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Chapter 4
Chemical
Reactions
Contents and Concepts
Ions in Aqueous Solution
Explore how molecular and ionic substances
behave when they dissolve in water to form
solutions.
1. Ionic Theory of Solutions and Solubility Rules
2. Molecular and Ionic Equations
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Types of Chemical Reactions
Investigate several important types of reactions
that typically occur in aqueous solution:
precipitation reactions, acid–base reactions, and
oxidation–reduction reactions.
3. Precipitation Reactions
4. Acid–Base Reactions
5. Oxidation–Reduction Reactions
6. Balancing Simple Oxidation–Reduction
Equations
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Working with Solutions
Now that we have looked at how substances behave
in solution, it is time to quantitatively describe these
solutions using concentration.
7. Molar Concentration
8. Diluting Solutions
Quantitative Analysis
Using chemical reactions in aqueous solution,
determine the amount of substance or species
present in materials.
9. Gravimetric Analysis
10. Volumetric Analysis
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Learning Objectives
Ions in Aqueous Solution
1. Ionic Theory of Solutions and Solubility Rules
a. Describe how an ionic substance can form
ions in aqueous solution.
b. Explain how an electrolyte makes a
solution electrically conductive.
c. Give examples of substances that are
electrolytes.
d. Define nonelectrolyte and provide an
example of a molecular substance that is a
nonelectrolyte.
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e. Compare the properties of solutions that
contain strong electrolytes and weak
electrolytes.
f. Learn the solubility rules for ionic
compounds.
g. Use the solubility rules.
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2. Molecular and Ionic Equations
a. Write the molecular equation of a chemical
reaction.
b. From the molecular equations for both
strong electrolytes and weak electrolytes,
determine the complete ionic equation.
c. From the complete ionic equation, write
the net ionic equation.
d. Write net ionic equations.
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Types of Chemical Reactions
3. Precipitation Reactions
a. Recognize precipitation (exchange)
reactions.
b. Write molecular, complete ionic, and net
ionic equations for precipitation reactions.
c. Decide whether a precipitation reaction will
occur.
d. Determine the product of a precipitation
reaction.
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4. Acid–Base Reactions
a. Understand how an acid–base indicator is
used to determine whether a solution is
acidic or basic.
b. Define Arrhenius acid and Arrhenius base.
c. Write the chemical equation of an
Arrhenius base in aqueous solution.
d. Define Brønsted–Lowry acid and
Brønsted–Lowry base.
e. Write the chemical equation of a
Brønsted–Lowry base in aqueous solution
f. Write the chemical equation of an acid in
aqueous solution using the hydronium ion.
g. Learn the common strong acids and strong
bases.
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h. Distinguish between a strong acid and a
weak acid and the solutions they form.
i. Distinguish between a strong base and a
weak base and the solutions they form.
j. Classify acids and bases as strong or
weak.
k. Recognize neutralization reactions.
l. Write an equation for a neutralization
reaction.
m. Write the reaction equations for a
polyprotic acid in aqueous solution.
n. Recognize acid–base reactions that lead
to gas formation.
o. Write an equation for a reaction with gas
formation.
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5. Oxidation–Reduction Reactions
a. Define oxidation–reduction reaction.
b. Learn the oxidation−number rules.
c. Assign oxidation numbers.
d. Write the half−reactions of an oxidation–
reduction reaction.
e. Determine the species undergoing
oxidation and reduction.
f. Recognize combination reactions,
decomposition reactions, displacement
reactions, and combustion reactions.
g. Use the activity series to predict when
displacement reactions will occur.
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6. Balancing Simple Oxidation–Reduction
Equations
a. Balance simple oxidation–reduction
reactions by the half−reaction method.
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Working with Solutions
7. Molar Concentration
a. Define molarity or molar concentration of a
solution.
b. Calculate the molarity from mass and volume.
c. Use molarity as a conversion factor.
8. Diluting Solutions
a. Describe what happens to the concentration
of a solution when it is diluted.
b. Perform calculations associated with dilution.
c. Describe the process for diluting a solution.
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Quantitative Analysis
9. Gravimetric Analysis
a. Determine the amount of a species by
gravimetric analysis.
