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Topic 4
Chemical Reactions
Chemical reactions are the heart of chemistry. Chemical
reactions involve a change from reactant substances to
product substances. The products have physical and
chemical properties different from those of the reactants.
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Ions in Aqueous Solution
Many ionic compounds (soluble salts) dissociate
into independent ions when dissolved in water.
H 2O


NaCl(s )  Na (aq )  Cl (aq )
• These compounds that “freely” dissociate into
independent ions in aqueous solution are called
electrolytes.
• Their aqueous solutions are capable of conducting an
electric current because they have charged ions capable
of carrying a current.
• Generally, ionic solids that dissolve (soluble) in water are
electrolytes. Not all ionic compounds are soluble;
depends on the charge and size of the ions involved in the
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ionic bond.
Ions in Aqueous Solution
Not all electrolytes are ionic compounds.
Some molecular compounds (mostly acids)
dissociate into ions.


HCl(aq )  H (aq )  Cl (aq )
• The resulting solution is electrically conducting,
and so we say that the molecular substance is an
electrolyte.
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Ions in Aqueous Solution
Most molecular compounds (except acids)
dissolve but do not dissociate into ions;
soluble but no ions formed.
C 6 H 12O 6 (s ) ( glucose )  C 6 H 12O 6 (aq )
H 2O
– These compounds are referred to as
nonelectrolytes. They dissolve in water to give a
nonconducting solution.
– Covalent bonds are stronger than ionic bonds and
do not dissociate in water resulting in neutral
dissolved species.
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Ions in Aqueous Solution
Electrolytes dissolve in water to produce ions, but do so
at varying extents and dictates their conductivity; higher
# ions, higher the conductivity.
A strong electrolyte is an electrolyte that exists in
solution entirely as ions (100% dissociation to saturation
point).


NaCl(s )  Na (aq )  Cl (aq )
H 2O
Most ionic solids that dissolve in water do so
almost completely as ions, so they are strong
electrolytes.
strong electrolytes – strong acids, strong bases,
soluble salts
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Ions in Aqueous Solution
A weak electrolyte is an electrolyte that dissolves in
water to give a relatively small percentage of ions.
HC N ( aq )  H 2 O (l)


