Download chem 100 class notes - Louisiana Tech University

CHEM 100 CLASS NOTES
Prof. Upali Siriwardane,
Chemistry Program, Louisiana Tech University, Ruston, LA 71272
CHAPTER 5. Chemical Reactions
KEY CONCEPTS
Solution Chemistry
Solubilty of chemical
compounds in water
Write molecular equation
Common types of chemical
reactions
Neutralization reactions
Gas-forming Reactions
Oxidizin/reducing agents
Single displacement
Solution Concentrations
Dilutions
Solvent and solute
Solubilities of chemical
compounds
Write complete ionic equation
Precipitation Reactions
Solubility rules
Common acids and bases
Oxidation-reduction reactions
Assigning oxidation numbers
Half reactions
Molarity
Solution stoichiometry
Aqueous solutions
Types of chemical equations
Write a net ionic equation
Acid/base Reactions
Strong/weak acids/bzses
Redox reactions
Activity series
Number electrons transferred
Preparing solutions
Titrations
Solution Chemistry
Solutions occur commonly in nature and many chemical reactions occur in solutions. By
learning about the nature of solutes, solvents and solutions, students can better understand both these
natural phenomena and the chemistry of solutions.
A soultionnis defined as a homogeneous mixture.
Many materials exist as homogeneous mixtures–air (g), lake water (l), gasoline (l), stainless steel (s)
Mixtures include suspensions, colloids, and solutions. Only soultions are considered true solutions.
Why is it necessary to carry out chemical reactions in solutions, especially in
water.
Faster chemical reactions occur in gas phase compared to reactions of solids and liquids.
Some solids when mixed do not show a reaction. However, when they are dissolved in a solvent,
an immediate reaction is usually observed. The molecules or the ions of solids are broken down
to individual particles when they go into solution. These particles or ions move around the
solution and have sufficient kinetic energy to initiate a chemical reaction when they collide with
particles of other chemicals.
There are many solvents that are available for chemical reactions. However, water is the
most abundant solvent dissolving many chemicals. Water can interact with both cations and
anions making it the best solvent for ionic and polar compounds.
Solvent and solute
1
Solution - homogeneous mixture of of two or more substances; often comprised of solvent and
solute, e.g., when gases or solids are dissolved in a liquid
Solvent - component with same phase as solution; substance present in excess in liquid-liquid
mixtures
Solute - minor component mixed with solvent
Aqueous solutions
Solubilty of chemical compounds in water
Degree of mixing of solute and a solvent
Miscible :When two liquids are soluble in all proportion.
Immiscible - When two liquids are not soluble in each other form layers upon mixing
Solubility - maximal amount of solute accommodated by solvent at specified temperature
Solubilities of chemical compounds
Solubility is defined as the maximum amount of a substance that can dissolve in a given amount
of solvent at a specific temperature.
Factors Affecting Solubility
What causes different solubilities?
General Rule: “Like dissolves like”
The definition of “like” involves the nature of the bonding and structure of solute and solvent:
ionic or polar covalent vs. non-polar covalent.
– Polar solvents dissolve polar solutes.
– Nonpolar solvents dissolve nonpolar solutes.
– Oil and water don’t mix.
- Water molecules are more attracted to each other than oil molecules.
The solution process involves making and breaking ionic bonds and/or intermolecular forces
(IMF). When attractions between solute molecules and solvent molecules are strong, solutions
are able to form. Solute molecules are surrounded by solvent molecules. (solvent cage). Process
is called solvation or dissolution.
Solubilites of ionic copounds are summerized as solubility rules
Solubility Rules
1. All common compounds of Group I and ammonium ions are soluble.
2. All nitrates, acetates, and chlorates are soluble.
3. All binary compounds of the halogens (other than F) with metals are soluble,
except those of Ag, Hg(I), and Pb. Pb halides are soluble in hot water.)
4. All sulfates are soluble, except those of barium, strontium, calcium, lead, silver,
and mercury (I). The latter three are slightly soluble.
5. Except for rule 1, carbonates, hydroxides, oxides, silicates, and phosphates are
insoluble.
6. Sulfides are insoluble except for calcium, barium, strontium, magnesium, sodium,
potassium, and ammonium.
2
Electrical conductivity of water solutions:
a) Strong electrolytes
b) Weak electrolytes
c) Nonelectrolytes
Strong electrolyte – a compound that dissolves nearly completely in water to give a solution
conducts electricity well
Weak electrolyte – a compound that dissolves to a small extent to give a solution that conducts
electricity weakly
Non-electrolyte – a compound that does not produce ions when dissolved in water
a) Strong Electrolytes: Strong electrolytes conduct electricity strongly when dissolved in water.
They are usually ionic compounds such as sodium chloride - NaCl. In the solution ionic
compounds break into cations and anions which are free to move in the solution and carry a
current through the solution.
b) Weak Electrolytes: Weak electrolytes conduct electricity weakly when dissolved in water.
They are normally covalent compounds which partially dissociate to form small amount of
cations and anions in water. This incomplete dissociation provides only few ions to carry a
electric current through the solution. Examples of weak electrolytes are organic acids, such as
acetic acid (HC2H3O2).
c) Nonelectrolytes: Nonelectrolytes do not conduct electricity when dissolved in water. They
are usually molecular or covalent compounds, which do not dissociate into ions in water.
Example of nonelectrolytes are alcohols such as ethanol (C2H5OH) and polyhydroxy alcohols
such as sugars.
Types of chemical equations
There are three types of chemical equations used to describe reactions in solution: (1) the
molecular equation, (2) the total ionic equation, and (3) the net ionic equation.
Molecular Equation
The molecular equation gives the overall reaction stoichiometry but not necessarily the actual
form of the reactants and products in solution. For example, when you mix HCl with aqueous
NaOH, a reaction occurs that forms water and the salt NaCl.
HCl(aq) + NaOH(aq) --> H2O(l) + NaCl(aq)
This is a molecular equation. One problem with molecular equations is that they don't give a
clear picture of what actually occurs in solution. As we learned earlier HCl, NaOH, and NaCl
will actually exist as separate ions in solution. A more accurate picture of the chemical reaction
that is taking place is given by the complete ionic equation.
