Unit 2
Types of Chemical Reactions and Solution Stoichiometry
Introduction
Many chemical reactions and virtually all biological processes take place in water. In
this chapter we will discuss three major categories of reactions that occur in aqueous
solutions.
4.1 Water, the Common Solvent
One of the most valuable properties of water is its ability to dissolve many substances.
Water is known as a polar substance, because it has a positive and negative side. The
molecule itself is bent; the oxygen obtains a partial negative charge (-) and the
hydrogens obtain a partial positive charge (+).
EXAMPLE: Draw a representation of a water molecule below.
The shape of a water molecule explains how it can dissolve an ionic substance. The
positive ends of the water molecules are attracted to the negatively charged anions and
the negative ends are attracted to the positively charged cations. This process is called
hydration. The strong forces present among the positive and negative ions of the solid
are replaced by strong water-ion interactions. When ionic substances dissolve in water,
they break up into individual ions. For instance, when NH4NO3 dissolves in water, the
resulting solution contains NH4+ and NO3- ions.
NH4NO3 (s)  NH4+ (aq) + NO3- (aq)
4.2 The Nature of Aqueous Solutions: Strong and Weak Electrolytes
A solution is a homogeneous mixture of two or more substances. The solute is the
substance present in smaller amount, and the solvent is the substance present in larger
amount.
Solutes fit into one of three categories: electrolytes, nonelectrolytes, and weak
electrolytes. Strong electrolytes are substances that conduct electricity in an aqueous
environment and includes all ionic compounds (such as NaCl and KBr), the strong acids
(such as HCl and HNO3) and the Group I metal bases (such as KOH and NaOH).
Nonelectrolytes do not conduct electricity and include many covalent compounds (such
as methanol, CH3OH). Weak electrolytes conduct electricity only slightly and include
weak acids (such as acetic acid, CH3COOH) and weak bases (like ammonia, NH3). The
characteristic of electrolytes that allows them to conduct electricity in water is the fact
that they ionize in an aqueous environment.
EXAMPLE:
Classify each of the following as an electrolyte or nonelectrolyte.
MgCl2
EXAMPLE:
CH2Cl2
HCl
C6H12O6
State what species (molecules or ions) are present when the following are
added to distilled water:
MgCl2 (s) →
C6H12O6(s) →
4.3 The Composition of Solutions
In general, the concept of concentration deals with the ratio of solute to solvent. In this
chapter, we will deal with the molar concentration of solutions, or the molarity, M.
M = molarity = moles of solute
L of solution
EXAMPLE:
What is the molarity of a solution which has 0.730 mol of glucose
dissolved in 500.0 mL of solution?
EXAMPLE:
Standard saline solution used in hospitals is 0.15 M NaCl. Calculate the
number of grams of NaCl present in 125 mL of standard saline solution.
EXAMPLE:
Calculate the molarity of all species present in a 1.0 M MgCl2 solution.
EXAMPLE:
Calculate the molarity of all species present when 250. mL of a 1.20 M
MgCl2 solution is mixed with 500. mL of a 1.50 M NaCl solution. Assume
the volumes are additive.
Dilution of Solutions
Whenever I make, say, 3.0 M HCl, I make it from a concentrated 12 M HCl “stock”
solution. (Of course, I always add the acid to the water.) If we dilute a solution, the
moles of solute remain the same. Therefore, we can write
MiVi = MfVf
Where the “i” and “f” subscripts refer to the initial and final conditions, respectively.
EXAMPLE:
Explain how you could make 250.0 mL of 3.00 M HCl starting from 12.0
M concentrated HCl.
4.4 Types of Chemical Reactions
We will organize reactions into three main types:
 Precipitation Reactions
 Acid-base Reactions
 Oxidation-reduction Reactions
4.5 Precipitation Reactions
One common type of reaction that occurs in aqueous solution is the precipitation reaction,
which results in the formation of an insoluble product, or precipitate. A precipitate is an
insoluble solid that separates from the solution.
AgNO3 (aq) + NaCl (aq)

NaNO3 (aq) + AgCl (s)
In order to predict when a precipitation reaction will occur, we need to know the
solubility rules. You will need to know these by memory.
