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Raffi Hovasapian

Raffi Hovasapian

Polyprotic Acids

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 (11)

1 answer

Last reply by: Professor Hovasapian
Fri Feb 26, 2016 1:07 AM

Post by RHS STUDENT on February 12, 2016

Sir, regarding to the first example, would you plz explain why the Hydrogen ion you found when dealing when the dissociation of H3PO4 can also be used in that of H2PO4- and HPO42-? Aren't the breakdown of H2PO4- and HPO42- also produce extra H+? Thx

1 answer

Last reply by: Professor Hovasapian
Thu Mar 13, 2014 7:57 PM

Post by Alexis Yates on March 12, 2014

Wonderful lesson (as always) Professor, thank you for all of the help!

0 answers

Post by William Dawson on December 14, 2013

You are defeinitely right about decimals being better when they are not unwieldy, and signifcant figures being overrated!

1 answer

Last reply by: Professor Hovasapian
Wed May 1, 2013 11:42 PM

Post by morgan franke on May 1, 2013

I am unsure which root to use in my h final equation. Can you help me!

1 answer

Last reply by: Professor Hovasapian
Wed May 1, 2013 5:41 AM

Post by Antie Chen on April 30, 2013

Hello Raffi, I cannot understand at the time about 30:00, "For solution...", can you explain for me again?
The negative pH in example 2 is because the solution is so concentrated?

1 answer

Last reply by: Professor Hovasapian
Tue Feb 26, 2013 3:36 PM

Post by kevin vaughn on February 26, 2013

very good,wish i found this site early. would have to take my chem class for a 3rd time at university ><

Related Articles:

Polyprotic Acids

  • Acids dissociate ONE hydrogen Ion at a time.
  • Sulfuric Acid is strong in its first dissociation, weak in its second.
  • Equilibrium problems involving solutions of Sulfuric Acid more dilute that 1.0 molar must use the entire expression without approximations ( no ignoring the “x”).

Polyprotic Acids

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
  • Polyprotic Acids 1:04
    • Acids Dissociation
    • Example 1
    • Example 2
    • Example 3

Transcription: Polyprotic Acids

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

Today, we are going to continue our discussion of acid-base equilibria, and we are going to talk about polyprotic acids.0004

If you remember, in the first lesson that we discussed acids in, we said that polyprotic acids are simply a fancy name for acids that have more than one hydrogen that they could possibly give up.0010

So, for example, sulfuric acid (H2SO4), carbonic acid (H2CO3)...probably the most important of the polyprotic acids is H3PO4, phosphoric acid.0021

It is how the inside of a cell actually maintains its pH; it is the buffer solution inside of the cell.0033

And the carbonic acid, the H2CO3, is how the blood maintains its pH; it is a buffer solution outside of the cell.0040

We are going to discuss polyprotic acids today and talk about they behave; and it is not altogether different than anything that you have seen already, as far as weak acids and strong acids.0048

We'll just jump right in, do a quick couple of definitions, and we'll start with the examples.0058

OK, so once again, the really important thing about all acids (polyprotic or otherwise--well, polyprotic especially) is that acids dissociate one hydrogen ion at a time.0065

So, in the case of, say, carbonic acid (let me actually write that out first: so acids dissociate one hydrogen ion at a time), which is H2CO3, it's in equilibrium, so it dissociates into H+ + HCO3-, and there is a Ka associated with this, an acid dissociation constant, 4.3x10-7.0076

And notice, we have a little 1 here, so Ka1, because it gives up one ion at a time.0112

Now, this HCO3- has another H that it can give up, and it also behaves like an acid; so it now shows up on this side, and it is in equilibrium with the proton that it gave up: CO32-.0118

And this is Ka2; this is the second acid dissociation constant, which is equal to 5.6x10-11.0135

One thing you want to notice here, which is a general trend for all polyprotic acids, is that the first Ka is going to be a certain number; the second Ka is one that is very, very small: 10-11...very tiny--certainly much smaller than this.0143

So basically, when it comes to polyprotic acids, it's really the first dissociation that counts the most.0160

The only exception to this case (which we will see in a little bit when we discuss sulfuric) is sulfuric acid, and the reason being that sulfuric acid is strong in its first dissociation--it behaves just like hydrochloric acid or hydrobromic acid--but in the second dissociation, it's actually a weak acid, so it's slightly different depending on the particular molarity of the solution that we are trying to find the pH of.0168

