Transcript Section 4.9 Oxidation–Reduction Reactions

Chapter 4
Chemical Quantities and Aqueous Reactions
Chapter 4
Chapter 4
Chemical Quantities and Aqueous Reactions
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
4.9
Climate Change and the Combustion of Fossil Fuels
Reaction Stoichiometry: How much Carbon Dioxide
Limiting Reactant, Theoretical Yield, and Percent
Yield
Solution Concentration and Solution Stoichiometry
Types of Aqueous Solutions and Solubility
Precipitation Reactions
Representing Aqueous Reactions: molecular, Ionic
and Complete Ionic Equations
Acid–Base and Gas – Evolution Reactions
Oxidation–Reduction Reactions
2
Section 4.1
Climate Change and the Combustion of Fossil Fuels
Greenhouse Gases
• The temperature on the surface of the earth
would be significantly cooler (~ 60°F) if not for
the presence of greenhouse gases.
– Water vapor, carbon dioxide, methane, nitrous oxide
(NO) and ozone (O3)
• Greenhouse gases trap heat energy from the
sun
3
Section 4.1
Climate Change and the Combustion of Fossil Fuels
The Greenhouse Effect
Fig 4.1
4
Section 4.1
Climate Change and the Combustion of Fossil Fuels
Fossil Fuels and Greenhouse Gases
• One of the greenhouse gases, carbon dioxide,
results from the combustion of fossil fuels.
• This is the balanced equation for the combustion
of octane (gasoline)
• 2 C8H18 (l) + 25 O2 (g) → 16 CO2 (g) + 18 H2O (g)
5
Section 4.2
Reaction Stoichiometry: How much Carbon Dioxide?
Reaction Stoichiometry
• Balanced chemical equations provide the exact
relationships between the amount of reactants
and products.
• 2 C8H18 (l) + 25 O2 (g) → 16 CO2 (g) + 18 H2O (g)
• For example 2 molecules of octane (gasoline)
react with 25 molecules of oxygen to produce 16
molecules of carbon dioxide gas and 18
molecules of water vapor
6
Section 4.2
Reaction Stoichiometry: How much Carbon Dioxide?
Reaction Stoichiometry
• But reactions don’t occur on this scale
• Molecules to moles
• 2 C8H18 (l) + 25 O2 (g) → 16 CO2 (g) + 18 H2O (g)
• 2 moles of octane (gasoline) react with 25 moles
of oxygen to produce 16 moles of carbon dioxide
gas and 18 moles of water vapor
7
Section 4.2
Reaction Stoichiometry: How much Carbon Dioxide?
Reaction Stoichiometry
• The coefficients in a balanced chemical
equation specify the relative amounts in moles
of each of the substance involved the the
reaction.
• 2 C8H18 (l) + 25 O2 (g) → 16 CO2 (g) + 18 H2O (g)
• Lets look at what this means and how we can use
it.
8
Section 4.2
Reaction Stoichiometry: How much Carbon Dioxide?
Making molecules: mol-to-mol Conversions
• 2 C8H18 (l) + 25 O2 (g) → 16 CO2 (g) + 18 H2O (g)
• 2 moles C8H18:16 moles CO2
• We can make 2 conversion factors from this
2 mol C 8H18
16 mol CO 2
16 mol CO 2
2 mol C 8H18
• How do we use them?
9
Section 4.2
Reaction Stoichiometry: How much Carbon Dioxide?
Making Molecules: mol-to-mol Conversions
• Say you are asked how much CO2 is produced
from the combustion of 15.0 moles of octane?
• 2 C8H18 (l) + 25 O2 (g) → 16 CO2 (g) + 18 H2O (g)
10
Section 4.2
Reaction Stoichiometry: How much Carbon Dioxide?
Making Molecules: mol-to-mol Conversions
• Say you are asked how much CO2 is produced
from the combustion of 15.0 moles of octane?
• 2 C8H18 (l) + 25 O2 (g) → 16 CO2 (g) + 18 H2O (g)
16 mol CO 2
15.0 mol C 8H18 x
 1.20 x 10 2 mol CO 2
2 mol C 8H18
• This is fine but we usually don’t think of quantities
in moles. We usually think and measure in
grams.
11
Section 4.2
Reaction Stoichiometry: How much Carbon Dioxide?
Making Molecules: mol-to-mol Conversions
• How many grams of CO2 is produced from the
combustion of 3.1 x 1015 grams of octane?
• 2 C8H18 (l) + 25 O2 (g) → 16 CO2 (g) + 18 H2O (g)
• We can’t go directly from grams to grams
• Use the mol-to-mol conversion.
12
Section 4.2
Reaction Stoichiometry: How much Carbon Dioxide?
Making Molecules: mol-to-mol Conversions
• How many grams of CO2 is produced from the
combustion of 3.1 x 1015 grams of octane?
• 2 C8H18 (l) + 25 O2 (g) → 16 CO2 (g) + 18 H2O (g)
• Grams of octane → moles of octane
• moles of octane → moles of CO2
• moles of CO2 → grams of CO2
13
Section 4.2
Reaction Stoichiometry: How much Carbon Dioxide?
Making Molecules: mol-to-mol Conversions
• How many grams of CO2 is produced from the
combustion of 3.5 x 1015 grams of octane?
• 2 C8H18 (l) + 25 O2 (g) → 16 CO2 (g) + 18 H2O (g)
3.5x1015 g C 8H18 x
1 mol C 8H18
16 mol CO 2
44.01 g CO 2
x
x
= 1.1 x 1016 g CO 2
114.22 g C 8H18 2 mol C 8H18
1 mol CO 2
14
Section 4.2
Reaction Stoichiometry: How much Carbon Dioxide?
