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Transcript
Homework Problems
Chapter 4 Homework Problems: 2, 8, 11, 14, 20, 22, 25, 28, 29, 30, 32,
42, 44, 48, 52, 54, 64, 68, 82, 84, 94, 124, 132, 150
CHAPTER 4
Reactions in Aqueous Solution
Solutions
A solution is a homogeneous mixture of two or more pure
chemical compounds. Solutions may be solids, liquids, or gases.
Solvent - The major component of a solution (if there is one)
Solute(s) - The minor component(s) of a solution
Example: A solution is prepared by adding a small amount of sodium
chloride (NaCl) to water.
solvent = water (major component)
solute = sodium chloride
Aqueous solution - Solution where water is the solvent.
Electrolytes and Nonelectrolytes
Many chemical substances dissolve in water. A substance that
forms ions when dissolved in water is called an electrolyte, while a
substance that does not form ions when dissolved in water is called a
nonelectrolyte.
Strong and Weak Electrolytes
Strong electrolyte - Completely dissociates (falls apart) when
added to solution, to form ions.
Weak electrolyte - Partially dissociates (falls apart) when added
to solution, to form ions.
Example:
HCl(aq)  H+(aq) + Cl-(aq)
strong electrolyte
NaBr(s)  Na+(aq) + Br-(aq)
strong electrolyte
HC2H3O2(aq)  H+(aq) + C2H3O2(aq)
weak electrolyte
HClO2(aq)  H+(aq) + ClO2-(aq)
weak electrolyte
C6H12O6(s)  C6H12O6(aq)
nonelectrolyte
Classification of Compounds
as Electrolytes (in Water)
We previously divided compounds into two general categories:
Ionic compounds - Collections of cations (positive ions) and
anions (negative ions) in a crystal structure. Usually form from a metal
(or cation group) and a nonmetal (or anion group).
Molecular compounds - Exist as individual molecules. Usually
form from two or more nonmetals.
We may use this as a starting point in classifying compounds as
strong electrolytes, weak electrolytes, and nonelectrolytes.
Ionic Compounds
Ionic compounds - All ionic compounds are strong electrolytes.
Examples:
NaCl(s)  Na+(aq) + Cl-(aq)
KOH(s)  K+(aq) + OH-(aq)
Ca(NO3)2(s)  Ca2+(aq) + 2 NO3-(aq)
Note that an insoluble ionic compound is considered a strong
electrolyte. This is because even though very little of the compound
dissolves in water, whatever does dissolve completely ionizes.
Molecular Compounds
Strong acids - Molecular compounds that are strong acids are also
strong electrolytes.
Examples:
HCl(aq)  H+(aq) + Cl-(aq)
HNO3(aq)  H+(aq) + NO3-(aq)
Weak acids and bases - Molecular compounds that are weak acids
or weak bases are also weak electrolytes. Note that an acid forms H+ ion
when added to water, and a base forms OH- ion.
Examples:
HF(aq)  H+(aq) + F-(aq)
NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq)
The double arrow “” indicates a reaction that proceeds in both the forward and reverse direction, establishing equilibrium.
Molecular Compounds (Continued)
Other molecular compounds - Molecular compounds that are
neither acids nor bases are nonelectrolytes.
Examples:
C6H12O6(s)  C6H12O6(aq)
CH3OH(l)  CH3OH(aq)
Note that water (H2O) is classified as a nonelectrolyte, even though a
small number of H+ and OH- ions are formed in pure water.
H2O(l)  H+(aq) + OH-(aq)
For pure water at T = 25 ° there are only ~ 2 x 10-7 M of ions present.
This is small enough that water is classified as a nonelectrolyte.
Flow Chart For Classifying Compounds
Precipitation Reactons
A precipitation reactions is a reaction that forms a solid (a
precipitate) out of soluble reactants.
Examples:
Pb(NO3)2(aq) + 2 KCl(aq)  PbCl2(s) + 2 KNO3(aq)
3 NaOH(aq) + FeBr3(aq)  Fe(OH)3(s) + 3 NaBr(aq)
Solubility (Water)
Solid compounds are divided into two general categories.
Soluble compound - a compound that will dissolve in water
Insoluble compound - a compound that will not dissolve in water
The above classification scheme is an oversimplification. Most
“insoluble” compounds will dissolve to a slight extent. For “soluble”
compounds there will be a limit as to the amount of compound that will
dissolve in a given amount of water.
We can also discuss solubility in other solvents.
Why Ionic Compounds Dissolve in Water
The forces holding an ionic compound together are the strong
electrical attraction that exists between cations and anions. It is therefore
somewhat surprising that ionic compounds will dissolve in water.
The reason some ionic compounds will dissolve in water is
because the water molecules have a partial negative charge on the
oxygen atom (-) and partial positive charges on the hydrogen atoms
(+), where “” indicates a small positive or negative charge. The reason
these partial charges exist will be discussed later in the semester.