Solubility Rules






Compounds containing alkali metal ions (Li+, Na+, K+, Rb+, Cs+) and the
ammonium ion (NH4+) are soluble
Nitrates (NO3-), bicarbonates (HCO3-), and chlorates (ClO3-) are soluble
Halides (Cl-, Br-, I-) are soluble. Exceptions: halides of Ag+, Hg2+2, and Pb+2
Sulfates are soluble. Exceptions: sulfates of Hg2+2, Ca+2, Ba+2, and Pb+2.
Carbonates (CO3-2), phosphates (PO4-3), chromates (CrO4-2) and sulfides (S-2) are
insoluble. Exceptions: compounds containing alkali metal ions and the
ammonium ion.
Hydroxides are insoluble. Exceptions: compounds containing alkali metal ions.
The compounds Ba(OH)2 and Ca(OH)2 are right on the border between soluble
and insoluble, and are sometimes referred to are marginally soluble.
(Note: All compounds are at least a little soluble in water, even the compounds that are
called “insoluble”. As a general rule, compounds are considered soluble if one can
dissolve 0.02 moles (or more) in 1.0 L of solution.)
EXAMPLE:
Classify the following as soluble or insoluble in water.
(a) BaSO4
(b) K2CO3
(c) Ag2S
(d) Pb(NO3)2
4.6 Describing Reactions in Solution
A molecular equation has the formulas of the compounds written as if they were whole
units. This type of equation is useful because it identifies the reagents . This type of
equation does not accurately describe what is happening. Since ionic compounds
dissolve and ionize in water, it is really the ions that are reacting. The ionic equation
shows the dissolved species as they really are – as dissolved ions. The net ionic equation
eliminates all spectator ions (ions which play no role in the reaction). For example, if we
consider the reaction of silver nitrate with sodium chloride, the equations become:
molecular:
AgNO3 (aq) + NaCl (aq)

NaNO3 (aq) + AgCl (s)
ionic: Ag+ (aq) + NO3- (aq) + Na+ (aq) + Cl- (aq)  Na+ (aq) + NO3- (aq) + AgCl (s)
net ionic:
Ag+ (aq) + Cl- (aq)  AgCl (s)
EXAMPLE:
Write the molecular, ionic, and net ionic equation for the reaction of
barium chloride with sodium sulfate.
4.7 Stoichiometry of Precipitation Reactions
Stoichiometric calculations can be done with solutions just as they are with massmass
calculations. You can back and forth between moles and volume by using the molarity.
The following example illustrates this point.
EXAMPLE:
Calculate the mass of solid NaCl that must be added to 1.50 L of a 0.100
M AgNO3 solution to precipitate all the Ag+ ions in the form of AgCl.
AgNO3 (aq) + NaCl(s)  AgCl (s) + NaNO3 (aq)
EXAMPLE:
Solid magnesium hydroxide can be reacted with hydrochloric acid to
produce magnesium chloride and water. How many mL of 3.00 M HCl
are required to react with 10.5 grams of solid magnesium hydroxide?
Mg(OH)2 (s) + 2HCl (aq)  MgCl2 (aq) + 2H2O (l)
EXAMPLE:
Suppose 100.0 mL of 2.40 M Pb(NO3)2 (aq) is mixed with 200.0 mL of
1.80 M KCl (aq).
(a) Write and balance the molecular equation.
(b) Calculate the mass of PbCl2 that precipitates.
(c) Calculate the molarity of all ions present in solution after the
reaction takes place.
4.8 Acid-Base Reactions
Acid-base chemistry plays a major role in chemistry. It is central to many industrial
processes and plays an essential role in sustaining biological systems. We begin our
discussion with some general comments on acids and bases.
General Properties of Acids
 The Arrhenius and Bronsted definition of an acid is a substance that ionizes in
water to produce H+. Acids are sometimes called proton donors.
 Acids have a sour taste, such as in vinegar and lemons.
 Acids react with strong bases (such as NaOH and KOH) to produce a salt and
water
HCl (aq) + NaOH (aq)  NaCl(aq) + H2O (l)

Acids react with certain metals (such as Zn, Mg and Fe) to produce hydrogen gas
2 HCl (aq) + Mg (s)  MgCl2 (aq) + H2 (g)

Acids react with carbonates and bicarbonates to produce a salt, water, and carbon
dioxide gas
2HCl (aq) + CaCO3 (s)  CaCl2 (aq) + H2O (l) + CO2 (g)


Aqueous solutions containing acids conduct electricity
The seven strong acids (HCl, HBr, HI, HNO3, HClO3, HClO4, H2SO4) ionize
100% in aqueous solution; all other acids are weak acids and ionize only slightly.
