ACE Chemistry
Chapter 4
IONS IN AQUEOUS SOLUTION
4.1 Ionic Theory of Solutions:
· Proposed by ____________________________ (1884)
· Certain substances produce freely moving _________________________ when they dissolve in
________________; the mobile ions conduct an ______________________________________.
Substances which are able to conduct electricity by ionic movement are called _________
_____________________
For example:
An aqueous solution of NaCl is a good conductor of electricity
Substances, which are not able to conduct electricity by ionic movement, are called_____________
____________________________
A particular substance must meet two conditions in order to be an electrolyte:
1. ______________________________
2. ___________________________________
Strong electrolytes-
Weak electrolytes-
Solubility rules:
4.2 Molecular and Ionic Equations:
Molecular equation-
Example:
Complete Ionic Equation-
Example:
Net ionic equation-
Example-
Spectator ions-
Problems:
Write the net ionic equation for each of the following molecular equations on page 133 of your text
exercise 4.2
a.
b.
Types of Chemical Equations
1.
2.
3.
Precipitate Reactions
Precipitate-
Exchange (metaathesis) reaction-
Example:
Problem: p. 136 exercise 4.3
You mix aqueous solutions of sodium iodide and lead (II) acetate. If a reaction occurs, write the
balanced molecular equation and the net ionic equation. If no reaction occurs, indicate this by writing
the formulas of the compounds and an arrow followed by NR
4.4 Acid Base Reactions
Acid base indicator-
Acids-
Bases-
Strong acids-
Weak acids-
Strong bases-
Weak bases-
p. 140 exercise 4.4
a.
b.
c.
d.
Neutralization Reactions-
Steps for Writing an equation for a neutralization reaction:
on page 142:
exercise 4.5:
Exercise 4.6
Acid Base Reactions with gas:
p. 144 exercise 4.7
4.5 Oxidations –Reduction Reactions
Oxidation number-
Oxidation-reduction reactions-
Rules For Assigning Oxidation Numbers:
p. 148 exercise 4.8
Describing Ox-red reactions:
Half reaction-
Oxidation-
Reduction-
Oxidizing agent-
Reducing agent-
Common Redox reactions:
1.
2.
3.
4.
Describe and give an example of each common redox reaction:
1.
2.
3.
4.
Balancing Simple Redox Rxn
4.7 Molar Concentrations
Molarity-
Exercise 4.11 and 4.12 page 156
4. 8 Diluting Solutions
Exercise 4.13 p. 157
4.9 Gravimetric Analysis
Exercise 4.14 p. 159
4.10 Volumetric Analysis
Exercise 4.16 p. 162
melting H2O
Na + Cl - (s) Na + Cl - (s) Na + (l)+Cl - (l) Na + Cl - (s) Na + (aq) + Cl (aq)
3
NOTE:
1. All soluble ionic compounds dissolved in water are STRONG ELECTROLYTES
2. The process by which a substance separates into ions (by melting or by dissolving in water) is
referred to, as DISSOCIATION.
3. SOLUBLE IONIC SUBSTANCES ARE COMPLETELY DISSOCIATED IN AQUEOUS SOLUTION
(all ions are separated and free to move).
EXAMPLES:
NaOH(aq) ¾¾¾¾> Na + (aq) + OH - (aq)
NOTE: 1 mole of NaOH produces by dissociation: 2 moles of ions: 1
mole of Na + ions, and
1
moles of OH - ions
MgCl2(aq) ¾¾¾¾> Mg 2+ (aq) + 2 Cl - (aq)
magnesium chloride magnesium ions chloride ions
NOTE: 1 mole of MgCl2 produces by dissociation: 3 moles of ions:1
mole of Mg 2+ ions, and
2
moles of Cl - ions
Na2SO4(aq) ¾¾¾¾> 2 Na + (aq) + SO4
2- (aq)
sodium sulfate sodium ions sulfate ions
NOTE: 1 mole of Na2SO4 produces by dissociation: 3 moles of ions: 2
moles of Na + ions, and
1
mole of SO4
2- ions
4. PARTLY SOLUBLE IONIC SUBSTANCES ARE PARTIALLY DISSOCIATED IN AQUEOUS
SOLUTION
(some, but not all, ions are separated and free to move)
Chemistry 101 Chapter 4
4
PURE MOLECULAR SUBSTANCES
SOLID STATE LIQUID STATE GASEOUS STATE
Solid Sucrose Liquid Water Gaseous Hydrogen Chloride
C12H22O11(s) H2O(l) HCl(g)
molecules only present molecules only present molecules
only present
Bulb is off Bulb is off Bulb is off
NO CONDUCTANCE NO CONDUCTANCE NO CONDUCTANCE
NONELECTROLYTE NONELECTROLYTE NONELECTROLYTE
CONCLUSION:
· Molecular Substances in pure form are NONELECTROLYTES, since they contain molecules
only (no ions)
Chemistry 101 Chapter 4
5
MOLECULAR SUBSTANCES IN AQUEOUS SOLUTION
Sucrose dissolved in water Acetic acid dissolved in water Hydrogen
chloride
dissolved in water
Bulb is off Bulb glows dimly Bulb glows brightly
No mobile ions present
(molecules only present)
A few mobile ions present
(mostly molecules)
Only mobile ions present
(no molecules)
Molecules separate
C12H22O11(s) ¾> C12H22O11(aq)
Molecules separate into ions:
HC2H3O2(aq) ¾> H + (aq) + C2H3O2
- (aq)
at the same time:
Ions combine and form molecules:
H + (aq) + C2H3O2
- (aq) ¾> HC2H3O2(aq)
Molecules separate into ions
HHCl(aq) ¾>H + (aq)+Cl - (aq)
No Ionization Partial Ionization Complete Ionization
No Conductance Weak Conductance Strong Conductance
NONELECTROLYTE WEAK ELECTROLYTE STRONG
ELECTROLYTE
Chemistry 101 Chapter 4
6
CONCLUSIONS:
1. Some molecular substances do not interact with water when they dissolve in it, and as such, they
do not form ions.
They exist in aqueous solution as MOLECULES ONLY, and are NONELECTROLYTES (NE).
Example: Sucrose
2. Some molecular substances interact with water and their molecules form ions.
This is referred to as IONIZATION. Ionization can be of two types:
(A) PARTIAL IONIZATION
· Some molecular substances interact only partially with water and a very few of their molecules
change into ions.
· In partial ionization, two processes take place at the same time, and a dynamic equilibrium is
established.
HC2H3O2 (aq) H + (aq) + C2H3O2
- (aq)
99 molecules 1 ion 1 ion
· The predominant species in solution are molecules.
· The weak conductance is due to the very few ions present (one out of every hundred molecules
ionizes)
· These substances exist in aqueous solution MOSTLY AS MOLECULES (a very few ions) and
are called WEAK ELECTROLYTES (WE)
Example:
An aqueous solution of ammonia:
NH3 (aq) + H2O (l) NH4
+ (aq) + OH - (aq)
Mostly molecules A very few ions
Chemistry 101 Chapter 4
7
(B) COMPLETE IONIZATION
§ For some molecular substances the interaction with the water molecules is essentially
complete, and all their molecules change into ions.
§ This is referred to as complete ionization.
