Analytical Chemistry...solutions. 3 2 General concept of acids and bases. 3 4 3 Strengths of acids...

Prepared by:

Dr. Mosad A. Elghamry

Analytical

Chemistry

First Year

2019/2018

Acknowledgments

This two-year curriculum was developed through a participatory and collaborative approach between the

Academic faculty staff affiliated to Egyptian Universities as Alexandria University, Ain Shams University,

Cairo University , Mansoura University, Al-Azhar University, Tanta University, Beni Souef University , Port

Said University, Suez Canal University and MTI University and the Ministry of Health and

Population(General Directorate of Technical Health Education (THE). The design of this course draws on

rich discussions through workshops. The outcome of the workshop was course specification with Indented

learning outcomes and the course contents, which served as a guide to the initial design.

We would like to thank Prof.Sabah Al- Sharkawi the General Coordinator of General Directorate of

Technical Health Education, Dr. Azza Dosoky the Head of Central Administration of HR Development, Dr.

Seada Farghly the General Director of THE and all share persons working at General Administration of the

THE for their time and critical feedback during the development of this course.

Special thanks to the Minister of Health and Population Dr. Hala Zayed and Former Minister of

Health Prof. Ahmed Emad Edin Rady for their decision to recognize and professionalize health education

by issuing a decree to develop and strengthen the technical health education curriculum for pre-service

training within the technical health institutes.

جىصيف مقشس دساسً تياوات المقشس -1

:اسم المقشس :الشمض الكىدي

Analytical Chemistry األول :المسحىي /الفشقة

:الحخصص

وظشي عملً : عذد الىحذات الذساسية

3 6

هذف -2

:المقشس Identify different methods of chemical analysis and their applications in life

:المسحهذف مه جذسيس المقشس -3

المعلىمات . ا

:والمفاهيم

- Identify general concept of acids and bases.

- Define the law of mass action.

- Describe acid- base Equilibria in water.

- Explain common ion effect.

- Identify different types of buffer solutions.

- Explain fractional precipitation.

- Identify different methods of expression of concentrations.

- Identify different methods of Precipitation reactions.

- Describe the complex formation reactions.

- Discuss the principle of gravimetric analysis.

- Identify technique of gravimetric analysis.

المهاسات -ب

:الزهىية

- Compare between acids and bases according to strengths.

- Apply Solubility product concept in precipitation of substances.

- Evaluate gravimetric factor in gravimetric analysis.

- Calculate the hydrogen ion exponent of acids and bases.

- Apply the titrimetric in determination of concentrations of

solutions.

- Solve the problems of titrimetric and gravimetric analyses

المهاسات -ج

المهىية الخاصة :تالمقشس

- measure the concentrations in human fluids using the different

methods of titrimetric titration.

- measure the quantity of precipitates of the analytes from solutions.

- measure the PH

value of solutions and fluids.

المهاسات -د

:العامة

- use problems solving skills

- use information technology

- apply the communication skills

مححىي -4

:المقشس

Theoretical Section:

Chapter 1: Fundamental theoretical principles 1.1. Electrolytic dissociation.

1.2. General concept of acids and bases.

1.3. The law of mass action.

1.4. Activity and activity coefficient.

1.5. Acid- base Equilibria in water.

1.6. Strengths of acids and bases.

1.7. Common ion effect.

1.8. Solubility product.

1.9. Fractional precipitation.

1.10. Complex ions.

1.11. The ionic product of water.

1.12. The hydrogen ion exponent.

1.13. The hydrolysis of salts.

1.14. Buffer solutions.

1.15. Methods of expression of concentrations of solutions.

Chapter 2: Titrimetric analysis. 2.1. Acid- Base titrations.

2.2. Redox titrations.

2.3. Precipitation reactions.

2. 4. Complex formation reactions.

3. Gravimetric analysis:

3.1. The principle of gravimetric analysis.

3.2. Technique of gravimetric analysis.

3.3. Calculations.

3.4. Problems.

Practical Section:

Titrimetric Method Experiments and some of their

applications. 1. Acid- Base titrations:

1.1. Standardization of HCl acid solution by Na2CO3 using the

methyl orange and phenolphthalein as indicator.

1.2. Standardization of NaOH solution using the oxalic acid

solution.

1.3. Analysis of a mixture of NaOH & Na2CO3

1.4. Analysis of a mixture of Na2CO3 & NaHCO3

2. Redox titrations:

2.1. Standardization of KMnO4 using the oxalic acid solution.

2.2. Analysis of a mixture of oxalic acid and sodium oxalate.

2.3. Standardization of Na2S2O3 solution.

2.4. Standardization of iodine solution using standard Na2S2O3

solution.

