Chapter 4
The Major Classes of
Chemical Reactions
4-1
Dr. Wolf’s CHM 101
The Major Classes of Chemical Reactions
4.1 The Role of Water as a Solvent
4.2 Writing Equations for Aqueous Ionic Reactions
4.3 Precipitation Reactions
4.4 Acid-Base Reactions
4.5 Oxidation-Reduction (Redox) Reactions
4.6 Elemental Substances in Redox Reactions
4.7 Reversible Reactions: An Introduction to Chemical
Equilibrium
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Dr. Wolf’s CHM 101
Water As a Solvent
Water will dissolve ionic compounds. The charged ions
become separated from each other and become
surrounded by water molecules.
One demonstration of this is the conduction of
electricity through an ionic solution.
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Dr. Wolf’s CHM 101
The electrical conductivity of ionic solutions
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Dr. Wolf’s CHM 101
Sample Problem 4.1
PROBLEM:
Determining Moles of Ions in Aqueous Ionic
Solutions
How many moles of each ion are in the following solutions?
(a) 5.0 mol of ammonium sulfate dissolved in water
(b) 78.5g of cesium bromide dissolved in water
(c) 7.42x1022 formula units of copper(II) nitrate dissolved in water
(d) 35mL of 0.84M zinc chloride
PLAN: We have to relate the information given and the number of moles of
ions present when the substance dissolves in water.
SOLUTION: (a) (NH4)2SO4(s)
2NH4+(aq) + SO42-(aq)
2mol NH4+
= 10.mol NH4+
5.0mol (NH4)2SO4
1mol (NH4)2SO4
5.0mol SO42-
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Dr. Wolf’s CHM 101
Sample Problem 4.1
Determining Moles of Ions in Aqueous Ionic
Solutions
continued
Cs+(aq) + Br-(aq)
(b) CsBr(s)
mol CsBr
78.5g CsBr
= 0.369mol CsBr
212.8g CsBr
(c) Cu(NO3)2(s)
= 0.369mol Cs+
= 0.369mol Br1-
Cu2+(aq) + 2NO3-(aq)
= 0.123mol Cu2+
mol Cu(NO3)2
7.42x1022 formula
= 0.123mol Cu(NO3)2
units Cu(NO3)2 6.022x1023 formula units
= 0.246mol NO
(d) ZnCl2(aq)
35mL ZnCl2
Zn2+(aq) + 2Cl-(aq)
1L
0.84mol ZnCl2
103mL
L
= 2.9x110-2mol Zn2+
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Dr. Wolf’s CHM 101
= 2.9x110-2mol ZnCl2
= 5.8x110-2mol Cl-
3
1-
Water is a Polar Solvent
The electron distribution in a water molecule is such that
an excess of electron density is on the oxygen atom and
an electron deficiency is on the hydrogen atoms.
The polarity of water is what allows it to separate charged
particles in ionic solutions.
It’s also what allows covalent compounds that are polar to
dissolve in water, e.g. sugar. Although these polar
compounds dissolve in water they are non-electrolytes.
They do not conduct electricity since they are not charged.
Water will also dissolve compounds with a polar bond to a H
atom. These compounds, called acids, dissociate to give
protons, H+, and anions.
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Dr. Wolf’s CHM 101
The Polarity of Water
Electron distribution in molecules of H2 and H2O
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The dissolution of an ionic compound
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Acids in water dissociate to give H+. The H+
bonds with a water molecule and becomes a
hydrated proton, H3O+ , a hydronium ion.
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Dr. Wolf’s CHM 101
Sample Problem 4.2
Determining the Molarity of H+ Ions in Aqueous
Solutions of Acids
PROBLEM: Nitric acid is a major chemical in the fertilizer and explosives
industries. In aqueous solution, each molecule dissociates and the
H becomes a solvated H+ ion. What is the molarity of H+(aq) in 1.4M
nitric acid?
PLAN:
Use the formula to find the molarity of H+.
SOLUTION: Nitrate is NO3 .
HNO3(l)
H+(aq) + NO3-(aq)
1.4M HNO3(aq) should have 1.4M H+(aq).
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Dr. Wolf’s CHM 101
Writing Equations for Aqueous Ionic Reactions
Molecular Equation - shows all reactants and products as if they
were undissociated compounds.