§ These substances exist in aqueous solution as IONS ONLY, and as such are
STRONG ELECTROLYTES (SE)
Examples:
HCl(aq), HNO3(aq), H2SO4(aq) (for the loss of one H + )
HCl(aq) ¾¾¾> H + (aq) + Cl - (aq)
HNO3(aq) ¾¾¾> H + (aq) + NO3
- (aq)
H2SO4(aq) ¾¾¾> H + (aq) + HSO4
- (aq)
Chemistry 101 Chapter 4
8
SUMMARY: ELECTROLYTES AND NONELECTROLYTES
IONIC SUBSTANCES MOLECULAR SUBSTANCES
Solid Liquid Aqueous Solution Solid Form Liquid Form Aqueous
Solution
Soluble Partly
Soluble
NaCl(s) NaCl(l) NaCl(aq) PbCl2(aq) C12H22O11(s) C12H22O11(l)
NH3(aq) HCl(aq)
NaOH(s) NaOH(l) NaOH(aq) HC2H3O2(s) HC2H3O2(l) HC2H3O2(aq)
HNO3(aq)
NE SE SE WE NE NE WE SE
Ions Ions Ions Few ions No ions No Ions Few ions Ions only
present present present and (molecules (molecules (mostly (no
but not and free and free and free only) only) molecules molecules)
to move to move to move to move free to move free to move
No Complete Complete Partial No No Partial Complete
Dissoc’n Dissoc’n Dissoc’n Dissoc’n Ionization Ionization Ionization
Ionization
Chemistry 101 Chapter 4
9
IONIC AND MOLECULAR EQUATIONS
· Chemical reactions which take place in aqueous solution are caused by the interactions between
ions
and are referred to as IONIC REACTIONS.
· Most IONIC REACTIONS are DOUBLE DISPLACEMENT REACTIONS (also called
METATHESIS REACTIONS)
A +X -+ B +Y -A +Y -+ B +X · Consider: the reaction between an aqueous solution of sodium sulfate and an aqueous solution of
barium
chloride forms a solid which is insoluble in water (in time, it settles as the bottom of the test tube and
is
referred to as A PRECIPITATE)
Na2SO4(aq) + BaCl2(aq) ¾¾> A PRECIPITATE
(soluble salt) (soluble salt) (opaque suspension)
spectator
ions
2 Na + (aq)
Na + (aq) Cl - (aq)
SO4
2- (aq) + Ba 2+ (aq) 2 Cl - (aq)
Na + (aq) Cl - (aq)
insoluble
solid
(precipitate)
BaSO4(s)
Chemistry 101 Chapter 4
10
Molecular Equation:
Na2SO4(aq) + BaCl2(aq) BaSO4(s) + 2NaCl(aq)
clear solution clear solution insoluble solid clear solution
(precipitate)
· This equation does not show which substances exist in ionic form.
Complete (Total) Ionic Equation:
· Strong electrolytes are written as separate ions in solution (completely dissociated)
2Na + (aq) + SO4
2- (aq) + Ba 2+ (aq) + 2 Cl - (aq) BaSO4(s) + 2 Na + (aq)+2 Cl - (aq)
Spectator Spectator Spectator Ions
· Spectator Ions: Ions in an ionic equation which do not take part in the reaction
SO4
2- (aq) + Ba 2+ (aq) BaSO4(s)
· This equation: is
called the NET IONIC EQUATION (focuses on the main event)
is
obtained by canceling out the spectator ions:
2Na
+ (aq)+ SO4
2- (aq) + Ba 2+ (aq) + 2 Cl - (aq) BaSO4(s) + 2 Na + (aq) + 2 Cl - (aq)
Spectator Spectator precipitate Spectator Ions
Net Ionic Equation: SO4
2- (aq) + Ba 2+ (aq) BaSO4(s)Chemistry 101 Chapter 4
11
TYPES OF CHEMICAL REACTION
I. METATHESIS REACTIONS (or DOUBLE DISPLACEMENT RXNS)
· In these reactions the ions of the reactants are exchanged:
A + B - (aq) + C + D - (aq) AD (?) + CB (?)
· DoubleDisplacement
Reactions can be further classified into:
1. Precipitation Reactions
2. Acid – Base Reactions
3. Reactions that form an unstable product
II. REDOX REACTIONS (Oxidation – Reduction Reactions)
· In these reactions an exchange of electrons occurs between the reactants.
· Redox Reactions can be further classified into:
1. Combination Reactions
2. Decomposition Reactions
3. Single Replacement Reactions
4. Combustion Reactions
Each type of these reactions will be discussed in detail.
Chemistry 101 Chapter 4
12
SOLUBILITY RULES
1. Precipitation Reactions
· In these reactions an insoluble solid (precipitate) forms.
· To better understand these reactions, a knowledge of solubility rules for ionic substances is
necessary.
· These solubility rules are summarized in solubility tables (See Table 4.1 in your textbook)
Solubility Rules:
· All compounds of group IA and (NH 4
+ ) are soluble.
· All nitrates, acetates, and most perchlorates are soluble.
· All chlorides, bromides, and iodides are soluble, except those of Ag + , Pb 2+ ,
Cu + and Hg 2
2+ .
· All sulfates are soluble, except those of Ca 2+ , Sr 2+ , Ba 2+ and Pb 2+ .
· All metal hydroxides are insoluble, except those of Group IA and larger
members of Group 2A.
· All carbonates and phosphates are insoluble, except those of Group IA and
(NH 4
+ ).
· All sulfides are insoluble, except those of Group IA, Group 2A and (NH 4
+ ).
Examples:
Use solubility table to determine if each of the following substances are soluble or insoluble:
CaCl2 ______________
PbSO4 ______________
Mg(OH)2 ______________
(NH4)2CO3 ______________
Chemistry 101 Chapter 4
13
PRECIPITATION REACTIONS
· A solution of silver nitrate is mixed with a solution of sodium chloride. A white precipitate is
formed. Write a NET IONIC EQUATION for this reaction.
AgNO3 (aq) + NaCl (aq) AgCl (?) + NaNO3 (?)
The solubility of each product must be known! Referring to the solubility rules, we find out that:
AgCl is insoluble in water (precipitate): AgCl(s)
NaNO3 is soluble in water (completely dissociated): NaNO3(aq)
· The Molecular Equation becomes:
AgNO3 (aq) + NaCl (aq) AgCl (s) + NaNO3 (aq)
· The Complete (Total) Ionic Equation is:
Ag + (aq) + NO3
- (aq) + Na + (aq) + Cl - (aq) AgCl(s) + Na + (aq) + NO3
- (aq)
ppt
· The Net Ionic Equation is obtained by canceling out the spectator ions:
Ag + (aq) + NO3
- (aq) + Na + (aq) + Cl - (aq) AgCl(s) + Na + (aq) + NO3
- (aq)
spectator spectator ppt spectator ions
ion ion
· The NET IONIC EQUATION is:
Ag + (aq) + Cl - (aq) AgCl(s)
Example:
Write balanced molecular and net ionic equations for the reaction of solutions of Pb(NO 3 ) 2 and KI.
Chemistry 101 Chapter 4
14
2. Acid – Base Reactions and Acid – Base Concepts
General Properties
ACIDS BASES
Taste sour bitter
Change color of
indicators:
Blue Litmus Red No change
Red Litmus No change Blue
Phenolphtalein Colorless Pink
Neutralization Reacts with bases to
produce salt and water
Reacts with acids to
produce salt and water
Arrhenius Concept of Acids and Bases
· This definition defines acids and bases in terms of the effect they have on water
ACIDS BASES
Substances that dissolve in water and increase the
concentration of hydronium ions (H3O + )
Substances that dissolve in water and
increase the concentration of hydroxide
ions (OH – )
Examples:
HCl (g) + H2O (l) ® H3O + (aq) + Cl – (aq)
Accepted simplification:
HCl (g) ¾H¾2 O¾ ® H+ (aq) + Cl – (aq)
Examples:
NaOH (s) ¾H¾2 O¾ ® Na+ (aq) + OH –
(aq)
HC2H3O2 (l) + H2O (l) ® H3O + (aq) + C2H3O2
– (aq)
Accepted simplification:
HC2H3O2 (l) ¾H¾2O ¾ ® H+ (aq) + C2H3O2
– (aq)
Ba(OH)2 (s) ¾H¾2O ¾ ® Ba2 + (aq) + 2 OH – (aq)
Limitations of Arrhenius definition:
1. Considers acidbase
reactions only in aqueous solutions.