10. Volumetric Analysis
a. Calculate the volume of reactant solution
needed to perform a reaction.
b. Understand how to perform a titration.
c. Calculate the quantity of substance in a
titrated solution.
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A strong electrolyte
dissolves to produce
ions. The ions, as
moving charges,
complete the circuit.
When a light bulb is
attached to the circuit, it
shines.
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A strong electrolyte is
an electrolyte that
exists in solution
almost entirely as ions.
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A weak electrolyte is an
electrolyte that dissolves
in water to give a
relatively small
percentage of ions. As a
result, the light bulb
shines weakly.
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Compounds that dissolve readily are said to be
soluble.
Compounds that dissolve very little are said to be
insoluble.
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Solubility Rules
1. Group IA and ammonium compounds are
soluble.
2. Acetates and nitrates are soluble.
3. Most chlorides, bromides, and iodides are
soluble.
Exceptions:
AgCl, Hg2Cl2, PbCl2;
AgBr, HgBr2, Hg2Br2, PbBr2;
AgI, HgI2, Hg2I2, PbI2
4. Most sulfates are soluble.
Exceptions:
CaSO4, SrSO4, BaSO4,
Ag2SO4, Hg2SO4, PbSO4
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5. Most carbonates are insoluble.
Exceptions:
Group IA carbonates and (NH4)2SO4
6. Most phosphates are insoluble.
Exceptions:
Group IA phosphates and (NH4)3PO4
7. Most sulfides are insoluble.
Exceptions:
Group IA sulfides and (NH4)2S
8. Most hydroxides are insoluble.
Exceptions:
Group IA hydroxides,
Ca(OH)2, Sr(OH)2, Ba(OH)2
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Molecular Equation
A chemical equation in which the reactants and
products are written as if they were molecular
substances, even though they may actually exist in
solution as ions.
State symbols are include: (s), (l), (g), (aq).
For example:
AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)
Although AgNO3, NaCl, and NaNO3 exist as ions in
aqueous solutions, they are written as compounds
in the molecular equation.
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Complete Ionic Equation
A chemical equation in which strong electrolytes
are written as separate ions in the solution. Other
reactants and products are written in molecular
form. State symbols are included: (s), (l), (g), (aq).
For example:
AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq)
In ionic form:
Ag+(aq) + NO3−(aq) + Na+(aq)Cl−(aq) 
AgCl(s) + Na+(aq) + NO3−(aq)
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Spectator Ion
An ion in an ionic equation that does not take part
in the reaction. It appears as both a reactant and a
product.
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Net Ionic Equation
A chemical equation in which spectator ions are
omitted. It shows the reaction that actually occurs
at the ionic level.
For example:
Ag+(aq) + NO3−(aq) + Na+(aq) + Cl−(aq) 
AgCl(s) + Na+(aq) + NO3−(aq)
In net ionic form:
Ag+(aq) + Cl−(aq)  AgCl(s)
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?
Decide whether the following reaction
occurs. If it does, write the molecular,
ionic, and net ionic equations.
KBr + MgSO4 
1. Determine the product formulas:
K+ and SO42− make K2SO4
Mg2+ and Br − make MgBr2
2. Determine whether the products are soluble:
K2SO4 is soluble
MgBr2 is soluble
KBr + MgSO4  no reaction
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?
Decide whether the following reaction
occurs. If it does, write the molecular,
ionic, and net ionic equations.
NaOH + MgCl2 
1. Determine the product formulas:
Na+ and Cl− make NaCl
Mg2+ and OH− make Mg(OH)2
2. Determine whether the products are soluble:
NaCl is soluble
Mg(OH)2 is insoluble
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Molecular Equation
(Balance the reaction and include state symbols)
2NaOH(aq) + MgCl2(aq) 
2NaCl(aq) + Mg(OH)2(s)
Ionic Equation
2Na+(aq) + 2OH−(aq) + Mg2+(aq) + 2Cl−(aq) 
2Na+(aq) + 2Cl−(aq) + Mg(OH)2(s)
Net Ionic Equation
2OH−(aq) + Mg2+(aq)  Mg(OH)2(s)
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?