2.5. Determination of copper in copper sulfate.

3. Precipitation titrations:

3.1. Standardization of AgNO3 solution against NaCl solution by:

a- Mohr method. b- Adsorption indicator method.

3.2. Standardization of NH4SCN solution against standard AgNO3

solution by Volhard method

3.3. Analysis of amixture of NaCl and HCl by indirect Volhard

method

4. Complex formation titrations:

4.1. Standardization of EDTA using standard ZnSO4 solution.

4.2. Determination of Cu2+

using the murexide indicator.

4.3. Determination of Co2+

using the xylenol orange indicator.

4.4. Determination of Fe3+

using the thiocyanate indicator.

4.5. Determination of Al3+

using the Eriochrome black T indicator

by back titration.

5. Some applications:

5.1. Determination of the concentration of antiacid tablets.

5.2. Examination of milk (Acidity test).

5.3. Estimation of urine acidity.

5.4. determination of vitamin C in fruit juice.

5.5. hydrolysis of starch by salivary amylase.

5.6. Estimation of chlorides in urine.

5.7. Determination of the percentage of iron in metal cans.

5.8. Determination of Total Hardness of water.

ية الحعليم والحعلمالأس -5

- Lecture. - Active learning. - Group work. - Power point presentation . - Clinical practical.

أسالية الحعليم والحعلم للطالب -6

روي القذسات المحذودة

:جقىيم الطالب -7

األسالية المسحخذمة -أ

a. Class work:

1. Quizzes

2. Midterm theoretical

3. Practical exam

4. Assignments

5. Participation

b. Final exam:

Written theoretical

الحىقيث -ب

a. Class work:

1. Quiz I (5th week) 5 marks

2. Attendance 5 marks

3. Midterm theoretical (7th week) 10 marks

4. Clinical work: 30 marks

b. Final exam

Practical exam (13th week) 10 marks

written theoretical exam (15th week) 90 marks

جىصيع الذسجات -ج

Case records and reports (5 marks) Quiz : 5 mark

Midterm: 10 marks

Attendance 5 marks

Clinical: 20 marks

Clinical exam:15 marks

Final written exam 90 marks.

Total percentage 150 mark

:قائمة الكحة الذساسية والمشاجع -8

مزكشات -أ

كحة ملضمة -ب

كحة مقحشحة -ج

Vogel Text book in: analytical chemistry(3th

Edition)1988

...... دوسيات علمية أو وششات -د

الخ

Analytical Chemistry Journal

Course Description ................................................................ vi

Chapter 1: Fundamental theoretical principles .............................. 1

Methods of expression of concentrations of solutions ................ 1

General concept of acids and bases ..................................... 11

Strengths of acids and bases .............................................. 20

Acid- base equilibria in water ............................................. 26

Autoionization of water .................................................... 34

The pH and the pH scale ................................................... 36

Hydrolysis of salts .......................................................... 41

Common ion effect .......................................................... 44

Buffer solutions .............................................................. 46

Solubility product ........................................................... 52

Chapter 2: Titrimetric analysis ................................................ 57

Titrimetry ..................................................................... 57

Acid- Base titrations ........................................................ 61

Redox titrations ............................................................ 73

Precipitation titrations .................................................... 89

Contents

iv

Gender and Social Norms

vi

Complex formation titrations ............................................ 98

Chapter 3: Gravimetric analysis ............................................. 109

The principle of gravimetric analysis .................................. 109

Technique of precipitation gravimetry ............................... 110

Gravimetric Calculations ................................................. 124

Problems .................................................................... 126

Practical analytical chemistry ................................................. 129

Basic tools and operations of analytical chemistry ................. 131

Acid- Base titrations ...................................................... 142

Redox titrations ............................................................ 155

Precipitation titrations ................................................... 168

Complexometric titrations ............................................... 184

References and Recommended Readings ................................... 203

حقىق الىشش والحأليف لىصاسة الصحة والسكان ويحزس تيعه

Analytical Chemistry

vii

This course will focus on provide the student a rigorous background in analytical

chemistry. and to develop in you the student an appreciation of the difficult task of

judging the accuracy and precision of experimental data and to show how these

judgments can be sharpened by the application of statistical methods. Also to introduce

the student too wide range of techniques of modern analytical chemistry and to teach the

laboratory skills that will give students competence in their ability to obtain high-quality

analytical data. In this course the student will learn how to apply the concepts of

chemical reactivity and equilibrium from general chemistry in a quantitative fashion to

the field of chemical analysis and how to design and implement a well- defined chemical

analysis that conveys the results with full scientific validity.