Stoichiometric Calculations
•
Chemical equations can be used to relate the
masses of reacting chemicals.
15
Section 4.2
Reaction Stoichiometry: How much Carbon Dioxide?
Stoichiometric Calculations
1. Balance the equation for the reaction.
2. Convert the known mass of the reactant
or product to moles of that substance.
3. Use the balanced equation to set up the
appropriate mol ratios.
4. Use the appropriate mol ratios to
calculate the number of moles of desired
reactant or product.
5. Convert from moles back to grams if
required by the problem.
16
Section 4.2
Reaction Stoichiometry: How much Carbon Dioxide?
Example
For the equation: 4 Cr (s) + 3 O2 (g) → 2 Cr2O3 (s)
How many grams of chromium(III) oxide can be
produced from 15.0 g of solid chromium and excess
oxygen gas?
grams of Cr to moles of Cr
moles of Cr to moles of Cr2O3
moles of Cr2O3 to grams of Cr2O3
grams to moles to moles to grams - GMMG
17
Section 4.2
Reaction Stoichiometry: How much Carbon Dioxide?
4 Cr (s) + 3 O2 (g) → 2 Cr2O3 (s)
15.0 g of Cr to moles of Cr to moles of Cr2O3 to g of Cr2O3
grams of Cr to moles of Cr.
moles of Cr to moles of Cr2O3
15.0 g Cr 
1 mol Cr
= 0.288 mol Cr
52.00 g Cr
2 mol Cr2O3
0.288 mol Cr 
4 mol Cr
moles of Cr2O3 to grams of Cr2O3. 0.144 mol Cr2O3 
Conversion string:
15.0 g Cr 
1 mol Cr
52.00 g Cr

2 mol Cr O
2
4 mol Cr
3

= 0.144 mol Cr2O3
152.00 g Cr2O3
1 mol Cr2O3
152.00 g Cr O
2
1 mol Cr O
2
3
= 21.9 g Cr2O3
= 21.9 g Cr O
2
3
3
18
Section 4.2
Reaction Stoichiometry: How much Carbon Dioxide?
Learning Check
Tin (II) fluoride is added to some dental
products to help prevent cavities. Tin (II)
fluoride is made according to the following
equation:
Sn (s) + 2 HF (aq) → SnF2 (aq) + H2 (g)
How many grams of tin (II) fluoride can be
made from 55.0 g of hydrogen fluoride if there
is plenty of tin available to react?
19
Section 4.2
Reaction Stoichiometry: How much Carbon Dioxide?
Learning Check
Consider the following reaction:
P4 (s) + 5 O2 (g) → 2 P2O5 (s)
If 6.25 g of phosphorus is burned, what mass
of oxygen does it combine with?
20
Section 4.3
Limiting Reactant, Theoretical Yield and Percent Yield
What is a limiting reactant?
• 1 crust + 5 oz sauce + 2 cups cheese → 1 pizza
• Say we have 4 crusts, 10 cups of cheese and 15 oz of sauce.
1 pizza
= 4 pizzas
crust
1 pizza
10 cups cheese x
= 5 pizzas
2 cups cheese
1 pizza
15 oz sauce x
= 3 pizzas
5 oz sauce
4 crusts x
• Even though we have crusts and cheese for 4 or 5 pizzas we
only have sauce for 3. Sauce is the limiting ingredient.
21
Section 4.3
Limiting Reactant, Theoretical Yield and Percent Yield
Limiting Reactants
•
•
•
Limiting reactant – the reactant that is
consumed first and therefore limits the
amounts of products that can be formed.
In our example sauce is the limiting
reactant
The other reactants are considered to be
in excess
22
Section 4.3
Limiting Reactant, Theoretical Yield and Percent Yield
Theoretical yield vs Actual Yield
•
•
So theoretically (in theory) we can make
3 pizzas. But lets say we drop one, or
burn one.
This is our actual yield. For this example
the actual yield would be 2.
23
Section 4.3
Limiting Reactant, Theoretical Yield and Percent Yield
Percent Yield
•
Percent Yield is the actual value over
the theoretical value
% Yield =
2 pizzas
x 100% = 67%
3 pizzas

24
Section 4.3
Limiting Reactant, Theoretical Yield and Percent Yield
Limiting Reactant, Theoretical Yield and
Percent Yield from Initial Reactant Masses
•
•
Balanced equation and masses of two
reactants.
From just this information it is possible to
determine
–
•
Limiting Reactant and Theoretical Yield
If the problem also gives an actual yield
–
You can determine Percent yield
25
Section 4.3
Limiting Reactant, Theoretical Yield and Percent Yield
Example
Urea is a common fertilizer synthesized from ammonia. How many grams
of Urea (CH4N2O) can be produced from the mixture of 10.00 g each of
ammonia and carbon dioxide? 2NH3 + CO2 → CH4N2O + H2O
First step is to determine which reactant makes the fewest moles of
product. This is the limiting reactant.
10.00 g NH3 x
1 mol NH3
17.03 g NH3
10.00 g CO2 x
x
1 mol CO2
44.01 g CO2
1 mol CH4 N 2O
x
2 mol NH3
 0.294 mol CH4 N 2O
1 mol CH4 N 2O
1 mol CO2
Then convert these moles to grams
 0.227 mol CH4 N 2O
0.227 mol x
60.06 g
 13.6 g CH 4 N 2O
mol
26
Section 4.3
Limiting Reactant, Theoretical Yield and Percent Yield
Example
If the actual yield for the previous problem was
10.5 g, calculate the percent yield. The theoretical
yield that we calculated was 13.6 g.
