 1 lets assume the ka was 1.8x10^-10 in this case would i use the weak acid or water?

1 answer

Last reply by: Professor Hovasapian
Tue Feb 4, 2014 1:17 AM

Post by Christian Fischer on February 3, 2014

Hi Raffi. What a great lecture as always!! I have a quick question regarding the chemistry taking place during this weak acid titration reaction. I'd really appreciate if you get time to answer it one day.

Is this correctly understood: Before we add NaOH we start off with a concentration of [H+]=1,34*10^-3 molar (at equilibrium). This means the solution only contains 1,34*10^-3M*50ml = 0.067 mmoles of H+ (at equilibrium) When we add 10,0ml NaOH we are adding 10ml*0,10M = 1mmoles of OH(-). Now we have more moles of OH(-) than H(+) and All the OH must react with H(+). Does this mean that
a) OH(-) attacks HA (the weak acid) directly?  OR  
b) Does OH(-) attack the 0.067 moles of H(+) in solution directly? If it attacks H(+) in solution directly there is not enought H(+) to meet the demand so does the equilibrium reaction between the weak acid in water HA + H2O = H(+) + A(-) shift to the right to produce more H+ when it senses OH- eating up the H(+)?

I want to thank you for explaining so well why pH can be different than 7 when when the same amount of strong base and weak acid have reacted 100%. I had a hard time understanding that until now! I'm very grateful :)  


Keep up the great work,
Christian
 

0 answers

Post by Edwin Wong on May 19, 2013

The pH should be 8.74 not 8.72 at 30:18.

1 answer

Last reply by: Professor Hovasapian
Sat May 4, 2013 4:01 AM

Post by morgan franke on May 3, 2013

when working on part 1 should you make the concentration of HAc = .1m/.5L = .2 because you added 50 ml of a .1 molar HAc? If not, please help me understand why sometimes the amount of solution is import in the ICE table and others it isn't.

1 answer

Last reply by: Professor Hovasapian
Mon Apr 29, 2013 1:41 AM

Post by ANURAAG PRAKASH KAMLE on April 29, 2013

Sir what are the other two zones of the Titration Curve called? the ones after the Buffering zone.

1 answer

Last reply by: Professor Hovasapian
Fri Mar 22, 2013 4:10 PM

Post by Kathryn Cosgrove on March 22, 2013

After figuring out the major species, how do you decided which species need to be the reactants and which are the products?

0 answers

Post by ARLENE ROOME on May 23, 2012

not sure how you got the PH = 5.35 when 50 ml was added

Related Articles:

Titrations: Weak Acid and Strong Base

  • At half-Equivalence the pH is equal to pKa, because the concentrations of Acid and conjugate base are equal, thus making the Log term in the Hen-Has equation zero.
  • pH at Equivalence for a Weak acid/strong base titration is greater than 7.
  • pH at Equivalence for a weak base/strong acid titration is less than 7.

Titrations: Weak Acid and Strong Base

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
  • Titrations: Weak Acid and Strong Base 0:43
    • Question
    • Part 1: No NaOH is Added
    • Part 2: 10.0 mL of NaOH is Added
    • Part 3: 25.0 mL of NaOH is Added
    • Part 4: 40.0 mL of NaOH is Added
    • Part 5: 50.0 mL (Total) of NaOH is Added
    • Part 6: 60.0 mL (Total) of NaOH is Added
    • Part 7: 75.0 mL (Total) of NaOH is Added
    • Titration Curve

Transcription: Titrations: Weak Acid and Strong Base

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

Last time, we talked about titrations; we started talking about them, and we did a strong acid-strong base titration.0004

Today, we are going to take the next step, and we are going to talk about a weak acid-strong base titration.0010

What I mean by that is: we are going to have a weak acid solution, and we are going to be adding a little bit of strong base to that weak acid solution, and we are going to be following the progress of the pH of that solution.0016

We are going to do the same thing: we are going to have a bunch of pHs, and we are going to take those pHs, and we are going to construct a titration curve, and it is going to look a lot like what it did last time--same general shape--except the equivalence point is going to be different.0027