Core Knowledge

At the completion of this course, the students are expected to develop the knowledge

and comprehension of the core concepts of analytical chemistry. By the end of this

course, students should be able to:

Identify different methods of expression of concentrations.

Identify general concept of acids and bases.

Describe acid-base equilibria in water.

Identify different types of salts.

Explain common ion effect.

Identify different types of buffer solutions.

Explain fractional precipitation.

Identify different methods of precipitation reactions.

Describe the complex formation reactions.

Discuss the principle of gravimetric analysis.

Identify technique of gravimetric analysis.

Course Description


Course Description vi

Core Skills

At the completion of this course, the students will have developed a set of

fundamental skills that can be applied to various analytical situations. By the end of

this course, students should be able to:

Compare between acids and bases according to strengths.

Apply solubility product concept in precipitation of substances.

Calculate the pH of acids and bases.

Apply the titrimetric methods in determination of concentrations of solutions.

Calculate the gravimetric factor in gravimetric analysis.

Solve the problems of titrimetric and gravimetric analyses.

Measure the concentrations in human fluids using the different methods of

titrimetric titration.

Estimate the quantity of precipitates of the analytes from solutions.

Measure the PH value of solutions and fluids.

Design and set up an experiment.

Collect and analyze data.

Use problems solving skills.

Estimate the solution to a problem.

Apply appropriate techniques to arrive at a solution.

Test the correctness of their solution.

Use information technology.

Apply the communication skills.

ix


Course Overview

Methods of Teaching/Training with

Number of Total Hours per Topic

ID

Topics

Inte

ract

ive

Lectu

re

Fie

ld W

ork

Cla

ss

Ass

ignm

ents

Rese

arc

h

Lab

1 Chapter 1: Fundamental theoretical principles.

Methods of expression of concentrations of solutions.

3

2 General concept of acids and bases.

3

4

3 Strengths of acids and bases. Acid- base equilibria in water. 3

4 Autoionization of water. The pH and the pH scale.

3

4

5 Hydrolysis of salts. Common ion effect.

3

4

6 Buffer solutions. Solubility product.

3

7 Chapter 2: Titrimetric analysis. Titrimetry. Acid- Base titrations.

3 12

8 Redox titrations. 3 4 12

9 Precipitation titrations. 3 12

10 Complex formation titrations. 2 4 12

11 Chapter 3: Gravimetric analysis. The principle of gravimetric analysis. Technique of precipitation gravimetry.

3

12 Technique of precipitation gravimetry (complement).

3

13 3.3. Gravimetric Calculations. 3.4. Problems.

1

4

TOTAL HOURS (108)

36

24

48

viii


Chapter 1 1

Objectives

Identify different methods of expression of concentrations.

Identify general concept of acids and bases.

Describe acid- base equilibria in water.

Compare between acids and bases according to strengths.

Calculate the pH of acids and bases.

Measure the PH value of solutions and fluids.

Identify different types of salts.

Explain common ion effect.

Identify different types of buffer solutions.

Apply solubility product concept in precipitation of substances.

Methods of expression of concentrations of solutions

A solution is a homogeneous mixture of two or more chemically nonreacting pure

substances.

The solution is consisting of solute (substance with minor amount) and solvent (substance

with major amount). In a solution, the solute is dispersed uniformly throughout the solvent.

The intermolecular forces between solute and solvent particles must be strong enough to

compete with those between solute particles and those between solvent particles.

As a solution forms, the solvent pulls solute particles apart and surrounds, or solvates, them

Fundamental theoretical principles

1 Chapter


2

The concentration of a solution is a measure of the amount of solute that is dissolved in a

given quantity of solvent.

A dilute solution is one that contains a small amount of solute.

A concentrated solution contains a large amount of solute.

There are many methods for expression of concentrations of solutions. Here, we concerned

with four basic ways to express concentration:

1) Molarity (M)

is the number of moles of solute dissolved in one liter of solution.

How do you calculate the molarity of a solution?

To calculate the molarity of a solution, divide the moles of solute by the volume of the

solution.

To make a 0.5-molar (0.5M) solution:

a) first add 0.5 mol of solute to a 1-L volumetric flask half filled with distilled water.

b) Swirl the flask carefully to dissolve the solute.

c) Fill the flask with water exactly to the 1-L mark.