As such, weak acids are essentially unionized in solution.
General Properties of Bases
 The Arrhenius definition of bases defines them as anything that ionizes in water
to release hydroxide, such as NaOH, or Ca(OH)2. The Bronsted definition says
that bases are proton acceptors, such as ammonia. (NH3 + H2O  NH4+ + OH-)
 Bases have a bitter taste
 Bases feel slippery. Many soaps contain bases.
 Aqueous solutions containing bases conduct electricity.
 Strong bases are those with group I and II metals with hydroxide. Weak,
Bronsted bases contain ammonia, or its derivatives (-NH2, -NH- groups). The
weak bases act by accepting a proton (H+) from water or an acid. Also, most
anions, except the anions of strong acids, are weak bases.
In the reaction below, a Bronsted acid (HCl) donates a proton to the Bronsted base (H2O),
forming a hydronium ion (H3O+) in the process.
HCl (aq) + H2O (l)  H3O+ (aq) + Cl- (aq)
Acid-Base Neutralization
A neutralization reaction is a reaction between an acid and a base. Generally, water and a
salt (an ionic compound) are the products.
acid + base  salt + water
The reaction between HCl and NaOH is an example of a neutralization reaction.
HCl (aq) + NaOH (aq)  NaCl (aq) + H2O (l)
EXAMPLE:
Write the ionic and net ionic equation for the reaction of HCl(aq) and
NaOH(aq)
EXAMPLE:
Write the molecular, ionic, and net ionic equation of an aqueous solution
of nitrous acid (a weak acid) reacting with an aqueous solution of
potassium hydroxide.
EXAMPLE:
Write the molecular, ionic and net ionic equation of hydrobromic acid
reacting with methanamine (CH3NH2) which is a weak base
EXAMPLE:
Write the molecular, ionic and net ionic equation for the reaction of HCl
(aq) with KNO2 (aq).
Acid-Base Titrations
In a titration, a solution of accurately known concentration (called a standard solution) is
added gradually to another solution of unknown concentration, until the chemical
reaction between the two solutions is complete.
To make a standard solution of NaOH, we must standardize it with a primary standard,
usually a weak solid acid nicknamed “KHP” (the actual formula is KHC8H4O4, molar
mass = 204.23 g). KHP is a monoprotic acid that reacts with NaOH in a 1:1 molar ratio.
In simple monoprotic acids and bases with only 1 hydroxide unit per formula unit,
titrations reach the equivalent point when the moles of acid equals the moles of base.
Moles Acid = Moles Base
Since moles equals molarity times liters, we have:
MaVa = MbVb
EXAMPLE:
How many mL of 2.00 M NaOH is required to neutralize 12.5 mL of a
4.50 M HCl solution?
EXAMPLE:
In a titration experiment, a student finds that 0.5468 g of KHP is
completely neutralized by 23.48 mL of a NaOH solution. What is the
concentration (molarity) of the NaOH solution? The molar mass of KHP
Is 204.23 g/mol.
4.9 Oxidation-Reduction Reactions
Acid-base reactions are proton transfer processes. Similarly, oxidation-reduction
(“redox”) reactions are electron transfer processes. Consider the reaction of calcium with
oxygen in the atmosphere:
Ca (s) + S (s)  CaS (s)
Calcium oxide is a compound made of ions (Ca+2 and O-2). Each calcium had to give up
two electrons, and each oxygen accepted two electrons, involving the exchange of four
electrons total. The half-reactions are given below:
Ca  Ca+2 + 2eS + 2e-  S-2
oxidation half-reaction
reduction half-reaction
Each of the half-reactions explicitly show the electrons involved in a “redox” reaction.
The sum of the half-reactions gives the overall reaction:
Ca + S + 2e-  Ca+2 + S-2 + 2eAnd if we cancel the electrons from both sides and combine the calcium and oxide ions
we get:
Ca + S  Ca+2 + S-2  CaS
The term oxidation reaction refers to the half-reaction that involves loss of electrons. A
reduction reaction is the half reaction that involves the gain of electrons. A handy way to
remember this distinction is the old saying “LEO the lion goes GER”.
