Workbook for General Chemistry II

GENERAL CHEMISTRY II
SOLVED PROBLEMS IN
CHEMISTRY
VON GODWIN C. JEQUINTO

2
TABLE OF CONTENTS
CHAPTER 1: CHEMICAL EQUILIBRIUM.....................................4
1.1 The Chemical Equilibrium Condition ...........................7
1.2 Writing the Reaction Quotient / Equilibrium Constant
Expression...............................................................................8
1.3 Predicting the Direction of a Reaction.......................10
1.4 Significance of the Equilibrium Constant...................11
1.5 Le Chatelier’s Principle..............................................12
WORKSHEET #1.........................................................................13
CHAPTER 2: ACID-BASE EQUILIBRIA AND SALT EQUILIBRIA .......21
2.1 The Chemical Equilibrium Condition .........................25
2.2 The Acid-Base Properties of Water ...........................26
2.3 pH- A Measure of Acidity ..........................................27
2.4 Strength of Acids and Bases......................................28
2.5 Weak Acids/Weak Bases ...........................................29
2.6 Relationship between the Ionization Constants of Acids
and their Conjugate Bases.....................................................30
2.7 The Common Ion Effect ............................................31
2.8 Buffer Solutions ........................................................32
2.9 Solubility Equilibria....................................................33
WORKSHEET #2.........................................................................34

3
CHAPTER 3: ELECTROCHEMISTRY..........................................37
3.1 Redox Reactions........................................................39
3.2 Galvanic Cells ............................................................40
3.3 Standard Reduction Potentials ..................................42
3.4 Spontaneity of Redox Reactions................................44
3.5 Batteries....................................................................46
3.6 Corrosion...................................................................47
3.7 Electrolysis ................................................................48
WORKSHEET #3.........................................................................49
REFERENCES..............................................................................51

4
CHAPTER 1: CHEMICAL EQUILIBRIUM
Chemical Equilibrium describes the state in
which the rates of forward and reverse reactions
are equal and the concentrations of the reactants
and products remain unchanged with time. This
state of dynamic equilibrium is characterized by an
equilibrium constant. Depending on the nature of
reacting species, the equilibrium constant can be
expressed in terms of molarities (for solutions) or
partial pressures (for gases). The Equilibrium
constant provides information about the net
direction of a reversible reaction and the
concentrations of the equilibrium mixture.

5
Key Points:
• A reversible reaction can proceed in both
the forward and backward directions
• Equilibrium is when the rate of the forward
reaction equals the rate of the reverse
reaction. All reactant and product
concentrations are constant at equilibrium.
• For reactions that are not at equilibrium, we
can write a similar expression called the
reaction quotient Q, which is equal to K.
• Kc and Q can be used to determine if a
reaction is at equilibrium, to calculate
concentrations at equilibrium, and to
estimate whether a reaction favors products
or reactants at equilibrium.

6
The General Equilibrium Expression for a Reaction:
aA+bB⇋cC+dD
CHAPTER OUTLINE
1.1 The Chemical Equilibrium Condition
The Concept of Equilibrium
1.2 Writing the Reaction Quotient /
Equilibrium Constant Expression
Homogeneous Equilibria • Equilibrium Constants
and Units •Heterogeneous Equilibria • The Form
of K and the Equilibrium Equation • Summary of
Rules for Writing Equilibrium Constant
Expressions
1.3 Predicting the Direction of a Reaction
Reaction Quotient
1.4 Significance of the Equilibrium Constant
Significance of Equilibrium Concentrations
1.5 Le Chatelier’s Principle
The Equilibrium Law

7
A reversible reaction can proceed in both the
forward and backward directions. Most reactions are
theoretically reversible in a closed system, though some
can be considered to be irreversible if they heavily favor the
formation of reactants or products. The double half-arrow
sign we use when writing reversible reaction equations, ⇋
, is a good visual reminder that these reactions can go either
forward to create products, or backward to create
reactants.
EXAMPLE:
The formation of
Nitrogen dioxide (NO2)
from Dinitrogen
tetroxide (N2O4):
N2O4 (g) ⇋ 2NO2 (g)
Imagine we added some colorless N2O4 (g) to an
evacuated glass container at room temperature. If we kept
our eye on the vial over time, we would observe the gas in
the ampoule changing to a yellowish orange color and
gradually getting darker until the color stayed constant. We
can graph the concentration of NO2 and N2O4 over time
for this process, as you can see in the graph.

8
1.2 Writing the Reaction Quotient / Equilibrium
Constant Expression
The equilibrium constant expression is the ratio of
the concentrations of a reaction at equilibrium. Each
equilibrium constant expression has a constant value
known as K, the equilibrium constant. When dealing with
partial pressures, Kp is used, whereas when dealing with
concentrations (molarity), Kc is employed as the
equilibrium constant. Reactions containing pure solids and
liquids results in heterogeneous reactions in which the
concentrations of the solids and liquids are not considered
when writing out the equilibrium constant expressions.
𝐾𝑐 =
[C]c
[D]d
[A]a
[B]b 𝐾𝑝 =
𝑃 𝐶
𝑐
𝑃 𝐷
𝑑
𝑃 𝐴
𝑎 𝑃 𝐵
𝑏
There are two different types of equilibria (homogeneous
and heterogeneous) separately, because the equilibrium
constants are defined differently.
• A homogeneous equilibrium has everything
present in the same phase.
• A heterogeneous equilibrium has things present in
more than one phase.

9
If the reaction involved partial pressures instead of
concentrations, then the equilibrium constant, now Kp,
would follow the same formula except with the pressures
of the gases written in. In that case, Kp is only applied to
reactions involving gases. Recall the equation relating Kp
and Kc:
Kp = Kc(RT)Δn
Where:
• R is the Ideal Gas Constant (0.0821 L atm
mol-1 K-1),
• T is the temperature in Kelvins, and
• Δn is the Sum of Coefficients of gaseous
Products - Sum of Coefficients of gaseous
Reactants.
Keep in mind the following facts:
1. The KP expression applies only to gaseous
reactions and
2. The concentration of solvent (usually water) does
not appear in the equilibrium constant expression.
3. If there are no gases present, KP does not apply
and we have only Kc.