2. Singles out the OH - ion as the source of base character; (other species can play a similar role)
Chemistry 101 Chapter 4
15
BronstedLowry
Concept of Acids and Bases
· This definition defines acids and bases in terms of H + (proton) transfer.
ACIDS BASES
Acids are H + (proton donors) Bases are H + (protons acceptors
NEUTRALIZATION: A reaction in which a H + (proton) is transfered
Examples of BronstedLowry
neutralizations :
1. HCl(g) + H2O(l) H3O + (aq) + Cl - (aq)
H+
HCl(aq) + H2O(l) H3O + (aq) + Cl - (aq)
acid base
2. NH3(g) + H2O(l) NH4
+ (aq) + OH - (aq)
H+
NH3(g) + H2O(l) NH4
+ (aq) + OH - (aq)
base acid
NOTE: H2O
gains a H + and acts as a base with HCl(g)
H2O
loses a H + and acts as an acid with NH3(g)
· A substance that can behave both as a base or an acid depending on the chemical environment
is called an amphiprotic species.
· H2O is an amphiprotic species.
Chemistry 101 Chapter 4
16
3. HC2H3O2(l) + H2O(l) H3O + (aq) + C2H3O2
- (aq)
acetic acid acetate ion
H+
HC2H3O2(l) + H2O(l) H3O + (aq) + C2H3O2
- (aq)
acid base
H2O is an amphiprotic species.
The BronstedLowry
Concept of Acids & Bases is more general than the Arrhenius concept:
The BronstedLowry
Concept introduces additional points of view:
1. A base is a species that accepts H + ions (protons). (OH - is only one example of a base)
2. Acids and Bases can be ions as well as molecular substances.
3. AcidBase
reactions are not restricted to aqueous solutions.
4. Amphiprotic species (like H2O) can act as either acids or bases, depending on what the
other reactant is.
SUMMARY OF ACID – BASE CONCEPTS
ACID BASE ACIDBASE
REACTION
ARRHENIUS
(less general)
Produces H + when
dissolved in water
Produces OH – when
dissolved in water H + + OH – ® H2O
BRONSTEDLOWRY
(more general) Proton donor Proton acceptor Proton transfer
Chemistry 101 Chapter 4
17
CHEMICAL PROPERTIES OF ACIDS AND BASES
· The most important property of ACIDS and BASES is their reaction with each other, called
NEUTRALIZATION:
ACID + BASE SALT + WATER
Ionic Substance
In General:
HA + BOH B + A - + H2O
any cation any anion
(except H + ) (except OH - )
Examples:
1. HCl (aq) + NaOH (aq) NaCl (aq) + H2O (l)
2. HCl (aq) + Ba(OH)2 (aq) BaCl2 (aq) + H2O (l)
unbalanced
Balanced: 2 HCl (aq) + Ba(OH)2 (aq) BaCl2 (aq) + 2 H2O (l)
3. HC2H3O2(aq) + NaOH(aq)
4. H2SO4(aq) + NaOH(aq)
unbalanced
Balanced: H2SO4(aq) + NaOH(aq)
Chemistry 101 Chapter 4
18
· Note that all SALTS are derived from an ACID and a BASE:
Salt Base from which
derived
Acid from which
Derived
Fe (NO3)3
iron (III) nitrate
Fe (OH)3
iron (III) hydroxide
HNO3
nitric acid
CaCl2
calcium chloride
Ca(OH)2
calcium hydroxide
HCl
hydrochloric acid
Na2CO3
sodium carbonate
(NH4)2SO4
ammonium sulfate
Na2S
sodium sulfide
NaC2H3O2
sodium acetate
K3PO4
(potassium phosphate)
Chemistry 101 Chapter 4
19
STRONG AND WEAK ACIDS AND BASES
· Acids and Bases can be classified according to their ability to ionize or dissociate in aqueous
solution:
ACIDS BASES
STRONG ACIDS WEAK
ACIDS STRONG BASES WEAK
BASES
Electrolyte
strength Strong electrolytes Weak
electrolytes Strong electrolytes Weak
electrolytes
Extent of
Ionization/
Dissociation
100%
Complete Ionization
Less than 100%
Partial
Ionization
100%
Complete
Ionization
Less than 100%
Partial
Ionization
Symbols used to
show extent of
ionization/
dissociation
Particles present
in aqueous
solution
Ions only
Mostly
molecules
(a few ions)
Ions only
Mostly
molecules
(a few ions)
Examples
H + (aq) + Cl – (aq)
H + (aq) + NO3
– (aq)
H + (aq) + HSO4
– (aq)
HC2H3O2 (aq)
HF (aq)
H2CO3 (aq)
Na + (aq) + OH – (aq)
K + (aq) + OH – (aq)
Ba 2+ (aq) + 2 OH – (aq)
NH4OH (aq)
also written
NH3 (aq) + H2O
Chemistry 101 Chapter 4
20
SUMMARY OF ACID AND BASE STRENGTHS
· In order to write net ionic equations for acidbase
(neutralization reactions), a knowledge of the strength
of acids and bases is essential.
I. ACIDS
1. Strong Acids
· are Strong Electrolytes
· are Acids that are completely ionized (100%) in aqueous solution and produce H 3O + (H + )
ions and an anion.