EXAMPLE:
The reaction of zinc with an aqueous solution of copper(II) sulfate is given
below. Write the ionic and net ionic reactions. Then write both the halfreactions. Determine what element is being reduced and what element is
being oxidized.
Zn(s) + CuSO4(aq)  ZnSO4(aq) + Cu(s)
Assigning Oxidation Numbers
An oxidation number is useful in redox reactions. This number signifies the charge an
atom would have if it transferred its electrons completely. It allows us to tell at a glance
what element is being oxidized, and what element is being reduced. The element being
oxidized experiences and increase in oxidation number, while the substance being
reduced experiences a decrease in oxidation number. We use the following rules to
assign oxidation numbers.
1) In free elements (the uncombined state), each atom has an oxidation number of zero.
So each atom in H2, Na, Be, O2, and P4 has the same oxidation number: zero.
2) For ions composed of only one atom, the oxidation number is equal to the charge on
the ion. So Li+ has an oxidation number of +1; Ba+2 has an oxidation number of +2;
O-2 has an oxidation number of –2 and so on. All alkali metals have an oxidation
number of +1 and all alkaline earth metals have an oxidation number of +2 in their
compounds. Aluminum has an oxidation number of +3 in all its compounds.
3) The oxidation number of oxygen in most compounds is –2. The exceptions: in
hydrogen peroxide (H2O2) and the peroxide ion (O2-2) it is –1.
4) The oxidation number of hydrogen is +1, except when it is bound to a metal in binary
compounds, such a NaH, in which case it is –1.
5) Fluorine has an oxidation number of –1 in all its compounds. Other halogens have
negative oxidation numbers when they occur as halide ions in their compounds.
When combined with oxygen they have a positive oxidation number.
6) In a neutral molecule, the sum of the oxidation numbers of all the atoms must be zero.
In a polyatomic ion, the sum of oxidation numbers of all the elements in the ion must
be equal to the net charge of the ion.
7) Oxidation numbers do not have to be integers.
EXAMPLE:
Assign oxidation numbers to each atom in the following compounds.
(a) Li2O
(b) HNO3
(c) Cr2O7-2
(d) PF3
(e) MnO4-
(f) H2
(g) H3PO3
(h) NaHCO3
(i) P4
(j) H3PO2
(k) LiH
(l) K2O2
(m) CH4
(n) NH4+
(o) Fe3O4
Oxidizing Agents and Reducing Agents
In a redox reaction, the substance that is being reduced is causing the other substance to
be oxidized. Therefore, the substance that is reduced is called the oxidizing agent. The
substance that is oxidized caused the other substance to be reduced and is called the
reducing agent. (The agents often include the entire compound; the following example
will show you what I mean.)
Example: Assign oxidation numbers to each of the following. Determine what is being
reduced, what is being oxidized, the oxidizing agent, and the reducing agent.
(1) 2FeBr3 + 3Cl2  2FeCl3 + 3Br2
(2) WO3 + 3H2  W + 3H2O
Types of Oxidation Reduction Reactions
1) Hydrogen Displacement
2Na + 2H2O  2NaOH + H2
2) Metal Displacement
Li + NaCl  LiCl + Na
3) Halogen Displacement
Cl2 + 2KBr  2KCl + Br2
4) Disproportionation
Cl2 + 2OH-  ClO- + Cl- + H2O
5) Combustion
CH4 + O2 → CO2 + H2O
6) Decomposition
Ag2O → Ag + O2
7) Combination or Synthesis
Mg + O2 → MgO
EXAMPLE: Write the balanced net ionic reactions for the following
(a) a sample of mercury(II) oxide is heated in a decomposition reaction
(b) a strip of silver metal is added to a solution of copper(II) sulfate
(c) a piece of potassium metal is added to water
(d) a piece of magnesium metal is added to a solution of hydrochloric acid
(e) liquid bromine is added to a solution of potassium iodide
(f) a iron filings are placed in a container with hot chlorine gas
(g) hexanol (C6H14O) is burned in air
4.10 Balancing Oxidation-Reduction Reactions
Assigning oxidation numbers can be helpful in balancing redox reactions. Both halfreactions should be written, and the electrons lost should equal the electrons gained. The
following example illustrates how to balance equations using redox half-reactions.
EX: Balance the following reaction using redox half-reactions:
HCl + Al  AlCl3 + H2
What makes redox reactions very difficult to balance sometimes is that both the atoms
and charges must balance. Redox half-reactions can make balancing these much easier.