Calculating the molarity of a solution

Example 1:


Chapter 1 3

Finding the moles of solute in a solution

Example 2:

What effect does dilution have on the total moles of solute in a solution?

Diluting a solution reduces the number of moles of solute per unit volume, but the total

number of moles of solute in solution does not change, so you can write this equation:

M1 and V1 are the molarity and volume of the initial solution, and M2 and V2 are the molarity

and volume of the diluted solution.


4

Preparing a dilute solution

Making a dilute solution

For example: To prepare 100 ml of 0.40M MgSO4 from a stock solution of 2.0M MgSO4:

a) first measures 20 mL of the stock solution with a 20-mL pipet.

b) then transfers the 20 mL to a 100-mL volumetric flask.

c) Finally she carefully adds water to the mark to make 100 mL of

solution.


Chapter 1 5

Example 3:

2) Percent solutions

What are two ways to express the percent concentration of a solution?

The concentration of a solution in percent can be expressed in two ways: as the ratio of the

volume of the solute to the volume of the solution or as the ratio of the mass of the solute

to the mass of the solution.

Concentration in percent (volume/volume)

For example: Isopropyl alcohol (2-propanol) is sold as a 91%

solution. This solution consist of 91 mL of isopropyl alcohol

mixed with enough water to make 100 mL of solution.


6

Calculating percent (volume/volume)

Example 4:

What volume of acetic acid is present in a bottle containing 350.0 mL of a

solution which measures 5 % (v/v)?

Concentration in percent (mass/mass)

Calculating percent (mass/mass)

Example 5:

Find the percent by mass in which 41.0 g of NaCl is dissolved in 331 g of water.

Solution:

Percent by mass (% m/m) = mass of solute

mass of solution x 100%


Chapter 1 7

mass of the solution = mass of the solute + mass of the solvent

= mass of NaCl + mass of water

= 41.0 g + 331 g = 372 g

% (m/m) = 41.0 g

372 g x 100% = 11.0%

Find the percent by mass in which 3.25 g of Ba(NO3)2 is dissolved in 85 g of

water.

3) Concentration in parts per million (ppm)

Parts per million (ppm): is the number of milligrams of solute per kg of solution.

Parts per million (ppm) = mass of solute(mg) mass

of solution(kg)

Assuming the density of water is 1.00 g/mL, 1 liter of solution = 1 kg and hence,

1 ppm = 1 mg/L

This is generally true for fresh water and other dilute aqueous solutions.

Parts per million (ppm) = mass of solute(mg)

volume of solution(L)

Parts per million concentrations are essentially mass ratios (solute to solution) x a million

(106). In this sense, they are similar to percent by mass (% m/m), which could be thought of

as parts per hundred.

Other variant on this theme is:

Parts per billion (ppb): is the number of micrograms of solute per kg of solution.

mass of solute(µg) Parts per billion (ppb) =

Parts per billion (ppb) =

mass of solution(kg)

mass of solute(µg)


1 ppb = 1 µg/L

x 109

mass of solute(g)

mass of solution(g) Parts per billion (ppb) =

x 106

mass of solute(g)

mass of solution(g) Parts per million (ppm) =


8

The units ppm or ppb are used to express trace concentrations. These are weight or volume

based, rather than mole based.

To convert concentrations in mg/L (or ppm in dilute solution) to molarity, divide by the

molar mass of the solute to convert mass into a corresponding number of moles.

Example 6:

What is the molarity of a 6.2 ppm solution of O2(aq)?

Solution:

Molarity (M) = moles of solute

volume of solution (L)

Molarity (M) = mass of solute(g)

molar mass of solute(g/��) x volume of solution (L)

M of O2 solution = mass of O2(g)

molar mass of O2(g/��) x volume of solution (L)

6.2 x 0.001 g =

32 g/�� x 1 L = 1.94 x 10-4 M

To convert from molarity to mg/L (or ppm in dilute solution), multiply by the molar mass of

the solute to convert number of moles into corresponding mass.

The maximum acceptable concentration (MAC) of Pb in drinking water is 10 ppb.

If a sample has concentration of 5.5 x 10-8 M, does it exceed the MAC?

4) Normality (N)

is the number of equivalents of solute dissolved in one liter of solution.

