Transcript Chapter 4

Chapter 4
Types of Chemical
Reactions and Solution
Stoichiometry
Chapter 4
Table of Contents
•
•
•
Precipitation Reactions
Acid–Base Reactions
Oxidation–Reduction Reactions
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3
Section 4.1
Water, the Common Solvent
•
•
•
One of the most
important substances
on Earth.
Can dissolve many
different substances.
A polar molecule
because of its unequal
charge distribution.
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Section 4.1
Water, the Common Solvent
Dissolution of a Solid in a Liquid
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Section 4.2
The Nature of Aqueous Solutions: Strong and Weak Electrolytes
Nature of Aqueous Solutions
•
•
•
Solute – substance being dissolved.
Solvent – liquid water.
Electrolyte – substance that when dissolved in
water produces a solution that can conduct
electricity.
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Section 4.2
The Nature of Aqueous Solutions: Strong and Weak Electrolytes
Nonelectrolytes:
-dissolves in water, produces no ions
-most molecular compounds don’t dissociate in water.
-no current flows
-”like dissolves like”
Ex: sugar, methanol
Weak electrolytes:
-weak acids and bases have some dissociation
-conduct only a small current
Ex: CH3COOH(aq)↔ H+(aq) + CH3COO-(aq)
NH3(aq) + H2O(aq) ↔ NH4+(aq) + OH-(aq)
Strong electrolytes:
-soluble salts, strong acids, & strong bases have complete dissociation
-conduct current very efficiently
Ex: NaCl(aq) ↔ Na+(aq) + Cl-(aq)
HCl(aq) ↔ H+(aq) + Cl-(aq)
NaOH(aq) ↔ Na+(aq) + OH-(aq)
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Section 4.2
The Nature of Aqueous Solutions: Strong and Weak Electrolytes
Good Time to Review Strong Acids……
STRONG
ACIDS
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Section 4.2
The Nature of Aqueous Solutions: Strong and Weak Electrolytes
Lets Review Strong Bases Now….
Notice
Group 1A
are all
soluble
with OHIn
Group 2A
are all
soluble
except
Mg+2
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Section 4.2
The Nature of Aqueous Solutions: Strong and Weak Electrolytes
Electrolyte Behavior
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Section 4.2
The Nature of Aqueous Solutions: Strong and Weak Electrolytes
Electrolyte Behavior
•
Smaller ions have stronger electric field so
they drag more water molecules around with
them.
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Section 4.3
The Composition of Solutions
Chemical Reactions of Solutions
We must know:
 The nature of the reaction.
 Is it acid-base rxn, ppt rxn, etc.
 The amounts of chemicals present in
the solutions.
 How much solute is in the solvent.
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Section 4.3
The Composition of Solutions
Chemistry Humor……
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Section 4.3
The Composition of Solutions
Molarity
•
Molarity (M) = moles of solute per
volume of solution in liters:
M = Molarity =
moles of solute
liters of solution
6 moles of HCl
3 M HCl =
2 liters of solution
mass/molar mass
M = Molarity =
liters of solution
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Section 4.3
The Composition of Solutions
Molarity
•
Intensive property: Independent on amount
In my prep room I have 1 liters of a 6M HCl solution
If I take 50ml of this solution, it is still 6M
•
Temperature dependent
Solutions expand & contract w/temp.
So, volume of the solution will change
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Section 4.3
The Composition of Solutions
Concentration of Ions
•
For a 0.25 M CaCl2 solution:
CaCl2 → Ca2+ + 2Cl–
 Ca2+: 1 × 0.25 M = 0.25 M Ca2+
 Cl–: 2 × 0.25 M = 0.50 M Cl–.
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Section 4.3
The Composition of Solutions
Concept Check
Which of the following solutions contains
the greatest number of ions?
a)
b)
c)
d)
400.0 mL of 0.10 M NaCl.
300.0 mL of 0.10 M CaCl2.
200.0 mL of 0.10 M FeCl3.
800.0 mL of 0.10 M sucrose.
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Section 4.3
The Composition of Solutions
Notice
•
The solution with the greatest number of
ions is not necessarily the one in which:
 the volume of the solution is the
largest.
 the formula unit has the greatest
number of ions.
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Section 4.3
The Composition of Solutions
Dilution
•
•
•
The process of adding water to a
concentrated or stock solution to achieve
the molarity desired for a particular
solution.
Dilution with water does not alter the
numbers of moles of solute present.
Moles of solute before dilution = moles of
solute after dilution
M1V1 = M2V2
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Section 4.4
Types of Chemical Reactions
•
•
•
Precipitation Reactions
Acid–Base Reactions
Oxidation–Reduction Reactions
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Section 4.5
Precipitation Reactions
Precipitation Reaction
•
A double displacement reaction in which
a solid forms and separates from the
solution.
 When ionic compounds dissolve in
water, the resulting solution contains
the separated ions.
 Precipitate – the solid that forms.
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Section 4.5
Precipitation Reactions
The Reaction of K2CrO4(aq) and Ba(NO3)2(aq)
•
Ba2+(aq) + CrO42–(aq) → BaCrO4(s)
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Section 4.5
Precipitation Reactions
Precipitation of Silver Chloride
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Section 4.5
Precipitation Reactions
Precipitates
•
•
•
Soluble – solid dissolves in solution; (aq)
is used in reaction.
Insoluble – solid does not dissolve in
solution; (s) is used in reaction.
Insoluble and slightly soluble are often
used interchangeably.
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Section 4.5
Precipitation Reactions
Simple Rules for Solubility
1. Most nitrate (NO3) salts are soluble.
2. Most alkali metal (group 1A) salts and NH4+ are
soluble.
3. Most Cl, Br, and I salts are soluble (except Ag+,
Pb2+, Hg22+).
4. Most sulfate salts are soluble (except BaSO4, PbSO4,
Hg2SO4, CaSO4).
5. Most OH are only slightly soluble (NaOH, KOH are
soluble, Ba(OH)2, Ca(OH)2 are marginally soluble).
6. Most S2, CO32, CrO42, PO43 salts are only slightly
soluble, except for those containing the cations in
Rule 2.
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Section 4.5
Precipitation Reactions
Concept Check
Which of the following ions form compounds
with Pb2+ that are generally soluble in water?
a)
b)
c)
d)
e)
S2–
Cl–
NO3–
SO42–
Na+
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Section 4.6
Describing Reactions in Solution
Formula Equation (Molecular Equation)
•
•
•
Gives the overall reaction stoichiometry
but not necessarily the actual forms of
the reactants and products in solution.
Reactants and products generally shown
as compounds.
Use solubility rules to determine which
compounds are aqueous and which
compounds are solids.
AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)
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Section 4.6
Describing Reactions in Solution
Complete Ionic Equation
•
Represents as ions all reactants and
products that are strong electrolytes.
Ag+(aq) + NO3(aq) + Na+(aq) + Cl(aq) 
AgCl(s) + Na+(aq) + NO3(aq)
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Section 4.6
Describing Reactions in Solution
Net Ionic Equation
•
Includes only those solution components
undergoing a change.