Total Ionic Equation
The total ionic equation represents all reactants and products that are strong electrolytes as ions.
For example,
H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) --> H2O(l) + Na+(aq) + Cl-(aq)
This is a better representation of the actual form of the reactants and products. In a total ionic
equation all substances that are strong electrolytes are represented as ions.
3
From the total ionic equation we see that the Na+ and Cl- ions don't really participate in the
reaction. The Na+ and Cl- ions are called Spectator Ions because they only watch the H+ and
OH- ions react to form water. If we focus on those ions that are actually involved in the chemical
change and eliminate the spectator ions from our equation then we are writing the net ionic
equation.
Spectator Ions: Ions in a solution that do not participate in a chemical reaction.
Net Ionic Equation
The net ionic equation includes only those species undergoing a chemical change, that is,
spectator ions are not included. For example,
H+(aq) + OH-(aq) --> H2O(l)
is the net ionic equation for the molecular equation above.
It is also possible to mix substances together in solution and have no net ionic equation. For
example, consider the following molecular equation:
CaCl2 (aq) + 2 HNO3 (aq) --> Ca (NO3 )2 (aq) + 2 HCl(aq)
In this example the total ionic equation is:
Ca2+(aq) + 2 Cl- (aq) + 2 H+(aq) + 2 NO3- (aq) --> Ca2+(aq) + 2 NO3- (aq) + 2 H+(aq) + 2 Cl- (aq)
Clearly, all the ions are spectator ions, so there is no net ionic equation in this example.
Write the ionic, net ionic equations for the following molecular equations and
figure out the spectator ions:
NaCl(aq) + AgNO3(aq)
HCl(aq) + NaOH(aq)
NaOH(aq) + HC2H3O2(l)
----> AgCl(s) + NaNO3(aq)
----> NaCl(aq) + H2O(l)
----> NaC2H3O2(aq) + H2O(l)
Molecular equation. Equation with formula, correct stoichiometric coefficients and physical
form written within parenthesis.
Total ionic equation: All the ionic compounds soluble in water are separated into ions written
with their ionic charge and (aq).
Net Ionic equation: Ionic equation with all spectator ions removed from both sides.
Spectator Ions: Ions appearing on both side of an ionic equation.
a) NaCl(aq) + AgNO3(aq) ----> AgCl(s) + NaNO3(aq)
Molecular equation:
NaCl (aq) + AgNO3 (aq) --> AgCl (s) + NaNO3 (aq)
Total ionic Equation:
Na+ (aq) + Cl- (aq) + Ag+ (aq) + NO3- (aq)
--> AgCl (s) + Na+ (aq) + NO3- (aq)
Spectator Ions:
Na+ (aq)
and
NO-3 (aq)
Net Ionic Equation:
Cl- (aq) + Ag+ (aq) --> AgCl (s)
b) HCl(aq) + NaOH(aq) ----> NaCl(aq) + H2O(l) Molecular Equation:
4
HCl (aq) + NaOH (aq) --> NaCl (aq) + H2O (l)
Total ionic Equation:
H+ (aq) + Cl- (aq) + Na+ (aq) + OH- aq)
--> Na+ (aq) + Cl- (aq) + H2O (l)
Spectator Ion:
Na+ (aq)
and
Cl- (aq)
Net Ionic Equation:
H+ (aq) + OH- (aq) --> H2O (l)
c) NaOH(aq) + HC2H3O2(l) -----> NaC2H3O2(aq) + H2O(l)
Molecular Equation:
NaOH (aq) + HC2H3O2 (l) -----> NaC2H3O2 (aq) + H2O (l)
Total ionic Equation:
Na+ (aq) + OH- (aq) + HC2H3O2 (l) --> Na+ (aq) + C2H3O-2 (aq) + H2O (l)
Spectator Ions:
Na+ (aq)
Net Ionic Equation:
OH- (aq) + HC2H3O2 (l) --> C2H3O-2 (aq) + H2O (l)
Classification of compounds as acids, bases and salts
Most of the chemical compounds can be classified either as an acid, a base or a salt.
Acids: Acids are compounds that increase the hydrogen ion concentration when dissolved in
water. They usually have at least one removable hydrogen atom attached to an oxygen atom.
These hydrogen are called acidic hydrogen because they come out as H+ or protons to give an
acidic pH (< 7) for the solution. Acidic protons are written at the beginning of formula e.g.
HC2H3O3. Some metal oxides such as SO2 and CO2, which produce protons in water, are also
considered as acids.
Bases: Bases are compounds that decrease the hydrogen ion concentration by adding OH - ions
(hydroxide) when added to water. They usually have OH-. These OH- ions (hydroxide) are
written after the metal in the formula of the base. e.g. NaOH. The pH of the solution increase
because OH- ions shift the H2O <==> H+ + OH- to left removing H+ from the solution. Some
compounds which accepts protons such as NH3 are also considered as bases. Some metal oxides
such as CaO and K2O, which
produce OH- ions in water, are also considered as bases.
Salts: Salts are formed when an acid reacts with a base. e.g. HCl(acid) + NaOH(base) --> NaCl
(salt) + H2O. Salts normally don't have acidic protons or OH - ions. One can get the parent acid
and base from a salt by breaking the compound into metal (cation) and anion (non-metal or
polyatomic anions) and adding H+ to anion and OH- to metal.
Identify acids bases and salts among the foloowing list:
5
AgNO3, NaCl, H3PO4, NaOH, HCl, NaNO3, H2SO4, HC2H3O2, KNO3, HNO3,
MgO, K2O, SO3, NH3, NH4OH.
Salts: AgNO3(Salt), NaCl(Salt), NaNO3(salt), KNO3(salt),
Acids: H3PO4(acid), HCl(acid), H2SO4(acid), HC2H3O2(acid), HNO3(acid), SO3(acid),
Bases: MgO(base), NaOH(base), K2O(base), NH3(base), NH4OH(base).