Normality(N) = ��

�� (�)

Equivalents of solute = weight (W) of solute

equivalent weight (E.W.) of solute

Equivalent weight (E.W.) of solute = Molecular weight (M.W.) of solute K

Equivalent weight (E.W.) also known as gram equivalent

Equivalents of solute = W of solute

X K M.W. of solute

Where, K = equivalents per mole = M.W. of solute

E.W. of solute


Chapter 1 9

K is an integer constant ≥ 1

Normality (N) = W of solute

X K M.W. of solute x volume of solution(L)

Normality (N) = moles of solute

x K volume of solution(L)

Hence,

or;

K for a particular species is defined by the reaction type and the balanced chemical

reaction.

For acid/base reactions: K is the number of moles of H+ ions produced or neutralized per

mole of acid or base supplied.

For salts: K is the number of cations (or anions) x oxidation number of cation (or anion).

Thus,

Acid/base K M.W. E.W.

HCl 1 36.5 36.5

H2SO4 2 98 49.0

CaCO3 2 100 50.0

Na2CO3 2 106 53.0

NaOH 1 40 40.0

Al(OH)3 3 78.0 26.0

For oxidation/reduction reactions: K is the number of moles of e- transferred per mole of

oxidant or reductant in the balanced half-reaction.

Balanced half reaction

K

Fe3+ + 3e- → Fe 3

I2 + 2e- → 2 I- 2

2S2O32- - 2e- →

S4O 2-

6

1

MnO4 + 5e → Mn - - 2+ 5

N = M x K Normality = Molarity x K


10

Calculating normality (N)

Example 7:

When 25.0 mL of NaOH solution was titrated, 23.4 mL of 0.572 N H2SO4 were required to

reach the end point. Find the normality of the NaOH.

Solution:

Method 1:

Normality (N) = number of equivalents


number of equivalents = N x V(L)

number of equivalents of H2SO4 = 0.572 x 23.4 x 10-3 = 0.0133848

number of equivalents of analyte = number of equivalents of titrant

number of equivalents of NaOH = number of equivalents of H2SO4

number of equivalents of NaOH = 0.0133848

[N x V(L)] for NaOH = 0.0133848

N x 25 x 10-3 = 0.0133848

N of NaOH = 0.535392 g.equiv./L

Method 2:

H2SO4 + 2NaOH → Na2SO4 + 2H2O

N = M x K

For H2SO4, 0.573 = M x 2

M of H2SO4 = 0.286 mol/L

number of moles M =


number of moles = M x V(L)

number of moles of H2SO4 = 0.286 x 23.4 x 10-3 = 0.0066924 mol

from the neutralization reaction equation;

1mol of H2SO4 = 2 mol of NaOH

0.0066924 mol of H2SO4 = X mol of NaOH


Chapter 1 11

X(number of moles of NaOH) = 0.0133848

M of NaOH =

N = M x K

0.0133848mol

25 x 0.001L = 0.535392 mol/L

For NaOH, N = 0.535392 g.equiv./L

What is the weight of Na2CO3 which required to prepare 500 mL of a solution (0.2

N).

General concept of acids and bases.

Acids and bases are among the most familiar and important of all chemical compounds. You

encounter them each and everyday. Vinegar- acteic acid, lemon juice - citric acid, milk of

magnesium - magnesium hydroxide, there is even HCl in your stomach to digest food.

In the early days of chemistry chemists were organizing physical and chemical properties of

substances. They discovered that many substances could be placed in two different

property categories:

The actual cause of acidity and basicity was ultimately explained in terms of the effect

these compounds have on water by Arrhenius in 1884. Arrhenius was the first person to

suggest a reason why substances are in A or B due to their ionization in water.

Arrhenius concept of acids and bases

The Swedish chemist Svante Arrhenius proposed the first definition of acids and bases.

According to the Arrhenius model, substances A and B became known as acids and bases.

Substance A

1. Sour taste

2. Reacts with carbonates to

make CO2

3. Reacts with metals to

produce H2

4. Turns blue litmus pink

5. Reacts with B substances to

make salt and water

Substance B

1. Bitter taste

2. Reacts with fats to make

soaps

3. Do not react with metals

4. Turns red litmus blue

5. Reacts with A substances to

make salt and water


12

An acid is a substance that, when dissolved in water, increases the concentration of

hydronium ion H3O+ (produces H+ ions).

HCl + H2O → H3O+ + Cl-

HCl → H+ + Cl-

Remember that, free H+ ions do not exist in water. They covalently react with water to

produce hydronium ions, H3O+.

or:

H+(aq) + H2O(l) →’ H3O+(aq)

This new bond is called a coordinate covalent bond since both new bonding electrons come

from the same atom. Hydronium ion is the name for H3O+ and is often times abbreviated as

H+(aq) they both mean the same thing.

The H3O+ is shown here bonded to three water molecules.