Because cations and anions can strongly interact with the
negative and positive partial charges on several water molecules at the
same time, it is possible for the ions in an ionic compound to be pulled
out of the solid and go into solution.
Hydration and Solvation
The process of surrounding a cation or anion with water
molecules to stabilize them in aqueous solution is called hydration.
Often these ions will be surrounded by a definite number of water
molecules, called a hydration shell.
The more general process of surrounding ions or molecules with
solvent molecules in a solution is called solvation. Water is unusual
because the interaction of ions with water molecules is much stronger
than the interactions that occur in most other solvents.
hydration of anion
hydration of cation
Solubility Rules
a) All compounds with group IA cations, or NH4+ (ammonium
ion), are soluble.
b) All compounds containing NO3- (nitrate ion), ClO4(perchlorate ion), ClO3- (chlorate ion), or C2H3O2- (acetate ion) are
soluble.
c) Most compounds containing Cl-, Br-, and I- are soluble.
(EXCEPTIONS: Cations Ag+, Pb2+, Hg22+).
d) Most compounds containing SO42- (sulfate ion) are soluble.
(EXCEPTIONS: Cations Ag+, Pb2+, Hg22+, Ca2+, Sr2+, Ba2+).
e) Most compounds containing OH- (hydroxide ion) are
insoluble. (EXCEPTIONS: Cations from group 1A, Ca2+, Sr2+, Ba2+).
f) Most other compounds are insoluble. (EXCEPTIONS:
Compounds covered by rules a-e).
Use of Solubility Rules
KClO4
(NH4)2SO4
CuCl2
PbI2
NiSO4
Mg(OH)2
Zn(IO3)2
Use of Solubility Rules
KClO4
soluble
Rules a and b
(NH4)2SO4
soluble
Rule a
CuCl2
soluble
Rule c
PbI2
insoluble
Rule c
NiSO4
soluble
Rule d
Mg(OH)2
insoluble
Rule e
Zn(IO3)2
insoluble
Rule f
Determining Whether a Precipitation Reaction Will
Occur
When solutions of two soluble substances mix we will not always
form a precipitate. We may use the solubility rules to decide if a
precipitate will form.
1) Write the ions formed from the starting compounds.
2) Look at all possible combinations of cations and anions.
a) If there is a combination of ions that forms an insoluble
compound, a precipitate will form and a precipitation reaction occurs.
b) If there is not a combination of ions that forms an insoluble
compound, then there will be no precipitation reaction.
Examples: For each of the following mixtures of two solutions
will a precipitate form? Note that all of the solutes are themselves
soluble.
Solution A
solution B
1)
NaCl(aq)
AgNO3(aq)
2)
NaCl(aq)
CuBr2(aq)
precipitate(?)
1)
Solution A
solution B
precipitate(?)
NaCl(aq)
AgNO3(aq)
yes, AgCl(s)
Na+ Cl-
Ag+ NO3-
NaCl(aq) + AgNO3(aq)  NaNO3(aq) + AgCl(s)
2)
NaCl(aq)
CuBr2(aq)
Na+ Cl-
Cu2+ Br-
No reaction.
no
Net Ionic Equations
To this point we have written chemical equations in terms of
neutral species (molecules, or formula units). However, in aqueous
solution it is often more appropriate to write equations in terms of ions.
There are three ways we normally write chemical equations:
1) Molecular equation. All reactants and products are written in
terms of their chemical formula.
2) Total ionic equation. Strong electrolytes (strong acids, strong
soluble bases, and soluble ionic compounds) are written as ions.
3) Net ionic equation. The total ionic equation after removing
spectator ions.
A spectator ion is an ion that appears on both the reactant and
product side of the equation, and which does not participate directly in
the chemical reaction.
Net Ionic Equation (Example)
We may show the various ways of writing chemical reactions as
follows:
Molecular
AgNO3(aq) + NaCl(aq)  NaNO3(aq) + AgCl(s)
Total Ionic
Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq)
 Na+(aq) + NO3-(aq) + AgCl(s)
Net Ionic
Ag+(aq) + Cl-(aq)  AgCl(s)
Finding the Total and New Ionic Equations
We find the total and net ionic equations as follows.
1) Begin with a balanced molecular equation.
2) Break apart strong electrolytes into ions to obtain the total
ionic equation.
3) Cancel ions that appear on both sides of the equation to obtain
the net ionic equation.