H 3 O ( aq )  C N ( aq )
Most soluble molecular compounds are either
nonelectrolytes or weak electrolytes.
Weak electrolytes - weak acids, weak bases,
insoluble salts
Solutions of weak electrolytes contain only a small
percentage of ions. We denote this situation by writing the
equation with a double arrow.
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We’ve stated the terms strong acid, strong base, soluble salt,
insoluble salt, but we haven’t describe how to determine which
species fall under these terms.
To be able to write chemical reactions correctly, we will need to
understand solubility and how strong and weak species
dissociate in water.
The first thing we will cover is the solubility rules for ionic
compounds. You must know the solubility rules to distinguish
between soluble and insoluble salts. Soluble salts (aq)
dissociate 100% into ions up to their saturation point while
insoluble salts (s) have very little dissociation.
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Solubility Rules for Ionic Compounds (Dissociates 100%)
1.) All compounds containing alkali metal cations (group I) and the
ammonium ion (NH4+) are soluble.
2.) All compounds containing NO3-, ClO4-, ClO3-, and C2H3O2- anions are
soluble.
3.) All chlorides, bromides, and iodides are soluble except those containing
Ag+, Pb2+, or Hg22+.
4.) All sulfates are soluble except those containing Hg22+, Pb2+, Ba2+, Sr2+,
or Ca2+. Ag2SO4 is slightly soluble.
5.) All hydroxides are insoluble except compounds of the alkali metals and
Ca2+, Sr2+, and Ba2+ are slightly soluble.
6.) All other compounds containing PO43-, S2-, CO32-, CrO42-, SO32- and
most other anions are insoluble except those that also contain alkali
metals or NH4+.
Generally, compounds that dissolve > 0.10 M are considered soluble (aq)
< 0.01 M are considered insoluble (s)
in between are slightly soluble
Soluble or insoluble?
Hg2Cl2 (s)
KI (aq)
Pb(NO3)2 (aq)
insoluble
soluble
soluble
(this class we will assume slightly soluble as soluble)
HW 26
code: soluble
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Strong Acids (Ionizes 100%)
HCl, HBr, HI, HClO4, HNO3, H2SO4 Memorize
these
Strong Bases (Dissociates 100%)
NaOH, KOH, LiOH, Ba(OH)2, Ca(OH)2,
Sr(OH)2
Rest of acids and bases (including acidic
cations and basic anions) are weak.
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Ions in Aqueous Solution
A molecular/formula unit equation is one in which the reactants and
products are written as if they were molecules/formula units, even
though they may actually exist in solution as ions.
Many ionic compounds undergo a displacement reaction between the cation
of one species with the anion of another. Displacement involves switching
atoms or ions between species, balancing charges between ions by adding
subscripts, balancing atoms in reaction by changing coefficients on
substances, and adding states (s, l, aq, g).
Calcium hydroxide + sodium carbonate
M.E.
Ca(OH)2 (aq) + Na2CO3 (aq)  CaCO3 (s) + 2 NaOH (aq)
strong base
soluble salt
insoluble salt
strong base
s solid
l liquid
aq aqueous (strong and weak acid/bases and soluble salts dissolved in water)
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g gases
Ions in Aqueous Solution
An total ionic equation represents strong electrolytes as separate
independent ions. This is a more accurate representation of the way
electrolytes behave in solution.
– A total ionic equation is a chemical equation in which strong
electrolytes (such as soluble ionic compounds, strong acids/bases) are
written as separate ions in solution. (note: g, l, insoluble salts (s), weak
acid/bases do not break up into ions)
M.E.
Ca(OH)2 (aq) +
strong base
Na2CO3 (aq)  CaCO3 (s) + 2 NaOH (aq)
soluble salt
insoluble salt
strong base
Total ionic
Ca2+ (aq) + 2OH- (aq)
+ 2Na+ (aq) + CO32- (aq) 
CaCO3 (s)
+ 2Na+ (aq) + 2OH- (aq)
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Net ionic equations.
– A net ionic equation is a chemical equation from
which the spectator ions have been removed.
– A spectator ion is an ion in an ionic equation that does
not take part in the reaction. Species does not change
form from reactant to product (Na+(aq) and OH-(aq) are
M.E. unchanged in this reaction and are spectators).
Ca(OH)2 (aq) +
Na2CO3 (aq)  CaCO3 (s) + 2 NaOH (aq)
Total Ionic
Ca2+ (aq) + 2OH- (aq) + 2Na+ (aq) + CO32- (aq)  CaCO3 (s) + 2Na+ (aq) + 2OH- (aq)
Net
Ca2+ (aq) + CO32- (aq)
 CaCO3 (s)
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Let’s try an example. First, we start with a molecular equation:
switch atoms or ions between species, balance charges between
ions by adding subscripts, balance atoms in reaction by
changing coefficients on substances, and add states (s, l, aq, g).
molecular equation
2 HNO3 (aq) + Mg(OH)2 (s)  Mg (NO3 )2(aq) + 2H2O (l)
strong acid
insoluble salt
soluble salt
Total ionic Break up strong acid/bases and soluble salts.
2 H+(aq) + 2NO3-(aq) + Mg(OH)2 (s)  Mg2+(aq) + 2NO3-(aq) + 2H2O(l)
Net Remove spectator ions.
2 H+(aq) + Mg(OH)2 (s)  Mg2+(aq) + 2H2O(l)
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Ions in Aqueous Solution
Molecular and Ionic Equations
HCN (aq) + NaOH (aq) 
weak acid
strong base
NaCN(aq) + H2O (l)
soluble salt
Total ionic
HCN (aq) + Na+(aq) + OH-(aq) 
Na+(aq) + CN-(aq) + H2O(l)
Net
HCN(aq) + OH-(aq)  CN-(aq) + H2O(l)
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Types of Chemical Reactions
Most of the reactions we will study fall into one
of the following categories
– Precipitation (solid) Reactions
– Acid-Base Reactions
– Oxidation-Reduction Reactions (transfer of
electrons)
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Types of Chemical Reactions
Precipitation Reactions
A precipitation reaction occurs in aqueous
solution because one product is insoluble.
– A precipitate (ppt) is an insoluble solid compound
formed during a chemical reaction in solution.
– For example, the reaction of sodium chloride with
silver nitrate forms AgCl(s), an insoluble precipitate.
NaCl(aq )  AgNO 3 (aq )  AgCl(s )   NaNO 3 (aq )
Is this a ppt rxn?
NaCl (aq) + Fe(NO3)2 (aq)  FeCl2 (aq) + 2 NaNO3(aq)
HW 27
No ppt is formed; therefore, not a ppt rxn - all aq ions; code: ppt
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basically, no reaction
Types of Chemical Reactions
Acid-Base Reactions
– Acids and bases are some of the most important
electrolytes.
– They can cause color changes in certain dyes
called acid-base indicators.
– Many household acids and bases encountered
everyday.
– Red cabbage juice is used as an acid-base
indicator.
– Vitamin C – ascorbic acid
– Aspirin – acetyl salicylic acid
– Milk of Magnesia
– Vinegar – acetic acid
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Types of Chemical Reactions
Different ways to define acids/bases
The Arrhenius Concept
– The Arrhenius concept defines acids as
substances that produce hydrogen ions, H+,
(also known as hydronium ions, H3O+) when
dissolved in water.
– An example is nitric acid, HNO3, a molecular
substance that dissolves in water to give H+ and
NO3-.