A hydronium ion (H3O

+)

is the ion that forms when a water molecule gains a hydrogen ion.


Chapter 1 13

4

4 4

Acids vary in the number of hydrogens they contain that can form hydrogen ions.

Some Common Acids

Name Formula

Hydrochloric acid HCl

Nitric acid HNO3

Sulfuric acid H2SO4

Phosphoric acid H3PO4

Ethanoic acid CH3COOH

Carbonic acid H2CO3

Monoprotic and polyprotic acids

A hydrogen atom that can form a hydrogen ion is described as ionizable.

Monoprotic acids

Acids that contain one ionizable hydrogen, such as nitric acid (HNO3), are called monoprotic

acids. The prefix mono- means ―one,‖ and the stem protic reflects the fact that a hydrogen

ion is a proton.

Polyprotic acids

Acids with more than one ionizable hydrogen ion

Acids that contain two ionizable hydrogens, such as sulfuric acid (H2SO4), are called

diprotic acids.

Acids that contain three ionizable hydrogens, such as phosphoric acid (H3PO4), are

called triprotic acids.

Their hydrogen ionizes in stages, for example, phosphoric acid (H3PO4) ionizes in three

stages

H3PO4(aq) + H2O ⇄ H3O+(aq) + H2PO -(aq)

H2PO -(aq) + H2O ⇄ H3O+(aq) + HPO 2-(aq)

HPO42-(aq) + H2O ⇄ H3O

+(aq) + PO43-(aq)


14

Not all compounds that contain hydrogen are acids. Only a hydrogen that is bonded to a very

electronegative element can be released as an ion. Such bonds are highly polar.

When a compound that contains such bonds dissolves in water, it releases hydrogen ions.

Methane (CH4) is an example of a hydrogen-containing compound that is not an acid.

The four hydrogen atoms in methane are attached to the central carbon atom by

weakly polar C—H bonds.

Methane has no ionizable hydrogens and is not an acid.

Ethanoic acid (CH3COOH), which is commonly called acetic acid, is an example of a molecule

that contains both hydrogens that do not ionize and a hydrogen that does ionize.

Although its molecules contain four hydrogens, ethanoic acid is a monoprotic acid.

The three hydrogens attached to a carbon atom are in weakly polar bonds, they do

not ionize. Only the hydrogen bonded to the highly electronegative oxygen can be

ionized.

A base, in the Arrhenius concept, is a substance that, when dissolved in water, increases the

concentration of hydroxide ion, OH-(aq) (produces OH- ions).


Chapter 1 15

Some Common Bases

Name Formula Solubility in

Water

Sodium hydroxide NaOH High

Potassium

hydroxide KOH High

Calcium hydroxide Ca(OH)2 Very low

Magnesium

hydroxide Mg(OH)2 Very low

Sodium hydroxide (NaOH) is a base known as lye. Sodium hydroxide is extremely

caustic. A caustic substance can burn or eat away materials with which it comes in

contact. This property is the reason that sodium hydroxide is a major component of

products that are used to clean clogged drains.

Potassium hydroxide (KOH) is another base. It dissociates to produce potassium ions

and hydroxide ions in aqueous solution.

Sodium hydroxide and potassium hydroxide are very soluble in water. The solutions

would typically have the bitter taste and slippery feel of a base, but you would not

want to test these properties. The solutions are extremely caustic to the skin. They

can cause deep, painful, slow-healing wounds if not immediately washed off.

→

Visitors to Bracken Cave wear

protective gear to keep ammonia gas out

of their eyes and respiratory tracts.

Think about the properties of bases.

Why are high levels of ammonia

harmful?

Ammonia is a

base, and bases

are caustic in high

concentrations.


16

Calcium hydroxide, Ca(OH)2, and magnesium hydroxide,

Mg(OH)2, are not very soluble in water. Their solutions

are always very dilute, even when saturated. The low

solubility of magnesium hydroxide makes the suspension

safe to consume. Some people use this suspension as an

antacid.

The Arrhenius concept is limited in that it looks at acids and bases in aqueous solutions only

involving H+ and OH-. There are many substances with acid/base properties that do not

involve these and cannot be classified by this theory. In addition, it singles out the OH- ion

as the source of base character, when other species can play a similar role. Sodium

carbonate (Na2CO3) and ammonia (NH3) act as bases when they form aqueous solutions.

Neither of these compounds is a hydroxide-containing compound, so neither would be

classified as a base by the Arrhenius definition.

There are broader definitions of acids and bases which we will cover.