Example:
molecular
Pb(NO3)2(aq) + 2 KI(aq)  PbI2(s) + 2 KNO3(aq)
molecular
Pb(NO3)2(aq) + 2 KI(aq)  PbI2(s) + 2 KNO3(aq)
total ionic
Pb2+(aq) + 2 NO3-(aq) + 2 K+(aq) + 2 I-(aq)
 PbI2(s) + 2 K+(aq) + 2 NO3-(aq)
net ionic
Pb2+(aq) + 2 I-(aq)  PbI2(s)
By writing the net ionic equation we are able to focus on the
important parts of a chemical reaction and observe similarities in
reactions.
mol.
AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)
total
Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq)
 AgCl(s) + Na+(aq) + NO3-(aq)
net
Ag+(aq) + Cl-(aq)  AgCl(s)
mol.
AgNO3(aq) + KCl(aq)  AgCl(s) + KNO3(aq)
total
Ag+(aq) + NO3-(aq) + K+(aq) + Cl-(aq)
 AgCl(s) + K+(aq) + NO3-(aq)
net
Ag+(aq) + Cl-(aq)  AgCl(s)
Net Ionic Equation When No Precipitate Forms
Writing the net ionic equation makes it clear when precipitation
reactions do or do not occur.
mol.
2 KBr(aq) + Zn(NO3)2(aq)  2 KNO3(aq) + ZnBr2(aq)
K+ Br-
total
Zn2+ NO3-
2 K+(aq) + 2 Br-(aq) + Zn2+(aq) + 2 NO3-(aq)
 2 K+(aq) + 2 NO3-(aq) + Zn2+(aq) + 2 Br-(aq)
All the ions are spectator ions, and so no reaction occurs.
Acids and Bases (Arrhenius)
Arrhenius definition
acid - A substance which, when added to water, forms H+ ion
base - A substance which, when added to water, forms OH- ion
Examples:
HCl(aq)  H+(aq) + Cl-(aq)
HF(aq)  H+(aq) + F-(aq)
NaOH(s)  Na+(aq) + OH-(aq)
NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq)
The advantage of the Arrhenius definition is that it is simple and
easy to implement. The disadvantage is that it is tied in to a particular
solvent (water) and is not a general definition.
Strong and Weak Acids
An acid is a substance that forms H+ ion when added to water
(Arrhenius definition). A strong acid forms one H+ ion for each acid
molecule. A weak acid forms much less than one H+ ion for each acid
molecule.
Examples:
HClO4(aq)  H+(aq) + ClO4-(aq)
strong
HNO2(aq)  H+(aq) + NO2-(aq)
weak
There are 7 common strong acids
binary
HCl
HI
HBr
ternary
HClO3
HNO3
HClO4
H2SO4 (1st proton)*
All other acids can be assumed to be weak acids.
*H
2SO4(aq)
 H+(aq) + HSO4-(aq)
Strong and Weak Bases
A base is a substance that forms OH- ion when added to water
(Arrhenius definition). A strong soluble base is a soluble hydroxide
compound that completely dissociates when added to water. An
insoluble base is an insoluble hydroxide compound. There are also a few
substances that act as weak bases in solution. These substances form
much less than one OH- ion per base molecule.
Examples:
KOH(s)  K+(aq) + OH-(aq)
strong soluble base
Fe(OH)3(s)  no reaction
insoluble base
NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq)
There are 8 common strong soluble bases
Group 1A - LiOH, NaOH, KOH, RbOH, CsOH
Group 2A - Ca(OH)2, Sr(OH)2, Ba(OH)2
All other hydroxide compounds are insoluble bases.
weak base
Acids and Bases (Bronsted)
Bronsted definition
acid - a proton (H+) donor
base - a proton (H+) acceptor
HF(aq) + H2O(l)  H3O+(aq) + F-(aq)
acid
base
In the Bronsted theory, in an acid-base reaction an acid donates a
proton, while a base accepts a proton. In addition, in Bronsted theory
acids form hydronium ion (H3O+ ion) instead of hydrogen ion (H+ ion)
when added to water.
The Bronsted definition of an acid and a base is more general
than the Arrhenius definition.
Examples of Acids (Bronsted)
Strong acid
HCl(aq) + H2O(l)  H3O+(aq) + Cl-(aq)
acid:
HCl
base: H2O
Weak acid
HC2H3O2(aq) + H2O(l)  H3O+(aq) + C2H3O2-(aq)
acid:
HC2H3O2
base: H2O
The second reaction goes in both directions.
Examples of Bases (Bronsted)
Strong base
NaOH(s)  Na+(aq) + OH-(aq)
Considered an ionization reaction in Bronsted theory, not an acidbase reaction.
Weak base
NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq)
base: NH3
acid:
H2O
The second reaction goes in both directions.
Example
Chlorous acid (HClO2) is a weak acid. Pyridine (C5H5N) is a
weak base. Indicate the behavior of these two substances when added to
water, according to Bronsted theory.
Example
Chlorous acid (HClO2) is a weak acid. Pyridine (C5H5N) is a
weak base. Indicate the behavior of these two substances when added to
water, according to Bronsted theory.