Let's go ahead and get started and see what we can do.0040

We want to consider...let us consider the titration of (oops, let's make the 50 a little bit more clear than this) 50.0 milliliters of a 0.10 Molar acetic acid solution (which, if you remember, we abbreviated as HAc, because I don't like writing out all of the C's, O's, and H's...it just hurts my eyes--it's too many atoms floating around), and we are going to titrate it with a 0.10 Molar NaOH.0046

Our titrant is the same as before: it is a strong base, sodium hydroxide, but this time we are titrating a weak acid, not a strong acid.0093

Now, because we are dealing with a weak acid, you can pretty much guarantee that the Ka is going to show up somewhere.0100

The Ka of acetic acid is 1.8x10-5; OK, so let's go ahead and get started.0108

The first thing we want to do is: we want to calculate the pH of a solution before any NaOH is added--so, no NaOH added yet.0116

Well, you know how to do this; this is a typical weak acid problem: there is no NaOH that had been added; you basically have 50 milliliters; you have a solution that is .1 Molar acetic acid; how do we calculate the pH?0130

Well, we are going to do an ICE chart; so, this is a typical weak acid problem, which I guess is probably pretty good--we'll do a little bit of quick review here.0142

Let's see what our major species are in solution; you will definitely want to get into the habit of doing this as a conditioned response.0157

Any kind of aqueous chemistry--you need to find out what is actually in that solution, to decide what chemistry is going to take place.0165

Our major species are going to be HAc (our acetic acid, which is a weak acid, which is why it shows up as the non-dissociated form), and we have H2O; that is it--those are the only things floating around in solution.0173

Well, between the two, 1.0x10-14 is the Ka of water; HAc has a Ka of 1.8x10-5; it's going to dominate the equilibrium.0186

The equilibrium that is going to take place here, as you drop it into solution, what is going to happen?--what is going to happen is the following.0199

The HAc, standard acid dissociation equilibrium: H+ + Ac-; it's always the same.0206

The acid loses a proton to its conjugate base--always the same.0215

We have an initial; we have a change; we have an equilibrium; when we are dealing with equilibrium issues, remember, ICE is for equilibrium; we deal in concentrations.0221

So, the concentration of HAc is 0.10 Molar; there is no this; and there is none of that.0230

Some of this is going to disappear; some of this is going to show up; and this is going to show up, the amount that disappeared, right?0238

We get 0.10-x; we have x; and we have x; OK.0245

Let's go ahead and do (I'm actually going to do it over here, off to the side, I think): the Ka is 1.8x10-5; it is equal to the hydrogen ion concentration, which is x, times the acetate concentration, which is x, all over 0.10-x, which is the HAc concentration.0252

Well, this is approximately equal to x2 over 0.10, because x is going to be pretty small; it's going to be pretty small because it's very little dissociation.0274

You can check it with the 5% rule if you want, but you can take my word for it.0284

We get x equals the H+ concentration, is going to equal 1.34x10 to the negative 1, 2...it looks like 3, which implies that our pH is going to be 2.87.0288

We are going to start off with this pH of 2.87, before any base has been added to the solution.0309

OK, so now, our next step is going to be: Add 10 milliliters of the NaOH; OK.0315

Now that we have added 10 milliliters of the NaOH, we do what we always do; what are the major species in solution, and what reaction is going to take place?0329

So, the major species are...well, HAc is in solution; H2O is in solution; you have dropped in sodium hydroxide, which is a strong base--it dissociates completely.0336

You are going to have free sodium, and you are going to have free hydroxide.0350

OK, anytime you have hydroxide in a solution, it is going to do one thing: it is going to seek out a source of protons that it can pull away to become water--that is it--that is all OH does.0354

Well, in this case, it has two sources of protons: OH- can pull H from H2O (which doesn't really change much, because it turns into H2O, but it leaves an OH- behind); so this is going to pull from that--this is the reaction that is always going to take place, always--weak acid-strong base.0364

So, the reaction that takes place is the following: OH- is going to react with the HAc, and it is going to produce H2O; well, A-, because this is going to pull off that H, plus H2O.0382