If the actual yield is 3.16 g then percent yield is
10.5 g
% Yield =
x 100% = 77.2%
13.6 g

27
Section 4.3
Limiting Reactant, Theoretical Yield and Percent Yield
Learning Check
Determine the theoretical yield of urea (CH4N2O) produced from 15.0 kg of
ammonia and 25.0 kg of carbon dioxide? 2NH3 + CO2 → CH4N2O + H2O
28
Section 4.3
Limiting Reactant, Theoretical Yield and Percent Yield
Definitions
• Limiting Reactant – reactant that is completely
consumed and limits amount of product
• Reactant in excess – reactant present in
greater quantity than limiting reactant
• Theoretical Yield – amount of product made
based on consumption of all the limiting reactant
• Actual Yield – amount of product actually
produced
• Percent Yield – (actual/theoretical) x 100%
29
Section 4.4
Solution Concentration and Solution Stoichiometry
Aqueous Solutions
• Homogeneous mixture of two substances =
solution:
– Solvent = substance that is the majority component
– Solute = substances that is the minority component
• Chemical reactions dissolved in water are the
most common.
• Aqueous solution
– Solvent = water
– Solute = chemicals dissolved in water
30
Section 4.4
Solution Concentration and Solution Stoichiometry
Solute Concentration
•
•
Concentration tells us how much solute is
dissolved in a given amount of solvent
There are lots of ways to express concentration
−
−
−
mg/mL
g/L
For now we are only going to consider Molarity
31
Section 4.4
Solution Concentration and Solution Stoichiometry
Molarity
moles
of
solute
M = Molarity =
liters of solution
• Notice Molarity is moles per liter of solution not
liter of solvent
• What is the difference?
32
Section 4.4
Solution Concentration and Solution Stoichiometry
Preparing a 1M NaCl Solution
• Weigh out
1.00 mol NaCl
• Add water up
to 1 L
• Do not add 1
L of water
• What is the
difference?
33
Section 4.4
Solution Concentration and Solution Stoichiometry
Solving Molarity Problems
• Basically two types
• 1. Given a mass (or moles) and a volume and
asked for molarity.
• 2. Given a molarity and volume and asked what
mass is required to make this solution.
34
Section 4.4
Solution Concentration and Solution Stoichiometry
Example
0.375 moles of potassium phosphate is
dissolved in enough water to make 1.50 L
of solution. What is the molarity of the
solution?
0.375 mol
= 0.250 M
1.50 L

35
Section 4.4
Solution Concentration and Solution Stoichiometry
Learning Check
1.17 moles of sodium nitrate is dissolved in
enough water to make 12.0 L of solution.
What is the molarity of the solution?
36
Section 4.4
Solution Concentration and Solution Stoichiometry
Example
A 250.0-g sample of potassium phosphate
is dissolved in enough water to make 1.50 L
of solution. What is the molarity of the
solution?
250.0 g K 3PO4 x 1mole K 3PO4 = 0.785 M
1.50 L
212.27 g K 3PO4
37
Section 4.4
Solution Concentration and Solution Stoichiometry
Learning Check
Consider a solution made by dissolving 200.0
g of KOH in a total 350.0 mL of solution.
Calculate the molarity of this solution
38
Section 4.4
Solution Concentration and Solution Stoichiometry
Determining Mass to make a certain
Molarity
A more typical calculation is where you are told to make a
certain volume of a solution of a certain molarity. So you
need to determine the grams you need to weight out.
39
Section 4.4
Solution Concentration and Solution Stoichiometry
Determining Mass to make a certain
Molarity
How many grams of Na3PO4 are needed to make
500.0 mL of a 0.250M solution of Na3PO4.
500.0 mL x
1L
0.250 mol 163.94 g Na 3 PO 4
x
x
= 20.5 g
1000 mL
L
mol
Notice here you start with the volume and end
with the mass.
Molarity is a conversion factor in this problem.
40
Section 4.4
Solution Concentration and Solution Stoichiometry
Learning Check
How many grams of MgCl2 do you need to weigh
out to make 250.0 mL of a 0.500 M solution of
MgCl2.
41
Section 4.4
Solution Concentration and Solution Stoichiometry
Solution Dilution
•
Solutions are often stored as concentrated
stocks.
–
Just like concentrated orange juice
•
To make a reagent we just dilute the
concentrated stock.
• There is a simple relationship for dilution problems
• M1V1 = M2V2
– Where M1 and V1 are the initial molarity and volume
– M2 and V2 are the final molarity and volume
• Lets solve a dilution problem with this formula.
42
Section 4.4
Solution Concentration and Solution Stoichiometry
Solving Dilution Problems
• Lets say we take 15.0 mL of a 10.0 M stock
solution and dilute it to 3.00 L.
–
–
–
–
M1 = 10.0 M
V1 = 15.0 mL = 0.0150 L (units have to be the same)
M2 = ?
V2 = 3.00L
M1V1  M 2V2 so M 2
M1V1
=
V2
10.0 M x 0.0150 L
M2 
= 0.0500 M
3.00 L
43
Section 4.4
Solution Concentration and Solution Stoichiometry
Learning Check
What volume of a 12.0 M HCl solution is
needed to make 750.0 mL of a 1.00 M HCl
solution?