Common types of chemical reactions
Classification of chemical reactions in solution based on their driving force:
a) Precipitation reactions
b) Acid-base reactions
c) Gas forming reactions
d) Oxidation-reduction (REDOX) reactions
a) Precipitation Reactions: They are double displacement reactions of ionic compounds where
an insoluble salt is formed when two aqueous salt solutions are mixed. The only reason or
driving force for these reactions to occur is the removal of cations and anions from the solution
as a precipitate. When two solutions of ionic compounds are mixed there will be no reaction if
the combinations of cations and anions in the mixture does not lead to an insoluble salt.
Foramtion of insoluble salt drives the reaction to completion.
E.g.
i) AgNO3 (aq) + NaCl (aq) = AgCl (s) + NaNO3 (aq)
Precipitation Reaction
occurs.
ii) NaCl (aq) + KNO3 (aq) = NaNO3 (aq) + KCl (aq)
No precipitation Reaction.
b) Acid-base Reactions: They are reactions of acids (compounds that produce H+ ion in water)
and bases (compound that produce OH- ions in water). The driving force for these reactions is
the formation of water H2O from the H+ and OH- ions in the solution. They are similar to
precipitation reactions except the insoluble salt or the product is water H2O. These reaction goes
to completion since formation of water is instantaneous once H+ and OH- are together.
E.g. NaOH (aq) + HCl (aq) ---> NaCl (aq) + H2O
c) Gas Forming Reactions
Gas is removed from the reaction mixture completing reaction to products. Some salts react
with acids to form gases. Carbonates and bicarbonates produce CO2.
This was done in last week’s lab. It works with any carbonate or bicarbonate. It is actually a
two-part rxn.
CaCO3(s) + 2 HCl(aq)  CaCl2(aq) + H2CO3(aq)
H2CO3(aq)  CO2(g) + H2O(l)
Sulfites and bisulfites produce SO2.
6
Sulfides produce H2S.
Halides produce hydrogen halides (with nonvolatile acids such as H2SO4 and H3PO4).
d) Redox (Reduction - Oxidation) Reactions
Redox reactions are chemical reactions where electrons are transferred from one compound to
another resulting in a chemical change.
E.g. Zn + H+Cl-(aq) --->
Zn2+Cl2-1 (aq) + H2
Zn loses two electrons to H+ to form H2 gas. The oxidation state (charge on the metal or nonmetal) of atoms on the left side of the chemical equation is changed when they react to form
products during redox reactions.
Precipitation Reactions
A precipitation reaction is a reaction in which soluble ions in separate solutions are mixed together to form an
insoluble compound that settles out of solution as a solid. That insoluble compound is called a precipitate.
Solubility Rules
1) All alkali metal (lithium, sodium, potassium, rubidium, and cesium) and ammonium
compounds are soluble.
2) All acetate, perchlorate, chlorate, and nitrate compounds are soluble.
3) Silver, lead, and mercury(I) compounds are insoluble.
4) Chlorides, bromides, and iodides are soluble
5) Carbonates, hydroxides, oxides, phosphates, silicates, and sulfides are insoluble. (6) Sulfates
are soluble except for calcium and barium.
Soluble compounds
•all salts of alkali metal ions and the NH4+ ion
•all salts of NO3 , CH3CO2 , ClO3 , ClO4 ions
•most salts of Cl , Br , and I ions
•except Ag+, Pb+2, and Hg2+2 salts
•most salts of the SO42 ion
•except Sr+2, Ba+2, and Pb+2 salts
Insoluble compounds
•most salts of CO3 2, PO4 3, OH , O 2, and S 2 ions
•except alkali metal and NH4+ salts
Solubility Rules for Ionic Compounds
Solubility rules are useful summaries of information about which ionic compounds (or
combinations of ions) are soluble in water and which are not. They are also important tools for
making predictions about whether certain ions will react with one another to form a precipitate.
In addition, they are useful for figuring out what ions might be involved when a precipitation
reaction has been observed.
Based on solubility rules, figure out whether following precipitation reactions
are likely or unlikely to occur
a) MgI2 + NaNO3
b) Ba(NO3)2 + Na2SO4
c) AgCl + NaNO3
= 2 NaI + Mg(NO3)2
= BaSO4 + 2 NaNO3
= AgNO3 + NaCl
7
An insoluble salt should be formed as a product if a precipitation reaction is to occur.
a) MgI2 + NaNO3
= 2 NaI + Mg(NO3)2
MgI (soluble)
NaNO3 (soluble)
NaI (soluble)
Mg(NO3)2 (soluble)
From solubility Rules
No precipitation Reaction.
b) Ba(NO3)2 (aq) + Na2SO4 (aq)
= BaSO4 (s) + 2NaNO3 (aq)
Ba(NO3)2 (soluble)
Na2SO4 (soluble)
From solubility
BaSO4 (insoluble)
Rules
NaNO3 (soluble)
A precipitation
Reaction.
c) AgCl + NaNO3
= AgNO3 + NaCl
AgCl (insoluble)
From solubility
Rules
No insoluble salts
are formed as
products
No precipitation
NaNO3 (soluble)
AgNO3 (soluble)
NaCl (soluble)
Acid/base Reactions
When an acid and a base are placed together, they react to neutralize the acid and base
properties, producing a salt. The H+ cation of the acid combines with the OH- anion of the base
to form water. The compound formed by the cation of the base and the anion of the acid is
called a salt. The combination of hydrochloric acid and sodium hydroxide produces common
table salt, NaCl:
Assign the parent acid and base of the following salts: AgNO3, NaCl, , AgCl,
KNO3, NaNO3, KCl, NaC2H3O2, NH4Cl.
Acid (A) + Base(B) = Salt + water (H2O)
HA + BOH = BA + H2O
E.g. LiNO3
B (Li)
A (NO3)
8
AgNO3=
NaCl
AgCl
KNO3
NaNO3
KCl
NaC2H3O2
NH4Cl
OH
H
BOH(LiOH)
HA(HNO3)
Base
Acid
AgOH + HNO3
=
NaOH + HCl
=
AgOH + HCl
=
KOH + HNO3
=
NaOH + HNO3
=
KOH + HCl
=
NaOH + HC2H3O2
=
NH4OH + HCl
Neutralization reactions
A chemical reaction in which an acid and a base react to form salt and water is called
Neutralization.