Brønsted Lowry concept of acids and bases

Johannes Brønsted and Thomas Lowry revised Arrhenius‘s acid-base theory to include this

behavior. They defined acids and bases as follows:

An acid is a hydrogen containing species that donates a proton (a proton donor) in a

reaction.

A base is any substance that accepts a proton (a proton acceptor) in a reaction.

notice this theory is only looking at proton not even looking at hydroxide. In any reversible

acid-base reaction, both forward and reverse reactions involve proton transfer.

You can use the Brønsted-Lowry theory to understand why ammonia is a base.

When ammonia dissolves in water, hydrogen ions (H+ ions) are transferred from water to

ammonia to form ammonium ions and hydroxide ions.

Ammonia is a Brønsted-Lowry base because it accepts hydrogen ions (protons). Water is a

Brønsted-Lowry acid because it donates protons.


Chapter 1 17

4

4 3 2

4

4 3

When the temperature of an aqueous solution of ammonia is increased, ammonia gas is

released.

NH + reacts with OH– to form more NH and H O.

In the reverse reaction, ammonium ions donate hydrogen ions to hydroxide ions.

NH + (the proton donor) acts as a Brønsted-Lowry acid, and OH− (the proton acceptor) acts

as a Brønsted-Lowry base.

In essence, the reversible reaction of ammonia and water has two acids and two bases.

A conjugate acid is the ion or molecule formed when a base gains a proton.

NH + is the conjugate acid of the base NH .

A conjugate base is the ion or molecule that remains after an acid loses a hydrogen ion.

OH– is the conjugate base of the acid H2O.

Conjugate acids are always paired with a base, and conjugate bases are always paired with

an acid.

A conjugate acid-base pair consists of two ions or molecules related by the loss or gain of

one proton.

The ammonia molecule (NH3) and the ammonium ion (NH +) are a conjugate acid-

base pair.

The water molecule (H2O) and the hydroxide ion (OH-) are also a conjugate acid-base

pair.


18

A second example: In the following reaction:

hydrogen chloride is the proton donor and is by definition a Brønsted-Lowry acid. Water is

the proton acceptor and a Brønsted-Lowry base.

The chloride ion is the conjugate base of the acid HCl. The hydronium ion is the conjugate

acid of the water base.

The figure below shows the reaction that takes place when sulfuric acid dissolves in water.

The products are hydronium ions and hydrogen sulfate ions. Use the figure to identify the

two conjugate acid-base pairs.

Some Conjugate Acid-Base Pairs

Acid Base

HCl Cl– Amphoteric Substances

Note that water appears in both the list of acids

and the list of bases.

Sometimes water accepts a hydrogen ion.

At other times, it donates a hydrogen ion.

How water behaves depends on the other

reactant.

H2SO4

–

HSO4

H3O+ H2O

HSO – 4 SO 2–

4

CH3COOH CH3COO–

H2CO3

−

HCO3

–

HCO3

2–

CO3

+ NH4 NH3


Chapter 1 19

H2O OH–


20

An amphoteric substance is a substance that can act as either an acid or a base (it can gain

or lose a proton).

Water is amphoteric:

In the reaction with hydrochloric acid, water accepts a proton and is therefore a

base.

In the reaction with ammonia, water donates a proton and is therefore an acid.

Lewis concept of acids and bases

According to Gilbert Lewis,

An acid is an electron pair acceptor and a base is an electron pair donor during a reaction.

This definition is more general than those offered by Arrhenius or by Brønsted and Lowry.

This concept broadened the scope of acid-base theory to include reactions that did not

involve H+.

The Lewis concept embraces many reactions that we might not think of as acid-base

reactions.

Consider the reaction of H+ and OH–:

The hydrogen ion donates itself to the hydroxide ion.

H+ is a Brønsted-Lowry acid, and OH− is a Brønsted-Lowry base.

The hydroxide ion can bond to the hydrogen ion because it has an unshared pair of

electrons.

OH− is also a Lewis base, and H+, which accepts the pair of electrons, is a Lewis acid.

A second example: of a reaction between a Lewis acid and a Lewis base is what happens

when ammonia dissolves in water.


Chapter 1 21

Hydrogen ions from the dissociation of water are the electron-pair acceptor and the

Lewis acid.

Ammonia is the electron-pair donor and the Lewis base.

A third example: of a reaction between a Lewis acid and a Lewis base is the reaction

between ammonia and boron trifluoride.

NH3 + BF3 → NH3BF3

Ammonia has an unshared pair of electrons to donate, so ammonia is the Lewis base.