44
Section 4.4
Solution Concentration and Solution Stoichiometry
Learning Check
What is the final concentration when 60.0 mL of a
2.00 M NaOH is used to prepare 0.150 L of NaOH
solution?
45
Section 4.4
Solution Concentration and Solution Stoichiometry
Solution Stoichiometry
• We have already seen how the coefficients in a
balanced equation can be used as conversion
factors between moles of reactants and moles of
products.
• grams reactant → moles reactant → moles
product → grams product
• To convert from grams to moles we used
molecular weight (g/mol) as a conversion factor
46
Section 4.4
Solution Concentration and Solution Stoichiometry
Solution Stoichiometry
• For reactions in solution we start with volumes
• A typical problem might ask what volume of a
solution is required to fully react with another
solution
• Volume reactant → moles reactant → moles
product → volume product
• Volume to moles to moles to volume
• VMMV
• To convert from volume to moles we use
molarity (mol/L) as a conversion factor
47
Section 4.4
Solution Concentration and Solution Stoichiometry
Example
Calculate the volume of 0.250 M AgNO3 that must
be added to completely react with 1.50 L of a
0.100 M NaCl solution according to the following
equation.
AgNO3 (aq) + NaCl (aq) → AgCl (s) + NaNO3 (aq)
0.100 mol NaCl 1 mol AgNO 3
1L
1.50 L x
x
x
 0.600 L
L
1 mol NaCl 0.250 mol AgNO 3
48
Section 4.4
Solution Concentration and Solution Stoichiometry
Learning Check
• What volume of 0.200 M Pb(NO3)2 is required to completely
react with 200. mL of 0.100 M Na3PO4 according to the
following equation.
2 Na3PO4 (aq) + 3 Pb(NO3)2 (aq) → 6 NaNO3 (aq) + Pb3(PO4) (s).
49
Section 4.5
Types of Aqueous Solutions and Solubility
The Polarity of Water
•
Water is one of the most important substances
on Earth.
–
•
Can dissolve many different substances.
The ability of water to dissolve so many
different substances is largely due to the
polarity of water
50
Section 4.5
Types of Aqueous Solutions and Solubility
The Polarity of Water
Fig 4.8
•
Uneven distribution
of electrons
51
Section 4.5
Types of Aqueous Solutions and Solubility
The Polarity of Water
•
Water molecule is
polar.
•
•
It has 2 poles
+ and –
•
Like a battery
Fig 4.8
52
Section 4.5
Types of Aqueous Solutions and Solubility
The Polarity of Water
•
•
•
How does being a polar molecule affect the
ability of water to dissolve solids?
Lets look at two different types of solids
Table salt – sodium chloride (NaCl)
–
•
Ionic compound
Table sugar – sucrose (C12H22O11)
–
molecular compound
53
Section 4.5
Types of Aqueous Solutions and Solubility
Dissolution of an Ionic Solid in Water
Water molecules “pull” ionic solids
apart
Fig 4-10
Negative end (oxygen) surrounds
(hydrates) cations
Positive end (hydrogens) surrounds
(hydrates) anions
54
Section 4.5
Types of Aqueous Solutions and Solubility
Dissolution of an Ionic Solid in Water
Fig 4.7
55
Section 4.5
Types of Aqueous Solutions and Solubility
Dissolution of an Ionic Solid in Water
• In a salt solution
the attraction of the
water molecules
for the ions
overcomes the
attraction of the
ions for each other.
Fig 4-10
• Solution is more
stable
56
Section 4.5
Types of Aqueous Solutions and Solubility
Dissolution of an Molecular Solid in Water
• Sucrose is a molecular solid. Not composed of
ions.
• How does is dissolve in water?
• Sugar molecules have areas that are polar – just
like water
• The partial positive and negative charges on the
sugar interact with the water molecules.
57
Section 4.5
Types of Aqueous Solutions and Solubility
Dissolution of an Molecular Solid in Water
• Polar water
molecules
interact with
polar regions of
a sugar
molecule.
Fig 4-12
58
Section 4.5
Types of Aqueous Solutions and Solubility
Dissolution of an Molecular Solid in Water
• In a sugar solution the
attraction of the water
molecules for the polar
parts of the sugar
molecules overcomes the
attraction of the sugar
molecules for each other.
Fig 4-13
• Solution is more stable
59
Section 4.5
Types of Aqueous Solutions and Solubility
Electrolyte and Nonelectrolyte Solutions
•
•
•
One useful property for characterizing
solutions is electrical conductivity
Whether or not the solution can conduct an
electrical current
Salt solutions can conduct electricity
–
•
Dissolved ions have positive and negative charge
Sugar solutions cannot
–
Dissolved molecules are uncharged
60
Section 4.5
Types of Aqueous Solutions and Solubility
Electrolyte and Nonelectrolyte Solutions
•
•
Substances that dissolve in water to form
solutions that can conduct electricity are called
electrolytes
Substances that dissolve in water to form
solutions but cannot conduct electricity are
called nonelectrolytes
61
Section 4.5
Types of Aqueous Solutions and Solubility
Strong vs Weak Electrolytes
•
Strong electrolytes dissociate 100% in solution
–
–
–
•
Salts
Strong acids
Strong Bases
NaCl, KNO3, Na2SO4
HCl, HNO3
NaOH, KOH
Weak electrolytes dissociate less than 100%
–
Weak acids
HC2H3O2 (acetic acid – vinegar)
62
Section 4.5
Types of Aqueous Solutions and Solubility
The Solubility of Ionic Compounds
•
•
Not all ionic compounds (salts) are soluble in
water
Silver chloride (AgCl) in water just sinks to the
bottom of the beaker
63
Section 4.5
Types of Aqueous Solutions and Solubility
The Solubility of Ionic Compounds
•
•
Why is NaCl soluble and AgCl not?
