Modules-Rev

MODULE IN
CHEMICAL ENGINEERING
CHE 1221
lec
Department of Chemical Engineering
SCHOOL OF ENGINEERING AND ARCHITECTURE

Prepared by: LRRAMOS
Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any
means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited. 1
REF SEA-BSCHE-CHEM1221-
2020
Table of Contents
COURSE LEARNING OUTCOMES ..................................................................................................................... 3
COURSE INTRODUCTION .................................................................................................................................. 4
MODULE AND UNIT TOPICS .............................................................................................................................. 4
COURSE STUDY GUIDE ...................................................................................................................................... 5
STUDY SCHEDULE .............................................................................................................................................. 7
FORMATIVE ASSESSMENT ACTIVITIES ............................................................................................10

SUMMATIVE ASSESSMENT ACTIVITIES ...........................................................................................11
GRADING SYSTEM .....................................................................................................................11
FACILITATOR CONTACT INFORMATION: ....................................................................................................... 12
MODULE 1:INTRODUCTION ............................................................................................................................ 12
MODULE 2: CHEMICAL EQUATIONS: FORMULATION AND BALANCING ..................................................... 14
MODULE 3: CONCENTRATIONS OF SOLUTIONS ............................................................................................ 34
MODULE 4: RATES AND EQUILIBRIUM CONSTANTS ...................................................................................... 77
MODULE 5: IONIZATION AND IONIZATION CONSTANTS .............................................................................. 93
MODULE 6: VOLUMETRIC METHODS OF ANALYSIS:UNIT 3: REDOX METHODS .......................................... 165
MODULE 7: GRAVIMETRIC ANALYSIS .......................................................................................................... 212
REFERENCES .................................................................................................................................................. 219

CHEM 1221
COURSE LEARNING OUTCOMES

At the end of the module you should be able
to:
1. Demonstrate formulation and

balancing of different types of
chemical reactions.
2. Apply skills in Mathematics, Chemistry

and Physical Science
3. Execute calculations involving
chemical equilibrium ,equilibrium
constants and problems related to it
4. Demonstrate a keen awareness of
contemporary issues such as water
ANALYTICAL
and air pollution and soil
contamination that may need the
application of chemical concepts and
CHEMISTRY chemical analysis
5. Use appropriate analytical techniques,
skills and tools for Chemical
Engineering practice

CHEm 1221 lec:
Analytical chemistry
COURSE INTRODUCTION
This is a 4-unit course, in Chemical Engineering curriculum th
at offers a profound study of the basic principles of qualitative and quantitative chemical
analyses.
In this course, the would – be Chemical Engineers learn the applications of analytical
concepts on chemical calculations especially regarding volumetric and gravimetric
methods. It also includes a study of the fundamental concepts about electrometric and
instrumental methods of analysis which at this point in industrial revolution are some of the
most commonly employed methods in various chemical industries
Module and Unit Topics
To ensure that you will demonstrate the above cited course learning outcome at the end
of the semester , this course designed into seven modules. Each module contains the oarts
of analytical chemistry. Each module is designed using the 5E constructivist model of
learning, developed by Roger Bybee, that encourages students to engage, explore,
explain, elaborate, and evaluate their knowledge of topics covered therein. It means that
at the end of each unit, each module, and the course as a whole, you will be assessed on
your progress in attaining the course learning outcomes. Outcomes based education
dictates that only when you can demonstrate the course learning outcomes by the end of
this course, can you be given a passing mark. The modules that form the building blocks to
help you attain the course learning outcomes are as follows:

Module 1: Introduction
Unit 1: Importance and significance of the study
Unit 2: Objectives, scope and overall view of the entire field of analytical chemistry
as applied to chemical engineering
Module 2 : Chemical Equations: Formulation and Balancing

Unit 1: Definition and Importance of chemical equations
Unit 2: Types of chemical equations
Unit 3: Redox and Non-Redox Reactions
Module 3: Concentrations of Solutions

Unit 1: Expression of Concentration
Unit 2: Calculations involving Concentrations
Unit 3:Colligative Properties of Solutions
Module 4: Chemical Reactions: Rates and Equilibrium Constants

Unit 1: Reaction Rates
Unit 2: Chemical Equilibrium
Module 5: Ionization and Ionization Constants

Unit 1: Ion Product Constant of water
Unit 2: Ionization of Weak acids and bases
Unit 3 :Common – ion Effect: Buffered Solutions
Unit 4: Dissociation Constant of Complex ions
Unit 5: Solubility Product Constant
Module 6: Volumetric Methods of Analysis

Unit 1: Introduction: Titration Concepts
Unit 2: Neutralization Methods
Unit 3: Redox Methods
Unit 4: Precipitation Methods
Module 7: Gravimetric Analysis

Unit 1: Introduction to Gravimetry
Unit 2: Law of Definite Proportions
Unit 3: Chemical Factor Derivation
Unit 4: Factor – weight Calculations
Course Study Guide

The key to successfully finish this online course relies heavily on your self-discipline
and time management skills. This module was prepared for you to learn diligently,
intelligently, and independently. Keeping yourself motivated to follow the schedules
specified in the learning plan, maintaining excellence in the expected student outputs,
and mastering the different technologies and procedures required in the delivery and
feedback for this course, will instill in you important qualities you will need in the future as an
engineer practicing your profession. The following course guides and house rules are
designed for you to practice decorum consistent with standards expected within a formal

academic environment. These guides shall lay the groundwork for consistency,
coherence, cooperation, and clear communication among learners and instructors
throughout the conduct of this course:
1. MANAGE YOUR MINUTES. Create a study routine and stick to it. Keep requirement
deadlines and study schedules always in mind by providing visual cues posted in your
place of study or listed in your reminders (electronically, online, or on paper).
Remember that there are other daily activities that take up your time, not to mention
other courses you may be concurrently taking. Choose a time of day when you are
most likely to maximize learning. Communicate your schedule to other members of
your household so they could help you keep it. It would also help to prepare a
dedicated space in your residence conducive for learning.
2. MIND YOUR MANNERS. Treat the distance learning environment as an academic

space not too different from a physical classroom. Do not do in the distance learning
environment, acts you would not normally do in a face-to-face classroom set up.
Avoid asking questions that have already been answered in the lessons or in the
instructions previously discussed or provided. This reflects your poor focus and
uninspired preparation for this course. Practice Conscientious Habitual Etiquette in
group chats, open forums, and similar electronic venues.
a. Use appropriate language and tone, correct grammar and spelling, and
complete sentences acceptable in an academic forum. Avoid text-speak,
slang, and all caps in your posts.
b. Express your opinions politely and do not dominate the conversation.
c. Avoid lengthy as well as offensive posts by sticking to the topic of the

discussion.
d. Take time to understand the salient points of the discussion, and provide
meaningful and well-thought responses to the posts of other participants.
e. For a live meeting or video/voice conferencing set-up, mute your microphone

when you are not speaking to keep the focus on the main speaker.
3. MASTER THE MEDIUM. The distance learning courses will be delivered making use of
the institutional Google Suite account of Saint Louis University. It would be worthwhile
on your part to devote some time and effort to learn the applications you will need
to access your course materials, interact with me and your classmates, and submit
course requirements. Applications of note are Google Classroom, Google Drive, and
Google Meet. There are also available alternatives to Microsoft Office tools you might
want to explore. Certain requirements will require you to take a video on your smart
phone, save it, and submit it electronically. Work on this skill as well. If you are offline,
identify the most convenient means for express mail correspondence and inform me
as early as possible so we can make the necessary arrangements ahead of time.

4. MAKE MASTERPIECES. Go beyond minimum requirements. The course learning
outcomes will serve as a guide to the minimum expected competencies you are to
acquire at the end of this course. It does not limit you from performing beyond it.
Keep in mind that the quality of your work reflects the amount of thought and care
you put into the process of completing it. It provides a very tangible measure of how
much of the competencies you have developed and fully obtained throughout this
course.
5. CONNECT CONSTANTLY. There are more than sufficient online and offline modes to
ensure that you are well informed and provided on time with the needed learning
materials, instructions, requirements, and feedback either from me or from your
classmates. Exhaust all means possible to keep in touch and updated. My contact
details can be found at the latter part of this document and will be made available
and widely disseminated to enrolees of this course.
6. OBSERVE ORIGINALITY. Your course outputs will largely be submitted in electronic

form. It is going to have a highly traceable and comparable digital footprint that can
be easily checked for originality. Cite your sources properly for referenced statements
you decide to use in your own work. Attribute statements by persons other than you
by using terms like according to, he said/she said, and the like.
7. INSTIGATE INDEPENDENCE. You are the focus of this course. Nobody else. All
assessment and evaluation tools in this course are designed to measure your
competence and not anybody else’s. You may use all resources at your disposal,
and ask other people for advice. In the end however, it is going to be your
independent work that will be judged against the standards set for this course. The
only way for you to maximize this course to your advantage is to learn as much from
it as an individual. Make it count.
Additional Guidelines for Offline Students:

 If you are a student opting for the offline mode of distance learning, you will be tasked
to send back the accomplished requirements at given stages of the course through
express mail correspondence on or before the scheduled date to me. Make sure you
will follow it up with me through text or any other media available for you.
 While waiting for my feedback of your accomplished requirements, continue doing
the task in the succeeding units of the module. If needed, do not hesitate to keep in
touch with me through any available means. Remember, if there is a will, there is a
way.
Study Schedule
Below is the complete weekly schedule for the attainment of the topic learning outcomes
vis-a-vis the activities. This contains also the schedule of the deadlines of the submission of
the accomplished course requirements or assignments and the examination.

Dates TOPIC LEARNING OUTCOMES ACTIVITIES
MODULE 1 Introduction
3 hrs TLO 1: : Illustrate different Engage: What is analytical chemistry

applications of analytical Explore: State and discuss the
chemistry with basic concepts of differences of qualitative and
chemistry as foundation quantitative chemistry
Explain/Elaborate :Give and discuss
the importance and significance of
the study, objectives, scope and
overall view of the entire field of
analytical chemistry as applied to
chemical engineering
Evaluate: Graded assignment
Module 2 Chemical Equations: Formulation and Balancing
5 hrs TLO 2: Demonstrate formulation Engage: Recall principles involved in

and balancing of different types chemical nomenclature
of chemical reactions Explore: The different types of
chemical reactions
Explain/Elaborate: Balancing of
Redox and Non-redox reactions
Graded Quiz 1
MODULE 3 Concentrations of Solutions
9hrs TLO 3: Solve problems on Engage: different types of solutions

random variables and their Explore: Methods of expressing
probability distributions, concentrations
cumulative distribution Explain: Calculations involving
functions, expected values of concentrations
random variables, the binomial Elaborate: Colligative properties of
distribution and the Poisson Solutions
distribution Evaluate: Graded Assignment
Evaluative Assessment
Graded Quiz 2
MODULE 4 Chemical Reactions: Rates and Equilibrium Constants
6 hrs TLO 4: : Evaluate chemical Engage: Different factors

reactions according to type affecting reaction rates
and solve problems involving Explore:Concepts on chemical equilibrium
chemical reactions and Explain/Elaborate: solve problems on
governing equilibrium chemical equilibrium

Evaluate: graded assignment
Module 4 Ionization and Ionization constants
10 hrs TLO 5: Execute Engage: Point out the distinction

calculations involving between strong and weak
chemical equilibrium, electrolytes.
ionization constants Explore: concepts on the ionization
of weak acids and bases
and problems related
Explain/Elaborate: Solve problems
to it on
ionization of weak acids and
bases ,ionization of water
solubility product principle
Evaluate; Graded Assignment
Midterm exam
Module 5 Volumetric Methods of Analysis
9 hrs TLO 6:Apply the basic Engage: Introduction to titration

concepts of chemical concepts
equilibrium, equilibrium Explore:Neutralization methods on
constants and reaction rates acidimetry and alkalimetry
in solving problems involving Explain/Elaborate: Illustrative
volumetric methods. example problems on neutralization
methods, Evaluate: graded
assignment
Quiz 2; Module 5: Units1& 2 andMidterm Exam

UNIT 3 Redox Methods
TLO 7: Apply reduction – Engage: Recall principles
oxidation concepts and involved in balancing by redox
chemical equilibrium in Explore: Different Redox
calculating problems involving Methods
Redox and precipitation Explain: Discuss
methods. permanganometry, Iodimetry
and different precipitation
Elaborate: Solve problems on
permanganometry, Iodimetry
and precipitation
Evaluate: graded assignment

Final Quiz 1
Module 6 Gravimetric Analysis
TLO 8 : Apply basic concepts of Engage: Introduction to gravimetry

chemistry, chemical equilibrium Explore: Law of definite proportions
and titration methods in Explain: derivation of chemical factor
gravimetric methods of analysis Elaborate: solve problems on
gravimetric method
Comprehensive Final Exam

a. Multiple choice
b. Problems solving
c. Essay (submit video recording; question to be assigned earlier)
V. Evaluation
The course modules rely on formative and summative assessments to determine the
progress of your learning in each module. To obtain a passing grade in this course, you
must:
1. Read all course materials and answer the pre-assessment quizzes, self-assessment
activities, and/or reflection questions.
2. Participate in online discussion forums.
3. Submit all assignments and graded quizzes.
4. Take the Midterm and Final Examination
Formative Assessment Activities

Formative assessments for this course are applied to ungraded activities that are used
to monitor your learning experience and provide feedback to improve both your
learning approach as well as my instructional approach.
• You are required to answer the pre-assessment quizzes, self-assessment activities,

and reflection questions but your scores in these activities will not be included in
the computation of your final grade.
• The reflection questions are designed to help you to critically analyse the course
readings for better understanding while the pre-assessment quizzes and self-
assessment activities are designed as a review management tool to prepare you
for the graded quizzes and examinations.
• Successfully answering formative activity questions and requirements will serve as
prompts to tell you if you need to study further or if you may already move forward
to the next unit of the module.
• The completeness of your answers to the pre-assessment quizzes, self-assessment
activities, and reflection questions will still be checked and will still be part of your
course completion. Hence, these must be answered.

• In doing your formative assessment activities, you can always ask the help of your
family and friends.
• The pre-assessment quizzes, self-assessment activities, and reflection questions are
required so you can take it anytime within the scheduled days assigned for each
unit.
Summative Assessment Activities

The evaluative assessments are graded activities designed to determine if your
acquisition of learning and performance in tests is at par with standards set at certain
milestones in this course.
A. Quizzes, Examinations, and Assignments
Graded quizzes, examinations, and assignments are essential to determine

whether your performance as a student is at par with standards/goals that need
to be achieved in this course. The scores obtained from each of the graded
activities will contribute to your final grade, the weights of which are presented
in the grading system described in the succeeding sections of this text. Direct
scoring can be used on straightforward requirements like short answers and
multiple choice responses, while scoring rubrics will be provided for answers that
are typically lengthy and involve a more complex level of thinking on your part.
B. Final Course Requirement
You will need to accomplish a comprehensive exam as a final requirement for the
course. And submit a position paper ( summary of what you learn from the subject)
on how you can apply what you learn in industry or in the field of chemical
engineering
Grading System
Activity Weight
MIDTERM GRADE (MG)
CS (online discussion, assignments, quiz) 50%
Examination 50%
Total 100%
TENTATIVE FINAL GRADE (TFG)

CS (online discussion, assignments, quiz) 50%
Examination 50%
Total 100%
Final Grade: MG 50% + TFG 50% = 100%

VI. Technological Tools
To be able to accomplish all the tasks in this course, you will need a computer or a
laptop with the following software applications: Word Processing, Presentation, and
Publication for requirements that do not require online access. A smart phone with video
recording and editing features will also be used for activities that will require you to record
videos for saving and submission.
If you are a student online, access to the institutional Google Classroom will be
provided through your institutional account. An invitation to join the Google Classroom will
be sent to you through the SLU Student Portal and your institutional email account, so make
sure to activate your institutional email account. It is equally important that you check your
SLU Student Portal account at least twice a week and turn your Gmail Notifications on your
mobile phone and computer.
If you are a student offline, the delivery of instructions and requirements will be primarily
through express mail correspondence of printed modules and saved digital content on a
USB flash drive. Feedback and clarifications will be facilitated through text messaging and
voice calls; hence, you need to have regular access to a cell phone. If you need to call, or
you want to talk to me, send me a message first and wait for me to respond. Do not give
my CP number to anybody. I will not entertain messages or calls from numbers that are not
registered in my phone. Hence, use only the CP number you submitted to me.
FACILITATOR CONTACT INFORMATION:

Engr. Lilibeth R. Ramos
CHEM 1221 Course Facilitator
Cellphone : 0915xxxxxx
SLU local extension number: Chemical Engineering local 391
Institutional email address: lramos@slu.edu.ph
MODULE 1:INTRODUCTION
At the end of the module the student should be able to:

 Illustrate different applications of analytical chemistry with basic concepts of
chemistry as foundation
CHEMISTRY - is a branch of science which deals with the study of:
a) composition and properties of matter

b) changes that matter undergoes
c) energy which accompany these changes
d) laws and principles which govern these changes
Branches of Chemistry:
1. General Chemistry - general survey of the entire field of chemistry with

particular emphasis placed on the fundamental concepts and elementary
laws.
2. Inorganic Chemistry – concerned with the preparation and properties of the

elements and their compounds. C – compounds are not included.
3. Organic Chemistry - study of C and its compounds
4. Analytical Chemistry – concerned with the detection and determination of

substances and their constituents.
a) Qualitative Chemistry – deals with the analysis of the component parts.

( What is/are present?)
b) Quantitative Chemistry – deals with the determination of the amount

of component parts. ( How much of it is /are present?)
5. Physical or Theoretical Chemistry – concerned with the laws and

generalizations underlying chemical changes wherever these laws are
expressed mathematically
6. Biochemistry – study of compounds and chemical changes that are brought

about by living processes
7. Radiochemistry or Nuclear Chemistry – study of the changes that take place

in atomic nuclei when bombarded with suitable projectile.
8. Industrial Chemistry – deals with the transformation of raw materials of nature

into finished products.
Applications of Analytical Chemistry
People engaged in the field of analytical chemistry do the following:
- perform qualitative and quantitative analysis

- use the science of sampling, defining, isolating, concentrating and preserving
samples
- set error limits
- validate and verify results through calibration and standardization
- perform separations based on differential chemical properties
- create new ways to make measurements, interpret data in proper context and
communicate results
SCIENTIFIC METHOD
- rational explanation of facts and laws designed to suggest why or how

something happen as it does
- logical approach to the solution of any problem which lends itself to
investigation
Steps:
1. Recognizing a phenomenon and stating it as a problem

2. Gathering of data
3. Form a hypothesis
4. Conduct experiment
5. Draw conclusion
6. Theory
7. Stating a law or generalization
MODULE 2: CHEMICAL EQUATIONS: FORMULATION AND BALANCING
CHEMICAL NOMENCLATURE
LANGUAGE OF CHEMISTRY
In olden times, medieval alchemists used symbols to stand for elements such as:
☉ ☾ ♂ ☿ ♀
Gold silver iron mercury copper

Crescent - for silver, symbolic of the silvery color of the moon
Circle - for gold, symbolic of the golden sun and of purity
John Dalton - an English teacher who made up other symbols.
- he pictured the atom as spheres and since atoms of different

elements were unlike, he represented his atoms with spheres with
distinguishing marks on them
Letters as symbols:
- J.J. Berzelius (Swedish Chemist)
1.) using the first letter of the name of the element as symbol
e.g. H for Hydrogen
 O for Oxygen
 C for Carbon
2.) the first and another characteristic letter of the name of
the element
- because the names of the several elements begin with

the same letter , Berzelius found it convenient to use 2
letters in some symbols
e.g Copper – Cu ; Calcium- Ca ; Chlorine- Cl ;
Cobalt- Co
- Latin name:
Silver - Ag: Argentum
Antimony - Sb: stibium
Iron - Fe: Ferrum
Potassium - K: Kalium
Copper - Cu: Cuprum
Mercury - Hg: hydrargyrum
Sodium - Na: natrium
Tungsten - Wolfranium
- The names of the elements came from many sources such as follows:
1. the discoverer
Curium- Curie Lawrencium- Lawrence
2. A scientist

Einsteintinium – Albert Einstein
Nobelium- Alfred Nobel
Mendelevium- Dmitri Mendeleev
Fermium- Enrico Fermi
3. the place or site of discovery
Americium- America Berkelium- Berkely
4. the country of the discoverer
Polonium- Poland
5. name of where it was found
Germanium- Germany
DEFINITION OF IMPORTANT TERMS
Chemical Nomenclature - a system of writing and naming of compounds
Chemical Symbol - the symbolic representation of elements; consists of either one or
two letters
ex. H, N, O, Na Ca
Chemical Formula - a combination of symbols of elements constituting a
compound; shows the number and kind of atoms present
- shorthand representation of a chemical compound
Ex. NaCl, Al2(SO4)3 , C12H22O11
Oxidation Number - a positive or negative whole number assigned to an element in

a molecule or ion on the basis of a set of formal rules. To some
degree, it reflects the positive or negative character of the atom. It is
also referred to as oxidation state.
Ex. Na +1, Ca +2, O-2 , Cl -1 oxidation no.
Radical - group of atoms with a corresponding charge
- polyatomic ion
Ex. NO3 -1, NH4+1, O2 -2
Subscript - are small numerals which are placed immediately after and a little
below the symbols of the elements concerned.
Al2(SO4)3

subscript
Chemical Reaction - a process in which one or more substances are converted into
other substances ; also called chemical change.
Chemical Equation - a representation of a chemical reaction using the chemical

formulas of the reactants and the products. A balanced chemical
equation contains equal numbers of atoms of each element on both
sides of the equation.
Stoichiometric Coefficients - the coefficients used to balance an equation.
RULES IN WRITING A CHEMICAL FORMULA:
1. A compound consists of a positive part and negative part.
The following combinations are possible:
POSITIVE NEGATIVE EXAMPLE
a. element element NaCl
b. element radical CaSO4
c. radical element (NH4)2O
d. radical radical NH4OH
2. The convention is to write the symbol of the positive part before the symbol of the
negative part. Write the symbols of the elements with the valences (oxidation
numbers) on top of them.
Ex. Na +1 Cl -1
Except: P -3 H + 1
3. Crisscross the valences. The valence of the positive part becomes the subscript
of the second and the valence of the second becomes the subscript of the first. A
compound is always neutral.
Ex. Al +3 O -2
Al2O3
4. When the subscript is supposed to be 1, it is no longer written.
Na1Cl1 x
NaCl √
5. Since a compound shows the simplest ratio in which the positive and the negative
part are combined:

a. omit writing subscripts whenever subscripts (by rule 3) are numerically
equal
Mg2O2 x
MgO √
Except for : O2 -2 as in the case of H2O2
b. whenever possible, reduce the subscripts to the simplest whole number

ratio. Be sure not to alter the formula of the radical.
Si +4 O -2 → Si 2/2 O 4/2 → SiO2
6. Whenever a radical needs a subscript, enclose it first with a parenthesis. If the original
already contains a parenthesis, enclose with a bracket.
Ex: (NH4)2SO4
Ca3[Fe(CN)6]2
RULES IN NAMING COMPOUNDS:
1. BINARY COMPOUNDS - contains 2 different elements
A. METAL + NON-METAL
To name: give the name of the positive element and change the last letters
of the negative element to IDE
NaCl - sodium chloride K2O - potassium oxide
B. COMPOUNDS WHOSE METALS ARE MULTIVALENT
- metals with variable valences or oxidation numbers
- can exist in more than 1 oxidation state
a. Old Method:
Suffix OUS - element exist in a lower oxidation state
IC - element exists in a higher oxidation state
b. Stock System/New Method

- give the name of the positive element .Indicate its
valence by a Roman Numeral enclosed in parenthesis.
Change the last letter of the negative element to IDE
Ex:
OLD METHOD STOCK SYSTEM
Hg2O mercurous oxide mercury (I) oxide
HgO mercuric oxide mercury (II) oxide

FeCl2 ferrous chloride iron (II) chloride
FeCl3 ferric chloride iron (III) chloride
PbO plumbous oxide lead (II) oxide
PbO2 plumbic oxide lead (IV) oxide
C. STOICHIOMETRIC PROPORTIONS
NON-METAL + NON-METAL
Prefix Table
prefix number prefix number
mono- 1 hexa- 6
di- 2 hepta- 7
tri- 3 octa- 8
tetra- 4 nona- 9
penta- 5 deca- 10
- the elements are named preceded by the prefix indicating the number
of atoms of each element. If the first element has only one atom there
is no need to give the prefix mono
CO - carbon monoxide P2O5 - diphosphorus pentoxide
CO2 - carbon dioxide N2O4 - dinitrogen tetorxide
CCl4 - carbon terachloride
2. ACIDS
a. non- oxy acids
- acid composed of hydrogen and non-metal
- to name , use the prefix HYDRO and the suffix IC for the negative
element followed by the term ACID
hydro + root Ic + acid
Formula Name of Compound Name of Acid
HI hydrogen iodide hydroiodic acid

HBr hydrogen bromide hydrobromic acid
HCl hydrogen chloride hydrochloric acid
b. oxyacid
- acid composed of hydrogen and a radical
- acid containing oxygen. To name, use the suffix IC for the acid forming
element or radical followed by the word acid.
Eg. H2CO3 - carbonic acid H2C2O4 - oxalic acid
c. oxyacids whose acid forming element is multivalent
- if radical ends in ITE use the suffix OUS and if it ends in ATE use
the suffix IC then follow with the word ACID
eg.
HNO2 - nitrous acid
HNO3 - nitric acid
H2SO3 - sulfurous acid
d. If more than two acids can be formed from the same elements because of
multivalency, use the prefix HYPO for the lowest oxidation state and PER for the
highest oxidation state followed by the word ACID
HClO - hypochlorous acid HClO3 – chloric acid
HClO2 - chlorous acid HClO4 - perchloric acid
3. Salts of Oxyacids
- To name, give the name of the positive element or radical and follow
with the name of the negative radical
Eg.
K3PO4 - potassium phosphate
(NH4)2Cr2O7 – ammonium dichromate
4. BASES
- compound composed of element/radical and hydroxide
- to name, give the name of the positive element or radical followed by the
term hydroxide

eg.
NaOH - sodium hydroxide NH4OH – ammonium hydroxide
Ternary Compounds - contain more than 2 elements in a chemical combination
oxy acids - hydrogen, non-metal, oxygen
- hydrogen with a polyatomic ion or radical

per (root) ic acid +7 oxidation state
(root)ic acid +6, +5 oxidation sates
(root)ous acid +4 +3 oxidation states
Hypo(root)ous acid +1 oxidation state
 metal ion together with a polyatomic ion or radical

Ion - an atom or group of atoms carrying a + or a - electric charge.
RULES IN ASSIGNING OXIDATION NUMBERS:

1. The oxidation state of an uncombined element is zero (in the free state).
Ex.Oxidation no.:
Na 0 + Cl0 → NaCl
H2 0 + O2 0 → H2O
2. In a compound, the more electronegative element is assigned a - oxidation state
and the less electronegative element a + oxidation state.
Ex.
Na+1Cl-1
3. In a formula of a compound the sum of the - oxidation state is equal to the sum of the
+ oxidation states.
Ex. Na2O
For Na : the oxidation state ( charge, valence, oxidation number) is +1, the
subscript of Na is 2, + 1(2) = +2
For O : the oxidation state ( charge, valence, oxidation number) is -2, the
subscript of O is 1, -2( 1) = -2
4. The algebraic sum of oxidation numbers of all atoms in the formula for a neutral
compound is equal to zero.
Ex. Na2O
+1(2) -2(1)
+2 -2 = 0
Al2S3 : the charge /oxdn number of Al is +3 , subscript is 2, +3(2) = +6

the charge /oxdn number of S is -2 , subscript is 3, -2(3) = -6
+6 -6 = 0
Ex. Determine the oxidation state/number (OS)of the underlined element:

a. H2SO4
to solve for the oxdn number of S, the sum of the oxdn numbers of all the
atoms = 0
2(+1) + S + 4(-2) = 0 (charge of H = +1 and the subscript is 2, charge of O =-2,
subscript is 1)
S = +6
b. H2SO3
2(+1) + S + 3(-2) = 0
S = +4
c. HClO
1(+1) + Cl + 1(-2) = 0
Cl = +1
d. HClO2
1(+1) + Cl + 2(-2) = 0
Cl = +3
e. HClO3
1(+1) + Cl + 3(-2) = 0
Cl = +5
f. HClO4
1(+1) + Cl + 4(-2) = 0
Cl = +7
 You can also use this in naming oxy acids

 In ex. C, Cl has a charge of +1 ,use Hypo(root)ous acid if the oxidation state is
+1, therefore the name would be:
Hypochlorous acid
For HClO2 , Cl has a charge of +3, use (root)ous acid if oxidation state is
+4 or +3 , therefore the name would be:
chlorous acid
For HClO3 , Cl has a charge of +5, use (root)ic acid if oxidation state is
+5 or +6 , therefore the name would be:
chloric acid
For HClO4 , Cl has a charge of +7, useper (root)ic acid if oxidation state is
+7 , therefore the name would be:
perchloric acid
5. The algebraic sum of oxidation numbers of all atoms in the formula of a polyatomic ion
equals the charge of the ion. ( Polyatomic ions – ions involving more than one element)
Ex. Determine the oxidation state/number (OS)of the underlined element:

a. NO3 -1
1N + 3(-2) = -1
N = +5

b. SO4 -2
1S + 4(-2) = -2
S = +6
c. C2O4-2
2C + 4(-2) = -2
C = +3
CHEMICAL EQUATION
- an expression which shows by the use of symbols and formulas, the changes
in the arrangement of the atoms which occur during a chemical reaction
REACTANTS → PRODUCTS
Reactants - original substances in a chemical reaction
Products - substances produced
CHEMICAL REACTION
- is the process by which one or more substances are changed into one or
more substances
- maybe represented by an equation
TYPES OF CHEMICAL REACTION
1. DIRECT COMBINATION (COMBINATION) OR SYNTHESIS REACTION

2 reactants combine to give one product
A + B → AB
Forms of direct combination reaction:
a. metal + oxygen → metal oxide

ex. Mg + O2 → MgO ( unbalanced)
In this equation, there are two O atoms on the left side of the arrow and one
O atom on the right side. We can increase the number of O atoms by
placing a coefficient of 2 on the product side
Mg + O2 → 2MgO ( unbalanced)
Now there are two O atoms and 2 Mg atoms on the right. Placing a
coefficient 2 in front of Mg brings both the Mg and O atoms into balance:
2Mg + O2 → 2MgO ( balanced)
Ex.2 Al + O2 → Al2O3 ( unbalanced)

In this equation, there are two O atoms on the left side of the arrow and 3 O
atoms on the right side. We can increase the number of O atoms by placing
a coefficient of 2 on the product side and 3 on the reactant side
Al + 3O2 → 2Al2O3 ( unbalanced)

Now there are 6 O atoms and 4 Al atoms on the right. Placing a coefficient 4
in front of Al brings both the Al and O atoms into balance:
4Al + 3O2 → 2Al2O3 ( balanced)
b. non – metal + oxygen → non metal oxide

C + O2 → CO2
S + O2 → SO2
c. metal + non metal → salt

2Na + Cl2 → 2NaCl
Mg + Cl2 → MgCl2
d. metal oxide + water → base ( OH ) metal hydroxide

MgO + H2O → Mg(OH)2
Na2O + H2O → 2NaOH
e. non – metal oxide + water → oxy acid

CO2 + H2O → H2CO3
SO2 + H2O → H2SO3
f. metal oxide + non metal oxide → salt

MgO + SO2 → MgSO3
Other examples of combination reaction:

N2 + 3H2 → 2NH3
2. Decomposition
- a single substance is decomposed to form 2 or more products:
AB → A + B
EXAMPLES:
a. metal oxide → metal + oxygen
2HgO → 2Hg + O2
2PbO → 2Pb + O2
b. metal carbonate → metal oxide + carbon dioxide

MgCO3 → MgO + CO2
PbCO3 → PbO + CO2
c. metal bicarbonates → metal carbonate + carbon dioxide +water

2NaHCO3 → Na2CO3 + CO2 + H2O
d. metal chlorates → metal chloride + oxygen

2KClO3 → 2KCl + 3O2
e. metal nitrate → metal nitrite + oxygen

2NaNO3 → 2NaNO2 + O2

3. Single replacement
- an element and a compound reacted to form a new element and a new
compound
A + BC → B + AC (A is metal)
A + BC → AB + C (A is non metal)
EXAMPLES:
a. sodium + water→ sodium hydroxide + hydrogen

2Na + 2H2O → 2NaOH + H2
refer to the activity series: Na is before H therefore it is more reactive,it will
replace/substitute H in H2O
b. magnesium + sulfuric acid→ magnesium sulfate +

hydrogen
Mg + H2SO4 → MgSO4 + H2
c. iron + cupric sulfate

Fe + CuSO4 → Fe SO4 + Cu
 Fe is more reactive than Cu ( refer to the activity series)
c. chlorine + sodium iodide → sodium chloride+ iodine

Cl2 + 2NaI → 2NaCl + I 2
 Cl is more reactive than I
Activity series:
Metals: Li K Ba Ca Na Mg Al Zn Fe Ni Sn Pb H Cu Hg Ag Pt Au
Non metals: F Cl Br I
4. Double Replacement/ Decomposition
The rxn b/n 2 compounds to form new compounds
AB + CD → AD + CB
A. acid + base → salt + water

HCl + NaOH → NaCl + H2O
a. formation of insoluble precipitate

NaCl + AgNO3 → NaNO3 + AgCl
Methods of Balancing Chemical Equations:
1. Inspection Method - trial and error

2. Algebraic Method
Steps:
1. Assign literal coefficients to all reactants and products

2. Form algebraic equations by doing balance on each element in the
chemical equation
3. Choose one literal coefficient to be equated to one. This is done because the
number of equation is less than the number of unknowns or literal coefficients.
4. Solve the rest of the algebraic equation.
5. Finally, replace all the literal coefficients by their equivalents in the chemical
equation
6. If fraction appears, eliminate them by multiplying the whole equation by the
LCD of all fractions.
Ex:
CaCO3 + H3PO4 → Ca3(PO4)2 + H2CO3
1. You first write the equation using letter variables for the coefficients:
aCaCO3 + bH3PO4 → cCa3(PO4)2 + d H2CO3
2. Then you set up a series of simultaneous equations, one for each element.
Ca bal:mlamml= 3c eqn 1
C bal:ml lammll=mmmd eqn 2
O bal :m 3a + 4b = 8c + 3d eqn 3
H bal:mm 3b =mmll2d eqn 4
P bal :mmlb = 2c eqn 5
3. Now you solve the five simultaneous equations.
Let's set c=1
Then substitute value of c in eqn 1:ia = 3 and in eqn 2:
Eld = a = 3 substitute value of c in eqn 5:
b = 2(1) = 2
So a=3; b=2; c=1; d=3
The balanced equation is
3CaCO3 + 2H3PO4 → Ca3(PO4)2 + 3 H2CO3
HNO3 + H2S → S + NO + H2O
aHNO3 + bH2S → cS + dNO + eH2O
H bal: a + 2b = 2e eqn 1
N bal: a = d eqn 2
O bal: 3a = d + e eqn 3
S bal: b = c eqn 4
Let a = 1
In eqn 1:
a =d ; d =1

in eqn 3:
3a = d + e
3(1) = 1 + e
e = 2
in eqn 1
a + 2b = 2e
1 + 2b = 2(2)
b = 3/2
in eqn 4
b =c
c = 3/2
2[HNO3 + 3/2 H2S → 3/2S + NO + 2H2O]
2HNO3 + 3H2S → 3S + 2NO + 4H2O
MnO + PbO2 + HNO3 → HMnO4 + Pb(NO3)2 + H2O
3. Redox Method
Oxidation - refers to a reaction in which an element increases in oxidation state

due to loss of electrons
Reduction - refers to a reaction in which an element decreases in oxidation state

due to gain of electrons
Oxidizing agent - substance that decrease in oxidation number
- substance responsible for oxidation
- particles which accepts electrons
Reducing agent - substance that increase in oxidation number
- substance responsible for reduction

- Particles which donates electrons
Steps:
1. Assign the oxidation number of all elements in the chemical equation

2. Look for those elements that changed in their oxidation number. There
must be at least two elements, one undergoes reduction, the other one
goes oxidation.
3. Write the partial ionic equations for those elements referred to in step 2.
4. Balance the ionic equations with respect to:
a) number of atoms; and
b) number of charges ( you can only add electron)
5. Balance the number of e- in both ionic equations.