But, we will get to that in a minute.0191

Another example would be phosphoric acid, like we mentioned, and it has three dissociations, because it has three protons to give up.0193

H3PO4 is in equilibrium with H+ + H2PO4-, and the first acid dissociation constant is 7.5x10-3.0200

And then, this H2PO4- dissociates into H+ + HPO42-; so make sure you keep track of the charges.0214

It's very, very important, especially with these polyprotic acids; they all sort of look alike--the H2PO4, the HPO4, the PO4, H3PO4; it gets very, very confusing, so definitely work very carefully, very, very slowly.0228

Chemistry is very symbol-heavy, and it's easy to get lost in the symbols and make silly mistakes, simply for that--as opposed to conceptual mistakes.0241

The second dissociation constant, the Ka2, is 6.2x10-8.0251

And then, this species, HPO4, the hydrogen phosphate (which is 2-), becomes H+ + PO43-.0258

Now, the phosphate is finally alone, and the third dissociation constant is equal to 4.8x10-13; so clearly, 10-3; 10-8; 10-13: you see the pattern.0270

At is dissociates more, it becomes harder and harder to actually pull off that second and third hydrogen.0285

Let's just go ahead and do an example, and I think everything will make sense.0291

Example 1: We want to calculate the pH of a 4.0 Molar H3PO4 solution, as well as the equilibrium concentrations of H3PO4, H2PO4-, HPO42-, and PO43-.0297

So, not only do we want the hydrogen ion concentration (in other words, the pH, when we actually take the negative log of it), but we want to find the concentration of all of the species in solution, once you take that phosphoric acid and drop it into water to create a 4 Molar solution.0349

What is the concentration of each of those species, in addition to the pH?0365

OK, well, let's just go ahead and start.0368

The first thing we do is the same thing we always do: we check to see the major species in the water before anything happens.0371

That is the whole idea; we want to see what is in there before the situation comes to equilibrium, before the system comes to equilibrium.0377

We want to check out the chemistry; we want to make some decisions about what is going to dominate--what chemistry is going to dominate.0384

There is always going to be one that dominates.0390

OK, so major species: well, phosphoric acid is a weak acid; you notice, Ka1 is 7.5x10-3; that is kind of small, so it's a weak acid.0392

A weak acid means that it's not going to dissociate very much.0405

So, most of it is going to be in this form: H3PO4.0409

That H3PO4 is just going to be floating around in solution, not very dissociated.0413

The major species in the solution: well, you have an H3PO4, and your other major species is just the water.0418

Well, both of these contribute hydrogen ions; well, the Ka of, or the Kw of, water (which happens to be the Ka of water) is 1x10-14.0425

Well, the Ka of this one is 7.5x10-3, 11 orders of magnitude bigger; so we can ignore water's contribution to the hydrogen ion concentration; this is the dominant species.0435

Because that is the dominant species, that is the equilibrium that we are going to work with.0447

The same thing we have always done: major species; decide which is dominant, which is going to control the chemistry; that is the one you concentrate on--ignore everything else.0451

So, let's go ahead and write our equilibrium expression and do our ICE chart.0459

Let's see...yes, that is fine.0465

OK, so we have: H3PO4 that dissociates into H+ + H2PO4-.0470

We have our initial concentration, our change, and our equilibrium.0482

Our initial concentration is 4.00 molarity, and this is 00, so this is before anything has had a chance to come to equilibrium.0486

A certain amount of this H3PO4 is going to dissociate, and for each amount that dissociates, 1:1, 1:1 ratio--that is the amount that shows up in solution of the other species.0495

Therefore, our equilibrium concentration is 4.00-x; this is x; this is x; and now that we have our equilibrium concentrations in terms of x, and we know what the Ka is, we set it equal to each other.0505

7.5x10-3 is the Ka; it is the x, the hydrogen ion concentration, times x, the dihydrogen phosphate concentration, divided by the phosphoric acid concentration, 4.00-x.0521

We do our normal approximation, which we can check the validity of; so this is equal to x2 over 4.00, just to make our math a little bit easier.0540

