Quiz on Wednesday

Study:

• Difference between aliphatic and aromatic
• Difference between alkanes, alkenes, alkynes
• What is a organic compound
• What is unique about carbon
• Nomenclature

Hydrocarbons

What are hydrocarbons?

• The simplest aliphatic and aromatic compounds
• Only contain hydrogen and carbon
• Can contain single, double, or triple bonds
• Classified by the types of bonds.

Alkanes

• Alkanes are aliphatic hydrocarbons that only contain single bonds.
• The general formula for alkanes is C_nH_2n+2
• Example: CH₄, C₂H₆, C₃H₈
• Considered saturated because carbon is surrounded by the max number of hydrogen.
• Methane is the simples alkane (CH4)
• Each consecutive alkane adds a carbon and its respective hydrogens.
• Methane: 1 carbon
• Ethane: 2 carbons
• Propane: 3 carbon
• Butane: 4 carbons
• Properties of Alkanes
• Very low melting and boiling points. They rise as carbons are added.
• Ex. methane
• Melting point (C):  -183
• Boiling point (C): -164
• Non-polar

Alkenes

• Hydrocarbons that contain double bonds between carbon atoms
• Contain the prefix -ene.
• The smallest is ethene. Why not methene?
• Considered unsaturated, because the double bond prevents the max number of hydrogen from bonding.
• Naming of alkenes requires numbering the carbons to identify the place where the double bond is.
• You start at the carbon that will give you the smallest number.
• This allows us to know where the double bond is. 
• Properties of Alkenes:
• Slightly higher melting/boiling points
• The first couple are gases at room temperature
• Relatively non polar

Alkynes

• Hydrocarbons that contain triple bonds between the carbon atoms
• Uses the prefix -yne
• The simplest is the most common, ethyne (acetylene).

Cyclic Aliphatic Compounds

• Not all hydrocarbons are open chains of carbon atoms.
• Some form a ring.
• 5 and 6 alkane rings are most abundant. 
• Some can have more than one double bond.

Aromatic Structures

• All contain a form of a molecule benzene
• They are called aromatic because they often smell good
• C₆H₆ is the simplest aromatic compound known
• It was hard to figure out the structure:
• Behaves like an alkane, but they knew from the molecular weight that it had several double and triple bonds.
• When they measured the bond length, the found that it should contain 1.5 bond lengths.
• Showed that carbon was in a ring and all had identical bonds
• In 1865, August Kekule proposed the structure. 
• He said that it was a dynamic equilibrium of the two.
• The double bonds were not “tied dow”, but are more or less shared.

Organic Chemistry Introduction

• Organic Compounds: covalently bonded carbon compounds, with the exception of carbonates, carbon oxides, and carbides.
• Biochemistry: the study of complex reactions taking place between organic compounds within living organisms.

Unique Carbon Atom

• Carbon has some unique properties that enable it to form hundreds of thousands of compounds.
• Carbon has 4 valence electrons, requiring 4 bonds to obtain an octet
• Carbon forms strong chemical bonds with other carbon atoms
• Carbon forms stable, almost non polar bonds with hydrogen
• Carbon atoms can bond to a wide variety of atoms
• H, P, O, N, S, the halogens, and even metal atoms.
• Bonds can be straight, branched, and in various lengths.
• They can even form rings
• Can form double and triple bonds

Structural Forumlas

• Structural formulas are used a lot in organic chemistry because molecular formulas can mean various compounds. 
• C₂H₆O can mean ethanol or dimethyl ether
• See page 446 in your books

Classification

• There are approx. 300,000 new organic compounds synthesized for the first time every year.
• It is important to have some categories:
• Aliphatic compounds: without a benzene ring
• Aromatic compounds: with a benzene ring

Redox Reactions

• Oxidation-reduction reactions (redox reactions):
• Reactions involving transfers or shifts of electrons 
• Oxidation:
• A loss of electrons which makes the oxidation number go up
• Occurs mainly in metals
• Occurs in some covalently bonded substances
• Does not require oxygen (that is not what oxidation means!)
• Reduction:
• A gain of electrons that makes the oxidation number go down (reduced)
• Occurs mainly in nonmetals that gain electrons by taking them from metals
• Review of oxidation numbers:
• Rule 1: free atoms = 0
• Rule 2: ion charge = oxidation number
• Rule 3: compound sum = 0
• Rule 4A: Group 1 = +1
• Rule 4B: Group 2 = +2
• Rule 4C: H = +1 or -1
• Rule 4D: O = -2 or -1
• Rule 4E: Group 17 = -1
• Rule 5: sum of ONs in polyatomic ion = charge
• Practice: Assign Oxidation Numbers
• H₂CO₃
• H: +1, O: -2, C: +4