HNO 3 (aq ) 
 H (aq )  NO 3 (aq )
H 2O
HNO3(aq) + H2O(l)  H3O+(aq) + NO3-(aq)
Note: You will encounter acid ionization equations written both ways:
as a dissociation and an ionization.
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Types of Chemical Reactions
Different ways to define acids/bases
The Arrhenius Concept
– The Arrhenius concept defines bases as
substances that produce hydroxide ions, OH-,
when dissolved in water.
– An example is sodium hydroxide, NaOH, an ionic
substance that dissolves in water to give sodium
ions and hydroxide ions.


NaOH(s ) 
 Na (aq )  OH (aq )
H 2O
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Types of Chemical Reactions
Different ways to define acids/bases
By Arrhenius concept is ammonia an acid or
base?




NH 3 (aq )  H 2O ( l )  NH 4 (aq )  OH (aq )
By Arrhenius concept, ammonia produces OH-;
therefore, it is a base. However, this concept does
not cover all acids/bases that exist which gets us to
a more useful acid/base concept especially when
dealing with aqueous solutions.
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Types of Chemical Reactions
Different ways to define acids/bases
The Brønsted-Lowry Concept
– The Brønsted-Lowry concept of acids and bases
involves the transfer of a proton (H+) from the acid
to the base.
– In this view, acid-base reactions are protontransfer reactions.
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Types of Chemical Reactions
Different ways to define acids/bases
The Brønsted-Lowry Concept
– The Brønsted-Lowry concept defines an acid as
the species (molecule or ion) that donates a
proton (H+) to another species in a protontransfer reaction.
– A base is defined as the species (molecule or
ion) that accepts the proton (H+) in a protontransfer reaction.
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Types of Chemical Reactions
Different ways to define acids/bases
The Brønsted-Lowry Concept
NH 3 (aq)  H 2O (l )
base
acid
H+


NH 4 (aq)  OH (aq)
We obtained the same conclusion that NH3 is
a base using this theory; however, we didn’t
look at the production of OH- to make the
decision. It was based on NH3 accepting a
proton in the aqueous rxn.
The H2O molecule is the acid because it donates a proton. The NH3
molecule is a base, because it accepts a proton. Using Brønsted-Lowry
concept makes it easy to write acid-base reactions. You remove a
proton off the acid component and place it on the base component
giving you the products. Water is amphiprotic meaning it can act as an
acid or base. Since NH3 is a better base, water is the acid in this
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reaction.
Types of Chemical Reactions
Different ways to define acids/bases
If you learn the acids and bases in water, it will inform you how water
reacts in the reaction. We know that nitric acid is an acid in water;
therefore, water must be the base and proton acceptor in the reaction.


HNO 3 (aq )  H 2O ( l )  NO 3 (aq )  H 3O (aq )
H+
where HNO3 is an acid (proton donor) and H2O is a
base (proton acceptor).
A consequence of the Bronsted-Lowry concept is that on the product side of
the acid/base reaction there are also acid/base products. This leads us to the
term of conjugate acid/base pairs. These are species that differ by a proton
with the species containing more protons being the acid.
acid/base
acid/base
acid/base
HCN/CN-
H3O+/H2O
H2SO4/HSO4-
acid/base
H2PO3-/HPO3224
Types of Chemical Reactions
Different ways to define acids/bases
In summary,
– The Arrhenius concept
acid: proton (H+) producer
base: hydroxide ion (OH-) producer
– The Brønsted-Lowry concept
acid: proton (H+) donor
base: proton (H+) acceptor
HW 28
code: acid
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Types of Chemical Reactions
Acid-Base Reactions
Strong and Weak Acids and Bases
– A strong acid is an acid that ionizes completely
in water; it is a strong electrolyte.
100%


HNO 3 (aq )  H 2O ( l )  NO 3 (aq )  H 3O (aq )
100%


HCl (aq )  H 2O ( l )  Cl (aq )  H 3O (aq )
none left; all
ionized
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Types of Chemical Reactions
Acid-Base Reactions
Strong and Weak Acids and Bases
– A weak acid is an acid that only partially ionizes
in water; it is a weak electrolyte.
< 5%



HCN(aq )  H 2O ( l )  CN (aq )  H 3O (aq )
bulk does
not ionize
far less hydronium
formed as compared to
a strong acid of the
same concentration
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Types of Chemical Reactions
Acid-Base Reactions
Strong and Weak Acids and Bases
– A strong base is a base that is present entirely
as ions, one of which is OH-; it is a strong
electrolyte.
100%
H O
2


NaOH(s )  Na (aq )  OH (aq )
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Types of Chemical Reactions
Acid-Base Reactions
Strong and Weak Acids and Bases
– A weak base is a base that is only partially
ionized in water; it is a weak electrolyte.
< 5%