10
1.3 Predicting the Direction of a Reaction
When we deal with reactions that are reversible in
nature, it is important to figure out the direction of the
reaction at a given instance. For example, in the formation
of ammonia at commercial scale in an industry from
nitrogen and hydrogen, we need to optimize the process
for efficient production. Thus, it is important to predict the
state of the reaction at any given instant.
For any chemical reaction,
• If Qc>Kc, the reaction goes from right to
left.
• If Qc<Kc, the reaction goes from left to
right.
• If Qc=Kc, the reaction is at equilibrium.

11
1.4 Significance of the Equilibrium Constant
For any given temperature, there is only one value
for the equilibrium constant. Kc only changes if the
temperature at which the reaction occurs changes. You can
make some predictions about the chemical reaction based
on whether the equilibrium constant is large or small.
If the value for Kc is very large, then the equilibrium
favors the reaction to the right and there are more products
than reactants. The reaction may be said to be "complete"
or "quantitative."
If the value for the equilibrium constant is small,
then the equilibrium favors the reaction to the left and
there are more reactants than products. If the value of Kc
approaches zero the reaction may be considered not to
occur.
If the values for the equilibrium constant for the
forward and reverse reaction are nearly the same then the
reaction is about as likely to proceed in one direction and
the other and the amounts of reactants and products will
be nearly equal. This type of reaction is considered to be
reversible.

12
1.5 Le Chatelier’s Principle
A statement of Le Chatelier's Principle:
Le Chatlier's principle (also known as "Chatelier's
principle" or "The Equilibrium Law") states that when a
system experiences a disturbance (such as concentration,
temperature, or pressure changes), it will respond to
restore a new equilibrium state. For example, if more
reactants are added to a system, Le Chatlier's principle
predicts that the reaction will generate more products to
offset the change and restore equilibrium.
Le Chatelier's principle states that if a dynamic
equilibrium is disturbed by changing the conditions, the
position of equilibrium shifts to counteract the change to
reestablish an equilibrium. If a chemical reaction is at
equilibrium and experiences a change in pressure,
temperature, or concentration of products or reactants,
the equilibrium shifts in the opposite direction to offset the
change. This page covers changes to the position of
equilibrium due to such changes and discusses briefly why
catalysts have no effect on the equilibrium position.

13
WORKSHEET #1
1. Calculate the equilibrium Constant of the following
reaction: 2NO (g) + 2H2(g) ⇌ N2(g) + 2H2O(g).
Where: NO= 0.789M, H2= 0.016M, N2=
0.214M, H2O= 0.14M
reaction: C2H5OH(l) + O2(g) ⇌ 2CO2(g) + H2O(g)
Where: C2H5OH= 0.189M, O2= 0.19M, CO2=
0.712M, H2O= 0.18M
reaction: C2H4 (g) + 2O2 (g) ⇌ CO2(g) + 2H2O(g).
Where: C2H4= 0.28M, O2= .0085M, CO2=
2.12M, H2O= 0.12M
4. If 9.0 grams of hydrogen peroxide is placed within a
sealed 750 milliliter container at 618 K. What is the
pressure of the oxygen gas produced in
atmospheres?
2H2O2 (g) ⇌ 2 H2O (g) +O2 (g)
5. The equilibrium constant Kp for the reaction:
2CO2(g) ⇌ 2CO(g)+O2(g) is 217 at 273 K. Calculate
PO2 if PCO2 = 0.200 atm ang PCO = 0.370 atm.
6. For the reaction 2N2(g)+6H2(g) ⇌ 4NH3(g), Kp is 2.73
x 10-5 at 372K. Calculate Kc for the reaction at this
temperature.

14
7. For the reaction 2H2(g)+CO(g) ⇌ CH3OH(g), The Kc
of this reaction is 6.18 at 703K. What is the value of
KP at this temperature?
8. At the start of a reaction, there are 0.249 mol H2,
0.00618 mol Cl2, and 0.61816 mol HCl in a 1.50L
reaction vessel at 275oC. If the equilibrium constant
(Kc) for the reaction H2(g)+Cl2(g) ⇌ 2HCl(g) is 284.78
at this temperature, determine if the system is at
equilibrium. If not, predict which way the net
reaction will proceed.
9. At the start of a reaction, there are 0.718 mol H2,
0.82 mol I2, and 0.0.0052 mol HI in a 4.2L reaction
vessel at 212oC. If the equilibrium constant (Kc) for
the reaction H2(g)+I2(g) ⇌ 2HI(g) is 2.35 at this
temperature, determine if the system is at
equilibrium. If not, predict which way the net
reaction will proceed.
10. At the start of a reaction, NO= 0.789M, H2= 0.016M,
N2= 0.214M, H2O= 0.14M in a 1.8L reaction vessel
at 178oC. If the equilibrium constant (Kc) for the
reaction:
2NO (g) + 2H2 (g) ⇌ N2 (g) + 2H2O (g)
is 10.18 at this temperature, determine if the
system is at equilibrium. If not, predict which way
the net reaction will proceed.