· are molecular substances in pure form Ex: HCl(g)
Example: A solution of 0.10 M HCl(aq)
HCl (aq) H + (aq) + Cl - (aq)
0 M 0.1 M 0.1 M
In General: HA(aq) H + (aq) + A - (aq)
2. Weak Acids
§ are Weak Electrolytes
§ are Acids that are partially ionized (less than 100%) in aqueous solution
§ molecular substances in pure form Ex: HC2H3O2(l)
Example: A solution of 0.10 M HC2H3O2(aq)
HC2H3O2 (aq) H + (aq) + C2H3O2
- (aq)
» 0.099 M » 0.001 M » 0.001 M
In General: HA(aq) H + (aq) + A - (aq)
Chemistry 101 Chapter 4
21
II. BASES
1. Strong Bases
§ are Strong Electrolytes
§ are Bases that are completely dissociated (100%) in aqueous solution and produce a
metallic cation and OH - ions
§ are ionic substances in pure form(all soluble metallic hydroxides) Ex: Na + OH - (s)
K + OH - (s)
Example: A solution of 0.10 M NaOH(aq)
NaOH(aq) Na + (aq) + OH - (aq)
0 M 0.1 M 0.1 M
Example: A solution of 0.10 M Ba(OH)2(aq)
Ba(OH)2(aq) Ba 2+ (aq) + 2OH - (aq)
0 M 0.1 M 0.2 M
In General: BOH(aq) B+(aq) + OH - (aq)
2. Weak Bases
§ are Weak Electrolytes
§ are Bases that are partially ionized (less than 100%) in aqueous solution
§ molecular substances in pure form Ex: NH3(g)
Example: A solution of 0.10 M NH3(aq)
NH3(aq) + H2O(l) NH4
+ (aq) + OH » 0.099 M » 0.001 M » 0.001 M
In General: B(aq) + H2O(l) BH + (aq) + OH - (aq)
Chemistry 101 Chapter 4
22
COMMON ACIDS AND BASES
Strong Acids
· Completely ionized and written in their ionic forms
HI(aq) ¾¾® H + (aq) + I - (aq)
HBr(aq) ¾¾® H + (aq) + Br - (aq)
HCl(aq) ¾¾® H + (aq) + Cl - (aq)
HNO3(aq) ¾¾®H + (aq) + NO3
- (aq)
H2SO4(aq) ¾¾® 2H + (aq) + SO4
2- (aq)
HClO4(aq) ¾¾® H + (aq) + ClO4
- (aq)
HClO3(aq) ¾¾® H + (aq) + ClO3
- (aq)
Weak Acids
· Partially ionized and written in their molecular forms
H2C2H3O2(aq)
HF(aq)
H2S (aq)
HCN(aq)
Strong Bases
· Completely ionized and written in their ionic forms
LiOH(aq) ¾¾® Li + (aq) + OH - (aq)
NaOH(aq) ¾¾® Na + (aq) + OH - (aq)
KOH(aq) ¾¾® K + (aq) + OH - (aq)
Ba(OH)2(aq) ¾¾® Ba 2+ (aq) + 2OH - (aq)
Sr(OH)2(aq) ¾¾® Sr 2+ (aq) + 2OH - (aq)
Ca(OH)2(aq) ¾¾® Ca 2+ (aq) + 2OH - (aq)
Weak Bases
· Partially ionized and written in their molecular forms
NH3 or NH4OH
CH3NH2
CO(NH2)2
Chemistry 101 Chapter 4
23
WRITING NET IONIC EQUATIONS FOR NEUTRALIZATION REACTIONS
Recall:
neutralization
ACID+ BASE SALT + WATER
· Neutralization Reactions can be classified according to the type of acid and base (strong or
weak) reacting with each other.
(A) Reaction of a Strong Acid with a Strong Base
Hydrochloric acid + Sodium hydroxide Sodium chloride + Water
Molecular Equation:
HCl (aq) + NaOH (aq) NaCl (aq) + H2O (l)
Strong Electrolyte Strong Electrolyte Strong Electrolyte NonElectrolyte
(completely ionized) (completely ionized) (completely ionized) (unionized)
Ions only Ions only Ions only Molecules only
Total Ionic Equation:
H + (aq) + Cl - (aq) + Na + (aq) + OH - (aq) Na + (aq) + Cl - (aq) + H2O(l)
Net Ionic Equation:
is obtained after canceling out the spectator ions from the Total Ionic Equation:
H + (aq) + Cl - (aq)+ Na + (aq) + OH - (aq) Na + (aq) + Cl - (aq) + H2O(l)
spectator ions spectator ions
H + (aq) + OH - (aq) H2O(l)
Example:
Write net ionic equation for the reaction of nitric acid and barium hydroxide.
Chemistry 101 Chapter 4
24
(B) Reaction of a Weak Acid with a Strong Base
Acetic Acid + Sodium hydroxide Sodium Acetate + Water
Molecular Equation:
HC2H3O2(aq) + NaOH(aq) NaC2H3O2(aq) + H2O(l)
Weak Electrolyte Strong Electrolyte Strong Electrolyte NonElectrolyte
(partially ionized) (completely ionized) (completely ionized) (unionized)
Mostly molecules Ions only Ions only Molecules only
Total Ionic Equation:
HC2H3O2(aq) + Na + (aq) + OH - (aq) Na + (aq) + C2H3O2(aq) - (aq) + H2O(l)
HC2H3O2(aq) + Na + (aq) + OH - (aq) Na + (aq) + C2H3O2(aq) - + H2O(l)
spectator ion spectator ion
Net Ionic Equation:
HC2H3O2(aq) + OH - (aq) C2H3O2(aq) - + H2O(l)
Example:
Write net ionic equation for the reaction of hydrosulfuric acid and potassium hydroxide.
Chemistry 101 Chapter 4
25
(C) Reaction of a Strong Acid with a Weak Base
Hydrochloric Acid + Ammonium hydroxide Ammonium Chloride + Water
Molecular Equation:
HCl(aq) + NH4OH(aq) NH4Cl(aq) + H2O(l)
Strong Electrolyte Weak Electrolyte Strong Electrolyte NonElectrolyte
(completely ionized) (partially ionized) (completely ionized) (unionized)
Ions only Mostly molecules Ions only Molecules only
Total Ionic Equation:
H + (aq) + Cl - (aq) + NH4OH(aq) NH4
+ (aq)
+ Cl(aq) - (aq) + H2O(l)
also written: NH3(aq) + H2O(l)
H + (aq) + Cl - (aq) + NH4OH(aq) NH4
+ (aq) + Cl - (aq) - + H2O(l)
(NH3(aq) + H2O(l)
spectator ion spectator ion
Net Ionic Equation:
H + (aq) + NH4OH(aq) NH4
+ (aq) + H2O(l)
OR
H + (aq) + NH3(aq) NH4
+ (aq)
Example:
Write net ionic equation for the reaction of sulfuric acid and ammonia.
Chemistry 101 Chapter 4
26
(D) Reaction of a Weak Acid with a Weak Base
Acetic Acid + Ammonium hydroxide Ammonium acetate + Water
Molecular Equation:
HC2H3O2(aq) + NH4OH(aq) NH4C2H3O2(aq) + H2O(l)
Weak Electrolyte Weak Electrolyte Strong Electrolyte NonElectrolyte
(partially ionized) (partially ionized) (completely ionized) (unionized)
Mostly molecules Mostly molecules Ions only Molecules only
Total Ionic Equation:
HC2H3O2(aq) + NH4OH(aq) NH4
+ (aq) + C2H3O2(aq) - + H2O(l)
also written: NH3(aq) + H2O(l)
NO SPECTATOR IONS ARE PRESENT!
Net Ionic Equation:
HC2H3O2(aq) + NH4OH(aq) NH4
+ (aq) + C2H3O2(aq) - + H2O(l)
Note: this is the same as the Total Ionic Equation
OR
HC2H3O2(aq) + NH3(aq) NH4
+ (aq) + C2H3O2
– (aq)
Example:
Write net ionic equation for the reaction of hydrofluoric acid and ammonia.
Chemistry 101 Chapter 4
27
3. Reactions that form an unstable product
· Some chemical reactions produce gas because one of the products formed in the reaction is
unstable.
· Three such substances that readily decompose are:
H2CO3(aq) CO2(g)+ H2O(l)
(carbonic acid)
H2SO3(aq) SO2(g) + H2O(l)
(sulfurous acid)
NH4OH(aq) NH3(g)+ H2O (l)
ammonium ammonia
hydroxide
· When any of these products appears in a chemical reaction, they should be replaced with their
decomposition products.