Remember what happens when we put a salt in
water
–
–
•
•
Water molecules pull the ions apart.
The solution of ions surrounded by water molecules
is more stable than the ions attracted to each other
This is not always true.
Ag+ and Cl– are more stable interacting with
each other than interacting with water molecules
64
Section 4.5
Types of Aqueous Solutions and Solubility
What are the rules that determine solubility?
•
•
•
There are a set of general guidelines that
determine whether or not an ionic compound will
be soluble
Don’t memorize them!
Know how to use them.
65
Section 4.5
Types of Aqueous Solutions and Solubility
Solubility Rules for Ionic Compounds in Water
66
Section 4.5
Types of Aqueous Solutions and Solubility
Concept Check
Indicate if the following compounds are
soluble or insoluble in water?
a)
b
c)
d)
e)
Li2S
LiCl
LiNO3
BaSO4
Mg(OH)2
67
Section 4.5
Types of Aqueous Solutions and Solubility
Learning Check
Which of the following ions form compounds
with Pb2+ that are soluble in water?
a)
b)
c)
d)
e)
S2–
Cl–
NO3–
SO42–
Mg2+
68
Section 4..6 – 4.9
Classification of Chemical Reactions
•
Classification of Chemical Reactions
– Precipitation Reactions
– Acid–Base Reactions
– Oxidation–Reduction Reactions
•
This is just one way to classify reactions
69
Section 4.6
Precipitation Reactions
Precipitation Reaction
•
•
A reaction that occurs when two
solutions are mixed and a solid product
is formed
Precipitate – the solid that forms.
70
Section 4.6
Precipitation Reactions
Precipitation Reaction
•
Aqueous solution of
potassium iodide, KI
– soluble
• Aqueous solution of lead
nitrate, Pb(NO3)2
– soluble
• When these solutions are
mixed, a yellow (insoluble)
solid forms.
71
Section 4.6
Precipitation Reactions
Precipitation Reaction
• The reactants consist of two soluble compounds
• Potassium iodide KI – soluble
– The two ions in solution are K+ and I–
• Lead (II) nitrate Pb(NO3)2 – soluble
– The two ions in solution are Pb2+ and NO3–
72
Section 4.6
Precipitation Reactions
Precipitation Reaction
• Two products form
• Potassium nitrate KNO3 – soluble
– The two ions in solution are K+ and NO3 –
• Lead (II) iodide PbI2 – insoluble
– Yellow precipitate
73
Section 4.6
Precipitation Reactions
Precipitation Reaction
•
What is the equation that describes this
chemical change?
•
•
The cations and anions exchange partners
Then use solubility rules to determine if the
products are soluble or insoluble.
74
Section 4.6
Precipitation Reactions
Precipitation Reaction
•
What happens if two solutions are mixed and
none of the possible products precipitate?
75
Section 4.6
Precipitation Reactions
Precipitation Reaction
•
Aqueous solution of
potassium iodide, KI (aq)
– Soluble – K+ and I–
• Aqueous solution of
sodium chloride, NaCl (aq)
– Soluble – Na+ and Cl–
• When these solutions are
mixed, nothing happens.
• All the ions stay in solution.
76
Section 4.6
Precipitation Reactions
Precipitation Reaction
•
•
Write the equation for the following precipitation
reaction
Na2CO3 (aq) + NiCl2 (aq) →
–
–
•
•
Exchange the cations and anions for the products
Na2CO3 (aq) + NiCl2 (aq) → NaCl + NiCO3
–
•
The ions for sodium carbonate are Na+ and CO32–
The ions for nickel (II) chloride are Ni2+ and Cl–
NaCl is soluble, NiCO3 is insoluble
Na2CO3 (aq) + NiCl2 (aq) → 2 NaCl (aq) +
NiCO3(s)
77
Section 4.6
Precipitation Reactions
Learning Check
•
•
Write the equation for the following precipitation
reaction
K2SO4 (aq) +Ba(NO3)2 (aq) →
78
Section 4.6
Precipitation Reactions
Concept Check
Use the solubility rules to predict the products
when aqueous solutions of silver nitrate
(AgNO3) and sodium chloride (NaCl) are
mixed
79
Section 4.7
Representing Aqueous Reactions: molecular, Ionic and Complete Ionic Equations
Molecular Equation
•
•
Chemical equation showing the complete,
neutral formula for every compound in a
reaction.
Reactants and products generally shown as
compounds.
AgNO3 (aq) + NaCl (aq) → AgCl (s) + NaNO3 (aq)
80
Section 4.7
Representing Aqueous Reactions: molecular, Ionic and Complete Ionic Equations
Complete Ionic Equation
•
Chemical equation showing all of the
species as they are actually present in
solution.
Ag+ (aq) + NO3- (aq) + Na+ (aq) + Cl- (aq) →
AgCl (s) + Na+ (aq) + NO3- (aq)
81
Section 4.7
Representing Aqueous Reactions: molecular, Ionic and Complete Ionic Equations
Net Ionic Equation
•
Includes only those species that actually
change during the reaction

Show only components that actually react.