15
SOLUTION:
1. Given: 2NO + 2H2 ⇌ N2 + 2H2O
Where: NO= 0.789M, H2= 0.016M, N2= 0.214M, H2O=
0.14M
Solution: 𝐾𝑐 =
[C]c[D]d
[A]a[B]b
𝐾𝑐 =
[0.214][0.14]
[0.789]2[0.016]2
𝐾𝑐 = 187.996
Answer : The equilibrium Constant Kc = 187.996
2. Given: C2H5OH(l) + O2(g) ⇌ 2CO2(g) + H2O(g)
Where: C2H5OH= 0.189M, O2= 0.19M, CO2= 0.712M,
H2O= 0.18M
[C]c[D]d
[A]a[B]b
𝐾𝑐 =
[0.712]2
[0.18]
[0.19]
𝐾𝑐 = 0.48

16
3. Given: C2H4 (g) + 2O2 (g) ⇌ CO2(g) + 2H2O(g)
Where: C2H4= 0.28M, O2= .0085M, CO2= 2.12M, H2O=
0.12M
[C]c[D]d
[A]a[B]b
𝐾𝑐 =
[2.12][0.12]2
[0.28][0.0085]2
𝐾𝑐 = 1509.046
4. 2H2O2 ⇌ 2 H2O +O2
Given: V= 750 m, T= 618 K, grams H2O2
Find: Pressure of the O2 produced in atmospheres
a. First, we need to determine the moles of O2
produced.
(9g H2O2 ) × (
1 𝑚𝑜𝑙 H2O2
34𝑔 H2O2
) × (
1 𝑚𝑜𝑙 O2
2 𝑚𝑜𝑙 H2O2
)
= 0.13 𝑚𝑜𝑙 O2
b. With the moles of Oxygen determined, we can now use
the Ideal Gas Law to determine the pressure.
PV = nRT

17
c. The Volume (750 mL = 0.75 L) and temperature (618
K) are already given, and R (0.0821 Latm mol-1K-1) is a
constant.
𝑃 =
𝑛𝑅𝑇
𝑉
𝑃 =
(0.13 𝑚𝑜𝑙 O2) × (0.0821 L atm mol−1
K−1 ) × (618K)
0.75𝐿
∴ 𝑷 = 8.79 atm
5. 2CO2(g) ⇌ 2CO(g)+O2(g)
Given: Kp= 217 PCO = 0.370
atm
PCO2 = 0.200 atm
Solution: 𝐾𝑝 =
𝑷PCO 𝑷 𝑶𝟐
𝑷 𝑪𝑶𝟐
𝑃𝑂2 =
𝑲𝒑 𝑷 𝑪𝑶𝟐
𝑷PCO
𝑃𝑂2 =
(217)(0.200 𝑎𝑡𝑚)
0.370 𝑎𝑡𝑚
∴ 𝑷 𝑶𝟐 = 𝟏𝟏𝟕. 𝟐𝟗𝟕𝟐𝟗𝟕𝟑 𝒐𝒓 𝟏𝟏𝟕. 𝟑𝟎
6. 2N2(g)+6H2(g) ⇌ 4NH3(g)
Given: Kp = 2.73 x 10-5 T= 372K
Kp = Kc (0.0821xT)Δn

18
2.73 x 10
−5
= Kc (0.0821 L atm mol−1
k−1
× 372K)−4
2.73 x 10−5
(0.0821×372K)−4
= Kc
∴ 𝐊𝐜 = 𝟐𝟑. 𝟕𝟓
7. 2H2(g)+CO(g) ⇌ CH3OH(g)
Given: Kc = 6.18 T= 703K
Kp = Kc(RT)Δn
Kp = 6.18 × (0.0821 L atm mol−1
k−1
× 703K)−2
∴ 𝐊𝐩 = 1.86 x 10-3
8. H2(g)+Cl2(g) ⇌2HCl(g)
Given: 0.249 mol H2 0.00618 mol Cl2
Kc= 284.78
0.61816 mol HCl V= 1.5 L
The initial concentrations of the reacting species
H2 =
0.249
1.5
= 0.166 M

19
Cl2 =
0.00618
1.5
= 4.12 x 10−3
M
HCl =
0.61816
1.5
= 0.41 M
Solving Qc
𝑄 𝑐 =
[0.41]2
[0.166][4.12 × 10−3]
𝑄 𝑐 = 245.79
Because Qc is smaller than Kc, the system is not at
equilibrium. The net result will be an increase in the
concentration of HCl and a decrease in the concentrations
of H2 and Cl2. The net reaction will proceed from left to right
until equilibrium is reached.
9. H2(g)+I2(g) ⇌ 2HI(g)
Given: 0.718 mol H2 0.82 mol I2 Kc=
2.35 V= 4.2L
0.0052 mol HI
The initial concentrations of the reacting species
H2 =
0.718
4.2
= 0.17 M HI =
0.0052
4.2
= 1.24 x 10−3
M
I2 =
0.82
4.2
= 0.2 M

20
Solving Qc:
𝑄 𝑐 =
[1.24 x 10−3
]2
[0.17][0.2]
𝑄 𝑐 = 0.04
Because Qc is smaller than Kc, the system is not at
concentration of HI and a decrease in the concentrations of
H2 and I2. The net reaction will proceed from left to right
until equilibrium is reached.
10. 2NO (g) + 2H2 (g) ⇌ N2 (g) + 2H2O (g)
Given: NO= 0.789M N2= 0.214M Kc= 10.18
H2= 0.016M H2O= 0.14M
Solve Qc:
𝑄𝑐 =
[0.214𝑀][0.98𝑀]2
[0.789𝑀]2[0.16]2
𝑄𝑐 = 12.9
Because Qc is bigger than Kc, the system is not at
concentrations of NO and H2 and a decrease in the
concentrations of N2 and H2O. The net reaction will proceed
from left to right until equilibrium is reached.

21
CHAPTER 2: ACID-BASE EQUILIBRIA AND SALT
EQUILIBRIA
On first encounter, the study of acid-base equilibria
is a little like a strange land with seemingly confusing trails
that make passage difficult. In fact, there is a road map that,
once understood, allows us to navigate acid-base equilibria
with confident precision and so become masters of its
domain. Here is an overview of this road map.
Fundamentally, aqueous acid-base equilibria are just a
particular example of the ideas and techniques we have
already learned in the study of gas phase chemical
equilibria.
However, there are two aspects that complicate the
application of these ideas. First, because the autoionization
of water is always present in aqueous solution, the analysis
of aqueous acid-base equilibria must always take into
account at least two competing equilibria, the acid or base
ionization and the water autoionization. Second, because
we will be interested in how acid-base equilibria respond to
changes in the system (typically by adding additional base
or acid), we need also to be able to separate the chemical
reactions that take place when things are combined from
the subsequent equilibration of the reaction product. The
key idea is to let what are combined react 100% as a limiting
reagent problem as a preliminary step done before
equilibration.