Now, we have to do the stoichiometry first: when you add this base, it is going to react with this until this is used up; any OH- that is added is going to react to completion--notice this single arrow; that means all of it reacts until there is no OH- left.0405

And then, when there is HAc left over, then we do the equilibrium problem.0426

You remember: this is no different than a buffer solution; in fact, each one of these steps that we are going to do is just a buffer problem, just like the last couple of lessons, a few lessons ago.0430

That is all that is happening here.0441

OK, so we have a Before; we have a Change; and we have an After.0444

Well, how much OH- did we actually end up adding?0450

Well, we added 10 milliliters, and the molarity is .10 Molar; OK, let me actually just write something here: this is the (let me do this in red) stoichiometry.0455

OK, so for a weak acid-strong base titration, we do the stoichiometry first; then, we do the equilibrium.0470

We didn't have to this for the strong acid-strong base titration, because it was just stoichiometry--there is no equilibrium to come to.0477

But, this is a weak acid, so there is still an equilibrium to deal with; here, we have to do the stoichiometry first, and then, we do the equilibrium.0484

When we do stoichiometry, we deal in moles.0492

So, this is going to be...I have added 10 milliliters...so 10 milliliters, times 0.10 molarity; that gives me 1 millimole of OH-.0495

Well, the HAc...I have 50 milliliters of a .1, so 50 milliliters, times 0.1 molarity; that gives me 5.0 millimoles.0514

Well, here: 1 millimole, 5 millimoles--this is a 1:1 ratio; this is the limiting reactant, so this is going to disappear.0529

1 millimole: that is going to leave 0 OH- (oops, we don't want these stray lines floating around); the change is going to be 1 millimole of HAc is going to be reacted with; 1 millimole of H+ is going to be pulled off, leaving 4.0 millimoles of HAc.0537

We are going to produce...let me see, we actually had 0 of that to start, and we are going to produce 1.0 millimole of that.0563

So now, at the end of our reaction, we have 1 millimole of acetate (oops, I'm so used to using...let me go back to blue here...this is acetate, not just a generic acid) floating around in solution; we have 4 millimoles of HAc floating around in solution; so now, let's check our major species to see where we are after the reaction.0575

Major species after the reaction: so, we have some HAc floating around, 4 millimoles, it looks like; we have Ac- floating around, 1 millimole; we have H2O, and we have Na+.0616

The OH- has been used up; what is going to dominate the reaction now is that.0633

We have a weak acid, and we have its conjugate base; there is going to be an equilibrium established here; so now, we do the equilibrium problem.0640

When we work in equilibrium, we work in molarity.0652

This is going to be: HAc is going to be in equilibrium with H+ + Ac-; and this is just an ICE chart, and we do this over and over and over again; that is all we are doing.0656

So now, let me do this one in blue.0674

So now, we deal in molarities; so we have 4 millimoles (let me write this a little farther to the left--I need a little bit more room, because I'm going to make this right); so let me write this as I, as C, as E.0677

I have 4 millimoles floating around, right?--4 millimoles of this in solution, that I just calculated from the stoichiometry.0696

It is floating around in 60 milliliters, because it's the 50, plus the 10 milliliters of solution that I added.0705

60 milliliters; so this is 0.067.0711

There is none of this; the Ac-, well...we had 1 millimole, and it is floating around in 60 milliliters; so we have 0.0167.0717

This is going to diminish by x; this is going to augment by x; augment by x; you end up with 0.067-x; this is going to be just x; and this is going to be 0.0167+x.0730

And now, we do the Ka, which is 1.8x10-5, is equal to x, times 0.0167+x, over 0.067-x; we can...the x is going to be pretty small, so let's just go ahead and write this as x, times 0.0167, divided by 0.067; and when we do this, we get a hydrogen ion concentration of 7.22x10-5, which implies that we have a pH equal to 4.14.0754

So again, we have added some hydroxide to this acid solution; we have to do the stoichiometry first--the base that we have added, the hydroxide, is going to react with the weak acid--it's going to pull protons off until the base runs out, until that limiting reactant actually runs out.0803