Acid + Base  Salt + Water
EXAMPLE:
HCl + NaOH  NaCl + H2O
HNO3 + NaOH  NaNO3 + H2O
Common acids and bases
Acid Name
Formula
Acetic acid
Hydrogen chloride
Nitric acid
Phosphoric acid
Sulfuric acid
Base Name
HC2H3O2
HCl
HNO3
H3PO4
H2SO4
Formula
Sodium hydroxide
Barium oxide
Calcium oxide
NaOH
BaO
CaO
Name
Sodium salt
Name
Sodium acetate
Sodium chloride
Sodium nitrate
Sodium phosphate
Sodium sulfate
Name
Name
Sodium chloride
Barium chloride
Calcium chloride
Strong/weak acids/bases
Acids
strong – completely ionized
weak – partially ionized
Strong Acids(strong electrolytes)
HCl, HBr, HI, H2SO4, HNO3, HClO4 (all others are weak)
Weak acids( weak electrolytes)
CH3COOH, HF, HCN, H3PO4, HCOOH, H2CO3
9
Formula
NaC2H3O2
NaNO3
Na3PO4
Na2SO4
Chloride salt
Formula
NaCl
BaCl2
CaCl2
Table 5.2 page 171.
Bases
Strong bases
Metal Hydroxides•(Group IA):Li, Na K. (Group IIA): Ca, Sr, Ba (all others metal
hydroxides are weak)
Weak bases
NH3, amines -CH3NH2•Table 5.2 page 171.
Gas-forming Reactions
It is very helpful to be able to recognize the formulas for those gases that may be used or
produced during the course of a chemical reaction. The way to indicate that a compound is a gas
when you write a chemical equation is to place (g) after the formula, such as NO2(g).
Here is a list of some of the more common gases:
CO2, SO2, H2S, NH3, ,SO3, CO, NO, NO2,
P2O3, P2O5, SiF4, HCl, HBr, HI, HF, N2O5, N2O3, N2O, F2, Cl2, H2, N2, O2
Sometimes in the solution reactions a gas will be involved as one of the reactants or products.
For example, consider the following reaction:
Formation of CO2
One way gases can form in solution is through the decomposition of weak electrolytes. For
example, H2CO3 readily decomposes into H2O and CO2 gas, that is:
H2CO3(aq)-->H2O(l) + CO2(g)
For example
HCl(aq) + NaHCO3(aq)-->NaCl(aq) + H2CO3(aq)
The product H2CO3, however, will decompose into water and carbon dioxide, giving
us a new equation:
HCl(aq) + NaHCO3(aq)-->NaCl(aq) + H2O(l) + CO2(g)
If we eliminate all the spectator ions we would write the net ionic equation as
H+(aq) + HCO3-(aq)-->H2O(l) + CO2(g)
Formation of SO2
Sulfites
Other substances that will decompose and form gases are H2SO3
10
CaSO3(s) + 2HCl(aq)  CaCl2(aq) + H2SO3(aq)
H2SO3(aq)-->H2O(l) + SO2(g)
Formation of H2S
Sulfides:
2HCl(aq) + Na2S(aq)-->H2S(gas) + 2NaCl(aq)
In this example the complete ionic equation would be:
2H+(aq) + 2Cl-(aq) + 2Na+(aq) + S2-(aq)-->H2S(g) + 2Na+(aq) + 2Cl-(aq)
Removing the spectator ions we obtain the net ionic equation:
2 H+(aq) + S2-(aq)-->H2S(g)
H2S is just one example of a gaseous substance that can form in a solution reaction.
Formation of Ammonia from NH4OH:
NH4Cl(aq)+ NaOH(aq) --> NaCl(aq) +
NH4OH(aq)
NH4OH(aq) -->H2O(l) + NH3(g)
Oxidation-reduction reactions(REDOX Reactions)
a) Oxidation
LEO The Lion Goes GER! Loss of electrons oxidation!
In old days oxidation was considered as reactions of metals with oxygen. The metal loses
electrons to oxygen forming metal a cation and an oxide ion. Taking electron transfer into
account, today oxidation is defined as loss of electrons from ions, atoms, elements, or
compounds.
E.g.
Na ----> Na+ + eb) Reduction
LEO The Lion Goes GER! Gain of electrons reduction!
In old days reduction was considered as reactions of hydrogen with metal oxide to
produce metal. The metal cations of the metal oxide gain electrons from hydrogen to form
neutral metal atoms or element. Taking electron transfer into account, today reduction is
defined as gain of electrons by ions, atoms, elements, or compounds.
E.g.
Cl2 + 2e- ----> 2Cl-
Redox reactions
Redox reactions primarily involve the transfer of electrons between two chemical species. The compound that loses
an electron is said to be oxidized, the one that gains an electron is said to be reduced. There are also specific terms
that describe the specific chemical species.
Reducing and oxidizing agents
11
A compound that is oxidized is refered to as a reducing agent, while a compound that is reduced is referred to as the
oxidizing agent. You should be able to visually determine when an oxidation-reduction reaction has occurred and
find oxidizing/reducing agents.
Reducing agent - the reactant that gives up electrons.
The reducing agent contains the element that is oxidized (looses electrons).
If a substance gives up electrons easily, it is said to be a strong reducing agent.
Oxidizing agent - the reactant that gains electrons.
The oxidizing agent contains the element that is reduced (gains electrons).
If a substance gains electrons easily, it is said to be a strong oxidizing agent.
Determination of oxidizing and reducing agents
Zinc metal is dissolved by hydroiodic acid. Bubbles of hydrogen gas are evolved.
Write the equation.
Zn + 2HCl ----> ZnCl2 + H2
Assign oxidation numbers to each atom on both sides.
Zn + 2HCl ----> ZnCl2 + H2
ON Zn= 0H= +1 Zn= +2 H=0
Cl= -1
Cl= -1
The oxidizing agent was the one reduced
Reduction Half Reaction(RHR):
2H+ +
2e----> 2H2
2H+ is the oxizidzing agent
The reducing agent was the reactant oxidized.