HClO2(aq) + H2O(l)  H3O+(aq) + ClO2-(aq)
C5H5N(aq) + H2O(l)  C5H5NH+(aq) + OH-(aq)
Notice that, depending on the reaction, water can act as either a Bronsted
base or a Bronsted acid. A substance that can act as either a Bronsted
acid or a Bronsted base is called amphoteric.
Polyprotic Acid
A polyprotic acid has two or more ionizable protons that can be
donated in an acid-base reaction.
Monoprotic (one ionizable proton)
HCl, HNO2, HC2H3O2
Diprotic (two ionizable protons)
H2CO3, H2SO4
Triprotic (three ionizable protons)
H3PO3
Example: H2SO4 (sulfuric acid)
1st proton H2SO4(aq) + H2O(l)  H3O+(aq) + HSO4-(aq)
2nd proton HSO4-(aq) + H2O(l)  H3O+(aq) + SO42-(aq)
Example: H2CO3 (carbonic acid)
1st proton H2CO3(aq) + H2O(l)  H3O+(aq) + HCO3-(aq)
2nd proton HCO3-(aq) + H2O(l)  H3O+(aq) + CO32-(aq)
Other Amphoteric Species
We previously pointed out that depending on the reaction taking
place water can act as either a Bronsted acid or a Bronsted base. Other
species have this same property.
Example: HCO3- ion
as acid
HCO3-(aq) + OH-(aq)  CO32- (aq) + H2O(l)
as base
HCO3-(aq) + H3O+(aq)  H2CO3(aq) + H2O(l)
Like water, HCO3- ion is classified as amphoteric. Note that the above
property of the hydrogen carbonate ion (and other amphoteric species)
can be useful in acid-base systems.
Dibasic Base
A dibasic base is a strong base that releases two OH- ions per
formula unit of base.
Example: Ba(OH)2 (barium hydroxide)
Ba(OH)2(s)  Ba2+(aq) +2 OH-(aq)
Acid-Base (Neutralization) Reactions
An acid-base reactions (or neutralization reaction) is the reaction
of an acid with a base. These reactions usually form a salt and water.
salt - An ionic compound that can be formed by the reaction of an
acid with a base. Note that if either the acid or the base is strong the
reaction will go completely from left to right.
Examples:
2 HClO4(aq) + Ca(OH)2(aq)  Ca(ClO4)2(aq) + 2 H2O(l)
HNO2(aq) + NaOH(aq)  NaNO2(aq) + H2O(l)
2 HBr(aq) + Cu(OH)2(s)  CuBr2(aq) + 2 H2O(l)
The last reaction above illustrates why insoluble hydroxide compounds
are considered insoluble bases. Note that ionic compounds other than
hydroxides (OH-) or oxides (O2-) are classified as salts, as they can be
formed from acid-base reactions.
Strong Acid + Strong Base
All strong acid + strong base reactions have the same net ionic
equation.
mol.
2 HClO4(aq) + Sr(OH)2(aq)  Sr(ClO4)2(aq) + 2 H2O()
total
2 H+(aq) + 2 ClO4-(aq) + Sr2+(aq) + 2 OH-(aq)
 Sr2+(aq) + 2 ClO4-(aq) + 2 H2O()
net
H+(aq) + OH-(aq)  H2O()
mol.
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O()
total
H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq)
 Na+(aq) + Cl-(aq) + H2O()
net
H+(aq) + OH-(aq)  H2O()
Net Ionic Equation for Weak
Acid or Base Reaction
When the acid or base in an acid base reaction is weak, the net
ionic equation will be slightly more complicated.
mol.
HNO2(aq) + KOH(aq)  KNO2(aq) + H2O()
mol.
HBr(aq) + NH3(aq)  NH4Br(aq)
mol.
HNO2(aq) + KOH(aq)  KNO2(aq) + H2O()
total
HNO2(aq) + K+(aq) + OH-(aq)  K+(aq) + NO2-(aq) + H2O()
net
HNO2(aq) + OH-(aq)  NO2-(aq) + H2O()
mol.
HBr(aq) + NH3(aq)  NH4Br(aq)
total
H+(aq) + Br-(aq) + NH3(aq)  NH4+(aq) + Br-(aq)
net
H+(aq) + NH3(aq)  NH4+(aq)
Notice that we do not break apart the weak acid or weak base, as
they are weak electrolytes, and therefore exist mainly as molecules rather
than ions.
Oxidation Number
An oxidation number is a number assigned to an atom in a
molecule or ion indicating whether that atom is electron rich (negative
oxidation number), electron poor (positive oxidation number), or neutral
(oxidation number of zero).
Oxidation numbers do not indicate charges of atoms, but changes
in oxidation number indicate whether atoms have gained or lost electron
density.