22
Key Points:
• Acid is a substance that contains hydrogen and
dissociates in water to yield a hydronium ion.
• A base is a substance that contains the hydroxyl group
and dissociates in water to yield : OH-
• Acid/Base theory and equilibria focus on the
relationship of these two ions and how they each affect
the equilibria of a whole class of compounds known as
acids and bases.
• Weak base is a chemical base that does not ionize fully
in an aqueous solution.
• The acid dissociation constant (Ka) quantifies the extent
of dissociation of a weak acid. The larger the value of
Ka, the stronger the acid, and vice versa.
• The base dissociation constant (or base ionization
constant) Kb quantifies the extent of ionization of a
weak base. The larger the value of Kb , the stronger the
base, and vice versa.
The General Weak Acid-Base Equilibria Formulas:
For a generic monoprotic weak acid HA with
conjugate base A- , the equilibrium constant has the
form: 𝐊 𝐚 =
[𝐇 𝟐 𝐎+][𝐀−]
[𝐇𝐀]
For a generic weak base B with conjugate acid BH+,
the equilibrium constant has the form: 𝐊 𝐛 =
[𝐁𝐇+][𝐎𝐇−]
[𝐁]

23
CHAPTER OUTLINE
2.1 Bronsted Acids and Bases
2.2 The Acid-Base Properties of Water
2.3 pH- A Measure of Acidity
2.4 Strength of Acids and Bases
Strong Acid–Strong Base Titrations •Weak Acid–
Strong Base Titrations •Strong Acid–Weak Base
Titrations
2.5 Weak Acids/Weak Bases
2.6 Relationship between the Ionization Constants of
Acids and their Conjugate Bases
2.7 The Common Ion Effect
2.8 Buffer Solutions
2.9 Solubility Equilibria
Solubility Product • Molar Solubility and Solubility
•Predicting Precipitation Reactions

24
ESSENTIAL CONCEPTS
 Buffer Solutions. A buffer solution contains a weak acid
and a salt derived from the acid. To maintain a
relatively constant pH, the acid and base components
of the buffer solution react with added acid or base.
Buffer solutions play an important role in many
chemical and biological processes.
 Acid-Base Titrations. The characteristics of an acid-
base titration depend on the strength of the acid and
base involved. Different indicators are used to
determine the end point of a titration.
 Solubility Equilibria. Another application of the
equilibrium concept is the solubility equilibria of
sparingly oluble salts, which are expressed as the
solubility product. The solubility of such a substance
can be affected by the resence of a common cation or
anion, or the pH. Complex-ion formation, an example
of the Lewis acid-base type eaction, increases the
solubility of an insoluble salt.

25
In 1923, chemists Johannes Nicolaus Brønsted and
Thomas Martin Lowry independently developed
definitions of acids and bases based on the compounds'
abilities to either donate or accept protons ( H+ ions). In
this theory, acids are defined as proton donors; whereas
bases are defined as proton acceptors. A compound that
acts as both a Brønsted-Lowry acid and base together is
called amphoteric.
The Bronsted-Lowry Theory of acids and bases states
that:
• An acid is a proton donor, any species that
donates an H+ ion. An acid must contain H in its
formula; HNO3 and H2PO4- are two examples, all
Arrhenius acids are Bronsted-Lowry acids.
• A base is a proton acceptor, any species
that accepts an H+ ion. A base must contain a
lone pair of electrons to bind the H+ ion; a few
examples are NH3, CO3-2, F -, as well as OH- .
Bronsted-Lowry bases are not Arrhenius bases,
but all Arrhenius bases contain the Bronsted-
Lowry base OH-.

26
2.2 The Acid-Base Properties of Water
Acids
Acids have long been recognized as a distinctive class of
compounds whose aqueous solutions exhibit the following
properties: A characteristic sour taste, changes the color of
litmus from blue to red, reacts with certain metals to
produce gaseous H2, and reacts with bases to form a salt
and water. Acidic solutions have a pH less than 7, with
lower pH values corresponding to increasing acidity.
Common examples of acids include acetic acid (in vinegar),
sulfuric acid (used in car batteries), and tartaric acid (used
in baking).
Bases
There are three common definitions of bases:
Arrhenius base: any compound that donates an hydroxide
ion (OH–) in solution, Brønsted-Lowry base: any compound
capable of accepting a proton, and Lewis base: any
compound capable of donating an electron pair. In water,
basic solutions will have a pH between 7-14.
H2O + H2O ⇌ H3O + +OH−
This is an example of autoprotolysis (meaning “self-
protonating”) and it exemplifies the amphoteric nature of
water (ability to act as both an acid and a base).
Autoprotolysis of water, the self-ionization of water
produces hydronium and hydroxide ions in solution.

27
2.3 pH- A Measure of Acidity
pH is a measure of hydrogen ion concentration; a
measure of the acidity or alkalinity of a solution. The pH
scale usually ranges from 0 to 14. Aqueous solutions at 25°C
with a pH less than seven are acidic, while those with a pH
greater than seven are basic or alkaline. A pH level of is 7.0
at 25°C is defined as 'neutral'. Very strong acids may have a
negative pH, while very strong bases may have a pH greater
than 14.
pH Equation: To calculate the pH of an aqueous
solution you need to know the concentration of the
hydronium ion in moles per liter (molarity). The pH is then
calculated using the expression:
pH = -log[H+]
where log is the base-10 logarithm and [H+] stands for the
hydrogen ion concentration in units of moles per liter
solution.
OH Equation: To calculate the pOH of a solution
you need to know the concentration of the hydroxide ion in
moles per liter (molarity). The pOH is then calculated using
the expression:
pOH = - log [OH-]
Where log is the base-10 logarithm and [H+] stands for the
hydrogen ion concentration in units of moles per liter
solution.