And then, you are going to have some acetate floating around, as well as some acetic acid floating around; so you have to...and then you go back, and because you have changed the volume of the solution, you calculate the molarity, and you calculate the pH.0821

OK, so let's do our next one, which is going to be 25 milliliters.0837

So now, add 25.0 milliliters total; and again, we always go back--we added 10; we are adding another 15; the total that we are adding is 25--we always go back to the beginning, in case we made a mistake, so it doesn't carry over.0846

NaOH added; OK, so major species before...let me write "major species before reaction": we have HAc; we have H2; we have Na+, and we have OH-.0867

This is going to react with that, and it is going to react according to the following.0895

It is going to be OH- + HAc; it is going to form Ac- + H2O.0899

Well, we have a Before; we have a Change; we have an After; now, we have added 25 milliliters of sodium hydroxide.0910

It is 25 milliliters, times 0.10 Molar, equals 2.5 millimoles; and here, we have our 50 milliliters, times a 0.10 Molar; this is Molar; equals 5.0 millimoles.0919

Well, this is 2.5; this is 5; this is going to run out first.0941

This is going to be -2.5 millimoles, leaving us with no hydroxide; it is going to react with 2.5 millimoles of that, leaving us with 2.5 millimoles of this; and it is going to...this is 0; this doesn't matter...this is going to add 2.5: because this is diminishing by 2.5, 2.5 Ac- is showing up.0945

This is going to be 2.5 millimoles; OK, so now what are our major species? (By the way, this was the stoichiometry part).0973

I keep forgetting to write that, but of course, hopefully you know that by now.0986

So now, we will go ahead and do the...actually, let me write down what the major species are first.0991

Let me go back to blue.0998

Major species after (this was before reaction; now, after reaction): I have, now, HAc floating around; I have Ac- floating around; I have H2O floating around; and I have Na+.1000

All of the OH- has been used up; that is what that number is.1018

This is going to dominate; so we are going to do our equilibrium; we are going to have (let me do this in red) HAc in equilibrium with Ac- + H+ (actually, let me write these in the order that I am accustomed to writing them in--otherwise, I think it's going to be confusing)...H+ + Ac-; Initial, Change, Equilibrium.1023

Well, HAc--we have 2.5 millimoles; and now, it's floating around in 50+25: it's floating around in 75 milliliters.1053

That concentration is 0.0333; there is none of that; the Ac-...I have 2.5 millimoles floating around in a total volume of 75 milliliters; it equals 0.033.1065

This is going to diminish by x; this is going to augment by x; this is going to augment by x; 0.033-x; this is x; 0.033+x.1090

Now, I do my Ka, which is 1.8x10-5; it is equal to x, times 0.033+x, divided by 0.033-x, which is approximately equal to x times 0.033, divided 0.033; those cancel; x equals the hydrogen ion concentration, of course, which equals 1.8x10-5, which implies that the pH is equal to 4.74.1105

OK, this is...let me see, where are we?--so we just did that; we calculated pH; OK.1145

This is a really, really important part; notice what happened, if you flip to the back of the last slide: we ended up with half of the original acetic acid being used up.1156

We started off with 5 millimoles; we used up half of it--we reacted half of it, leaving 2.5 millimoles.1171

Now, we also ended up producing 2.5 millimoles of acetate; so now, this is confirmed right here: the concentrations are the same, because they are in the same volume.1179

When the concentrations are the same, the pH of the solution happens to equal the pKa.1191

So, let me write this as: At this point, one-half of the HAc has reacted with the added OH-, producing equal amounts of HAc and Ac-.1198

Well, what does that remind you of, when you have equal amounts of a weak acid and its conjugate base?--you have the ideal buffer.1242

We have the ideal buffer.1252

In the process of titrating a weak acid, you end up creating a buffer solution, and you will see what the buffer solution looks like when you see the titration curve.1258

It is actually going to look like it's going to be very little change in the pH.1266

That is going to increase very, very slowly before that sudden jump.1270

OK, so a buffer...which is why pH equals pKa, because, if you remember, pH in the Henderson-Hasselbalch equation, plus the log of the base concentration over the acid concentration...well, here the base concentration and the acid concentration are the same, so this is 1.1275