A reaction where some of the oxidation numbers change in going
from reactants to products is called an oxidation-reduction (redox)
reaction.
Rules For Assigning Oxidation Numbers
1) The sum of the oxidation numbers of a species is equal to the
charge of the species.
2) Atoms in elemental forms have an oxidation number of 0.
3) Hydrogen has an oxidation number of +1 when bonded with
nonmetals (molecular compounds) and -1 when combined with metals
(ionic compounds).
4) Oxygen usually has an oxidation number of -2. Exceptions:
Elemental forms, peroxides (H2O2), OF2.
H2O2 H +1, O -1
OF2
F -1, O +2
5) Halogens. F in compounds is always -1. Other halogens are
often -1, but in compounds with oxygen or other halogens the oxidation
number may take on different values.
6) For assigning oxidation numbers in ionic compounds, it is
useful to break the compounds up into ions.
Examples:
Assign oxidation numbers for each atom in the following
substances.
H2O
Fe(NO3)2
SO42O3
HClO2
MnO4-
H2O
H +1; O -2
Fe(NO3)2
= Fe2+ and NO3so Fe +2, O -2, and so N +5
SO42-
O -2, and so S +6
O3
elemental form, so O = 0
HClO2
= H+ and ClO2- (even though molecular)
so H +1, O -2, and so Cl +3
MnO4-
O -2, and so Mn +7
Common Oxidation Numbers
Oxidation-Reduction Reactions
In an oxidation-reduction (redox) reaction the oxidation numbers
of some of the atoms involved in the reaction change when going from
reactants to products.
2 Fe2O3(s) + 3 C(s)  4 Fe(s) + 3 CO2(g)
+3 -2
0
0
+4 -2
We use the following terms in reference to redox reactions.
Oxidation - An increase in the value for the oxidation number.
Reduction - A decrease in the value for the oxidation number.
There will always be one oxidation and one reduction process per
reaction.
C
0 to +4, and so is oxidized.
Fe
+3 to 0, and so is reduced.
Oxidizing and Reducing Agents
We call the species that is oxidized in a redox reaction the
reducing agent, and the species that is reduced in a redox reaction an
oxidizing agent. This seems confusing, but is based on the idea that
there must be both an oxidation and a reduction taking place in a redox
reaction.
C(s) + O2(g)  CO2(g)
0
0
+4 -2
In this reaction carbon is oxidized (from 0 to +4) and so C is a
reducing agent, and oxygen is reduced (from 0 to -2) and so O2 is an
oxidizing agent.
SO2(g) + 2 Fe3+(aq) + 2 H2O(l)  2 Fe2+(aq) + SO42-(aq) + 4 H+(aq)
+4 -2
+3
+1 -2
+2
+6 -2
+1
In this reaction S is oxidized (from +4 to +6) and so SO2 is a
reducing agent, and Fe is reduced (from +3 to +2) and so Fe3+ is an
oxidizing agent.
Activity Series
The higher the element in the activity series, the easier it is to
oxidize. The lower the anion in the activity series, the easier it is to
reduce.
Active and Noble Metals
Some metals in the activity series (those at or near the top of the
series) are so reactive that they are never found in the metallic state in
nature. Such metals are called active metals. The alkali metals and
alkaline earth metals are active metals.
Metals at or near the bottom
of the activity series are not very
reactive, and so are often found in
the metallic state in nature. Such
metals are called noble metals.
Examples are gold (Au), platinum
(Pt), and mercury (Hg).
Use of the Activity Series
The activity series can be used to predict which metals will
reduce a particular ion, or which ions will oxidize a particular metal.
Example: A solution contains cobalt II ions (Co2+). Which of the
following metals would be expected to react if placed in the solution?
Cu, Fe, Ni, Zn
A solution contains cobalt II ions (Co2+). Which of the following metals
would be expected to react if placed in the solution?
Cu, Fe, Ni, Zn
Metals above cobalt in the activity series have a stronger tendency to
undergo oxidation and form ions. Those below have a weaker tendency
to form ions than cobalt.
So - Will react - Fe, Zn
Co2+(aq) + Fe(s)  Co(s) + Fe2+(aq)
Co2+(aq) + Zn(s)  Co(s) + Zn2+(aq)
Will not react - Cu, Ni
Hydrogen Ion
A solution containing hydrogen ion (H+) will, according to the
activity series, oxidize any metal above it (so Pb, Sn, Ni, …)
Sn(s) + 2 H+(aq)  Sn2+(aq) + H2(g)
Balancing Simple Oxidation-Reduction
Reactions
The following three step procedure can be used to balance simple
oxidation-reduction reactions. A variation of this method will be used
later to balance more complicated oxidation-reduction reactions.