Example 1:
Na2CO3(aq) + 2 HCl(aq) NaCl(aq) + H2CO3(aq)
unstable
Na2CO3(aq) + 2 HCl(aq) 2 NaCl(aq) + H2CO3(aq)
CO2(g) H2O(l)
Na2CO3(aq) + 2 HCl(aq) 2 NaCl(aq) + CO2(g) + H2O(l)
Total Ionic Equation:
2Na + (aq) + CO3
2- (aq) + 2H + (aq) + 2Cl - (aq) ® 2Na+(aq) + 2Cl - (aq) + CO2(g) + H2O(l)
Net Ionic Equation:
2Na + (aq) + CO3
2- (aq) + 2H + (aq) + 2Cl - (aq) ® 2Na+(aq) + 2Cl - (aq) + CO2(g) + H2O(l)
CO3
2- (aq) + 2H + (aq) CO2(g) + H2O(l)
Chemistry 101 Chapter 4
28
Example 2:
CaCO3(s) + 2 HNO3(aq) Ca(NO3)2(aq) + H2CO3(aq)
unstable
CaCO3(s) + 2 HNO3(aq) Ca(NO3)2(aq) + H2CO3(aq)
CO2(g) H2O(l)
CaCO3(s) + 2 HNO3(aq) Ca(NO3)2(aq) + CO2(g) + H2O(l)
Total Ionic Equation:
CaCO3(s) + 2H + (aq)+ 2NO3
- (aq) Ca 2+ (aq) + 2NO3
- (aq) + CO2(g) + H2O(l)
Net Ionic Equation:
CaCO3(s) + 2H + (aq) + 2NO3
- (aq) Ca 2+ (aq) + 2NO3
- (aq) + CO2(g) + H2O(l)
CaCO3(s) + 2H + (aq) Ca 2+ (aq) + CO2(g) + H2O(l)
Example 3:
Write a balanced net ionic equation for the reaction of sodium sulfite and hydrobromic acid.Chemistry
101 Chapter 4
30
SOLUTIONS
· Solutions are homogeneous mixture of two or more substances: (solute/s) dispersed throughout
another substance (solvent)
SOLUTION = SOLUTE(S) + SOLVENT
(homogeneous mixture) substance being substance doing
dissolved the dissolving
· AQUEOUS solutions are solutions in which the solvent is water
Examples:
Aqueous Solution = Solute + Water
Vinegar = Acetic Acid + Water
Salt Water = Salt + Water
Soda Water = CO2 + Water
Rubbing Alcohol = Isopropyl Alcohol + Water
Wine = Ethyl Alcohol + Water
(+ natural flavors
from grapes)
Chemistry 101 Chapter 4
31
CONCENTRATION OF SOLUTIONS
· Concentration expresses the relative amount of solute dissolved in a given amount of solution.
· Concentration may be expressed:
I. Qualitatively (no precise quantities are given)
· Solutions may be:
Concentrated Dilute
A relatively large amount of solute is
dissolved in a given amount of solution
A relatively small amount of solute is
dissolved in a given amount of solution
· Concentrated and Dilute are Relative Terms
II. Quantitatively
The ratio between a given amount of solute and a given amount of solution is given.
1. Mass Percent Solution (g solute/g solution)
2. Mass/Volume Percent (grams solute/mL solution)
3. Volume Percent (mL solute/mL solution)
4. Molar Concentration or Molarity (M)
Most common and useful in the Chemistry Laboratory
· Molarity is defined as moles of solute dissolved in one liter of solution
moles of solute moles
Molarity = M = ¾¾¾¾¾¾¾= ¾¾¾
L of solution L
Chemistry 101 Chapter 4
32
Examples:
1. You work in a lab and your job is to prepare 250.0 mL of 0.2000 M solution of copper (II) sulfate
pentahydrate (CuSO4 . 5 H2O). How many grams of CuSO4 . 5 H2O are needed?
0.2000 moles CuSO4.5H2O ? g
? g CuSO4.5 H2O = 250.0 mL solution¾¾¾¾¾¾¾¾¾¾¾¾x¾¾¾¾¾¾¾¾¾
1000 mL solution 1 mole CuSO4.5H2O
The mass of 1 mole of CuSO4.5H2O must be calculated:
1 Cu = 1 x 63.5 = 63.5
1 S = 1 x 32.1 = 32.1
4 O = 4 x 16.0 = 64.0
10 H = 10 x 1.0 = 10.0
5 O = 5 x 16.0 = 80.0
Mass of 1 mole = 249.6 g/mole
0.2000 moles CuSO4.5H2O 249.6 g
? g CuSO4.5 H2O = 250.0 mL solution¾¾¾¾¾¾¾¾¾¾¾¾x¾¾¾¾¾¾¾¾¾
1000 mL solution 1 mole CuSO4.5H2O
= 12.48 g CuSO4 . 5 H2O needed
250mL
mark
Chemistry 101 Chapter 4
33
Examples:
2. Calculate the number of moles of NaOH in 27.40 ml of 0.08543 M NaOH solution.
1 L NaOH solution 0.08543 moles NaOH
? moles NaOH = 27.40 ml solution x¾¾¾¾¾¾¾¾¾ x ¾¾¾¾¾¾¾¾¾
1000 mL solution 1 L NaOH solution
= 0.002341 moles NaOH
Examples:
3. Determine the molarity of a solution prepared by dissolving 32.0 g of NaOH in 185 mL of solution.
4. How many mL of 0.150 M AgNO3 solution contains 3.25 g of solute?
5. How many grams of solute are present in 225 mL of a 3.5% NaCl solution?
Conversion Between Molarity and Mass %
The molarity of a particular brand of vinegar (solution of acetic acid, HC2H3O2, in water) is 0.8527 M.
The
density of vinegar is 1.0052 g/mL. Calculate the mass percent of HC2H3O2 in vinegar.
g HC2H3O2 0.8527 moles HC2H3O2 60.06 g HC2H3O2 1 L vinegar 1 mL vinegar
? ¾¾¾¾ x 100 =¾¾¾¾¾¾¾¾¾ x ¾¾¾¾¾¾¾ x ¾¾¾¾¾ x ¾¾¾¾¾¾ x 100
g vinegar 1 L vinegar 1 mole HC2H3O2 1000 mL vinegar 1.0052g vinegar
= 5.095 % g HC2H3O2 / g vinegar
Chemistry 101 Chapter 4
34
DILUTING SOLUTIONS
· Suppose you are making orange juice from frozen concentrate:
++
1 can of frozen 2 cans of water 3 cans of orange juice
concentrate
NOTE: The diluted Orange Juice: has
three times the volume of the concentrate (3x)
one/
third the concentration of the concentrate (1/3)
Meaning: Volume and Concentration are inversely proportional
Chemistry 101 Chapter 4
35
· Suppose you want to prepare 100 mL, 3 M CuSO4 from 6 M CuSO4.
a add water
100 mL mark
100 mL
??? mL 6 M
12M
???? mL 100 mL
6M3M
Concentrated solution dilute solution
NOTE: The
concentration is halved (from 6 M to 3 M)
The
volume must have been doubled
Concentrated Solution Dilute Solution
Mc = 6 M Md = 3 M
Vc = ???? Vd = 100 mL
· Recall: Volumes and concentrations (Molarities are inversely proportional)
Mc Vd Md x Vd (3M) (100 mL)
¾¾ = ¾¾ or by cross multiplying: Mc x Vc = Md x Vd Vc = ¾¾¾¾ = ¾¾¾¾¾¾
Md Vc Mc 6 M
Vc = 50 mL
Chemistry 101 Chapter 4
36
General Dilution Formula
Mc x Vc = Md x Vd OR Mf x Vf = Mi x Vi OR M1 x V1 = M2 x V2
concentrated final Solution 1
dilute initial Solution 2
Examples:
1. 25.00 mL of a vinegar solution was diluted to 250.0 mL. The concentration of the diluted vinegar
solution was determined to be
0.08527 M. What was the concentration of the original vinegar ?
Conc’d Solution Dilute Solution Note: The
volume increased 10 times
Vc = 25.00 mL Vd = 250.0 mL (10fold
dilution)
Mc = ?????? Md = 0.08527 M The
concentration must have
decreased 10 times
(Mc = 0.8527 M)
Mathematically: Mc x Vc = Md x Vd
Md x Vd (0.08527 M) (250.0 mL)
Mc = ¾¾¾¾ = ¾¾¾¾¾¾¾¾¾¾ = 0.8527 M
Vc 25.00 mL
Chemistry 101 Chapter 4
37
2. 1.00 mL of a solution of 6.00 x 10 -4 M ferric chloride is diluted to 15.00 mL by addition of water.