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.
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Section 4.6
Describing Reactions in Solution
Concept Check
Write the correct formula equation, complete ionic equation,
and net ionic equation for the reaction between cobalt(II)
chloride and sodium hydroxide.
Formula Equation:
CoCl2(aq) + 2NaOH(aq)  Co(OH)2(s) + 2NaCl(aq)
Complete Ionic Equation:
Co2+(aq) + 2Cl(aq) + 2Na+(aq) + 2OH(aq) 
Co(OH)2(s) + 2Na+(aq) + 2Cl(aq)
Net Ionic Equation:
Co2+(aq) + 2Cl(aq)  Co(OH)2(s)
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Section 4.7
Stoichiometry of Precipitation Reactions
Solving Stoichiometry Problems for Reactions in Solution
1. Identify the species present in the combined
solution, and determine what reaction if any
occurs.
2. Write the balanced net ionic equation for the
reaction.
3. Calculate the moles of reactants.
4. Determine which reactant is limiting.
5. Calculate the moles of product(s), as required.
6. Convert to grams or other units, as required.
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Section 4.7
Stoichiometry of Precipitation Reactions
Concept Check (Part I)
10.0 mL of a 0.30 M sodium phosphate
solution reacts with 20.0 mL of a 0.20 M
lead(II) nitrate solution (assume no volume
change).
 What precipitate will form?
lead(II) phosphate, Pb3(PO4)2

What mass of precipitate will form?
1.1 g Pb3(PO4)2
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Section 4.7
Stoichiometry of Precipitation Reactions
Let’s Think About It
•
Where are we going?

•
To find the mass of solid Pb3(PO4)2 formed.
How do we get there?