1) Find the oxidation numbers for all reactants and products
2) Write a half-reaction for oxidation, and a half-reaction for
reduction. Use electrons for charge balance.
a) Oxidation - will produce electrons
b) Reduction - will consume electrons
3) Combine the two half reactions to get the net ionic equation.
This may mean multiplying one or both of the half reactions to get the
electrons to cancel.
Example
Balance the following oxidation-reduction reactions.
a) Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
b) Cr(s) + Pb2+(aq)  Cr3+(aq) + Pb(s)
a) Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
0
+2
+2
0
oxidation
Zn(s)  Zn2+(aq) + 2 e-
reduction
Cu2+(aq) + 2 e-  Cu(s)
NET
Zn(s) + Cu2+(aq)  Zn2+(aq) +Cu(s)
b) Cr(s) + Pb2+(aq)  Cr3+(aq) + Pb(s)
0
+2
+3
0
oxidation
(Cr(s)  Cr3+(aq) + 3 e- ) x 2
reduction
(Pb2+(aq) + 2 e-  Pb(s) ) x 3
NET
2 Cr(s) + 3 Pb2+(aq)  2 Cr3+(aq) +3 Pb(s)
Classifying Redox Reactions
Redox reactions can be classified into various categories.
Combination reaction - Two or more reactants form a simple product
Example:
H2(g) + Cl2(g)  2 HCl(g)
Decomposition reaction - One reactant forms two or more products
Example:
2 KClO3(s)  2KCl(s) + 3 O2(g)
Disproportionation reaction - A decomposition reaction where the same
element is both oxidized and reduced
Example:
2 H2O2(aq)  2 H2O(l) + O2(g)
Classifying Redox Reactions (continued)
Combustion reaction - A single substance reacts with oxygen to form
combustion products
Example:
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l)
In combustion reactions molecular oxygen (O2) is reduced from an
oxidation number of 0 to an oxidation number of -2.
Displacement reaction - A reaction where one species replaces another
species
Example:
Zn(s) + CuCl2(aq)  ZnCl2(aq) + Cu(s)
Br2(aq) + 2 KI(aq)  2 KBr(aq) + I2(aq)
Molarity
The most common unit of concentration for laboratory work is
molarity. By definition, the molarity of a solution (M) is
M = moles solute
liters of solution
Unfortunately, we use the same symbol, M, for molarity and for
molecular mass. The meaning of M must therefore be determined from
the context. To avoid confusion, we often use square brackets “[ ]” to
indicate molarity (for example, [KNO3] for the concentration of a
potassium nitrate solution, in moles/L).
Example: Find the molarity of NaCl in a solution prepared by
dissolving 5.000 g NaCl (M = 58.44 g/mol) in water to form a solution
with final volume V = 100.0 mL.
Solution Preparation
Solutions of known concentration can easily be prepared using an
analytical balance and standard laboratory glassware.
Example: Find the molarity of NaCl in a solution prepared by
dissolving 5.000 g NaCl (M = 58.44 g/mol) in water to form a solution
with final volume V = 100.0 mL.
moles NaCl = 5.000 g 1 mol = 0.08556 mol
58.44 g
liters of solution = 100.0 mL
1L
= 0.1000 L
1000.0 mL
So [NaCl] = 0.08556 mol NaCl = 0.8556 mol/L = 0.8556 M NaCl
0.1000 L soln
Concentration and Stoichiometry
When we give concentrations for solutions, we usually express
the concentration in terms of the substance added to the solution. For
strong electrolytes the species present will be ions whose concentrations
can be found based on the ionization taking place.
Example: What are the concentrations of Ca2+ ion and NO3- ion
in a 0.0240 M solution of calcium nitrate?
Example: What are the concentrations of Ca2+ ion and NO3- ion
in a 0.0240 M solution of calcium nitrate?
The equation for the ionization of calcium nitrate is
Ca(NO3)2(s)
H2O
Ca2+(aq) + 2 NO3-(aq)
[Ca2+] = 0.0240 mol Ca(NO3)2
1 mol Ca2+ = 0.0240 M Ca2+ ion
1 mol Ca(NO3)2
[NO3-] = 0.0240 mol Ca(NO3)2 2 mol NO3- = 0.0480 M NO3- ion
1 mol Ca(NO3)2
Note that for weak electrolytes finding the concentrations of species in
solution is more complicated. This will be discussed next semester.
Stock Solutions and Dilutions
A stock solution is a solution of high concentration used to
prepare more dilute solution by combining known amounts of the stock
solution with pure solvent. If the concentration of the stock solution is
known and the method used to carry out the preparation of the dilute
solution is also known, the concentration of the dilute solution may be
found.
Example: A solution is prepared by mixing 20.00 mL of the
0.8556 M NaCl solution discussed above with water to form a dilute
solution with a final volume equal to 500.0 mL. What is the concentration of this dilute solution?