Oxidation Half Reaction(OHR): Zn ----> Zn 2+ + 2eZn will be the reducing agent
Activity series
Among the elements there is a great range of chemical reactivity. Some, like sodium and
fluorine, are so reactive that they are never found in the free or uncombined state in nature.
Others, like xenon and platinum, are nearly inert and can be made to react with other elements
only under special conditions. The reactivity of an element is related to its tendency to lose or
gain electrons; that is, to be oxidized or reduced. In principle it is possible to arrange nearly all
the elements into a single series in order of their reactivities. A
series of this kind indicates which free elements are capable of
displacing other elements from their compounds. Such a list is
known as an activity or electromotive series. To illustrate the
preparation of an activity series we will experiment with a small
group of selected elements and their compounds.
A generalized single displacement reaction is represented in the
form
A + BC → B + AC
Element A is the more active element and replaces element B
from the compound BC. But if
element B is more active than element A, no reaction will occur.
Let’s look at two specific examples, using copper and mercury.
Example 1. A few drops of mercury metal are added to a
solution of copper(II) chloride(CuCl2)
12
Example 2. A strip of metallic copper is immersed in a solution of mercury(II) chloride (HgCl2)
In Example 1 no change is observed even after the solution has been standing for a prolonged
time, and we conclude that there is no reaction. In Example 2 the copper strip is soon coated
with metallic mercury and the solution becomes pale green. From this evidence we conclude that
mercury will not displace copper in copper compounds but copper will displace mercury in
mercury compounds.
Therefore copper is a more reactive metal than mercury and is above mercury in the
activity series. In terms of chemical equations these facts may be represented as
Example 1.
Hg + CuC12 → No reaction
Example 2.
Cu + HgC12 → Hg + CuCl2
The second equation shows that, in terms of oxidation numbers, the chloride ion remained
unchanged, mercury changed from +2 to 0, and copper changed from 0 to +2. The +2 oxidation
state of
copper is the one normally formed in solution.
Expressed another way, the actual reaction that occurred was the displacement of a mercury ion
by a copper atom. This can be expressed more simply in equation form:
Cu0 + Hg2+ → Cu2+ + Hg0
In contrast to double displacement reactions, single displacement reactions involve changes in
oxidation number and therefore are also classified as oxidation-reduction reactions.
Single displacement Reaction
Single displacement Reaction is a redox reaction. During single replacement, one element
replaces another element in a compound. There are two different possibilities:
1. One cation replaces another: single displacement. Written using generic symbols, it is:
A + BC → B + AC
Element A is the more active element and replaces element B from the compound BC. But if
element B is more active than element A, no reaction will occur.
Some examples are:
Cu + AgNO3 ---> Ag + Cu(NO3)2
Fe + Cu(NO3)2 ---> Fe(NO3)2 + Cu
Ca + H2O ---> Ca(OH)2 + H2
Zn + HCl ---> ZnCl2 + H2
Half reactions
There are two half-reactions in every redox reaction: one where the oxidation is talking place
and the other where reduction is taking place. The Oxidation Number (ON) to each atom in
chemicals is assigned and two half reaction for the redox reaction can be separated out.
E.g. 2 Na + Cl2 ----> 2NaCl
Oxidation Half Reaction(OHR): Na ----> Na+ + eReduction Half Reaction(RHR):Cl2 + 2e- ----> 2Cl-
Number electrons transferred
13
This electron refers to the definitions of oxidation-reduction. Oxidation is loss of electrons and
reduction, gain of electrons
E.g. 2 Na + Cl2 ----> 2NaCl
Oxidation Half Reaction(OHR): Na ----> Na+ + eElectron balanced half reaction: 2Na ----> 2Na+ + 2 eReduction Half Reaction(RHR):Cl2 + 2e- ----> 2ClElectron balanced Reaction(RHR):Cl2 + 2e- ----> 2ClIt is the number of electrons in the balanced half reactions. There are two electrons tranferred in
this redox reaction.
E. g. MnO2 + 4HBr ----> Br2 + MnBr2 + 2H2O
MnO2 + 4HBr ----> Br2 + MnBr2 + 2H2O
ON Mn=+4
H=+1 Br= 0 Mn=+2
H= +1
O= -2 Br= -1
Br= -1 O= -2
Mn+4
----> Mn 2+
+
2eThe half-reactions should include the species present in the solution, therefore, Mn+4 is included
as MnO2.
In order to balance the half-reaction and to account for 2O2- from MnO2 we have to add 4H+ ions
to the left and 2H2O to the right.
Reduction Half Reaction(RHR):
4H+ + MnO2 + 2e- ----> Mn2+ + 2H2O
Mn=+4
Mn=+2
Oxidation Half Reaction(OHR):
2Br- -----> Br2 + 2eBr=-1
Br=0
Electron balanced Reaction(OHR): 2Br- -----> Br2 + 2eElectron balanced Reaction(RHR): 4H+ + MnO2 + 2e- ---->
Mn2+ + 2H2O
It is the number of electrons in the balanced half reactions. There are two electrons tranferred in
this redox reaction.
Define following terms used in oxidation-reduction reactions.
a) Oxidation
d) Redox reactions
b) Reduction
e) Half redox reactions
c) Oxidation number
a) Oxidation
In old days oxidation was considered as reactions of metals with oxygen. The metal loses
electrons to oxygen forming metal a cation and an oxide ion. Taking electron transfer into
account, today oxidation is defined as loss of electrons from ions, atoms, elements, or
compounds.
E.g.
Na ----> Na+ + eb) Reduction
In old days reduction was considered as reactions of hydrogen with metal oxide to produce
metal. The metal cations of the metal oxide gain electrons from hydrogen to form neutral metal
14
atoms or element. Taking electron transfer into account, today reduction is defined as gain of
electrons by ions, atoms, elements, or compounds.
E.g.