28
2.4 Strength of Acids and Bases
All acids and bases do not ionize or dissociate to the
same extent. This leads to the statement that acids and
bases are not all of equal strength in producing H+ and OH-
ions in solution. The terms "strong" and "weak" give an
indication of the strength of an acid or base. The terms
strong and weak describe the ability of acid and base
solutions to conduct electricity. If the acid or base conducts
electricity strongly, it is a strong acid or base. If the acid or
base conducts electricity weakly, it is a weak acid or base.
The terms "strong" and "weak" give an indication
of the strength of an acid or base. The terms strong and
weak describe the ability of acid and base solutions to
conduct electricity. If the acid or base conducts electricity
strongly, it is a strong acid or base. If the acid or base
conducts electricity weakly, it is a weak acid or base. Acids
or bases with strong bonds exist predominately as
molecules in solutions and are called "weak" acids or
bases. Acids or bases with weak bonds easily dissociate
into ions and are called "strong" acids or bases.

29
2.5 Weak Acids/Weak Bases
Weak acids and bases are only partially ionized in
their solutions, whereas strong acids and bases are
completely ionized when dissolve in water. Some common
weak acids and bases are given here. Furthermore, weak
acids and bases are very common, and we encounter them
often both in the academic problems and in everyday life.
The ionization of weak acids and bases is a chemical
equilibrium phenomenon. The equilibrium principles are
essential for the understanding of equilibria of weak acids
and weak bases.
Because the acid is weak, an equilibrium expression
can be written. An acid ionization constant (Ka) is the
equilibrium constant for the ionization of an acid.
𝐾𝑎 =
[ 𝐻+][𝐴−]
[𝐻𝐴]
The numerical value of K_b is a reflection of the
strength of the base. Weak bases with relatively higher K_b
values are stronger than bases with relatively lower K_b
values.

30
2.6 Relationship between the Ionization Constants of
Acids and their Conjugate Bases
As with acids, bases can either be strong or weak,
depending on their extent of ionization.A weak base is a
base that ionizes only slightly in an aqueous solution.When
a weak base such as ammonia is dissolved in water, it
accepts an H + ion from water, forming the hydroxide ion
and the conjugate acid of the base, the ammonium ion.
𝑁𝐻3 + 𝐻2 𝑂 ⇆ 𝑁𝐻4
+
+ 𝑂𝐻−
An equilibrium expression can be written for the
reactions of weak bases with water. Because the
concentration of water is extremely large and virtually
constant, the water is not included in the expression. A base
ionization constant (𝐾𝑏) is the equilibrium constant for the
ionization of a base. For ammonia the expression is:
𝐾𝑏 =
[𝑁𝐻4
+
][𝑂𝐻−
]
[𝑁𝐻3]
The numerical value of 𝐾𝑏 is a reflection of the
strength of the base. Weak bases with relatively higher 𝐾𝑏
values are stronger than bases with relatively lower 𝐾𝑏
values.

31
2.7 The Common Ion Effect
The common ion effect states that in a chemical solution, if
the concentration of any one of the ions is increased, then,
according to Le Chatelier's principle, some of the ions in
excess should be removed from solution, by combining
with the oppositely charged ions. Some of the salt will be
precipitated until the ion product is equal to the solubility
product. In short, the common ion effect is the suppression
of the degree of dissociation of a weak electrolyte
containing a common ion.
The solubility products Ksp's are equilibrium constants in
hetergeneous equilibria (i.e., between two different
phases). If several salts are present in a system, they all
ionize in the solution. If the salts contain a common cation
or anion, these salts contribute to the concentration of the
common ion. Contributions from all salts must be included
in the calculation of concentration of the common ion. For
example, a solution containing sodium chloride and
potassium chloride will have the following relationship:
[Na+]+[K+]=[Cl−]
When NaCl and KCl are dissolved in the same
solution, the Cl− ions are common to both salts. In a system
containing NaCl and KCl, the Cl− ions are common ions.

32
2.8 Buffer Solutions
A buffer is an aqueous solution containing a weak acid and
its conjugate base or a weak base and its conjugate acid. A
buffer’s pH changes very little when a small amount of
strong acid or base is added to it. It is used to prevent any
change in the pH of a solution, regardless of solute. Buffer
solutions are used as a means of keeping pH at a nearly
constant value in a wide variety of chemical applications.
For example, blood in the human body is a buffer solution.
Buffer solutions are resistant to pH change because of the
presence of an equilibrium between the acid (HA) and its
conjugate base (A–). The balanced equation for this
reaction is: HA⇌H+
+A−
When some strong acid (more H+) is added to an
equilibrium mixture of the weak acid and its conjugate
base, the equilibrium is shifted to the left, in accordance
with Le Chatelier’s principle. This causes the hydrogen ion
(H+) concentration to increase by less than the amount
expected for the quantity of strong acid added. Similarly, if
a strong base is added to the mixture, the hydrogen ion
concentration decreases by less than the amount expected
for the quantity of base added. This is because the reaction
shifts to the right to accommodate for the loss of H+ in the
reaction with the base.

33
2.9 Solubility Equilibria
The solubility product is the equilibrium constant
representing the maximum amount of solid that can be
dissolved in aqueous solution. The general form of the
solubility product constant (Ksp) for the equation: aA (s) ⇌
bB (aq) + cC (aq) is Ksp = [B]b[C]c
The solubility product constant (Ksp) is the equilibrium
constant for a solid that dissolves in an aqueous solution.
All of the rules for determining equilibrium constants
continue to apply. An equilibrium constant is the ratio of
the concentration of the products of a reaction divided by
the concentration of the reactants once the reaction has
reached equilibrium. Consider this reaction:
𝐴𝑔𝐶𝑙(𝑠) → 𝐴𝑔+
(𝑎𝑞) + 𝐶𝑙−
(𝑎𝑞)
The equilibrium expression for the reaction is:
𝐾 =
[𝐴𝑔+][𝐶𝑙−
]
[𝐴𝑔𝐶𝑙]
Because the AgCl is a solid, its concentration before
and after the reaction is the same. The equilibrium
equation can therefore be rearranged as: 𝑲 𝒔𝒑 =
[𝑨𝒈+][𝑪𝒍−
]
For substances in which the ions are not in a 1:1 ratio,
the stoichiometric coefficients of the reaction become the
exponents for the ions in the solubility-product expression.
The general formula for a reaction 𝑎𝐴(𝑠) ⇌ 𝑏𝐵(𝑎𝑞) + 𝑐𝐶(𝑎𝑞)
is: 𝑲 𝒔𝒑 = [𝑩] 𝒃
[𝑪] 𝒄