Decide whether the following reaction
occurs. If it does, write the molecular,
ionic, and net ionic equations.
K3PO4 + CaCl2 
1. Determine the product formulas:
K+ and Cl− make KCl
Ca2+ and PO43− make Ca3(PO4)2
2. Determine whether the products are soluble:
KCl is soluble
Ca3(PO4)2 is insoluble
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Molecular Equation
(Balance the reaction and include state symbols)
2K3PO4(aq) + 3CaCl2(aq) 
6KCl(aq) + Ca3(PO4)2(s)
Ionic Equation
6K+(aq) + 2PO43−(aq) + 3Ca2+(aq) + 6Cl−(aq) 
6K+(aq) + 6Cl−(aq) + Ca3(PO4)2(s)
Net Ionic Equation
2PO43−(aq) + 3Ca2+(aq)  Ca3(PO4)2(s)
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Types of Chemical Reactions
1. Precipitation reactions: a solid ionic
substance forms from the mixture of two
solutions of ionic substances.
2. Acid–base reactions: reactions that involve
the transfer of a proton (H+) between reactants.
3. Oxidation–reduction reactions: reactions that
involve the transfer of electrons between
reactants.
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A precipitate is an insoluble solid compound
formed during a chemical reaction in solution.
Predicting Precipitation Reactions
1. Predict the products (exchange of parts).
2. Determine the state of each product: (s), (l),
(g), (aq).
3. If all products are aqueous (aq), no net
reaction occurred.
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Arrhenius Acid
A substance that produces hydrogen ions, H+,
when it dissolves in water.
Arrhenius Base
A substance that produces hydroxide ions, OH−,
when it dissolves in water.
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Brønsted–Lowry Acid
The species (molecule or ion) that donates a
proton to another species in a proton−transfer
reaction.
Brønsted–Lowry Base
The species (molecule or ion) that accepts a
proton from another species in a proton−transfer
reaction.
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Household Acids
and Bases
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Acid−Base Indicator
A dye used to distinguish between an acidic and
basic solution by means of the color changes it
undergoes in these solutions.
The sample beakers
show a red cabbage
indicator in beakers
varying in acidity from
highly acidic (left) to
highly basic (right).
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Strong Acid
An acid that ionizes completely in water. It is
present entirely as ions; it is a strong electrolyte.
Common strong acids:
HNO3
H2SO4
HCl
HBr
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HClO4
HI
4 | 36
Weak Acid
An acid that only partly ionizes in water. It is
present primarily as molecules and partly as ions;
it is a weak electrolyte.
If an acid is not strong, it is weak.
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In Figure A, a solution of
HCl (a strong acid)
illustrated on a
molecular/ionic level,
shows the acid as all
ions.
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In Figure B, a solution of
HF (a weak acid) also
illustrated on a
molecular/ionic level,
shows mostly molecules
with very few ions.
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Strong Base
A base that ionizes completely in water. It is
present entirely as ions; it is a strong electrolyte.
Common strong bases:
LiOH
NaOH
Ca(OH)2
Sr(OH)2
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KOH
Ba(OH)2
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Weak Base
A base that is only partly ionized in water. It is
present primarily as molecules and partly as ions;
it is a weak electrolyte. These are often nitrogen
bases such as NH3:
NH3(aq) + H2O(l)  NH4+(aq) + OH−(aq)
If a base is not strong, it is weak.
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?
Classify the following as strong or
weak acids or bases:
a. KOH
b. H2S
c. CH3NH2
d. HClO4
a.
b.
c.
d.
KOH is a strong base.
H2S is a weak acid.
CH3NH2 is a weak base.
HClO4 is a strong acid.
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Polyprotic Acid
An acid that results in two or more acidic
hydrogens per molecule.
For example:
H2SO4, sulfuric acid
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Neutralization Reaction
A reaction of an acid and a base that results in an
ionic compound (a salt) and possibly water.
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?
Write the molecular, ionic, and net ionic
equations for the neutralization of
sulfurous acid, H2SO3, by potassium
hydroxide, KOH.