A solution is prepared by mixing 20.00 mL of the 0.8556 M NaCl
solution discussed above with water to form a dilute solution with a final
volume equal to 500.0 mL. What is the concentration of this dilute
solution?
moles NaCl = 0.02000 L 0.8556 mol = 0.01711 mol NaCl
L stock soln
M dilute soln = 0.01711 mol NaCl = 0.03422 mol/L
0.5000L
General Relationship For Dilution of Stock Solutions
Based on the method used in the above calculation a general
relationship useful in preparing dilutions of stock solutions may be
found.
Let
Mc = molarity of initial (concentrated) solution
Lc = volume, in liters, of initial (concentrated) solution
Md = molarity of final (dilute) solution
Ld = volume, in liters, of final (dilute) solution
Then Mc Lc = Md Ld
This relationship is based on conservation of mass, in this case,
the mass (or number of moles) of solute in the solutions. It will work for
any volume unit, as long as the same unit is used for Lc and Ld.
We can check this relationship using our previous example.
A solution is prepared by mixing 20.00 mL of the 0.8556 M NaCl
solution discussed above with water to form a dilute solution with a final
volume equal to 500.0 mL. What is the concentration of this dilute
solution? We previously found M = 0.03422 mol/L.
Since Mc Lc = Md Ld, then if we divide both sides by Ld, we get
Md = Mc Lc = (0.8556 M) (20.00 mL) = 0.03422 mol/L
Ld
(500.0 mL)
the same as before.
One of the main uses of this relationship is to use stock solutions
of known concentration and volumetric glassware (pipettes, flasks) to
prepare more dilute solutions with particular concentrations.
Example: A student has a 0.4846 M aqueous stock solution of
glucose, and wishes to prepare 100.0 mL of a 0.1000 M glucose solution.
How can she do this?
A student has a 0.4846 M aqueous stock solution of glucose, and
wishes to prepare 100.0 mL of a 0.1000 M glucose solution. How can
she do this?
We could use the method used earlier, but is is faster and more
simple to use the relationship
Mc Lc = Md Ld
So
Mc = 0.4846 M
Md = 0.1000 M
Lc = ?
Ld = 100.0 mL
(0.4846 M) Lc = (0.1000 M) (100.0 mL)
Lc = (0.1000 M) (100.0 mL) = 20.6 mL
(0.4846 M)
So add 20.6 mL of the stock solution to a 100.0 mL volumetric
flask, and then add water to obtain the final volume of solution.
Serial Dilution
It is difficult to prepare very dilute solutions (concentration < 10-3
M) directly, as it would involve precisely weighing out a small mass of
solute. Such dilute solutions are prepared by the process of successive,
or serial, dilution of a stock solution.
Example: Starting with a 0.1000 M stock solution of KNO3,
prepare a dilute solution with a concentration of 1.00 x 10-5 M.
Example: Starting with a 0.1000 M stock solution of KNO3,
prepare a dilute solution with a concentration of 1.00 x 10-5 M.
Stock solution
0.1000 M
Take 5.00 mL of stock solution and dilute to a final volume of 500.0 mL.
Md = Mc Lc = (0.1000 M) (5.00 mL) = 0.001000 M
Ld
(500.0 mL)
Dilute this new solution in the same way as in the previous step
Md = Mc Lc = (0.00100 M) (5.00 mL) = 1.00 x 10-5 M
Ld
(500.0 mL)
Gravimetric Analysis
In gravimetric analysis, we use mass measurements to determine
composition or concentrations for unknowns.
One common gravimetric method is to add a compound to a
solution to form a precipitate, which can then be filtered out to determine
the mass of solid formed.
Example: An aqueous solution contains an unknown concentration of silver nitrate (AgNO3). Excess solid sodium chloride (NaCl) is
added to a 200.00 mL sample of the silver nitrate solution, and 0.184 g of
precipitate forms. What is the concentration of silver nitrate in the
original solution?
Example: An aqueous solution contains an unknown concentration of silver nitrate (AgNO3). Excess solid sodium chloride (NaCl) is
added to a 200.00 mL sample of the silver nitrate solution, and 0.184 g of
precipitate forms. What is the concentration of silver nitrate in the
original solution?
AgNO3(aq)  Ag+(aq) + NO3-(aq)
AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)
M(AgCl) = 143.32 g/mol
mol AgNO3 = 0.184 g AgCl 1 mol AgCl
143.32 g
= 0.00128 mol AgNO3
[AgNO3] = 0.00128 = 0.00642 M
0.2000L
1 mol AgNO3
1 mol AgCl
Titration
The term titration refers to a reaction between two substances
where one reactant is slowly added to a fixed amount of a second
reactant until complete reaction occurs.
Titrations are often used to find the concentrations of solutions.