6. Transfer the coefficient obtained in step 4 to the overall chemical
equation ( the original one). Balance the rest of the overall equation by
inspection with H & O the last ones being balanced.
Ex:
Using Change- in-Oxidation State/Number Method (Valence-Change

Method)
1. FeCl3 + SO2 + H2O → FeCl2 + HCl + H2SO4 ( unbalanced)

Soln:
Step 1: Assign oxidation numbers:
FeCl3 + SO2 + H2O → FeCl2 + HCl + H2SO4
+3 -1 +4 -2 +1 -2 +2 -1 +1 -1 +1 +6 -2
The oxidation numbers of Fe and S have changed, Fe from +3 to +2
and S from +4 to +6
Step2: Write the oxidation and reduction steps. Balance the number of
atoms and then balance the electrical charge using electrons:
 Select the atom of that element in the oxidizing agent whose oxidation
number is changed within the reaction, indicate in the equation the number
of electrons gained in this change
+3 reduction +3-(+2) = 1e- gained +2

( Fe gains 1 e- )
 Select the atom of that element in the reducing agent whose oxidation
number is changed within the reaction, indicate in the equation the number
of electrons lost in this change
+4 oxidation +6-(+4)= 2e- lost +6

( S loses 2 electrons )
Step 3: Adjust loss and gain of electrons so that they are equal. Determine the
smallest common denominator for electrons gained, and multiply the oxidizing and
reducing agents by suitable small numbers such that the total electrons gained and
lost are equal to ach other
 Multiply the oxidation step by 1 and the reduction step by 2

+3 reduction +3-(+2) = 1e- gained +2 X2
+4 oxidation +6-(+4)=2e- lost +6 X1
Step 4: Transfer the coefficients from the balanced redox equations into the original
equation . We need to use 2 FeCl 3 , 2 FeCl2, 1 SO2, 1 H2SO4
2 FeCl3 + SO2 + H2O → 2 FeCl2 + HCl + H2SO4 (unbalanced)
Step 5: Balance the remaining elements by inspection: (last to bal: H & O)

2 FeCl3 + SO2 + 2 H2O → 2 FeCl2 + 2HCl + H2SO4 (balanced)
Check: the final balanced equation contains:

Reactant side Product side
Fe: 2 atoms 2 atoms
Cl: 2x3 = 6 (2x2) + 2 = 6
S: 1 1
H: 4 2+2 = 4
O: 2+2 = 4 4
OR (ANOTHER SOLUTION):
FeCl3 + SO2 + H2O → FeCl2 + HCl + H2SO4 ( unbalanced)
Soln:
Step 1: Assign oxidation numbers:
+3 -1 +4 -2 +1 -2 +2 -1 +1 -1 +1 +6 -2
The oxidation numbers of Fe and S have changed, Fe from +3 to +2 and S from +4 to +6
Step2: Now write two new equations, using only the elements that change in oxidation
number. Then add electrons to bring the equations into electrical balance. One equation
represents the oxidation step; the other represents the reduction step. Remember:
Oxidation produces electrons; reduction uses electrons
1 Fe +3 + 1 e- → 1Fe +2 reduction
( Fe gains 1 electron )
+3
1S +4 → 1 S+6 + 2e- oxidation

(S loses 2 electrons)
+4

Step 3: Multiply the two equations by the smallest whole numbers that will make the
electrons lost by oxidation equal to the number of electrons gained by reduction. In
this reaction the oxidation step is multiplied by 1 and the reduction step by 2. The
equations become
2Fe +3 + 2 e- → 2Fe +2 reduction
( 2Fe gains 2 electron )
+3
S +4 → S+6 + 2e- oxidation

(S+4 loses 2 electrons)
 We have now established the ratio of the oxidizing to the reducing agent as being
2 atoms Fe to 1 atom of S
Step 4: Transfer the coefficients in front of each substance in the balanced redox
equations to the corresponding substance in the original equation . We need to use 2
FeCl3 , 2 FeCl2, 1 SO2, 1 H2SO4
2 FeCl3 + SO2 + H2O → 2 FeCl2 + HCl + H2SO4 (unbalanced)
Step 5: In the usual manner, balance the remaining elements that are not oxidized or
reduced by inspection: (last to bal: H & O) to give the final balanced equation:
2 FeCl3 + SO2 + 2 H2O → 2 FeCl2 + 2HCl + H2SO4 (balanced)
Check: the final balanced equation contains:

Reactant side Product side
Fe: 2 atoms 2 atoms
Cl: 2x3 = 6 (2x2) + 2 = 6
S: 1 1
H: 4 2+2 = 4
O: 2+2 = 4 4
Oxidizing Agent (O.A): FeCl 3

Reducing Agent (R.A): SO2
SOLVE:
2. MnO + PbO2 + HNO3 → HMnO4 + Pb(NO3)2 + H2O
3. Na2S2O3 + I 2 → NaI + Na2S4O6
Balancing a Redox Reaction that Occurs in Acidic Solution
Steps:
1. Divide the equation into two incomplete half-reactions, one for oxidation
and the other for reduction.

2. Balance each half – reaction.
a. First, balance the elements other than H and O.
b. Next, balance the O atoms by adding H2O.
c. Then, balance the H atoms by adding H +.
d. Finally, balance the charge by adding e- to the side with greater
overall positive charge.
3. Multiply each half-reaction by an integer so that the number of electrons
lost in one half-reaction equals the number gained in the other.
4. Add the two half-reactions and simplify where possible by canceling
species appearing on both sides of the equation.
5. Check the equation to make sure that there are the same number of
atoms of each kind and the same total charge on both sides.
Balance these equations using the ion-electron method:
1. Cr2O7 -2 (aq) + Cl – (aq) → Cr +3 ( aq) + Cl2 (g) (in acid solution)
+6 -2 -1 +3 0
Step 1: Write the two-half reactions , one containing the element being oxidized and
the other, the element being reduced ( use the entire molecule or ion):
Cr2O7 -2 → Cr+3 reduction
Cl – → Cl2 0 oxidation
Step 2: Balance the elements other than oxygen and hydrogen ( accomplished in
step 1: 2Cr and 2 Cl on each side)
Cr2O7 -2 → 2Cr+3 reduction
2Cl – → Cl2 0 oxidation
Step 3: Balance O and H. Remember that the solution is acidic. The oxidation
requires neither O and H, but the reduction needs 7 H2O on the right side and 14 H+
on the left:
14 H+ + Cr2O7 -2 → 2Cr+3 + 7 H2O reduction

2Cl – → Cl2 0 oxidation
Step 4: Balance each half - reaction electrically with electrons:
6e- g + 14 H+ + Cr2O7 -2 → 2Cr+3 + 7 H2O reduction

net charge = + 6 on each side
2Cl – → Cl2 0 + 2 e- l oxidation

net charge = -2 on each side
Step 5: Equalize loss and gain of electrons . In this case, multiply the oxidation by 3
and reduction equation by 1:
6e- g + 14 H+ + Cr2O7 -2 + → 2Cr+3 + 7 H2O

6Cl – → 3Cl2 0 + 6 e- l
Step 6: Add the two-half reactions together, canceling the 6e- from each side ,to
obtain the balanced equation:
6e- g + 14 H+ + Cr2O7 -2 → 2Cr+3 + 7 H2O
6Cl – → 3Cl2 0 + 6 e- l
14 H+ + Cr2O7 -2 + 6Cl – → 2Cr+3 + 3Cl2 0 + 7 H2O (balanced)
Check:
Charge: ( left side) : +14 -2 -6 = +6 right side: 2(+3) = 6
Number of atoms for each element:
Left right
Cr: 2 2
Cl: 6 6
H: 14 14
O: 7 7
Ex.2 KMnO4 + KCl + H2SO4 → MnSO4 + K2SO4 + H2O + Cl2 ( in acid sol’n)
Balancing Equations for Reactions Occurring in Basic Solutions
- if a redox reaction occurs in basic solution, the equation must be

completed by using OH- and H2O rather than H+ andH2O. The half-reactions
can be balanced initially as if they occurred in acidic solution. The H + ions
can then be “ neutralized” by adding an equal number of OH - ions to both
sides of the equation and canceling, where appropriate , the resulting water
molecules.
Balance these equations:
1. Al + NO2-1 → Al(OH)4-1 + NH3 (basic solution)
2. CN- (aq) + MnO4- (aq) → CNO- (aq) + MnO2 (s) (basic solution)
Ex.1
1. Al + NO2-1 → Al(OH)4-1 + NH3 (basic solution)
Assign Oxidation states:
Al + NO2-1 → Al(OH)4 -1 + NH3
0 +3 -2 +3 -2 +1 -3 +1
Solution: Step 1: Write the two half-reactions:
Al 0 → Al(OH)4 -1 oxidation

NO2-1 → NH3 reduction
Step 2: Balance elements other than H and O (accomplished in Step 1)
Step 3: Remember the solution is basic. Balance the O and H as though the solution
were acidic. Use H2O and H+. To balance O and H in the oxidation equation,
add 4H2O on the left and 4H+ on the right side:
Al 0 + 4H2O → Al(OH)4 -1 + 4H+
Because H+ does not exist in any appreciable concentration in basic solution,

we remove it from the equation by adding an appropriate amount of OH - to
both sides of the equation to neutralize the 4H +. The 4OH- and 4H+ form 4H2O
Al 0 + 4H2O + 4OH- → Al(OH)4 -1 + 4H+ + 4OH-
Combine 4H+ and 4OH- as 4H2O and rewrite, canceling H2O on each side:
Al 0 + 4H2O + 4OH- → Al(OH)4 -1 + 4H2O
Al 0 + 4OH- → Al(OH)4 -1 oxidation
To balance O and H in the reduction equation, add 2H2O on the right and
7H+ on the left side:
NO2-1 + 7H+ → NH3 + 2H2O
Add 7 OH- to each side:

NO2-1 + 7H+ + 7 OH- → NH3 + 2H2O + 7 OH-
Combine 7H+ + 7 OH- → 7H2O :
NO2-1 + 7H2O → NH3 + 2H2O + 7 OH-
Rewrite canceling 2H2O from each side:

NO2-1 + 5H2O → NH3 + 7 OH- reduction
Step 4: Balance each half-reaction electrically with electrons:

Al 0 + 4OH- → Al(OH)4 -1 + 3 e- g (balanced oxidation equation)
6 e- l + NO2-1 + 5H2O → NH3 + 7 OH- (balanced reduction equation)
Step 5: Equalize the loss and gain of electrons. Multiply the oxidation reaction by 2:
2Al 0 + 8OH- → 2Al(OH)4 -1 + 6 e- g
6 e- l + NO2-1 + 5H2O → NH3 + 7 OH-
Step 6: Add the two half- reactions together, calceling the 6 e- and 7 OH- from each side of
the equation:
2Al 0 + 8OH- → 2Al(OH)4 -1 + 6 e- g

6 e- l + NO2-1 + 5H2O → NH3 + 7 OH-
2Al 0 + NO2-1 + 5H2O + OH- → 2Al(OH)4 + NH3 balanced equation
ASSIGNMENT: BALANCE THE FOLLOWING EQUATIONS BY:
A. ALGEBRAIC AND CHANGE IN OXIDATION STATE METHOD
1. MnO + PbO2 + HNO3 → HMnO4 + Pb(NO3)2 + H2O

2. Na2S2O3 + I 2 → NaI + Na2S4O6
B. ION ELECTRON METHOD

3. CN- (aq) + MnO4- (aq) → CNO- (aq) + MnO2 (s) (basic solution)
4. 2 KMnO4 + KCl + H2SO4 → MnSO4 + K2SO4 + H2O + Cl2 ( in acid sol’n)
MODULE 3: CONCENTRATIONS OF SOLUTIONS

SOLUTION: HOMOGENEOUS MIXTURE
INTRODUCTION:
A solution is a homogeneous mixture of 2 or more substances, these substances

being of molecular dimensions, whose composition may be varied up to certain limits
A. COMPONENTS OF A SOLUTION:
1. solvent - dissolving medium of a solution
- it is normally the component of a solution present in the greater

amount
2. solute- substance dissolved in a solvent to form a solution
- it is normally the component of a solution present in the smaller

amount
B. CLASSIFICATION OF SOLUTIONS:
1. BASED ON THE PHYSICAL STATE OF THE SOLVENT:

SOLUTION TYPE SOLUTE SOLVENT EXAMPLE
Gaseous solution Solid gas dust in air
Liquid gas H2O vapor in air
Gas gas O2 in air
Liquid solution Solid liquid salt solution
Liquid liquid alcohol solution

Gas liquid carbonated
drinks
Solid solution Solid solid Zn in Cu (brass)
Liquid solid Hg in Cu
gas solid H2 in Pt

2. BASED ON SOLUBILITY:
SOLUBILITY OF SOLUTE - the amount of substance that dissolves in a given
quantity of solvent at a given temperature to form a saturated solution
Factors affecting solubility of solute:
1. Nature of solvent
2. Surface area exposed if the solute is solid
3. State of the solution when the solute is added
4. Temperature
a. An increase in temperature increases the solubility of most solid solutes
dissolved in liquid solvents
b. An increase in temperature decreases the solubility of gaseous solutes
dissolved in liquid solvents.
5. Pressure
a. For solid and liquid solutes, the solubility is not greatly affected by pressure
b. For gases, the greater the pressure, the higher the solubility
a. Unsaturated Solution -
- one where less solute than the possible amount is dissolved in the
solvent

b. Saturated Solution
- contains the maximum amount of solute that can be dissolved under

the conditions at which the solution exists
- contains the amount of dissolved solute necessary for the existence of

an equilibrium between dissolved and undissolved solute
c. Supersaturated solution
- contains more than the maximum amount of the solute that could
normally dissolve in the given amount of solvent at a specified
temperature
- more concentrated than a saturated solution

3. BASED ON THE AMOUNT OF SOLUTE DISSOLVED:
a. concentrated solution - solution that contains relatively large amount of
solute relative to the amount that could dissolve
b. dilute solution - solution where a small amount of solute is present in a

solution relative to the amount that could dissolve

C. SOLUBILITY CHARACTERISTICS ASSOCIATED WITH LIQUIDS:
1. miscible substances - substances which dissolve in any amount in each

other
2. partially miscible - substances which have limited solubility in each other
3. immiscible - substances which do not dissolve in each other and forms 2
layers upon mixing
 AQUEOUS SOLUTION – a solution in which water is the solvent

D. UNIT OF EXPRESSING SOLUBILITY:
- grams of solute per 100 g solvent
- the temperature of the solvent must be specified
QUALITATIVE SOLUBILITY TERMS:
Solute solubility Qualitative solubility

description
g solute / 100 g solvent
less than 0.1 insoluble
0.1 – 1.0 slightly soluble
1.0 – 10 soluble
greater than 10 very soluble
E. WHY SUBSTANCES DISSOLVE: SOLUTION FORMATION
In a solution solute particles are uniformly dispersed throughout the solvent.

Considering what happens in the molecular level during the solution process
will help us understand how this is being achieved.
Attractions which must be overcome for a solute to dissolve in a solvent:
1. Solute – solute attraction – attraction between solute particles

2. Solvent – solvent attraction – attraction between solvent particles
Driving force for solution formation:
- Solute solvent attraction – attraction between solute and solvent particles

UNIT 2: CALCULATIONS INVOLVING CONCENTRATIONS
METHODS OF EXPRESSING CONCENTRATION OF SLUTIONS:
Concentration - amount of solute present in a specified amount of

solvent or specified amount of solution
A. PHYSICAL METHODS- done on the basis of the solute

1. PERCENTAGE OF SOLUTE:
The concentration of a solution is often specified in terms of the
percentage of solute to the total amount of solution. Since the amount
of solute and solution present may be stated in terms of either mass or
volume, different types of percentage units exists.
a. Percent by Weight- Weight (% W /W)
or parts per hundred (pph)
- percentage unit most frequently used by chemists
- gives the number of grams of solute per 100 grams of solution
% w/w = mass of solute X 100
mass of solution
 Parts per thousand (ppt) = mass of solute X 1000

mass of solution
Parts per million ( ppm) = mass of solute X 106
mass of solution
B. Percent by weight – volume (% W /V)

- often encountered in hospitals and industrial settings
- convenient when working with a solid solute and liquid solvent
% w/v = wt. solute X 100
vol of solution
C. Percent by volume – volume(% V /V)

- used as concentration unit in situations where both solute and solvents
are liquids
- gives directly mL of solute per 100 mL of solution

%v/v = volume of solute X 100
volume of solution
B. CHEMICAL METHODS – done on the basis of moles of solute

2. MOLE FRACTION (X)
- ratio of the number of moles of solute to the number of moles of
solution
Mole (n)- amount of a substance
X solute = moles solute
moles solution
X solvent = moles solvent
moles solution
n solution = n solute + n solvent
3. MOLARITY (M)
- number of gram- molecular weights of solute per liter of solution
- concentration most often used in chemical laboratories
- ratio of the number of moles of solute per liter of solution
M = no.of moles solute
liter solution
gram-solute
= MW solute
liter solution

4. FORMALITY (F)
- Number of gram formula weights of solute per liter of solution
F = moles solute
liter solution
5. NORMALITY (N)
- most often encountered in situations that involve reactions of acids

and bases
- ratio of the number of gram-equivalents of solute per liter of solution
N = no. of gram-equivalent weights
liter solution
= no. of equivalents solute
Liter solution
weight in g solute
equivalent weight
= liter of solution
6. MOLALITY (m)
- finds use in experimental situations where changes in temperature are

concerned
- for dilute solution, aqueous solutions, M and m are identical
- ratio of the number of moles of solute per Kg of solvent
m = no. of moles of solute
kg of solvent

W
= MW
kg solvent
 Notes:
a. Specific gravity of a substance is numerically equal to density of substances
in units gram/mL
b. Density of dilute aqueous solution is approximately that of water which is
1gram/mL
c. ppm of solute in dilute aqueous solution = milligrams solute/liter of solution
. DILUTION CONCEPT:
A common problem encountered when working with solutions in the

laboratory is that of diluting a solution of known concentration (usually a
stock solution ) to a lower concentration.
DILUTION - a process in which more solvent is added to a solution in order

to lower the concentration of the solution. The amount of solute is
present but is now distributed in a larger amount of solvent.
Formula: V1C1 = V2C2
- amount of dissolved solute before dilution = amount of dissolved solute after

dilution
-
SUPPLEMENTARY PROBLEMS:
1. What is the percent by weight concentration of Na 2SO4 in a solution made by

dissolving 7.60 g Na2SO4 in enough water to give 87.3 g of solution?
2. In the treatment of illnesses in the human body, a 0.92% w/v% NaCl solution is to be
administered intravenously. How many grams of NaCl are required to prepare 345
mL of this solution?
3. The density of a solution of 5.0 gram toluene and 22.5 grams of benzene is 8.76
g/mL. Calculate :
a. molarity of the solution
b. % by weight of benzene
c. mole % of benzene
4. A 2.5 gram sample of ground water was found to contain 5.4 microgram of Zn +2,
what is this concentration in parts per million.
5. Determine the molarities of the following solutions:
a. 2.37 moles of KNO3 are dissolved in enough water to give 650 mL of solution.

b. 25.0 g of NaOH are dissolved in enough water to give 2.50 L of solution.
6. Express in formality the concentration of each of the following solutions:
a. 28.4% NH3 by weight, having a density of 0.808 g/mL
b. 36.0 % HCl by weight, having a density of 1.19 g/mL
7. How many grams of solute are in 500 mL of 6F HCl?
8. Calculate the molality of a solution made by dissolving 10.0 g AgNO 3 (MW = 169.9) in
275 g of water.
9. A 4.10 molal solution of H2SO4 has a density of 1.21 g/mL. What is the molar
concentration of the solution? What is the normality?
10. Five grams calcium hydroxide is dissolved in 950 mL of water. Determine the ff:
a. % solute by weight e. molarity
b. % solute by mole f. ppm of solute
c. molarity g. specific gravity
d. normality h. density
11. Calculate the equivalent weight of the following solutes:

a. Na3PO4 b. H2SO4 c. Al(OH)3
12. Calculate the normality of the solution that results when 4.00 g of Al(NO 3)3 (MW =
213.0) is dissolved in enough water to give 250.0 mL of solution.
13. A solution contains 10.0 g of glucose (MW= 180) and 85.0 g of water. What is the
mole fraction of glucose in the solution? What is the mole percent glucose in the
solution?
14. What are the mole fractions of ethyl alcohol, C2H5OH and water, in a solution made
by dissolving 11.5 g of ethanol in 27 g water?
15. What volume of concentrated HCl should be used to prepare 500.0 mL of a 3.00 M
HCl solution? Concentrated HCl solution is 12.0 M
16. A solution of H2SO4 containing 50.0% H2SO4 by mass has a density of 1.40 g/mL.
Express its concentration in terms of mole fraction, mole percent, molarity, molality
and normality.
17. How many grams of solute are required to prepare each of the following aqueous
solutions?
a. 500.00 mL of a 0.500M urea solution , CO (NH 2)2
b. 500.0 mL of 1.50 N H3PO4 solution
c. 1.0 liter of a 2.o M CaBr2 solution
d. 1.0 kg of a 5.0 % by mass glucose, C6H12O6
18. What is the normality of each of the following solutions as an acid or as a base?
a. 2.0 liters of 18M H2SO4 diluted to 100 liters
b. 140 mL of a 12 M HCl diluted to 2.0 liters

19. How many grams of ethanol are there in a liter-volume of wine (D = 0.934 g/mL)
that is 12.0% by mass ethanol, C2H5OH? 12% by volume ethanol?( D of pure ethanol
= 0.789 g /mL).
20. What is the molarity of concentrated HBr solution if the acid is 48.0% by mass HBr? (
D= 1.50 g/ mL)
21. What volume of concentrated HNO3 should be used to prepare 1.0 liter of 0.15M
HNO3? Concentrated HNO3 is 70.0 % by mass HNO3 with a density of 1.42 g/mL .
22. A 100mL sample of concentrated H2SO4 is diluted with water until the final volume is
500 mL. Concentrated H2SO4 is 96.0% by mass with a density of 1.84 g/mL. What is
the normality and molarity of the new solution?
23. Commercial aqueous nitric acid has a specific gravity of 1.42 and is 16M. Calculate
the % HNO3 by mass in the solution.
24. Propylene glycol, C3H6(OH)2, is sometimes used in automobile antifreeze solutions. If

an aqueous solution has a mole fraction of propylene glycol equal to 0.10,
calculate : a) % propylene glycol by mass; b) molality of the solution.
25. Caffeine, C8H10N4O2, is a stimulant found in coffee and tea. If a solution of caffeine
in chloroform , CHCl 3, as a solvent has a concentration of 0.085 molal, calculate: a)
% caffeine by mass, b) mole fraction of caffeine.
SOLUTIONS TO PROBLEMS:
1. What is the percent by weight concentration of Na 2SO4 in a solution made by dissolving

7.60 g Na2SO4 in enough water to give 87.3 g of solution?
Given:
W Na2SO4 = 7.60 g
W soln = 8.3 g
Reqd: % w-w
Soln:
𝑊 𝑠𝑜𝑙𝑢𝑡𝑒
%𝑤 = × 100
𝑊 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
7.60 𝑔
%𝑤 = × 100
87.7 𝑔
% 𝑊 = 8.71%
2. In the treatment of illnesses in the human body, a 0.92% w/v% NaCl solution is to be
administered intravenously. How many grams of NaCl are required to prepare 345
mL of this solution?

Given:
% w-v = 0.92%
V soln = 345 mL
Reqd: W NaCl
Soln:
%𝑤−𝑉 = × 100
𝑉 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
0.92 % = × 100
345 𝑚𝐿
𝑊 𝑠𝑜𝑙𝑢𝑡𝑒 = 3.2 𝑔
3. The density of a solution of 5.0 gram toluene and 22.5 grams of benzene is 0.876
g/mL. Calculate :
a. molarity of the solution
b. % by weight of toluene
c. mole % of toluene
Given:
W toluene = 5.0 g MW = 92 g/mol
W benzene = 22.5 g MW = 78 g/mol
D(𝜌) = 0.876 g/mL
Reqd: a. molarity of the solution

b. % by weight of toluene
c. mole % of toluene
Soln:
𝑛 𝑠𝑜𝑙𝑢𝑡𝑒
𝑀= ; 𝑀= 𝑀𝑊 𝑠𝑜𝑙𝑢𝑡𝑒
𝑉𝑠𝑜𝑙𝑛 𝑖𝑛 𝐿
𝑉𝑠𝑜𝑙𝑛 𝑖𝑛 𝐿
 The solute is toluene ( present in lesser amount)

 To solve for the M of the solution , we first solve the volume of the solution in L by using
the density of the solution.
 The weight of the solution is equal to the wt of toluene + wt of benzene
𝜌 = 𝑉 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
5.0𝑔 +22.5
𝑉 𝑠𝑜𝑙𝑛 = × 100
0.876𝑔/𝑚𝐿
𝑉𝑠𝑜𝑙𝑛 = 31.3927 𝑚𝐿

 Convert the V soln in liters : 1L = 1000 mL
 Thus the Molarity is
𝑊𝑠𝑜𝑙𝑢𝑡𝑒
𝑀 =
𝑀𝑊 𝑠𝑜𝑙𝑢𝑡𝑒 × 𝑉𝑠𝑜𝑙𝑛 𝑖𝑛 𝐿
5𝑔
𝑀 = 𝑔 1𝐿
92 × 31.3927 𝑚𝐿 ×
𝑚𝑜𝑙 1000𝑚𝐿
𝑀 = 1.7023 𝑚𝑜𝑙𝑒𝑠 /𝐿
b. %𝑤 = × 100
5.00𝑔
%𝑤 = × 100
27.5 𝑔
% 𝑤 𝑡𝑜𝑙𝑢𝑒𝑛𝑒 = 18.1818 %
c. 𝑚𝑜𝑙𝑒 % = 𝑚𝑜𝑙𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 × 100

 We first solve the mole fraction of toluene :
𝑊 𝑡𝑜𝑙𝑢𝑒𝑛𝑒
𝑋 𝑡𝑜𝑙𝑢𝑒𝑛𝑒 = 𝑀𝑊 𝑡𝑜𝑙𝑢𝑒𝑛𝑒
𝑊 𝑡𝑜𝑙𝑢𝑒𝑛𝑒 𝑊 𝑏𝑒𝑛𝑧𝑒𝑛𝑒
+
𝑀𝑊 𝑡𝑜𝑙𝑢𝑒𝑛𝑒 𝑀𝑊 𝑏𝑒𝑛𝑧𝑒𝑛𝑒
5.0𝑔
92 𝑔/𝑚𝑜𝑙
𝑋 𝑡𝑜𝑙𝑢𝑒𝑛𝑒 =
5.0 𝑔 22.5 𝑔
+
92𝑔/𝑚𝑜𝑙 78𝑔/𝑚𝑜𝑙
𝑋 𝑡𝑜𝑙𝑢𝑒𝑛𝑒 = 0.1585
𝑚𝑜𝑙𝑒 % = 0.1585 × 100
𝑚𝑜𝑙𝑒 % = 15.8537 %

10. Five grams calcium hydroxide is dissolved in 950 mL of water. Determine the ff:
a. % solute by weight e. molality
Given:
W Ca(OH)2 = 5.0 g
V H2O = 950 mL
Reqd: a. % solute by weight e. molality

Soln:
a. % solute by weight
The Volume of water = V soln
%𝑤 = × 100
W soln = wt solute + wt solvent
To solve for the wt of solvent (H2O), we make use of the density of water which is 1 g/mL
𝑊
𝜌 =
𝑉
𝑊
1 𝑔/𝑚𝐿 =
950 𝑚𝐿
𝑊 𝐻2 𝑂 = 950 𝑔
Therefore, the wt of soln = 950 + 5 = 955 g
5.0 𝑔
%𝑤 Ca(OH)2 = × 100
955 𝑔
% 𝑊 = 0.5236%
b. % solute by mole
𝑛 Ca (OH)2
% Ca(OH)2 𝑏𝑦 𝑚𝑜𝑙𝑒 = 𝑛 𝑠𝑜𝑙𝑛
𝑊 Ca(OH)2
𝑀𝑊 Ca(OH)2
% Ca(OH)2 𝑏𝑦 𝑚𝑜𝑙𝑒 = 𝑊 Ca(OH)2 𝑊𝐻 𝑂 𝑥 100
+ 𝑀𝑊 𝐻2 𝑂
𝑀𝑊 Ca(OH)2 2

5𝑔
74𝑔/𝑚𝑜𝑙
% Ca(OH)2 𝑏𝑦 𝑚𝑜𝑙𝑒 = 𝑥 100
5𝑔 950 𝑔
+
74 𝑔/𝑚𝑜𝑙 18 𝑔/𝑚𝑜𝑙
% Ca(OH)2 𝑏𝑦 𝑚𝑜𝑙𝑒 = 0.1279%
c. molarity
𝑛 𝑠𝑜𝑙𝑢𝑡𝑒
𝑀= ; 𝑀= 𝑀𝑊 𝑠𝑜𝑙𝑢𝑡𝑒
𝑉𝑠𝑜𝑙𝑛 𝑖𝑛 𝐿 𝑉𝑠𝑜𝑙𝑛 𝑖𝑛 𝐿
𝑀 =
𝑀𝑊 𝑠𝑜𝑙𝑢𝑡𝑒 × 𝑉𝑠𝑜𝑙𝑛 𝑖𝑛 𝐿
5𝑔
𝑀 = 𝑔 1𝐿
74 × 950 𝑚𝐿 ×
𝑚𝑜𝑙 1000𝑚𝐿
𝑀 = 0.0711 𝑚𝑜𝑙𝑒𝑠 /𝐿
d. normality
to solve for the N of the solution, we first determine the factor, f of Ca(OH) 2 , for a
base , the factor would be the number of replaceable OH -,so the factor is 2
𝑊 𝑠𝑜𝑙𝑢𝑡𝑒 ×𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑁= 𝑀𝑤 𝑠𝑜𝑙𝑢𝑡𝑒 ×𝑉 𝑠𝑜𝑙𝑛 𝑖𝑛 𝐿
5𝑔 × 2 𝑒𝑞/𝑚𝑜𝑙
𝑁 = 𝑔
74𝑚𝑜𝑙 × 0.950 𝐿
𝑁 = 0.1422 𝑁
𝑚𝑜𝑙𝑒𝑠 𝑒𝑞
OR: 𝑁 = 𝑀 ×𝑓 ; 𝑁= 0.0711 ×2 = 0.1422 𝑁
𝐿 𝑚𝑜𝑙
e. molality

𝑚 =
𝑀𝑊 𝑠𝑜𝑙𝑢𝑡𝑒 × 𝑘𝑔 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
5𝑔
𝑚 =
𝑔 1𝑘𝑔
74 × 950 𝑔 × 1000 𝑔
𝑚𝑜𝑙
𝑚𝑜𝑙𝑒𝑠
𝑚 = 0.0711
𝑘𝑔
f. ppm of solute
𝑝𝑝𝑚 = × 106
OR:
𝑝𝑝𝑚 = % 𝑤 × 104
𝑝𝑝𝑚 = 0.5236 × 104
𝑝𝑝𝑚 = 5236 𝑝𝑝𝑚
g. specific gravity
To solve for the specific gravity ,SG, we must solve first the density of Ca(OH) 2 ,
since SG = is equal to the ratio of the density of the subst to the density of
water
𝑊
𝜌 =
𝑉
𝜌 𝑠𝑢𝑏𝑠𝑡
𝑆𝐺 =
𝜌 𝑤𝑎𝑡𝑒𝑟
955𝑔
𝜌 𝐶𝑎(𝑂𝐻)2 = = 1.0053 𝑔/𝑚𝐿
950 𝑚𝐿
𝜌 𝑠𝑢𝑏𝑠𝑡
𝑆𝐺 =
𝜌 𝑤𝑎𝑡𝑒𝑟
𝜌 𝑤𝑎𝑡𝑒𝑟 = 1 𝑔/𝑚𝐿

1.0053 𝑔/𝑚𝐿
𝑆𝐺 = = 1.0053
1 𝑔/𝑚𝐿
h. density of water :  = 1.0053 𝑔/𝑚𝐿
6. Express in formality the concentration of each of the following solutions:

a. 28.4% NH3 by weight, having a density of 0.808 g/mL
b. 36.0 % HCl by weight, having a density of 1.19 g/mL
Given: a. % wt NH3 = 28.4 %  = 0.808 g/mL

Reqd: F of NH3 soln
Solution:
𝐹 =
𝐹𝑊 𝑠𝑜𝑙𝑢𝑡𝑒 × 𝑉𝑠𝑜𝑙𝑛 𝑖𝑛 𝐿
From the data , we are given, only the density of soln and % W ,the weight of NH3 and
volume of solution are unknown. Therefore we make can make a basis of 100 g solution
since we know the % by wt of soln or we can also assume a volume of 1L soln, since by
definition formality = formula weights per liter
Basis: 100 g of soln

 To solve for the weight of solute:
 %𝑤 = 𝑊 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
× 100
% 𝑤 × 𝑊 𝑠𝑜𝑙𝑛
𝑊 𝑠𝑜𝑙𝑢𝑡𝑒 =
100
28.4 % × 100𝑔
100
 To solve for the volume of solution:
𝑊
 𝜌 = 𝑉
100𝑔
𝑉 =
0.808 𝑔/𝑚𝐿
𝑉 = 123.7624 𝑚𝐿

28.4 𝑔
𝐹 =
𝑔 1𝐿
17 × 123.7624 𝑚𝐿 × 1000𝑚𝐿
𝑚𝑜𝑙
𝐹 = 13.4984 𝐹
Another Solution:
Basis: 1L of soln
 To solve for the weight of solution:
𝑊
𝜌 =
𝑉
𝑔 1000𝑚𝑙
𝑊 𝑠𝑜𝑙𝑛 = 0.808 × 1𝐿 ×
𝑚𝐿 1𝐿
𝑊 𝑠𝑜𝑙𝑛 = 808 𝑔
To solve for the weight of solute:
 %𝑤 = × 100
100
28.4 % × 808𝑔
100
 :
229.472 𝑔
𝐹 = 𝑔
17 × 1𝐿
𝑚𝑜𝑙
𝐹 = 13.4984 𝐹
DILUTION
18. What is the normality of each of the following solutions as an acid or as a base?
a. 2.0 liters of 18M H2SO4 diluted to 100 liters
b. 140 mL of a 12 M HCl diluted to 2.0 liters

Given: V1 = 2.0 L V2 = 100 L
C1 = 18M C2 = ?
Let V1, V2 = volume of solutions 1 and 2, respectively
Let C1, C2 = concentration of solutions 1 and 2, respectively (either M or N)

V2 = V1 + V solvent added
Illustration: V H2O
V1= 2.0 L H2SO4

C1 = 18M H2SO4 Mixing or dilution Diluted soln
V2 = 100 liters
1 C2 = ? N 2
To solve for the final concentration in N, we have to convert the initial concentration from M
to N
N= M × f
𝑚𝑜𝑙𝑠 2 eq
N = 18 × = 36 𝑁
𝐿 𝑚𝑜𝑙
C1 = 36N
𝑉1 𝐶1 = 𝑉2 𝐶2
2.0𝐿( 36𝑁) = 100𝐿 𝐶2
𝐶2 = 1.3889 𝑁
21. What volume of concentrated HNO3 should be used to prepare 1.0 liter of 0.15M
HNO3? Concentrated HNO3 is 70.0 % by mass HNO3 with a density of 1.42 g/mL .
Given:
V1= ?
C1 : Mixing or dilution Diluted soln
70.0 % by mass V2 = 1 liter
 = 1.42 g/mL C2 = 0.15M

Soln:
𝑉1 𝐶1 = 𝑉2 𝐶2
To solve for the initial volume of HNO3, we must solve for the initial concentration of HNO 3 in
M since the unit of the final conc is in M
To solve for C1 we are given only the % W and density of the concentrated HNO 3 soln ,
therefore we assume a value. We can make a basis of 100 g soln or 1L soln
Basis: 100 g of soln

 To solve for the weight of solute:
 %𝑤 = × 100
100
70.0 % × 100𝑔
100
 To solve for the volume of solution:
𝑊
 𝜌 = 𝑉
100𝑔
𝑉 =
1.42 𝑔/𝑚𝐿
𝑉 = 70.4225 𝑚𝐿
70.0 𝑔
𝑀 =
𝑔 1𝐿
63 × 70.4225 𝑚𝐿 × 1000𝑚𝐿
𝑚𝑜𝑙
𝑀 = 15.7778𝑀 = 𝐶1
𝑉1 𝐶1 = 𝑉2 𝐶2
𝑉1 ( 15.7778𝑀) = 1𝐿(0.15𝑀)
𝑉1 = 9.5070 𝑚𝐿
ASSIGNMENT; SOLVE THE REMAINING SUPPLEMENTARY PROBLEMS: 4,5,6,7,8,9,11-25

MODULE 3: CONCENTRATIONS OF SOLUTIONS
UNIT 3: COLLIGATIVE PROPERTIES OF SOLUTIONS
COLLIGATIVE PROPERTIES OF SOLUTIONS

When a solute is added to a solvent, changes in the properties of solvent take place
such that the solution exhibits properties different from those of pure solvent. These solution
properties are called COLLIGATIVE PROPERTIES
They are dependent on the number of solute particles that are dissolved in a given
quantity of solvent.
Properties of solutions that depend only on the concentration of the solute rather than
its nature.
Required conditions:
1. solute has no appreciable vapor pressure
- nonvolatile : vapor pressure is so low that we assume it to be zero
2. solute does not form ions
- non – electrolyte
COLLIGATIVE PROPERTY LAW

- The freezing point, boiling point and vapor pressure of a solution differ from
those of the pure solvent by amounts which are directly proportional to the molal
concentration of the solute
4 Types:
A. Freezing Point Lowering/ Depression
- The addition of a non-volatile solute decreases the freezing point of the pure
solvent
- the freezing point of a solution is always lower than the freezing point of a pure
solvent
Freezing Point- temperature at which solid and liquid phases are in equilibrium
∆𝑇𝑓 = 𝑘𝑓 m
∆𝑇𝑓 = 𝑇𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 − 𝑇𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
Where:
∆𝑇𝑓 = freezing point depression in ℃

𝑘𝑓 = freezing point depression constant
℃−𝑘𝑔
 H2O : 𝑘𝑓 = 1.86 ℃ / m or
𝑚𝑜𝑙
m = molality of solution = moles solute / kg solvent
Example 1: Ethylene glycol (EG), CH2OH CH2OH , is a common automobile antifreeze.