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Molecular Equation
(Balance the reaction and include state symbols)
H2SO3(aq) + 2KOH(aq)  2H2O(l) + K2SO3(aq)
Ionic Equation
H2SO3(aq) + 2K+(aq) + 2OH−(aq) 
2H2O(l) + 2K+(aq) + SO32−(aq)
Net Ionic Equation
H2SO3(aq) + 2OH−(aq)  2H2O(l) + SO32−(aq)
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Acid−Base Reaction with Gas
Formation
Some salts, when treated with an
acid, produce a gas. Typically
sulfides, sulfites, and carbonates
behave in this way producing
hydrogen sulfide, sulfur trioxide, and
carbon dioxide, respectively.
The photo to the right shows
baking soda (sodium
hydrogen carbonate) reacting
with acetic acid in vinegar to
give bubbles of carbon
dioxide.
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Acid−base reactions with gas−formation
Sulfides, carbonates, sulfites react with acid to
form a gas.
Na2S(aq) + 2HCl(aq)  2NaCl(aq) + H2S(g)
Na2CO3(aq) + 2HCl(aq) 
2NaCl(aq) + H2O(l) + CO2(g)
Na2SO3(aq) + 2HCl(aq) 
2NaCl(aq) + H2O(l) + SO2(g)
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?
Write the molecular, ionic, and net ionic
equations for the reaction of copper(II)
carbonate with hydrochloric acid.
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Molecular Equation
(Balance the reaction and include state symbols)
CuCO3(s) + 2HCl(aq) 
CuCl2(aq) + H2O(l) + CO2(g)
Ionic Equation
CuCO3(s) + 2H+(aq) + 2Cl−(aq) 
Cu2+(aq) + 2Cl−(aq) + H2O(l) + CO2(g)
Net Ionic Equation
CuCO3(s) + 2H+(aq)  Cu2+(aq) + H2O(l) + CO2(g)
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Oxidation Number
For a monatomic ion, the actual charge of the
atom or a hypothetical charge assigned to the
atom in the substance using simple rules.
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Rules for Assigning Oxidation Numbers
1. Elements: The oxidation number of an atom in
an element is zero.
2. Monatomic ions: The oxidation number of an
atom in a monatomic ion equals the charge on
the ion.
3. Oxygen: The oxidation number of oxygen is −2
in most of its compounds. (An exception is O in
H2O2 and other peroxides, where the oxidation
number is −1.)
4. Hydrogen: The oxidation number of hydrogen
is +1 in most of its compounds. (The oxidation
number of hydrogen is −1 in binary compounds
with a metal such as CaH2.)
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5. Halogens: The oxidation number of fluorine is
−1. Each of the other halogens (Cl, Br, I) has
an oxidation number of −1 in binary
compounds, except when the other element is
another halogen above it in the periodic table
or the other element is oxygen.
6. Compounds and ions: The sum of the
oxidation numbers of the atoms in a compound
is zero. The sum of the oxidation numbers of
the atoms in a polyatomic ion equals the
charge on the ion.
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?
Potassium permanganate, KMnO4, is a
purple−colored compound; potassium
manganate, K2MnO4, is a
green−colored compound. Obtain the
oxidation numbers of the manganese in
these compounds.
K
Mn
O
1(+1) + 1(oxidation number of Mn) + 4(−2) = 0
1 + 1(oxidation number of Mn) + (−8) = 0
(−7) + (oxidation number of Mn) = 0
Oxidation number of Mn = +7
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K
Mn
O
2(+1) + 1(oxidation number of Mn) + 4(−2) = 0
2 + 1(oxidation number of Mn) + (−8) = 0
(−6) + (oxidation number of Mn) = 0
Oxidation number of Mn = +6
In KMnO4, the oxidation number of Mn is +7.
In K2MnO4, the oxidation number of Mn is +6.
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?
What is the oxidation number of Cr in
dichromate, Cr2O72−?
Cr
O
2(oxidation number of Cr) + 7(−2) = −2
2(oxidation number of Cr) + (−14) = −2
2(oxidation number of Cr) = +12
Oxidation number of Cr = +6
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Half−reaction
One of two parts of an oxidation–reduction
reaction, one part of which involves a loss of
electrons (or increase in oxidation number) and the
other part of which involves a gain of electrons (or
decrease in oxidation number).