Cl2 + 2e- ----> 2Clc) Oxidation number
Oxidation number (sometimes called oxidation state) is number assigned to an atom in
compounds, ions and polyatomic ions to show the number of electrons relative to an atom in the
element, There are specific rules to assign oxidation numbers to atoms as given below. Read
problem 9 for more detail.
d) Redox reactions
Redox reaction are (red-stands for reduction, ox-stands for oxidation) chemical reactions
where at least one element of the reactants gain electrons (reduction) and another element lose
electrons (oxidation). The total reaction is called redox to show the involvement of both
reduction and oxidation in the process. As you can see yourself, reduction and oxidation cannot
take place independently, because there has to be an electron giver (reducing agent) and
electron acceptor (oxidizing agent). As a rule, the oxidation number of an atom increases
during reduction and oxidation number of an atom decreases during oxidation.
E.g.
2 Na + Cl2 ----> 2NaCl
ON 0
0
+1 -1
+
Oxidation Na ----> Na + eNa increase ON, 0 ----> +1
Reduction Cl2 + 2e- ----> 2ClCl decrease ON, 0 ----> -1
e) Half redox reactions
There are two half-reactions in every redox reaction: one where the oxidation is talking place
and the other where reduction is taking place. The Oxidation Number (ON) to each atom in
chemicals is assigned and two half reaction for the redox reaction can be separated out.
E.g. 2 Na + Cl2 ----> 2NaCl
Oxidation Half Reaction(OHR): Na ----> Na+ + eReduction Half Reaction(RHR):Cl2 + 2e- ----> 2Cl-
Assigning oxidation numbers
Assign the oxidation states to each atom in NaCl, O2, CBr4, S8, MnO2, KMnO4
and K2Cr2O7.
Oxidation number
Oxidation number (sometimes called oxidation state) is number assigned to an atom in
compounds, ions and polyatomic ions to show the number of electrons relative to an atom in the
element, There are specific rules to assign oxidation numbers to atoms as given below. Read
problem 9 for more detail.
a) Oxidation number of atoms in an element is zero (0). e.g. O2 , oxidation state of each
oxygen atom in O2 is zero.
15
b) Sum of the oxidation numbers in an element, compound is equal to zero. Sum of the
oxidation numbers in an ion, cation or anion is equal to the ionic charge
E.g. H2O, oxidation number (ON) of H is equal to +1 and oxidation number of O is equal to -2
making the ON= 2x1(2H) + (-2)= 0. Notice that before you add ON numbers you have to
multiply ON by the subscript for that atom.
E.g. SO42-, oxidation number (ON) of O is equal to -2 and oxidation number of S is equal to +6
making the ON= +6(S) + 4 (-2)= -2. Notices when you are adding you have to multiply ON by
the subscript for that atom.
c) As a rule ONs of H =+1, and O=-2 almost most of the time.
Cl (or halogens, group VIIB) = -1, E.g. Br = -1
alkali metal ( group IA)= +1, E.g. K = +1
alkaline earth metals (group IIA) = +2 ,E.g. Ca = +2
The group number in the periodic table could be used for main group elements (p and s blocks).
d and f block elements show variable ONs E.g. Fe shows either +3 or +2.
Examples:
NaCl: NaCl is an ionic compound made up of Na+ and Cl- ions.
For Na+ and Cl-, monoatomic ions, the oxidation number is equal to their charge.
Na+ ClON
+1
-1
O2: Oxygen is an element. The oxidation number of an atom in an element is 0.
O
ON
0
CBr4: CBr4 is a covalent compound (non-metal + non-metal). Br is a halogen. Halogens
normally have ON of -1. ON of C should be +4 since total of ON of a neutral compound such as
CBr4 is 0.
C
4Br
+4
+ 4 (-1) = 0
S8: Sulfur is an element. The oxidation number of an atom in an element is 0.
S
ON
0
MnO2:: MnO2 is an ionic compound composed of Mn 4+ and 2O2- ions. Oxygen always have an
oxidation number -2. MN should have an ON of +4 since total of ON of a neutral compound
such as MnO2 is 0.
ON
ON
Mn 4+ 2O2+4
+
Mn O
+4
-2
2(-2) = 0
KMnO4: KMnO4 is an ionic compound composed of K+ and MnO4- ions. MnO4- ion could be
considered as made up of Mn 7+ and 4O2-. The charge on Mn 7+ ion is obtained because oxygen
16
always have an oxidation number -2 and total of ON of atoms in MnO4- should add up to -1
which is the charge on the ion. Mn should have a ON of +7 and K+ have ON of +1 since total
of ONs of a neutral compound such asKMnO4 is 0.
K+
MnO4ON
+1 + +7 + [4(-2)] = 0
K
Mn
O
ON
+1
+7
-2
K2Cr2O7: K2Cr2O7 is an ionic compound composed of 2K+ and Cr2O72- ions. Cr2O72- ion could
be considered as made up of 2Cr+6 and 7O2-. The charge on each Cr+6 ion is obtained because
oxygen always have an oxidation number -2 and total of ON of atoms in Cr2O72- should add up
to -2 which is the charge on the ion. Each Cr should have an ON of +6 and K + have ON of +1
since total of ONs of a neutral compound such as K2Cr2O7 is 0.
2K+ Cr2O72ON
2(+1) +
2(+6) + [7(-2)] = 0
K
Cr
O
ON
+1
+6
-2
Which of the following reactions are redox?
a) NaCl + AgNO3 ----> AgCl + NaNO3
b) NaOH + HCl ----> NaCl + H2O
c) Zn + 2HCl ----> ZnCl2 + H2
d) 2Cr + 6HCl ----> 2CrCl3 + 3H2
e) MnO2 + 4HBr ----> Br2 + MnBr2 + 2H2O
a) NaCl + AgNO3 ----> AgCl + NaNO3
NaCl + AgNO3 ----> AgCl + NaNO3
ON Na= +1 Ag= +1
Ag= +1 Na= +1
Cl= -1
N= +5 Cl= -1
N= +5
O= -2
O= -2
Since the oxidation numbers (ON) in atoms of either reactants or products are not
changed , definitely this is not a redox reaction. If you remember the possible types for
chemical reactions, you should have figured out this reaction as a precipitation reaction because
an insoluble salt AgCl is formed as one of the products.
b) NaOH + HCl ----> NaCl + H2O
NaOH + HCl ----> NaCl + H2O
ON Na= +1
H= +1
Na= +1 H= +1
O= -2 Cl= -1 Cl= -1 O= -2
H= +1
Since the oxidation numbers (ON) in atoms of either reactants or products are not
changed , definitely this is not a redox reaction. If you remember the possible types for
chemical reactions, you should have figured out this reaction as an acid/base reaction because an
salt NaCl and water (H2O) are formed as products.