Total Ionic Equation - shows any compounds that are soluble ionic
compounds dissociated into ions.
Net Ionic Equation - eliminates spectator ions showing only actual
chemical reaction taking place.
Precipitation Reactions - Reactant ions form an insoluble
product which precipitates from solution.
Acid-Base Reactions - (neutralization) - A reaction
between H+ ions from an acid dissolved in water and OHions from a base dissolved in water. The product is a
molecule of water.
Redox Reactions - (later in chapter)
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Writing equations for aqueous ionic reactions.
A precipitation reaction and its equation.
©
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Table 4.1 Solubility Rules For Ionic Compounds in Water
Soluble Ionic Compounds
1. All common compounds of Group 1A(1) ions (Li+, Na+, K+, etc.) and
ammonium ion (NH4+) are soluble.
2. All common nitrates (NO3-), acetates (CH3COO- or C2H3O2-) and most
perchlorates (ClO4-) are soluble.
3. All common chlorides (Cl-), bromides (Br-) and iodides (I-) are soluble,
except those of Ag+, Pb2+, Cu+, and Hg22+.
4. All common sulfates (SO42-) are soluble,
except those of Ca2+ , Sr2+ , Ba2+ , and Pb2+ .
Insoluble Ionic Compounds
1. All common metal hydroxides are insoluble, except those of Group 1A(1)
and the larger members of Group 2A(2)(beginning with Ca2+).
2. All common carbonates (CO32-) and phosphates (PO43-) are insoluble,
except those of Group 1A(1) and NH4+.
3. All common sulfides are insoluble except those of Group 1A(1), Group
2A(2) and NH4+.
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Dr. Wolf’s CHM 101
Sample Problem 4.3
Predicting Whether a Precipitation Reaction
Occurs; Writing Ionic Equations
PROBLEM: Predict whether a reaction occurs when each of the following pairs of
solutions are mixed. If a reaction does occur, write balanced
molecular, total ionic, and net ionic equations, and identify the
spectator ions.
(a) sodium sulfate(aq) + strontium nitrate(aq)
(b) ammonium perchlorate(aq) + sodium bromide(aq)
PLAN:
write ions
SOLUTION:
(a) Na2SO4(aq) + Sr(NO3)2 (aq)
2NaNO3(aq) + SrSO4(s)
2Na+(aq) +SO42-(aq)+ Sr2+(aq)+2NO3-(aq)
2Na+(aq) +2NO3-(aq)+ SrSO4(s)
combine anions & cations
check for insolubility
Table 4.1
eliminate spectator ions
for net ionic equation
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Dr. Wolf’s CHM 101
SO42-(aq)+ Sr2+(aq)
(b) NH4ClO4(aq) + NaBr (aq)
SrSO4(s)
NH4Br (aq) + NaClO4(aq)
All reactants and products are soluble so no reaction
occurs.
Strong / Weak Acids and Bases
Strong acids and bases dissociate completely into ions in water.
Weak acids and bases only slightly dissociate when dissolved in water.
Acids
Bases
Strong
Strong
hydrochloric acid, HCl
sodium hydroxide, NaOH
hydrobromic acid, HBr
potassium hydroxide, KOH
hydroiodic acid, HI
calcium hydroxide, Ca(OH)2
nitric acid, HNO3
strontium hydroxide, Sr(OH)2
sulfuric acid, H2SO4
barium hydroxide, Ba(OH)2
perchloric acid, HClO4
Weak
hydrofluoric acid, HF
phosphoric acid, H3PO4
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acetic acid, CH3COOH (or
HC2H3O2)
Dr. Wolf’s CHM 101
Weak
ammonia, NH3
Sample Problem 4.4
PROBLEM:
Writing Ionic Equations for Acid-Base Reactions
Write balanced molecular, total ionic, and net ionic equations for
each of the following acid-base reactions and identify the spectator
ions.