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Oxidation
The half−reaction in which there is a loss of
electrons by a species (or an increase in oxidation
number).
Reduction
The half−reaction in which there is a gain of
electrons by a species (or a decrease in oxidation
number).
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Oxidizing Agent
A species that oxidizes another species; it is itself
reduced.
Reducing Agent
A species that reduces another species; it is itself
oxidized.
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Common Oxidation–Reduction Reactions
1.
2.
3.
4.
Combination reaction
Decomposition reaction
Displacement reaction
Combustion reaction
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Combination Reaction
A reaction in which two substances combine to
form a third substance.
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For example:
2Na(s) + Cl2(g)  2NaCl(s)
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Decomposition Reaction
A reaction in which a single
compound reacts to give two or
more substances.
For example:
2HgO(s)  2Hg(l) + O2(g)
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Displacement Reaction
A reaction in which an element
reacts with a compound, displacing
another element from it.
For example:
Zn(s) + 2HCl(aq) 
H2(g) + ZnCl2(aq)
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Combustion Reaction
A reaction in which a substance reacts with
oxygen, usually with the rapid release of heat to
produce a flame.
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For example:
4Fe(s) + 3O2(g)  2Fe2O3(s)
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Balancing Simple Oxidation−Reduction
Reactions: Half−Reaction Method
First, identify what is oxidized and what is reduced
by determining oxidation numbers.
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For the reaction
Zn(s) + Ag+(aq)  Zn2+(aq) + Ag(s)
0
+1
+2
0
Zn is oxidized from 0 to +2.
Ag+ is reduced from +1 to 0.
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Next, write the unbalanced half−reactions.
Zn(s)  Zn2+(aq)
(oxidation)
Ag+(aq)  Ag(s)
(reduction)
Now, balance the charge in each half reaction by
adding electrons.
Zn(s)  Zn2+(aq) + 2e− (oxidation)
e− + Ag+(aq)  Ag(s)
(reduction)
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Since the electrons lost in oxidation are the same
as those gained in reduction, we need each
half−reaction to have the same number of
electrons. To do this, multiply each half−reaction
by a factor so that when the half−reactions are
added, the electrons cancel.
Zn(s)  Zn2+(aq) + 2e−
2e− + 2Ag+(aq)  2Ag(s)
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(oxidation)
(reduction)
4 | 69
Lastly, add the two half−reactions together.
Zn(s) + 2Ag+(aq)  Zn2+(aq) + 2Ag(s)
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?
Balance the following
oxidation−reduction reaction:
FeI3(aq) + Mg(s)  Fe(s) + MgI2(aq)
The oxidation numbers are given below the
reaction.
FeI3(aq) + Mg(s)  Fe(s) + MgI2(s)
+3 −1
0
0
+2 −1
Now, write the half−reactions. Since Iodide is a
spectator ion it is omitted at this point.
Mg(s)  Mg2+(aq)
(oxidation)
Fe3+(aq)  Fe(s)
(reduction)
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Balancing the half−reactions:
Mg(s)  Mg2+(aq) + 2e−
Fe3+(aq) + 3e−  Fe(s)
(oxidation)
(reduction)
Multiply the oxidation half−reaction by 3 and the
reduction half−reaction by 2.
3Mg(s)  3Mg2+(aq) + 6e− (oxidation)
2Fe3+(aq) + 6e−  2Fe(s)
(reduction)
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Add the half−reactions together.
2Fe3+(aq) + 3Mg(s)  2Fe(s) + 3Mg2+(aq)
Now, return the spectator ion, I−.
2FeI3(aq) + 3Mg(s)  2Fe(s) + 3MgI2(aq)
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Molar Concentration, Molarity, (M)
Moles of solute per liter of solution
moles of solute
Molarity (M ) 
liters of solution
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To prepare a solution, add the measured amount
of solute to a volumetric flask, then add water to
bring the solution to the mark on the flask.
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Dilution
When a higher concentration solution is used to
make a less−concentration solution, the moles of
solute are determined by the amount of the
higher−concentration solution. The number of
moles of solute remains constant.