What are the ions present in the combined solution?
What is the balanced net ionic equation for the
reaction?
What are the moles of reactants present in the
solution?
Which reactant is limiting?
What moles of Pb3(PO4)2 will be formed?
What mass of Pb3(PO4)2 will be formed?
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Section 4.7
Stoichiometry of Precipitation Reactions
Concept Check (Part II)
10.0 mL of a 0.30 M sodium phosphate
solution reacts with 20.0 mL of a 0.20 M
lead(II) nitrate solution (assume no volume
change).
 What is the concentration of nitrate ions
left in solution after the reaction is
complete?
0.27 M
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Section 4.7
Stoichiometry of Precipitation Reactions
Let’s Think About It
•
Where are we going?

•
To find the concentration of nitrate ions left in
solution after the reaction is complete.
How do we get there?


What are the moles of nitrate ions present in the
combined solution?
What is the total volume of the combined
solution?
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Section 4.7
Stoichiometry of Precipitation Reactions
Concept Check (Part III)
10.0 mL of a 0.30 M sodium phosphate
solution reacts with 20.0 mL of a 0.20 M
lead(II) nitrate solution (assume no volume
change).
 What is the concentration of phosphate
ions left in solution after the reaction is
complete?
0.011 M
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Section 4.7
Stoichiometry of Precipitation Reactions
Let’s Think About It
•
Where are we going?

•
To find the concentration of phosphate ions left in
solution after the reaction is complete.
How do we get there?




What are the moles of phosphate ions present in
the solution at the start of the reaction?
How many moles of phosphate ions were used
up in the reaction to make the solid Pb3(PO4)2?
How many moles of phosphate ions are left over
after the reaction is complete?
What is the total volume of the combined
solution?
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41
Section 4.8
Acid–Base Reactions
Acid–Base Reactions (Brønsted–Lowry)
•
•
•
Acid—proton donor
Base—proton acceptor
For a strong acid and base reaction:
H+(aq) + OH–(aq)  H2O(l)
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Section 4.8
Acid–Base Reactions
Neutralization of a Strong Acid by a Strong Base
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Section 4.8
Acid–Base Reactions
Performing Calculations for Acid–Base Reactions
1. List the species present in the combined
solution before any reaction occurs, and decide
what reaction will occur.
2. Write the balanced net ionic equation for this
reaction.
3. Calculate moles of reactants.
4. Determine the limiting reactant, where
appropriate.
5. Calculate the moles of the required reactant or
product.
6. Convert to grams or volume (of solution), as
required.
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44
Section 4.8
Acid–Base Reactions
Acid–Base Titrations
•
•
•
Titration – delivery of a measured volume of
a solution of known concentration (the titrant)
into a solution containing the substance
being analyzed (the analyte).
Equivalence point – enough titrant added to
react exactly with the analyte.
Endpoint – the indicator changes color so
you can tell the equivalence point has been
reached.
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Section 4.8
Acid–Base Reactions
Concept Check
For the titration of sulfuric acid (H2SO4) with
sodium hydroxide (NaOH), how many moles
of sodium hydroxide would be required to
react with 1.00 L of 0.500 M sulfuric acid to
reach the endpoint?
1.00 mol NaOH
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Section 4.8
Acid–Base Reactions
Let’s Think About It
•
Where are we going?

•
To find the moles of NaOH required for the
reaction.
How do we get there?