Calculate the freezing point of a solution containing 651 g of EG in 2505 g of water. Would
you keep the substance in your car radiator during the summer? The molar mass of EG is
62.01 g/mol. Kf = 1.86 oC/m and Kb = 0.52 oC for water
Given:
WEG = 651 g Kf = 1.86 ℃/m
Wwater = 2505 g MW of compound = 62.01 g/mol
Reqd: FP soln or 𝑇𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
SOLUTION:
To calculate the freezing point of the solution, we can first calculate for the ∆𝑇𝑓 , use the
equation

𝑠𝑖𝑛𝑐𝑒 𝑡ℎ𝑒 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 𝑖𝑠 𝑤𝑎𝑡𝑒𝑟, 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑡ℎ𝑒 𝑓𝑟𝑒𝑒𝑧𝑖𝑛𝑔 𝑝𝑜𝑖𝑛𝑡 𝑖𝑠 0℃ ∶ 𝑇𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 = 0 ℃
OR Use this equation:
𝑊𝐸𝐺
∆𝑇𝑓 = 𝑘𝑓 (𝑀𝑊 )
𝐸𝐺 × 𝑘𝑔𝑠𝑜𝑙𝑣𝑒𝑛𝑡
℃−𝑘𝑔 651
0℃ − 𝑇𝑓𝑠𝑜𝑙𝑛 = 1.86 ( 1𝑘𝑔 )
𝑚𝑜𝑙 62.01× 2505 𝑔 × 1000 𝑔
𝑇𝑓𝑠𝑜𝑙𝑛 = − 7.795 ℃
 the solution will freeze at − 7.795 ℃
Example 2:
2.00 g of some unknown compound reduces the freezing point of 75.00 g of benzene
from 5.53 to 4.90 ∘C
. What is the molar mass of the compound?
Given:
Wunknown cpd = 2.00 g Tf solvent = 5.53 0C
Wbenzene = 75.00 g Tf soln = 4 .9 0C
Reqd: MW of compound
SOLUTION
First we must compute the molality of the benzene solution, which will allow us to find
the number of moles of solute dissolved.
∆𝑇𝑓
𝑚 = 𝑘𝑓
𝑘𝑓𝑏𝑒𝑛𝑧𝑒𝑛𝑒 = 5.12 ℃/m ( freezing point depression constant
for benzene)
∆𝑇𝑓 = 𝑇𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 − 𝑇𝑓𝑠𝑜𝑙𝑛

∆𝑇𝑓 = 5.53℃ − 4.90℃ = 0.63 ℃
∆𝑇𝑓
𝑚 = 𝑘𝑓
0.63℃
𝑚 =
℃
5.12 𝑚
𝑚 = 0.1230 𝑚 𝑜𝑟 𝑚𝑜𝑙𝑒𝑠/𝑘𝑔
We can now find the molecular weight of the unknown compound by using this
formula:
𝑚 =
𝑀𝑊𝑠𝑜𝑙𝑢𝑡𝑒 × 𝑘𝑔𝑠𝑜𝑙𝑣𝑒𝑛𝑡
𝑚𝑜𝑙𝑒𝑠 2.00 𝑔
0.1230 =
𝑘𝑔 1𝑘𝑔
𝑀𝑊𝑠𝑜𝑙𝑢𝑡𝑒 × 75.00 𝑔 × 1000 𝑔
Molecular Weight =216.8022 g/mol
The freezing point depression is especially vital to aquatic life. Since saltwater will
freeze at colder temperatures, organisms can survive in these bodies of water.
B. Boiling Point Elevation

- The addition of a nonvolatile solute increases the boiling point of the
pure solvent
- The temperature at which a solution boils is higher than that of the pure solvent
if the solute is nonvolatile
 Boiling Point - temperature at which the vapor

pressure of a liquid equals the prevailing atmospheric
pressure

∆𝑇𝑏 = 𝑘𝑏 m
∆𝑇𝑏 = 𝑇𝑏 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 − 𝑇𝑏 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
Where:
𝑇𝑏 = boiling point elevation in ℃
𝑘𝑏 = boiling point elevation constant or ebullioscopic
constant
 H2O: 𝑘𝑏 = 0.52 ℃ / m
m = molality of solution

Example 3: Calculate the boiling point elevation of the solution in Example 1 . Would you
keep the substance in your car radiator during the summer?
Given:
WEG = 651 g Kb = 0.52 ℃/m
Wwater = 2505 g MW of compound = 62.01 g/mol
Reqd: BP soln or 𝑇𝑏 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
SOLUTION:
The boiling point elevation can be calculated in the same way.
∆𝑇𝑏 = 𝑇𝑏 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 − 𝑇𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
𝑠𝑖𝑛𝑐𝑒 𝑡ℎ𝑒 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 𝑖𝑠 𝑤𝑎𝑡𝑒𝑟, 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑡ℎ𝑒 𝑏𝑜𝑖𝑙𝑖𝑛 𝑝𝑜𝑖𝑛𝑡 𝑖𝑠 100℃ ∶ 𝑇𝑏 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 = 100 ℃
𝑊𝐸𝐺
∆𝑇𝑏 = 𝑘𝑏 (𝑀𝑊 × 𝑘𝑔𝑠𝑜𝑙𝑣𝑒𝑛𝑡
)
𝐸𝐺
℃−𝑘𝑔 651
𝑇𝑏𝑠𝑜𝑙𝑛 − 100℃ = 0.52 ( 1𝑘𝑔 )
𝑚𝑜𝑙 62.01× 2505 𝑔 × 1000 𝑔
𝑇𝑏𝑠𝑜𝑙𝑛 = 102.1793 ℃
 the solution will boil at 102.1793 ℃
Because the solution will boil at 102.1793 ℃, it would be preferable to leave the antifreeze
in the car radiator in summer to prevent the solution from boiling.

C. Vapor Pressure Lowering
RAOULT - studied extensively the vapor pressure of solutions at

various concentrations
 He found out that the lowering of vapor pressure of a solvent
is proportional to the mole fraction of the solute or the solution
vapor pressure is proportional to the mole fraction of the
solvent, P0
Vapor Pressure - is the pressure exerted by the gaseous molecules in

equilibrium with the liquid at a given temperature
Pressure Units:
1 atm = 760 mm Hg
= 760 torr
 H2O: P0 at 25 0C = 23.8 mm Hg
∆𝑉𝑃 = 𝑉𝑃 of pore solvent x mole fraction of solute

∆𝑉𝑃 = 𝑉𝑃 of pure solvent - 𝑉𝑃 of solution

𝑉𝑃 of solution = 𝑉𝑃 of pure solvent x mole fraction of solvent

Ex.5
Ex. 6: The vapor pressure of pure benzene (C6H6) is 100. torr at 26.1 ℃. Calculate the vapor
pressure of a solution containing 24.6 g of camphor (C 10H16O) dissolved in 100. mL of
benzene. The density of benzene is 0.877 g/mL.
Given:
P0 C6H6 = 100.0 torr W C10H16O = 24.6 g
V = 100 mL T = 26.1 ℃.
 C6H6 = 0.877 g/mL.
Reqd: 𝑉𝑃 of solution (P0 soln)

Soln:
𝑉𝑃 of solution = 𝑉𝑃 of pure solvent x mole fraction of solvent
𝑉𝑃 of solution = 𝑉𝑃𝐶6 𝐻6 x 𝑋 𝐶6 𝐻6
We need to determine the mole fraction of benzene , using this formula:
𝑛𝐶6 𝐻6
𝜒𝐶6 𝐻6 =
𝑛𝐶6 𝐻6 + 𝑛𝐶10 𝐻6 𝑂
The weight of benzene is not given , however we are given the volume and density of
benzene.

; 𝑚 𝑜𝑟 𝑊 = 𝜌 × 𝑉 ;
𝑔
𝑊 = 0.877 × 100 mL = 87.7 g ; 𝑊 = 87. 7 𝑔
𝑚𝐿
molar mass or molecular weight ,MW C6H6 = 78.1 g/mol

molar mass or molecular weight ,MW C10H16O = 152.2 g/mol
substituting the values of solute and solvent in the formula:

87.7 𝑔
𝑔
78.1
𝜒𝐶6 𝐻6 = 𝑚𝑜𝑙 = 0.8742
87.7 𝑔 24.6 𝑔
𝑔 + 𝑔
78.1 152.2
𝑚𝑜𝑙 𝑚𝑜𝑙
𝑉𝑃 of solution = 𝑉𝑃𝐶6 𝐻6 x 𝑋 𝐶6 𝐻6
𝑉𝑃 of solution = 100 torr x 0.8742
𝑉𝑃 of solution = 87.4 torr
Ex .7 The vapor pressure of water at 80C is 355 torr.Calculate the vapor pressure of an
aqueous solution made by dissolving 50262 grams of ethylene glycol in 50 grams of water.
What is the vapor pressure lowering of water in this solution?
D. Osmotic Pressure - pressure that must be applied to prevent osmosis

* Osmosis - Greek word meaning “ push”
- a process in which solvent molecules moves through a
semi- permeable membrane from a solution of lower
solute to a higher solute concentration

* semi – permeable membrane - selective membrane
𝜋 = MRT
Where: 𝜋 = osmotic pressure in atm
M = molarity of solution = moles solute / L soln
R = universal gas constant = 0.08205 L-atm / mole-K
T = absolute temperature ( K) = 0C + 273
𝑛
𝜋 = (𝑉 ) 𝑅𝑇

Types of osmotic solution:
1. Isotonic or Isoosmotic - two solutions that have the same solute
concentration or osmotic pressure
ex. 0.9% NaCl solution ( 0.15 M ) – physiological saline solution
5.0% glucose solution ( 0.29 M )

2. Hypotonic or Hypoosmotic - a solution that contains a lower solute
concentration or osmotic pressure than that of another solution.
ex. < 0.9 % NaCl solution

3. Hypertonic or Hyperosmotic - a solution that contains a higher solute
concentration or osmotic pressure than that of another solution
ex. > 0.9% NaCl solution

Ex. 8; What is the osmotic pressure at 0 0C of an aqueous solution containing 46.0 g of
glycerine (C3H8O3) per liter ?
Given:
Wt C3H8O3 = 46.0 g T = 00C
V soln = 1L
Reqd: 𝜋
Soln:
The temperature , (T = 00C ) , mass of C3H8O3 ( 46.0 g ) and volume of soln(V soln = 1L)
are given and we know the value of R, so we can use the equation below to calculate the
osmotic pressure , 𝜋
𝜋 = MRT
In doing so,
1. we must convert temperature from 0C to K and the osmotic pressure from
torr to atm.
2. We then calculate the molarity, M. use the weight of C3H8O3 and the
molar mass(MW) to determine the number of moles of solute
𝝅 = 𝑴𝑹𝑻
𝑊
𝜋= ( ) 𝑅𝑇
𝑀𝑊

46 𝑔 𝐿−𝑎𝑡𝑚
𝜋= ( 𝑔 ) (. 08205 ) (0 ℃ + 273)𝐾
92 𝑚𝑜𝑙 𝑚𝑜𝑙−𝐾
𝜋= 11.1998 𝑎𝑡𝑚
Ex.9: The osmotic pressure of an aqueous solution of a certain protein was measured in
order to determine its molar mass. The solution contained 3.50 mg of protein dissolved in
sufficient water to form 5.00 mL of solution. The osmotic pressure of the solution at 25 0C was
found to be 1.54 torr. Calculate the molar mass of the protein
Given:
Wt protein = 3.50 mg T = 250C
V soln = 5.00 mL 𝜋 = 1.54 𝑡𝑜𝑟𝑟
Reqd: molar mass (MW) of the protein
Soln:
The temperature , (T = 250C ) and osmotic pressure , (𝜋 = 1.54 𝑡𝑜𝑟𝑟) are given and we
know the value of R, so we can use the equation below to calculate the molarity of the
solution , M
𝜋 = MRT
In doing so,
3. we must convert temperature from 0C to K and the osmotic pressure from
torr to atm.
4. We then use the molarity and the volume of the solution (V soln = 5.00 mL)
to determine the number of moles of solute
5. Finally, we obtain the molar mass(MW) by dividing the mass of solute (Wt
protein = 3.50 mg) by the number of the moles of solute
Solve:
T = 25 0C + 273 = 298 K
𝜋
𝑀 = 𝑅𝑇
1𝑎𝑡𝑚
1.54 𝑡𝑜𝑟𝑟 (760 𝑡𝑜𝑟𝑟) 𝑚𝑜𝑙
𝑀 = 𝐿−𝑎𝑡𝑚 = 8.2873 × 10−5
(0.08205 𝑚𝑜𝑙−𝐾)(298 𝐾 ) 𝐿
𝑛
𝑀 = 𝑉
𝑚𝑜𝑙 1𝐿
𝑛 = 𝑀 × 𝑉 = 8.2873 × 10−5 × 5. 𝑜𝑜 𝑚𝐿 × 1000𝑚𝐿
𝐿

𝑛 = 4.1436 × 10−7 𝑚𝑜𝑙
𝑊
𝑀𝑊 = 𝑛
1𝑔
3.50 𝑚𝑔 ×1000 𝑚𝑔
𝑀𝑊 = 4.1436 ×10−7 𝑚𝑜𝑙
𝑀𝑊 = 8446.6745 𝑔/𝑚𝑜𝑙
PROBLEMS:
1. Calculate the freezing point and boiling point of a solution made by dissolving 5
grams of sugar, C12H22O11, in 100 grams of water.
Given:
W C12H22O11 = 5.0 g
Wwater = 100 g
Reqd: a. 𝑇𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
b. 𝑇𝑏 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
SOLUTION:
a. To calculate the freezing point of the solution, we can first calculate for the ∆𝑇𝑓 , use
the equation
𝑠𝑖𝑛𝑐𝑒 𝑡ℎ𝑒 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 𝑖𝑠 𝑤𝑎𝑡𝑒𝑟, 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑡ℎ𝑒 𝑓𝑟𝑒𝑒𝑧𝑖𝑛𝑔 𝑝𝑜𝑖𝑛𝑡 𝑖𝑠 0℃ ∶ 𝑇𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 = 0 ℃
And Kf = 1.86 0C-kg/mol-K
𝑊𝐸𝐺
∆𝑇𝑓 = 𝑘𝑓 (𝑀𝑊 )
℃−𝑘𝑔 5.0 9
0℃ − 𝑇𝑓𝑠𝑜𝑙𝑛 = 1.86 ( 𝑔 1𝑘𝑔 )
𝑚𝑜𝑙 342 𝑚𝑜𝑙× 100 𝑔 × 1000 𝑔
𝑇𝑓𝑠𝑜𝑙𝑛 = − 0.2719 ℃
 the solution will freeze at − 0.2719 ℃

b. To calculate the boiling point of the solution, we can first calculate for the ∆𝑇𝑏 , use
the equation
∆𝑇𝑏 = 𝑇𝑏 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 − 𝑇𝑏 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
𝑠𝑖𝑛𝑐𝑒 𝑡ℎ𝑒 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 𝑖𝑠 𝑤𝑎𝑡𝑒𝑟, 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑡ℎ𝑒 𝑏𝑜𝑖𝑙𝑖𝑛𝑔 𝑝𝑜𝑖𝑛𝑡 𝑖𝑠 100℃ ∶ 𝑇𝑏 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 = 100 ℃
And Kb = 0.52 0C-kg/mol-K
𝑊𝐸𝐺
∆𝑇𝑏 = 𝑘𝑏 (𝑀𝑊 )
℃−𝑘𝑔 5.0 9
𝑇𝑏 𝑠𝑜𝑙𝑛 − 100℃ = 0.52 ( 𝑔 1𝑘𝑔 )
𝑚𝑜𝑙 342 𝑚𝑜𝑙× 100 𝑔 × 1000 𝑔
𝑇𝑏𝑠𝑜𝑙𝑛 = 100.0760 ℃
the solution will boil at 100.0760 ℃
ASSIGNMENT:
ADDITIONAL PROBLEMS:
1. How many rams of ethylene glycol, C2H4(OH)2 , must be dissolved in 200 grams of
water to produce a solution which freezes at -2.10C?..
2. A solution made from water and a non-dissociating solute, X, contains 10.0 g of
solute X and 800 g of water. The freezing point of the solution is -0.310C. What is the
molecular weight (MW) of X?
3. The freezing point of a solution containing 65 g of a compound in 100 g of

benzene is 3.80C. What is the MW of the compound if the freezing point of benzene
is 5.50C and kf is 4.90C / molal?
4. Calculate the boiling point of a solution that contains 1.5 g of glycerin, C 3H8O3, in
30 g of water
5. A solution of 1.04 g 0f unknown solute, A, in 25.3 g of benzene has a boiling point
of 80.78 0C. Under the same conditions, pure benzene boiled at 80.06 0C
.Calculate the MW of A. (kb of benzene = 2.53 0C/ molal)
6. The P0 of water at 280C is 28.35 mmHg. Compute the vapor pressure at 280C of a
solution containing 68 g of cane sugar in 1000 g of water .
7. What would be the osmotic pressure at 17 0C of an aqueous solution containing
1.75 sucrose per 150 mL solution ?
8. An aqueous solution containing 1.00 g of solute X per liter has an osmotic pressure
of 3.1 torr at 250C. Find the MW of solute X.
9. A solution is prepared by dissolving in 3.00 g of naphthalene ( C 10H8 ) in 20.0 g of
benzene ( C6H6).
a. What is the freezing point of the solution? (kf of benzene=4.90C/molal)

b. What is the mole fraction of benzene in the solution?
c. What is the boilingpoint of the solution? (kb of benzene=2.530C/molal)
10. How many grams of glucose , C6H12O6, are required to lower the freezing point of
150 g of water by 0.750C? What will be the boiling point of the solution?
11. If 8.0 L of ethylene glycol , C2H4(OH)2, density = 1.113 g/mL, is placed in an
automobile radiator and diluted with 32 L of water, density = 1.0 g/mL, what is the
approximate freezing point of the solution?
12. Calculate approximately how much methanol , CH 3OH, would be needed for an
automobile radiator holding 4.0 gal to keep the radiator solution from freezing until
a temperature of – 10.00F is reached?
13.Which solution has the lower freezing point?

a. 10.0 g of CH3OH in 100.0 g H2O or 20.0 g CH3CH2OH in 200.0 g of H2O
b. 10.0 g of H2O in 1.00 kg of CH3OH or 10.0 g CH3CH2OH in 1.00 kg of CH3OH
14. Which solution has the higher boiling point?

a. 35.0 g of C3H8O3 in 250 g ethanol or 35.0 g of C2H6O2 in 250.0 g ethanol
b. 20.0 g C2H6O2 in 0.50 kg of H2O or 20.0 g of NaCl in 0.50 kg of H2O
15. Calculate the vapor pressure of a solution of 44.0 g of glycerol (C 3H6O3) in

500.0g of water at 250C. The vapor pressure of water at 250C is 23.76 torr.
16. Calculate the vapor pressure of a solution of 0.39 mol of cholesterol in 5.4 mol of
toluene at 320C. Pure toluene has a vapor pressure of 41 torr at 32 0C.
17. What is the freezing point of 0.111 m of urea in water?
18. What is the boiling point of 0.200 m lactose in water?
19. The boiling point of ethanol (C2H5OH) is 78.50C. What is the boiling point of
3.4 g of vanillin (MW = 152.14 g/mol) in 50.0 g of ethanol (k b of ethanol
= 1.22 0C/m)
20. What is the minimum mass of ethylene glycol (C2H6O2) that must be
dissolved in 14.5 kg of water to prevent the solution from freezing at
-10.0oF?
21. Wastewater discharged into a stream by a sugar refinery contains sucrose

(C12H22O11) as its main impurity. The solution contains 3.42 g of sucrose/L. A
government industry project is designed to test the feasibility of removing
the sugar by reverse osmosis. What pressure must be applied to the apparatus
at 200C to produce pure water?
22. In a study designed to prepare new gasoline-resistant coatings, a polymer chemist
dissolves 6.053 g of poly(vinyl alcohol) in enough water to make 100.0 mL of
solution. At 250C, the osmotic pressure of this solution is 0.272 atm. What is the
molar mass of the polymer sample?

CHEm 1221 lec:
Analytical chemistry
MODULE 4: RATES AND EQUILIBRIUM CONSTANTS
At the end of this module the student should be able to:

 Evaluate chemical reactions according to type and solve problems involving
chemical reactions and governing equilibrium
A. CHEMICAL REACTION - chemical change involving interaction between

substances of the electrons in their atomic, molecular or ionic structure.
2 Types of Chemical Reaction:

1. Reversible reaction - reaction that involves the formation of reactants
from their original products
- does not go to completion
A + B ⇌ C + D
2 opposing reactions:
- one proceeding to the products
- the other, the products recombining to produce the initial reactants
2. Irreversible reaction - the products are not capable of forming the

reactants
- goes to completion
A + B → C + D
Evidences:
a. when a gas is formed / formation of bubbles
b. when water droplets are formed
c. when an insoluble precipitate is formed

B. REACTION RATE - magnitude of change in the concentration of reactants or
products per unit time
- rate or speed at which reactants are consumed or products are
formed
- also rate of chemical reaction or velocity of a chemical reaction
- the amount of chemical change which takes place in a given
interval of time usually expressed in unit of moles of substance
used up or formed / liter of solution•unit time like:
mole/ li . sec or M /sec
Ex. A + 2B → 3C
t=0 5moles/L 3moles/L 0
t = 5min 1mole/L 2moles/L 3moles/L
What is the reaction rate of the above reaction in terms of A? of B? of C?
Factors Which Influence the Speed of Reactions:

1. nature of reacting substances- the more reactive the reactants , the faster
is the reaction rate
ex.a.) Active metals displace hydrogen vigorously and rapidly from acids,
while less active metals act slowly
b.) Nitrogen is inert; it combines very slowly with other elements
c.) halogens combine with most of the other elements readily
2. concentration of reactants – increases in the concentration of reactants,

increases the rate of formation of products
3. temperature – increase in temperature will speed up an endothermic

reaction and slows down an exothermic reaction
4. presence of catalyst/s
a. positive catalyst- increase the rate of a reaction
b. negative catalyst – reduces the rate of reaction
5. pressure – affects only gaseous systems
C. CHEMICAL EQUILIBRIUM - state at which two opposing reactions proceed at the

same rate
- only reversible reactions attain chemical equilibrium
- a state of reversible reaction where the rates of forward and backward
reactions are equal
A + B ⇌ C + D
at chemical equilibrium: rate of forward reaction = rate of backward reaction

Rf = Rb
C.1 Law of Mass Action ( Guldberg and Waage) - states that the rate of a
chemical reaction is directly proportional to the molar concentration of the
reactants each raised to a number equivalent to the corresponding
coefficient in the balanced chemical equation
aA + bB ⇌ cC + dD
Rf ∝ [ A ] a [ B ] b Rf = ka [ A ] a [ B ] b
Rb ∝ [ C ] c [ D ] d Rb = kb [ C ] c [ D] d
C.2 Law of Chemical Equilibrium – states that for a reversible chemical reaction,
at a fixed temperature and in a state of equilibrium, the product of the formula
– weight concentrations of substances formed in the reaction divided by the
product of the formula – weight concentrations of the reactants, each raised
to the power indicated by the number of molecules or ions in the balanced
equation is equal to a CONSTANT.
Ex. aA + bB ⇌ cC + dD
[𝐶]𝑐 [𝐷]𝑑
keq = [𝐴]𝑎 [𝐵]𝑏
D. LE CHATELIER’S PRINCIPLE – states that when a stress is applied to a system in

equilibrium whereby the equilibrium is altered, the equilibrium will shift in such
a manner as to relieve or neutralize the effect of the added stress.
N2 + 3H2 ⇌ 2 NH3
∆𝐻 = −22080 𝑐𝑎𝑙 𝑎𝑡 25℃
Since ∆𝐻 is negative , it is exothermic rxn

Forward Reaction – production of NH3
Backward Reaction - production of N2 & H2
stress – change in temp ( inc) supply of heat to the system

 result Rb > Rf

E. FACTORS AFFECTING A SYSTEM AT CHEMICAL EQUILIBRIUM
1. Concentration
- An increase in concentration of the reactants will cause the equilibrium
to shift to the right; it is to the left when the concentration of the
reactants are decreased
H2 + Cl2 ⇌ 2 HCl
Increase [ H2] = forward(FW) reaction is favored
Increase [ HCl] = backward (BW) reaction is favored
Decrease [ HCl] = FW reaction is favored
Decrease [ H2] = BW reaction is favored
2. Temperature

- For endothermic reactions, an increase in temperature will cause the
equilibrium to shift to the right, a decrease in temperature will produce
a shift to the left. The opposite is expected for exothermic reactions
Ex. N2 + 3H2 ⇌ 2 NH3 + heat

Exothermic: increase temp: BW reaction is favored
decrease temp : FW reaction is favored
N2 + 3H2 + heat ⇌ 2 NH3

Endothermic: increase temp: FW reaction is favored
decrease temp : BW reaction is favored

3. Pressure
- applicable only to gaseous systems ( mole = volume)
N2 + 3H2 ⇌ 2 NH3
increase P ( favors side occupying lower volume ) = FW reaction is favored

decrease P ( favors side occupying greater volume ) = BW reaction is
favored
 Catalyst – a substance that influences the speed of a chemical reaction

without itself undergoing a permanent change.
- A catalyst cannot change the numerical value of the equilibrium
constant and hence the relative amounts of reactants and products
present at equilibrium. However, it may greatly reduce the time
necessary for the establishment of equilibrium ( therefore , only the value
of the rate of chemical reaction is affected by a catalyst)
2 types:
a. Positive catalyst ( true catalyst)
o Speeds up the rate of chemical reaction
𝑀𝑛𝑂2
Ex. 2KClO3 → 2KCl + 3O2
b. Negative catalyst ( inhibitor)

- Slows down or retards the rate of chemical reaction
𝐶𝑢𝑆𝑂4
Ex. Mg + 2HCl → MgCl2 + H2
 Since the catalyst will have the same effect on forward and backward reactions,
there is no actual effect
EQUILIBRIUM CONSTANTS
- It is the product of the molar concentrations of the product for a chemical reaction,
each raised to the power of the respective coefficient in the equation, divided by the
product of the molar concentration of the reactants each raised to the power of the
respective coefficient in the equation.

Equilibrium the numbers in front of each subst
“c” shows that (coefficient) in the cheml eqn
it is in terms of the brackets show
concentration concentration in mol/L
[ ] – represents Molar concentration (moles/L)
- Note: In writing equilibrium constant expressions, only concentrations of

gases and substances in aqueous solution are written in an equilibrium
constant expression. The reason for this convention is that , other
substances (pure solids and pure liquid) have constant concentrations.
WRITING THE EQUILIBRIUM-CONSTANT EXPRESSSION:

d. 2NO (g) + 5H2 (g) ⇌ 2NH3 (g) + 2H2O (g)
[𝑁𝐻3]2 [𝐻2𝑂]2
𝐾𝑐 = [𝑁𝑂]2 [𝐻2]5
e. NH4Cl (s) ⇌ NH3 (g) + HCl (g)
𝐾𝑐 = [𝑁𝐻3 ][𝐻𝐶𝑙 ]
 NOTE: NH4Cl is not included in the equilibrium constant expression since it is in
solid form
COMPUTING EQUILIBRIUM CONSTANTS AND EQUILIBRIUM CONCENTRATIONS:
Examples:
1. The following concentrations were measured from equilibrium mixture at 500K

contains [ N2 ] = 3.0 x 10 -2 M, [ H2 ] = 3.7 x 10 -2 M, [NH3] = 1.6 x 10 -2 M. Calculate
the equilibrium constant for the reaction: N2 (g) + 3H2 (g) ⇌ 2 NH3 (g)
Solution:
Analyze: We are given the balanced equilibrium equation and equilibrium
concentrations and are asked to determine the equilibrium constant
Plan: Using the balanced equation, we write the equilibrium-constant expression.
We then substitute the equilibrium concentrations into the expression and solve for K c.
Solve:
[𝑁𝐻3 ]2
𝐾𝑐 =
[𝑁2 ][𝐻2 ]3
(1.6 × 10−2 )2
𝐾𝑐 =
(3.0 × 10−2 )(3.7 × 10−2 )3
𝐾𝑐 = 168
2. An equilibrium mixture of gaseous oxygen, nitrogen oxide and nitrogen dioxide at

500K contains 1.0 x 10 -3 M O2 and 5.0 x 10 -2 M NO2 at this temperature, the
equilibrium constant is 6.9 x 10 5. Calculate for [NO]

2NO (g) + O2 (g) ⇆ 2NO2 (g)
Solution:
concentrations of O2 and NO2 and the equilibrium constant , and we are asked to
determine concentration of NO
We then substitute the equilibrium concentrations into the expression ,K c and solve for
concentration of [NO]
Solve:
[𝑁𝑂2 ]2
𝐾𝑐 =
[𝑂2 ][𝑁𝑂]2
(5.0 × 10−2 )2
6.9 × 105 =
(1.0 × 10−3 )[𝑁𝑂]2
[𝑁𝑂] = 1.9035 × 10−3 M or mol/L

3. Nitryl chloride, NO2Cl, is in equilibrium in a closed container with NO 2 and Cl2, at
equilibrium, the concentrations of the substances are: [NO 2Cl] = 0.00106 M, [NO2]
= 0.0108 M, and [Cl2] = 0.00538 M. from these data calculate the equilibrium
constant.
2NO2Cl (g) ⇆ 2NO2 (g) + Cl2 (g)
Solution:
concentrations and are asked to determine the equilibrium constant
We then substitute the equilibrium concentrations into the expression and solve for K c.
Solve:
[𝑁𝑂2 ]2 [𝐶𝑙2 ]
𝐾𝑐 =
[𝑁𝑂2 𝐶𝑙 ]2
(0.0108)2 (0.00538)
𝐾𝑐 =
(0.00106)2
𝐾𝑐 = 0.5585

COMPUTING EQUILIBRIUM CONCENTRATIONS FROM ORIGINAL CONCENTRATIONS:
We often don’t know the equilibrium concentrations of all chemical species in an

equilibrium. If we know the equilibrium concentration of at least one species, however, we
can generally use the stoichiometry of the reaction to deduce the equilibrium
concentrations of the other species in the chemical equation. We will use the following
procedure to do this:
1. Tabulate the known initial and equilibrium concentrations of all species in the
equilibrium-constant expression.
2. For those species for which both initial and equilibrium concentrations are known,
calculate the change in concentration that occurs as the system reaches
equilibrium.
3. use the stoichiometry of the reaction ( that is, use the coefficients in the balanced
chemical equation) to calculate the changes in concentration for all the other
species in the equilibrium
4. From the initial concentrations and the changes in concentration ,calculate the
equilibrium concentrations. These are used to evaluate the equilibrium constant.
Sample Exercise 1:
A mixture of 5.00 x 10 -3 mol H2 and 1.0 x 10 -2 mol I 2 is placed in a 5.00 L container at

4480C and allowed to come to equilibrium. Analysis of the equilibrium mixture shows
that the concentration of HI is 1.87 x 10 -3M. Calculate Kc at this temperature for the
reaction:
H2 (g) + I 2 (g) ⇆ 2HI (g)
Solution:
Analyze: We are given starting amounts in mol of H2 and I2 and an equilibrium concentration
of the product and we are asked to determine the value of the equilibrium constant for the
formation of HI
Plan: We construct a table to find equilibrium concentrations of all species and use the
equilibrium concentrations to calculate the equilibrium constant.
Solve: First , we calculate for the initial concentrations in Molar, since we are given the initial
amounts of the substances in mol and we are given the volume . We use the formula, M =
mols/ V in L.
5.00×10−3 𝑚𝑜𝑙
[𝐻2 ] = = 1.0 × 10−3 𝑀
5𝐿
1.00×10−2 𝑚𝑜𝑙
[𝐼2 ] = = 2.0 × 10−3 𝑀
5𝐿
Second , we tabulate the known initial concentrations of all the species in the equilibrium
constant expression. We also provide space in our table for listing the changes in

concentrations. As shown, it is convenient to use the chemical equation as the heading for
the table.
H2 (g) + I2 (g) ⇆ 2HI (g)
Initial , I: 1.0 × 10 𝑀−3 −3
2.0 × 10 𝑀 0
Change, C:
Equilibrium, E: 1.87 × 10−3 𝑀
Third, we calculate the change in concentration of HI , using the initial and equilibrium
values. The change is the difference between the equilibrium and initial values ,
1.87 x 10 -3 M
Fourth, we use the stoichiometry of the reaction to calculate the changes in the other
species. The balanced chemical equation indicates that for 1 mol of H 2 that reacts ,1mol of
I2 is also consumed and 2 moles of HI are produced. Thus the amount of H 2 consumed is
1.87×10−3
M
2
1.87×10−3
the amount of I 2 consumed is M
2
Fifth, we calculate the equilibrium concentrations , using the initial concentrations and the
changes. The equilibrium concentrations of H 2 and I2 are the initial concentration minus
that consumed
1.87×10−3
[𝐻2 ] = 1.0 × 10−3 𝑀 − = 6.5 × 10−3 𝑀
2
1.87×10−3
[𝐼2 ] = 2.0 × 10−3 𝑀 − = 1.605 × 10−3 𝑀
2
Likewise, the equilibrium concentration of HI is
[𝐻𝐼 ] = 0 + 1.87 × 10−3 𝑀 = 1.87 × 10−3 𝑀
The completed table now looks like the following :
H2 (g) + I2 (g) ⇆ 2HI (g)

Initial , I: 1.0 × 10 −3
𝑀 2.0 × 10 −3
𝑀 0
1.87×10−3 1.87×10−3
Change, C: −
2
− + 1.87 × 10−3 𝑀
2
Equilibrium, E: 6.5 × 10−3 𝑀 1.605 × 10−3 𝑀 1.87 × 10−3 𝑀
Finally, now that we know the equilibrium concentration of each reactant and product, we
can use the equilibrium-constant expression to calculate the equilibrium constant

[𝐻𝐼 ]2
𝐾𝑐 = [𝐻2 ][𝐼2 ]
(1.87 ×10−3 𝑀)2

𝐾𝑐 = (6.5 ×10−3 𝑀)(1.605 ×10−3 𝑀)
= 50.5
Sample Exercise 2:
Five moles of HBr were placed in a 2.0L flask and the flask was heated to 1025 0C
where the equilibrium HBr (g) ⇆ H2 (g) + Br2 (g) was established. Assuming no
volume change of the flask at the high temp, what would be the concentrations of
H2 and Br2 at equilibrium? The Kc for HBr at 10250C is 7.32 × 10−8
Given: V = 2.0 L T = 10250C

HBr = 5 mols Kc = 7.32 × 10−8
Reqd: [ H2], [Br2]
Analyze: We are given a volume, an equilibrium constant. and starting mol amount of
reactant for an equilibrium and we are asked calculate equilibrium concentrations of H 2
and Br2
Plan: We construct a table to find equilibrium concentrations of all species .

Solve: First , we calculate for the initial concentrations in Molar, since we are given the initial
amounts of the substances in mol and we are given the volume . We use the formula, M =
mols/ V in L.
5𝑚𝑜𝑙
[𝐻𝐵𝑟] = = 2.5 𝑀
2𝐿
Second , we tabulate the known initial concentrations of all the species in the equilibrium
constant expression. We also provide space in our table for listing the changes in
concentrations. As shown, it is convenient to use the balanced chemical equation as the
heading for the table.
2HBr (g) ⇆ H2 (g) + Br2 (g)

Initial , I: 2.5 𝑀 0 0
Change, C:
Equilibrium, E:
Third, we use the stoichiometry of the reaction to calculate the changes in concentrations
that occur as the reaction proceeds to equilibrium. Let’s represent the change in
concentration of HBr by the variable x. the balanced chemical equation tells us the
relationship between the changes in concentration of the three gases

For each x mol H2 and Br2 produced , 2x mole HBr must be consumed
Fourth, we use the initial concentration of HBr and the changes in concentrations as
dictated by stoichiometry , to express the equilibrium concentrations. With all of our entries,
we now have the following:
2HBr (g) ⇆ H2 (g) + Br2 (g)

Initial , I: 2.5 𝑀 0 0
Change, C: −2𝑥 +𝑥 +𝑥
Equilibrium, E: 2.5 𝑀 − 2𝑥 𝑥 𝑥
Finally, we substitute the equilibrium concentrations into the equilibrium-constant expression

and solve for the single unknown , 𝑥
[𝐻2 ][𝐵𝑟2 ]
𝐾𝑐 = [𝐻𝐵𝑟 ]2
(𝑥)(𝑥)
7.32 × 10−8 = (2.5 𝑀−2𝑥 )2
if you have an equation solving calculator, you can solve this equation directly for x if not,
get the square root of both sides since it is a perfect square. ( in some problems that are not
perfect square use quadratic equation in solving X )
√7.32 × 10−8 = (𝑥)(𝑥)

√ 2
(2.5 𝑀 − 2𝑥)
√7.32 × 10−8 𝑥
=
2.5 𝑀 − 2𝑥
Simplifying:
𝑥
𝑥 =
2.5 𝑀 − 2𝑥
𝑥 = 6.7602 × 10−4 𝑀 = [𝐻2 ] = [𝐵𝑟2 ]
ASSIGNMENT:
SUPPLEMENTARY PROBLEMS:
1. Write an equilibrium constant expression, Kc, for the following reactions:

a. NO (g) + O2 (g) ⇆ NO2 (g)
b. CS2 (g) + H2 (g) ⇆ CH4 (g) + H2S (g)
c. NH3 (g) + O2 (g) ⇆ NO (g) + H2O (g)
d. NO2 (g) + H2 (g) ⇆ NH3 (g) + H2O (g)
e. CaCO3 (s) ⇆ CaO (s) + CO2 (g)
f. (CH3)2CO (l) ⇆ (CH3)2CO (g)
g. Na2CO3 (s) + 2C (s) + N2 (g) ⇆ NaCN (s) + CO (g)
2. What is the value of the equilibrium constant for the reaction:
N2 (g) + O2 (g) ⇆ NO2 (g)

If the concentrations of each species at equilibrium are as follows:
[N2] = 0.0013M, [O2] = 0.0042 M, [NO2] = 0.00065 M
3. At 250 0C , 1.10 mol PCl5 (g) was introduced into a 1.0 L container, equilibrium was
established. PCl5(g) ⇆ PCl3(g) + Cl2 (g)
At equilibrium, the concentration of PCl 3 (g) was 0.050 M
a) What were the equilibrium concentrations of Cl 2 (g) and PCl5 (g)?
b) What is the value of Kc at 250 0C?
4. A mixture of 0.0080 mol of SO2 (g) and 0.0056 mol O2 (g) is placed in a 1.0 –liter
container. When equilibrium is established, 0.0040 mol of SO 3 (g) is present.
SO2 (g) + O2 (g) ⇆ SO3 (g)
a) What are the equilibrium concentrations of SO2 (g) and O2 (g) ?
b) What is the Kc value?
5. For the reaction : H2 (g) + CO2 (g) ⇆ H2O (g) + CO (g), KC is 0.771 at 750 0C, if
0.0100 mol of H2 and 0.0100 mol of CO2 (g) are mixed in a 1.0- L container at 7500C.
What are the concentrations of all substances present at equilibrium?
6. If 0.025 mol of COCl2 (g) is placed in a 1.0L container at 4000C, 16.0% of the COCl 2 (g)
is dissociated when equilibrium is established. Calculate Kc value for the equilibrium
at 4000C
COCl2 ⇔ CO (g) + Cl2 (g)

MODULE 5: IONIZATION AND IONIZATION CONSTANTS
IONIZATION OF WEAK ELECTROLYTES
Certain covalent compounds such as HC2H3O2, NH4OH, etc undergo partial

ionization when in water solutions. Equilibrium is thus established between ionized and non-
ionized molecules. The equilibrium constant principle may therefore be applied to such
ionized equation.
Ionization - process by which a compound in its molten (dissolved in medium) is broken

down in its individual ionic components
Some Postulates of Modern View of Ionization
 Electrolytes, when dissolved in water or any other ionizing medium produced

electrically charged particles called ions.
2 types of ions:
a. cation - positively charged ion
b. anion - negatively charged ion

 Electrolytes are compounds whose water solutions conduct electricity due to the
presence of ions.
2 kinds of electrolytes
a. strong electrolytes - substance whose water solution is completely ionized
(100%)
- they have more ions in water than weak electrolytes
- good conductor of electricity

Examples: strong acids
strong bases
salts like NaCl, KCl, CaCl 2, Na2SO4
all nitrates like NaNO3, AgNO3
b. weak electrolytes - substances whose water solution are partially ( or

incompletely ionized in water solution
- they have few ions in water
- poor conductor of electricity
Examples: weak acids

weak bases
slightly soluble salts

Note:The production of ions from a weak electrolyte is a reversible reaction to which
the law of chemical equilibrium is applicable. Ionization of strong electrolyte
are irreversible and thus, not covered by the chemical equilibrium law.