MiVi = MfVf
Note:
The units on Vi and Vf must match.
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Diluting a
solution
quantitatively
requires
specific
glassware.
The photo at
the right shows
a volumetric
flask used in
dilution.
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?
You place a 1.52−g of potassium
dichromate, K2Cr2O7, into a 50.0−mL
volumetric flask. You then add water to
bring the solution up to the mark on the
neck of the flask. What is the molarity
of K2Cr2O7 in the solution?
Molar mass of K2Cr2O7 is 294 g.
1 mol
1.52 g
294 g
 0.103 M
3
50.0  10 L
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?
A solution of sodium chloride used for
intravenous transfusion (physiological
saline solution) has a concentration of
0.154 M NaCl. How many moles of
NaCl are contained in 500.−mL of
physiological saline? How many grams
of NaCl are in the 500.−mL of solution?
mol  M  L
 0.154 M  0.500 L
 0.0770 mol NaCl
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Molar mass NaCl  58.4 g
58.4 g
0.0770 mol
1 mol
 4.50 g NaCl
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?
A saturated stock solution of NaCl is
6.00 M. How much of this stock
solution is needed to prepare 1.00−L of
physiological saline soluiton (0.154 M)?
M iVi  M fVf
M fVf
Vi 
Mi
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(0.154 M )(1.00 L)
Vi 
6.00 M
Vi  0.0257 L or 25.7 mL
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Quantitative Analysis
The determination of the amount of a substance or
species present in a material.
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Gravimetric Analysis
A type of quantitative analysis in which the amount
of a species in a material is determined by
converting the species to a product that can be
isolated completely and weighed.
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The figure on the
right shows the
reaction of Ba(NO3)2
with K2CrO4 forming
the yellow BaCrO4
precipitate.
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The BaCrO4
precipitate is being
filtered in the figure
on the right. It can
then be dried and
weighed.
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?
A soluble silver compound was
analyzed for the percentage of silver by
adding sodium chloride solution to
precipitate the silver ion as silver
chloride. If 1.583 g of silver compound
gave 1.788 g of silver chloride, what is
the mass percent of silver in the
compound?
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Molar mass of silver chloride (AgCl) = 143.32 g
1mol AgCl
1mol Ag 107.9 g Ag
1.788 g AgCl 


143.32 g AgCl 1mol AgCl 1mol Ag
= 1.346 g Ag in the compound
1.346 g Ag
100%
1.583 g silver compound
= 85.03% Ag
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Titration
A procedure for determining the amount of
substance A by adding a carefully measured
volume with a known concentration of B until the
reaction of A and B is just complete.
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In the titration above, the indicator changes color
to indicate when the reaction is just complete.
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Volumetric Analysis
A type of quantitative analysis based on titration.
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?
Zinc sulfide reacts with hydrochloric acid
to produce hydrogen sulfide gas:
ZnS(s) + 2HCl(aq)  ZnCl2(aq) + H2S(g)
How many milliliters of 0.0512 M HCl are
required to react with 0.392 g ZnS?
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Molar mass of ZnS = 97.45 g
1mol ZnS
2 mol HCl
1L solution
0.392 g ZnS 


97.45 g ZnS 1mol ZnS 0.0512 mol HCl
= 0.157 L = 157 mL HCl solution
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?
A dilute solution of hydrogen peroxide
is sold in drugstores as a mild
antiseptic. A typical solution was
analyzed for the percentage of
hydrogen peroxide by titrating it with
potassium permanganate:
5H2O2(aq) + 2KMnO4(aq) + 6H+(aq) 
8H2O(l) + 5O2(g) + 2K+(aq) + 2Mn2+(aq)
What is the mass percent of H2O2 in a
solution if 57.5 g of solution required 38.9
mL of 0.534 M KMnO4 for its titration?
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Molar mass of H2O2 = 34.01 g
38.9  103 L 
0.534 mol KMnO4
5 mol H2O2
34.01g H2O2


1L
2 mol KMnO4
1mol H2O2
= 1.77 g H2O2
1.77 g H2 O 2
100%
57.5 g solution
= 3.07% H2O2
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