34
WORKSHEET #2
1. Find the pH of a 0.0125 M HCl solution. The HCl is a
strong acid and is 100% ionized in water. The
hydronium ion concentration is 0.0025 M.
2. Calculate the pH for a specific [H+]. Calculate pH given
[H+] = 1.8 x 10-5 M
3. Calculate [H+] from a known pH. Find [H+] if pH = 7.8
4. What is the pOH of a solution that has a hydroxide ion
concentration of 6.18 x 10-5 M?
5. What is the hydroxide ion concentration in a solution
that has a pOH of 6.90?
[H+] = 6 M
[H+] = 2.89 M
8. Given the H+ concentration to be 6.18×106, what is the
pH? Identify if it is acidic, basic or neutral.
9. Express the solubility equilibrium constant of the
following dissociation equation: NaCl⇌ Na+ + Cl-
following dissociation equation: Fe2(SO4)3 ⇌ 2Fe3+ +
3SO4
2-

35
SOLUTION:
1. Given: 0.0025M HCl
Solution: pH = - log (0.0125)
= - ( - 2.90) = 2.90
Answer : pH= 2.90
2. Given: [H+] = 1.8 x 10-5 M
Solution: pH = - log (1.8 x 10-5)
= - ( - 4.74) = 4.74
Answer : pH= 4.74
3. Given: pH = 7.8
Solution: [H+] = 10-pH
[H+] = 10-7.8
[H+] = 1.58 x 10-8
Answer : [H+] = 1.58 x 10-8
4. Given: 6.18 x 10-5 M
pOH = - log [6.18 x 10-5]
= - ( - 4.22) = 4.21
Answer : pOH= 4.21
5. Given: pOH of 6.90
Solution: [OH-] = 10-pOH
[OH-] = 10-6.90
[OH-] = 1.26 x 10-7
Answer : [OH-] = 1.26 x 10-7
6. Given: [H+] = 6 M
Solution: pH = - log (6)
= - (-0.78) = 0.78

36
Answer : pH= 0.78
7. Given: [H+] = 2.89 M
Solution: pH = - log (2.89)
= - ( - 0.46) = 0.46
Answer : pH= 0.46
8. Given: [H+] = 6.18×106M
Solution: pH = - log (6.18×106)
= - ( - 6.79) = 6.79
Answer : pH= 6.79 and therefore ACIDIC
following dissociation equation. NaCl⇌ Na+ + Cl-
Answer : 𝐾𝑠𝑝 = [𝑁𝑎+][𝐶𝑙−
]2
following dissociation equation. Fe2(SO4)3 ⇌ 2Fe3+
+ 3SO4
2-
Answer : 𝐾𝑠𝑝 = [𝐹𝑒3+]2
[𝑆𝑂42
2−
]
3

37
CHAPTER 3: ELECTROCHEMISTRY
We encounter electrochemical cells in all facets of
our everyday lives from the disposable AA batteries in our
remote controls and the lithium-ion batteries in our
iPhones to the nerve cells strewn throughout our bodies.
There are two types of electrochemical cells: galvanic, also
called Voltaic, and electrolytic. Galvanic cells derives its
energy from spontaneous redox reactions, while
electrolytic cells involve non-spontaneous reactions and
thus require an external electron source like a DC battery
or an AC power source. Both galvanic and electrolytic cells
will consist of two electrodes (an anode and a cathode),
which can be made of the same or different metals, and an
electrolyte in which the two electrodes are immersed
Electrochemistry is the study of chemical processes
that cause electrons to move. This movement of electrons
is called electricity, which can be generated by movements
of electrons from one element to another in a reaction
known as an oxidation-reduction ("redox") reaction.
Key Points:
 Electrolytic processes are reactions in which
chemicalchanges occur on the passage of an
electrical current
 Galvanic or Voltaic processes are chemical reactions
that result in the production of electrical energy

38
CHAPTER OUTLINE
3.1 Redox Reactions
3.2 Galvanic Cells
3.3 Standard Reduction Potentials
3.4 Spontaneity of Redox Reactions
3.5 Batteries
3.6 Corrosion
3.7 Electrolysis

39
3.1 Redox Reactions
A redox reaction is a reaction that involves a change in
oxidation state of one or more elements. When a substance
loses an electron, its oxidation state increases; thus, it is
oxidized. When a substance gains an electron, its oxidation
state decreases, thus being reduced.
𝐻2 + 𝐹2 → 2𝐻𝐹
Can be rewritten as follows:
Oxidation reaction: 𝑯 𝟐 → 𝟐𝑯+
+ 𝟐𝒆−
Reduction reaction: 𝑭 𝟐 + 𝟐𝒆−
→ 𝟐𝑭−
Overall reaction: 𝑯 𝟐 + 𝑭 𝟐 → 𝟐𝑯+
+ 𝟐𝑭−
Oxidation is the loss of electrons, whereas reduction
refers to the acquisition of electrons, as illustrated in the
respective reactions above. The species being oxidized is
also known as the reducing agent or reductant, and the
species being reduced is called the oxidizing agent or
oxidant. In this case, 𝐻2 is being oxidized (and is the
reducing agent), while 𝐹2 is being reduced (and is the
oxidizing agent).