CONJUGATE ACID-BASE PAIRS
Examples:
0R:

remove H+
HC2H3O2 + H2O ⇌ H3O+ + C2H3O2-

Acid base conjugate conjugate
Acid base
add H+
Ionization Constant , Ki - equilibrium constant describing the ionization of a weak

acid or base
Degree of Ionization, ∝ - a measure of the extent of ionization
∝ = concentration of ions
concentration of solution
% ionization = concentration of ions X 100

concentration of solution

I. Ionization of Weak Acids ( Ka)
Acid - defined as a proton donor ( Bronsted Lowry Theory )
Kinds of acids:
1. monoprotic weak acid - those that yield only one H+ upon ionization
ex. HC2H3O2 , HF, HC7H5O2, HNO2
2. polyprotic weak acids - those that yield more than one H + upon
ionization
ex. H2S , H3PO4 , H2C2O4 , H2CrO4
A. Ionization of a weak monoprotic acid:
HC2H3O2 + H2O ⇌ H3O + + C2H3O2 –
Ke = [ H3O+ ] [ C2H3O2-]
[HC2H3O2][H2O]
but [H2O] = 55.5 F

and Ke x 55.5 F = Ka
Ka = [ H3O+ ] [ C2H3O2-]
[HC2H3O2]
General Representation:
HA + H2O ⇌ H3O + + A–
HA - general weak acid
Ka = [ H3 O +] [ A- ]
[ HA ]
B. Ionization of weak base
General representation:
B + H2O <==> BH+ + OH-
Kb = [BH+][OH-]
[B]
OR

BOH <==> B+ + OH-
Kb = [B+][OH-]
[BOH]
Ex. NH3 + H2O ⇌ NH4+ + OH -
Kb = [ NH4+ ] [ OH-]
[NH3]
In partially ionized solution, equilibrium exists between dissociated ions and
undissociated ions – thus, the greater the dilution, the higher is the degree of ionization
SOLUTIONS TO PROBLEMS ON PP.81-82 ( QUANTITATIVE ANALYSIS BY GILREATH)
1/81] Nitrous acid, in a 0.1F solution, is 6.5% ionized. Calculate the ionization constant for this
acid.
Given: % ionization = 6.5 %
Reqd: Ka
Solution:
 We are given the formal concentration of an aqueous solution of weak acid and
the percentage ionization and we are asked to determine the value of K a for the
acid.
 Although we are dealing specifically with the ionization of weak acid, this problem is
very similar to the equilibrium problems we encountered in the previous module. We
can solve it starting with the chemical reaction and a tabulation of initial equilibrium
concentrations.
∝ = 6.5 %/100 = 0.065, therefore only 0.065 x initial conc of HNO2

is being ionized or dissociated into H3O + and NO2-
HNO2 + H2O ⇄ H3O + + NO2-

Init conc or I: 0.1 F 0 0
During reaction or C: – 0.1(0.065) + 0.1(0.065) + 0.1(0.065)
_______________________________________________________________________
At eqbm or E : 0.1 – 0.1(0.065) 0.1(0.065) 0.1(0.065)
Note: there are no entries in the column beneath water because water is a solvent and
does not appear in the ionization constant expression
Ka = [H3O +][NO2-] = (0.0065) (0.0065)

[ HNO2 ] ( 0.1 – 0.0065)
Ka = 4.52 x 10-4

2/81 Compute the formula-weight concentration of NH3 in a solution in which it is known
to be 4.0 % ionized.
Given: % ionization = 4 %
Reqd: FW conc
Solution:
Let C = initial concentration of [NH3]
∝ = 4%/100 = 0.04, therefore only .04 x initial conc of NH3 is being
ionized or dissociated into NH4 +1 and OH-1
NH3 + H2O ⇄ NH4 +1 + OH-1

Init conc or I: C 0 0
During reaction or C: -.04 C +.04 C +.04 C
______________________________________________________________________________
At eqbm or E : C- .04 C .04 C .04 C
[𝑁𝐻4 +1 ][𝑂𝐻−1 ]
K NH3 =
[𝑁𝐻3 ]
−5 (0.04𝐶)(0.04𝐶)
1.8 × 10 = K NH3 = 1.8 x 10-5 p.269 (Gilreath)
(𝐶−.04𝐶)
 Since Kb ≤ 10−4 ,neglect the value 0.04C in the denominator .From a mathematical
viewpoint, ,it maybe discarded only if it is a very small value in comparison the value
from which it is subtracted
(0.04𝐶)(0.04𝐶)
Therefore: 1.8 × 10−5 = (𝐶)
C = 0.01F (gram-formula weight conc is F)
5/81) What concn of hydroxide ions exists in a 0.009F of NH3 at 250C?
Given: [NH3] = 0.009F

Reqd: conc of OH-
Soln:
Let X = amount of NH3 that ionized
 We want to find the equilibrium value for [OH -].let’s call this quantity x. the
concentration of NH3 before any of it ionizes is 0.009 F. The chemical equation tells
us that for each molecule of NH3 that ionizes, one OH- and one NH4 +1 are formed.
Consequently ,if x moles per liter of OH- form at equilibrium, x moles per liter of NH4 +1
must also form and x moles per liter of NH3 must be ionized.

NH3 + H2O ⇄ NH4 +1 + OH-1
Init conc or I: 0.009 0 0
During reaction or C: - X +X + X
_______________________________________________________________________
At eqbm or E : 0.009- X X X
 We need to substitute the equilibriumconcentrations into the equilibrium-constant

expression
[𝑁𝐻4 +1 ][𝑂𝐻 −1 ]
K NH3 = [𝑁𝐻3 ]
(𝑋)(𝑋)
1.8 × 10−5 = K NH3 = 1.8 x 10-5 p.269 (Gilreath
(0.009−𝑋)
 This expression leads to a quadratic equation in x, which we can solve by using the
quadratic formula. We can also simplify the problem, however by noting that the
value of Kb is quite small.
 Since Kb ≤ 10−4 ,neglect the value of X in the denominator .From a mathematical

viewpoint, ,it maybe discarded only if it is a very small value in comparison the value
from which it is subtracted, in this particular problem, X can be ignored bec it does
not represent a significant number as compared with the concentration of
ammonia , given as 0.009F
 OR: use the RULE OF 5%
= Ka/Kb ÷ init conc x 100

= 1x10-5/ 0.009 x 100
=0 .2%
X = [OH-1] = 4.02 X 10-4 F
6/81) What is the hydronium –ion conc of a 0.05F HCN solution?
Given: [HCN] = 0.05F

Reqd: conc of H3O+
Soln:
Let X = amount of HCN that ionized
HCN + H2O ⇄ H3O +1 + CN-1

Init conc or I: 0.05F 0 0
_______________________________________________________________________

[𝐻3 𝑂+ ][𝐶𝑁−1 ]
K HCN = [𝐻𝐶𝑁]
(𝑋)(𝑋)
7.2 × 10−10 = K HCN = 7.2 x 10-10 p.269 (Gilreath
(0.05−𝑋)
 Since Ka ≤ 10−4 ,neglect the value of X in the denominator .

 OR: use the RULE OF 5%
X = [H3O+] = 6 X 10 -6 F
9/82) What weight of lactic acid must be dissolved in a liter of solution to produce a
hydronium- ion concentration of 1x10-3F?
Given: [H3O+] = 1x10-3F V = 1lL

Reqd: g of HC3H5O3
Soln:
Let C = initial concentration of HC3H5O3
HC3H5O3 + H2O ⇄ H3O +1 + C3H5O3-1

During reaction or C: - .001 +.001 +.001
_______________________________________________________________________
At eqbm or E : C- . 001 0.001 0.001
NOTE: at eqbm the conc of [H3O+] is 1x10-3 F, therefore the amt of HC3H5O3 ionized is also
1x10-3F
[𝐻3 𝑂+ ][C3H5O3−1 ]
K HC3H5O3 = [HC3H5O3]
(0.001 )(0.001 )
1.39 × 10−4 =
(C− .001)
 Since Ka ≤ 10−4 ,neglect 0.001 in the denominator .

C = [HC3H5O3] = 7.19 X 10 -3 F
𝑊
Since weight is unknown: 𝐹 = (𝐹𝑊)(𝑉 𝑖𝑛 𝐿)
𝑊
7.19 𝑥 10−3 = (90𝑔/𝑚𝑜𝑙)(1𝐿) W= 0.65 g

10/82) What is the hydronium –ion concentration of 1g acetic acid in 100 mL of water?
Given: HC2H3O2 = 1 g V = 100 mL

Reqd: [H3O+]
Soln:
Let X = amount of HC2H3O2 that ionized
𝑊
𝐹 = (𝐹𝑊)(𝑉 𝑖𝑛 𝐿)
1𝑔
= (60𝑔/𝑚𝑜𝑙)(0.1𝐿)
= 0.167F
HC2H3O2 + H2O ⇄ H3O +1 + C2H3O2-1

_______________________________________________________________________
[𝐻3 𝑂+ ][ C2H3O2−1 ]
K HC2H3O2=
[HC2H3O2]
(X )(X )
1.75 × 10−5 = (0.167− X)
 Since Ka ≤ 10−4 ,neglect X in the denominator . (refer to explanation on previous

problems)
X = [H3O+] = 1.7X 10 -3 F
AUTOPROTOLYSIS OF WATER (IONIZATION OF WATER)
water is an amphoteric/amphiprotic solvent (can either act as an acid or a base)

- can undergo self ionization (autoprotolysis)

H2O + H2O <==> H3O+ + OH-
at eqbm: c-x x x
the equilibrium constant:

KC = [H3O+][OH-] [H2O] = 55.5M
[H2O]2
KC [H2O] = [H3O+][OH-]
2
KW = [H3O+][OH-]
or KW = [H+][OH-]
Kw - ion-product constant
1 × 10−14 =[H3O+][OH-]
1 × 10−14 = (𝑥)(𝑥)
X = 1x10-7F = [H3O+] =[OH-]
in pure water at 25oC and in exactly neutral solutions:
[H3O +] = [OH-] = 1 x 10-7 M
KW = [H3O+][OH-] = [1 x 10-7]2
KW = 1 x 10-14
1 x 10-14 = [H3O+][OH-]
 in non-neutral solution [H3O +] ≠ [OH-]
Potential of Hydrogen (pH)

- pH concept was developed in 1909 by S.P.L. Sorensen, a Danish Biochemist
- pH is a convenient way of expressing the hydrogen ion concentration
- measures the relative acidity
- negative logarithm to the base 10 of hydrogen ion
p expression – expression of small numbers in terms of its negative logarithm
forms:
1. pKi = - log Ki
2. pKa = - log Ka
3. pKb = - log Kb
4. pH = – log [H+] = – log [H3O+]
5. pOH = - log [OH-]
6. pKw = - log Kw
[H3O+] = antilog –pH

[OH-] = antilog –pOH

pH + pOH = pKW pKW = - log KW
pH + pOH = 14
 In neutral solution pH = pOH = 7

 in non-neutral solution 𝒑𝑯 ≠ 𝒑𝑶𝑯, however the sum of pH & pOH should be
equal to 14
pH > 7 =basic
pH < 7 = acidic
pH = 7 neutral solution
0 𝟏𝒙𝟏𝟎−𝟏 𝟏𝒙𝟏𝟎−𝟐 𝟏𝒙𝟏𝟎−𝟕 𝟏𝒙𝟏𝟎−𝟖 𝟏𝒙𝟏𝟎−𝟏𝟒
pt of neutrality
pH 1 2 7 8 14
ACIDIC BASIC

pH calculations:
Calculate the pH of each of the ff solutions:

a) 0.0010M HCl c) 0.0050N HNO3
b) 0.0010 N NaOH d) 0.0050M Ba(OH)2 assuming they are 100% ionized.
Relationship between Ionization of weak acids and weak bases:
KW = Kb x Ka
pKW = – log KW
pKa = – log Ka
pKb = – log Kb
pKa = – log Ka and

HA <==> H+ + A-
Ka = [H+][A-]
substituting [HA]
pKa = – log [H+][A-] = – log [H+] – log [A-]
[HA] [HA]
= pH – log [A-]
[HA]
• Guidelines in determining strength of acids and bases using Ka and Kb values :
Ka or Kb > 103 very strong acid or base

Ka or Kb = 10-2 to 102 strong acid or base
Ka or Kb = 10-7 to 10-3 weak acid or base
Ka or Kb < 10-7 very weak acid or base
For Ka / Kb calculations:
1- write the balanced equation
2- write the known initial and eqbm concns
3- write changes in concn at eqbm
4- from initial concns and changes in concns, calc eqbm concns
5- write Ka / Kb expression
6-substitute eqbm concns to Ka / Kb expression
a. HCl + H2O → H3O + + Cl-
init conc :t =0 0.001 0 0
during rxn/t rxn: - 0.001 0.001 0.001
after rxn: 0 0.001 0.001
 Note: HCl is strong acid, therefore it will ionize completely
pH = – log [H+] = – log [H3O+]

pH = -log (0.001)

pH= 3 (strong acid)
b. M = N/f
0.001 𝑒𝑞/𝐿
𝑀= 1𝑒𝑞/𝑚𝑜𝑙
𝑀 = 0.001𝑀
NaOH + H2O → Na + + OH-

init conc :t =0 0.001 0 0
during rxn/t rxn: - 0.001 0.001 0.001
after rxn: 0 0.001 0.001
pOH = – log [OH-] pOH = -log (0.001)

pOH= 3
pH = 14-pOH
pH = 14 – 3 = 11 (strong base)
d. Ba(OH)2 → Ba +2 + 2OH-
init conc :t =0 0.005 0 0
during rxn/t rxn: - 0.005 0.005 2(0.005)
after rxn: 0 0.005 2(0.005)
pOH = – log [OH-] pOH = -log (2x0.005)

pOH= 2
pH = 14-pOH
pH = 14 – 2 = 12 (strong base)
Weak acids and weak bases:
ex. What is the [H +] of a solution of acetic acid that is 0.05F? What is its pH?
Given: [HC2H3O2] = 0.05F
Reqd: conc of H3O+ , pH
Soln: Formality = molarity
HC2H3O2 + H2O ⇄ H3O +1 + C2H3O2-1

During reaction or C: -X +X + X
_______________________________________________________________________
At eqbm or E : 0.05F - X X X
[𝐻3 𝑂+ ][𝐶2𝐻3 𝑂2−1 ]

𝐾 𝐻𝐶2 𝐻3 𝑂2 = [𝐻𝐶2 𝐻3 𝑂2 ]

(X )(X )
1.75 × 10−5 = (0.05− X)
 Since Ka ≤ 10−4 ,neglect X in the denominator . (refer to explanation on previous

problems)
X = [H3O+] = 9.3541X 10 -5 F
pH = – log [H3O+]
pH = – log [9.3541X 10 -5]
pH = 4.0290
27/82] What concentration of nitrous acid is required to produce a solution with a pH of

2.6?
Given: pH = 2.6
Reqd: init conc [HNO2]
Soln:
[H3O+] = antilog –pH
[H3O+] = antilog –2.6

[H3O+] = 2.5119 x10-3 F
Let C = initial concentration of HNO2

HNO2 + H2O ⇄ H3O +1 + NO2-1
During reaction or C: -2.5119 x10 -3 + 2.5119 x10-3 + 2.5119 x10-3
_______________________________________________________________________
At eqbm or E : C - 2.5119 x10-3 2.5119 x10-3 2.5119 x10-3
[𝐻3 𝑂+ ][𝑁𝑂2 −1 ]
𝐾 𝐻𝑁𝑂2 =
[𝐻𝑁𝑂2 ]
(2.5119 x−3 )(2.5119 x−3 )
4 × 10−4 = ( C − 2.5119 x−3 )
 Since Ka ≤ 10−4 ,neglect 2.5119 x10-3 in the denominator . (refer to explanation on

previous problems)
C = [HNO2] = 0.0158 F

COMMON – ION EFFECT
- phenomenon wherein the dissociation of a weak electrolyte is decreased by adding to

the solution a strong electrolyte that has an ion common with the weak electrolyte
- the effect upon the addition of a substance ( strong electrolyte- salt solution) which
furnishes an ion that is common on what is produced on the ionization of a weak
electrolyte
examples: a) if sodium acetate, NaC2H3O2 is added to acetic acid, HC2H3O2

NaC2H3O2  Na+ + C2H3O2-
spectator ion common ion
HC2H3O2 <=> H+ + C2H3O2-
- the addition of C2H3O2- from NaC2H3O2 causes the equilibrium to shift to the left thus,
decreasing the H+ concentration (Le Chatelier’s Principle)
 NOTE: NaC2H3O2 dissociates completely in aqueous soln

inc [C2H3O2-] & inc [ H3O+] ⇨ BW rxn is favored
inc [HAc] & inc [H2O] ⇨ FW rxn is favored

example: What is the pH of a 0.2 F acetic acid, HC2H3O2 ? What is the new pH of this
solution if 2.0 g of sodium acetate, NaC2H3O2 were added to 250 mL of this acetic acid
solution?
HC2H3O2 + H2O <=> H3O+ + C2H3O2-

I: 0.2F 0 0
C: -X +X +X
E: 0.2- X X X
[𝐻3 𝑂+ ][𝐶2𝐻3 𝑂2− ]

𝑲𝒂 = [𝑯𝑪𝟐 𝑯𝟑 𝑶𝟐 ]
(𝑿)(𝑋)
𝟏. 𝟕𝟓𝑿𝟏𝟎−𝟓 = (𝟎.𝟐−𝑿)
X in the denominator = 0 (refer to previous notes)
𝑿 = [𝐻3 𝑂+ ] = 1.8708 𝑋 10−3 𝐹

pH = -log (1.8708 𝑋 10−3 )
= 2.7280
𝑊
b) 𝐹 = (𝐹𝑊)(𝑉)
2𝑔
𝐹= 𝑔
(82𝑚𝑜𝑙)(0.25𝐿)
F = 0.0976 F

100%
t=0: I 0.0976 0 0
t=rxn: C - 0.0976 0.0976 0.0976
after: E 0 0.0976 0.0976
finally:
HC2H3O2 + H2O <=> H3O+ + C2H3O2-

I: 0.2F 0 0.0976
C: -X +X +X
E: 0.2- X X 0.0976 + X
[𝐻3 𝑂+ ][𝐶2 𝐻3 𝑂2− ]

𝑲𝒂 =
[𝑯𝑪𝟐 𝑯𝟑 𝑶𝟐 ]
(𝑿)(𝟎.𝟎𝟗𝟕𝟔+𝑋)
𝟏. 𝟕𝟓𝑿𝟏𝟎−𝟓 =
(𝟎.𝟐−𝑿)
X in the denominator and x in the numerator = 0 ( the value of x is
very, very small compared to the value 0.0976)
(refer to previous notes)
𝑿 = [𝐻3 𝑂 + ] = 3.5861 𝑋 10−5 𝐹

pH = -log ( 3.5861 𝑋 10−5 )
= 4.45
There was a decrease in the [𝐻3 𝑂 + ] and an increase in pH after the addition of
NaC2H3O2
11/82) Calculate the hydroxide concentration of a solution that is 0.1F with NH3, and 0.25 F
in respect to (NH4)2SO4?
 Follow up question: What happens to the [OH-],pOH & pH of the 0.1F with NH3
after the addition of (NH4)2SO4
Given:
(NH4)2SO4
NH3 or 0.25F (NH4)2SO4

NH4OH 0.1F NH3
Reqd: [OH-] of the mixture

Soln:

100%
(NH4)2SO4  2NH4+ + SO4-2
common ion spectator ion
t=0: I 0.25F 0 0
t=rxn: C - 0.25 2 (.25) 0.25
after: E 0 0.5 0.25
finally:
NH3 + H2O <=> NH4+ + OH-

I: 0.1F 0.5 0
C: -X +X +X
E: 0.1- X 0.5 + X X
[𝑵𝑯𝟒+ ][𝑂𝐻 − ]
𝑲𝒃 = [𝑵𝑯𝟑 ]
(𝑿)(𝟎.𝟓+𝑋)
𝟏. 𝟖𝑿𝟏𝟎−𝟓 =
(𝟎.𝟏−𝑿)
Neglect value of x in NH4+ and x in NH3 since Kb ≤ 10-4 (refer to prev probs)
𝑿 = [𝑂𝐻 + ] = 3.6 𝑋 10−6 𝐹

pOH = -log ( 3.6 𝑋 10−6 )
= 5.54
pH = 14- 5.54 = 8.46
- The original pure 0.1F NH3 has an [OH-] of 1.33x10-3 F and has a pOH of 2.88
- There was a decrease in pH in the NH3 after the addition of (NH4)2SO4
12/82] What concentration of NaC2H3O2 must be added to 0.1F HC2H3O2 to give a

hydronium –ion concentration of 2x10-6gfw/liter?
Given:
? F NaC2H3O2
[H3O+] = 2x10-6F
0.1F NaC2H3O2
HC2H3O2 + 0.1 HC2H3O2
Reqd: [NaC2H3O2]

Soln:
Let x = initial conc of NaC2H3O2
100%
t=0: I x 0 0
t=rxn: C -x x x
after: E 0 x x
finally:
HC2H3O2 + H2O <=> H3O+ + C2H3O2-

I: 0.1F 0 x
C: -2x10-6 +2x10-6 + 2x10-6
E: 0.1- 2x10-6 2x10-6 X+ 2x10-6
[𝐻3 𝑂+ ][𝐶2𝐻3 𝑂2− ]

𝑲𝒂 = [𝑯𝑪𝟐 𝑯𝟑 𝑶𝟐 ]
(2x10−6)(2x10−6+𝑋)
𝟏. 𝟕𝟓𝑿𝟏𝟎−𝟓 = (𝟎.𝟏−2x10−6)
Neglect 2x10-6 in C2H3O2-and 2x10-6 in HC2H3O2(refer to prev probs)
X = [NaC2H3O2] = 0.875 F

BUFFERED SOLUTION
- solution that resists a change in pH upon adding small amounts of acid or base
- mixture of weak electrolyte and its salt which by common-ion effect maintains a nearly
constant hydronium ion concentration when a strong acid or strong base is added
- The principle involved in buffer solutions is the common ion effect
Buffer capacity – amount of acid or base the buffer can neutralize before pH begins to
change
Two types of Buffer:

1. Acid buffer
2. Base buffer

examples:
a) buffer solution composed of weak acid and its salt

- acetic acid, HC2H3O2 and ammonium acetate NH4C2H3O2
- acetic acid, HC2H3O2 and sodium acetate NaC2H3O2
- hydrocyanic acid, HCN and potassium cyanide, KCN
- carbonic acid, H2C2O3 and NaHCO3 , sodium bicarbonate
-potassium dihydrogen phosphate, KH2PO4 and dipotassium hydrogen phosphate, , K 2HPO4
b) buffer solution composed of weak base & its salt

- ammonium hydroxide, NH4OH and ammonium acetate, NH4C2H3O2
- ammonium hydroxide and ammonium chloride, NH4Cl
HC2H3O2 – NaC2H3O2 Buffer Solution may be prepared by any of the ff:
a) adding directly sodium acetate to acetic acid
b) reacting acetic acid with sodium hydroxide (strong base) in such a way that some of
acetic acid remain unreacted
HC2H3O2 + NaOH  NaC2H3O2 + H2O
after mixing: still present all consumed present present
Note: there is a common ion effect between the unreacted HAc and NaAc produced
c) reacting sodium acetate with hydrochloric acid (strong acid) in such a way as to leave
some part of sodium acetate, NaC2H3O2 unreacted
HCl + NaC2H3O2  Na+ + Cl- + HC2H3O2


Note: there is a common ion effect between the unreacted NaAc and HAc produced
NH4OH - NH4Cl Buffer Solution may be prepared by any of the ff:

a) adding ammonium chloride directly to ammonium hydroxide
b) reacting ammonium hydroxide and hydrochloric acid (strong acid) in such a way that
some of ammonium hydroxide remain unreacted
NH4OH + HCl  NH4Cl + HOH
c) reacting sodium hydroxide (strong base) and ammonium chloride in such a way as to
leave part of ammonium chloride (NH4Cl) unreacted
NaOH + NH4Cl  Na+ + Cl- + NH4OH
after mixing: still present all consumed present
Uses of Buffered Solutions:

1. Buffered solutions are used extensively in analytical chemistry, biochemistry and
bacteriology, as well as in photography and the leather and dye industries. In each of
these areas , particularly in biochemistry and bacteriology, certain rather narrow pH ranges
may be required for optimum (best) results. If during the course of a chemical reaction, the
concentration of acids ( or bases) is allowed to increase , an undesirable action may occur
or the desired reaction may be inhibited. The activity of enzymes, the growth of bacterial
cultures, and other biochemical processes depend upon the control of the pH by buffered
systems.
2. Buffered solutions in the body: Intracellular and extra cellular fluids in living organisms
contain conjugate acid-base pairs that function as buffers at the pH of the fluids. The major
intracellular buffer is the dihydrogen phosphate- monohydrogen phosphate, H2PO4- - HPO4-
2, conjugate acid-base pair. The major extra cellular buffer is the carbonic acid-
bicarbonate, H2CO3 – HCO3- , conjugate acid-base pair. The latter buffered system helps
maintain the pH of the blood at nearly constant value, close to 7.4, even though acidic
and basic substances continually pass into the bloodstream. If the pH- regulating
mechanisms of the body fail, as may happen during illness, and if the pH of the blood falls
below 7.0 or rises above 7.8, irreparable damage may result.
Notes:
a. The catalytic activity of enzymes is extremely sensitive to small changes in pH. Their
activity declines sharply on high or low side of 7.4. A change in [H +] of as little as 2.5
times ( say, from 7.4- 7.0) can be fatal, thus buffer solutions are useful for enzymes to
maintain required pH.
b. Blood, milk, digestive juices, and other fluids, which are produced or used in living
tissues are highly buffered solutions.
Examples:
1. A 1 L buffer solution is made up of 1F HAc and 1F NaAc
a) What is the pH of the original solution
b) What is the pH that results from adding 0.1F HCl solution
c) What is the pH that results from adding 0.1FNaOH solution

(assume no vol change with the addition of HCl and NaOH)
2. A 500 mL buffer solution is made up of 1F HAc and 2F NaAc
a) What is the pH of the original solution
b What is the pH that results from adding 10 mL of 0.3F HCl solution
c What is the pH that results from adding 10 mL of 0.3FNaOH solution
1. a. Common-ion effect:
1 F NaC2H3O2
pH?
1F 1FNaC2H3O2
HC2H3O2 + 1F HC2H3O2
Reqd: [NaC2H3O2]
Soln:
100%
t=0: I 1F 0 0
t=rxn: C -1F 1F 1F
after: E 0 1F 1F
finally:
HC2H3O2 + H2O <=> H3O+ + C2H3O2-

I: 1F 0 1F
C: -X X X
E: 1- X X 1+ X
[𝐻3 𝑂+ ][𝐶2𝐻3 𝑂2− ]𝑡𝑜𝑡𝑎𝑙

𝑲𝒂 = [𝑯𝑪𝟐 𝑯𝟑 𝑶𝟐 ]
[C2H3O2-]Total = 1 + X

(X)(1+𝑋)
𝟏. 𝟕𝟓𝑿𝟏𝟎−𝟓 = (𝟏−X)
X = [H3O+] = 1.75 x 10 -5 F
X in the denominator and x in the numerator = 0 ( the value of x is very, very small
compared to the value 1)
pH = - log [H3O+]
pH = - log ( 1.75 X 10 -5)
pH = 4.76
b.
0.1F HCl
pH?
1FNaC2H3O2
+ 1F HC2H3O2
Before reaction:
Available moles HC2H3O2 (HAc): F= n/V
n = 1mol/L (1L) = 1mol reactant in excess
Available moles NaC2H3O2 (NaAc): n = 1mol/L (1L) = 1mol reactant in excess

Available moles HCl: n = 0.1mol/L (1L) = 0.1mol limiting reactant
During reaction: (there is a chemical reaction between NaAc and HCl)
NaAc + HCl  HAc + NaCl

t=0: I 1 mol 0.1 mol 1mol
t=rxn: C - 0.1 mol -0.1 mol + 0.1 mol
after: E 0.9 mol 0 1.I mol
After reaction: New post reaction concentrations:
[NaAc] F= n / V = 0.9 mol / 1L = 0.9F

[HAc] = 1.1 mol / 1L = 1.1 F

100%
NaAc  Na+ + Ac-
t=0: I 0.9 0 0
t=rxn: C - 0.9F 0.9F 0.9F
after: E 0 0.9F 0.9F
finally:
HAc + H2O <=> H3O+ + Ac-

I: 1.1 0 0.9 F
C: -X +X +X
E: 1.1- X X 0.9+ X

𝑲𝒂 = [𝑯𝑪𝟐 𝑯𝟑 𝑶𝟐 ]
[C2H3O2-]Total = 0.9 + X
(X)(0.9+𝑋)
𝟏. 𝟕𝟓𝑿𝟏𝟎−𝟓 = (𝟏.𝟏−X)
X = [H3O+] = 2.1389 x 10 -5 F
compared to the value 1.1 and 0.9), or use Rule of 5%
pH = - log [H3O+]
pH = - log ( pH = - log [H3O+]
pH = - log (2.1389 x 10 -5 F)
pH = 4.670
 conclusion: there is no great change in pH after the addition of strong acid, therefore
it is a buffer solution
c.
0.1F NaOH
pH?
1FNaC2H3O2
+ 1F HC2H3O2

Before reaction:

Available moles NaOH : n = 0.1mol/L (1L) = 0.1mol limiting reactant
During reaction: (there is a chemical reaction between HAc and NaOH)
HAc + NaOH  NaAc + H2O

t=rxn: C - 0.1 mol -0.1 mol + 0.1 mol
after: E 0.9 mol 0 1.1 mol
[HAc] F= n / V = 0.9 mol / 1L = 0.9F

[NaAc] = 1.1 mol / 1L = 1.1 F
100%
NaAc  Na+ + Ac-
t=0: I 1.1F 0 0
t=rxn: C - 1.1F 1.1F 1.1F
finally:
HAc + H2O <=> H3O+ + Ac-

I: 0.9 0 1.1 F
C: -X +X +X
E: 0.9- X X 1.1+ X

𝑲𝒂 = [𝑯𝑪𝟐 𝑯𝟑 𝑶𝟐 ]
[C2H3O2-]Total = 1.1 + X
(X)(1.1+𝑋)
𝟏. 𝟕𝟓𝑿𝟏𝟎−𝟓 =
(𝟎.𝟗−X)
X = [H3O+] = 1.4318 x 10 -5 F

pH = - log [H3O+]
pH = - log (1.4318 x 10 -5 F)
pH = 4.844
 conclusion: there is no great change in pH after the addition of strong base,

therefore it is a buffer solution
HENDERSON-HASSELBALCH EQUATION:
buffer pair of weak acid and its corresponding base:
HA <==> H+ + A-
acid ca cb
at equilibrium: Ka = [H+][A-]
[HA]
solving for [H+]: [H+] = Ka [HA]

[A-]
x eqn (– log)
– log [H+] = – log {Ka [HA] }
[A-]
= – {log Ka + log [HA] }
[A-]
Henderson-Hasselbalch Equation:
– log [H+] = – log Ka – log [HA]

[A-]
pH = pKa – log [HA]
[A-]
OR:
Acid Buffer: pH = pKa + log [salt]
[WA]
Base Buffer: pOH = pKa + log [salt]
[WB]
pH = pKa + log [base] HHE

[acid]
Solve the previous example using HHE.

pH = pKa – log [acid]
[base]
Acid-Base concepts:
Arrhenius concept:
an acid when dissolved in water, increases the H3O+ concentration
a base when dissolved in water, increases the OH- concentration
Bronsted-Lowry concept:
- an acid is the specie donating a proton in a proton transfer reaction
- a base is the specie accepting a proton in a proton transfer reaction
note: a conjugate acid-base pair consists of two species in an acid-base reaction, one
acid and one base that differ by the loss or gain of a proton
SOLUTION TO # 2 ON BUFFER
#2.
a. Common-ion effect:
2 F NaC2H3O2
pH?
1F 1FNaC2H3O2
HC2H3O2 + 1F HC2H3O2
Reqd: [NaC2H3O2]
Soln:
100%
t=0: I 2F 0 0
t=rxn: C -2F 2F 2F
after: E 0 2F 2F
finally:
HC2H3O2 + H2O <=> H3O+ + C2H3O2-

I: 1F 0 2F
C: -X +X + X
E: 1- X X 2+ X

𝑲𝒂 = [𝑯𝑪𝟐 𝑯𝟑 𝑶𝟐 ]
[C2H3O2-]Total = 1 + X
(X)(2+𝑋)
𝟏. 𝟕𝟓𝑿𝟏𝟎−𝟓 = (𝟏−X)
X = [H3O+] = 8.75 x 10 -6 F
compared to the value 1)
pH = - log [H3O+]
pH = - log ( 8.75 X 10 -6)
pH = 5.0580
c. What is the pH that results from adding 10 mL of 0.3FNaOH solution
10mL
0.3F NaOH
pH?
1FNaC2H3O2
+ 1F HC2H3O2
Before reaction:
n = 1mol/L (0.5L) = 0.5mol reactant in excess
Available moles NaC2H3O2 (NaAc): n = 2mol/L (0.5L) = 1mol reactant in excess

Available moles NaOH: n = 0.3mol/L (0.010L) = 3 X 10 -3 mol limiting reactant

t=0: I 0.5 mol 3 X 10 -3 mol 1mol
t=rxn: C - 3 X 10 -3 -3 X 10 -3 +3 X 10 -3
after: E 0.497 mol 0 1.003mol

[NaAc] F= n / V = 1.003 mol / 0.510L = 1.9667F

[HAc] = 0.497 mol / 0.510L = 0.9745 F
100%
NaAc  Na+ + Ac-
t=0: I 1.9667 0 0
t=rxn: C - 1.9667 +1.9667 +1.9667
after: E 0 1.9667 1.9667
finally:
HAc + H2O <=> H3O+ + Ac-

I: 0.9745 0 1.9667
C: -X X X
E: 0.9745- X X 1.9667+ X

𝑲𝒂 = [𝑯𝑪𝟐 𝑯𝟑 𝑶𝟐 ]
[C2H3O2-]Total = 1.9667+ X
(X)(1.9667 + 𝑋)
𝟏. 𝟕𝟓𝑿𝟏𝟎−𝟓 =
(𝟎. 𝟗𝟕𝟒𝟓 − X)
X = [H3O+] = 8.6713 x 10 -6 F
pH = - log [H3O+]
pH = - log [H3O+]
pH = - log (8.6713 x 10 -6 F)
pH = 5.062

Henderson-Hasselbalch Equation:
Acid Buffer: pH = pKa + log [salt]

[WA]
pH = - log ( 1.75 X 10 ) + log (1.9667)
-5
pH = 5.062 (0.9745)

OR: ANOTHER SOLUTION:
[HAc] 510 mL ; V1C1 = V2C2
500 mL( 1F) = 510mL(C2)

C2 = 0.98 F
[NaAc] = 500 mL(2F) = 1.961 F

510 mL
[NaOH] = 10 mL(0.3F) = 5. 882 X 10-3 F

510 mL

t=0: I 0.98 F 5.882 X 10 -3 F 1.961F
t=rxn: C - 5.882 X 10 -3 F 5.882 X 10 -3 F 5.882 X 10 -3 F
after: E 0.974 F 0 1.967 F
common ion effect
NaAc  Na+ + Ac-

t=0: I 1.967 0 0
t=rxn: C - 1.967 1.967 1.967
after: E 0 1.967 1.967
finally:
HAc + H2O <=> H3O+ + Ac-

I: 0.974 0 1.967
C: -X X X
E: 0.974- X X 1.967+ X

𝑲𝒂 =
[𝑯𝑪𝟐 𝑯𝟑 𝑶𝟐 ]
[C2H3O2-]Total = 1.967+ X
(X)(1.967 + 𝑋)
𝟏. 𝟕𝟓𝑿𝟏𝟎−𝟓 =
(𝟎. 𝟗𝟕𝟒 − X)
X = [H3O+] = 8.6713 x 10 -6 F

pH = - log [H3O+]
pH = - log [H3O+]
pH = - log (8.6713 x 10 -6 F)
pH = 5.062
0.1F HCl
pH?
1FNaC2H3O2
+ 1F HC2H3O2
Before reaction:

Available moles HCl: n = 0.1mol/L (1L) = 0.1mol limiting reactant
During reaction: (there is a chemical reaction between NaAc and HCl)
NaAc + HCl  HAc + NaCl

t=rxn: C - 0.1 mol 0.1 mol + 0.1 mol
after: E 0.9 mol 0 1.I mol
[NaAc] F= n / V = 0.9 mol / 1L = 0.9F

[HAc] = 1.1 mol / 1L = 1.1 F
100%
NaAc  Na+ + Ac-
t=0: I 0.9 0 0
t=rxn: C - 0.9F 0.9F 0.9F

finally:
HAc + H2O <=> H3O+ + Ac-

I: 1.1 0 0.9 F
C: -X X X
E: 1.1- X X 0.9+ X

𝑲𝒂 = [𝑯𝑪𝟐 𝑯𝟑 𝑶𝟐 ]
[C2H3O2-]Total = 0.9 + X
(X)(0.9+𝑋)
𝟏. 𝟕𝟓𝑿𝟏𝟎−𝟓 = (𝟏.𝟏−X)
X = [H3O+] = 2.1389 x 10 -5 F
pH = - log [H3O+]
pH = - log (2.1389 x 10 -5 F)
pH = 4.670
 conclusion: there is no great change in pH after the addition of strong acid, therefore
it is a buffer solution
c.
0.1F NaOH
pH?
1FNaC2H3O2
+ 1F HC2H3O2
Before reaction:

Available moles NaOH : n = 0.1mol/L (1L) = 0.1mol limiting reactant


t=rxn: C - 0.1 mol -0.1 mol + 0.1 mol
after: E 0.9 mol 0 1.1 mol
[HAc] F= n / V = 0.9 mol / 1L = 0.9F

[NaAc] = 1.1 mol / 1L = 1.1 F
100%
NaAc  Na+ + Ac-
t=0: I 1.1F 0 0
t=rxn: C - 1.1F 1.1F 1.1F
finally:
HAc + H2O <=> H3O+ + Ac-

I: 0.9 0 1.1 F
C: -X +X +X
E: 0.9- X X 1.1+ X

𝑲𝒂 = [𝑯𝑪𝟐 𝑯𝟑 𝑶𝟐 ]
[C2H3O2-]Total = 1.1 + X
(X)(1.1+𝑋)
𝟏. 𝟕𝟓𝑿𝟏𝟎−𝟓 = (𝟎.𝟗−X)
X = [H3O+] = 1.4318 x 10 -5 F
pH = - log [H3O+]
pH = - log (1.4318 x 10 -5 F)
pH = 4.844