17
c) Zn + 2HCl ----> ZnCl2 + H2
Zn + 2HCl ---->
ZnCl2 + H2
ON Zn= 0
H= +1 Zn= +2
H=0
Cl= -1
Cl= -1
Since the oxidation numbers (ON) in atoms of either reactants or products are changed ,
definitely this is a redox reaction.
d) 2Cr + 6HCl ----> 2CrCl3 + 3H2
2Cr + 6HCl ----> 2CrCl3 + 3H2
ON Cr= 0
H= +1
Cr= +3
Cl= -1
Cl= -1
H=0
e) MnO2 + 4HBr ----> Br2 + MnBr2 + 2H2O
MnO2 + 4HBr ----> Br2 + MnBr2 + 2H2O
ON Mn=+4
H= +1
Br= 0 Mn=+2
O= -2 Br= -1
Br= -1 O= -2
H= +1
Separate the reduction and oxidation half reactions in following chemical equations.
a) Zn + 2HCl ----> ZnCl2 + H2
b) MnO2 + 4HBr ----> Br2 + MnBr2 + 2H2O
c) 10K + 2KNO3 ----> N2 + 6K2O
a) Zn + 2HCl
----> ZnCl2 +
2+
Mn+4
----> Mn
H2
+
2eZn +
2HCl ----> ZnCl2 + H2
The half-reactions should include the
ON
Zn= 0 H= +1
Zn=
species present in the solution, therefore,
+2
H=0
Mn+4 is included as MnO2. In order to
Cl= -1 Cl= -1
balance the half-reaction and to account for
Oxidation Half Reaction(OHR):
2O2- from MnO2 we have to add 4H+ ions
2+
Zn ----> Zn
+ 2e
to the left and 2H2O to the right.
Reduction Half Reaction(RHR):
Reduction Half Reaction(RHR):
2H+ +
2e----> 2H2
4H+ + MnO2 + 2e- ----> Mn2+
+
2H2O
Mn=+4
Mn=+2
b) MnO2 + 4HBr ----> Br2 + MnBr2 +
Oxidation Half Reaction(OHR):
2H2O
MnO2 + 4HBr ----> Br2 + MnBr2
2Br- -----> Br2 + 2e+ 2H2O
Br=-1
Br=0
ON Mn=+4 H=+1
Br= 0
Mn=+2
H= +1
2
O= -2 Br= -1
Br=
-1
B
O= -2
r
18
>
B
r
=
B
r
2
0
+
2
e
c) 10K + 2KNO3 ----> N2 + 6K2O
10K + 2KNO3 ----> N2 + 6K2O
ON K= 0 K= +1 N= 0 K= +1
N= +5
O= -2
O= -2
Oxidation Half Reaction(OHR):
-
B
r
=
10K ----> 10K + +
10eK=0
K= +1
Reduction Half Reaction(RHR):
2NO3- + 10e- ---->
N2 + 6O-2
N=+5
N=0
O= -2
1
Solution Concentration Units
a) Molarity (M)
Molarity (M)
=
moles of solute
------------------Liters of solution
b) mole fraction (a)
moles of solute (substance)
a
= -------------------------------------------------moles of solute + moles of solvent
c) Mass percent (% weight)
Mass Solute
Mass percent = -----------------------Mass of solution
d Volume percent (% volume)
Volume Solute
Volume percent = -----------------------Volume of solution
e) "Proof"
Proof = Volume percent x 2
f) ppm and ppb
19
ppm =
ppb
=
mass of solute
------------------------ x 106
mass of solvent
mass of solute
------------------------ x 109
mass of solvent
Molarity
Describe and define the following term used to express solution composition:
a) Molarity(M)
moles of solute
Molarity (M)
=
------------------Liters of solution
Why is molarity an important conversion factor in chemical stoichiometry?
Molarity of solution is defined as number of solutes moles per liter of solution:
moles of solute
Molarity (M)
=
---------------------Liters of solution
The moles of compounds participating in a chemical reaction have to be calculated from the volume of the
solution, when solutions are mixed to do chemical reactions.
Molarity of the solutions provides a convenient conversion factor in all stoichiometric calculation carried
out during solution chemistry.
Calculate the molarity of a solution prepared by dissolving 200.0 g of K2SO4 in enough water to make 500.0
mL solution.
moles of solute
500.0 mL = ? Liters of solution
Molarity(M) = ---------------------Liters of solution
500.0 mL
1L
1000 mL
solute = K2SO4; m.w. = 174.27 g/mol
= 0.5 L
moles of K2SO4 = ?
1.148 mol K2SO4
200 g K2SO4
1 mol K2SO4
Molarity of K2SO4 solution = -----------------------174.27 g K2SO4
0.5 Liters of solution
= 1.148 mol K2SO4
= 2.30 mole/Liter = 2.30 M (M = moles/liters)
How many grams of KNO3 are contained in 500 mL of a 0.500 M solution of potassium nitrate?
0.500 mol KNO3
Molarity (M) of 0.500 M KNO3 solution means = -----------------------1.0 Liters of solution
M.W. of KNO3 = 101.11 g/mol
First find out the moles and then calculate KNO3 grams.
500 mL
1L
0.500 mol KNO3
1000 mL
mL --> L
conversion
= 25.28 g KNO3
1L
L --> mol
conversion
20
101.11 g KNO3
1 mol KNO3
mol --> gram
conversion
Preparing Solutions
There are two ways solutions can be prepared: from a pure solute and from another solution.