NH 3 (aq )  H 2O ( l )  NH 4 (aq )  OH  (aq )
bulk does
not ionize
far less hydroxide
formed as compared to
a strong base of the
same concentration
The main question is what are the weak bases?
Weak acids are easier because if you learn the 6 strong
acids, you know all the rest are weak.
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Weak bases are NH3, NH2-, NH- compounds and some anions.
Salts may be acidic/basic/neutral and are composed of cations
(positive ions) and anions (negative ions).
cation anion
acidic or neutral
basic or neutral
cations of strong bases
are neutral; otherwise,
cation contributes acidity
anions of monoprotic strong
acids are neutral; otherwise,
anion contributes basicity
Na+, K+, Li+, Ca2+, Sr2+, Ba2+
are neutral; rest of cations are
acidic to salt
Cl-, Br-, I-, NO3-, ClO4- are
neutral; rest of anions are
basic to salt
Learn which cations and anions are neutral and the rest of
the cations are acidic and the rest of the anions are basic.
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Predicting Whether a Salt is Acidic,
Basic, or Neutral
To predict the acidity or basicity of a salt,
you must examine the acidity or basicity
of the ions composing the salt.
Depending on the two components (cation/anion)
the overall salt will be acidic/neutral/basic :
neutral cation neutral anion  neutral salt
acidic cation
neutral anion  acidic salt
neutral cation basic anion  basic salt
acidic cation
basic anion  depends on which is
larger Ka or Kb
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Predicting Whether a Salt is Acidic,
Basic, or Neutral


N H 4 C l ( s )  N H 4 ( aq )  C l ( aq )
neutral  acidic salt
pH < 7
acidic


N aC N ( s )  
  N a ( aq )  C N ( aq )
neutral
basic  basic salt
pH > 7
KC2H3O2(s)  K+(aq) + C2H3O2-(aq)
neutral
basic