IONIZATION OF WEAK POLYPROTIC ACIDS:

- acids having more than 1 H+ per molecule
- occurs in several steps where 1 H+ is removed at a time
- acids w/c react with water to give more than 1 H 3O+ per molecule of the original acid
- dissociation or ionization occurs in stages
- in a soln of polyprotic acid 1 proton at a time dissociates from the acid molecule and
each dissociation step has a different Ka
2 types of polyprotic acids:

a. diprotic acid
- yield 2 H3O+
ex: H2S , H2CO3, H2SO3
b. triprotic acid
- yield 3 H3O+
ex: H3PO3, H3PO4
H2CO3 + H2O <=> H3O+ + HCO3 -1 (1st /primary ionization) Ka1 = [H3O+][HCO3-]
[H2CO3]
HCO3 - + H2O <=> H3O+ + CO3 -2 (2nd /secondary ionization) Ka2 = [H3O+][CO3-2]
[HCO3 -]
 NOTE: IN SOLN ALL THE [H3O+] WILL COME FROM THE 1ST DISSOCIATION

0
 USEFUL APPROXIMATION : [H3O ]TOTAL = [H3O ]1ST + [H3O ]SUCCEEDING CONC
+ + +
Ka1
H3PO4 + H2O <=> H3O + + H2PO4- Ka1 = [H3O+][H2PO4-]
[H3PO4]
Ka2
H2PO4- 1 + H2O <=> H3O + + HPO4 2- Ka2 = [H3O +][HPO42-]
[H2PO4-]
Ka3
HPO4 2- + H2O <=> H3O + + PO4 3- Ka3 = [H3O +][PO43-]
[HPO42-]

Over-all reaction:
H3PO4 + 3 H2O <=> 3 H3O + + PO43-
Ka (over all ionization constant) = Ka1 X Ka2 X Ka3

Ka1 = 7.5 x 10-3
Ka2 = 6.2 x 10-8
Ka3 = 4.8 x 10-13
Ka = 2.232 x 10-22 = [H3O +]3 [PO43-]

[H3PO4]
note: Ka1 >>> Ka2 >>> Ka3
it is always easier to remove the first proton (H 3O +) from a polyprotic acid than the
second
100%
ex . H2SO4 + H2O ⇨ H3O + + HSO4 –
HSO4 – + H2O <=> H3O + SO4 -2
ONLY K2 ( p.269 Gilreath)
Ka2 = [H3O +][SO42-]

[HSO4-]
Ka2 = 1.2 x 10-2 ( p.269 Gilreath)

 Sulfuric acid is a strong acid with respect to the removal of the 1st proton , complete
ionization
HSO4 – is a weak acid
DERIVATION OF H2S OVERALL

H2S + H2O <=> H3O + + HS-
HS- + H2O <=> H3O+ + S=
Ka1 = [H3O+][HS-]
[H2S]
Ka2 = [H3O +][S2-]

[HS-]
OVERALL IONIZATION:
K H2S OVERALL :
Ka1 X Ka2 = [H3O+] [HS-] X [H3O +][S2-]
[H2S] [HS-]

[𝐻3 𝑂 + ]𝟐 [𝑆 = ]
(𝟓. 𝟕 𝑿 𝟏𝟎−𝟖 )(𝟏. 𝟐 𝑿 𝟏𝟎−𝟏𝟓 ) =
[𝑯𝟐 𝑺]
H2S OVERALL:
[𝐻3 𝑂 + ]𝟐 [𝑆 = ]
𝟔. 𝟖𝟒 𝑿 𝟏𝟎−𝟐𝟑 =
[𝑯𝟐 𝑺]
 For a system involving H2S:

To obtain [H3O+] - use K1 expression
To obtain [S ] -
= use overall expression
Over-all reaction:
H2S + 2H2O <=> 2H3O + + S2-
 NOTE: ANY SOLN SATURATED WITH H2S AT RM TEMP (250C); [ H2S] = 0.1 F
Ex: 1. Determine the pH of :a) 0.5F H3PO4 b.) 2F H2S soln
a. Solution: let X = amt of H3PO4 that ionized
H3PO4 + H2O <=> H3O+ + H2PO4 -1

I: 0.5F 0 0
C: -X +X +X
E: 0.5 - X X X
[𝐻3 𝑂+ ][𝐻2 𝑃𝑂4−1 ]

𝑲𝒂 = [𝑯𝟑 𝑷𝑶𝟒 ]
(X)(𝑋)
𝟕. 𝟓𝑿𝟏𝟎−𝟑 = (𝟎.𝟓−X) Ka > 10-4
USE QUADRATIC FORMULA:

−𝑏 ± √𝑏 2 − 4𝑎𝑐
𝑥=
2𝑎
𝑿𝟐 = 𝟕. 𝟓 𝑿 𝟏𝟎 −𝟑 (𝟎. 𝟓 − 𝑿)
𝑿𝟐 = 𝟑. 𝟕𝟓 𝑿𝟏𝟎 −𝟑 − 𝟕. 𝟓 𝑿𝟏𝟎 −𝟑 𝑿
𝑿𝟐 + 𝟕. 𝟓 𝑿𝟏𝟎 –𝟑 𝑿 − 𝟑. 𝟕𝟓 𝑿𝟏𝟎 –𝟑 = 𝟎
−𝟕.𝟓 𝑿𝟏𝟎 −𝟑 ±√(𝟕.𝟓 𝑿𝟏𝟎 −𝟑 )2 −4(1)(− 𝟑.𝟕𝟓 𝑿𝟏𝟎 −𝟑 )

𝑥= 2( 1)
X = [H3O+] = 0.0576 F

[H3O+] 1st ≈ [H3O+] total
pH = - log (0.0576 )
pH = 1.24
b. 2F H2S
Solution: let X = amt of H2S that ionized
H2S + H2O <=> H3O+ + HS -1

I: 2.0 F 0 0
C: -X +X +X
E: 2.0 - X X X
[𝐻3 𝑂+ ][𝐻𝑆 −1 ]
𝑲𝒂𝟏 = [𝑯𝟐 𝑺]
(X)(𝑋)
𝟓. 𝟕𝑿𝟏𝟎−𝟖 = Ka ≤ 10-4
(𝟐−X)
X = [H3O+] = 3.376 X 10-4 F
pH = - log (3.376 X 10-4 )

pH = 3.47
2. What is the phosphate ion conc [PO4-3] in 0.5 F H3PO4 ?
Solution: let X = amt of H3PO4 that ionized
H3PO4 + H2O <=> H3O+ + H2PO4 -1

I: 0.5F 0 0
C: -X +X +X
E: 0.5 - X X X
[𝐻3 𝑂+ ][𝐻2 𝑃𝑂4−1 ]

𝑲𝒂 = [𝑯𝟑 𝑷𝑶𝟒 ]
(X)(𝑋)
𝟕. 𝟓𝑿𝟏𝟎−𝟑 = (𝟎.𝟓−X) Ka > 10-4 USE QUADRATIC FORMULA
X = [H3O+] = 0.0576 F

OVERALL REACTION:
H3PO4 + 3H2O <=> 3 H3O + + PO43-

I: 0.5F 0 0
C: - 0.0576 /3 0.0576 X
E: 0.4808 0.0576 X
[𝐻3 𝑂+ ]𝟑 [𝑃𝑂4−3 ]
𝑲𝒐𝒗𝒆𝒓𝒂𝒍𝒍 =
[𝑯𝟑 𝑷𝑶𝟒 ]
(0.05760)𝟑 [𝑃𝑂4−3 ]
𝟐. 𝟐𝟑𝟐𝑿𝟏𝟎−𝟐𝟐 =
𝟎.𝟒𝟖𝟎𝟖
[PO4-3] = 5.616 X 10-19 F
3. How many grams of carbonic acid should be dissolved in 500 mL water to have a
pH of 5?
Soln: let X = initial concentration of H2CO3
pH = -log [H3O+ ]
[H3O+ ] = antilog – pH
[H3O+ ] = antilog – 5
[H3O+ ] = 1 x 10 -5
H2CO3 + H2O <=> H3O+ + HCO3 -1

I: X 0 0
C: -1 x 10 -5 1 x 10 -5 1 x 10 -5
E: X - 1 x 10 -5 1 x 10 -5 1 x 10 -5
[𝐻3 𝑂+ ][𝐻𝐶𝑂3−1 ]
𝑲𝒂 = [𝑯𝟐 𝑪𝑶𝟑 ]
(𝟏 𝑿 𝟏𝟎 −𝟓 )𝟐
𝟒. 𝟑 𝑿 𝟏𝟎 −𝟕 =
(𝑿− 𝟏 𝑿𝟏𝟎−𝟓 )
X = [H2CO3] = 2.326 X 10-4 F
Grams of H2CO3 = 2.326 X 10-4 mol/L x 62 g/mol = 7.2 x 10-3 g

19/82] calculate the sulfide – ion conc of a 0.5 F HCl sol which has been satd with H 2S at
250C
Given: [HCl] = 0.5 F
[H2S] = 0.1 F
Reqd: [ S=]
Soln: let x = amt of [ S=]
100%
HCl + H2O  H3O+ + Cl-
t=0: I 0.5 F 0 0
t=rxn: C - 0.5 +0.5 + 0.5
after: E 0 0.5 0.5
Over-all reaction:
H2S + 2H2O <=> 2H3O + + S=
At E: 0.1 0.5 X
[𝐻3 𝑂+ ]𝟐 [𝑆 −2 ]
𝑲 𝑯𝟐𝑺 𝒐𝒗𝒆𝒓𝒂𝒍𝒍 = [𝑯𝟐 𝑺]
(0.5)𝟐 [𝑆 −2 ]
𝟔. 𝟖𝟒 𝑿𝟏𝟎−𝟐𝟑 = 𝟎.𝟏
[S-2] = 2.74 X 10-23 F
24/82] To 500 mL of 0.05F H2S solution is added 50 mL 3F HAc solution. What is the sulfide ion
conc of the resulting soln?
Given: [HAc] = 3.0 F V = 50 mL

[H2S] = 0.05 F V = 500 mL
Reqd: [ S=]
Soln: V1C1 = V2C2
[H2S] : 500 mL( 0.05 F) = 550 mL(C 2)

C2 = [H2S] = 0.045 F
[HAc] : 50 mL( 3 F) = 550 mL(C2)

C2 = [HAc] = 0.27 F
HC2H3O2 + H2O ⇄ H3O +1 + C2H3O2-1
During reaction or C : - X +X + X
_______________________________________________________________________
At eqbm or E : 0.27F - X X X

[𝐻3 𝑂+ ][𝐶2𝐻3 𝑂2−1 ]
𝐾 𝐻𝐶2 𝐻3 𝑂2 = [𝐻𝐶2 𝐻3 𝑂2 ]
(X )(X )
1.75 × 10−5 = (0.27− X)
X = [H3O+] = 2.17 x 10 -3 F
Over-all reaction:
H2S + 2H2O <=> 2H3O + + S=
At E: 0.045 2.17 x 10 -3 X
[𝐻3 𝑂+ ]𝟐 [𝑆 −2 ]
𝑲 𝑯𝟐𝑺 𝒐𝒗𝒆𝒓𝒂𝒍𝒍 =
[𝑯𝟐 𝑺]
𝟐
(𝟐.𝟏𝟕 𝒙𝟏𝟎−𝟑 ) [𝑆 −2 ]
𝟔. 𝟖𝟒 𝑿𝟏𝟎−𝟐𝟑 =
𝟎.𝟎𝟒𝟓
[S-2] = 6.54 X 10-19 F
ASSIGNMENT:
1. Calculate the degree of ionization of a 0.05F solution of lactic acid.
2. Compute the ionization constant of hydrofluoric acid, if a 0.1F solution is 8.13
percent ionized.
3. What concentration of nitrous acid has a hydronium –ion concentration of 0.01 F?
4. If a weak monoprotic acid is 3.8 percent ionized in a 0.2F solution, calculate its
ionization constant. What is the percentage ionization of this acid in a 0.01F
solution?
5. The pOH of a solution, which is 0.5F with respect to a weak monoprotic acid is 9.2.
calculate the ionization constant of the acid.
6. What concentration of NH3 is required to make a solution with a pH of 10.2?
7. If 3 g ammonium chloride and 2 g of ammonia are dissolved in sufficient water to
produce 500 mL of solution, calculate the hydroxide – ion concentration of the
solution. What is the pH of the solution?
8. What is the Hydrogen ion concentration in 500 mL of a 0.100 M solution of acetic
acid at 25oC if the solution contains an additional 2.00 g of acetate ions added in
the form of sodium acetate (Ka = 1.86 x 10-5 at 25OC). What is the Hydrogen ion
concentration b) if 4.00 millimoles of NaOH and c) if 4.00 millimoles of HCl are
introduced into the buffered solution? What is the pH value in each of the three
cases?
9. Compute the sulfide-ion concentration of a 0.5F HC2H3O2 solution which has been
saturated with H2S at 250C.
10. Calculate the pH of a 0.1 F solution of NaHSO4

REF SEA-BSCHE-CHE526-2020
MODULE 5: UNIT 5: SOLUBILITY PRODUCT PRINCIPLE
SOLUBILITY PRODUCT PRINCIPLE
SOLUBILITY PRODUCT PRINCIPLE
- is an application of the law of chemical equilibrium to heterogeneous equilibria

produced by solids in equilibrium with ions in solution, or more specifically, to equilibria
produced by saturated solutions of difficultly / slightly soluble electrolytes .
- involves the ionization of the precipitates formed during chemical reactions
DEFINITION: SOLUBILITY PRODUCT PRINCIPLE

- In a saturated solution of difficultly soluble electrolyte, the product of the formal
concentration of the ions, each raised to the power equal to the number of times it
occurs in the formula, is a constant at a given temperature.
Saturated Solution
- one which contains or dissolves the maximum amount of solute that can be dissolved
in the solvent; addition of further solute will produce super saturated solution
In a saturated solution of difficultly / slightly soluble electrolyte; there is

equilibrium with solid electrolyte with its ions in solution. This is a heterogeneous
system in which the chemical equilibrium constant takes a new name,which is
solubility product constant or Ksp
APPLICATIONS OF SOLUBILITY PRODUCT PRINCIPLE:
1. predict solubility of solution

solubility - is the amount of solute that can be dissolved in a given weight or
volume of solvent at a specific temperature
2. predict if precipitate will form or not when two solutions are mixed

- when the Ksp of the salt is exceeded in a solution, the salt is precipitated
3. predict the amount of a substance needed to precipitate another
4. predict the maximum concentration of a substance that can co-exist with another
without causing precipitation
5. predict the maximum concentration of a substance that has to be present in order to

initiate precipitation
6. predict order of solubility

- arranging in order of increasing/decreasing solubility from a list of substances
ex. a. BaCO3 Ksp= 8.1 x 10 -9
b. BaSO4 Ksp= 1.1 x 10 – 10
c. PbS Ksp= 3.4 x 10 – 28
d. AgCl Ksp= 1.56 x 10 – 10
 the lower the Ksp the more insoluble is the substance and the more it readily
precipitates
increasing solubility
7. predict order of precipitation

ex. arranging from a list of substances, beginning the substance with higher
tendency to precipitate
Influence of a Common Ion on Solubility
- the effect of common ion upon a precipitate in equilibrium with its ions is the decrease
in the solubility of the solid phase, thus more precipitate is formed.
Fractional Precipitation
- process of precipitating one at a time several ions present in one solution by the
gradual addition of the same precipitating agent
- possible only if Ksp of ions precipitated have wide difference in their Ksp values
ex. H2S precipitates both group II and group III cations based upon a careful
control of the concentration of the sulphide ion
 SOLUBILITY RULES (IN WATER)
1. All common salts of the Family IA elements and NH 4+ are soluble.
2. All common acetates and nitrates are soluble.
3. All binary compounds of Family VIIA elements (other than fluorine) with metals are
soluble except those of Ag, Hg, Pb (ex. PbI 2).
4. All sulfates are soluble except those of barium, strontium,lead, calcium,silver and
mercury
5. Except for those in Rule I, CO3, OH’s, oxides and phosphates are insoluble
 Mainly Water Insoluble:

S-2 except IA and IIA elements, (NH4)2S
CO3-2 except those of IA elements and (NH4)2CO3
SO3-2 except those of IA elements and (NH4)2SO3
PO4-3 except those of IA elements and (NH4)4PO4
OH- except those of IA elements and Ba(OH)2, Sr(OH)2, Ca(OH)2,
 Mainly Water Soluble:

NO3- all nitrates are soluble
C2H3O2- all acetates are soluble
ClO3 - all chlorates are soluble
Cl- all chlorides are soluble except AgCl, Hg2Cl2 and PbCl2
Br - all bromides are soluble except AgBr, Hg2Br2 and PbBr2
I - all iodides are soluble except AgI, Hg2I2 and PbI 2
SO4-2 all sulfates are soluble except CaSO4, SrSO4, BaSO4,PbSO4, Ag2SO4
DERIVATION OF Ksp EXPRESSION:
Ex. Write the expression for the solubility-product constant for the ff:
1. AgCl
2. Bi2S3
3. MgNH4PO4
We are asked to write an equilibrium- constant expression

For the process by which the following compounds dissolve in water. We apply the
same rules for writing any equilibrium-constant expression, making sure to exclude the
solid reactant from the expression. We assume that these compounds dissociate
completely into its component ions.
1. AgCl ⇌ Ag+ + Cl-
Ksp = [Ag+][Cl-]
2. 𝐵𝑖2 𝑆3 ⇋ 2 𝐵𝑖 +3 + 3𝑆 −2
𝐾𝑠𝑝 = [𝐵𝑖 +3 ]2 [𝑆 −2 ]3
3. 𝑀𝑔𝑁𝐻4 𝑃𝑂4 ⇋ 𝑀𝑔+2 + 𝑁𝐻4+2 + 𝑃𝑂4−3
𝐾𝑠𝑝 = [𝑀𝑔+2 ][𝑁𝐻4+1 ][𝑃𝑂4−3 ]

I. CALCULATION OF Ksp FROM SOLUBILITY
1/102] Gilreath
𝑔
The solubility of AgI is 2.9 × 10−6 . Calculate its solubility product constant.
𝐿
Given:
𝑔
solubility of AgI = 2.9 × 10−6
𝐿
Reqd: Ksp
Soln:
First step is to calculate the Formal solubility in mols per liter(gfw/L) since we are given the
𝑔 1
solubility of AgI in g/L by using the formula : ×
𝐿 𝑀𝑊
𝑔
MW (gram-formula wt) = 235
𝑚𝑜𝑙

𝑔 1𝑚𝑜𝑙 𝑚𝑜𝑙
Formal solubility of 𝐴𝑔𝐼 = 2.9 × 10−6 × = 1.234 × 10−8 ,𝐹
𝐿 235 𝑔 𝐿
−8 𝑚𝑜𝑙
 From our calculation, 1.234 × 10 (𝐹) of AgI to dissociate completely at
𝐿
equilibrium
 The stoichiometry of the equilibrium dictates that x moles/L of 𝐴𝑔+1 and x moles of
𝐼 −1 are produced for each x moles/L AgI that dissolve.
The equilibrium equation and the equation for Ksp are :
𝐴𝑔𝐼 ⇌ 𝐴𝑔+1 + 𝐼 −1
1.234 × 10−8 𝐹 1.234 × 10−8 𝐹 1.234 × 10−8 𝐹
𝐾𝑠𝑝 = [𝐴𝑔+1 ][𝐼 −1 ]

𝐾𝑠𝑝 = (1.234 × 10−8 𝐹 )(1.234 × 10−8 𝐹 )
𝐾𝑠𝑝 = 1.5228 × 10−16 .
2/102] A saturated solution of BaSO4 is 1.1 × 10−5 𝐹 at room temperature. Calculate the
solubility product constant of barium sulfate.
Given:
solubility of BaSO4 = 1.1 × 10−5 𝐹
Reqd: Ksp
Soln:
we are given the formal solubility of BaSO4 = 1.1 × 10−5 𝐹
The equilibrium equation and the equation for Ksp are :
𝐵𝑎𝑆𝑂4 ⇌ 𝐵𝑎 +2 + 𝑆𝑂4 −2
XF XF XF
Eqbm: 1.1 × 10−5 𝐹 1.1 × 10−5 𝐹 1.1 × 10−5 𝐹
𝐾𝑠𝑝 = [𝐵𝑎 +2 ][𝑆𝑂4 −2 ]

𝐾𝑠𝑝 = (1.1 × 10−5𝐹 )(1.1 × 10−5 𝐹 )
𝐾𝑠𝑝 = 1.21 × 10−10 .

Ex. What is the solubility product constant and pKsp of silver sulfate, Ag 2SO4 (MW = 312
g/mol) , if the solubility of the salt is 5.7 x 10-3 g/mL
Given:
solby of = 5.7 x 10-3 g/mL
Reqd: Ksp, pKsp

Soln:
𝑔 1𝑚𝑜𝑙 1000𝑚𝐿
𝐹 = 5.7 𝑥 10−3 𝑚𝐿 × ×
312 𝑔 𝐿
F = 0.01827 mol /L
H2O
Ag2SO4 ⇄ 2Ag+ + SO4-2
At eqbm: 0.01827 F 2(0.01827 ) F 0.01827 F
Ksp = [Ag+]2 eqbm [ SO4-2] eqbm

Ksp = ( 2 x 0.01827)2(0.01827)
Ksp = 2.43 x 10-5
pKsp = -log Ksp

= - log (2.43 x 10-5)
pKsp= 4.61

II. CALCULATION OF SOLUBILITY FROM Ksp
8/102] The solubility product constant for AgBr at 250C is 7.7 x 10 -13( table A-9 pp 271-272
Gilreath) . What is the solubility of AgBr in grams per liter at 250C?
Given:
Ksp = 7.7 x 10 -13
Reqd: solby of AgBr in g/L
Soln:
Let X = moles AgBr dissolved/ liter of soln
H2O
AgBr ⇄ Ag+ + Br-
At eqbm: XF XF XF
Ksp AgBr = [ Ag+]eqbm [Br-1] eqbm
7.7 x 10 -13 = (X)(X)

𝑋2 = 7.7 𝑥 10−13
𝑋 = 8.77 𝑥 10−7 F = [AgBr] = [ Ag+]eqbm = [Br-1] eqbm
𝑔 𝑚𝑜𝑙 𝑔 𝑔
𝑠𝑜𝑙𝑏𝑦 𝐴𝑔𝐵𝑟 𝑖𝑛 = 8.77 𝑥 10 −7 × 188 = 1.65 𝑥 10−4
𝐿 𝐿 𝑚𝑜𝑙 𝐿
9./102] The solubility product constant for cupric iodate is 1.4 x 10 -7 at 250C. Calculate its
formal solubility
Given:
Ksp Cu(IO3)2 = 1.4 x 10-7
Reqd: solby of Cu(IO3)2 in F
Soln:
Let X = moles Cu(IO3)2 dissolved/ liter of soln
H2O
Cu(IO3)2 ⇄ Cu +2 + 2 IO3-1
At eqbm: XF XF 2X F
𝐾𝑠𝑝𝐶𝑢(𝐼𝑂3)2 = [𝐶𝑢+2 ]𝑒𝑞𝑏𝑚 [𝐼𝑂3−1 ]2 𝑒𝑞𝑏𝑚
1.4 𝑥 10−7 = (𝑥 ) (2𝑥)2
𝑥 = 3.27 𝑥 10−3 𝐹
Ex: Determine the solubility in g/L solution of the following slightly soluble salts and identify
which one is the most soluble in H2O at room temp.
a. Bi2S3
b. Al(OH)3
c. Cu(IO3)2

 Note for Ksp Values if not given , refer to T-9 pp 271-272 (Gilreath)
Soln:
a. Bi2S3
Let X = moles Bi2S3 dissolved/ liter of soln
H2O
Bi2S3 ⇄ 2Bi +3 + 3S-2
At eqbm: X F 2X F 3X F
Ksp Bi2S3 = [Bi+3]2 eqbm [ S-2]3eqbm
1.6 𝑥 10−72 = (2𝑥 )2 (3𝑥)2

−72
1.6 𝑥 10 = 108𝑥 5
𝑚𝑜𝑙𝑒𝑠 𝐵𝑖2 𝑆3
𝑥 = 1.71 𝑥 10 −5
𝐿 𝐻2 𝑂
g Bi2S3 dissolved / L of H2O = ( 1.71 x 10-5 moles/L)( 514 g/mol)
= 8.79 x 10-13 g/L H2O
b. Al(OH)3
Let X = moles Al(OH)3 dissolved/ liter of soln
H2O
Al(OH)3 ⇄ Al +3 + 3OH-1
At eqbm: X F XF 3X F
Ksp Al(OH)3 = [Al +3]eqbm [OH-1]3 eqbm
1.9 𝑥 10−33 = (𝑥)(3𝑥)3 = 27 𝑥 4
𝑚𝑜𝑙𝑒𝑠 𝐴𝑙(𝑂𝐻)3
𝑥 = 2.9 𝑥 10 −9
𝐿 𝐻2 𝑂
g Al(OH)3 dissolved / L of H2O = ( 2.9 x 10-9 moles/L)( 78 g/mol)
= 2.26 x 10-7 g/L H2O

c. From the previous prob
𝑥 = 3.27 𝑥 10−3 𝐹
𝑚𝑜𝑙𝑒𝑠 𝐶𝑢(𝐼𝑂3 )2
X = 3.27 𝑥 10 −3 𝐿 𝐻2 𝑂
g Cu(IO3)2 dissolved / L of H2O = ( 3.27 x 10-3 moles/L)( 413.55 g/mol)
= 1.352 g/L H2O
 The most soluble in H2O at room temperature is Cu(IO3)2

 The least soluble in H2O at room temperature is Bi2S3
III. INFLUENCE OF COMMON ION ON SOLUBILITY

-
13/102] Calculate the solubility in grams per liter of silver chloride in 0.01F hydrochloric acid.
(Assume that no complex ion is formed).
Given:
0.01 F HCl
AgCl solby AgCl ?

AgCl soln 0.01 F HCl
Soln:
100%
HCl + H2O → H3O + + Cl-
init conc :t =0 0.01 0 0
during rxn/t rxn: - 0.01 0.01 0.01
after rxn: 0 0.01 0.01
Let X = moles AgCl dissolved/ liter of soln
H2O
AgCl ⇄ Ag+ + Cl-
At eqbm: XF XF XF
Ksp AgCl = [ Ag+]eqbm [Cl-1] Total @ eqbm
[Cl-]TOTAL = [Cl-1] AgCl + [Cl-1] HCl

≈0
[Cl-]TOTAL = X + 0.01 value of X is very small compared to 0.01
1.56 x 10 -10 = ( X ) ( X +.01 )
𝑚𝑜𝑙
𝑋 = 1.56𝑥 10−10 F = [AgCl]
𝐿
𝑔 𝑚𝑜𝑙 𝑔
𝑠𝑜𝑙𝑏𝑦 𝐴𝑔𝐶𝑙 𝑖𝑛 = 1.56 𝑥 10 −10 × 143
𝐿 𝐿 𝑚𝑜𝑙
= 2.23 x10-6 g/L
16/102] What is the solubility in grams per liter of ferric hydroxide in 0.2 F KOH solution? Ferric
hydroxide has a Ksp of 1.1 x 10 -36
Given:
0.2 F KOH
Fe(OH)3 solby Fe(OH)3?

Fe(OH)3 soln 0.2 F KOH
Soln:
100%
KOH → K+ + OH-
init conc :t =0 0.2 0 0
during rxn/t rxn: - 0.2 0.2 0.2
after rxn: 0 0.2 0.2
Let X = moles Fe(OH)3 dissolved/ liter of soln
H2O
Fe(OH)3 ⇄ Fe+3 + 3OH-
At eqbm: XF XF 3X F
Ksp Fe(OH)3 = [Fe+3 ]eqbm [OH-1]3 Total @ eqbm
[OH-]TOTAL = [OH-1] Fe(OH)3 + [OH-1] KOH
[OH-]TOTAL = 3X + 0.2 value of X is very small compared to 0.2

1.1 x 10 -36 = ( X ) ( 3X +.2 )3
𝑚𝑜𝑙
𝑋 = 1.38𝑥 10−34 F = [Fe(OH)3]
𝐿
𝑔 𝑚𝑜𝑙 𝑔
𝑠𝑜𝑙𝑏𝑦 𝐹𝑒(𝑂𝐻)3 𝑖𝑛 = 1.38 𝑥 10 −34 × 107
𝐿 𝐿 𝑚𝑜𝑙
= 1.47 x10-32 g/L
18./102] The Ksp of PbI2 is 1.39 x 10-8 . Calculate the solubility in grams per 100 mL in a 0.1F
solution CaI 2.
Given:
100ml soln 0.1FCaI2
solby PbI 2?
100mLsoln
PbI2 soln 0.1FCaI 2
PbI2
Soln:
100%
CaI 2 → Ca+2 + 2I-
init conc :t =0 0.1 0 0
during rxn/t rxn: - 0.1 0.1 2( 0.1)
Let X = moles PbI 2 dissolved/ liter of soln
H2O
PbI2 ⇄ Pb+2 + 2I-
At eqbm: XF XF 2X F
Ksp PbI2 = [Pb+2 ]eqbm [I -1]2 Total @ eqbm
[I-]TOTAL = [I-1] PbI2 + [I -1] CaI2
[OH-]TOTAL = 2X + 0.2 value of X is very small compared to 0.2
1.39 x 10 -8 = ( X ) ( 2X +.2 )2
𝑚𝑜𝑙
𝑋 = 3.48𝑥 10−7 = [PbI2]
𝐿
𝑔 𝑚𝑜𝑙 𝑔 1𝐿
𝑠𝑜𝑙𝑏𝑦 𝑃𝑏𝐼2 𝑖𝑛 = 3.48 𝑥 10 −7 × 461 × 100 𝑚𝐿 ×
𝐿 𝐿 𝑚𝑜𝑙 1000 𝑚𝐿
= 1.6 x10-5 g

IV. FRACTIONAL PRECIPITATION
PREDICTING WHETHER A PPT WILL FORM OR NOT
- In predicting whether a ppt will form or not when 2 or more solutions are mixed,
compare actual ion product (aip), which is the Ksp with known Ksp from table
 If actual ion product (Ksp) < Ksp table

- the soln is unsatd
- ∴ no ppt is formed
 If actual ion product (Ksp) = Ksp table
- the soln is satd
- ∴ no ppt is formed
 If actual ion product (Ksp) > Ksp table

- the soln is supersatd
- ∴ ppt is formed
22/103] If 100 mL of a solution containing 0.01 g of Fe +3 is mixed with 1 mL of 0.1 F NH 3.

Calculate whether or not a ppt will form
Given:
1mLof 0.1 F NH3 soln
Will a ppt form?

100 mL soln 101 mLsoln
0.01gFe+3
Soln:
 If ever there is a ppt formed , it is Fe(OH)3 ; try to solve actual Ksp of Fe(OH)3 in the 101
mL mixture
0.01 𝑔
[𝐹𝑒 +3 ] = 56 𝑔 = 1.77 𝑥 10−3 F
× 0.101 𝐿
𝑚𝑜𝑙
[NH3]101 mL : V1C1 = V2C2

(1 mL)(0.1F) = 101 mL(C 2)
= 9.9 x 10-4 F
Solve for the [OH-] in NH3:
Let X = amount of NH3 that ionized

NH3 + H2O ⇄ NH4 +1 + OH-1
Init conc or I: 9.9 x 10 F
-4 0 0
_______________________________________________________________________
At eqbm or E : 9.9 x 10-4 - X X X
[𝑁𝐻4 +1 ][𝑂𝐻 −1 ]
K NH3 =
[𝑁𝐻3 ]
(𝑋)(𝑋)
1.8 × 10−5 = K NH3 = 1.8 x 10-5 p.269 (Gilreath
(9.9 𝑥 10−4 −𝑋)
X = [OH-] = 1.33 x 10-4 F
H2O
Fe(OH)3 ⇄ Fe+3 + 3OH-
Actual Ksp Fe(OH)3 = [Fe+3 ] [OH-1]3

= (1.77 x 10-3)( 1.33 x 10-4)
= 4. 21 x 10 -15
Ksp theoretical ( fr T-9 p. 271) = 6 x 10-38
Ksp actual > Ksp theoretical

∴ a ppt of Fe(OH)3 will form in the 101 mL mixture
Ex. Calculate whether or not precipitate will be formed in each case

a. 100 mL of 0.2 F BaCl2 + 75 mL of 0.1 F K2CO3
b. 85 mL of 0.7 F lead nitrate + 150 mL NH3
c. 1L of 1.5 F H2SO4 + 800 mL of 0.9 F SrCl2
d. 5 g of MgCl2 in 85 mL H2O + 50 mL 1.2 F HF
Given:
75mLof 0.1 F K2CO3 soln
Will a ppt form?

100 mL soln 175 mLsoln
0.2F BaCl2
Soln:
 If ever there is a ppt formed , it is BaCO3 ; try to solve actual Ksp of BaCO3in the 175
mL mixture

[BaCl2]175 mL : V1C1 = V2C2
(100 mL)(0.2F) = 175 mL(C 2)
= 0.114 F
[K2CO3]175 mL : V1C1 = V2C2

(75 mL)(0.01F) = 175 mL(C 2)
= 4.286 x 10-3 F
100%
BaCl2 → Ba+2 + 2Cl-

0.114 F 0.114 F 2(0.114) F
100%
K2CO3 → 2K+1 + CO3-2
4.286 x 10 F
-3 2(4.286 x 10-3) F 4.286 x 10-3 F
H2O
BaCO3 ⇄ Ba+2 + CO3-2
Actual Ksp BaCO3 = [Ba+2 ] [CO3-2]

= (0.114 )( 4.286 x 10-3)
= 4. 886 x 10 -4
Ksp theoretical ( fr T-9 p. 271) = 8.1 x 10-9
Ksp actual > Ksp theoretical

∴ a ppt of BaCO3 will form in the 175 mL mixture
ORDER OF PRECIPITATION
a. The lower the Ksp value, the more insoluble, the more it readily precipitates
ex. Arrange in the order of increasing solubility and precipitation

a. BaCO3 Ksp= 8.1 x 10 -9
b. BaSO4 Ksp= 1.1 x 10 – 10
c. PbS Ksp= 3.4 x 10 – 28
d. AgCl Ksp= 1.56 x 10 – 10
ans: c – b- d –a
incg solby in water
a- d- b – c
incg precipitation

b. A solution contg 2 or more ions that can be precipitated by a common pptng rgt, to
predict the order or sequence of pptn, calculate the conc of the pptng rgt required
by each ion, the ion requiring the lesser pptng rgt precipitates first
24/102] If K2CrO4 is added to a solution contg 10 g Ag + and 1 g Pb++ per liter, potassium
dichromate is added dropwise. Which ion will ppt first?
Given:
K2CrO4 soln
10g Ag+ which ion will ppt first?

1g Pb++
1L soln
Soln:
 The precipitates which will be formed are Ag2CrO4 and PbCrO4
 Determine the [CrO4-2] requirement of each ion from K sp , the one with the lesser [CrO4-
2]
will precipitate first because if the [CrO4-2] theoretical is exceeded , the salt is
precipitated
10 𝑔
[𝐴𝑔+1 ] = 108 𝑔 = 0.09 F
× 1𝐿
𝑚𝑜𝑙
1.0 𝑔
[𝑃𝑏 +2 ] = 207 𝑔 = 4.83 𝑥 10−3 F
× 1𝐿
𝑚𝑜𝑙
H2O
Ag2CrO4 ⇄ 2Ag+1 + CrO4-2
At eqbm: X 0.093 F x
Ksp Ag2CrO4 = [Ag+1]2 [CrO4-2]
9 𝑥 10−12 = (0.093)2 (𝑥)
X = 1.04 x 10-9 F = [CrO4-2] = [Ag2CrO4]
H2O
PbCrO4 ⇄ Pb+2 + CrO4-2
At eqbm: X 4.83 x 10-3 F x
Ksp PbCrO4 = [Pb+2] [CrO4-2]
2 𝑥 10−14 = (4.83 𝑥 10−3 )(𝑥)

X = 4.17 x 10-12 F = [CrO4-2] = [PbCrO4]
[CrO4-2] in PbCrO4 < [CrO4-2] in Ag2CrO4
∴ Pb++ will precipitate first
Ex. A solution contains 0.010 mol of KI and 0.010 mol of KCl per liter. AgNO 3 is gradually
added to this solution. Which will be precipitated first?
Given:
AgNO3 soln
0.01 mol KI which ion will ppt first?

0.01 mol KCl
1L soln
Soln:
 The precipitates which will be formed are AgI and AgCl
 Determine the [Ag+] requirement of each ion from Ksp , the one with the lesser [Ag+]
will precipitate first because if the [Ag+] theoretical is exceeded , the salt is
precipitated
F = mol /L = 0.01mol / 1L = 0.01 F

100%
KI → K+ + I-
0.01 0.01 0.01
H2O
AgI ⇄ Ag+1 + I-
At eqbm: X x 0.01
Ksp AgI = [Ag+1] [ I-]
1.5𝑥 10−16 =(𝑥)(0.01)
X = 1.5 x 10-14 F = [Ag+] = [AgI]
100%
KCl → K+ + Cl-
0.01 0.01 0.01
H2O
AgCl ⇄ Ag+1 + Cl-
At eqbm: X x 0.01
Ksp AgCl = [Ag+1] [ Cl-]

1.56𝑥 10−10 =(𝑥)(0.01)
X = 1.56 x 10-8 F = [Ag+] = [AgCl]
[Ag+] in AgI < [Ag+] in AgCl
∴ AgI will precipitate first
H2S AS PRECIPITATING AGENT
28/103] A solution is 0.01 F with respect to Co ++ and 0.3F in H3O+ ions. If the solution is
saturated with H2S, indicate calculation as to whether or not a precipitate of CoS will be
formed?
Given:
Satd with H2S soln ( 0.1F)
0.01F Co++ will ppt form?