Well, the logarithm of 1 is 0; what you are left with is pH=pKa.1299

So, at the halfway point of the titration of a weak acid with a strong base, what you have created is the perfect buffer.1304

That is it; we just want you to know that.1312

OK, now, #4: now we have 40 milliliters of NaOH added.1314

Well, proceeding as before, just like we did the last few that we did (you should probably check this for yourself): proceeding as before, we get a pH equal to 5.35.1326

All right, now let's do 50.0 milliliters total of NaOH added; so this one, we are definitely going to do.1344

Let's check our major species (this one we are going to treat just like we did before): we have HAc; we have H2O; we have OH-; and we have Na+.1361

Well, OH- is going to react with HAc, and you are going to get the following.1375

I am not going to do all of the calculations here, but they are just like before, so you should recognize where my numbers are coming from; they are not just dropping out of the sky.1383

The OH- is going to react with the HAc; it is going to produce acetate, and it is going to produce water, because this is an acid-base neutralization.1392

We have a Before; we have a Change; we have an After.1404

Well, 50 milliliters of a .1 Molar NaOH produces 5.0 millimoles1408

Well, 50 milliliters of a .1 Molar acetic acid solution produces 5.0 millimoles; there is none of this, and there is none of that.1417

Equal amounts here--we have equal amounts, so that means they are both going to vanish--they are going to react with each other.1428

OH- is going to take this, produce that, produce water; so it is going to be -5.0, -5.0, +5.0, water doesn't matter; at the end, we have no hydroxide floating around; we have no acetic acid floating around; we have all acetate ion floating around.1435

Here is where we are now; so now, the major species in solution after reaction--major species after reaction: you end up with, well--we have Ac-, H2, and Na+.1459

There is no hydroxide left--that is what this says.1486

There is no acetic acid (there is no HAc) left--that is what this says.1490

All I am left with is the acetate, the water, and that.1494

OK, now we do our equilibrium problem.1499

However, our equilibrium problem is going to be slightly different now, and this is what you need to recognize; and this is why it's important to know the chemistry and watch what happens.1502

Notice what you have in solution: we have acetate; we have water; and we have sodium ion--sodium ion isn't going to do anything; water isn't going to do anything.1511

But this is the conjugate base of a weak acid; so what it is going to do: it is actually going to react with the water to create hydroxide ion, producing a basic solution.1520

But now...well, let me just write that out: so again, you don't have any HAc; so because there is no HAc, there is no acid dissociation equilibrium to write down; now, what you are going to have, because you only have a base and water, and this is the conjugate base of a weak acid (therefore, it's a relatively strong base): it is going to pull hydrogen ions off of the only source it has, which is water.1534

So now, the equilibrium that you are going to do is going to be a weak base equilibrium.1562

It is going to be the equilibrium of a base reacting with water to produce hydroxide ion.1567

It is going to be the base association equilibrium, and you remember this one--it looks like this: A base, plus water (I'll write it as HOH) is going to be in equilibrium with HAc, plus OH-.1575

It is going to end up producing hydroxide ion, because now this is the only thing in solution.1593

The reaction took place; nothing was left over except the Ac-; now the Ac-, when we allow the system to come to equilibrium (which we are doing right now)--it is going to follow this reaction.1598

It's really, really important that you recognize this.1609

Initial, Change, Equilibrium: well, the initial concentration of Ac- is going to be 5.0 millimoles, divided by the total volume, which is now 100 milliliters; you have added 50 milliliters of solution to 50 milliliters of solution; this is 100 milliliters.1612

You are going to end up with a 0.050; water doesn't matter; there is none of this, and there is none of this.1634

Well, some of this is going to diminish; water doesn't matter; some of this is going to show up, and some of this is going to show up.1643

Here we have 0.050-x; this is x; this is x; OK.1653

Now, here is where it's interesting: we must use the Kb, not Ka.1661

The Ka represents the acid dissociation constant, HAc going to H+ + Ac-.1679

The Kb represents the base association constant: Ac- + H2O goes to HAc + OH-.1688