When we do this, we get x2 (actually, you know what, I'm just going to skip this line altogether; and I'm just going to go ahead)...and you multiply by the 4; you take the square root; you end up with x is equal to 0.173.0548

That is equal to the hydrogen ion concentration; therefore, the pH is the negative log of the hydrogen ion concentration--you get a pH of 0.76.0563

That is our first part: 0.76 is our pH; let's check the validity of what we have; by checking the validity, that means...0574

So let's do it over here; let's say "check validity of our approximation"--I mean, we know it's good, but it's good to just sort of check it.0583

0.173, which is x, divided by 4.0, times 100; you end up with 4.3%.0592

It's close to 5, but it's still under the 5% rule, so we're still actually pretty good; it's not a problem.0600

A perfectly good, valid approximation: we don't need to do the quadratic equation; this is a good concentration; this is the pH.0606

Now let's see where we stand: we found the pH, but we also wanted to find the concentrations of all the other species in solution.0613

Let's see what we have; we have the H+ concentration; that is equal to x, which is 0.173 Molar.0620

Well, the H concentration also happens to be the H2PO4- concentration.0629

The H2PO4- concentration is the same thing, 0.173 Molar.0634

Well, since we have x, we also have the phosphoric acid concentration (H3PO4), which is equal to 4.00-0173, and I'll just go ahead and...no, we'll just do 3.8 Molar (I'll just go ahead and round it up a little bit; sorry about that).0641

So, that is going to be 3.8 Molar; so now, the only thing that we are actually missing is: we are missing the HPO4- concentration and the PO43- (this is 2-; see, again, I am making the same mistakes; I have symbols and charges all over the place).0667

So now, we need the concentration of the hydrogen phosphate, and we need the concentration of the phosphate ion.0688

Well, like we said, an equilibrium is established; it doesn't matter where these species come from (the phosphate, the hydrogen...); the Ka that we have for a given species, for a given equilibrium; if it involves the species that we want, that is good enough.0693

The equilibrium that we do have, that involves the HPO42-, is the following.0713

We have the H2PO4 concentration; well, that is in equilibrium with H+, plus the HPO42-, which is the species that we are looking for.0721

Well, we have this concentration, and we have this concentration; that is this and this; and we also have the second Ka, the second dissociation constant.0734

That is what this is here: this is the second dissociation of phosphoric acid, and the Ka for this one, Ka2, is equal to 6.2x10-8.0742

So now, we can just basically rearrange the equilibrium expression and solve for HPO42-.0757

And then, for the PO43-, we do the same thing with Ka3, the third dissociation constant.0764

Let's go ahead and write everything out, so that we see it.0769

Ka2, based on the equation that was just written (the dissociation of the H2PO4-), becomes the concentration of H+, times the concentration of HPO42-, over the concentration of H2PO4- (oh, this is crazy--all kinds of charges and symbols going on).0774

Therefore, I rearrange this to solve for this species, because I have this, I have this, and I have this; it's a simple algebra problem.0797

We have HPO42-, the concentration--moles per liter--is equal to the Ka2 times the H2PO4- concentration, divided by the H+ concentration.0806

We get: that is equal to (I'm going to move this over here) 6.2x10-8, which is our Ka, times 0.173, which was our concentration of H2PO4; it also happens to be the concentration of our H+.0825

So, these cancel, of course, leaving us with 6.2x10-8 molarity for...that is the HPO42- concentration--that is one of the last things that we needed.0849

Notice, it's very, very, very small.0866

Well, now we want to know the PO43- concentration; that is the final concentration--the final species whose concentration we want.0870

OK, well, now that we have found the HPO42- concentration (it's this), we have an equilibrium.0880

The third equilibrium is (the third dissociation of phosphoric acid--excuse me): HPO42- (oh, wow, you see what I mean--all of these charges, all of these symbols floating around; you're bound to make mistakes--you have to go very carefully) is in equilibrium--it dissociates into H+ + PO43-.0886

We can write a Ka for this: the third dissociation constant is equal to the H+ concentration, times the PO43- concentration, over the HPO42- concentration (which we just found).0912

We rearrange this, and you get: the PO43- concentration is equal to the Ka3, times the HPO42- concentration, divided by the H+ concentration.0929

So, the Ka3 is going to be (let me see: what is the Ka3?) 4.8x10-13, times the HPO4 concentration, which we just found, which is 6.2x10-8, over the hydrogen ion concentration, which we found in the first step: .173.0953