40
3.2 Galvanic Cells
Galvanic cells traditionally are used as sources of DC
electrical power. A simple galvanic cell may contain only
one electrolyte separated by a semi-porous membrane,
while a more complex version involves two separate half-
cells connected by a salt bridge. The salt bridge contains an
inert electrolyte like potassium sulfate whose ions will
diffuse into the separate half-cells to balance the building
charges at the electrodes. According to the mnemonic “Red
Cat An Ox”, oxidation occurs at the anode and reduction
occurs at the cathode. Since the reaction at the anode is the
source of electrons for the current, the anode is the
negative terminal for the galvanic cell.
In a cell with zinc sulfate anode and copper sulfate
cathode, Zn(s) is continuously oxidized, producing aqueous
Zn2+:
𝑍𝑛(𝑠) → 𝑍𝑛2+
(𝑎𝑞) + 2𝑒−
Conversely, in the cathode, Cu2+ is reduced and
continuously deposits onto the copper bar:
𝐶𝑢2+
(𝑎𝑞) + 2𝑒−
→ 𝐶𝑢(𝑠)
The cell diagram is a shorthand notation to represent
the redox reactions of an electrical cell. For the cell
described, the cell diagram is as follows:
𝑍𝑛(𝑠)|𝑍𝑛2+
(𝑎𝑞)‖𝐶𝑢2+
(𝑎𝑞)|𝐶𝑢(𝑠)

41
 A double vertical line (||) is used to separate the
anode half reaction from the cathode half reaction.
This represents the salt bridge.
 The anode (where oxidation occurs) is placed on the
left side of the ||.
 The cathode (where reduction occurs) is placed on
the right side of the ||.
 A single vertical line (|) is used to separate different
states of matter on the same side, and a comma is
used to separate like states of matter on the same
side.
These are the required parts for a galvanic cell:
• Two half cells
• Two electrodes
• One electrically conductive wire
• One salt bridge
• One device, usually an ammeter or a
voltmeter
For an oxidation-reduction reaction to occur, the
two substances in each respective half-cell are connected
by a closed circuit such that electrons can flow from the
reducing agent to the oxidizing agent. A salt bridge is also
required to maintain electrical neutrality and allow the
reaction to continue.

42
3.3 Standard Reduction Potentials
The standard reduction potential is the tendency for
a chemical species to be reduced, and is measured in volts
at standard conditions. The more positive the potential is
the more likely it will be reduced. The standard reduction
potential is in a category known as the standard cell
potentials or standard electrode potentials. The standard
cell potential is the potential difference between the
cathode and anode. For more information view Cell
Potentials. The standard potentials are all measured at 298
K, 1 atm, and with 1 M solutions.
The standard potentials are all measured at 298 K, 1
atm, and with 1 M solutions.It is written in the form of a
reduction half reaction. An example can be seen below
where "A" is a generic element and C is the charge. 𝑨 𝑪
+
+𝑪𝒆−
→ 𝑨
The standard oxidation potential is much like the
standard reduction potential. It is the tendency for a
species to be oxidized at standard conditions. It is also
written in the form of a half reaction, and an example is
shown: 𝑨(𝒔) → 𝑨 𝒄+
+ 𝑪𝒆−
The standard oxidation potential and the standard
reduction potential are opposite in sign to each other for
the same chemical species.

43
Copper's Standard Oxidation Potential
Cu(s)→Cu2++2e−
Eo
0 (SOP)= −0.34V
The standard oxidation potential and the standard
reduction potential are opposite in sign to each other for
the same chemical species.
Relation Between Standard Reduction Potential (SRP) and
the Standard Oxidation Potential (SOP)
Eo
0 (SRP)=−Eo0(SOP)
Standard reduction potentials are used to determine the
standard cell potential. The standard reduction cell
potential and the standard oxidation cell potential can be
combined to determine the overall cell potentials of a
galvanic cell. The equations that relate these three
potentials are shown below:
Eo
cell=Eo
reduction of reaction at cathode +
Eo
oxidation of reaction at anode
or alternatively
Eo
cell=Eo
reduction of reaction at cathode−Eo
reduction of
reaction at anode

44
3.4 Spontaneity of Redox Reactions
A positive voltage that forms across the electrodes of
a voltaic cell indicates that the oxidation-reduction reaction
is a spontaneous reaction for reduction at the cathode and
oxidation at the anode.
 Conversely, if the potentials of the half-cells are known,
then it is possible to predict whether a given redox
reaction will be spontaneous (i.e. result in a positive
voltage)
 A negative voltage indicates that the reverse reaction is
spontaneous (i.e. oxidation at the cathode, and
reduction at the anode; by convention you would need
to swap the labels on the electrodes)
 Basically, any current flow indicates that there is a
spontaneous redox reaction occurring in a voltaic cell.
The sign of the voltage indicates at which electrode the
reduction or oxidation is occurring. (electrons flow
towards the half-cell where reduction is occurring - by
convention, the cathode)

45
As long as we can identify the actual reduction and
oxidation processes that will occur in a redox reaction, the
general description of the standard reduction potential for
any redox reaction (and not just one occurring in a voltaic
cell) would be:
E0 = E0
red (reduction process) - E0 = E0
red (oxidation
process)
 Thus, E0 will be positive for the case where the
reaction is spontaneous
 E0 will be zero for a redox reaction at equilibrium
 E0 will be negative for the case where the reaction is
spontaneous in the reverse direction
 Note that there is no reference here to what is the
cathode and what is the anode

46
3.5 Batteries
Chemistry is the driving force behind the magics of
batteries.
A battery is a package of one or more galvanic cells used for
the production and storage of electric energy by chemical
means. A galvanic cell consists of at least two half cells, a
reduction cell and an oxidation cell. Chemical reactions in
the two half cells provide the energy for the galvanic cell
operations.
Each half cell consists of an electrode and an electrolyte
solution. Usually the solution contains ions derived from
the electrode by oxidation or reduction reaction.
We will make this introduction using a typical setup as
depicted here. The picture shows a copper zinc galvanic cell
(battery).
A galvanic cell is also called a voltaic cell. The spontaneous
reactions in it provide the electric energy or current.
Two half cells can be put together to form an electrolytic
cell, which is used for electrolysis. In this case, electric
energy is used to force nonsponaneous chemical reactions.