0.3F H3O+
soln
Soln:
 If there is a precipitate, it will be CoS
Actual Ksp = [ Co++] [S-2]
Get this from sat’d sol’n of H2S & 0.3 F H3O+
H2S + 2H2O ⇄ 2 H3O+ + S-2

I: 0.1 F 0.3 F 0
C: -X 2X X
E: 0.1-X 0.3 + 2X X
[𝐻3 𝑂 + ]𝟐 [𝑆 −2 ]
𝑲 𝑯𝟐𝑺 𝒐𝒗𝒆𝒓𝒂𝒍𝒍 =
[𝑯𝟐 𝑺]
(03+2X)𝟐 [𝑆 −2 ]
𝟔. 𝟖𝟒 𝑿𝟏𝟎−𝟐𝟑 =
𝟎.𝟏−𝑿
[S-2] = 7.6 X 10-23 F

H2O
CoS ⇄ Co+2 + S-2
At eqbm: X 0.01 7.6 X 10-23
actual KSP = ( 0.01)( 7.6 X 10-23)
= 7.6 X 10-25

Theoretical Ksp = 7 x 10-22
∴ no precipitate of CoS will form
ASSIGNMENT:
1. The solubility of strontium fluoride at room temperature is 2.3 x 10 -1g/L. Calculate its
solubility product constant.
2. The solubility of silver chromate is 0.0014 g/100 mL. What is the solubility product
constant of this salt?
3. The solubility product constant of silver carbonate at 250C is 6.2 x 10-12. Calculate the
solubility of silver carbonate in g/L
4. The solubility product constant of MgNH 4PO4 is 2.5 x 10-13.Calculate its solubility in
gram-formula weights per liter.
5. What would be the sulfide-ion formality in a saturated solution of Bi2S3 if the Ksp of this
compound is 1.6 x 10-72 ?
6. A solution is 0.01F with respect to Mg ++ ions. What concentration of hydroxide ions in
formula weights per liter will be needed to precipicate Mg(OH)2? The Ksp of Mg(OH)2
is 1.2 x10-11
7. The solubility of Mn(OH)2 in water is 0.0002 g/100 mL. What is the solubility in gram-
formula weights per liter in a solution which is 0.01F with respect to the hydroxide ions?
8. What is the maximum concentration of Mg++, in formula weights per liter that can exist
in a solution which is 0.5F with respect to NH4Cl and 0.1 F with respect to NH3.
9. Aa solution containing 10 g each of Ba ++ and Pb++ per liter, potassium chromate is
added dropwise, which will precipitate first?
10. Calculate whether or not precipitate will be formed in each case
a. 85 mL of 0.7 F lead nitrate + 150 mL NH3
b. 1L of 1.5 F H2SO4 + 800 mL of 0.9 F SrCl2
c. 5 g of MgCl2 in 85 mL H2O + 50 mL 1.2 F HF
HYDROLYSIS
- a chemical reaction or process in which a molecule splits into two parts by
reacting with a water molecule – one part gets the H+ from the water molecule and the
other part gets the OH- - this causes a change in the pH
- a reaction between water and ions of a salt to form a solution that is either acidic or
basic
- reaction when H+ or OH- of water to form a weak electrolyte
CLASSIFICATION OF SALTS:
1. SALT WHOSE CATION AND ANION ARE APROTIC

- Salts of Strong Acid and Strong Base
- no hydrolysis , no gain or of electrons
- no alteration of equilibrium concentration in the solvent
eg. NaCl → Na+ + Cl-
2. SALTS WHOSE ANIONS ARE BASIC

- salts of strong base and weak acid - hydrolyze

Eg. KNO2, KC2H3O2, NaC2H3O2
3. SALTS WHOSE CATIONS ARE ACIDIC

- salts of strong acid and weak base- hydrolyze
Eg. NH4Cl
4. SALTS WHOSE CATIONS ARE ACIDIC AND WHOSE ANIONS ARE BASIC
- salts of weak acid and weak base- hydrolyze
Eg. NH4C2H3O2
CASE 1: Hydrolysis of anions that are bases (or salts of Strong base (SB) and weak acid
(WA)
example: hydrolysis of sodium acetate
NaC2H3O2  Na+ + C2H3O2- 1st
hydrolysis of C2H3O2- 2nd

(Na+ is aprotic ion)
C2H3O2 + HOH  HC2H3O2
- + OH-
KC = [HC2H3O2][OH-]
[C2H3O2-][HOH]
KC [HOH] = Kh = [HC2H3O2][OH-]
[C2H3O2-]
Kh = [HC2H3O2][OH-] x [H+] = KW
[C2H3O2-] [H+] Ka
the Kh of any anion base is equal to Kw divided by Ka of the conjugate acid formed when
the anion is ionized
[HC2H3O2] = [OH-] and [C2H3O2-] ≈ c
Kh = KW = [OH-]2 = [OH-]2
Ka [C2H3O2-] c
 c= concn of salt(sodium acetate)
[OH-] = KW c [H+] = KW
Ka [OH-]
[H+] = KW
c(KW /Ka)

pH = ½ pKW + ½ pKa + ½ log c
Suppose 0.20 gfw of sodium acetate is dissolved in sufficient water to produce a liter of
solution, calculate the [H+] and the pH of the solution. Also compute the degree of
hydrolysis of the acetate ion in this concentration. Ka for acetic acid = 1.75 x 10-5
KW = 1 x 10-14
KW = 1 x 10-14
[H+] = KW = 1 x 10-14
KW Ka 1 x 10-14 (1.75 x 10-5)
√ c √ 0.20
= 1.06 x 10-5
pH = – log [H+] = 4.97 = 7 - 2.378 + 0.35
deg of hydrolysis = 1.06 x 10-5 / 0.2 = 5.3 x 10-5
percent hydrolysis = 5.3 x 10-5 (100) = 0.0053%
Ex.2 Calculate the concentration of hydroxide ion in 0.1F solution of potassium nitrite.
Soln: does not hydrolyze
Will hydrolyze
KNO2 ⟶ K+ + NO2-
t=0 0.1F 0 0
t rxn: - 0.1 0.1 0.1
Next: Hydrolysis reaction

NO2- + H2O ⇌ HNO2 + OH-
I: 0.1 0 0
C: - X +X +X
E: 0.1 –X X X
𝐾𝑤 [𝐻𝑁𝑂2 ] [𝑂𝐻 −1 ]
= 𝐾ℎ =
𝐾𝑁𝑂2 [𝑁𝑂2−1 ]
1 × 10−14 (𝑋)2
=
1 × 10−4 (0.1 − 𝑋)
𝑿 = [𝑶𝑯−𝟏 ] = 𝟏. 𝟓𝟖 × 𝟏𝟎−𝟔 F
CASE 2: Hydrolysis of cations that are acids (or salts of weak base (WB) and strong acid
(SA)
example: hydrolysis of ammonium chloride
NH4Cl  NH4+ + Cl- 1st

(Cl- is aprotic ion)
hydrolysis of NH4+ 2nd

NH4+ + HOH  NH4OH + H+
or NH4+ + HOH  NH3 + H3O+
KC = [NH4OH ][H+] Kh = KC [HOH]

[NH4+][HOH]
Kh = [NH4OH ][H+] x [OH-] = KW

[NH4+] [OH-] Kb
[NH4OH ] = [H+]
Kh = [NH4OH ][H+] = KW
[NH4+] Kb
and [NH4+] ≈ c
c = concn of salt (NH4Cl)
kh = [H+]2 = KW [H+]2 = KW
[NH4+] kb c Kb
[H+] = sq rt of KW c
kb
pH = – ½ log KW + ½ log Kb – ½ log c
= ½ pKW – ½ pKb – ½ log c
Example3: What is the Kb for an aqueous solution of ammonia if 0.1 F ammonium chloride
has a pH of 5.13?
pH = – log [H+] [H+] = inv log – 5.13
= 7.413 x 10-6
[H+]2 = KW (7.413 x 10-6)2 = 1 x 10-14

[NH4Cl] Kb 0.1 Kb
Kb = 1.812 x 10
EX.4: Ammonium chloride in 0.001 F solution is 0.076% hydrolyzed. Calculate the ionization
constant of NH3 from these values.
Soln:
𝑎𝑚𝑡 𝑜𝑓 𝑁𝐻4 𝐶𝑙 ℎ𝑦𝑑𝑟𝑜𝑙𝑦𝑧𝑒𝑑 𝑋
% ℎ𝑦𝑑𝑟𝑜𝑙𝑦𝑠𝑖𝑠 = 𝑎𝑚𝑡 𝑜𝑓 𝑁𝐻 × 100 = × 100
4 𝐶𝑙 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙𝑙𝑦 𝑝𝑟𝑒𝑠𝑒𝑛𝑡 0.001
% ℎ𝑦𝑑𝑟𝑜𝑙𝑦𝑠𝑖𝑠 = 7.6 × 10−7

100%
𝑁𝐻4 𝐶𝑙 → 𝑁𝐻4+1 + 𝐶𝑙−
t=0 0.001F 0 0
t rxn: - 0.001 0.001 0.001
after rxn: 0 0.001 0.001
Next: Hydrolysis reaction

NH4+ + H2O ⇌ NH3 + H3O+
I: 0.001 0 0
C: - 7.6 X10-7 +7.6 X10-7 +7.6 X10-7
E: 0.001 –7.6 X10-7 7.6 X10-7 7.6 X10-7
𝐾𝑤 [𝑁𝐻3 ] [𝐻3 𝑂+1 ]

= 𝐾ℎ =
𝐾𝑁𝐻3 [𝑁𝐻4+1 ]
1 × 10−14 (7.6 × 10−7 )2

=
𝐾𝑁𝐻3 (0.001 − 7.6 × 10−7 )
𝐾𝑁𝐻3 = 𝟏. 𝟕𝟑 × 𝟏𝟎−𝟓 F
EX.5 Calculate the pH of 1F solution of methylammonium chloride
[ 𝐶𝐻3 𝑁𝐻3 𝐶𝑙] = 1 𝐹 → 𝑝𝐻?

100%
𝐶𝐻3 𝑁𝐻3 𝐶𝑙 → 𝐶𝐻3 𝑁𝐻3+1 + 𝐶𝑙−
t=0 1F 0 0
t rxn: - 1 1 1
after rxn: 0 1 1
CH3NH3+ + H2O ⇌ CH3NH2 + H3O+

I: 1 0 0
C: - X +X +X
E: 1–X X X
𝐾𝑤 [𝐶𝐻3 𝑁𝐻2 ] [𝐻3 𝑂+1 ]
= 𝐾ℎ =
𝐾 𝐶𝐻3 𝑁𝐻3 [𝐶𝐻3 𝑁𝐻3+1 ]
1 × 10−14 (𝑋)2
=
4.38 × 10−4 (1 − 𝑋)
𝑋 = [𝑯𝟑 𝑶+ ] = 𝟒. 𝟕𝟕𝟖𝟐 × 𝟏𝟎−𝟔 𝑭

𝑝𝐻 = log 𝟒. 𝟕𝟕𝟖𝟐 × 𝟏𝟎−𝟔 𝑭
𝑝𝐻 = 5.32
CASE 3: Hydrolysis of salts whose cations are acids and whose anions are bases [or salts of
weak acid (WA) and weak base (WB)]
example: hydrolysis of ammonium acetate
NH4C2H3O2  NH4+ + C2H3O2- 1

hydrolysis of NH4+ and C2H3O2-
2nd
NH4+ + C2H3O2- + HOH  NH4OH + HC2H3O2
Kh = KC [HOH] = [NH4OH ][HC2H3O2] x [H+][OH-]

[NH4+] [C2H3O2-] [H+] [OH-]
Kh = KW
Ka Kb
[NH4OH ] = [HC2H3O2]
KW = [HC2H3O2]2 = [HC2H3O2]2
Ka Kb [NH4+] [C2H3O2-] c2
and [NH4+] = [C2H3O2-] ≈ c

c = concn of salt
(NH4acetate)
[HC2H3O2]2 = KW c2 or [H+]2 [C2H3O2-]2 = KW c2

Ka Kb Ka2 Ka Kb
[H+]2 (c2) = KW c2
Ka2 Ka Kb
[H ] = sq rt KW Ka
+
Kb
pH = ½ pKW + ½ pKa – ½ pKb
Example: What is the pH of a solution which is 0.1 F with respect to NH 4CN
Kb ammonia = 1.8 x 10-5
Ka HCN = 7.2 x 10-10
[H+] = KW Ka = sq rt 1 x 10-14(7.2 x 10-10)

Kb 1.8 x 10-5

= 6.32 x 10-10
pH = 9.20
Or:
100%
𝑁𝐻4 𝐶𝑁 → 𝑁𝐻4+1 + 𝐶𝑁 −
t=0 0.1F 0 0
t rxn: - 0.1 0.1 0. 1
NH4+ + CN- + H2O ⇌ NH4OH + CN-

I: 0.1 0.1 0 0
C: - X -X +X +X
E: 0. 1–X 0.1-X X X
𝐾𝑤 [ 𝑁𝐻4 𝑂𝐻] [𝐻𝐶𝑁]
= 𝐾ℎ =
𝐾 𝑁𝐻4 𝑂𝐻 × 𝐾𝐻𝐶𝑁 [𝑁𝐻4+1 ][𝐶𝑁 − ]
1 × 10−14 (𝑋)2
=
(1.8 × 10−5 )(7.2 × 10−10 ) (0.1 − 𝑋)
 do not neglect X since Kh = 0.77 (large value)

SOLVING FOR X
𝑋 = [𝑵𝑯𝟒 𝑶𝑯] = [𝑯𝑪𝑵] = 𝟎. 𝟎𝟒𝟔𝟕𝟔 𝑭
HCN + H2O ⇌ H3O++ CN-

I: 0.04676 0 0.1-X = 0.1-0.4676= 0.05324
C: - y +y +y
E: 1–y y 0.05324 + y
[𝐻3 𝑂 +1 ] [𝐶𝑁 −1 ]
𝐾𝑎 =
[𝐻𝐶𝑁]
(𝑦)(0.05324 + 𝑦)
7.2 × 10−10 =
(0.04676 − 𝑦)
𝑦 = [𝑯𝟑 𝑶+ ] = 𝟔. 𝟑𝟐𝟒 × 𝟏𝟎−𝟏𝟎 𝑭
𝑝𝐻 = log 𝟔. 𝟑𝟐𝟒 × 𝟏𝟎−𝟏𝟎 𝑭
𝑝𝐻 = 9.2

CASE 4: Hydrolysis of salts whose ions are aprotic [or salts of strong acid (SA) and sttrong
base (SB)]
NO hydrolysis takes place between the ions of this type of salt and water
NaCl  Na+ + Cl- 1st
hydrolysis of Na+ and Cl- 2nd
Na+ + Cl - + HOH  NaOH + HCl
no Kh value because no hydrolysis take place
𝐾𝑤 [𝐻𝑁𝑂2 ] [𝑂𝐻 −1 ]
= 𝐾ℎ =
𝐾𝑁𝑂2 [𝑁𝑂2−1 ]
ASSIGNMENT:
1. calculate the pH of a 2F solution of sodium formate.
2. What are the pH value and percentage hydrolysis at 250C in a o.1F solution of
sodium cyanide?
3. What concentration of potassium acetate has a pH of 8.9?
4. Calculate the grams of ammonium nitrate which must be added to a liter of
aqueous solution to produce a pH of 5.4
5. Calculate the percentage hydrolysis and pH of 0.1F solution of NH 4CL

MODULE 6: VOLUMETRIC METHODS OF ANALYSIS:UNIT 1: ACIDIMETRY AND

ALKALIMETRY
UNIT 1: INTRODUCTION: TITRATION CONCEPTS
QUANTITATIVE ANALYSIS
DIVISIONS OF ANALYTICAL CHEMISTRY:
a. Qualitative Analysis - nature of the constituents of a given material

– identification of constituents present in a sample
b. Quantitative Analysis - proportion in which these constituents are present
-measures the mass of a substance chemically related to the

analyte
DIVISIONS OF QUANTITATIVE ANALYSIS:

1. Volumetric Analysis - measures the volume of a standard reacting reagent with
the desired constituent in a definite chemical reaction
- commonly known as “TITRATION”
2. Gravimetric Analysis - based entirely upon weight, the original substance

is
weighed and from it the constituent to be determined is isolated as an element or as
a compound of definite composition

3. Colorimetric Method - the substance to be determined is converted to
some compound which imparts to its solution a distinct color, the intensity of which
varies in proportion to the amount of a compound in a solution
4. Other physicochemical Methods - quantities of constituents from

measurements of such properties as the turbidity of suspensions of semi colloidal
precipitate
5. Gas Volumetric Methods - determination of a mixture of gases
VOLUMETRIC ANALYSIS
Divisions:
1. Neutralization methods – acidimetry (acid determination) and alkalimetry (alkali

determination)
2. Oxidation and Reduction – “redox” – oxidimetry and reductimetry
3. Precipitation methods – precipitimetry
4. Complex formation methods - compleximetry
Definition of terms:
 Titration – determination of the concentration (unknown) of a solution by the

gradual
addition of a standard (known) solution
 titrant – solution of known concentration (reagent solution)
 analyte – substance under study (substance being titrated)
 Standard solution – solution whose concentration is accurately known

a. standard acid solutions:
HCl, HNO3, H2SO4, H2C2O4
b. standard alkali solutions:

NaOH, KOH, Ba(OH)2
 Standardization – process of determining the exact concentration of a titrant
 Primary standard – a substance of highest purity used for standardization
characteristics: highest purity, soluble in water, stable towards air, high temp
and humidity
example: potassium hydrogen phthalate (KHP) or KHC8H4O4 ; sodium

carbonate, sulfuric acid
 Indicator – a substance (colored dye) that is sensitive hydrogen ion concentration,

changes in color once the end point is reached (from acidity to alkalinity and vice
versa)
example: phenolphthalein, methyl orange, congo red, bromocresol

green, litmus paper
• Endpoint – point in titration where the color of the indicator changes which
coincides with the EQUIVALENCE POINT
• Equivalence point – the point in titration in which the chemical reaction is

completed or
the point of complete neutralization
- point where the acid and the base added in amounts are
equivalent to
each other

• Normality = equivalents of solute / liter of solution
• equivalents, E = weight of solute .
equivalent weight of solute
also: equivalents,E = N X V in liters
Note: N = eq/ L = meq/mL
NEUTRALIZATION - a reaction between a base and an acid to form salt and water.
base + acid → salt + water
ex. NaOH + HCl → NaCl + water
Equivalent weights in Neutralization Methods
fundamental reaction of acidimetry and alkalimetry is:
H+ + OH-  H2O
- the neutralization of an acid by a base or the neutralization of a base by an acid
• equivalent weight of a substance acting as an acid is that weight of it which is

equivalent in total neutralizing power to one mole of Hydrogen as H +
• Equivalent weight of a substance acting as a base is that weight of it which will

neutralize one mole of hydrogen ions (or equivalent in total neutralizing power to
one mole of hydroxyl ions, OH-)
equivalent weight of solute = molecular weight
factor
Recall: factor determination:
acid – the number of H+ replaceable
HCl f=1 H2SO4 f=2
HC2H3O2 f=1 H3PO4 f=3
base – the number of OH- replaceable
NaOH f=1 Al(OH)3 f=3
Ca(OH)2 f=2
Others: needed in precipitimetry, redox, compleximetry
salt – the total positive charge

NaCl f=1 Na3PO4 f=3
Na2SO4 f=2
redox – change in oxidation state
Mn+2  Mn+7 f=5

S+6  S-2 f=8
DETERMINATION OF NORMALITY OF SOLUTION
g solute
N = eq. wt solute
liter solution
eq. wt = MW solute / f
or N = W (g) solute x f solute

MW solute x L solution
WORKING FORMULA:
equivalents of acid = equivalents of base
or milliequivalents of acid = milliequivalents of base
meq a = meq b
 me liquid = V mL x N , meq/mL
 me solid = w (g) solid

MW
f(1000)
equivalents = N in eq/L X V in L
3 CONDITIONS IN TITRATION:
(1) 1 LIQUID / 1 LIQUID

meq a = meq b
V a x Na = Vb x Nb
(2) 1 LIQUID / 1 SOLID

meq a = meq b
V a x Na = w(g) solid
MW
f(1000)

3. 1 SOLID / 1 SOLID
meq a = meq b
w(g) a = w(g) b
MW a MW b
fa(1000) fb(1000)
Ex. How many grams of each of the following solutions constitute the g-equiv wt as
an acid assuming complete neutralization unless otherwise stated
a. HNO3
b. KHSO4
c. H2SO3
d. H2C2O4.2H2O
e. CH3COOH
f. P2O5 – forming H2PO4-1
g. SO3 (acid fr H2SO4)
Soln:
Equivalents = g - equivalents
g-equiv wt = equiv wt
 Whatever will be the equiv –wt will be the gram ( stated fr the problem)
Basis: a g-equiv wt of a subst acting as an acid is that number of g of the subst which in
neutralization process furnishes one g-atom(1.0089) of replaceable hydrogen:
𝑀𝑊𝑎𝑐𝑖𝑑
𝑔 − 𝑒𝑞𝑢𝑖𝑣 𝑤𝑡 =
𝑓𝑎𝑐𝑖𝑑
𝑔
63.02𝑚𝑜𝑙 𝑔
a. 𝐻𝑁𝑂3 = 𝑒𝑞 = 63.02 = 63.02 𝑔
1𝑚𝑜𝑙 𝑒𝑞
𝑔
136.17 𝑔
𝑚𝑜𝑙
b. 𝐾𝑆𝐻𝑂4 = 𝑒𝑞 = 136.17 = 136.17 𝑔
𝑔
82.08𝑚𝑜𝑙 𝑔
c. 𝐻2 𝑆𝑂4 = 𝑒𝑞 = 41.04 = 41.04 𝑔
ASSIGNMENT: 1. d,e,f,g
Ex.2 What is the millieqquivalent wt of each of the ff subst acting as bases with complete
neutralization in each case
a. 𝐶𝑎 (𝑂𝐻)2 f. 𝑍𝑛𝑂
b. 𝐵𝑎𝑂 g. 𝐶𝑎𝐶𝑂3
c. 𝐾𝐻𝐶𝑂3 h. 𝐹𝑒2 𝑂3

d. 𝑁𝑎2 𝑂
e. 𝑁𝑎2 𝑂2
𝑀𝑊 𝑔 − 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑤𝑡 𝑜𝑓 𝑏𝑎𝑠𝑒
𝑚𝑖𝑙𝑙𝑖𝑒𝑞𝑢𝑖𝑣 𝑤𝑡(𝑚𝑒 𝑤𝑡) = =
𝑓(1000) 1000 𝑚𝐿
𝑔
74.1𝑚𝑜𝑙
a. 𝐶𝑎 (𝑂𝐻)2 = 𝑒𝑞 𝑚𝑒𝑞 = 0.03705
2𝑚𝑜𝑙(1000 )
𝑒𝑞
𝑔
153.34𝑚𝑜𝑙
b. 𝐵𝑎𝑂 = 𝑒𝑞 𝑚𝑒𝑞 = 0.07667
2𝑚𝑜𝑙(1000 )
𝑒𝑞
𝑔
100.12𝑚𝑜𝑙
c. 𝐾𝐻𝐶𝑂3 = 𝑒𝑞 𝑚𝑒𝑞 = 0.1002
1𝑚𝑜𝑙(1000 )
𝑒𝑞
ASSIGNMENT 2: d,e,f,g,h
MIXING PROCESSES:
CASE I: solution B (acid)
Solution A (acid) Resulting solution (R.S )
(acid)
Solution C (acid)
meq A (acid) + meq B (acid) + meq C (acid) = meq R.S. (acid)
A, B, C & R.S. could be bases too..
Note: meq pure water = 0
CASE II:
- Solutions mixed are different (1.e. acids and bases)

Solution B (base)
Solution A (acid) Resulting solution

(either acidic or basic depending on
total meq acid & total meq base)
Solution C (base)
R.S is acidic if total meq acid > total meq base

R.S is basic if total meq acid < total meq base
Example: If 3.00 g of solid KOH and 5.00 g of solid NaOH are mixed, dissolved in water, and
the solution made up to 1500 mL, what is the normality of the solution as a base?
Given: W KOH = 3.00 g V soln = 1500 mL

W NaOH = 5.00 g
Reqd: N
Soln:
Tot eq of base = eq KOH + eq NaOH

N = eq/L
eq of KOH = 3.00 g KOH = 0.0535 eq

(56.1 g/mol)/1(eq/mol)
eq of NaOH = 5.00 g NaOH = 0.125 eq

(40.0 g/mol)/ 1(eq/mol)
N= (0.0535 + 0.125) = 0.1189 eq/L

1500 mL (1L/1000mL)
Adjusting a Solution to a desired Normality

- dilution concept
Example: To what volume must 750 mL of 2.400 N solution be diluted in order to make it
1.760 N?
Given:
V1 = 750 mL
C1 = 2.400 N C2 = 1.760 N
Reqd: V2
Soln:
V 1 C1 = V2 C2
(750 mL)(2.400 meq/mL) = V2 (1.760 meq/mL)
V2 = 1023 mL

*add 273 mL of water to make 1023 mL of a 1.76N
Soln
Example: How much 0.600 N base must be added to 750 mL of a 0.200 N base in order for
the solution to be 0.300N?
Given:
Solution A(base) Resulting solution

0.600 N (ei 0.300N
Solution B (base)
750 mL ,0.200 N
V1C1 + V2C2 = VmCm

V1(0.6meq/mL) + (750ml x 0.2meq/mL) = (750 + V1)(0.3meq/mL)
V1 = 250 mL
Example: A solution containing 25.3 mL of 0.1065 N HCl is added to one containing 92.2
mL of 0.2715 M H2SO4 and 50.0 mL of 1.00 N KOH are added. Is the solution acid or alkaline?
What volume of 0.100N acid or alkali must be added to neutralize the soln?
50mL KOH
1N
25.3 mL
0.1065 N HCl
Resulting solution , RS
92.2 mL
0.2715 M H2SO4
Soln:
a. if the total me acid> tot me base ; the RS is acid
b. . if the total me acid< tot me base ; the RS is base
𝑡𝑜𝑡𝑎𝑙 𝑚𝑒𝑞 𝑎𝑐𝑖𝑑𝑠 = 𝑚𝑒𝑞 𝐻𝐶𝑙 + 𝑚𝑒𝑞 𝐻2𝑆𝑂4
𝑡𝑜𝑡𝑎𝑙 𝑚𝑒𝑞 𝑎𝑐𝑖𝑑𝑠 = 𝑉𝐻𝐶𝑙 𝑁𝐻𝐶𝑙 + 𝑉𝐻2 𝑆𝑂4 𝑁𝐻2 𝑆𝑂4

For the unit of Normality, N it can be eq/L or me/mL
Since we are given the concentration of H2SO4 in M , convert it to N by using the

formula : N = M x f
N = 0.2715 mol/L X 2 eq/mol = 0.543 eq/L or meq/mL
𝑚𝑒𝑞 𝑚𝑒𝑞
𝑡𝑜𝑡𝑎𝑙 𝑚𝑒 𝑎𝑐𝑖𝑑𝑠 = (25.3 𝑚𝐿) (0.1065 ) + (92.2 𝑚𝐿) (0.543 ) = 52.76 𝑚𝑒
𝑚𝐿 𝑚𝐿
𝑡𝑜𝑡𝑎𝑙 𝑚𝑒 𝑏𝑎𝑠𝑒 = 𝑚𝑒 𝐾𝑂𝐻

𝑚𝑒𝑞
𝑡𝑜𝑡𝑎𝑙 𝑚𝑒 𝑏𝑎𝑠𝑒 = (50 𝑚𝐿) (1 ) = 50 𝑚𝑒
𝑚𝐿
52.76 𝑚𝑒𝑞 > 50𝑚𝑒𝑞 ∴ 𝑅𝑆 𝑖𝑠 𝑎𝑐𝑖𝑑𝑖𝑐
b. for neutralizing the RS

𝑛𝑒𝑡 𝑚𝑒𝑞 𝑅𝑆 𝑎𝑠 𝑎𝑐𝑖𝑑 = 𝑚𝑒𝑞 𝑏𝑎𝑠𝑒
OR:
𝑡𝑜𝑡 𝑚𝑒𝑞 𝑎𝑐𝑖𝑑 = 𝑚𝑒𝑞 𝑏𝑎𝑠𝑒

Na Va = Nb Vb
(25.3 mL)(0.1065 N) + (92.2mL)(0.543N) = (V)(0.1N) + (50mL)(1N)
V= 27.60 mL
Neutralization – the reaction between an acid and a base to form water and salt
Ex. NaOH + HCl ⟶ HOH + NaCl
Example: Given the ff data: 1.00 mL of NaOH ≎ 1.117 mL HCl, the HCl is 0.4876 N. How
much water should be added to 100 mL of the alkali to make it 0.500 N
Given: V NaOH = 1.00mL V HCl = 1.117 mL

N HCl = 0.4876 N
Reqd; V H2O
Soln:
me NaOH = me HCl
(𝑁𝑁𝑎𝑂𝐻 )(𝑉𝑁𝑎𝑂𝐻 ) = (𝑁𝐻𝐶𝑙 )(𝑉𝐻𝐶𝑙 )

(𝑁𝑁𝑎𝑂𝐻 )(1.00𝑚𝐿) = (0.4876 𝑁)(1.117𝑚𝐿)
𝑁𝑁𝑎𝑂𝐻 = 0.5446 𝑁

(𝑉1 )(𝐶1 ) = (𝑉2 )(𝐶2 )
(100𝑚𝐿)(0.5446𝑁) = (𝑉2 )(0.5𝑁)

𝑉2 = 108.92 𝑚𝐿
𝑉𝐻2 𝑂 = 𝑉2 − 𝑉1 = 108.92 − 100 = 8.92 𝑚𝐿
Example: If 30.00 g KHC2O4. H2C2O4 ( potassium tetroxalate, Ktet) are dissolved ,diluted to
1000 mL, and it is found that 40.00 mL are neutralized by 20.0 mL of a solution of KOH, what
is the normality of the alkali solution?
Given: W Ktet = 30.00 g V Ktet = 40 mL

V soln = 1000 mL, V KOH = 20.00 mL
Reqd: N KOH
Soln:
We are given the wt and volume of potassium tetroxalate solution, so we can solve for the
concentration in N by using the formula:
𝑊 ×𝑓
𝑁 = 𝑀𝑊 × 𝑉
𝑒𝑞
30.0 𝑔 ×3 𝑚𝑜𝑙
𝑁 𝐾𝑡𝑒𝑡 = 𝑔 = 0.354 𝑁
242.2𝑚𝑜𝑙 × 1𝐿
In neutralization process , we are given the volume of Ktet to neutralize KOH
meq acid = meq base

𝑚𝑒𝑞 KHC2O4.H2C2O4 = 𝑚𝑒𝑞 𝐾𝑂𝐻
(𝑁 × 𝑉 )𝐾𝑡𝑒𝑡 = (𝑁 × 𝑉 )𝐾𝑂𝐻
(0.354 𝑁)(40𝑚𝐿) = (𝑁)(20𝑚𝐿)
𝑁𝐾𝑂𝐻 = 0.708 𝑁
Example: a sample of pure oxalic acid, H2C2O4.2H2O weighs 0.2000 g and requires 30.12 mL
of KOH solution for complete neutralization. What is the N of the KOH solution?
Given:
W H2C2O4.2H2O = 0.2000 g V KOH = 30.12 mL
Reqd: N KOH
Soln:
Eq acid = eq base
Eq H2C2O4.2H2O = eq KOH
weight oxalic x factoroxalic = NKOH VKOH

MWoxalic acid

0.2000 g x 2 eq/mol = 30.12 mL X NKOH
126.07 g/mol 1000 mL/L
NKOH = 0.1053 eq/L
Example: What is the normality of a solution of HCl if 20.00 mL is required to neutralize the
NH3 that can be liberated from 4.00 millimoles of (NH4)2SO4?
Given:
V HCl = 20.00 mL
(NH4)2SO4 = 4.00 millimoles
Reqd: N HCl
Soln:
Eq Base = Eq acid
Eq NH3 = Eq HCl
weightb x factorb = NHCl VHCl

MWb
molb x factorb = NHCl VHCl
4.00 mmol (NH4)2SO4 x 2mmol NH3 = 8 mmols NH3

1mmol (NH4)2SO4
8 mmols NH3 x 1 meq = 20.00 mL (NHCl)

mmol
NHCl = 0.400 meq/mL
STANDARDIZATION OF ACID AND BASE
Standard acid solutions

- Prepared from HCl, HNO3 , H2SO4 , oxalic acid
Standard alkali solutions
- NaOH, KOH, Ba(OH)2
 the standardization of a solution of an acid

- determine the volume of an acid which is equivalent to a known weight of a
pure basic substance
 the standardization of a solution of a base
- determine experimentally the volume of a base which is equivalent to a known
weight of a pure acid

ex. In standardizing solution of HCl it is found that 47.26 mL of the acid are
equivalent to 1.216 g of pure Na2CO3( methyl orange indicator).What is the
Normality of the acid?
Given:
V HCl = 47.26 mL
W Na2CO3 = 1.216 g
Reqd: N HCl
Soln;
meq acid = meq base

𝑚𝑒𝑞 HCl = 𝑚𝑒𝑞 𝑁𝑎2 𝐶𝑂3
(𝑊𝑁𝑎2 𝐶𝑂3 )
(𝑁 × 𝑉 )𝐻𝐶𝑙 = 𝑀𝑊
𝑓 ×1000
(1.216 𝑔)
( 𝑁)(47.26𝑚𝐿) =
105.99 𝑔/𝑚𝑜𝑙
2 𝑒𝑞/𝑚𝑜𝑙 × 1000
𝑁𝑁𝑎2 𝐶𝑂3 = 0.4856 𝑁
BACKTITRATION
– when there is over addition of a titrant at the endpoint, a standard solution

(backtitrant) is added to neutralize the over added titrant until a color change is
indicated
Example:
A solution of sulphuric acid is standardized against a sample which has been previously
found to contain 92.44% CaCO3 and no other basic material. The sample weighing
0.7423g is treated with 42.42mL of the acid then required 11.22 mL of NaOH solution.
If 1.00 mL of the acid is equivalent to 0.9976 mL of the NaOH, what is the normality of
each?
Standardization of acid: 1mL base ≎ X ml base

1mL H2SO4 ≎ 0.9976 mL NaOH
1 𝑚𝐿 𝐻2 𝑆𝑂4 0.9976 𝑚𝐿 𝑁𝑎𝑂𝐻
=
𝑥 𝑚𝐿 𝐻2 𝑆𝑂4 11.22 𝑚𝐿 𝑁𝑎𝑂𝐻
mL 𝐻2 𝑆𝑂4 = 11.2469
net Volume H2SO4 = 42.42 - 11.2469 = 31.17301 mL
𝑚𝑒 𝐶𝑎𝐶𝑂3 = 𝑚𝑒 𝐻2 𝑆𝑂4

(𝑊𝐶𝑎𝐶𝑂3 )
𝑀𝑊 = 𝑁 𝐻2 𝑆𝑂4 × 𝑉𝐻2 𝑆𝑂4
𝑓 ×1000
𝑤𝑡 𝑠𝑢𝑏𝑠𝑡
To calculate for the wt of CaCO3: % 𝑤𝑡 = × 100
𝑤𝑡 𝑠𝑎𝑚𝑝𝑙𝑒
92.44%
(0.7423 𝑔)( )
100
100.08 𝑔/𝑚𝑜𝑙 = 𝑁 𝐻2 𝑆𝑂4 × 31.17 𝑚𝐿
2 𝑒𝑞/𝑚𝑜𝑙 ×1000
𝑁 𝐻2 𝑆𝑂4 = 0.4399𝑁
𝑚𝑒 𝑁𝑎𝑂𝐻 = 𝑚𝑒 𝐻2 𝑆𝑂4
1mL H2SO4 ≎ 0.9976 mL NaOH

Therefore in 1 mL NaOH ≎ 𝑥 𝑚𝐿 𝐻2 𝑆𝑂4
1𝑚𝐿
𝑚𝐿 𝐻2 𝑆𝑂4 = = 1.0024 𝑚𝐿
0.9976
(1𝑚𝐿 )(𝑁𝑁𝑎𝑂𝐻 ) = (1.0024 𝑚𝑙)(0.4399𝑁)
𝑁𝑁𝑎𝑂𝐻 = 0.4409 𝑁
Example: When CaCO3 is used as a standard for a strong acid, it is necessary to dissolve it
in an excess of the acid and back titrate with NaOH solution. In such a standardization, a
water suspension of 1.000 g of CaCO3 was used. A volume of 49.85 mL of HCl was added
from a buret, and warming the solution to remove any
dissolved CO2, the solution required 6.32 mL NaOH to reach an endpoint. If a separate 50.0
mL pipetful of the HCl required 48.95 mL of the NaOH for neutralization, what is the N of the
HCl and of the NaOH?
eq CaCO3 = eq HCL – eq NaOH

wt CaCO3 x factor CaCO3 = NHCl VHCl – NNaOH VNaOH
MW CaCO3
1.000g x 2 eq/mol = 49.85mL NHCl – 6.32mL NNaOH
100.08 g/mol
 equation 1
But me HCl = me NaOH
50.0ml NHCl = 48.95mL NNaOH  equation 2
NHCl = 48.95mL NNaOH
50.0 mL
NHCl = 0.4605 N NNaOH = 0.4704 N
Assignment:
3. How many mL of 0.1421N KOH are required to neutralize 13.72 mL of 0.06850 M
H2SO4?