Well, we get a very, very tiny number: we get 1.7x10-19 molarity; that equals the PO43- concentration.0976

There you have it.0990

So, a polyprotic acid: you handle it in the exact same way--most polyprotic acids are going to be weak acids.0993

The only exception is sulfuric; sulfuric is strong in its first dissociation, weak in its second dissociation; we are going to handle a problem in just a minute.0998

But aside from that, you handle it in exactly the same way.1007

And then, if you happen to need the concentrations of the other species--a further dissociation (so, phosphoric acid, dihydrogen phosphate, hydrogen phosphate, phosphate), you use the concentrations that you found, and then you use the next dissociation, and the next dissociation, with the appropriate Ka, to find the concentrations of all the species that you want.1010

About the only difficult problem here, as you saw, was just keeping track of the symbols and not losing your way in the symbology.1035

That is going to be the hardest problem, which, as far as I am concerned, that is the problem you always want--you don't want conceptual problems; you want mechanical problems; those are easy to deal with.1042

OK, so let's do another example; this time, we are going to deal with sulfuric acid, and we are going to do a couple of variations of it.1050

So, let's just start Example 2: We want to calculate the pH of a 1.2 Molar H2SO4.1057

OK, well, so sulfuric acid: let's write down its dissociations.1074

Its first dissociation is: H2SO4 dissociates into H+ + HSO4- (the hydrogen sulfate ion, also called the bisulfate ion).1079

This Ka is large: we said it's a strong acid in its first dissociation; it's large--it doesn't even have a number.1091

You remember, strong acids don't have Kas; they are huge.1097

The second dissociation is: HSO4- dissociates into H+...1101

Actually, I'm not even going to give the equilibrium sign here; this is going to be a one-sided arrow pointing in one direction--full dissociation; there is no H2SO4 left--very, very important to remember that for strong acids.1107

For strong acids, there is an arrow pointing to the right, and that is it; it's not in equilibrium.1119

We have HSO4 going to H+ + SO42-, and the Ka (it's small, but it's actually still pretty large, relatively speaking): 1.2x10-2.1125

So, a little larger than the other weak acids, but it still behaves as a weak acid; there is an equilibrium here; it is still less than 1.1138

OK, we want to calculate the pH; well, let's do what we always do--let's see what the major species are.1145

Be careful here: the major species in solution--you have taken sulfuric acid; you have dropped it into water; sulfuric acid is strong in its first dissociation.1152

Therefore, a solution of sulfuric acid is entirely composed of free hydrogen ion, free hydrogen sulfate ion, and (the other species in water is) H2O.1163

There you go; now, again, we are calculating a pH here; once again, strong dissociation; it's a strong acid in its first dissociation; that means in species, before any equilibrium is reached, it's this and this and this.1179

"Strong acid" means full dissociation--it breaks up completely: there is none of this left.1197

It is all that in solution: this floating around, this floating around, this floating around; what is the hydrogen ion concentration?1202

Well, the question is: Since it's a strong acid in its first dissociation, can we treat it just like any other strong acid and just take the molarity and just take the negative log of it?1211

Can we just do -log of 1.2?1220

Because 1.2 moles per liter produced 1.2 moles per liter of H and 1.2 moles per liter of that, and this is actually pretty weak, it's probably not going to contribute too much.1223

However, we need to make sure; so let's ask our question: Can we just treat this like any other strong acid problem?1235

The answer is maybe; so, we have to check.1263

The reason we have to check is, we need to see if this actually does contribute anything.1269

It is small, but it's not so small (like on the order of 10-5 or 10-6, 10-7; it's 10-2.1274

When you get to the 10-2 range, you're going to have to be careful; and we'll see in a minute what a general rule of thumb is.1284

A general rule of thumb is: when you have second-dissociation constants, such as 10-2, if you are below about 1 molarity, you can't really ignore it; you have to include it.1291

So, not only do you have to take the 1.2 Molar, but some of this HSO4, because it also dissociates, is going to produce a little bit more H+; so you might have to include that; that is what we are going to check.1302

We are going to check to see if the second dissociation is significant--produces enough H+ so that we have to include it into the 1.2.1317