What
is the concentration of the diluted solution ?
Concentrated Solution Dilute Solution
V1 = 1.00 mL V2 = 15.00 mL
M1 = 6.00 x 10 -4 M M2 = ???
M1 x V1 = M2 x V2
11
2
2
M = M V = (6.00x10
V
4 M)(1.00 mL) = 4.00x10
15.00 mL
5
3. What volume of 0.73M solution must be used to prepare 1.36 L of a 0.20M solution?
Concentrated Solution Dilute Solution
V1 = ??? V2 = 1.36 L
M1 = 0.73 M M2 = 0.20 M
M1 x V1 = M2 x V2
22
1
1
V = M V = (0.20 M)(1.36 L) = 0.37 L
M 0.73 M
4. How much water must be added to 60.0 mL of 0.150M solution of HCl to prepare a 0.100 M
solution?
Chemistry 101 Chapter 4
38
STOICHIOMETRY OF AQUEOUS SOLUTIONS
(VOLUMETRIC ANALYSIS)
· Stoichiometry is the calculation of quantities of reactants and products in a chemical reaction.
· Stoichiometry is based on mole ratio between amounts of substances in a balanced chemical
reaction.
grams grams
given moles/g g/moles calculated
MOLE
RATIO
moles given moles calculated
· In aqueous solutions, the amounts of substances are commonly given based on volume and
molarity.
· As a result, mass measurements are replaced with volume measurements.
Volume moles MOLE L Volume
given L RATIO moles calculated
molarity molarity
is used is used
for this for this
conversion conversion
Chemistry 101 Chapter 4
39
Examples:
1. When aqueous solutions of Na 2 SO 4 and Pb(NO 3 ) 2 are mixed, PbSO 4 precipitates. What
mass of
PbSO 4 is formed when 1.25 L of 0.0500 M Pb(NO 3 ) 2 and 2.00 L of 0.0250 M Na 2 SO 4 are
mixed?
Solution Plan:
· Write a balanced equation.
· Calculate moles of each reactant from volume and concentration.
· Calculate moles and mass of product using molar ratios and molar mass.
· Write a balanced equation:
Na2SO4 (aq) + Pb(NO3)2 (aq) ®
· Calculate moles of each reactant from volume and concentration:
Moles Na2SO4 =
Moles Pb(NO3)2 =
· Calculate moles and mass of product formed using molar rations and molar mass:
Chemistry 101 Chapter 4
40
Examples:
2. A beaker contains 35.0 mL of 0.175 M H2SO4. How many milliliters of 0.250 M NaOH must be
added
to completely neutralize the sulfuric acid?
Solution Plan:
· Write a balanced equation.
· Calculate moles of acid from volume and concentration.
· Calculate moles of base using molar ratios.
· Calculate volume of base using moles and concentration.
· Write a balanced equation:
H2SO4(aq) + 2 NaOH(aq) ® Na2SO4(aq) + 2 H2O(l)
· Calculate moles of acid from volume and concentration:
Moles H2SO4 =
· Calculate moles of base using molar ratios:
Moles NaOH =
· Calculate volume of base using moles and concentration
· Alternate solution:
Chemistry 101 Chapter 4
41
TITRATION
· Titration is a laboratory procedure that uses the reaction between two substances to determine the
concentration of one
substance.
· Titration is based on the balanced chemical equation that represents the reaction.
aA + bB
VA (known volume) VB (known volume)
MA (known molarity) MB (unknown molarity)
a
The MOLE RATIO: ¾ must be known
b
ACID – BASE TITRATION
· Uses the neutralization reaction between an acid and a base to determine the concentration of the
acid or the base in a solution.
Example: Find the concentration of an aqueous solution of HC2H3O2(aq), acetic acid.
Available: An aqueous solution of NaOH(aq) of known molarity (0.07776 M) (this solution is referred
to as the TITRANT)
Procedure: 1. An exact volume of acetic acid, HC2H3O2(aq) (for example 20.00 mL) is measured into
an Erlenmeyer flask.
2. Phenolphtalein (indicator) is added. There is no color change (colorless)
3. NaOH(aq) is added dropwise
from a buret until the solution in the Erlenmeyer flask just turns faint pink
(experimental end point).
4. The volume of the NaOH needed to reach the end point is accurately recorded.
Chemistry 101 Chapter 4
42
buret reading
28.35 mL
buret reading
NaOH(aq) 46.44 mL 46.50 mL
0.07776 M
20.00 mL HC2H3O2(aq) NaOH was added to the solution Common
error:
+ in the flask until a faint pink The addition of several drops
23
drops phenolphtalein color was reached, marking the of NaOH
solution beyond the
experimental end point of the end point gives a deep pink
titration. color
(NaOH is in very slight excess) (NaOH is in great excess)
Volume of NaOH added = 46.44 mL – 28.35 ml = 18.09 mL (Buret
is read in reverse)
Chemistry 101 Chapter 4
43
Calculations:
1 HC2H3O2(aq) + 1 NaOH(aq) NaC2H3O2(aq) + H2O(l)
20.00 mL 18.09 mL
?????? M 0.07776 M
0.07776 moles
0.01809 L x¾¾¾¾¾¾
1L
1.407 x 10 -3 moles NaOH
1.407 x 10 -3 moles NaOH reacts exactly with 1.407 x 10 -3 moles
HC2H3O2 (1:1 Mole Ratio)
moles of HC2H3O2 1.407 x 10 -3 moles
Molarity of HC2H3O2 = ¾¾¾¾¾¾¾¾ =¾¾¾¾¾¾¾¾ = 0.07035
M
L of HC2H3O2 0.02000 LChemistry 101 Chapter 4
43
OXIDATIONREDUCTION
REACTIONS
· Oxidationreduction
(redox) reactions are reactions that involve transfer of electrons from one
species to another.
· As an example:
Fe (s) + CuSO4 (aq) ® FeSO4 (aq) + Cu (s)
Net ionic equation:
Fe (s) + Cu 2+ (aq) ® Fe 2+ (aq) + Cu (s)
· In this reaction, the electrons are transferred from iron atoms to copper ions to form iron ions and
copper atoms.
· To better understand oxidationreduction
reactions, the concept of oxidation numbers was
developed as a simple way of keeping track of electrons in a reaction.
· Using oxidation numbers, one can determine if electrons have been transferred between
substances during a reaction.
· If transfer of electrons has occurred during the reaction, an oxidationreduction
reaction has
occurred.
Chemistry 101 Chapter 4
44
OXIDATION NUMBERS
· The oxidation number of an atom is the number of electrons lost, gained or unequally shared by an
atom. Oxidation numbers can be zero, positive or negative.
· Oxidation number of a monatomic ion is equal to the charge of the ion. For example:
NaCl
Na + = sodium has lost an electron O.N. = +1
Cl – = chloride has gained an electron O.N. = –1
· In covalent compounds, the oxidation numbers are based on relative electronegativities.