basic salt
pH > 7
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Predicting Whether a Salt is Acidic, Basic, or Neutral
acidic neutral  acidic
• AlCl 3
• Zn(NO3)2 acidic neutral  acidic
• KClO4
neutral neutral  neutral
• Na3PO4
neutral basic  basic
• LiF
neutral basic  basic
• NH4F (Ka = 5.6x10-10 > Kb = 1.4x10-11)
acidic basic  acidic
• NH4ClO (Ka = 5.6x10-10 < Kb = 3.6x10-7)
acidic basic  basic
• NH4C2H3O2 (Ka =
33
5.6x10-10
= Kb =
5.6x10-10)
acidic basic  neutral
Types of Chemical Reactions
Acid-Base Reactions
Neutralization Reactions
– One of the chemical properties of acids and bases is
that they neutralize one another.
– A neutralization reaction is a reaction of an acid
and a base that results in an ionic compound and
water.
– The ionic compound that is the product of a
neutralization reaction is called a salt.
Is the salt product acidic, basic, or neutral?
HCN (aq )  KOH (aq )  KCN (aq )  H 2O ( l )
acid
base
salt
Based on the cation being
neutral and the anion being
basic, this is a basic salt.
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During a neutralization reaction an acid and base are mixed together to make a
solution, the resulting solution will contain a salt and water
Acid + Base  salt + water
One of the chemical properties of acids and bases is that they neutralize one another;
however, that doesn’t mean that the product will be neutral.
The misconception is that if the acid and base are in stoich proportions that the
resulting solution is neutral. This is not true. The salt formed may be a acidic, basic,
or neutral salt and will dictate the pH of the solution. Common sense can help you
predict the pH of a mixture.
SA + SB  neutral salt
only true neutralization pH =7
WA + SB 
basic salt
original acid and base neutralize but
product has basic properties and basic pH
SA + WB 
acidic salt
original acid and base neutralize but
product has acidic properties and acidic pH
HCN (aq )  KOH (aq )  KCN (aq )  H 2O ( l )
weak acid
strong base
basic salt
strong acid
strong base
neutral salt
HCl (aq)  KOH (aq)  KCl (aq)  H 2O (l )
HW 29-30
code: neu
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Types of Chemical Reactions
Acid-Base Reactions
Acid-Base Reactions with Gas Formation with certain salts
– Carbonates react with acids to form CO2, carbon
dioxide gas.
H2CO3(aq)
H2O(l) + CO2(g)
Na 2CO 3  2 HCl  2 NaCl  H 2O  CO 2 
– Sulfites react with acids to form SO2, sulfur dioxide
gas.
H SO (aq)
H O + SO
2
3
2
(l)
Na 2SO 3  2 HCl  2 NaCl  H 2O  SO 2 
2(g)
– Sulfides react with acids to form H2S, hydrogen
sulfide gas.
Na 2S  2 HCl  2 NaCl  H 2S 
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Types of Chemical Reactions
Acid-Base Reactions
Other Acid-Base reactions
Nonmetal oxides (or acid oxides or acid anhydrides)
react with water to form acids
SO2 + H2O  H2SO3
Metal oxides (or basic oxides or basic anhydrides)
react with water to form bases
Na2O + H2O  2NaOH
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Types of Chemical Reactions
Oxidation-Reduction Reactions (Redox rxn)
– Oxidation-reduction reactions involve the
transfer of electrons from one species to another.
– Oxidation is defined as the loss of electrons
(higher oxidation state).
– Reduction is defined as the gain of electrons
(lower oxidation state).
– Oxidation and reduction always occur
simultaneously.
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Redox reactions – transfer of ereduction – oxidation reactions
Reduction – gain of e- / gain of H / lost of O
Fe3+ + 1e-  Fe2+
(lower ox state)
note: must balance atoms and charges;
electrons are always on reactants side.
39
Oxidation - loss of e- / loss of H / gain of O
Fe2+
 Fe3+ + 1e-
We must calculate charge on Br
to decide if it is oxidized or
reduced in charge (O: 2-; ion: 1-)
(higher ox state)
Br + 4(-2) = -1
Br = -1 +8 = +7
ox: H2O + BrO3-  BrO4- + 2H+ + 2eBr + 3(-2) = -1
(Br oxidized: charge goes 5+  7+)
Br = -1 +6 = +5
red: 2H+ + 2e-  H2
(H reduced: charge 1+  0)
Oxidizing agent is species that undergoes reduction.
Reducing agent is species that undergoes oxidation.
Note: need both for reaction to happen; can’t have
something being reduced unless something else is being
oxidized. Electrons are always on the product side for ox.40
Types of Chemical Reactions
Oxidation-Reduction Reactions
Copper prefers to be reduced while iron prefers to be
oxidized when present together. Lets write a net ionic
reaction for iron and copper (II) sulfate.
Fe(s )  CuSO 4 (aq )  FeSO 4 (aq )  Cu(s )
soluble salt
soluble salt
total ionic: Fe(s) + Cu2+(aq) + SO42-(aq)  Fe2+(aq) + SO42-(aq) + Cu(s)
net ionic: Fe(s) + Cu2+(aq)  Fe2+(aq) + Cu(s)
41
Types of Chemical Reactions
In this oxidation-reduction reaction, Feo lost
2e- and was oxidized to Fe2+ while Cu2+ gained
2e- and was reduced to Cuo. the charge is basically an inventory of
decreased by 2e- - oxidation
2
electrons as compared to protons: +#
means neutral species lost e- while -#
means neutral species gained e-
2
Fe(s )  Cu (aq )  Fe (aq )  Cu(s )
Reducing
agent
Oxidizing
agent
increased by 2e- - reduction
Notice that not only the atoms are balanced on both sides, the charges are
balanced as well
reactants: 0 + 2+ = 2+
products: 2+ + 0 = 2+ which gets us to the topic of oxidation numbers.
42
Types of Chemical Reactions
Oxidation-Reduction Reactions
Oxidation Numbers
– The concept of oxidation numbers is a simple way
of keeping track of electrons in a reaction.