(a) strontium hydroxide(aq) + perchloric acid(aq)
(b) barium hydroxide(aq) + sulfuric acid(aq)
PLAN:
SOLUTION:
reactants are strong acids and (a) Sr(OH)2(aq)+2HClO4(aq) 2H2O(l)+Sr(ClO4)2(aq)
bases and therefore completely
Sr2+(aq) + 2OH-(aq)+ 2H++(aq)+ 2ClO4-(aq)
ionized in water
2H2O(l)+Sr2+(aq)+2ClO4-(aq)
products are
2OH-(aq)+ 2H+(aq)
2H2O(l)
water
spectator ions
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(b) Ba(OH)2(aq) + H2SO4(aq)
Ba2+(aq) + 2OH-(aq)+ 2H+(aq)+ SO42-(aq)
2H2O(l)+Ba2+(aq)+SO42-(aq)
2OH-(aq)+ 2H+(aq)
Dr. Wolf’s CHM 101
2H2O(l) + BaSO4(aq)
2H2O(l)
An aqueous strong acid-strong base reaction on
the atomic scale.
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An acid-base titration
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Sample Problem 4.5
PROBLEM:
Finding the Concentration of Acid from an
Acid-Base Titration
You perform an acid-base titration to standardize an HCl solution by
placing 50.00mL of HCl in a flask with a few drops of indicator
solution. You put 0.1524M NaOH into the buret, and the initial
reading is 0.55mL. At the end point, the buret reading is 33.87mL.
What is the concentration of the HCl solution?
PLAN:
volume(L) of base
multiply by M of base
mol of base
SOLUTION:
NaOH(aq) + HCl(aq)
(33.87-0.55)mL x
NaCl(aq) + H2O(l)
1L
= 0.03332L
103mL
molar ratio
0.03332L X 0.1524M
mol of acid
divide by L of acid
Molar ratio is 1:1
5.078x10-3mol HCl
M of acid
0.050L
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Dr. Wolf’s CHM 101
= 5.078x10-3mol
NaOH
= 0.1016M HCl
Writing Equations for Aqueous Ionic Reactions
Molecular Equation - shows all reactants and products as if they
were undissociated compounds.
Total Ionic Equation - shows any compounds that are soluble ionic
compounds dissociated into ions.
Net Ionic Equation - eliminates spectator ions showing only actual
chemical reaction taking place.
Oxidation-Reduction (Redox) Reactions Reactions where there is movement of electrons
from one reactant to another. It can be in the form
of a transfer of electrons from one atom to another
or in the case of many covalent compounds, an
uneven distribution of electron density in the
covalent bond.
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The redox process in compound formation
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To keep track of the transfer of electrons, atoms are
assigned oxidation numbers.
Table 4.3 Rules for Assigning an Oxidation Number (O.N.)
General rules
1. For an atom in its elemental form (Na, O2, Cl2, etc.): O.N. = 0
2. For a monoatomic ion: O.N. = ion charge
3. The sum of O.N. values for the atoms in a compound equals zero. The
sum of O.N. values for the atoms in a polyatomic ion equals the ion’s charge.
Rules for specific atoms or periodic table groups
1. For Group 1A(1):
O.N. = +1 in all compounds
2. For Group 2A(2):
O.N. = +2 in all compounds
3. For hydrogen:
O.N. = +1 in combination with nonmetals
4. For fluorine:
O.N. = -1 in combination with metals and boron
5. For oxygen:
O.N. = -1 in peroxides
O.N. = -2 in all other compounds(except with F)
O.N. = -1 in combination with metals, nonmetals
(except O), and other halogens lower in the group
6. For Group 7A(17):
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Dr. Wolf’s CHM 101
Sample Problem 4.6
PROBLEM:
PLAN:
Determining the Oxidation Number of an Element
Determine the oxidation number (O.N.) of each element in these
compounds:
(a) zinc chloride (b) sulfur trioxide (c) nitric acid
The O.N.s of the ions in a polyatomic ion add up to the charge of the
ion and the O.N.s of the ions in the compound add up to zero.
SOLUTION:
(a) ZnCl2. The O.N. for zinc is +2 and that for chloride is -1.
(b) SO3. Each oxygen is an oxide with an O.N. of -2. Therefore the
O.N. of sulfur must be +6.
(c) HNO3. H has an O.N. of +1 and each oxygen is -2. Therefore
the N must have an O.N. of +5.