Ag+ (aq) + Cl- (aq) → AgCl (s)
•
Spectator ions are not included (ions that
do not participate directly in the reaction).

Na+ and NO3 are spectator ions.
82
Section 4.7
Representing Aqueous Reactions: molecular, Ionic and Complete Ionic Equations
Molecular Equation:
AgNO3 (aq) + NaCl (aq) → AgCl (s) + NaNO3 (aq)
Complete Ionic Equation:
Ag+ (aq) + NO3- (aq) + Na+ (aq) + Cl- (aq) → AgCl (s) + Na+ (aq) + NO3- (aq)
Net Ionic Equation:
Ag+ (aq) + Cl- (aq) → AgCl (s)
Notice the states of the reactants are indicated for every species.
83
Section 4.7
Representing Aqueous Reactions: molecular, Ionic and Complete Ionic Equations
Concept Check
Write the correct molecular, complete ionic, and net
ionic equation for the reaction between cobalt(II)
chloride and sodium hydroxide.
Don’t forget to balance and indicate state for each
species.
84
Section 4.8
Acid–Base and Gas – Evolution Reactions
Acid-Base Reactions
• Acid – Substance that Produces H+ ions in
aqueous solution
• Base – Substance that produces OH– ions in
aqueous solution
• Arrhenius Definition of acids and bases
– Learn more about acid/base definitions in Ch. 15
85
Section 4.8
Acid–Base and Gas – Evolution Reactions
Acid-Base Reactions
• HCl is an acid because it produces protons in
aqueous solution
– HCl (aq) → H+ (aq) + Cl– (aq)
• H+ in solution normally associate with a water
molecule to form a hydronium ion.
– H+ (aq) + H2O (l) → H3O+ (aq)
86
Section 4.8
Acid–Base and Gas – Evolution Reactions
Acid-Base Reactions
• HCl dissolved in water is written this way
• HCl (aq) + H2O (l) → H3O+ (aq) + Cl– (aq)
87
Section 4.8
Acid–Base and Gas – Evolution Reactions
Acid-Base Reactions
• Mono vs polyprotic acids
– HCl – monoprotic acid – one ionizable H+
– H2SO4 – diprotic – two ionizable H+
88
Section 4.8
Acid–Base and Gas – Evolution Reactions
Ionization of HCl and H2SO4
• When these acids are dissolved in water
• HCl completely ionizes into H+ and Cl–
• The first hydrogen on H2SO4 completely ionizes
– Behaves like a strong acid
• The second hydrogen from H2SO4 only partially
ionizes
– behaves like a weak acid.
– Chapter 15 will go into why the first and second H+ on
diprotic acids behave differently.
89
Section 4.8
Acid–Base and Gas – Evolution Reactions
Acid-Base Reactions
• Base – Substance that produces OH– ions in
aqueous solution
• Strong bases (metal hydroxides)
– NaOH – 1 mol OH per mol NaOH
– Ca(OH)2 – 2 moles OH per mol Ca(OH)2
• Weak base
– Ammonia (NH3) is a weak base
90
Section 4.8
Acid–Base and Gas – Evolution Reactions
Acid Base reaction
• 1 HCl combines with 1 NaOH
• NaOH (s) + HCl (l) → NaCl (aq) + H2O(l)
• 1 H2SO4 combines with 2 NaOH
• 2NaOH (aq) H2SO4 (aq) + → Na2SO4 (aq) + 2H2O (l)
91
Section 4.8
Acid–Base and Gas – Evolution Reactions
Ionization of Acids
• Ionization of Acids vs Dissociation of Bases
• When these acids are dissolved in water the
hydrogens ionize (ions are formed and get
released)
– Acids are molecular/covalent compounds
• When strong bases are dissolved in water the
ions dissociate (ions already exist and come
apart)
– Bases are ionic compounds
92
Section 4.8
Acid–Base and Gas – Evolution Reactions
Acid-Base Reactions
• There are only 6 acids that are strong acids (all the rest
are weak) and one type of base (OH base) that is a
strong base (all other bases are weak bases.)
93
Section 4.8
Acid–Base and Gas – Evolution Reactions
Acid-Base Reactions
•
Aqueous solution of
Hydrochloric Acid, HCl (aq)
– Ions – H+ and Cl–
• Aqueous solution of sodium
hydroxide, NaOH (aq)
– Ions – Na+ and OH–
• H+ ion from the acid
combines with the OH– from
the base to form liquid water.
• The other two ions form a
soluble salt.
94
Section 4.8
Acid–Base and Gas – Evolution Reactions
Acid-Base Reactions
•
What is the equation that describes this
chemical change?
•
The products of the reaction between a strong
acid and a strong base are always water and
a salt.
These reactions are also called neutralization
reactions
•
95
Section 4.8
Acid–Base and Gas – Evolution Reactions
Acid-Base Reactions
•
•
Net ionic equation for the reaction of a strong acid
and strong base
Molecular Equation
NaOH (s) + HCl (l) → NaCl (aq) + H2O(l)
•
Complete ionic equation
Na+ (aq) + OH– (aq) + H+ (aq) + Cl– (aq) → Na+ (aq) + Cl – (aq) + H2O (l)
•
Net ionic equation
H+ (aq) + OH− (aq) → H2O(l)
96
Section 4.8
Acid–Base and Gas – Evolution Reactions
Concept Check
Write the correct molecular, complete ionic, and net
ionic equation for the reaction between sulfuric acid
and potassium hydroxide.