For instance, try balancing the following reaction, both in atom numbers and in charge:
Cu +
H+ + NO3-
 Cu+2 + NO + H2O
Redox Reactions in Acidic and Basic Solutions
Many redox reactions occur in aqueous solutions where the H+ and OH- have an effect
on the oxidizing and reducing processes. As a result, the balancing of the redox equation
is more difficult. In this section, I will show you how to balancer a number of redox
equations in acidic and basic solutions.
The steps you take to balance redox reactions in acidic solutions are:
1) Write the two half-reactions.
2) For each half reaction:
Balance all atoms exclusive of hydrogen and oxygen.
Balance each excess oxygen with one H2O.
+
Balance each hydrogen with one H .
Balance charge with electrons.
3) If necessary, multiply each half-reaction by an appropriate factor to make the total
electrons gained in the reduction half-reaction equal to the total electrons lost in the
oxidation half-reaction.
4) Add the half reactions, and cancel identical species.
5) Check that the elements and charges are balanced.
EX: Apply the previous steps to the balancing of the following redox reaction in acidic
solution:
Cr2O7-2(aq) + C2H6O(l)  Cr+3(aq) + CO2(g)
EX:
(acidic solution)
Balance the following redox reacton, which occurs in an acidic environment:
Cr2O7-2 + Cl-  Cl2 +Cr+3
(acidic solution)
In basic solutions, the steps are:
1) Use the half-reaction method as specified for acidic solutions to obtain the final
balanced equation as if H+ ions were present.
2) To both sides of the equation obtained above, add a number of OH- ions that is equal
to the number of H+ ions (we want to eliminate H+ by forming water).
3) Form H2O on the side containing both H+ and OH- ions, and eliminate the number of
H2O molecules that appear on both sides of the equation.
4) Check that elements and charges are balanced.
EX: Apply the above steps to balance the following redox reaction, assuming it occurred
in basic solution:
I2 → I– + IO3–
(basic solution)
Redox Reactions and Net Ionic Equations
In order to write the net ionic reactions for redox reactions, you need no know some of
the most important oxidizing agents and reducing agents and the products that form. The
list below summarizes these agents.
Important Oxidizing Agents
MnO4- (acidic solution)
MnO4- (basic solution)
Cr2O72- (acidic solution)
CrO42- (basic solution)
HNO3, concentrated
HNO3, dilute
H2SO4, hot, concentrated
Metal ions
Halogens
H2O2
Formed in Reaction
Mn2+
MnO2
Cr3+
Cr3+
NO2
NO
SO2
Metal or metal ions of lower charge
halide ions
H2O, O2
Important Reducing Agents
Halide ions
Metals
Metal ions
Sulfite ions
Halogens (dilute, basic)
Halogens (concentrated, basic)
Formed in Reaction
halogens
metal ions of most stable charge
Metal ions of higher charge
sulfate ions
hypohalite (ex: hypochlorite) ions
halite (ex: chlorite) ions
(Last but not least, don’t forget to balance oxygens with water and hydrogens with H+ in
acidic solution and OH- in basic solution!)
EXAMPLE:
Write the unbalanced net ionic equations for the following:
(a) a solution of acidic potassium permanganate is added to a solution of
iron(II) sulfate.
(b) A piece of copper wire is placed in a solution of silver nitrate.
(c) A piece of iodine is added to a dilute potassium hydroxide solution in a
disproportination reaction
(d) Some hydrogen peroxide is catalytically decomposed by light
(e) An acidified solution of potassium dichromate is added to a solution of
sodium bromide
(f) A piece of copper metal is dropped into dilute nitric acid
(g) A piece of zinc metal is added to a basic solution of sodium chromate
Redox Titrations
Redox titrations require the same type of calculations (based on moles) that many other
stioichiometric calculations use. Many times the equations and the stoichiometry can be
fairly complex. The following example illustrates this point.
EXAMPLE:
A 16.42 mL volume of 0.1327 M KMnO4 solution is needed to oxidize
20.00 mL of a Fe+2 solution to Fe+3. The reaction takes place in an
acidic environment.
(a) Write the net ionic reaction that occurs.
(b) Balance the reaction.
(c) What is the original concentration of the Fe+2 solution?
(d) How many mL of 3.00 M HNO3 solution would have been necessary to
complete this reaction?