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Dr. Wolf’s CHM 101
Highest and lowest
oxidation numbers of
reactive main-group
elements
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A summary of terminology for oxidationreduction (redox) reactions
e-
X
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transfer
Y
or shift of
electrons
X loses electron(s)
Y gains electron(s)
X is oxidized
Y is reduced
X is the reducing agent
Y is the oxidizing agent
X increases its
oxidation number
Y decreases its
oxidation number
Dr. Wolf’s CHM 101
Sample Problem 4.7
Recognizing Oxidizing and Reducing Agents
PROBLEM: Identify the oxidizing agent and reducing agent in each of the following:
(a) 2Al(s) + 3H2SO4(aq)
Al2(SO4)3(aq) + 3H2(g)
(b) PbO(s) + CO(g)
(c) 2H2(g) + O2(g)
Pb(s) + CO2(g)
2H2O(g)
PLAN: Assign an O.N. for each atom and see which gained and which lost
electrons in going from reactants to products.
An increase in O.N. means the species was oxidized (and is the
reducing agent) and a decrease in O.N. means the species was
reduced (is the oxidizing agent).
SOLUTION:
0
+1 +6 -2
(a) 2Al(s) + 3H2SO4(aq)
+3 +6 -2
0
Al2(SO4)3(aq) + 3H2(g)
The O.N. of Al increases; it is oxidized; it is the reducing agent.
The O.N. of H decreases; it is reduced; it is the oxidizing agent.
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Dr. Wolf’s CHM 101
Sample Problem 4.7
continued
+2 -2
Recognizing Oxidizing and Reducing Agents
+2 -2
(b) PbO(s) + CO(g)
0
+4 -2
Pb(s) + CO2(g)
The O.N. of C increases; it is oxidized; it is the reducing agent.
The O.N. of Pb decreases; it is reduced; it is the oxidizing agent.
0
0
(c) 2H2(g) + O2(g)
+1 -2
2H2O(g)
The O.N. of H increases; it is oxidized; it is the reducing agent.
The O.N. of O decreases; it is reduced; it is the oxidizing agent.
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Dr. Wolf’s CHM 101
Sample Problem 4.8
Balancing Redox Equations by the Oxidation
Number Method
PROBLEM: Use the oxidation number method to balance the following equations:
(a) Cu(s) + HNO3(aq)
Cu(NO3)2(aq) + NO2(g) + H2O(l)
(b) PbS(s) + O2(g)
SOLUTION:
0
+1 +5 -2
(a) Cu(s) + HNO3(aq)
PbO(s) + SO2(g)
+2 +5 -2
+4 -2
+1 -2
Cu(NO3)2(aq) + NO2(g) + H2O(l)
O.N. of Cu increases because it loses 2e-; it is oxidized and is the reducing agent.
O.N. of N decreases because it gains 1e-; it is reduced and is the oxidizing agent.
loses 2eCu(s) + HNO3(aq)
balance other ions
Cu(NO3)2(aq) + NO2(g) + H2O(l)
gains 1e-
x2 to balance e-
balance unchanged polyatomic ions
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Dr. Wolf’s CHM 101
Sample Problem 4.8
Balancing Redox Equations by the Oxidation
Number Method
continued
Cu(s) + 4
2 HNO3(aq)
Cu(NO3)2(aq) + 2 NO2(g) + 2 H2O(l)
2
+2 -2
0
+2 -2
(b) PbS(s) + O2(g)
+4 -2
PbO(s) + SO2(g)
loses 6ePbS(s) + 3/2 O2(g)
PbO(s) + SO2(g)
gains 2e- per O; need 3/2 O2 to make 3O2Multiply by 2 to have whole number coefficients.
2PbS(s) + 3O2(g)
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Dr. Wolf’s CHM 101
2PbO(s) + 2SO2(g)
A redox titration
permanganate ion / oxalate ion
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Dr. Wolf’s CHM 101
Sample Problem 4.9
Finding an Unknown Concentration by a Redox
Titration
2+
PROBLEM: Calcium ion (Ca ) is required for blood to clot and for many other
cell processes. An abnormal Ca2+ concentration is indicative of
disease. To measure the Ca2+ concentration, 1.00mL of human
blood was treated with Na2C2O4 solution. The resulting CaC2O4
precipitate was filtered and dissolved in dilute H2SO4. This solution
required 2.05mL of 4.88x10-4M KMnO4 to reach the end point. The
unbalanced equation is
KMnO4(aq) + CaC2O4(s) + H2SO4(aq)
MnSO4(aq) + K2SO4(aq) + CaSO4(s) + CO2(g) + H2O(l)
(a) Calculate the amount (mol) of Ca2+.