Don’t forget to balance and indicate state for each
species.
97
Section 4.8
Acid–Base and Gas – Evolution Reactions
Acid Base Titrations
•
•
•
Can apply the principles of acid-base
neutralization and reaction stoichiometry to
titration
In a titration there is a solution of a know
concentration (called a standard or the titrant)
added to a solution of an unknown
concentration (called the unknown or the
analyte)
The purpose is to determine the concentration
of the unknown.
98
Section 4.8
Acid–Base and Gas – Evolution Reactions
Acid Base Titrations
•
•
•
So lets say the standard (titrant) is the base
and the unknown (analyte) is an acid.
At the equivalence point, the number of moles
of OH– equal the moles of H+, the titration is
complete.
The equivalence point is usually signaled by an
indicator – a dye that changes color
depending on acidity/basicity
99
Section 4.8
Acid–Base and Gas – Evolution Reactions
Acid Base Titrations
Fig 4-19
100
Section 4.8
Acid–Base and Gas – Evolution Reactions
Acid Base Titrations
Fig 4-20
101
Section 4.8
Acid–Base and Gas – Evolution Reactions
Example
The titration of 25.0 mL of an HCl solution of unknown
concentration requires 55.0 mL of 0.250 M NaOH. What is the
concentration of the HCl?
Molecular equation
NaOH (s) + HCl (l) → NaCl (aq) + H2O(l)
Determine moles of standard (known - titrant)
1L
0.250 mol NaOH
55.0 mL NaOH x
x
= 0.0138 mol NaOH
1000 mL
L
Determine moles of unknown (anylate)
1 mol HCl
0.0138 mol NaOH x
= 0.0138 mol HCl
1 mol NaOH
Determine concentration of unknown
0.0138 mol HCl 1000 mL
x
= 0.552 M HCl
25.0 mL
L
102
Section 4.8
Acid–Base and Gas – Evolution Reactions
Concept Check
The titration of 35.0 mL of an H2SO4 solution of
unknown concentration requires 15.0 mL of 0.150
M NaOH. What is the concentration of the H2SO4?
103
Section 4.8
Acid–Base and Gas – Evolution Reactions
Example
What volume of a 0.100 M HCl solution is needed
to neutralize 25.0 mL of 0.350 M NaOH?
Molecular equation
NaOH (s) + HCl (l) → NaCl (aq) + H2O (l)
Calculate moles of reactant
25.0 mL x
1L
0.350 mol NaOH
x
= 0.00875 mol NaOH
1000 mL
L
Determine volume of HCl to neutralize NaOH.
0.00875 mol NaOH x
1 mol HCl
1L
x x
= 0.0875 L HCl
1 mol NaOH
0.100 mol HCl
104
Section 4.8
Acid–Base and Gas – Evolution Reactions
Concept Check
What volume of a 0.200 M H2SO4 solution is
needed to neutralize 55.0 mL of 0.150 M NaOH?
105
Section 4.9
Oxidation–Reduction Reactions
Definition of Redox
• Many redox reactions involve the reaction of a
substance with oxygen
• Rusting of iron
• 4 Fe (s) + 3 O2 (g) → 2 Fe2O3 (s)
• Combustion of octane
• 2 C8H18 (l) + 25 O2 (g) → 16 CO2 (g) + 18 H2O (g)
• Combustion of hydrogen
• 2 H2 (g) + O2 (g) → 2 H2O (g)
106
Section 4.9
Oxidation–Reduction Reactions
Definition of Redox
• Redox reactions don’t have to involve oxygen
• Formation of sodium chloride from solid sodium
and gaseous chlorine is a redox reaction
• 2 Na (s) + Cl2 (g) → 2 NaCl (g)
107
Section 4.9
Oxidation–Reduction Reactions
Definition of Redox
• What all redox reactions have in common is the
transfer of electrons from one reactant to
another.
• Can be the transfer of entire electrons
(formation of an ionic compound)
– 2 Na (s) + Cl2 (g) → 2 NaCl (g)
– 2 Na+ + 2 Cl – → 2 NaCl
– Electrons moves from Na to Cl
108
Section 4.9
Oxidation–Reduction Reactions
Definition of Redox
• In a Redox reaction
electrons transfer from one
reactant to the other
• Can be the transfer of
electron density
(formation of molecular
compound)
• H2 (g) + Cl2 (g) →2 HCl (g)
109
Section 4.9
Oxidation–Reduction Reactions
Definition of Redox
• Oxidation – the atom that loses
electrons or electron density is said to
be oxidized in a redox reaction
• Reduction – the atom that gains
electrons or electron density is said to
be reduced in a redox reaction
110
Section 4.9
Oxidation–Reduction Reactions
Definition of Redox
• 2 Na (s) + Cl2 (g) → 2 NaCl (g)
− Sodium loses electrons – is oxidized
− Chlorine gains electrons – is reduced
• H2 (g) + Cl2(g) →2 HCl (g)
− Hydrogen loses electron density – is oxidized
− Chlorine gains electron density – is reduced
• Metals tend to lose electrons – get oxidized
• Nonmetals tend to gain electrons – get reduced
− in a struggle between 2 nonmetals the more electronegative (Ch
8) wins
111
Section 4.9
Oxidation–Reduction Reactions
Oxidation States
•
•
When we look at a reaction how do we determine
if it is a redox reaction or not?
When an ionic compound forms it is fairly
straightforward.
–
•
Because we know ionic compounds are formed from
ions we know entire electrons had to move from one
reactant to another
What about molecular compounds – reactions
between non metals?