In an acid-base titration, for example,
base is slowly added to a fixed amount of acid
until complete reaction occurs (the titration can
also be carried out by the slow addition of acid
to a fixed amount of base). An indicator (a
molecule that has a different color in acid and
base solution) is used to visually estimate the
equivalence point of the titration (the point
where the reaction is complete). This should
occur as close as possible to the end point of the
titration, the point where the indicator color
change occurs.
Calculations Involving Titrations
Since titrations are chemical reactions, we may carry out
calculations associated with titrations in the same way as other
calculations involving chemical reactions. We require a balanced
chemical equation and information about the amounts of the acid and
base solutions used in the titration and the concentration of one of the
solutions (or the number of moles of one of the reactants).
Example: A 20.00 mL sample of an HCl solution of unknown
concentration is titrated with a 0.3145 M NaOH solution. The
equivalence point for the titration occurs after the addition of 28.33 mL
of the NaOH solution. What is the concentration of the HCl solution (in
units of mol/L)?
A 20.00 mL sample of an HCl solution of unknown
concentration is titrated with a 0.3145 M NaOH solution. The
equivalence point for the titration occurs after the addition of 28.33 mL
of the NaOH solution. What is the concentration of the HCl solution (in
units of mol/L)?
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
Strategy:
1) Find the number of moles of NaOH (the solution where you
know both the volume and concentration).
2) Use the balanced chemical equation to find the number of
moles of HCl.
3) Find the concentration of the HCl solution, using the number
of moles (found in step 2) and the volume of HCl solution (given in the
problem).
A 20.00 mL sample of an HCl solution of unknown
concentration is titrated with a 0.3145 M NaOH solution. The
equivalence point for the titration occurs after the addition of 28.33 mL
of the NaOH solution. What is the concentration of the HCl solution (in
units of mol/L)?
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O()
n(NaOH) = 0.02833 L 0.3145 mol = 8.910 x 10-3 mol NaOH
L soln
n(HCl) = 8.910 x 10-3 mol NaOH 1 mol HCl = 8.910 x 10-3 mol HCl
1 mol NaOH
M(HCl) = 8.910 x 10-3 mol HCl = 0.4455 mol/L = 0.4455 M
0.02000 L soln
Oxidation-Reduction (Redox) Titration
An oxidation-reduction titration (redox titration) is a titration
based on an oxidation-reduction reaction. Calculations are done in the
same way as in acid-base titrations.
Example: A 20.00 mL sample of a solution containing Fe2+ ion is
titrated with a 0.02418 M solution of BrO3- ion. The equivalence point
of the titration occurs after the addition of 38.67 mL of the BrO3solution. What is the concentration of the Fe2+ solution?
6 Fe2+(aq) + BrO3-(aq) + 6 H+(aq)  6 Fe3+(aq) + Br-(aq) + 3 H2O()
Example: A 20.00 mL sample of a solution containing Fe2+ ion is
titrated with a 0.02418 M solution of BrO3- ion. The equivalence point
of the titration occurs after the addition of 38.67 mL of the BrO3solution. What is the concentration of the Fe2+ solution?
6 Fe2+(aq) + BrO3-(aq) + 6 H+(aq)  6 Fe3+(aq) + Br-(aq) + 3 H2O()
Strategy:
1) Find the number of moles of BrO3- (the solution where you
know both the volume and concentration).
2) Use the balanced chemical equation to find the number of
moles of Fe2+.
3) Find the concentration of the Fe2+ solution, using the number
of moles (found in step 2) and the volume of Fe2+ solution (given in the
problem).
6 Fe2+(aq) + BrO3-(aq) + 6 H+(aq)  6 Fe3+(aq) + Br-(aq) + 3 H2O()
20.00 mL Fe2+ soln
38.67 mL BrO3- soln
0.02418 M BrO3-
moles BrO3- = 0.03867 L soln 0.02418 mol = 9.350 x 10-4 mol BrO3L soln
moles Fe2+ = 9.350 x 10-4 mol BrO3- 6 mol Fe2+ = 5.610 x 10-3 mol Fe2+
1 mol BrO3molarity of Fe2+ soln =
5.610 x 10-3 mol Fe2+ = 0.2805 M
0.02000L soln
End of Chapter 4
A physicist, a biologist and a chemist went to the ocean for the
first time. The physicist saw the ocean and was fascinated by the waves.
He said he wanted to do some research on the fluid dynamics of the waves
and walked into the ocean. He was drowned and never returned.
The biologist saw the ocean and said that he wanted to do research
on the marine flora and fauna he saw there. He walked into the ocean and
also never returned.
The chemist waited for a long time and then wrote down the
following observation. “Physicists and biologists are soluble in ocean
water”. - anonymous
Two questions to distinguish chemists from non-chemists:
(1) How do you pronounce “unionized”?
(2) What is a mole?