What are the ions present in the combined
solution? What is the reaction?
What is the balanced net ionic equation for the
reaction?
What are the moles of H+ present in the solution?
How much OH– is required to react with all of the
H+ present?
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Section 4.9
Oxidation–Reduction Reactions
Redox Reactions
•
Reactions in which one or more electrons
are transferred.
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Section 4.9
Oxidation–Reduction Reactions
Reaction of Sodium and Chlorine
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Section 4.9
Oxidation–Reduction Reactions
Rules for Assigning Oxidation States
1. Oxidation state of an atom in an element = 0
2. Oxidation state of monatomic ion = charge of
the ion
3. Oxygen = 2 in covalent compounds (except in
peroxides where it = 1)
4. Hydrogen = +1 in covalent compounds
5. Fluorine = 1 in compounds
6. Sum of oxidation states = 0 in compounds
7. Sum of oxidation states = charge of the ion in
ions
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Section 4.9
Oxidation–Reduction Reactions
Exercise
Find the oxidation states for each of the
elements in each of the following
compounds:
•
•
•
•
•
K2Cr2O7
CO32MnO2
PCl5
SF4
K = +1; Cr = +6; O = –2
C = +4; O = –2
Mn = +4; O = –2
P = +5; Cl = –1
S = +4; F = –1
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Section 4.9
Oxidation–Reduction Reactions
Redox Characteristics
•
•
•
•
Transfer of electrons
Transfer may occur to form ions
Oxidation – increase in oxidation state
(loss of electrons); reducing agent
Reduction – decrease in oxidation state
(gain of electrons); oxidizing agent
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Section 4.9
Oxidation–Reduction Reactions
Concept Check
Which of the following are oxidation-reduction
reactions? Identify the oxidizing agent and the
reducing agent.
a)Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g)
b)Cr2O72-(aq) + 2OH-(aq)  2CrO42-(aq) + H2O(l)
c)2CuCl(aq)  CuCl2(aq) + Cu(s)
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Section 4.10
Balancing Oxidation–Reduction Equations
Balancing Oxidation–Reduction Reactions by Oxidation States
1. Write the unbalanced equation.
2. Determine the oxidation states of all atoms in
the reactants and products.
3. Show electrons gained and lost using “tie
lines.”
4. Use coefficients to equalize the electrons
gained and lost.
5. Balance the rest of the equation by
inspection.
6. Add appropriate states.
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Section 4.10
Balancing Oxidation–Reduction Equations
•
Balance the reaction between solid zinc
and aqueous hydrochloric acid to
produce aqueous zinc(II) chloride and
hydrogen gas.
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Section 4.10
Balancing Oxidation–Reduction Equations
1. What is the unbalanced equation?
•
Zn(s) + HCl(aq)  Zn2+(aq) + Cl–(aq) + H2(g)
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Section 4.10
Balancing Oxidation–Reduction Equations
2. What are the oxidation states for each atom?
•
Zn(s) + HCl(aq)  Zn2+(aq) + Cl–(aq) + H2(g)
0
+1 –1
+2
–1
0
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Section 4.10
Balancing Oxidation–Reduction Equations
3. How are electrons gained and lost?
1 e– gained (each atom)
•
Zn(s) + HCl(aq)  Zn2+(aq) + Cl–(aq) + H2(g)
0
+1 –1
+2
–1
0
2 e– lost
•
The oxidation state of chlorine remains unchanged.
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Section 4.10
Balancing Oxidation–Reduction Equations
4. What coefficients are needed to equalize the
electrons gained and lost?
1 e– gained (each atom) × 2
•
Zn(s) + HCl(aq)  Zn2+(aq) + Cl–(aq) + H2(g)
0
+1 –1
+2
–1
0
2 e– lost
•
Zn(s) + 2HCl(aq)  Zn2+(aq) + Cl–(aq) + H2(g)
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Section 4.10
Balancing Oxidation–Reduction Equations
5. What coefficients are needed to balance the
remaining elements?
•
Zn(s) + 2HCl(aq)  Zn2+(aq) + 2Cl–(aq) + H2(g)
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Section 4.10
Balancing Oxidation–Reduction Equations
Chemical Reactions:
Reactions can actually be categorized in the following
3 main groupings:
1)
The Precipitation
2)
The Redox
3)
The Acid/Base
Synthesis and Decomposition (special case acid/base OR redox)
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Section 4.10
The Precipitation
Balancing Oxidation–Reduction Equations
1
2
3
Look
at
Cation
Anion
Anion
Is it a(an):