• N2
• N: 0
• Zn(OH)₄^-2
• Zn: +2, H: +1, O: -2

Redox

• Short for reduction­-oxidation
• Pronounced “REE-docs”
• Must occur together (An element cannot take electrons without another one losing them.)
• LEO the GERm
• Lose Electrons Oxidation
• Gain Electrons Reduction
• Determine which element is oxidized and which is reduced?
• Zn + 2H⁺ ➝ Zn²⁺ + H₂
• Zn is oxidized (ON: 0 ➝ +2)
• H⁺ is reduced (ON: +1 ➝ 0)
• 3Hg²⁺ + 2 Fe(s) ➝ 3Hg + 2Fe³⁺
• Hg²⁺: reduced 
• Fe: oxidized
• Oxidizing and Reducing Agents:
• Reducing agent is a substance used to reduce another substance.
• If a substance is oxidized it is the reducing agent.
• Oxidizing agent is a substance used to oxidize another substance.
• If a substance is reduced it is the oxidizing agent.
• Example:
• Zn + 2H⁺ ➝ Zn²⁺ + H₂
• Zn is oxidized (ON: 0 ➝ +2) [REDUCING AGENT]
• H⁺ is reduced (ON: +1 ➝ 0) [OXIDIZING AGENT]
• 3Hg²⁺ + 2 Fe(s) ➝ 3Hg + 2Fe³⁺
• Hg²⁺: reduced [OXIDIZING AGENT]
• Fe: oxidized [REDUCING AGENT]

Balancing Redox Reactions

See practice for the procedure, but here are the steps:

1. Assign oxidation numbers
2. Make half reactions (oxidized reaction and reduced reaction) [forget everything else for now]
3. Balance electrons
4. Add everything back in and balance traditionally [see Chapter 8 for procedure]

Answers to Chapter 17 Practice 28/03

Here are the answers and more or less how to do the process. If you have any questions, you can email me or ask on Monday.

Oxidized/Reducing Agent: Na 

Reduced/Oxidizing Agent: H₂O 

Oxidized/Reducing Agent: HCl 

Reduced/Oxidizing Agent: HNO₃

Oxidized/Reducing Agent: Fe

Reduced/Oxidizing Agent: SnCl₄

Oxidized/Reducing Agent: CO

Reduced/Oxidizing Agent: I₂O₅

Oxidized/Reducing Agent:  Fe²⁺

Reduced/Oxidizing Agent: MnO₄^-

Oxidized/Reducing Agent: Fe

Reduced/Oxidizing Agent: Cu⁺

Review for Test

1. Definitions of acids/bases.
• Properties
• Arrhenius defnition
• Bronsted-Lowry definiton
• Conjugate pairs
• Lewis definition
2. Acid Base Equilibria
3. Self-Ionization of Water (K_w)
4. pH, pOH, [H₃O⁺], [OH^-]
5. pH scale and values
6. Acid-Base Strength (Concentration vs. Strength)
• Ka and Kb
7. Polyprotic Acids
8. Titration
9. Buffers
10. Important Equations:
• Kw = [H₃O⁺][OH^-] = 1.0 x 10^-14
• pH = -log[H₃O⁺]
• [H₃O⁺] = 10^-pH
• pOH = -log[OH^-]
• [OH^-] = 10^-pOH
• pH + pOH = 14
• (M_K)(V_K)=(M_U)(V_U)

1. In the following chemical reactions, identify the acid, base, conjugate acid, and conjugate base.
• HC₂H₃O₂ + H₂O ⇌ H₃O⁺ + C₂H₃O₂^-
• NH₃ + H₂O ⇌ NH₄⁺ + OH^-
• HNO₃ + H₂O ⇌ NO₃^- + H₃O⁺
2. Find the [H₃O⁺] and pH of the following:
• 0.025 M HNO₃
• 3.4 x 10^-4 M H₂SO₄
3. Calculate the [H₃O⁺] and pOH of the following solutions:
• pH = 3.2
• pH = 5.0
• [OH^-] = 8.2 x 10^-9
• pH = 12.4
4. Find the [OH^-] of the following solutions:
• [H₃O⁺] = 3.9 x 10^-6 M
• [H₃O⁺] = 0.0014 M
• pH = 4.2
5. If 35.2 mL of 12M HCl is used to titrate 50.0 mL of KOH, what is the concentration of the KOH?
6. What is the volume of 2.5 M NaOH needed to titrate 25.0 mL of 1.0 M HCl?
7. Given the following equation, find the concentration of NaOH when 24.09 mL of 1.605 M H₂SO₄ is needed to titrate 50.0 mL of NaOH.      H₂SO₄ + 2NaOH ➝ Na₂SO₄ + 2H₂O
8. What is the salt that is formed during the following neutralization reactions?
• Mg(OH)₂ + HCl ➝ ? + H₂O
• H₂SO₄ + 2NH₄O ➝ ? + 2H₂O
• Ni(OH)₂ + 2HClO₄ ➝ ? + 2H₂O
• Mg(OH)₂ + H₂SO₄ ➝ ? + 2H₂O