Remember, now it is acting as a base; there is no acid around to get in equilibrium with--now it's completely Ac-, so we have to use the Kb, not the Ka.1700

OK, now, we say Kb is equal to the concentration of OH-, times the HAc, over the Ac-.1713

Well, we know the Ka of the acetic acid; well, how do we get Kb?1730

Remember, Kb is equal to Kw, over Ka.1737

It is equal to 1.0x10-14, divided by 1.8x10-5.1742

When we list Ka values for weak acids, we list Ka values; we don't list Kb values for their conjugate bases; but we can calculate Kb values for the conjugate bases if the situation comes up (which it does).1750

In this case, the Kb ends up being 5.6x10-10.1763

We have 5.6x10-10 is equal to x times x, over 0.050-x, which is approximately equal to x squared over 0.050.1770

Well, we get that x, which equals the hydroxide ion concentration, is equal to 5.3x10-6; that means that the pOH is equal to 5.26, which means that the pH is 14 minus the 5.26; we get a pH of 8.72.1788

There we go; now, here is the important part: the pH at equivalence is not 7.1810

This is equivalence; the amount of hydroxide that we added reacted with all of the acetic acid that was present, that gave up its proton; the OH- pulled off the H+ and produced water, leaving just acetate ion.1828

Equal amounts: 5 millimole reacted with 5 millimole; but the pH of the solution that we calculated is 8.72.1845

It is a basic solution; and the reason that it is basic is because the base that was left over, the Ac-, ended up reacting with water in a base association to produce hydroxide.1852

So, we saw that in a strong acid-strong base titration, the pH at equivalence was 7; with a weak acid-strong base titration, at equivalence, the pH is bigger than 7--in this case, 8.72.1863

The number itself is relevant only as far as this problem is concerned, because it is acetic acid; but other weak acids--the idea is: you are going to get...that equivalence of pH is going to be greater than 7, not 7.1878

It is the stoichiometry that defines equivalence, not the pH.1891

OK, now we will keep going, adding some more: 60 milliliters total of NaOH is added; OK.1896

Major species: we have HAc, OH-...1919

I know this is getting a little tedious, but believe me; when you come to a point where you are sick of a problem, that means you understand it so well that you never have to think about it again; so getting sick of a problem is good.1923

If it is becoming tedious, that is a good sign; that means that you know what you are doing.1934

60 milliliters total of sodium hydroxide is added; our major species are the (let me write them again on this page; I think it's a good idea)...major species: now, we have again HAc, OH-; we have H2O.1938

We run our stoichiometry again; our stoichiometry is going to be the OH- is going to react with the HAc (I might put a little more room here); it is going to produce acetate ion, plus water.1959

We have a Before; we have a Change; we have an After.1981

Here, again, 60 milliliters times .1 moles per liter; we have 6.0 millimoles; here we have the 5.0 millimoles (hope you know where that came from; that is the 50 times .1); 0; water doesn't matter.1986

Here, because now this is the limiting reactant, this is going to be -5.0 millimole, -5.0 millimole; this is going to be +5.0 millimole; water doesn't matter.2005

We are left with 1.0 millimole of OH- in solution; no HAc in solution; and 5.0 millimoles of Ac in solution.2020

Let's check our major species after reaction--we have: OH- in solution; we have H2O; we have Ac-; and we have Na+.2032

Na+ is not going to do anything; that is not going to do anything; ah, we have two bases--we have OH-, and we have Ac-.2050

Between the two, just for your information, hydroxide is a very strong base.2060

We saw, from the previous calculation that we did, that the hydroxide ion concentration was 5.3x10-6.2068

This was with absolutely no hydroxide in solution, other than just the hydroxide that shows up from this reacting with water.2076

Here, we actually have a whole millimole of hydroxide ion; so, for all practical purposes, any contribution that this Ac- could make to the hydroxide ion concentration is completely negligible.2084

This is the only thing that matters--this is the only thing that matters.2097

We are not even worried about that.2103

So, our hydroxide ion concentration (let me write this) is going to be 1 millimole, divided by the total volume, which is now 50 milliliters + 60 milliliters, so 110 milliliters.2105