The first method involves measuring an amount of a pure solute. An example of this would be
the sucrose solution described earlier in this module. If you dissolve 21.7 grams of pure sucrose
(C12 H22O11; M.W. 342 g/mol) in water until the total volume of the solution was 500 mL. The
concentration of this solution will be be (21.7/342)/ 0.5 = 0.127 M.
Dilutions
A similiar solution could be prepared using the second method. Suppose a 1.0 M sucrose
solution was available to you instead of pure, solid sucrose. In using this method, it is important
to realize that adding water to a solution does not change the number of moles of solute that is
present. Since it increases the total volume of the solution, the concentration will decrease
(since concentration is expressed in moles of solute per liters of solution). For example, if you
took one liter of the 1.0 M sucrose solution and added water until the total volume was equal to
two liters, the resulting solution would still contain one mole of sucrose. However, since the
volume of solution has increased to two liters, the concentration would now be 0.50 M. This can
be expressed mathematically:
initial moles = final moles
ni = nf
ni = MiVi and nf = MfVf
MiVi = MfVf
Mi = initial molarity
Vi = initial volume
Mi = 1.00
Mf = final molarity
Vi = 1L
Vf = final molarity
Mf = ?L
Vf = 2.00
Mf = MiVi/ Vf = (1.00 x 1)/2.0 = 0.5 M
If you wanted to prepare a 500 mL of a sucrose solution similiar to the solution
prepared previously (concentration = 0.127 M), what volume of a 1.0 M sucrose
solution would you use?
Mi = 1.00
Vi = ? L
Mf = 0.127 L
Vf = 0.5 L
MiVi = MfVf
MfVf / Mi
= (0.127 x 0.5)/1.0 = 0.0635 L or 63.5 mL
You need to add 436.5 mL of water to 3.5 mL of 1.0 M sucrose solution to a final
voluvem of 500 mL.
Dilutions
21
How many mL of 2.00 M solution of HNO3 are required with water to make a 250 mL of
1.50 M nitric acid solution?
This is a dilution problem. You have to add water to certain volume of HNO3 and make
a 250 mL solution with the molarity, 1.50 M HNO3.
Equation is: MiVi = MfVf
Mi = initial molarity
Vi = initial volume
Mi = 2.00
Mf = final molarity
Vi = ?
Vf = final molarity
Mf = 1.50
250
Vf = ------------= 0.25 Liters
1000
Vi =
MfVf
--------------Mi
1.50 x 0.25
= ---------------- = 0.1875 L
2.00
0.1875 L = 187.5 mL of 2.00 M solution.
Solution stoichiometry
Same idea as for stoichimetry of pure substances: Balanced equation! Moles!
However need to use concentrations and volumes to find numbers of moles.
Remember that the key to stoichiometry is to find moles! In solutions, we can find moles by
using the molarity (M), the most common expression for solution concentration. Molarity is
defined as follows:
solution of Liters solute moles Molarity (in words “molarity equals moles per liter”)
For solutions, then ! moles of solute = Molarity x Liters (i.e. mol = M x L)
How many mLs of 0.100 M BaCl2 are required to react completely with 25 mL
of 0.200 M Fe2(SO4)3?
This is a stoichiometric calculation involving a precipitation reaction. First, write equation and
balance.
3 BaCl2(aq) + Fe2(SO4)3(aq) ------> 3 BaSO4(s) + 2 Fe Cl3(aq)
3 BaCl2 = 1 Fe2(SO4)3
Then calculate the moles of Fe2(SO4)3 in 25 mL of 0.200 M solution:
25 mL
1L
0.200 mol Fe2(SO4)3
1000 mL
1L
= 0.005 mole Fe2(SO4)3
22
Then convert Fe2(SO4)3 to BaCl2 mole, BaCl2 moles to liters and liters to mL.
0.005 mol Fe2(SO4)3
3 mol BaCl2
1 mol
Fe2(SO4)3
= 150 mL of BaCl2
23
1L
1000 mL
0.100 mol BaCl2
1L
Titrations
How can we know the concentration of some solution of interest? One answer to this
problem lies in the method of titration. In titration we will make use of a second solution
known as a standard solution which has the following characteristics:
1.The second solution contains a chemical which reacts in a defined way, with known
stoichiometry, with the
solute of the first solution
2.The concentration of the solute in this second solution is known.
Classic titrations include so-called acid-base titrations. In these experiments a solution of
an acid with an unknown concentration is titrated with a solution of known concentration
of base (or vice versa). For example, we may have a solution of hydrochloric acid (HCl)
of unknown concentration and a standard solution of NaOH. To a fixed amount of the
HCl solution is added incremental amounts of the NaOH solution until the acid is
completely neutralized - i.e. a
stoichiometrically equivalent quantity of HCl and NaOH have been combined. This is
known as the equivalence point in the titration. By knowing the concentration of the
standard solution, and the amount added to achieve stoichiometric equivalency, we can
determine the amount of moles of HCl in the original sample volume.
How do we know when we have reached the equivalence point in such a titration
experiment?
In this type of acid-base titration, so called indicator-dyes are used. For example
phenolphthalein is colorless in acidic solutions and turns red in basic solutions. Thus, in
the above experiment we will add a small amount of this indicator-dye and add base until
we barely begin to see a color change to red.
Acid–base indicator
An acid or base which exhibits a colour change on neutralization by the
basic or acidic titrant at or near the equivalence point of a titration.
How many mLs of 0.300 M NaOH are required to react with 500 mL of
0.170 M H2SO4 completely?
Calculate the moles of H2SO4 in 500 mL of 0.170 M solution. 2 NaOH = 1 H2SO4 from
the chemical equation
500 mL
1L
.170 mol H2SO4
1000 mL
1L
= 0.085 mole
H2SO4
Then convert, H2SO4 moles to NaOH moles, NaOH moles to liters of NaOH
solution and finally to mL of NaOH solution.
0.85 mol H2SO4
2 mol NaOH
1L
24
1000 mL
1 mol H2SO4
0.300 mol NaOH
1L
H2SO4 mole to NaOH mole NaOH mol to NaOH liters
= 566.7 mL
25
26