Answers:

1. Identifying the acid, base, conjugate acid, and conjugate base.
• Acid: HC₂H₃O₂Base: H₂OC. Acid: H₃O⁺C. Base: C₂H₃O₂^-
• Acid: H₂OBase: NH₃C. Acid: OH^-C. Base: NH₄⁺ 
• Acid: HNO₃Base: H₂OC. Acid: H₃O⁺C. Base: NO₃^- 
2. Find the [H₃O⁺] and pH of the following:
• [H₃O⁺] = 0.025 & pH = 1.6
• [H₃O⁺] = 6.8 x 10^-4 & pH = 3.1
3. Calculate the [H₃O⁺] and pOH of the following solutions:
• [H₃O⁺] = 6.0 x 10^-4 & pOH = 10.8
• [H₃O⁺] = 1.0 x 10^-5 & pOH = 9
• [H₃O⁺] = 1.2 x 10^-6 & pOH = 8.1
• [H₃O⁺] = 4.0 x 10^-13 & pOH = 1.6
4. Find the [OH^-] of the following solutions:
• [OH^-] = 2.5 x 10^-9
• [OH^-] = 7.1 x 10^-12
• [OH^-] = 1.6 x 10^-10
5. 8.4 M KOH
6. 10.0 mL of NaOH
7. 1.5 M NaOH
8. What is the salt that is formed during the following neutralization reactions?
• MgCl₂
• (NH₄)₂SO₄
• Ni(ClO₄)₂
• MgSO₄

Acid Base Equilibrium

• Why can we eat and drink some acids and bases and not others?
• The degree to which they release or accept protons
• Equilibrium constants describe how readily acids deprotonate and bases protonate.

Self-Ionization of Water

Water can react with itself.

One molecule accepts and one donates a proton. 

This is called self-ionization

• This reaction is very important. Because it gives us a constant for acid base equilibrium. Kw is the equilibrium constant for water. 
• Kw = [OH^-][H₃O⁺] = 1.0 x 10^-14 M
• Note: M stands for molarity which is moles/L
• Whether a solution is acidic, basic or neutral, the product of the [H3O+] and [OH-] is always equal to Kw.
• Example problem: 
• The [H₃O⁺] in a mild acid is found to be 5 x 10^-7 mol/L. What is the concentration (molarity) of hydroxide ions?
• [OH^-][H₃O⁺] = 1.0 x 10^-14 M
• [OH^-] = 1.0 x 10^-14 M/[H₃O⁺]
• [OH^-] = 1.0 x 10^-14 M/5.0 x 10^-7 M
• [OH^-] = 2.0 x 10^-8 M

pH Scale

• pH stands for “power of hydronium”
• pH is the negative logarithm of the [H₃O⁺]
• pH = -log [H₃O⁺]
• Examples:
• If [H₃O⁺] = 0.0025 M
• pH = -log(0.0025) = 2.6
• If [H₃O⁺] = 4.57 x 10^-9 M
• pH = -log(4.57 x 10^-9 M) = 8.34

pH Values

• A pH of 7 means neutral.
• A pH of 0-7 is an acid.
• A pH of 7-14 is basic

pH Scale

• Example 1:
• The [H₃O⁺] in a shampoo is 2.0 x 10^-5 M. What is the pH of this shampoo?
• pH = -log [H₃O⁺]
• pH = -log(2.0 x 10^-5) = 4.7
• Example 2:
• What is the pH of an aqueous solution of 0.40 g of HI dissolved in 500 mL of water?
• First, convert grams of HI to moles of HI.
• 0.4 g HI x (1 mol HI/127.9 g HI) = 0.00031 mol HI
• Next calculate molarity (M).
• (0.0031 mol HI/500 mL) x (1000 mL/1L) = 6.3 x 10^-3 M 
• Solution:
• pH = -log[H₃O⁺
•       = -log(6.3 x 10^-3 M)
•       =  2.2
• Example 3
• Find the pH of a solution whose [H₃O⁺] equals 9.5 x 10^-8.
• pH = -log[H₃O⁺]
•       = -log(9.5 x 10^-8 M)
•       =  7.02