47
3.6 Corrosion
Corrosion is a process through which metals in
manufactured states return to their natural oxidation
states. This process is a reduction-oxidation reaction in
which the metal is being oxidized by its surroundings, often
the oxygen in air. This reaction is both spontaneous and
electrochemically favored. Corrosion is essentially the
creation of voltaic, or galvanic, cells where the metal in
question acts as an anode and generally deteriorates or
loses functional stability.
Corrosion can be defined as the deterioration of
materials by chemical processes. Of these, the most
important by far is electrochemical corrosion of metals, in
which the oxidation process M → M+ + e– is facilitated by
the presence of a suitable electron acceptor, sometimes
referred to in corrosion science as a depolarizer.
In a sense, corrosion can be viewed as the
spontaneous return of metals to their ores; the huge
quantities of energy that were consumed in mining,
refining, and manufacturing metals into useful objects is
dissipated by a variety of different routes.

48
3.7 Electrolysis
The use of electric current to stimulate a non-
spontaneous reaction. Electrolysis can be used to separate
a substance into its original components/elements and it
was through this process that a number of elements have
been discovered and are still produced in today's industry.
In Electrolysis, an electric current it sent through an
electrolyte and into solution in order to stimulate the flow
of ions necessary to run an otherwise non-spontaneous
reaction. Processes involving electrolysis include: electro-
refining, electro-synthesis, and the chloro-alkali process.
Faraday's Constant
The amount of electric charge associated with one mole of
electrons.
Faraday's Constant: 1 mole e- = 96,485 C (where
C easures Coulombs)
To find the amount of moles of electrons that have been
involved in an electrolysis reaction use the following
equation:
Charge (C)= current (C/s) x time (s) x (1 mole e-/96,485 C)

49
WORKSHEET #3
1. Express the oxidation half-reaction of the following
redox reaction: Cd+2 + Fe  Fe +2 + Cd
Answer: Cd +2 + 2 e -  Cd
2. Express the oxidation half-oxidation of the following
redox reaction: Zn + 2H+ → Zn2+ + H2
Answer: Zn → Zn2+ + 2 e -
3. Express the reduction half-reaction of the
following redox reaction: Fe +3 + H2 →
Fe +2 + H+
Answer: Fe +3 + 3e- → Fe +2
4. Determine which element is oxidized and which
element is reduced in the Zn + 2H+ → Zn2+ + H2
reaction.
Answer: Zn is oxidized (Oxidation number: 0 → +2);
H+ is reduced (Oxidation number: +1 → 0)
element is reduced in the 2Al + 3Cu2+→2Al3+ +3Cu
reaction
Answer: Al is oxidized (Oxidation number: 0 → +3);
Cu2+ is reduced (+2 → 0)
element is reduced in the CO3
2- + 2H+→ CO2 + H2O
reaction.

50
Answer: This is not a redox reaction because each
element has the same oxidation number in both
reactants and products: O= -2, H= +1, C= +4.
7. Determine the Oxidation State of the bold element
Na3PO3
Answer: The oxidation numbers of Na and O are +1
and -2. Because sodium phosphite is neutral, the
sum of the oxidation numbers must be zero. Letting
x be the oxidation number of phosphorus, 0= 3(+1) +
x + 3(-2). x=oxidation number of P= +3.
8. Determine the Oxidation State of the bold element
H2PO4
-
Answer: Hydrogen and oxygen have oxidation
numbers of +1 and -2. The ion has a charge of -1, so
the sum of the oxidation numbers must be -1. Letting
y be the oxidation number of phosphorus, -1= y +
2(+1) +4(-2), y= oxidation number of P= +5.
9. Determine what is oxidized in the Zn+2H+→Zn2++ H2
reaction.
Answer: The oxidation state of H changes from +1 to
0, and the oxidation state of Zn changes from 0 to +2.
Hence, Zn is oxidized and acts as the reducing agent.
10. What is reduced species in this reaction? Zn+
2H+→Zn2++ H2
Answer: The oxidation state of H changes from +1 to
0, and the oxidation state of Zn changes from 0 to +2.
Hence, H+ ion is reduced and acts as the oxidizing
agent.

51
REFERENCES
 Fidentify-solution-neutral-base-acidic-
8346.html&h=ATOmL5lCAi2-y_pOxAiOmvOTFi-
LsYrHSsAJesZu3mmUN12-
YsmbNsKPJCcP2vYK5MQEbzId9ql6WJhfiEx3XCEGEz
c-F0xxSkXk3gxUE6KvnpO-ecCA
 http://web.gccaz.edu/~lisys52871/00152note/16n
otes.pdf
 https://chem.libretexts.org/Core/Analytical_Chemi
stry/Electrochemistry/Redox_Chemistry/Oxidation-
Reduction_Reactions
 https://chem.libretexts.org/Core/Physical_and_Th
eoretical_Chemistry/Equilibria/Solubilty/Common_
Ion_Effect
eoretical_Chemistry/Acids_and_Bases/Ionization_
Constants/Weak_Acids_and_Bases
eoretical_Chemistry/Acids_and_Bases/Acid/Overvi
ew_of_Acids_and_Bases
eoretical_Chemistry/Acids_and_Bases/Acid/Overvi
ew_of_Acids_and_Bases
 https://courses.lumenlearning.com/boundless-
chemistry/chapter/acids-and-bases/
 https://courses.lumenlearning.com/boundless-
chemistry/chapter/solubility-equilibria/
 https://courses.lumenlearning.com/cheminter/cha
pter/acid-and-base-ionization-constants/

52
 https://sciencing.com/identify-solution-neutral-
base-acidic-8346.html
 https://www.chem.purdue.edu/gchelp/howtosolve
it/Equilibrium/Calculating_pHandpOH.htm
 https://www.chemguide.co.uk/physical/acidbasee
qia/buffers.html
 https://www.khanacademy.org/science/chemistry/
acids-and-bases-topic
 https://www.thoughtco.com/how-to-calculate-ph-
quick-review-606089

Workbook for General Chemistry II

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Similar to Workbook for General Chemistry II

Similar to Workbook for General Chemistry II (20)

More from Mariz Pascua

More from Mariz Pascua (17)

Recently uploaded

Recently uploaded (20)

Workbook for General Chemistry II