– The oxidation number (or oxidation state) of an
atom in a substance is the actual charge of the
atom if it exists as a monatomic ion.
– Alternatively, it is hypothetical charge assigned
to the atom in the substance by simple rules.
43
Types of Chemical Reactions
Oxidation-Reduction Reactions
Oxidation Number Rules
Rule Applies to
Statement
1
Elements
The oxidation number of an atom in an element
is zero. i.e. Fe (s), H2 (g), Br2 (l)
2
Monatomic ions
The oxidation number of an atom in a
monatomic ion equals the charge of 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, O22-, where the oxidation
number is 1-).
44
Oxidation-Reduction Reactions
Rule Applies to
Statement
4
Hydrogen
The oxidation number of hydrogen is 1+ in
most of its compounds (exception metal
hydrides, i.e. CaH2).
5
Halogens
Fluorine is 1- in all its compounds. The other
halogens are 1- unless the other element is
another halogen or oxygen (oxoanions).
6
Compounds
and ions
The sum of the oxidation numbers of the
atoms in a compound is zero. The sum in a
polyatomic ion equals the charge on the ion.
Na2SO4 has Na+, O 2- , S 6+
SO422 (+1) + S + 4 (-2) = 0
S + 4 (-2) = -2
2 + S -8 = 0
S -8 = -2
S = 8 -2 = 6+
S = 8 -2 = 6+
Cl2
Cl : 0
ClO2
ClO2-
+4
+3
Cl2O5
+5
HW 31
code: number
You can determine charges based on whole compound
being equal to zero and using the rules to determine
charges with calculating unknown charges or you can do it based on the ions separately.
45
Types of Chemical Reactions
Oxidation-Reduction Reactions
Some Common Oxidation-Reduction Reactions
– Most of the oxidation-reduction reactions fall into one of
the following simple categories:
– Combination Reaction
– Decomposition Reactions (breaks up)
– Displacement Reactions (switches)
– Combustion Reactions (reacts with O2)
46
Types of Chemical Reactions
Oxidation-Reduction Reactions
Combination Reactions
– A combination reaction is a reaction in which two
substances combine to form a third substance.
R1 + R2  P
2 Sb (s)  3 C l 2 (g)  2 SbC l 3 (l)
47
Types of Chemical Reactions
Oxidation-Reduction Reactions
Decomposition Reactions
– A decomposition reaction is a reaction in which
a single compound reacts to give two or more
substances.
R  P1 + P2
2 HgO (s)  2 Hg (l)  O 2 (g)
48
Types of Chemical Reactions
Oxidation-Reduction Reactions
Displacement Reactions
– A displacement reaction (also called a singlereplacement reaction) is a reaction in which an
element reacts with a compound, displacing an
element from it.
Zn(s )  2 HCl(aq )  ZnCl 2 (aq )  H 2 ( g )
49
Types of Chemical Reactions
Oxidation-Reduction Reactions
Combustion Reactions
– A combustion reaction is a reaction in which a
substance reacts with oxygen, usually with the
rapid release of heat to produce a flame.
4 Fe (s) + 3 O 2 (g)  2 Fe 2 O 3 (s)
Common combustion reactions involve hydrocarbons
Hydrocarbon + O2  CO2 + H2O
HW 32
code: rxns
50
Types of Chemical Reactions
Oxidation-Reduction Reactions
How do we balance Oxidation-Reduction Reactions?
Look again at the net reaction of iron with copper(II)
sulfate.
2
2
Fe(s )  Cu (aq )  Fe (aq )  Cu(s )
We can write this reaction in terms of two halfreactions.
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Types of Chemical Reactions
Oxidation-Reduction Reactions
A half-reaction is one of the two parts of an oxidation-reduction reaction.
One involves the loss of electrons (oxidation) and the other involves the gain
of electrons (reduction).
To write a balanced redox reaction, you must write the half-reactions, balance
the atoms, balance the charges, and finally balance the number of electrons
transferred.
2
2
Fe ( s )  Cu ( aq )  Fe ( aq )  Cu( s )
oxidation half-reaction
reduction half-reaction
Notice: atoms
and charges are
balanced.
2
Fe ( s )  Fe ( aq ) + 2 e2
2 + C u ( aq )  C u ( s )
______________________________
e-
2
2
Fe ( s )  Cu ( aq )  Fe ( aq )  Cu ( s )
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Balancing Redox Equations
- Must know charges (oxidation numbers) of
species including polyatomic ions
- Must know strong/weak acids and bases
- Must know the solubility rules
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Rules to balance redox
1.) Convert to net ionic form if equation is originally in molecular form
(eliminate spectator ions).
2.) Write half reactions.
3.) Balance atoms using H+ / OH- / H2O as needed:
– acidic: H+ / H2O put water on side that needs O or H
– basic: OH- / H2O put water on side that needs H but if there is no H
involved then put OH- on the side that needs the O in a 2:1 ratio;
basically balance O with OH, double OH, add 1/2 water to
other side.
4.) Balance charges for half rxn using e-.
5.) Balance transfer/accept number of electron in whole reaction.
6.) Convert equation back to molecular form if necessary (re-apply
spectator ions).
Notes: H+ and OH- on same side equals H2O;
If only need H+ or OH- , just use it by itself.
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Let’s look at a simple example to demonstrate the procedure. First,
we will need to find the net ionic equation (remove spectator ions).
Zn(s) + AgNO3(aq)  Zn(NO3)2(aq) + Ag(s)
Total ionic:
Zn(s) + Ag+(aq) + NO3-(aq)  Zn2+(aq) + 2NO3-(aq) + Ag(s)
Net ionic:
Zn(s) + Ag+(aq)  Zn2+(aq) + Ag(s)
Next, we will need to write the half-reactions.
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Zn(s) + Ag+(aq)  Zn2+(aq) + Ag(s)
Net:
0
Oxidation:

2+ + 2e- = 0
Zn(s)  Zn2+(aq) + 2e0 = 1e- + 1+
Reduction:

0
[ 1e- + Ag+(aq)  Ag(s) ] 2
Multiplying the silver reaction by 2 will give
both half reactions 2e- that will cancel out.
Balanced net:
Now, we must balance
atoms (already balanced in
this reaction) and charges;
we can only add e- (-1) to
balance the charge.
Next, we need to balance
the number of electrons
transferred.
Zn(s) + 2 Ag+(aq)  Zn2+(aq) + 2 Ag(s)
Balanced eq:
Zn(s) + 2 AgNO3(aq)
 Zn(NO3)2(aq) + 2 Ag(s)
Lastly, we put back the spectators making sure that we keep the atoms
balanced. If correctly done, charges and atoms should be balanced.
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Let’s look at a net ionic example. First, write the half-reactions.
this is an acidic solution with
H+ Indicates
H /H O available.
2+
Net:
MnO4 (aq) + Fe (aq)  Mn2+(aq) + Fe3+(aq)
+
2
Now, we must balance atoms; however, in this example the reduction half-reaction will need to use
H+/H2O (acidic) to balance the reaction. We place the H2O on the side that needs O.
Next, decide how many H2O you need based on O needed to balance the sides.
Next, we need to balance
Now place the number of H+ needed on the opposite side to balance the sides.
the number of electrons
2+
3+
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transferred.
Fe (aq)  Fe (aq) + 1e
[
Ox:
]
Next, we must balance the charges; we can only add e- (-1) to balance the charge.
Red:
5e- + 8 H+(aq) + MnO4-(aq)  Mn2+(aq) + 4 H2O(l)
Balanced net:
Finally, the balanced net ionic equation (charges & atoms)
is as follows
8 H+(aq) + MnO4-(aq) + 5 Fe2+(aq)  Mn2+(aq) + 5 Fe3+(aq) + 4 H2O(l)
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Let’s look at a molecular equation example. First, we need the net eq.
KMnO4(aq) + NaNO2(aq) + HCl(aq)  NaNO3(aq) + MnCl2(aq) + KCl(aq) + H2O(l)
Net: MnO4-(aq) + NO2-(aq) + H+(aq)  NO3-(aq) + Mn2+(aq) + H2O(l)
Noticed that after we removed the spectators, KCl is not part of the net. We must remember to
put it back at the end.
H2O(l) + NO2-(aq)  NO3-(aq) + 2 H+(aq) + 2 e- ] 5
Ox:
[
Red:
[ 5 e- + 8 H+(aq) + MnO4-(aq)  Mn2+(aq) + 4 H2O(l) ] 2
Balanced net:
Equation can be reduced since it has common species on both sides.
2 MnO4-(aq) + 5 NO2-(aq) + 16 H+(aq) + 5 H2O(l)  2Mn2+(aq) + 8 H2O(l) + 5 NO3-(aq) +10 H+(aq)
2 MnO4-(aq) + 5 NO2-(aq) + 6 H+(aq)  2Mn2+(aq) + 3 H2O(l) + 5 NO3-(aq)
Balanced eq:
2 KMnO4(aq) + 5 NaNO2(aq) + 6 HCl(aq)  2MnCl2(aq)+ 3 H2O(l)+ 5 NaNO3(aq) + 2 KCl
Lastly, we put back the spectators making sure that we keep the atoms balanced.
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Note: we must add back KCl and balanced K and Cl for the entire reaction can be balanced.
Let’s look at a net ionic example (basic).
OH-
Net:
CrI3 (s) + Cl2 (g)  CrO42-(aq) + IO4-(aq) + Cl-(aq)
Note: Cr (3+  6+) and I (1-  7+)
are being oxidized in same half-rxn.
Ox:
[ 32 OH-(aq) + CrI3(s)  CrO42-(aq) + 3 IO4-(aq) + 16 H2O(l) + 27 e- ] 2
Since we only need to balance the O and we don’t want to mess up H as we balance the O (H not
present), we will double the amount of OH- on the side that needs O (needs 16 but will put 32) and put
half as many H2O on the other side (2OH-:1H2O ratio).
Red:
[ 2 e- + Cl2(g)  2 Cl-(aq) ] 27
Balanced net:
64 OH-(aq) + 2 CrI3(s) + 27 Cl2(g)  2 CrO42-(aq) + 6 IO4-(aq) + 54 Cl-(aq) + 32 H2O(l)
HW 33
code: redox
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Working with Solutions
The majority of chemical reactions discussed
in general chemistry occur in aqueous
solution (water as solvent).
When you run reactions in liquid solutions, it is
convenient to dispense the amounts of reactants by
measuring out volumes of reactant solutions.
Note: volumes are temperature dependent and in
some instances are not useful; mass is not
temperature dependent.
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Working with Solutions
Molar Concentration
When we dissolve a substance in a liquid, we
call the substance the solute (species
dissolved) and the liquid the solvent (species
doing the dissolving).
Coffee:
coffee is the solute while water is the solvent.
The general term concentration refers to the
quantity of solute in a standard quantity of solution
(solute + solvent).
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Working with Solutions
Molar Concentration
Molar concentration, or molarity (M), is
defined as the moles of solute dissolved in
one liter (cubic decimeter) of solution.
moles of solute
Molarity (M) 
liters of solution
L solution = L solute + L solvent
Since volumes are not additive and the solution accounts for both
the volume of the solute and the solvent, we typically make
solutions by mixing them in glassware with a specific volume
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indicator; i.e., volumetric glassware.
Working with Solutions
Molar Concentration
A sample of NaNO3 weighing 0.38 g is placed in a
50.0 mL volumetric flask. The flask is filled with
H2O to the mark on the neck, dissolving the solid.
What is the molarity of the solution?
Goal:
We start with a ratio between the mass, g, and
volume, mL, and convert the units to the desired
units of mols and L (M).
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Working with Solutions
Molar Concentration
An experiment calls for the addition of 0.184 g NaOH in an
aqueous solution to reaction vessel. How many milliliters
of 0.150 M NaOH should be added? g NaOH  mL soln
We want to determine the volume of NaOH that has
0.184g NaOH in it from a 0.150 M solution. We need to