It is going to give us 9.1x10-3 molarity.2123

That implies that the pOH of this solution is 2.04, which implies that the pH of this solution is 11.96 (very, very, very basic).2130

OK, and now, our final, just to round things out: 75 milliliters total NaOH added; proceeding as we did right here, we end up with pH=12.30.2144

OK, now, we want to take a look at our titration curve to get a pictorial view of all of these calculations that we did.2167

Titration curve; it's going to look like this: once again, this is the volume of OH- titrant added, and it's a unit of milliliters; this is going to be the pH.2181

Let's go to 25; let's go to 50; let's do 1, 2, 3, 4, 5, 6, 7, 8, 9; let's mark off our 7; 10, 11, 12, 13, 14; we are going to start off somewhere in the neighborhood of about 3, and at 50 milliliters, we are going to be at .009.2201

I just want to mark off this equivalence point; so it is going to look something like this.2229

There we go; so again, you notice that it looks the same as it did for the strong acid-strong base, but the numbers are different.2240

So, this right here--this is the equivalence point; and again, the equivalence point is where enough hydroxide has been added to react precisely with the amount of acid available; total neutralization; pure water.2249

However, it is not pure water, because, in this particular case, we dealt with a weak acid.2265

A weak acid has a strong conjugate base; so, once that Ac- shows up, it is going to start reacting with the water present to produce a little bit more hydroxide, which is why our pH at equivalence is roughly around 9--at 8.72.2271

So again, for a weak acid-strong base titration, at equivalence, the pH is above 7; it is a basic solution.2289

Or, think of strong base-weak acid; they are not balanced out; it is going to be more basic (strong base); that is another way of thinking about it.2299

OK, a couple of other things to consider here: remember, at 25 milliliters, that was the halfway point to the titration; that was when we reacted half of the acetic acid with a certain amount...and what we produced was equal concentrations of acetic acid and acetate ion.2306

In other words, we created the perfect buffer.2327

This region right here--this is called the buffering zone: basically, you have--in solution, you have a weak acid, plus its conjugate base, floating around simultaneously; that is a buffer--that is the definition of a buffer.2329

This zone is called the buffering zone; and you can tell that it's a buffering zone, because you notice you are adding all of this base, but there is very, very little change in pH.2352

As you get close to the equivalence point, it jumps up.2360

So, once again, the take-home lesson for this particular graph, qualitatively, is that equivalence is defined by stoichiometry, not pH.2364

So, on a test question (and you are going to get this question; it always shows up): acetic acid is titrated with sodium hydroxide; the pH at equivalence is 7--true or false?2386

The answer is false, because what you have is a weak acid titrated with a strong base; the pH is bigger than 7.2405

Equivalence is defined by stoichiometry; equal amounts of base and acid have reacted to produce neutralization.2411

It is not defined by the pH; the pH is only 7 for a strong acid-strong base titration.2420

OK, so there you have it: so what we have done is: we have taken a weak acid; we have followed the progress of titrating that weak acid by additional volumes of sodium hydroxide.2428

We have showed you how to actually calculate the pH of that solution during this buffering zone, at equivalence, and after equivalence.2440

So, for a weak acid-strong base, you have three different regions: you have this region; you have this region; and you have this region--and all of those were dealt with, as far as solving this titration problem.2450

We essentially did 7 problems in one big, long problem here; a typical problem that you will see is something like: 30 milliliters of a .2 Molar acetic acid solution is titrated with 17 milliliters of a .5 Molar sodium hydroxide; what is the pH of that solution?"2466

Well, you are dealing with a weak acid and a strong base; you do the stoichiometry first; after the stoichiometry is taken care of, you check the major species, and then you do the equilibrium part, depending on where you are--depending on what has been used up and what is actually floating around in solution.2485

I hope this has helped.2502

Next time, we will talk a little bit more about titrations, and we will begin to talk about acid-base indicators--how we actually know, in a laboratory, when we have reached the equivalence point--as opposed to doing it analytically (in other words, calculations).2505

Thank you for joining us here at Educator.com.2520

We'll see you next time; goodbye.2522

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