Let's write: first of all, maybe, so we have to check; so now, mind you, our final H+ concentration that we are going to take the negative log of to find the pH is going to equal, in this case, the 1.2+x.1327

This x is going to be any hydrogen ion that comes from this dissociation.1345

It's strong in this first dissociation, so I know that, before anything happens, I know that at least 1.2 moles per liter of hydrogen ion is floating around in solution.1352

However, I need to know if the second dissociation also produces enough hydrogen ion; so I have to add it to this.1360

That is the thing; we are going to compare x and 1.2 to see what the real difference is; if it's small, we can ignore it, and just treat it as a regular strong acid problem; if it's not small, we have to include it before we take the final negative log of that final hydrogen ion concentration.1369

I hope that makes sense.1384

The ICE chart is going to look slightly different; OK.1386

Well, let's go ahead and do our ICE chart, then: so again, we know that we have the 1.2of that, so we don't have to worry about this.1389

We have to worry about this equilibrium: how much H+ is this HSO4 going to produce under these circumstances?1397

So, the equilibrium that we want to look at is the HSO4-: H+ + SO42-.1405

I hope that makes sense: major species, ICE charts...you do ICE charts with weak acids; you don't do them with strong acids.1414

I have at least 1.2 moles per liter of H+ floating around, because it's a strong acid in its first dissociation.1421

I need to check the second dissociation to see if it produces a significant amount of H+ to add to the 1.2.1428

So, Initial, Change, Equilibrium: our initial concentration of HSO4 is 1.2 Molar, because it is produced in the same way that this is produced.1436

When H2SO4 dissociates, it produces 1.2 Molar of this, 1.2 Molar of that.1448

Now, what is our initial H+ concentration?1453

Here is where we have to be careful, and where the ICE chart is different: our initial H+ concentration is also 1.2, because we have free H+ floating around and free HSO4 floating around; that is what these two are.1455

But, there is no SO42- yet, before anything else happens.1471

In solution, before the system comes to equilibrium, this and this are the same from the first dissociation.1475

Now, the change is going to -x here; this is going to be +x here; +x here; we add vertically down; we get 1.2-x for the HSO4; we get 1.2+x...like I said, 1.2+x is going to be our final hydrogen...and we get +x.1482

Now, we form our equilibrium; let's see, our Ka is (oops, here we go with the stray lines again; they show up in the most interesting places; OK--let me just do it off to the side here) 1.2x10-2, equals 1.2+x, times x, over 1.2-x.1504

So now, we have to solve this equation.1542

OK, here is what we are going to do: we are going to presume that x is small in order to do our approximation; then, we are going to check our approximation to see if it's valid.1544

So, when we take x small, watch what happens to this approximation.1556

OK, it's approximately equal to...that means this x disappears--not this x, this x.1561

So, it becomes 1.2 times x, over 1.2.1568

Well, these cancel, and I am left with: x is equal to 1.2x10-2, which is the same as 0.012.1579

I hope you'll forgive me; I actually prefer to work in decimals, as opposed to scientific notation.1590

I like scientific notation, but I just like the way that regular numbers look.1594

OK, now, we said that x is .012; now, we have to go up here: our final hydrogen ion concentration is going to be 1.2 (the original amount floating around), plus the x, plus the 0.012 that came from the second dissociation.1599

That equals 1.2.1627

It equals 1.212, but it's 1.2 to the correct number of significant figures.1632

So again, significant figures become a little bit of an issue; I personally don't care about significant figures all that much; a lot of people sort of criticize me for that, but you know what, it doesn't really matter; I'm more interested in concepts than I am in actual significant figures.1640

I think they are fine, and I think it's good to sort of use them, and in this context, yes, it sort of helps, because you notice: 1.2 plus .012--this is actually pretty small compared to this.1652

Since, working with significant figures, you still end up with 1.2, this says that you are actually justified in ignoring the second dissociation.1665

In other words, it doesn't really contribute all that much, in terms of hydrogen ion; you can just take the negative log of the 1.2.1674

So, here we have the 1.2; so the pH is equal to -log of 1.2, and you get a (let me see what...yes) -0.079 for a pH.1682

Yes, a pH can be negative; that means it's a really, really strong acid.1699

Now, if you ignore the significant figures, like I personally tend to do, I'll just go ahead and put that number down, too.1703