(b) Calculate the amount (mol) of Ca2+ ion concentration expressed
in units of mg Ca2+/100mL blood.
PLAN:
(a)
volume of KMnO4 soln
multiply by M
mol of KMnO4
mol of CaC2O4
molar ratio
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Dr. Wolf’s CHM 101
mol of Ca2+
ratio of elements in formula
Sample Problem 4.9
Finding an Unknown Concentration by a Redox
Titration
continued
2.05mL soln
SOLUTION:
L
4.88x10-4mol KMnO4
103 mL
1.00x10-6mol KMnO4
= 1.00x10-6mol KMnO4
L
5mol CaC2O4
= 2.50x10-6 mol CaC2O4
2mol KMnO4
2.50x10-6 mol CaC2O4
1mol Ca2+
= 2.50x10-6 mol Ca2+
1mol CaC2O4
PLAN:
(b)
mol Ca2+/1mL blood
multiply by 100
mol Ca2+/100mL blood
multiply by M
g Ca2+/100mL blood
SOLUTION:
2.50x10-6 mol Ca2+ x100 =2.50x10-4 mol Ca2+
1mL blood
100mL blood
2.50x10-4 mol Ca2+ 40.08g Ca2+ mg
100mL blood
mol Ca2+
10-3g
10-3g = 1mg
=10.0mg Ca2+/100mL blood
mg Ca2+/100mL blood
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Dr. Wolf’s CHM 101
Older Approach to Classify of Chemical Reactions
Combination Reactions - Where two or more reactants
form one product.
Decomposition Reactions - Where one reactant formed
two or more products.
Displacement Reactions - Where atoms or ions of
reactants exchange places in products.
Also - Combustion Reactions - Special redox reactions
with oxygen a reactant which serves as the oxidant.
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Combining elements to form an ionic compound
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Decomposing a compound to its elements
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Displacing one metal with another
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Sample Problem 4.10
PROBLEM:
Identifying the Type of Redox Reaction
Classify each of the following redox reactions as a combination,
decomposition, or displacement reaction, write a balanced
molecular equation for each, as well as total and net ionic
equations for part (c), and identify the oxidizing and reducing
agents:
(a) magnesium(s) + nitrogen(g)
(b) hydrogen peroxide(l)
magnesium nitride (aq)
water(l) + oxygen gas
(c) aluminum(s) + lead(II) nitrate(aq)
aluminum nitrate(aq) + lead(s)
PLAN: Combination reactions produce fewer products than reactants.
Decomposition reactions produce more products than reactants.
Displacement reactions have the same number of products and reactants.
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Sample Problem 4.10
continued
Identifying the Type of Redox Reaction
0
(a) Combination
3Mg(s) +
0
+2 -3
N2(g)
Mg3N2 (aq)
Mg is the reducing agent; N2 is the oxidizing agent.
+1 -1
(b) Decomposition
+1 -2
0
H2O2(l)
H2O(l) + 1/2 O2(g)
2 H2O2(l)
2 H2O(l) +
or
O2(g)
O is the oxidizing and reducing agent.
(c) Displacement
0
+2 +5 -2
Al(s) + Pb(NO3)2(aq)
+3 +5 -2
Al(NO3)3(aq) + Pb(s)
2Al(s) + 3Pb(NO3)2(aq)
Pb is the oxidizing and Al is the reducing agent.
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Dr. Wolf’s CHM 101
0
2Al(NO3)3(aq) + 3Pb(s)
Reversible Reactions
Chemical Equilibrium
For some reactions, which proceed from reactants to
products, can also undergo the reverse reaction of
“products” to “reactants”.
For example:
CaCO3
CaO + CO2
CaO + CO2
and
CaCO3
But both reactions are taking place simultaneously and an
equilibrium is reached. The equation is written:
CaCO3
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Dr. Wolf’s CHM 101
CaO + CO2
End of Chapter 4
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