4. Given the ff data: 1.00 mL≎ 1.117 mL HCl, the HCl is 0.4876 N. How much water
should be added to 100 mL of the alkali to make it 0.500 N?
5. A solution of H2SO4 is found to be 0.5172 N and 39.65 mL of it are equivalent to 21.74
mL of standard alkali solution. What is the N of the alkali and how many grams of
sulfamic acid (HSO3.NH2) will 1.oo mL of it neutralize?
6. In standardizing a solution of NaOH against 1.431g of KHC 8H4O4, the analyst uses
35.50 mL of the alkali and has to run back with 5.12 mL of acid ( 1mL ≎0.0031g Na2O)
What is the normality of NaOH?
7. A solution of H2SO4 is standardized against a sample of which has been previously
found to contain 92.44% CaCO3 and no other basic material. The sample weighing
0.7423 g is treated with 42.42 mL of the acid and the excess acid then requires 11.22
mL of NaOH solution. If 1.00 mL of the acid is equivalent to 0.9976 mL of the NaOH,
what is the N of each solution?
8. - If 44.97 mL of a solution of HCl are equivalent to 43.76 mL of a solution of NaOH,
and if 49.14 mL of the latter will neutralize 0.2162 g of KHC2O4.H2C2O4.2H2O, what
volume of water should be added to 100 mL of the HCl in order to prepare a mixture
which is 0.0500 N as an acid?
9. What is the equivalent weight of an organic acid if 44.00 mL NaOH solution (1.000
mL ≎ 1.100 mL HCl ≎ 0.01001 g CaCO3) are required to neutralize 0.5192 g of the
organic acid?
DETERMINATION OF TOTAL ACIDITY AND ALKALINE STRENGTH
Calculation of Percent Purity from Titration Values
Na V a = Nb V b
weight a x factora = Nb Vb 1
MWa
weight a = Nb Vb x MWa
factora
% component (a) = weighta x 100 2
weight of sample
% component (a) = Nb Vb x MWa x 1 x100
factora wt sample

Example: A sample of pearl ash (technical grade of K2CO3) weighing 2.000 g is titrated with
HCl and requires 25.00 mL. What is the alkaline strength of the ash in terms of percent K 2O if
20.00 mL of the HCl will just neutralize the NH3 that can be liberated from 4.000 mmols of
(NH4)2HPO4?
Given:
Wt sample = 2.000 g V HCl = 20.0 mL
V HCl = 25.0 mL (NH4)2HPO4= 4.000 mmols
Reqd: % K2O
Soln:
me K2O = me HCl
meq HCl = me (NH4)2HPO4
20.00 mL (NHCl) = 4 mmols x 2 meq/mmol
NHCl = 0.400 me/mL
weight K2O x 2 me/mmol = 25.00 mL(0.40 me/mL)

94.20 mg/mmol
Weight K2O = 471 mg or 0.471 g

% K2O = 0.471 g x 100 = 23.55%
2.00 g
Example: On ignition, rochelle salt, KNaC4H4O6.4H2O, is converted to KNaCO3. A sample of

the original salt weighs 0.9546 g, and the ignition product is titrated with H 2SO4. From the ff
data find the % purity of the sample: H2SO4 used = 41.72 mL; 10.27 mL H2SO4 ≈ 10.35 mL
NaOH; the NaOH is 0.1297N; NaOH used for back titration = 1.91 mL
me KNaC4H4O6.4H2O = me H2SO4 – me NaOH(back titration)

me H2SO4 = me NaOH
10.27mL (NH2SO4) = 10.35 mL (0.1297 N)
NH2SO4 = 0.1307 N
wt KNaC4H4O6.4H2O x 2 me/mmol
282.19 mg/mmol
= 0.1307N(41.72 mL) – 0.1297N(1.91 mL)
wt KNaC4H4O6.4H2O = 734.41 mg = 0.73441 g
% purity = (0.73441/ 0.9546) = 76.93%

Example: What weight of soda ash should be taken for analysis such that the percentage
of Na2O present can be found by multiplying by 2 the number of milliliters of 0.200 N acid
solution used in titration?
factora
% component (a) = weighta x 100
weight of sample
factora wt sample
% Na2O = 2 x (Vacid)
2 x (Vacid) = 62 mg/mmol x (0.200 me/ml)(Va) x 1 .
2 me/mmol wt sample
weight sample = 0.3100 g
Example: A sample of pearl ash (technical grade of K2CO3) weighing 2.000 g is titrated with
HCl and requires 25.00 mL. What is the alkaline strength of the ash in terms of percent K 2O if
20.00 mL of the HCl will just neutralize the NH3 that can be liberated from 4.000 mmols of
(NH4)2HPO4?
me K2O = me HCl
meq HCl = me (NH4)2HPO4
20.00 mL (NHCl) = 4 mmols x 2 meq/mmol
NHCl = 0.400 me/mL
weight K2O x 2 me/mmol = 25.00 mL(0.40 me/mL)
94.20 mg/mmol
Weight K2O = 471 mg or 0.471 g
% K2O = 0.471 g x 100 = 23.55%
2.00 g
Example: On ignition, rochelle salt, KNaC4H4O6.4H2O, is converted to KNaCO3. A sample of

the original salt weighs 0.9546 g, and the ignition product is titrated with H2SO4. From the ff
data find the % purity of the sample: H2SO4 used = 41.72 mL; 10.27 mL H2SO4 ≈ 10.35 mL
NaOH; the NaOH is 0.1297N; NaOH used for back titration = 1.91 mL
me KNaC4H4O6.4H2O = me H2SO4 – me NaOH(back titration)

me H2SO4 = me NaOH
10.27mL (NH2SO4) = 10.35 mL (0.1297 N)
NH2SO4 = 0.1307 N
wt KNaC4H4O6.4H2O x 2 me/mmol
282.19 mg/mmol
= 0.1307N(41.72 mL) – 0.1297N(1.91 mL)
wt KNaC4H4O6.4H2O = 734.41 mg = 0.73441 g
% purity = (0.73441/ 0.9546) = 76.93%

Example: What weight of soda ash should be taken for analysis such that the percentage
of Na2O present can be found by multiplying by 2 the number of milliliters of 0.200 N acid
solution used in titration?
factora
% component (a) = weighta x 100
weight of sample
factora wt sample
% Na2O = 2 x (Vacid)
2 x (Vacid) = 62 mg/mmol x (0.200 me/ml)(Va) x 1 .
2 me/mmol wt sample
weight sample = 0.3100 g
INDIRECT VOLUMETRIC METHOD

- calculation is made by determining the total number of milliequivalents of reagent
added and subtracting the number of milliequivalents used in titration – the difference in
the number is the number of milliequivalents of desired substance
substance reagent
determined added
+  products
A A
B
Titrated
amount
Example: A sample of meat scrap weighing 2.000 g is digested with concentrated H 2SO4
and a catalyst. The resulting solution is made alkaline with NaOH and the liberated
ammonia distilled into a 50.0 mL of 0.6700 N H 2SO4. The excess then requires 30.10 mL of
0.6520 N NaOH for neutralization. What is the percentage of nitrogen in the meat?
Example: A sample of meat scrap weighing 2.000 g is digested with concentrated H 2SO4
and a catalyst. The resulting solution is made alkaline with NaOH and the liberated
ammonia distilled into a 50.0 mL of 0.6700 N H 2SO4. The excess then requires 30.10 mL of
0.6520 N NaOH for neutralization. What is the percentage of nitrogen in the meat?
Given:
Wt sample = 2.000 g V NaOH = 30.10 mL
V H2SO4 = 50.0 mL N NaOH = 0.6520 N
N H2SO4= 0.6700 N
Reqd: %N

Soln:
Kjeldahl method – determination of N in organic matl
meq H2SO4 = 50.0 mL x 0.6700 meq/mL = 33.50 me

meq NaOH = 30.10 mL x 0.6520 meq/mL = 19.62 me
Difference in meq = 13.88 me

- this is the meq of H2SO4 that reacted with NH3 in the meat
weight a x factora = Nb V b
MWa
meq NH3 = meq H2SO4

weightNH3 x 1 me/mmol = 13.88 me
17 mg/mmol MW of NH3
At wt N
weight NH3 = 235.96 mg/1000mg/g = 0.23596 g
weight N = 0.23596 g NH3 x 1mol NH3 x 1mol N x 14gN = 0.19432 g
14g NH3 1mol NH3 1mol N
Or meq N = meq NH3 = meq H2SO4

weight N x 1 me/mmol = 13.88 me
14 me/mmol
wt N = 0.19432 g
%N = 0.19432 g x 100 = 9.716%

2.000 g
Example: When a direct current is passed through a solution of NaCl, using metallic Hg as a
cathode, a compound having the formula NaHg2 is formed as an amalgam in the Hg. It is
used as a powerful reducing agent. A sample of the material weighing 5.00 g is added to
water and after the evolution of H2 ceases, the resulting solution requires 40.75 mL of 0.1067
N HCl for titration. a) Write an equation for the action of the amalgam and b) calculate
the percentage of Na in the sample.
a) 2NaHg2 + 2H2O  2NaOH + 4Hg + H2
b) me Na = me NaHg2 = me HCl
weight Na x 2 me/mmol = 40.75 mL (0.1067 N)
2(23) mg/mmol
weight Na = 100 mg = 0.0100 g
% Na = (0.010 g /5.00 g) x 100 = 2.00%

Assignment:
16- If all the N in 10.0 mmols urea, CO(NH2)2, is converted to NH4HSO4, and if with excess
NaOH the NH3 is evolved and caught in 50.0 mL of HCl (1.00 mL ≈ 0.03000 g CaCO 3), what
volume of NaOH (1.00 mL ≈ 0.3465 g H2C2O4.2H2O) would be required for complete
titration?
17- The percentage of protein in meat products is determined by multiplying the %N as
determined by the kjeldahl method by the arbitrary factor 6.25. A sample of processed
meat scrap weighing 2.000 g is digested with concd H 2SO4 and Hg (catalyst) until the N
present has been converted to NH4HSO4. This is treated with excess NaOH, and the
liberated NH3 is caught in 50.0 mL pipetful H2SO4 (1.000 mL ≈ 0.01860 g Na2O). The excess
acid requires 28.80 mL NaOH (1.000 mL ≈ 0.1266 g KHP). Calculate % protein in the meat
scrap.
18- A 2.00 g sample of steel is burned in O2, and the evolved CO2 after passing through
appropriate purifying trains is caught in 100 mL Ba(OH)2 solution. The supernatant liquid
requires 75.0 mL HCl (1.00 mL ≈ 0.00626 g Na 2CO3 , 1.00 mL ≈ 1.12 ml Ba(OH)2 ). What is the
% Carbon in the steel?
Assignment:
19- A sample of limestone is titrated for its value as a neutralizing agent. A sample weighing
1.000 g is taken. What must be the normality of the titrating acid so that every 10 ml will
represent 4 ½ % of the neutralizing value expressed in terms of percentage of CaO.
20- What weight of soda ash must be taken for analysis so that by using 0.5000N HCl for
titrating, (a) buret reading will equal the percentage of Na 2O, (b) three times the buret
reading will equal the percentage of Na 2O, (c) every 3 mL will represent 1% Na2O (d) each
mL will represent 3% Na2O (e) the buret reading and the percentage of Na2O will be in the
respective ratio 2:3?
Example: When a direct current is passed through a solution of NaCl, using metallic Hg as a
cathode, a compound having the formula NaHg2 is formed as an amalgam in the Hg. It is
used as a powerful reducing agent. A sample of the material weighing 5.00 g is added to
water and after the evolution of H2 ceases, the resulting solution requires 40.75 mL of 0.1067
N HCl for titration. a) Write an equation for the action of the amalgam and b) calculate
the percentage of Na in the sample.
c) 2NaHg2 + 2H2O  2NaOH + 4Hg + H2
d) me Na = me NaHg2 = me HCl
weight Na x 2 me/mmol = 40.75 mL (0.1067 N)
2(23) mg/mmol
weight Na = 100 mg = 0.0100 g
% Na = (0.010 g /5.00 g) x 100 = 2.00%
Assignment:
16- If all the N in 10.0 mmols urea, CO(NH2)2, is converted to NH4HSO4, and if with excess
NaOH the NH3 is evolved and caught in 50.0 mL of HCl (1.00 mL ≈ 0.03000 g CaCO 3), what
volume of NaOH (1.00 mL ≈ 0.3465 g H2C2O4.2H2O) would be required for complete
titration?

17- The percentage of protein in meat products is determined by multiplying the %N as
determined by the kjeldahl method by the arbitrary factor 6.25. A sample of processed
meat scrap weighing 2.000 g is digested with concd H 2SO4 and Hg (catalyst) until the N
present has been converted to NH4HSO4. This is treated with excess NaOH, and the
liberated NH3 is caught in 50.0 mL pipetful H2SO4 (1.000 mL ≈ 0.01860 g Na2O). The excess
acid requires 28.80 mL NaOH (1.000 mL ≈ 0.1266 g KHP). Calculate % protein in the meat
scrap.
18- A 2.00 g sample of steel is burned in O2, and the evolved CO2 after passing through
appropriate purifying trains is caught in 100 mL Ba(OH)2 solution. The supernatant liquid
requires 75.0 mL HCl (1.00 mL ≈ 0.00626 g Na 2CO3 , 1.00 mL ≈ 1.12 ml Ba(OH)2 ). What is the
% Carbon in the steel?
Assignment:
19- A sample of limestone is titrated for its value as a neutralizing agent. A sample weighing
1.000 g is taken. What must be the normality of the titrating acid so that every 10 ml will
represent 4 ½ % of the neutralizing value expressed in terms of percentage of CaO.
20- What weight of soda ash must be taken for analysis so that by using 0.5000N HCl for
titrating, (a) buret reading will equal the percentage of Na 2O, (b) three times the buret
reading will equal the percentage of Na 2O, (c) every 3 mL will represent 1% Na2O (d) each
mL will represent 3% Na2O (e) the buret reading and the percentage of Na2O will be in the
respective ratio 2:3?
DOUBLE INDICATOR TITRATIONS
- The fact that certain indicators change color at different stages of a

neutralization is sometimes made use of in volumetric work to determine the
proportions of the components of certain mixtures by observing two endpoints in
a single titration.
 A practical application of the principle of a double indicator titration is in the

analysis of mixtures containing Na2CO3 and NaOH or Na2CO3 and NaHCO3.
The change of color of phenolphthalein is not sharp at the point of formation
of bicarbonate; hence, the precision of these double indicator titrations is not
great as in most alkalimetric titrations, but by keeping the solution ice-cold,
three significant accuracy can be obtained. The use of creso red plus thymol

blue in place of phenolphthalein, together with the use of bromphenol blue
in place of methyl orange gives greater precision.
Case 1: NaOH + Na2CO3 + inert matter

- Consider the case of a mixture containing Na2CO3 and NaOH ( and inert
matter). The titration with standard acid , using phenolphthalein , would give a
color change of pink to colorless only when the NaOH had been neutralized and
the Na2CO3 has been half-neutralized(i.e., converted to NaHCO3). Methyl
orange would give a yellow color to the solution, and the additional volume of
acid required to change the color to pink would be that required to complete
the reaction with NaHCO3
- Additional volume of acid required for methyl orange end point is less than
volume required for phenolphthalein end point
NaOH
HCl A, mL A + B mL - phenolphthalein endpt

changes color(pink to colorless)
NaCl + H2O Na2CO3
* NaOH- neutralized
HCl B, mL Na2CO3 – half neutralized
NaHCO3
HCl B, mL
Methyl orange end pt
CO2 + NaCl (yellow color to pink) to complete
reaction
Case 2: Na2CO3 + NaHCO3 + inert matter

- The titration with standard acid, using phenolphthalein , would produce a color
change of pink to colorless when the Na2CO3 had been converted to NaHCO3.
Methyl orange would give a yellow color to the solution, and the additional
volume of acid required to change the color of the methyl orange would be
that required to complete the reaction with the NaHCO 3 formed from the
Na2CO3, plus that required to react with the original NaHCO 3.
- _ Additional volume of acid required for methyl orange end point is greater
than volume required for phenolphthalein end point

Na2CO3 - phenolphthalein endpt
HCl A, mL when Na2CO3 is converted to NaHCO3
NaHCO3
A, mL A+B * A = original vol

B = addtl vol of acid
CO2 NaHCO3
HCl B, mL

reaction
ex. A sample consisting of Na2CO3, NaOH and inert matter weighs 1.179 g. It is titrated with
0.300 N HCl with phenolphthalein as the indicator, and the solution becomes colorless after
the addition of 48.16 mL. methyl orange is then added and 24.08 mL more of the acid are
needed for the color change. What is the percentage NaOH and Na 2CO3?
NaOH
HCl 48.16 mL -24.08 mL - phenolphthalein endpt

NaCl + H2O Na2CO3 48.16
* NaOH- neutralized
HCl 24.08 mL Na2CO3 – half neutralized
NaHCO3
HCl 24.08 mL
reaction
Soln:
 If the acid is added slowly, the stronger base (NaOH) is neutralized first. After
this reaction is complete, the Na2CO3 is converted to NaHCO3, at which the
phenolphthalein changes color. All this requires 30.00 mL of the acid. Since
the further reaction with the bicarbonate formed requires 5.00 mL of the acid,
of the total 35.00 mL used, 10.00 mL must have reacted with Na2CO3 , and
therefore, 25. 00 mL with the NaOH

𝑚𝑒𝑞 𝑁𝑎𝑂𝐻 = 𝑚𝑒𝑞 𝐻𝐶𝑙
𝑊𝑁𝑎𝑂𝐻
= 𝑁𝐻𝐶𝑙 × 𝑉𝐻𝐶𝑙
𝑀𝑊
𝑓 × 1000
𝑵𝑯𝑪𝒍 ×𝑽𝑯𝑪𝒍 × 𝑴𝑾𝑵𝒂𝑶𝑯

𝑾𝑵𝒂𝑶𝑯 =
𝒇𝑵𝒂𝑶𝑯 × 𝟏𝟎𝟎𝟎
𝒈
(𝟎. 𝟑 𝑵)(𝟐𝟒. 𝟎𝟖 𝒎𝑳) (𝟒𝟎 )
𝑾𝑵𝒂𝑶𝑯 = 𝒎𝒐𝒍
𝒆𝒒
𝟏 × 𝟏𝟎𝟎𝟎
𝒎𝒐𝒍
𝑾𝑵𝒂𝑶𝑯 = 𝟎. 𝟐𝟖𝟗 𝒈
𝑾𝑵𝒂𝑶𝑯
% 𝑵𝒂𝑶𝑯 = × 𝟏𝟎𝟎
𝑾𝒔𝒂𝒎𝒑𝒍𝒆
𝟎. 𝟐𝟖𝟗 𝒈
% 𝑵𝒂𝑶𝑯 = × 𝟏𝟎𝟎 = 𝟐𝟒. 𝟓𝟏%
𝟏. 𝟏𝟕𝟗 𝒈
𝑚𝑒𝑞 𝑁𝑎2 𝐶𝑂3 = 𝑚𝑒𝑞 𝐻𝐶𝑙
𝑊𝑁𝑎2 𝐶𝑂3
𝑀𝑊
𝑓 × 1000
𝑵𝑯𝑪𝒍 × 𝑽𝑯𝑪𝒍 × 𝑴𝑾𝑵𝒂𝟐 𝑪𝑶𝟑

𝑾𝑵𝒂𝟐 𝑪𝑶𝟑 =
𝒇𝑵𝒂𝟐 𝑪𝑶𝟑 × 𝟏𝟎𝟎𝟎
𝒈
(𝟎. 𝟑 𝑵)(𝟒𝟖. 𝟏𝟔. 𝒎𝑳) (𝟏𝟎𝟓. 𝟗𝟗 )
𝑾𝑵𝒂𝟐 𝑪𝑶𝟑 = 𝒎𝒐𝒍
𝒆𝒒
𝟐 × 𝟏𝟎𝟎𝟎
𝒎𝒐𝒍
𝑾𝑵𝒂𝟐 𝑪𝑶𝟑 = 𝟎. 𝟕𝟔𝟓𝟕 𝒈
𝑾𝑵𝒂𝟐 𝑪𝑶𝟑
% 𝑵𝒂𝟐 𝑪𝑶𝟑 = × 𝟏𝟎𝟎
𝟎. 𝟕𝟔𝟓𝟕 𝒈
% 𝑵𝒂𝟐 𝑪𝑶𝟑 = × 𝟏𝟎𝟎 = 𝟔𝟒. 𝟗𝟒%
𝟏. 𝟏𝟕𝟗 𝒈

Ex.2 A 1.200 g sample containing Na2CO3 and NaHCO3 and inert impurities is titrated cold
with 0.500N HCl with phenolphthalein as the indicator the solution turns colorless after the
addition of 15.0 mL of the acid . Methyl orange is then added and 22.mL more of the acid
are required to change the color of this indicator. What is the percentage Na 2CO3 and
NaHCO3 ?
Given:
Na2CO3 - phenolphthalein endpt
HCl 15, mL when Na2CO3 is converted to NaHCO3
NaHCO3
15, mL
22 mL
CO2 NaHCO3
HCl 22-15 mL

reaction
Reqd; a. % Na2CO3
b. % NaHCO3
Soln:
 Na2CO3 is first converted to NaHCO3, requiring 15.00 mL of the acid. Of the
22.0 mL of additional acid added, 15.00 mL must have been required to
complete the reaction with this NaHCO3 formed and 7.00mL to react with the
NaHCO3 originally present
a. 𝑚𝑒𝑞 𝑁𝑎2 𝐶𝑂3 = 𝑚𝑒𝑞 𝐻𝐶𝑙
𝑊𝑁𝑎2 𝐶𝑂3
𝑀𝑊
𝑓 × 1000
𝑵𝑯𝑪𝒍 × 𝑽𝑯𝑪𝒍 ×𝑴𝑾𝑵𝒂𝟐 𝑪𝑶𝟑

𝑾𝑵𝒂𝟐 𝑪𝑶𝟑 =
𝒇𝑵𝒂𝟐 𝑪𝑶𝟑 × 𝟏𝟎𝟎𝟎
𝒈
(𝟎. 𝟓 𝑵)(𝟏𝟓 + 𝟏𝟓 𝒎𝑳) (𝟏𝟎𝟓. 𝟗𝟗 )
𝑾𝑵𝒂𝟐 𝑪𝑶𝟑 = 𝒎𝒐𝒍
𝒆𝒒
𝟐 × 𝟏𝟎𝟎𝟎
𝒎𝒐𝒍
𝑾𝑵𝒂𝟐 𝑪𝑶𝟑 = 𝟎. 𝟕𝟗𝟒𝟗 𝒈

𝑾𝑵𝒂𝟐 𝑪𝑶𝟑
% 𝑵𝒂𝟐 𝑪𝑶𝟑 = × 𝟏𝟎𝟎
𝟎. 𝟕𝟗𝟒𝟗 𝒈
% 𝑵𝒂𝟐 𝑪𝑶𝟑 = × 𝟏𝟎𝟎 = 𝟔𝟔. 𝟐𝟒𝟏𝟕%
𝟏. 𝟐 𝒈
b.
𝑚𝑒𝑞 𝑁𝑎𝐻𝐶𝑂3 = 𝑚𝑒𝑞 𝐻𝐶𝑙
𝑊𝑁𝑎𝐻𝐶𝑂3
𝑀𝑊
𝑓 × 1000
𝑵𝑯𝑪𝒍 × 𝑽𝑯𝑪𝒍 ×𝑴𝑾𝑵𝒂𝑯𝑪𝑶𝟑

𝑾𝑵𝒂𝑯𝑪𝑶𝟑 =
𝒇𝑵𝒂𝑶𝑯 × 𝟏𝟎𝟎𝟎
𝒈
(𝟎. 𝟓𝑵)(𝟐𝟐 − 𝟏𝟓 𝒎𝑳) (𝟖𝟒. 𝟎𝟏 )
𝑾𝑵𝒂𝑯𝑪𝑶𝟑 = 𝒎𝒐𝒍
𝒆𝒒
𝟏 × 𝟏𝟎𝟎𝟎
𝒎𝒐𝒍
𝑾𝑵𝒂𝑯𝑪𝑶𝟑 = 𝟎. 𝟐𝟗𝟒𝟎 𝒈
𝑾𝑵𝒂𝑯𝑪𝑶𝟑
% 𝑵𝒂𝑯𝑪𝑶𝟑 = × 𝟏𝟎𝟎
𝟎. 𝟐𝟗𝟒𝟎 𝒈
% 𝑵𝒂𝑯𝑪𝑶 𝟑 = × 𝟏𝟎𝟎 = 𝟐𝟒. 𝟓𝟎𝟐𝟗 %
𝟏. 𝟐 𝒈
Determination of the Proportion in which Components are present in a mixture:
Example: If 0.5000 g of a mixture of CaCO3 and BaCO3 requires 30.00 mL of a 0.2500N HCl
for neutralization, What is the percentage of each component?
let x = g CaCO3 and y = g BaCO3

x + y = 0.5000  equation1
eq of CaCO3 present = x = x .
CaCO3/2 50.04
eq of BaCO3 present = y = y .
BaCO3/2 98.69

eq CaCO3 + eq BaCO3 = eq HCl
x + y = 30.0 mL (0.250eq/L)
50.04 98.69 (1000mL/L)
 equation2
Solving equations 1 and 2 simultaneously:
x = 0.247 g
y = 0.253 g
% CaCO3 = (0.247 / 0.500) x 100 = 49.4%
% BaCO3 = (0.253 / 0.500) x 100 = 50.6%
Example: The weight of combined LiOH, KOH and Ba(OH)2 in a mixture is 0.5000 g and
requires 25.43 mL of 0.500N acid for neutralization. The same weight of sample with CO 2
gives a precipitate of BaCO3 that requires 5.27 mL of the above acid for neutralization. Find
the weights of LiOH, KOH and Ba(OH)2 in the original mixture.
Let x = g LiOH and y = g KOH and z = g Ba(OH)2
x + y + z = 0.5000 1
eq LiOH = x eq Ba(OH)2 = z .
23.95/1 171.36/2
eq KOH = y .
56.11/1
eq LiOH + eq KOH + eq Ba(OH)2 = 25.43mL(0.5eq/L)
1000mL/L
 equation 2
eq Ba(OH)2 = eq BaCO3 = 5.27 mL (0.5 eq/L)
1000mL/L
wt Ba(OH)2 x 2 eq/mol = 2.635 x 10-3 eq
171.36 g/mol
wt Ba(OH)2 = 0.2258 g
x + y + 0.2258 = 0.5000 1
x + y = 0.2742
eq LiOH + eq KOH + eq Ba(OH)2 = 25.43mL(0.5eq/L)
1000mL/L
 equation 2
eq LiOH + eq KOH + 2.635 x 10-3 eq = 0.012715 eq
x + y = 0.01008
23.95/1 56.11/1
Solving equations 1 and 2 simultaneously:
x = 0.217 g
y = 0.0572 g
Assignment:
21- A mixture consisting entirely of Li2CO3 + BaCO3 weighs 1.000 g and requires 15.00 mL of
1.000 N HCl for neutralization. Find the percentage of BaCO 3 and of combined Li in the
sample.
22- A mixture of pure BaCO3 and pure Na2CO3 weighs 1.000 g and has the total neutralizing
power of 15.37 meq of CaCO3. Calculate the percentage of combined CO2 in the mixture
and the weight of Li2CO3 that has the same neutralizing power as 1.000 g of the above
mixture.

MODULE 6: VOLUMETRIC METHODS OF ANALYSIS:UNIT 3: REDOX TITRATION
QUANTITATIVE ANALYSIS -
REDOX
•Volumetric Analysis
Divisions:
1- Neutralization methods – acidimetry (acid determination) and alkalimetry (alkali
determination)
2- Oxidation and Reduction – “redox” – oxidimetry and reductimetry
3- Precipitation methods – precipitimetry
4- Complex ion formation methods - compleximetry
Volumetric Analysis is more commonly as

TITRATION
REDOX TITRATION
Redox titration is a division of volumetric analysis where the reactants are oxidizing
reducing agents
Oxidation - refers to a reaction in which an element increases in oxidation state due to
loss of electrons
– loss of electrons
- oxidation state tends to move to the positive side
oxidation numbers
– 8 … – 3 – 2 – 1 0 +1 +2 + 3 … + 8
   
substances that undergo these type of change in oxidation state are called reducing
agents
example: Mn+2  Mn+7
Mn+2 is a reducing agent and is capable of being oxidized to Mn +7

Reduction - refers to a reaction in which an element decreases in oxidation state due to
gain of electrons
– gain of electrons
- oxidation state tends to move to the negative side
oxidation numbers
– 8 … – 3 – 2 – 1 0 +1 +2 + 3 … + 8
   
substances that undergo these type of change in oxidation state are called oxidizing
agents
example: S+6  S–2

S+6 is an oxidizing agent and is capable of being reduced to S–2
Example: H+ + e-  H0
   
H is an oxidizing agent and is capable of being reduced to hydrogen
+
Zn + 2H+  Zn2+ + H2
   
free Hydrogen is a reducing agent and is capable of being oxidized to hydrogen
ion
Zinc ion is an oxidizing agent and is capable of being reduced to a zinc metal
Oxidizing agent - substance that decrease in oxidation number

- substance responsible for oxidation
- particles which accepts electrons
Reducing agent - substance that increase in oxidation number

- substance responsible for reduction
- Particles which donates electrons
Solutions used in titrations:

Oxidizing agents Reducing agents
K2Cr2O7 FeSO4
KMnO4 FeSO4.(NH4)2SO4.6H2O
K3Fe(CN)6 H2C2O4
I2 Na2C2O4
Ce(SO4)2 Na2S2O3
KIO3 SnCl2
KBrO3 Na3AsO3
Important Combinations of Oxidizing Agent and Reducing Agent in Redox Titrations:
1. K2Cr2O7 and ferrous solutions

2. KMnO4 and ferrous solutions
3. KMnO4 and Na2C2O4
4. I 2 and Na2S2O3
5. Hypochlorites (ex. HClO4) and arsenious acid
OXIDATION - REDUCTION EQUIVALENTS

g-equiv wt of O.A. AND R.A. = FW / tot. change in O.S of its constant element
= FW / f
Ex. Mn +7 → Mn +2
g-equiv wt = KMnO4
5
How many grams of the following reducing substances constitute the gram equivalent
weight in each case:
(a) FeSO4.7 H2O
(b) SnCl2
(c) H2C2O4.2H2O (oxalic acid)
(d) KHC2O4.H2O (potassium binoxalate)
(e) KHC2O4.H2C2O4.2H2O (potassium tetroxalate)
(f) H2S (oxidized to S)
(g) H2S (oxidized to H2SO4)
(h) Na2S2O3.5H2O (oxidized to Na2S4O6)
(i) H2O2
reducing substances – oxidation process takes place

(a) FeSO4.7 H2O
Fe2+ (+2)  Fe3+ (+3) + e- f=1
eq wt = FeSO4.7 H2O g/mol = 278 grams/eq
1 eq/mol
(b) SnCl2
Sn2+ (+2)  Sn4+ (+4) + 2e- f=2
eq wt = SnCl2 g/mol = 94.8 grams/eq
2 eq/mol
(c) H2C2O4.2H2O
C2O42- (C+3)  2CO2 (C+4) + 2e- f=2
eq wt = H2C2O4.2H2O g/mol = 63.03 grams/eq
2 eq/mol
(d) KHC2O4.H2O
C2O42- (C+3)  2CO2 (C+4) + 2e- f=2
eq wt = KHC2O4.2H2O g/mol = 73.07 grams/eq
2 eq/mol
(e) KHC2O4.H2C2O4.2H2O
2C2O4-2  4CO2 + 4e- f=4
eq wt = KHC2O4.H2C2O4.2H2O g/mol = 63.55 g/eq
4 eq/mol

(f) H2S (oxidized to S)
S2- (–2)  S0 (0) + 2e- f=2
eq wt = H2S g/mol = 17.04 grams/eq
2 eq/mol
(g) H2S (oxidized to H2SO4)
S2- (–2)  S6+ (+6) + 8e- f=8
eq wt = H2S g/mol = 4.26 grams/eq
8 eq/mol
(h) Na2S2O3.5H2O (oxidized to Na2S4O6)
2S2O32- (+ 2)  S4O62- (+2.5) + 1e- f=1
eq wt = Na2S2O3.5H2O g/mol = 248.2 g/eq
1 eq/mol
(i) H2O2 (as a reducing agent, oxidizes to free O)
H2O2 (–1)  O20 (0) + 2e- f=2
eq wt = H2O2 g/mol = 17.01 g/eq
2 eq/mol
How many grams of the following oxidizing substances constitute the gram equivalent
weight in each case:
(a) K3Fe(CN)6 (d) I2
(b) KMnO4 (e) K2Cr2O7
(c) KBrO3 (reduced to bromide) (f) H2O2
oxidizing substances – reduction process takes place

(a) K3Fe(CN)6
Fe(CN)63- (Fe+3) + 1e-  Fe(CN)64- (Fe+2) f = 1
eq wt = K3Fe(CN)6 g/mol = 0.3293 g/eq
1 eq/mol
(b) KMnO4
** when reduced in the presence of an acid, permanganate forms manganous ions
MnO4–1 (+7) + 5e-  Mn2+ (+2) f = 5

eq wt = KMnO4 g/mol = 31.61 g/eq
5 eq/mol
** in alkaline solution, permanganate is reduced to MnO2

MnO4–1 (+7) + 3e-  Mn2+ (+4) f=3
eq wt = KMnO4 g/mol = 52.68 g/eq
3 eq/mol
(c) KBrO3 (reduced to bromide)

KBrO3 (+5) + 6e-  Br- (–1) f=6
eq wt = KBrO3 g/mol = 27.84 g/eq
6 eq/mol
(d) I2
I2 (0) + 2e-  2I- (–1) f=2

eq wt = I2 g/mol = 126.9 g/eq
2 eq/mol
(e) K2Cr2O7
Cr2O72- (+6) + 6e-  2Cr3+ (+3) f=6
eq wt = K2Cr2O7 g/mol = 49.03 g/eq
6 eq/mol
(f) H2O2 (as an oxidizing agent, reduce to H2O)

H2O2 (–1) + 2e-  H2O (–2) f=2
eq wt = H2O2 g/mol = 17.01 g/eq
2 eq/mol
ASSIGNMENT
Exercises:
1. How to determine the factor of reducing agent and oxidizing agent:
a. Fe + HCl → FeCl3 + H2
b. HNO3 + H2S → S + NO + H2O
c. KMnO4 + LiCl + H2SO4 → Cl2 + MnSO4 + K2SO4 +Li2SO4 +H2O
d. K2Cr2O7 + KI + H3PO4 → I2 + CrPO4 + K3PO4 + H2O
e. K2Cr2O7 + FeSO4 + H2SO4 → Cr2(SO4)3 + Fe2(SO4)3 + K2SO4 + H2O
f. MnO4 - + Fe +2 + H+ → Mn++ + Fe3+ + H2O
g. MnO4 + C2O4 + H
- -2 + → Mn ++ + CO2 + H2O
2. What fraction of the formula weight of each of the ff compounds represents the
equivalent weight in a redox process in which the product formed is as indicated:
(a) Ce(SO4)2.2(NH4)2SO4.2H2O ( Ce3+)
(b) As2O5 ( As3+)
(c) KIO3 ( ICl32-)
(d) Na2SeO4 ( SeO32-)
(e) VOSO4 ( VO3-)
Mo2O3 ( H2MoO4
REDOX:OXIDATION- REDUCTION
OXIDATION - REDUCTION EQUIVALENTS
g-equiv wt of O.A. AND R.A.