If you ignore the significant figures, and take the final hydrogen ion concentration to equal 1.212--if I include that .012, there is a little bit of extra hydrogen ion in there--well, you know what--your pH is going to be slightly different.1711

This is .084; 0.079, 0.084...you know, it's a small difference; if you are doing analytical work, it's important--for normal work, it's not altogether that important.1737

But you are going to end up with a pH of .08; that is really what it comes down to.1749

So, as it turns out here, the rule of thumb ends up being the following.1754

We notice that, for a solution which is about 1.2 Molar, roughly 1 Molar, the contribution of the hydrogen sulfate dissociation can be ignored.1765

You can treat it like a strong acid problem.1776

Below about 1.0 Molar, 1 Molar, .9 Molar...you can't ignore it; the second dissociation does produce a significant amount of hydrogen ion that has to be included in the concentration that came from the first dissociation.1779

You have to add those two; so, the first dissociation and then a little bit of the second dissociation...that is when you have your final hydrogen ion concentration.1796

For solutions of H2SO4 more dilute than approximately 1 Molar, the full expression for the equilibrium constant must be used.1805

In other words, we can't approximate; we have to do the 1+x, times x, over 1-x; we have to solve the quadratic, which means the quadratic expression must be solved.1842

OK, so now, we'll do a quick example; I'm actually just going to set it up for you (I'm not actually going to solve it), just so you see what it looks like.1864

The Example 3 is: Calculate the pH of a 0.010 Molar H2SO4.1872

This is pretty dilute: .01 Molar--a lot less than 1.1889

So, we can't ignore the second dissociation; OK.1896

Let's take a look at what we have; we have our major species, and again, we have the H+ from the first dissociation; we have the HSO4 from the first dissociation; and we have water.1900

These two are going to dominate the equilibrium; we know what this is already--this is just .01 Molar, because it's fully dissociated (strong).1918

The final hydrogen ion concentration is going to be 0.010+x, and x is the hydrogen ion concentration we get from the dissociation of the HSO4-.1927

That is the equilibrium we want to look at, HSO4- goes to H+ + SO42-; we have Initial; we have Change; we have Equilibrium.1943

The initial concentration is 0.010, right?--and 0.010, strong, dissociates into this and this (first dissociation).1955

This is 0; the change is -x, +x, x; we get 0.010-x over here; we get 0.010+x over here; we get x over here, and we write Ka2, which is 1.2x10-2, is equal to 0.010+x, times x (oops, let me just take out this parentheses here), and then, that is going to be over 0.010-x.1965

That is it; we have to solve this equation.2003

Multiply through; get your x2; get your x term; get your...it's going to be ax2+bx+c=0.2007

b and c and a--they can be negative, if they need to be.2019

And then, you just plug it into the quadratic equation, or use your graphing utility (your TI-83, 84 calculator, your graphing calculator) to find the roots of this equation.2022

Now, you are going to get two roots; check to see what makes sense--one of the roots is completely not going to make sense.2031

The other root will make sense, and that is the whole idea.2037

So, your final hydrogen ion concentration (oops, let's make this a little bit...): when you get x, you are going to get x equal to some number; we'll just put a little box there.2041

Your final hydrogen ion concentration is going to equal the 0.010, plus this boxed number, and your pH is going to equal the negative log of your final hydrogen ion concentration.2054

And that is it: so, for polyprotic acids, treat them like any other equilibrium; usually, most polyprotic acids are weak acids; it is going to be the first dissociation that dominates.2071

You can pretty much ignore the second, third, fourth...I don't think...I don't even know if there is a...well, there is, but...2081

Weak acid--you can pretty much ignore the second and third dissociations.2086

For sulfuric acid, maybe; maybe not.2092

If you are talking about 1 Molar or above, you can ignore the second dissociation; if you are talking about more dilute than 1 Molar (.8, .7, .6...below that), the second dissociation is going to contribute a significant amount of hydrogen ion.2094

You have to take the amount that comes from the original concentration, the first dissociation; find the amount that comes from the second; add them; and then take the negative log of that.2110

I hope that helps; again, you see: it's pretty much all the same--slightly different when you are dealing with multiple dissociations in the case of sulfuric acid.2119

Thank you for joining us here at Educator.com.2128

We will see you next time to discuss the acid-base properties of salts.2130

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