· For nonpolar
covalent compounds each atom is assigned an oxidation number of 0 because they
have the same electronegativity. For example:
H¾H Cl¾Cl
EN 2.1 2.1 3.0 3.0
O.N. 0 0 0 0
· For polar covalent compounds, the element with the greater electronegativity is assigned the
negative oxidation number, and the element with the lower electronegativity is assigned the positive
oxidation number. For example:
H¾Cl
EN 2.1 3.0
O.N. +1 –1
· Many elements have multiple oxidation numbers. For example:
N2 N2O NO N2O3 NO2 N2O5 NO3
–
O.N.. of N 0 +1 +2 +3 +4 +5 +5
· The oxidation numbers for elements must be assigned based on a set of rules outlined next.
Chemistry 101 Chapter 4
45
RULES FOR ASSIGNING OXIDATION NUMBERS
1. All elements in their free states have oxidation number of zero.
000000
Cu Mg Ag H2 O2 N2
2. For monatomic ions, O.N. equals the charge of the ion.
MgS NaCl MgO AlCl3 Na2O SnO2
+2 –2 +1 –1 +2 –2 +3 –1 +1 –2 +4 2
MgS NaCl MgO AlCl3 Na2O SnO2
3. Oxygen is -2, except in peroxides, where it is -1 and OF2 , where it is +2.
2
2
1
1
H2O CO2 Exceptions: Peroxides: H2O2 Na2O2
4. In Molecular Compounds, the negative ON is assigned to the atom that is more to the right and
higher
up on the Periodic Table
SiC NH3 CO Negative O.N.
+4 4
3
+1 +2 2
SiC NH3 CO
5. Hydrogen is +1, except when combined with metals, where it is -1.
+1 +1 +1 +1
HCl H2O H2SO4 HNO3
+1 –1 +2 –1 +3 1
Exceptions: Metallic hydrides: NaH CaH2 AlH3
(Ionic Compounds)
Chemistry 101 Chapter 4
46
6. The algebraic sum of the oxidation number of elements in a compound is equal to zero.
+1 ? –2 +1? –2 +1? –2 +1? –2
H2SO4 H2SO3 HNO3 HNO2
+2 (?) – 8 = 0 +2 (?) – 6 = 0 +1 (?) – 6 = 0 +1 (?) – 4 = 0
O.N. of S = +6 O.N. of S = +4 O.N. of N = +5 O.N. of N = +3
7. The algebraic sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.
2
?2
CO3
-2 or CO3 ? – 6 = 2
O.N. of C is +4
3
?2
PO4
-3 or PO4 ? – 8 = 3
O.N. of P is ____
2
?2
Cr2O7
-2 or Cr2O7 ? – 14 = 2
O.N. of Cr is ____
Examples:
1. Calculate O.N. of Cl in the following Compounds:
HCl HClO HClO2 HClO3 HClO4
+1 ? +1 ? 2
+1? 2
+1 ? 2
+1 ? 2
HCl HClO HClO2 HClO3 HClO4
1
+1 ____ ____ ____
OXIDATION NUMBER INCREASES
2. Calculate O.N. of Mn in KMnO4
+1 ? 2
+1 ( ? ) – 8 = 0 O.N. of Mn = ______
KMnO4
Chemistry 101 Chapter 4
47
OXIDATIONREDUCTION
REACTIONS
(Also referred to as Redox Reactions)
· Redox reactions are rxns in which the O.N.’s of at least 2 elements change
2e
(oxidation)
0 +2
Mg + S ¾¾¾>MgS
0 +2e (reduction) 2
In General
An element that is oxidized: (Mg) An element that is reduced: (S)
· loses electrons
· its O.N. increases
· is the Reducing Agent
· gains electrons
· its O.N. is reduced
· is the Oxidizing Agent
OXIDATION (LOSS OF ELECTRONS)
6
5
4
3
2
1
0 +1 +2 +3 +4 +5 +6 +7
REDUCTION (GAIN OF ELECTRONS)
Chemistry 101 Chapter 4
48
Example:
–2e (is oxidized)
Copper 0 +2
Cu(s) + 2AgNO3(aq) ¾> Cu(NO3)2(aq) + 2Ag(s)
(clear) (green)
Silver crystals +1 0
(+1e)x2 = +2e (is reduced)
Aqueous solution
of silver nitrate
Total Ionic Equation:
2e
Cu 0 (s) + 2Ag(aq) + + 2NO3
- (aq) Cu 2+ (aq) + 2NO3
- (aq) + 2Ag 0 (s)
Net Ionic Equation:
2e
Cu 0 (s) + 2Ag(aq) + Cu 2+ (aq) + + 2Ag 0 (s)
· Redox Reactions are discussed (sometimes balanced) by writing two Half – Reactions:
Chemistry 101 Chapter 4
49
OXIDATION HALFREACTION
REDUCTION HALFREACTION
involves
loss of electrons involves
gain of electrons
increase
in Oxidation Number decrease
in Oxidation Number
Cu(s) Cu 2+ (aq) + 2 e – 2Ag + (aq) + 2e – 2Ag 0 (s)
NOTE:
Number of electrons lost in the Number of electrons gained in the
Oxidation Reaction = Reduction Reaction
CONCLUSION:
Oxidation Reduction
(Redox) Reactions
· are reactions in which the Oxidation Numbers of at least two elements change
· involve transfer of electrons:
Ø from : the element that is oxidized (called Reducing Agent)
Ø to: the element that is reduced (called Oxidizing Agent)
Chemistry 101 Chapter 4
50
CLASSIFICATION OF SOME COMMON REDOX REACTIONS
1. Combination Reactions
(Element 1 + Element 2 Compound)
Examples:
(a) Zn(s) + S(s) ZnS(s)
0 2e
(oxidation) +2
Zn(s) + S(s) ZnS(s)
0 +2e(reduction) –2
2e
Zn(s) + S(s) ZnS(s)
Oxidation HalfReaction
Reduction HalfReaction
Zn 0 ® Zn 2+ + 2 e – S 0 + 2e – ® S 2Zn is oxidized S is reduced
Zn is the Reducing Agent S is the Oxidizing Agent
Chemistry 101 Chapter 4
51
(b) Al(s) + I2(s) AlI3(s)
0 3e
(oxidation) +3
Al(s) + I2(s) AlI3(s)
0 2x (+1e) = + 2e (reduction) –1
NOTE: Number of electrons lost and gained must be equal:
0 2 x ( 3e
) = 6e
+3
2Al(s) + 3I2(s) 2 AlI3(s)
0 3 x (+ 2e) = + 6e –1
Balancing can also be done by using the halfReactions:
Oxidation HalfReaction
Reduction HalfReaction
Al ® Al 3+ + 3 e – I2 + 2e – ® 2I NOTE: Number of electrons lost and gained must be equal:
2 Al ® 2 Al 3+ + 6 e – 3 I2 + 6 e – ® 6 I Balanced Redox Reaction:
6e
2Al(s) + 3I2 2AlI3 (s)
Chemistry 101 Chapter 4
52
2. Decomposition Reactions
(Compound Element 1 + Element 2)
Examples:
(A) Red mercury(II) oxide decomposes when heated and produces silvery droplets of liquid mercury
and
oxygen gas.