112
Section 4.9
Oxidation–Reduction Reactions
Oxidation States
•
•
•
Oxidation states are a way to keep track of
electron density in molecular compounds
Oxidation state (oxidation number) is the charge
an atom in a compound would have if all the
shared electrons in the bond belonged entirely to
that atom.
What does this mean?
113
Section 4.9
Oxidation–Reduction Reactions
Oxidation States
•
•
•
•
Lets consider our HCl
example from a few slides
ago
In this example the Cl atom
attracts the electrons more
strongly than the H atom.
The Cl has gained the H
electron (charge of –1)
The H has lost its electron
(charge of +1)
114
Section 4.9
Oxidation–Reduction Reactions
Oxidation States
•
•
•
•
•
Do I have to figure out oxidation number
every time?
Nope
There are a handy (and short) set of rules
Don’t memorize them
Know how to use them.
115
Section 4.9
Oxidation–Reduction Reactions
Rules for Assigning Oxidation States
1. Oxidation state of an atom in a free
element = 0
2. Oxidation state of monatomic ion =
charge of the ion
3. The sum of oxidation states of all
atoms in a neutral molecule is 0 or
an ion is the charge of the ion.
4. In compounds metals have positive
oxidation states = group number
5. For covalent compounds non
metals are assigned oxidation
states according to the table.
116
Section 4.9
Oxidation–Reduction Reactions
Practice
•
CO2
Covalent O = – 2 so C = +4
•
Br2
Elemental form so 0
•
SF6
Covalent F = – 1 so S = + 6
•
NO3 –
O = – 2 but N = +5 so that overall charge
stays at – 1
•
NaNO3
This is an ionic compound Na is +1
because the polyatomic ion is – 1. O = – 2
and N = +5 just like above
•
H2SO4
O = – 2 and H = +1 which leaves +6 for S
117
Section 4.9
Oxidation–Reduction Reactions
Concept Check
Find the oxidation states for each of the
elements in each of the following
compounds:
•
•
•
•
•
CO32MnO2
PCl5
SF4
K2Cr2O7
118
Section 4.9
Oxidation–Reduction Reactions
Identifying Redox Reactions
•
•
•
•
Earlier we defined
Oxidation as loss of electrons
Reduction as gain of electrons
When we look at this reaction how can we tell
which element (if any) has been oxidized or
reduced?
•
C + 2S→CS2
119
Section 4.9
Oxidation–Reduction Reactions
Identifying Redox Reactions
•Oxidation state of C changes from 0 to +4
–Loss of 4 electrons
•Oxidation state of S changes from 0 to –2
–Gain of 2 electrons (x 2 S = –4)
•Since C lost electrons it was oxidized
•Since S gained electrons it was reduced
120
Section 4.9
Oxidation–Reduction Reactions
Learning Check
• Use oxidation numbers to identify which species
is oxidized and which species is reduced for the
combustion of methane
• CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (g)
121
Section 4.9
Oxidation–Reduction Reactions
Example
•
CH4 (g) + 2 O2(g) → CO2(g) + 2 H2O(g)
–4 +1
0
+4 –2
+1 –2
Reduction
Oxidation
C loses electrons
is oxidized - process is oxidation
is the reducing agent
Oxygen (O2) gains electrons
is reduced – process is reduction
is the oxidizing agent
122
Section 4.9
Oxidation–Reduction Reactions
Concept Check
Which of the following are redox reactions?
Identify which atom is oxidized and which is reduced.
Identify the oxidizing agent and the reducing agent.
a) HBr (aq) + KOH (aq) → H2O (l) + KBr (aq)
b) PbO (s) +CO (g) → Pb (s) +CO2 (g)
c) 2 CuCl (aq) → CuCl2 (aq) + Cu (s)
123
Section 4.9
Oxidation–Reduction Reactions
Conceptual Connection
• Which statement is true?
• (a) A redox reaction involves either the transfer of an
electron or a change in the oxidation state of an element
• (b) If any of the reactants or products in a reaction contain
oxygen, the reaction is a redox reaction
• (c) In a reaction, oxidation can occur independently of
reduction.
• (d) In a redox reaction, any increase in the oxidation state
of a reactant must be accompanied by a decrease in the
oxidation state of a reactant.
124
Section 4.9
Oxidation–Reduction Reactions
Conceptual Connection
• Explain why a, b and c are false.
• (a) A redox reaction involves either the transfer of an
electron or a change in the oxidation state of an element
• (b) If any of the reactants or products in a reaction contain
oxygen, the reaction is a redox reaction
• (c) In a reaction, oxidation can occur independently of
reduction.
125
Section 4.9
Oxidation–Reduction Reactions
Combustion Reactions
• Combustion reactions are a type of redox
reaction
• Oxygen is the oxidizing agent
– Readily accepts electrons
• And some type of fuel is the reducing agent
− Readily gives up electrons
126
Section 4.9
Oxidation–Reduction Reactions
Combustion Reactions
• Fuels that contain C and H always produce CO2
and H2O as products
– CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (g)
– C2H5OH (l) + 3 O2 (g) → 2 CO2 (g) + 3 H2O(l)
• Other combustion reactions are the reaction of
oxygen with only C or H
– C (s) + O2 (g) → CO2 (g)
– H2 (g) + O2 (g) → 2H2O (g)
127
Section 4.9
Oxidation–Reduction Reactions
Learning check
• Write a balanced equation for the combustion of
propane gas C3H8.
128