= FW or MW / tot. change in O.S of its constant element
𝐹𝑊
𝑔 − 𝑒𝑞𝑢𝑖𝑣 𝑤𝑡 = 𝑡𝑜𝑡 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑂.𝑆
Where: FW = formula weight

MW = molecular weight
O.S = oxidation state/number
Ex. Mn +7 → Mn +2
g-equiv wt = KMnO4
5
Fe +2 → Fe +3
f =1
g-equiv wt = Fe
5
𝑵𝑹.𝑨 𝑵𝒂𝒄𝒊𝒅
=
𝒇𝑹.𝑨 𝒇𝒂𝒄𝒊𝒅
𝑵𝑶.𝑨 𝑵𝒂𝒄𝒊𝒅
=
𝒇𝑶.𝑨 𝒇𝒂𝒄𝒊𝒅
Example: How much water must be added to 50.0 mL of a solution of HNO 3which is 2N as
an acid to make the resulting solution 2.0N as an oxidizing agent? Assume
reduction of the HNO3 to NO
HNO3 to NO
𝑓𝐻𝑁𝑂3 = |5 − 2| = 3 × 1 = 3
=
𝒇𝑶.𝑨 × 𝑵𝒂𝒄𝒊𝒅
𝑵𝑶𝑨 =
𝒇𝒂𝒄𝒊𝒅
𝒆𝒒
𝟑 𝒎𝒐𝒍 × 𝟐𝑵
𝑵𝑶𝑨 = 𝒆𝒒 = 𝟔𝑵 𝒂𝒔 𝑶. 𝑨
𝟏𝒎𝒐𝒍
𝑉1 𝐶1 = 𝑉2 𝐶2
As Oxidizing agent ,O.A

𝑉1 = 50 𝑚𝐿 V2= ?
C1 = 6N C2 = 2N
50mL(6N) = V2 ( 2N)
V2 = 150 mL
𝑉𝐻2𝑂 = 150 − 50 = 100 𝑚𝐿

Example: From the following data , find the ratio of the Normality of the HNO 3 as an
oxidizing agent, to the normality of the tetroxalate as a reducing agent.
1.00 mL HNO3 ≎ 1.246 mL NaOH
1.00 mL NaOH ≎ 1.743 mL KHC2O4.H2 C2O4.2H2O
The NaOH is 0.1200 N ( oxalate oxidized to CO2; nitrate is reduced to NO)
Given:
1.00 mL HNO3 ≎ 1.246 mL NaOH
1.00 mL NaOH ≎ 1.743 mL KHC2O4.H2 C2O4.2H2O
N NaOH = 0.1200 N
Reqd:
Ratio of N HNO3 as O.A to N Ktet as R.A
Soln:
[𝑂]
𝐶2 𝑂4−2 → 𝐶𝑂2−2 𝑓 = 4 𝑎𝑙𝑤𝑎𝑦𝑠 𝑅. 𝐴
[𝑅]
𝑁𝑂3−1 → 𝑁𝑂 𝑓 = 3
𝑚𝑒𝑞𝐻𝑁𝑂3 𝑎𝑠 𝑎𝑐𝑖𝑑 = 𝑚𝑒𝑞𝑁𝑎𝑂𝐻 𝑎𝑠 𝑏𝑎𝑠𝑒

𝑁𝑎 𝑉𝑎 = 𝑁𝑏 𝑉𝑏
(0.1200𝑁) (1.246 𝑚𝐿)
𝑁𝐻𝑁𝑂3 = = 0.1495 𝑁 𝑎𝑠 𝑎𝑛 𝑎𝑐𝑖𝑑
1𝑚𝐿
N as O.A
=
𝒇𝑶.𝑨 × 𝑵𝒂𝒄𝒊𝒅
𝑵𝑶𝑨 = 𝒇𝒂𝒄𝒊𝒅
𝒆𝒒
𝟑 𝒎𝒐𝒍 × 𝟎.𝟏𝟒𝟗𝟓𝑵
𝑵𝑶𝑨 = 𝒆𝒒 = 𝟎. 𝟒𝟒𝟖𝟓𝑵 𝒂𝒔 𝑶. 𝑨
𝟏
𝒎𝒐𝒍
𝑚𝑒𝑞𝑁𝑎𝑂𝐻 𝑎𝑠 𝑏𝑎𝑠𝑒 = 𝑚𝑒𝑞𝐾𝑡𝑒𝑡 𝑎𝑠 𝑎𝑐𝑖𝑑
 Note: base is always base
𝑁𝑏 𝑉𝑏 = 𝑁𝑎 𝑉𝑎
(0.1200𝑁) (1.0 𝑚𝐿)
𝑁𝐾𝑡𝑒𝑡 = = 0.0688 𝑁 𝑎𝑠 𝑎𝑛 𝑎𝑐𝑖𝑑
1.743 𝑚𝐿
𝑵𝑹.𝑨 𝑵𝒂𝒄𝒊𝒅
=
𝒇𝑹.𝑨 𝒇𝒂𝒄𝒊𝒅
𝒇𝑹.𝑨 × 𝑵𝒂𝒄𝒊𝒅
𝑵𝑹𝑨 = 𝒇𝒂𝒄𝒊𝒅

𝒆𝒒
𝟒 𝒎𝒐𝒍 × 𝟎.𝟎𝟔𝟖𝟖𝑵
𝑵𝑹𝑨 = 𝒆𝒒 = 𝟎. 𝟎𝟗𝟏𝟖 𝒂𝒔 𝑹. 𝑨
𝟑𝒎𝒐𝒍
𝑵𝑯𝑵𝑶𝟑 𝒂𝒔 𝑶.𝑨 𝟎. 𝟒𝟒𝟖𝟓 𝑵

𝑹𝒂𝒕𝒊𝒐 = = = 𝟒. 𝟖𝟖𝟔
𝑵𝑲𝒕𝒆𝒕 𝒂𝒔 𝑹.𝑨 𝟎. 𝟎𝟗𝟏𝟖 𝑵
Example: a ferrous solution contains 1.176 g of FeSO4.(NH4)2 SO4.6H2O in 30.00 ml; a

dichromate solution contains 0.2940 g of K2Cr2O7 in 20.00 mL. Find (a) the normality of the
ferrous solution as a reducing agent, (b) the normality of the dichromate solution as an
oxidizing agent, (c) the volume of the dichromate equivalent to 1.000 mL of the ferrous
soln.
NFeSO4.(NH4)2SO4.6H2O = (1.176 g)1 eq/mol(1000 mL/L)
(392.14 g/mol)(30.00 mL)
= 0.09996 N
NK2Cr2O7 = (0.2940 g)6 eq/mol(1000 mL/L) = 0.2998 N
(294.19 g/mol)(20.00 mL)
eqFeSO4.(NH4)2SO4.6H2O = eqK2Cr2O7
(0.09996N)(1.000 mL) = (0.2998N)(V K2Cr2O7)
VK2Cr2O7 = 0.3334 mL
ASSIGNMENT:
3. How many grams per milliliter does a soln of KNO2 contain if it is 0.100 N as a
reducing agent?How many grams of SO2 is contained in a liter of a solution of
H2SO3 which is 0.05860 N as a reducing agent?
4. What is the normality of a nitric acid solution to be used as an oxidizing agent
(reduced to NO) if it contains 55.50% by weight of HNO 3and has a specific
gravity of 1.350
5. In the reaction expressed by the equation: 13Pb 3O4(s) + 2Mn3O4(s) + 72H + 
6MnO4- + 39Pb2+ + 36H2O (a) What is the numerical value of the equivalent
weight of Pb3O4 as an oxidizing agent (b) the milliequivalent weight of Mn 3O4
as a reducing agent, and (c) the volume of 0.1500 N FeSO 4 solution required
to titrate the permanganate formed from 0.2000 mmol of Mn 3O4?
PERMANGANATE PROCESS
KMnO4 – used extensively as an oxidimetric standard – serves as its own indicator

3 ways to use KMnO4 solution:
1- it is used in the presence of acid in the direct titration of a number or oxidizable cations
and anions:
Substance Oxidized to Substance Oxidized to
Fe 2+ Fe 3+ H2O2 O2
Sn2+ Sn4+ Mo3+ MoO32–
VO2+ VO3– Ti3+ TiO2+

C2O42– CO2 U4+ UO22+
NO2 – NO3 – As 3+ AsO43–
SO3 2– SO4 2–
2- it is used in the presence of acid in the direct titration of a number of reducible

substances ( a measured amount of
reducing agent (ferrous salt or oxalate) is added and after reduction is complete, the
excess reducing agent is titrated with standard permanganate)
Substance Reduced to
MnO4 – Mn2+
Cr2O72 – Cr3+
MnO2, Mn3O4 Mn2+
PbO2, Pb2O3, Pb3O4 Pb2+
Ce 4+ Ce 3+
3- it is used in neutral or alkaline solution in the titration of few substances. In these case
permanganate is reduced to MnO2, w/c precipitates. The MnO4 – has an
oxidizing power 3/5 of what it has in the presence of acid
Substance Oxidized to
Mn 2+ MnO2
HCOOH (formic acid) CO2
Example: What is the N of a solution of potassium permanganate if 40.00 mL will oxidize that
weight of potassium tetroxalate, KHC2O4.H2C2O4.2H2O, which requires 30.00 mL of 0.5000 N
sodium hydroxide solution for its neutralization, and what is the value of 1.000 mL of the
KMnO4 in terms of grams As2O3 in the titration of As3+ to H3AsO4 in the presence of acid?
Neutralization process: eq tetroxalate = eq NaOH

weightKHC2O4.H2C2O4.2H2O x factortetrox = NNaOH VNaOH
MWtetrox
weightKHC2O4.H2C2O4.2H2O = 0.50N(30 mL) x 254.2 g/mol
3 eq/mol
weightKHC2O4.H2C2O4.2H2O = 1271 g
Redox process: eq tetroxalate = eq KMnO4 weightKHC2O4.H2C2O4.2H2O(factortetro)= NKMnO4VKMnO4
MWtetrox
1271 g (4 eq/mol ) = NKMnO4(40mL)
254.2 g/mol
NKMnO4 = 0.500 N
eq KMnO4 = eq As2O3
0.500 N(1.000mL) = weight As2O3 (4 eq/mol )
1000mL/L 197.84g/mol
weight As2O3 = 0.02473 g As2O3
Example: What is the percentage of iron in a sample of iron ore weighing 0.7100 g if, after
solution and reduction of the iron with amalgamated zinc, 48.06 mL of KMnO 4 (1.000 mL ≈
0.006700 g Na2C2O4) is required to oxidize the iron? How many grams of KMnO 4 are
contained in each milliliter of the solution?

eq KMnO4 = eq Na2C2O4
NKMnO4 (1.00 mL) = 0.006700 g Na2C2O4 (2 eq/mol )
1000mL/L 134 g/mol
NKMnO4 = 0.100 N
eq KMnO4 = eq Fe
0.100N(48.06 mL) = weight Fe (1 eq/mol )

1000mL/L 55.85 g/mol
weight Fe = 0.2684 g % Fe = 0.2684/0.71 = 37.80%
each mL of normal KMnO4 contains KMnO4

5000
wt KMnO4 = (0.100 eq)(158.04 g/mol)( 1 L ) = 0.003161 g
mL L 5 eq/mol 1000mL Ml
Example: How many grams of H2O2 are contained in a solution that requires for titration
14.05 mL of KMnO4 of which 1.000 mL ≈ 0.008378 g Fe (i.e., will oxidize that amount of iron
from the divalent to the trivalent state)? How many g and how many mL of oxygen
measured dry and under standard conditions are evolved during the titration?
eq KMnO4 = eq Fe
NKMnO4 (1.000 mL ) = 0.008378 g Fe (1 eq/mol )

1000mL/L 55.85g/mol
NKMnO4 = 0.1500 N
eq KMnO4 = eq H2O2
0.1500N( 14.05 mL ) = weight H2O2 (2 eq/mol )
1000mL/L 34.02g/mol
weight H2O2 = 0.03585 g H2O2
eq KMnO4 = eq O2
0.1500N( 14.05 mL ) = weight O2 (2 eq/mol )
1000mL/L 32g/mol
weight O2 = 0.03372 g O2
At STP, 1mol O2 occupies 22.4 L

volume O2 = 0.03372 g O2 (1.00 mol)(22400mL)
32 g 1 mol
= 23.60 mL
Example: What is the percentage of MnO2 in a pyrolusite ore if a sample weighing 0.4000 g
is treated with 0.6000 g of pure H2C2O4.2H2O and dilute H2SO4 and after reduction has
taken place (MnO2 + H2C2O4 + 2H+  Mn2+ + 2CO2 + 2H2O), the excess oxalic acid requires
26.26 mL of 0.1000 N KMnO4 for titration? If pure As2O3 were used instead of oxalic acid,

how many grams would be required in order for the other numerical data to remain the
same?
eq MnO2 = eq H2C2O4.2H2O – eq KMnO4

wt MnO2 (2 eq/mol ) =
86.94g/mol
0.600g(2eq/mol) – 26.26mL(0.1N)
126.07 g/mol 1000mL/L
weight MnO2 = 0.2996 g % MnO2 = 0.2996/0.4 = 74.9%
eq As2O3 = eq H2C2O4.2H2O
weight As2O3 ( 4 eq/mol ) = 0.600g(2eq/mol)
197.84 g/mol 126.07 g/mol
weight As2O3 = 0.4708 g
Example: What is the milliequivalent weight of Pb 3O4 and of Pb in the calculation analysis of
red lead (impure Pb3O4) Pb3O4(s) + H2C2O4 + 3SO42- + 6H+  3Pb2+ + CO2 + 4H2O
meq wt Pb3O4 = Pb3O4 mg/mmol = 0.3428

2000 meq/mmol
meq wt Pb = 3Pb mg/mmol = 0.3108

2000 meq/mmol
Example: A steel containing 0.90% Mn is analyzed by the three standard methods:
Bismuthate, Chlorate (Williams)and Volhard, in each case with a 2.50 g sample, 0.0833 N
KMnO4 and 0.100 N FeSO4 solutions. Calculate the volume of KMnO4 required in each case.
Bismuthate Method: Mn is oxidized to KMnO4, and after reduction with 25.0 mL of standard
FeSO4 (MnO4- + 5Fe2+ + 8H+  Mn2+ + 5Fe3+ + 4H2O) the excess ferrous ion is titrated
with the standard KMnO4
eq Mn = eq FeSO4 – eq KMnO4
wt Mn x 5 eq/mol = 0.100N(25.0 mL) – 0.0833NVKMnO4
54.94g/mol 1000mL/L 1000mL/L
Weight Mn = 0.90% (2.50 g) = 0.0225 g

100%
2.048 x 10-3 = 2.5 x 10-3 – 8.33 x 10-5 VKMnO4
VKMnO4 = 5.426 mL
Determination of Mn in carbon steel
Example: A steel containing 0.90% Mn is analyzed by the three standard methods:

Bismuthate, Chlorate (Williams)and Volhard, in each case with a 2.50 g
sample, 0.0833 N KMnO4 and 0.100 N FeSO4 solutions. Calculate the volume
of KMnO4 required in each case.

Bismuthate Method: Mn is oxidized to KMnO4, and after reduction with 25.0 mL of standard
FeSO4 (MnO4- + 5Fe2+ + 8H+  Mn2+ + 5Fe3+ + 4H2O) the excess ferrous ion
is titrated with the standard KMnO4
meq Mn = meq FeSO4 – meq KMnO4

W = NxV - NxV
MW
f(1000)
W Mn = 0.100N(25.0 mL) – 0.0833N xVKMnO4

54.94 g/mol
5eq/mol(1000)
Weight Mn = 0.90% (2.50 g) = 0.0225 g

100%
2.048 = 2.5 – 8.33 x 10-2 VKMnO4
VKMnO4 = 5.426 mL
Chlorate Method: Mn is oxidized with KClO3 to MnO2, which is filtered and dissolved in 25.0
mL of the standard FeSO4 (MnO2 + 2Fe2+ + 4H+  Mn2+ + 2Fe3+ + 2H2O). The
excess acid is titrated with the standard KMnO4.
MnO2 (Mn+4) + 2e-  Mn2+ (+2)

meq Mn = meq FeSO4 – meq KMnO4
W = NxV - NxV
MW
f(1000)
W Mn = 0.100N(25.0 mL) – 0.0833N xVKMnO4

54.94 g/mol
2eq/mol(1000)
Weight Mn = 0.90% (2.50 g) = 0.0225 g

100%
8.1908 x 10-1 = 2.5 – 8.33 x 10-2 VKMnO4
VKMnO4 = 20.18 mL = 20.2 mL
Volhard Method: Mn is titrated directly with KMnO4 in a solution kept neutral with ZnO
(3Mn2+ + 2MnO4- + 2ZnO  5MnO2 + 2Zn2+)
The oxidizing power of KMnO4 in neutral solution only three fifths (3/5) as great as it is in acid
solution
Mn2+ (+2)  MnO2 (Mn+4) + 2e-
Mn (+2) + MnO4 (Mn+7) & MnO4-(Mn+7)  MnO2(Mn+4)
2+ -

5 3
meq Mn = meq KMnO4
W = NxV
MW
f(1000)
W Mn = 0.0833Nacid (3/5) VKMnO4

54.94 g/mol
2eq/mol(1000)
Weight Mn = 0.90% (2.50 g) = 0.0225 g

100%
8.1908 x 10-1 = 4.998 x 10-2 VKMnO4
VKMnO4 = 16.388 mL = 16.4 mL
DICHROMATE AND CERIC PROCESSES
Dichromate and Processes are redox titrations where the oxidizing agents use are K2Cr2O7
and Ce(SO4)2, respectively. The most common reducing agents used in this
process are ferrous solutions
Advantages of Dichromate and Ceric Processes over Permanganate Process

- stable even at higher temperatures unlike permanganate which easily
decomposes
Disadvantages of Dichromate and Ceric Processes over Permanganate Process

-special indicator (known as redox indicator) is used in each case whereas
permanganate serves as its own indicator
- the sodium salt of diphenyl amine sulfonate is the most commonly used redox
indicator
whose change is from colorless to purple
Reactions:
K2Cr2O7 → Cr +3 factor of K2Cr2O7 = ( 6 -3)( 2) = 6
Cr2O7= → 2Cr+3
Ce(SO4)2 → Ce +3 factor of Ce(SO4)2 = ( 4 -3)(1) = 1
MnO4 -, Cr2O7=, Ce +3 : O.A

Fe + 2 → Fe in the presence of acid
+3
HCl
Applications of Dichromate/Ceric Process

1. Determination of Fe in Limonite in solutions containing Fe like FeCl 3 solution
2. Determination of Cr in Chromite, an ore whose composition is Fe(CrO2)2
Example:
1. In the standardization of a K2Cr2O7 solution against 99.85% pure Fe wire , 42.42
mL of the dichromate were added to the HCl solution of the wire. The wt of
the wire was 0.2200 g and 3.27 mL of FeSO4 soln ( 0.1011 N as reducing agent)
were required to complete the titration. Calculate the normality of the
dichromate as an oxidizing agent
me Fe = me K2Cr2O7 (O.A) – me FeSO4 (R.A)
W = NxV - NxV
MW
f(1000)
(0.2200 g)(0.9985) = (42.42 mL)(N K2Cr2O7) – (3.27mL x 0.1011 N)

55.85 g/mol
1eq/mol(1000)
N K2Cr2O7 = 0.1005 N
2. A sample of chromite contains 30.00% Cr 2O3. After fusion of a 0.200-g sample of

Na2O2 and dissolving in acid, how many grams of FeSO4.(NH4)2SO4.6H2O should be
added so that the excess ferrous ions will require 15.00 mL of a solution of K2Cr2O7
containing 0.200 millimole of Cr millilitre?

me Cr2O3 = me FAS - me K2Cr2O7
(0.3)(0.2 g) = W FAS -(15mL)(0.2 mmol/mL)(3 me/mol)

152 g/mol 392 g/mol
6(1000) 1 (1000)
WFAS = 4.458 g
3. In the analysis of a sample of limonite by titrating with a solution of K 2Cr2O7 of which 1.00
mL ≎ 0.01117 g Fe. What wt of sample should be taken so that the percentage of Fe 2O3
will be found by multiplying the buret reading by 4? How many grams of K 2Cr2O7 are in
each millilitre of the above dichromate?
a. me K2Cr2O7 = me Fe
(N K2Cr2O7)(1 mL) = 0.01117g
-----------
55.85 g/mol
-------------
(1eq/mol)(1000)
N K2Cr2O7 = 0.2N
b. me Fe2O3 = me K2Cr2O7
% Fe2O3 = N K2Cr2O7 x V K2Cr2O7 x me wt Fe2O3 x 100

W spl
4 V K2Cr2O7 = ( 0.2 N)( V K2Cr2O7 )(159.7 g/mol)

W spl x 2 eq/mol x 1000
W sample = 0.3992 g
c. N K2Cr2O7 = W x f
MW x V
W K2Cr2O7 = (0.2 eq/mol)(294.7 g/mol)(1ml/1000)

6 eq/mol
W K2Cr2O7 = 0.009807 g
4. A 0.500 g sample of chromite is fused with Na2O2 leached with water, and acidified.
The Cr is reduced by adding 2.78 g of FeSO4.7H2O crystals. The excess ferrous ions

then then require 10.0 mL K2Cr2O7 for oxidation , and 1.00mL K2Cr2O7 ≎ 0.1060 g
Fe2O3. What is the percentage of Cr in the chromite?
me K2Cr2O7 = me Fe2O3
(1mL)( N K2Cr2O7 ) = 0.0160 g

156.7 g/mol
2(1000)
N K2Cr2O7 = 0.2 N
me Cr = me FeSO4.7H2O - me K2Cr2O7
WCr = 2.78 g - (10mL)(0.2N)

52 g/mol 278.02g/mol
3(1000) 1(1000)
W Cr = 0.1387 g
% Cr = 0.1387 g x 100
0.5 g
OR:
% Cr = (me FeSO4.7H2O - me K2Cr2O7) x me wt Cr x 100
W sample
% Cr = 2.78 - (10mL)(0.2N) 52
278.02/1(1000) 3(1000)
X 100
0.5 g
% Cr = 27.7%
ASSIGNMENT:
16. What is the percentage of Fe2O3 in a sample of limonite ore if the iron from a
0.5000 g sample is reduced and titrated with 35.15 mL of K2Cr2O7 solution of
which 15.00 mL is equivalent in oxidizing power to 25.00 mL of KMnO 4 solution
which has an “iron value” of 0.004750 g?
-17. A solution contains 2.608 g KMnO4 per 750 mL. (a) What is the normality as an
oxidizing agent? And what is the value of each mL in terms of g of (b)
FeSO4.(NH4)2SO4.6H2O, (c) As2O3, (d) KHC2O4, (e) H2O2 and (f) U(SO4)2
(oxidized to UO22+)?
18. - A sample of pyrolusite weighing 0.6000 g is dissolved in a solution containing 5.00
mL of 6.00 N H2SO4 and 0.900 g of H2C2O4.2H2O. The excess oxalate then requires
24.00 mL of KMnO4 solution for titration. If each mL of the KMnO4 will oxidize the Fe(II)
in 0.03058 g FeSO4.7H2O, what is the oxidizing power of the sample in terms of MnO2?
84.33%

IODIMETRY
Iodimetry is the redox titration of iodine ( as an oxidizing agent) against sodium thiosulfate
with starch as the indicator. The end point color is deep blue
Reaction:
I2 + 2Na2S2O3 → 2NaI + Na2S4O6
Where:
Na2S2O3 is sodium thiosulfate
Na2S4O6 is sodium tetrathionate
The factor of I 2 = [ 0- (-1)][2] = 2

The factor of Na2S2O3 = [2.5-2][2] = 1
Applications of Iodimetry:
1. Analysis of Reducing Agents (such as H 2S, sulfites, arsenites, stannous salts)
Reducing Agent is directly titrated with I 2
me I 2 = me reducing agent
2. Analysis of Oxidizing agents ( such as permanganate, chromate, H 2O2)

The method makes use of the fact that all oxidizing agents oxidize iodide in dilute
acid solution to free iodine ( as I 2), which in turn can be titrated with standard
sodium thiosulfate
Oxidizing agent + excess KI free iodine in the form I 2 which in turn

is titrated with Na2S2O3
NaI + Na2S4O6
me oxidizing agent = me I 2 = me Na2S2O3
Note: When O.A are analyzed iodimetrically, it is important to titrate most of the
liberated iodine with the thiosulfate before adding the starch indicator.
Otherwise , so much of the blue-iodo starch compound is formed that the
thiosulfate reacts only very slowly with it
Standardization of Iodine:
Stoichiometry : H3AsO3 + I 2 + HCO3- H3AsO4 + 2I- + 2CO2 + H2O
As2O3 + 6OH- ⇔ 2 AsO3 -3 + 3H2O
me wt As2O3 =
MW As2O3
2(2X1000)
Antimony can be oxidized by iodine: Sb +3 → Sb+5

Standardization of Thiosulfate:
me wt KBrO3 = KBrO3 / 6(1000)

BrO3 - (+5) + 6e- → Br- (-1) f= 6
me wt KIO3 = KIO3 / 6 (1000)

IO3- (+5) + 6e- → I- (-1) f=6
me wt K2Cr2O7 = K2Cr2O7 /6(1000)

Cr2O7= + 6I- + 14H+ → 3I2 + 2 Cr+3 + 7H2O
me wt Cu = Cu / 1(1000)
2Cu++ + 4I- → 2 CuI + I2
Stoichiometry:
BrO3- + 6I- + 6H+ → Br- 3 I2 + 3H2O
IO3- + 5I - + 6H+ → 3 I2 + 3H2O
Cr2 O7= + 6I - + 14H+ → 2Cr+3 + 3 I 2 + 7H2O
2Cu++ + 4I - → 2CuI + I 2
1. What is the value in terms of grams of As2O3 of each milliliter of an I 2 soln of which 1.00
mL is equivalent to 0.0300 g of Na2S2O3 ?
a. me I 2 = me Na2S2O3
1mL(N I 2) = 0.03 g
158.11 g/mol
1 eq/mol(1000)
N I2 = 0.1897 N
b. me I 2 = me As2O3
1mL( 0.1897 N) = g As2O3

As+3 → As+5
197.84 g/mol
f = 5-3= 2x2 = 4
4eq/mol(1000)
W As2O3 = 0.009373 g
2. If 48.0 mL of a soln of thiosulfate are required to titrate the I 2 liberated from an excess of
KI by 0.300 g of KIO3 what is the normality of the thiosulfate and the value of
each milliliter of it in terms of grams of I2 ?

0.30 g KIO3 + x’s KI → I2 + Na2S2O3 → NaI
me S2O3= = me KIO3 KIO3 (+5) + 6e- → KI (-1)

f = +5 –(-1) = 6x1 = 6
48 mL(N S2O3= ) = 0.300 g

214.01 g/mol
6eq/mol(1000)
N S2O3= = 0.1752 N
me S2O3= = me I 2 I2 (0) +2e- →2 I- ( -1)

f = [0- (-1)] = 1x2 =2
(1mL)(0.1752N) = W I2
253.81g/mol
2eq/mol(1000)
3. A sample of stibnite containing 70.05% is given out for analysis. A student titrate it
with a soln of I 2 which 1.000 mL is equivalent to 0.004946 g of AS2O3.Due to an
error in standardization, the student’s analysis shows the sample to contain
70.32% Sb. Calc the true normality of the I 2 and the percentage error in the
analysis.
me I2 = me As2O3
1mL (N I 2) = 0.004946 g
197.84 g/mol
4eq/mol(1000)
N I2 = 0.10 N
% error = 70.32 - 70.05 x 100

70.05
= 0.385 %
4. Wt. of copper ore taken for analysis = 1.200 g; vol of Na 2S2O3 used= 40.00 mL; 1mL
Na2S2O3 ≎ 0.004715 g KBrO3. Calc. the copper content of the ore in terms of percentage
Cu2O.
me Cu++ → me Cu2O → me Na2S2O3
me Na2S2O3 → me KBrO3
1mL( N Na2S2O3) = 0.004175 g

167.01 g/mol
6 eq/mol(1000)

N Na2S2O3 = 0.15 N
% Cu2O = (mL Na2S2O3 x V Na2S2O3 x me wt Cu2O x 100

W sample
% Cu2O = (40 mL)(0.15N) 143.08g/mol
2eq/mol(1000) x 100
1.2 g
% Cu2O = 35.77 %
5. What is the percentage of Sb in a 0.2500-g sample of stibnite if 20.83 mL of I 2 are used

in the final titration and each milliliter of the I 2 is equivalent to 0.004495 g of As ?
me I 2 = me As
1mL( N I 2) = 0.004495 g
75 g/mol
2eq/mol(1000)
N I2 = 0.1198 N
% Sb = V I 2 x N I 2 x me wt Sb x 100
W spl
% Sb = ( 20.83 mL)(0.1198 N) 122 g/mol

2eq/mol(1000) x 100
0.25 g
= 60.87 %
6. What volume of thiosulfate soln ( 1.00 mL ≎ 0.004873 g of potassium biiodate) will be

required to titrate the I 2 liberated by the Cl2 evolved when a 0.450- g sample of red
lead containing 95. 0% Pb3O4 is boiled with HCl ?
me thiosulfate =me KIO3. HIO3

IO3 ( +5) → I- (-1) f= 5-(-1)= 6x2=12
1mL (N thio) = 0.004873 g

75 g/mol
12eq/mol(1000)
N thio = 0.15 N
0.45 g red lead + HCl → Cl2 + x’s KI → I2
(contg 95% Pb3O4) O.A + Na2S2O3

NaI + Na2S4O6
me Pb3O4 = me Cl2 = me I2 = me Na2S2O3

0.45g (0.95)
685.57 g/mol = 0.15 N ( V thio)
2eq/mol(1000)
Pb3O4 : PbO2 (+4) + 2e- =→ Pb ( +2) p.277
V thio = 8.31 mL
MODULE 7: GRAVIMETRIC ANALYSIS

GRAVIMETRIC ANALYSIS

• Qualitative Analysis – identification of constituents present in a sample
• Quantitative Analysis – determines the amount of the constituents present in a
sample
Methods of analyses:
1- volumetric analysis – measures the volume of the solution necessary to react
completely with analyte
2- gravimetric analysis – measures the mass of a substance chemically related to

the analyte
• Gravimetric Analysis
- based on the law of Definite Proportions:
states that a chemical compound always contains exactly the same proportion
of elements by mass
Law of Multiple Proportions:

states that if two elements form more than one compound between them, then the
ratios of the masses of the second element which combine with a fixed mass of the first
element will be ratios of small whole numbers.
GRAVIMETRIC
- Involves preparation and weighing of a stable substance of known composition
that contains that constituent to be determined
- the most common method of doing this is to cause the stable substance to ppt from
a solution leaving behind materials that might contaminate & alter the composition of the
substance
STOICHIOMETRY
- measurement of weight relative b/n constituents of substances and products of
reaction
GRAVIMETRIC METHOD:
- from a known weight of a sample , the weight of a constituent can be determined
by multiplying it by a factor
- represents that weight of a desired constituent equivalent to one unit weight of a
given substance
Steps involved in Gravimetric Methods

1. Weighing the sample
2. Dissolving the sample
3. Separation of some substance from solution containing, or bearing a definite
relation to, the constituent being measured, under conditions which render this
separation as complete as possible
4. Isolation of the separated substance, commonly by filtration
5. Determination of the weight of the isolated substance, or that of some derivative
formed from it on ignition

Divisions of Gravimetric analysis
1. Chemical Precipitation Methods
2. Electrolytic Deposition methods
Gravimetric Methods versus Volumetric methods

- Vlumetric methods are generally more rapid, requires less apparatus,
and are frequently capable of greater accuracy than gravimetric
methods
Example Calc the gravimetric factor for a) Sn in SnO2 b) MgO in Mg2P2O7 ; c) P2O5 in
Mg2P2O7
D) Fe in Fe2O3 e ) SO3 in BaSO4
a. Sn in SnO2
gf = 1mol Sn (At wt Sn)
1 mol SnO2(MW SnO2)
= 1mol Sn( 118.69 g/mol) = 118.69 g Sn
1mol SnO2 ( 150.69 g/mol) 150.69 g SnO2
= 0.7876
b. MgO in Mg2P2O7
gf = 2 mols MgO( 40.31 g/mol)
1mol (222.57 g/mol)
= 0.3622
c. P2O5 in Mg2P2O7
gf = 1mol P2O5( 141.95 g/mol)
1mol Mg2P2O7(222.57 g/mol)
= 0.637
d. Fe in Fe2O3
gf = 2 mol Fe(55.85g/mol)
1mol Fe2O3(159.70 g/mol)
= 0.6994
e. SO3 in BaSO4
gf = 1mol SO3( 80.06 g/mol)
1mol BaSO4( 233.40 g/mol)
= 0.3430

• Example: A sample of impure sodium chloride is dissolved in water, and the chloride
is precipitated with silver nitrate (Cl - + Ag+  AgCl) furnishing 1.000 g of silver
chloride. What is the weight of chlorine in the original sample?
• Example: What weight of Fe3O4 (magnetite) will furnish 0.5430 g of Fe2O3?(hematite)

2Fe3O4 + ½ O2  3Fe2O3
Wt of Fe3O4 = 0.5430 g Fe2O3 x MW Fe3O4 x ratio

MW Fe2O3
= 0.5430 g Fe2O3 x 231.54 g/mol Fe3O4 x 2mol Fe3O4

159.69 g/mol Fe2O3 3mol Fe2O3
= 0.5249 g Fe3O4
35.45 g/mol Cl x 1 mol Cl

143.32 g/mol AgCl 1 mol AgCl
and
231.54 g/mol Fe3O4 x 2mol Fe3O4
159.69 g/mol Fe2O3 3mol Fe2O3
can be shortened called gravimetric factor

Gravimetric factor or Chemical factor – weight of desired substance equivalent to a unit
weight of given substance
1 Cl 2 Fe3O4
1 AgCl 3 Fe2O3
Example. How many grams of Mn3O4 can be obtained from 1.00 g of MnO2?
g Mn3O4 = 1 g MnO2 x 228.82 g Mn3O4 X 1 mol Mn3O4
86.94 g MnO2 3 mols MnO2
= 0.8773 g Mn3O4
OR:
g Mn3O4 = 1 g MnO2 x 1 mol MnO2 x 1 mol Mn3O4 x 228.82 g Mn3O4
86.94 g MnO2 3 mols MnO2 1 mol Mn3O4
= 0.8773 g Mn3O4

• Calculation of Percentages
% constituent = wt of constituent in sample x 100
wt of sample
Example: If 2.000 g of impure sodium chloride is dissolved in water and, with an excess of
silver nitrate, 4.6280 g of silver chloride is precipitated, what is the percentage of chlorine in
the sample?
g Cl = 4.6280 g AgCl x 1 Cl = 4.6280 g x 35.45
1 AgCl 143.32
= 1.145 g
% Cl = wt of Cl = 1.145 x 100 = 57.25%
wt of sample 2.000
• Calculations Involving A Factor weight Sample

% constituent = wt of constituent in sample x 100
wt of sample
wt of constituent = wt known x g f (unknown)
(known)
Example: The gravimetric factor of a certain analysis is 0.3427. It is desired to regulate

the weight of sample taken so that a) each centigram of the precipitate obtained will
represent 1.00% of the desired constituent, b) the percentage will be twice the number of
centigrams of precipitate. What weight of sample should be taken in each case?
a) The relationship between the weight of precipitate and the percentage of constituent
is such that
0.01 g = 1.00% therefore
0.01 x 0.3427 x 100 = 1 x = 0.3427 g
x
a) 0.01 x 0.3427 x 100 = 2 x = 0.1714 g
x
• Calculation of the Volume of a Reagent required for a given reaction
Concentration is expressed as g solute
volume of solution
Example: How many mL of barium chloride solution containing 90.0 g of BaCl 2.2H2O
per liter are required to precipitate the sulfate as BaSO4 from 10.0 g of pure Na2SO4.10H2O?
Ba2+ + SO42-  BaSO4
1 mol BaCl2.2H2O 1 mol Na2SO4.10H2O
g BaCl2.2H2O = 10 g Na2SO4.10H2O x 1 BaCl2.2H2O
1 Na2SO4.10H2O
= 10.0 g x 244 = 7.58 g BaCl2.2H2O

322
7.58 g BaCl2.2H2O = 90.0 g BaCl2.2H2O
y (mL) 1 L x 1000mL/L
y = 84.2 mL

Example: How many mL of ammonia water of SG = 0.950 (containing 12.74% of NH 3
by mass) are required to precipitate iron from 0.800 g of pure ferrous ammonium sulfate,
FeSO4(NH4)2SO4.6H2O, after oxidation of the iron to ferric state?
Since, Fe3+ + 3 NH3 + 3H2O  Fe(OH)3 + 3NH4+
It follows to precipitate ferric ion: the wt of NH3
= 0.800 g FeSO4(NH4)2SO4.6H2O x 3 NH3 .

FeSO4(NH4)2SO4.6H2O
g NH3 = 0.800 g x 3(17) = 0.1042 g
392.1
From SG of NH3 : 0.950 ≈ 0.950 g / mL solution and contains 12.74% NH 3 , therefore
wt of NH3 in 1 mL solution = 0.950 x 0.1274
= 0.121 g
0.1042 g NH3 (reqd to ppt) = 0.121 g NH3
y (mL) mL solution
y = 0.861 mL
Example: How many mL of sulfuric acid (SG =1.135, 18.96% H 2SO4 by mass) are
required to neutralize 75.0 mL of ammonium hydroxide (SG = 0.960, containing 9.91% NH 3 by
mass)
H2SO4 + 2NH3 (2NH4OH)  (NH4)2SO4 + 2HOH
mass NH3 = ρV = 0.960 g x 0.0991 x 75 mL = 7.1352g
mL
mass H2SO4 = 7.1352 g NH3 x H2SO4 (98.08)
2NH3 2(17.03)
= 20. 5467 g
Volume of H2SO4 = 20. 5467 g x 1 mL x 1 =
1.135g 0.1896
= 95. 48 mL ≈ 95.0 mL
Concentration of a reagent mixture:

Example: What weight of water must be added to 100 mL of sulfuric acid containing
26.0% by mass H2SO4 (SG = 1.19) in order for the resulting solution to contain 12.3%
H2SO4 by mass?
Mass H2SO4 initially = Vρ = 100 mL ( 1.19 g ) = 119 g
(solution) mL
Mass pure H2SO4 (solute) = 119 g (0.26) = 30.94 g pure
% mass = mass solute x 100
mass solution (finally)
mass solution = 30.94 + mass H2O (x)

12. 3 = 30.94 x 100
119 + x
x = 132.5447 g
INDIRECT GRAVIMETRIC ANALYSIS:

Example: In the analysis of a 2.00 g sample of limestone, the weight of combined
oxides of iron and aluminum (Fe2O3 + Al2O3) is found to be 0.0812 g. by volumetric methods,

the percentage of the limestone of total iron calculated as FeO is found to be 1.50. What is
the percentage of Al2O3 in the sample?
mass of FeO = 2.00 g x (1.50/100) = 0.0300 g FeO

mass of Fe2O3 = 0.0300 g FeO x Fe2O3 159.69 =
2FeO 2(71.85)
= 0.0333 g
Mass of Al2O3 = 0.0812 g – 0.0333 = 0.0479 g
% Al2O3 = 0.0479 x 100 = 2.40%
2.00
example: In the analysis of a sample of feldspar weighing 0.4150 g, a mixture of KCl +

NaCl is obtained weighing 0.0715 g. From these chlorides, 0.1548 og of K 2PtCl6 is obtained.
Calculate percentage of Na2O in the sample
.
Let x = mass of NaCl in combined chlorides
0.0715 – x = mass of KCl
mass KCl x K2PtCl6 = 0.1548
2KCl
(0.0715 – x) [ 486.01 ] = 0.1548
2(74.56)
x = 0.0240 g NaCl
mass Na2O = 0.0240 g NaCl x Na2O [ 61.98 ]
2NaCl 2(58.45)
= 0.012725 g
% Na2O = 0.012725 g x 100 = 3.066%
0.4150 g

References
Brown, T. L., LeMay, E. H., Bursten, B. E., & Burdge, J. R. (2004). Chemistry The Central Science. New Jersey:
Pearson Education South Asia PTE LTD.
Hamilton, L. F., & Simpson, S. G. (1971). Quantitative Chemical Analysis. New York: MacMillan Publishing
Co., Inc.
S, G. E. (1964). Qualitative Analysis. New York: McGraw-Hill Book Company Inc.
FACILITATOR:
Engr. Lilibeth R. Ramos
CHEM 1221 Course Facilitator
Cellphone : 09433003149
SLU local extension number: Chemical Engineering local 391
Institutional email address : lramos@slu.edu.ph


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MODULE IN

Department of Chemical Engineering

SCHOOL OF ENGINEERING AND ARCHITECTURE

FORMATIVE ASSESSMENT ACTIVITIES ............................................................................................10

Prepared by: LRRAMOS

COURSE LEARNING OUTCOMES

1. Demonstrate formulation and

2. Apply skills in Mathematics, Chemistry

Prepared by: LRRAMOS

Module and Unit Topics

Prepared by: LRRAMOS

Module 2 : Chemical Equations: Formulation and Balancing

Module 3: Concentrations of Solutions

Module 4: Chemical Reactions: Rates and Equilibrium Constants

Module 5: Ionization and Ionization Constants

Module 6: Volumetric Methods of Analysis

Module 7: Gravimetric Analysis

Course Study Guide

Prepared by: LRRAMOS

2. MIND YOUR MANNERS. Treat the distance learning environment as an academic

b. Express your opinions politely and do not dominate the conversation.

c. Avoid lengthy as well as offensive posts by sticking to the topic of the

e. For a live meeting or video/voice conferencing set-up, mute your microphone

Prepared by: LRRAMOS

6. OBSERVE ORIGINALITY. Your course outputs will largely be submitted in electronic

Additional Guidelines for Offline Students:

Prepared by: LRRAMOS

3 hrs TLO 1: : Illustrate different Engage: What is analytical chemistry

Module 2 Chemical Equations: Formulation and Balancing

5 hrs TLO 2: Demonstrate formulation Engage: Recall principles involved in

9hrs TLO 3: Solve problems on Engage: different types of solutions

6 hrs TLO 4: : Evaluate chemical Engage: Different factors

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Module 4 Ionization and Ionization constants

10 hrs TLO 5: Execute Engage: Point out the distinction

Module 5 Volumetric Methods of Analysis

9 hrs TLO 6:Apply the basic Engage: Introduction to titration

Quiz 2; Module 5: Units1& 2 andMidterm Exam

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TLO 8 : Apply basic concepts of Engage: Introduction to gravimetry

Comprehensive Final Exam

Formative Assessment Activities

• You are required to answer the pre-assessment quizzes, self-assessment activities,

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Summative Assessment Activities

A. Quizzes, Examinations, and Assignments

Graded quizzes, examinations, and assignments are essential to determine

B. Final Course Requirement

TENTATIVE FINAL GRADE (TFG)

Final Grade: MG 50% + TFG 50% = 100%

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FACILITATOR CONTACT INFORMATION:

At the end of the module the student should be able to:

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CHEMISTRY - is a branch of science which deals with the study of:

a) composition and properties of matter

1. General Chemistry - general survey of the entire field of chemistry with

2. Inorganic Chemistry – concerned with the preparation and properties of the