change the mass term into mols and then use the ratio of
mols to L (the molarity) to determine the volume.
This means that 30.7 mL of 0.150 M NaOH soln has a
mass of 0.184 g.
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Working with Solutions
Molar Concentration
How many grams of NaCl should be put in a 50.0 mL
volumetric flask to make a 0.15 M NaCl solution
when the flask is filled to the mark with water?
We will convert L  g by using
the ratio of L to mol (molarity)
and then molar mass (mol to g).
This means if we take 0.44 g NaCl and dilute to 50.0 mL
with H2O, we will have a 0.15 M NaCl soln.
HW 34
code: solution
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Working with Solutions
Molar Concentration
The molarity of a solution and its volume are
inversely proportional. Therefore, adding
water makes the solution less concentrated.
This inverse relationship takes the form of:
M i  Vi  M f  V f
C c  Vc  C d  Vd
This equation can be used for any concentration term, C, where c –
concentrated and d – diluted. Only use for dilution calculations.
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Working with Solutions
Molar Concentration
Describe how you would make 100.0 mL of
1.00 M NH3 from a 14.8 M NH3 solution?
using Cc x Vc = Cd x Vd
= 6.76 mL
You must take 6.76 mL of 14.8 M NH3
soln and dilute to 100.0 mL.
HW 35
code: dilution
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Quantitative Analysis
Analytical chemistry deals with the
determination of composition of materialsthat is, the analysis of materials.
– Quantitative analysis involves the determination
of the amount of a substance or species (analyte)
present in a material.
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Quantitative Analysis
Gravimetric Analysis
Gravimetric analysis is a type of quantitative
analysis in which the amount of a species in a
material is determined by converting the
species into a product that can be isolated
and weighed.
– Precipitation reactions are often used in gravimetric
analysis.
– The precipitate from these reactions is then filtered,
dried, and weighed.
– If the end product is the analyte, you can determine
the quantity directly or if the end product is not the
analyte, mol-to-mol ratios can be used to indirectly
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determine the quantity of analyte.
Quantitative Analysis
Gravimetric Analysis
Consider the problem of determining the
amount of lead in a sample of drinking water.
– Adding sodium sulfate (Na2SO4) to the sample will
precipitate lead(II) sulfate.
Na 2SO 4 (aq )  Pb 2 (aq )  2 Na  (aq )  PbSO 4 (s )
– The PbSO4 can then be filtered, dried, and
weighed.
– The quantity of PbSO4 can be used to determine
the amount of Pb in the sample because all the Pb
in the sample should have precipitated with the
PbSO4.
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Quantitative Analysis
Gravimetric Analysis
Suppose a 1.00 L sample of polluted water was analyzed for
lead(II) ion, Pb2+, by adding an excess of sodium sulfate to it.
The mass of lead(II) sulfate that precipitated was 229.8 mg.
What is the mass of lead (mg) in the sample?
Na 2SO 4 (aq )  Pb 2 (aq )  2 Na  (aq )  PbSO 4 (s )
? mg
229.8 mg
m
m
We must convert the amount of PbSO4 precipitated to
the amount of Pb in the sample. We will need to convert
it first to mols and then use the mol-to-mol ratio
between Pb and PbSO4. After that, we will use molar
mass to obtain the mass (mols to g). Notice we used the
milli concept to avoid unnecessary conversions.
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Quantitative Analysis
Volumetric Analysis
An important method for determining the amount of a
particular substance is based on measuring the volume of
the reactant solution.
– Titration is a procedure for determining the
amount of unknown substance A by adding a
carefully measured volume of a solution with
known concentration of B until the reaction of A
and B is just complete (equivalence point: #mols
of titrant = #mols analyte by stoich).
– Volumetric analysis is a method of analysis
based on titration.
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Quantitative Analysis
Volumetric Analysis
Consider the reaction of sulfuric acid, H2SO4, with sodium hydroxide,
NaOH (note: must write a balanced equation to determine mol-tomol ratios):
Suppose a beaker contains 35.0 mL of 0.175 M
H2SO4. How many milliliters of 0.250 M NaOH must
be added to completely react with the sulfuric acid?
H2SO4 (aq) + 2 NaOH (aq)  Na2SO4 (aq) + 2 H2O (l)
35.0 mL 0.175 M
0.250 M ? mL
Note: volume times M gives us mols of
H2SO4 used. Next, we can do a mol-to-mol
ratio to NaOH and then determine volume
from molarity (mol to L ratio). We will take
advantage of the milli prefix once again.
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Quantitative Analysis
Volumetric Analysis
A flask contains a solution with an unknown amount of HCl.
This solution is titrated with 0.207 M NaOH. It takes 4.47 mL
of NaOH to complete the reaction. What is the mass (g) of HCl
in the sample? Note: Must write reaction before you work the
problem; need to know the mol-to-mol ratio.
HCl (aq) + NaOH (aq)  NaCl (aq) + 2 H2O (l)
?g
0.207 M
4.47 mL
Warning: Although this problem ended up with a 1:1
mol ratio, you should get in the habit of writing the
reaction before doing any calculations. Most of the
time, the reactions will not be 1:1.
HW 36
code: volume
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