Then the compound is
Alkali metal(Grp IA) or ammonium
Nitrate, acetate, or perchlorate
Halide
SOLUBLE
SOLUBLE
SOLUBLE
Except: Ag+, Hg22+,
Pb2+
SOLUBLE
Except: Ba2+, Hg2+,
Pb2+, Ca2+
INSOLUBLE
Except: Rule 1
INSOLUBLE
Except: Rule 1
Likely INSOLUBLE
4
Anion Sulfate
5
6
Anion Hydroxide
(Mg and Ca are slightly soluble)
Anion Sulfide, carbonate, phosphate
7
Cation If you get this far, the compound is
Ex: Solutions of copper(II) nitrate and sodium phosphate are mixed
3Cu2+ + 2PO 3-  Cu (PO )
4
3
42
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Section 4.10
The Precipitation: Now You Try
Balancing Oxidation–Reduction Equations
Aqueous solutions of silver nitrate and sodium chloride are mixed, resulting in the
formation of a precipitate in a solution.
Ag+ + Cl-  AgCl
Solutions of nickel(II) acetate and potassium hydroxide are mixed, resulting in
visible precipitation in the solution.
Ni2+ + 2OH-  Ni(OH)
2
Solutions of iron(III) chloride and lithium carbonate are mixed,
resulting in a visible precipitate in a solution.
2Fe3+ + 3CO 2-  Fe (CO )
3
2
33
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Section 4.10
Redox #1
Balancing Oxidation–Reduction Equations
1)
Single replacement
Magnesium metal is placed in a solution of iron(III)
chloride
3Mg + 2 Fe3+ 
3Mg2+
+ 2Fe
Solid potassium is added to distilled water
2K + 2H2O  2K+ + 2OH- + H2
Solutions of fluorine and sodium chloride are mixed
F2 + 2 Cl-  2F- + Cl2
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Section 4.10
Redox #1: Now You Try
Balancing Oxidation–Reduction Equations
A solution of silver nitrate is poured over copper metal
2Ag+ + Cu  2Ag + Cu2+
Cadmium metal is placed in a solution of tin(II) chloride
Sn2+ + Cd  Cd2+ + Sn
A solution of sodium iodide is place in bromine water
2I- + Br
2
 I + 2Br2
Solutions of tin(II) chloride and iron(III) chloride are mixed
Sn2+ + 2Fe3+  Sn4+ + 2Fe2+
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Section 4.10
Balancing Oxidation–Reduction Equations
Redox #2
Combustion
butanol (C4H10O) is burned in air
C4H10O + 6 O2 
4 CO2 + 5 H2O
calcium is burned in air
2Ca + O2  2CaO
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Section 4.10
Redox #2: Now You Try
Balancing Oxidation–Reduction Equations
Propane (C H ) is burned in air
3 8
C H + 5O  3 CO + 4H O
3 8
2
2
2
Ethyne (C H ) is burned in air
2 2
2C H + 5O  4CO + 2H O
2 2
2
2
2
Magnesium metal is burned in air
2Mg + O
2
 2MgO
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Section 4.10
Synthesis and Decomposition As Redox Now You Try
Balancing Oxidation–Reduction Equations
Hydrogen peroxide is left in light
2H O  O + 2H O
2 2
2
2
Lithium metal is strongly heated in nitrogen gas
3Li + N
 2Li N
2
3
Molten sodium chloride is electrolyzed
2Na+ + 2Cl-  2Na + Cl
2
Aqueous sodium iodide is electrolyzed
2H O + 2I-  H + 2OH- + I
2
2
2
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Section 4.10
Acid/Base:
Balancing Oxidation–Reduction Equations
Bronsted Lowry: H+ transfer
1) Traditional:
Solutions of ethanoic (CH COOH) acid and sodium hydroxide are mixed
3  H O + CH COOCH COOH + OH3
2
3
Look for stoichiometry
“excess” or “equal volumes of equimolar”
A) equal volumes of equimolar solutions of sodium phosphate and hydrochloric
acid are mixed
H+ + PO 3- 
HPO 24
4 to aqueous sodium phosphate
B) excess hydrochloric acid is added
3H+ + PO 3
H PO
4
3 4
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Acid/Base: Now You Try
Section 4.10
Balancing Oxidation–Reduction Equations
Solutions of hydrochloric acid and sodium hydroxide are mixed
H+ + OH-  H O
2
Hydroiodic acid is poured over solid calcium carbonate
2H+ + CaCO  Ca2+ + H O + CO
3
2
2
A solution of nitric acid was poured into an ammonia solution
H+ + NH
3
 NH +
4
Solutions of nitrous acid and potassium hydroxide are mixed
HNO + OH-  H O + NO 2
2
2
Solutions of hydrofluoric acid and methylamine are mixed
HF + CH NH  CH NH + + F3 2
3 3
20. mL of 0.10 M hydrochloric acid is mixed with 5.0 mL of 0.2 M sodium phosphate
2H+ + PO 3-  H PO 4
2 4
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Section 4.10
Final Thoughts
Balancing Oxidation–Reduction Equations
When information is given pertaining to reactivity,
a reaction ALWAYS occurs.
Use checklist approach to categorize reaction type
Remember unstable acids: H CO , H SO
2 3 2 3
and NH OH
4
Sulfuric acid appears as either:
H+ + HSO - (sometimes 2H+ + SO 2-)
4
4
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