Mercury
Mercury(II) oxide
heating
HgO (s) Hg (l) + O2(g)
2x (+2e) = + 4e
+2 0
2HgO(s) 2Hg(l) + O2(g)
2
0
2x (2e)
= 4e
4e
2 HgO(s) 2Hg(l) + O2(g)
Chemistry 101 Chapter 4
53
(B) Electrolysis of water: 2 H2O(l) 2 H2(g) + O2(g)
4 x (+1e) = +4e
+1 0
2H2O(l) 2H2(g) + O2(g)
2
0
2 x (2e)
= 4e
4e
2 H2O(l) 2H2(g) + O2(g)
reduced oxidized
(oxidizing (reducing
agent) agent)
3. Single Replacement Reactions
electron(s)
A + B - (aq) + C 0 (s) C + B - (aq) + A 0 (s)
ionic free ionic free
compound metal compound metal
Oxidation HalfReaction
Reduction HalfReaction
C 0 (s) ® C + (aq) + e – A + (aq) + e – ® A(s)
C is oxidized A + is reduced
C is the Reducing Agent A + is the Oxidizing Agent
Since metal C replaces A + from its compound:
Ø C is more active than A
Ø C loses electrons easier than A
Ø C is a stronger reducing agent than A
Chemistry 101 Chapter 4
54
Example 1
A copper strip dipped into a silver nitrate solution becomes coated with crystals of silver, while the
solution
turns greenishblue,
due to the presence of the copper(II) ion.
Cu(s) + 2AgNO3(aq) Cu(NO3)2(aq) + 2Ag(s)
copper strip
copper(II) nitrate solution
silver crystals
2e
(oxidation)
0 +2
Cu(s) + 2AgNO3(aq) ¾> Cu(NO3)2(aq) + 2 Ag(s)
+1 0
2(+1e) = +2e (reduction)
2e
N.I.E. Cu(s) + 2Ag + (aq) ® Cu +2 (aq) + 2Ag(s)
Oxidation Reduction
Cu® Cu 2+ + 2 e – 2Ag +1 + 2e – ¾> 2Ag
Since Cu replaces Ag + from its compound:
Ø Cu is more active than Ag
Ø Cu loses electrons easier than Ag
Ø Cu is a stronger reducing agent than Ag
It follows: Ag(s) + Cu(NO3)2(aq) ¾> No Reaction
Since Ag does not replace Cu 2+ from its compound:
Ø Ag is less active than Cu
Ø Ag does not lose electrons as easily as Cu
Ø Ag is a weaker reducing agent than Cu
Chemistry 101 Chapter 4
55
Example 2:
2e
(oxidation)
0 +2
Zn(s) + CuSO4(aq) ¾> ZnSO4(aq) + Cu(s)
+2 0
+2e (reduction)
2e
N.I.E. Zn(s) + Cu +2 (aq) ¾> Zn +2 (aq) + Cu(s)
Oxidation Reduction
Zn® Zn 2+ + 2 e – Cu +2 + 2e – ¾> Cu 0
Zn is oxidized Cu 2+ is reduced
Zn is the Reducing Agent Cu 2+ is the Oxidizing Agent
Since Zn replaces Cu 2+ from its compound:
Ø Zn is more active than Cu
Ø Zn loses electrons easier than Cu
Ø Zn is a stronger reducing agent than Cu
It follows: Cu(s) + ZnSO4(aq) ¾> No Reaction
Since Cu does not replace Zn 2+ from its compound:
Ø Cu is less active than Zn
Ø Cu does not lose electrons as easily as Zn
Ø Cu is a weaker reducing agent than Zn
Chemistry 101 Chapter 4
56
Example 3:
Zn(s) + H2SO4(aq) ¾> ZnSO4(aq) + H2(g)
Zinc strip
Sulfuric acid
Bubbles of hydrogen on zinc strip
2e
(oxidation)
0 +2
Zn(s) + H2SO4(aq) ¾> ZnSO4(aq) + H2(g)
+1 0
(+1e)x2 = +2e (reduction)
2e
N.I.E. Zn 0 (s) + 2H + (aq) ¾> Zn +2 (aq) + H2
0 (g)
Oxidation Reduction
Zn – 2e ¾> Zn 2+ 2H + + 2e ¾> H2
0 (g)
Zn is oxidized H + is reduced
Zn is the Reducing Agent H + is the Oxidizing Agent
Since Zn replaces H + from its compound:
Ø Zn is more active than H2
Ø Zn loses electrons easier than H2
Ø Zn is a stronger reducing agent than H2
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Summing up:
The elements studied above may be ranked according to their activity:
Example 1: Cu is more active than Ag Cu > Ag
Example 2: Zn is more active than Cu Zn > Cu
Example 3: Zn is more active than H2 Zn > H2
Zn > H2 > Cu > Ag
ease of oxidation increases
metallic activity increases
¬¾¾¾¾¾¾¾
· Such a listing is called “An Activity Series”
· A more complete “Activity Series” includes the majority of common metals:
Ease K Very Active: React with water to produce H2
of Ba Ex:
Oxidation Ca Na(s) + H2O(l) ¾> NaOH(aq) + H2(g)
increases Na
Mg Moderately Active: React with nonoxidizing acids to produce H2 gas
Al
Zn Ex:
Cr Zn(s) + H2SO4(aq) ¾> ZnSO4(aq) + H2(g)
Fe
Ni
Sn
Pb
H2
Cu Relatively inactive:
Ag Do not react with nonoxidizing acids to produce hydrogen gas.
Hg
Au
NOTE: A free metal can displace the ion of a second metal from solution if the free metal is
above the second metal in the activity series.
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4. Combustion Reactions
· Combustion reactions are those reactions in which a substance reacts with oxygen , with the rapid
release of heat to produce a flame.
Example 1:
Iron wool burns in air (reacts with oxygen) to produce iron (III) oxide.
4 Fe (s) + 3O2 (g) 2Fe2O3 (s)
4 x (3e)
= 12
e (oxidation)
0 +3
4 Fe(s) + 3O2(g) 2Fe2O3(s)
02–
6 x (+ 2e) = + 12 e (reduction)
12 e
4 Fe(s) + 3O2(g) 2Fe2O3(s)
Is oxidized Is reduced
Is the Reducing Agent is the Oxidizing Agent
NOTE: The above reaction can also be classified as a Combination Reaction.
Example 2:
Methane gas (CH4) burns in air (reacts with oxygen) to produce carbon dioxide and water vapor.
CH4(g) + 2O2(g) CO2(g) + 2H2O(g)
8e
(oxidation)
44+
CH4(g) + 2O2(g) CO2(g) + 2H2O(g)
0 24
x (+2e) = + 8e (reduction)
8e
CH4(g) + 2O2(g) CO2(g) + 2H2O(g)
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SUMMARY OF OXIDATION REDUCTION REACTIONS
· Redox reactions can be identified by checking the Oxidation Numbers (O.N.) of all the elements
involved.
· If the O.N.’s of at least 2 elements change, the reaction is an Oxidation – Reduction (Redox)
Reaction.
· In redox reactions an exchange of electrons occurs between the reactants
· Redox reactions can be classified as:
1. Combination Reactions: A 0 + B 0 AB
2. Decomposition Reactions: CD C 0 + D 0
3. Single Replacement Reactions:
A + B - (aq) + C 0 (s) C + B - (aq) + A 0 (s)
4. Combustion Reactions: Burning in the presence of oxygen gas
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SUMMARY OF TYPES OF CHEMICAL REACTION
I. DOUBLE DISPLACEMENT (METATHESIS) RXNS.
· The ions of the reactants are exchanged
· Oxidation Numbers of the elements do not change
(Charges of ions do not change from Reactant to Product side)
A + B - (aq) + C + D - (aq) AD(?) + CB(?)
DoubleDisplacement
Reactions can be further classified into:
1. Precipitation Reactions
2. Acid – Base Reactions
3. Reactions that form an unstable product
II. REDOX REACTIONS (Oxidation – Reduction Reactions)
· An exchange of electrons occurs between the reactants
· Oxidaton Numbers of at least 2 elements change
Redox Reactions can be further classified into:
1. Combination Reactions
2. Decomposition Reactions
3. Single Replacement Reactions
4. Combustion Reactions