The boron atom can accept the donated electrons, so boron trifluoride is the Lewis

acid.

The following table compares the definitions of acids and bases.

Acid-Base Definitions

Type Acid Base

Arrhenius H+ producer OH– producer

Brønsted-

Lowry H+ donor H+ acceptor

Lewis electron-pair

acceptor

electron-pair

donor

Strengths of acids and bases

Electrolytes and nonelectrolytes

An electrolyte is a substance that dissociates into ions when dissolved in water.

A nonelectrolyte may dissolve in water, but it does not dissociate into ions when it

does so.


22

Soluble ionic compounds tend to be electrolytes. Molecular compounds tend to be

nonelectrolytes, except for acids and bases.

A strong electrolyte dissociates completely when dissolved in water.

A weak electrolyte only dissociates partially when dissolved in water.

Acids and bases are classified as strong or weak based on the degree to which they ionize in

water.

A strong acid is an acid that ionizes completely (100% ionized) in aqueous solution to give

H3O+(aq) and an anion.

A single arrow → is used to represent the ionization of a strong acid.

For example:

HCl(aq) + H2O H3O+(aq) + Cl-(aq)

H2SO4(aq) + H2O H3O+(aq) + HSO4-(aq)

Complete ionization of hydrochloric acid


Chapter 1 23

A strong base is a base that dissociates completely (100% ionized) in aqueous solution to

give OH-(aq) and a cation.

For example:

NaOH(aq) Na+(aq) + OH-(aq)

Ba(OH)2(aq) Ba2+(aq) + 2 OH-(aq)

According to Brønsted Lowry concept of acids and bases

The stronger acids are those that lose their hydrogen ions more easily than other

acids; donate proton quicker. From this point of view, we can order acids by their

relative strength as hydrogen ion donors.

Similarly, the stronger bases are those that hold onto hydrogen ions more strongly

than other bases; harder to lose proton.

Easier give up proton, stronger the acid and harder to give up proton, stronger the

base.

Rest of acids and bases that you encounter are weak. They are not completely ionized and

exist in reversible reaction with the corresponding ions.

Weak acid is an acid that ionizes only partially in aqueous solution.

Double arrows ⇌ (Equilibrium) are used to represent weak acids.


24

For example:

Partial ionization of acetic acid

HF(aq) + H2O ⇌ H3O+(aq) + F-(aq)

HOCl(aq) + H2O ⇌ H3O+(aq) + ClO-(aq)

Note: At any given time only a small portion of the acid molecules are ionized and since

reactions are running in both directions the mixture composition stays the same (an

equilbrium).

Weak base is a base that ionizes only partially in aqueous solution:

For example:

NH3(aq) + H2O NH4+(aq) + OH-(aq)

PO43-(aq) + H2O HPO4

2-(aq) + OH-(aq)

Comparison between the extent of the dissociation of a strong acid and a weak acid.

Dissociation of an acid (HA) in water yields H3O+ and an anion, A–.

The extent of dissociation of strong acids (As a result, [H3O+] is high in an aqueous solution of strong acid)

Chapter 1 25


The extent of dissociation of weak acids (By contrast, weak acids remain largely undissociated)

Relative strength of acids and their conjugate bases

It is logical to assume that if an acid is considered strong, its conjugate base (that is,

its anion) would be weak, since it is unlikely to accept a proton. It wants to donate

proton as soon as it accepts a proton.

In other words, the stronger the acid, the weaker the conjugate base (can't be

strong acid and strong conjugate base). Which gets us to the following statement:

26


The stronger the conjugate acid is an acid, the weaker its conjugate base is a base.

The stronger the conj base is a base, the weaker its conjugate acid is an acid.

Strong acids lose protons very readily → weak conjugate bases; weak acids do not

lose protons very readily → strong conjugate bases.

HCl + H2O → H3O+ + Cl-

strong acid extremely weak base,

non-existent

HCN + H2O ↔ H3O+ + CN-

weak acid stronger base than Cl-

HCl stronger acid than HCN; therefore, CN- stronger base than Cl-

Relative strength of acids and their conjugate bases

Chapter 1 27


a

Acid- base equilibria in water

Strong acids and strong bases completely dissociate in water. Therefore they do not

create an equilibrium system.

Weak acids and weak bases disassociate to a far lesser degree in an aqueous

solution. They will create an equilibrium system. They have their own dissociation

constants.

The acid and base dissociation (ionization) constant, Ka & Kb

Acid dissociation constant (Ka)

Dissociation of an acid (HA) in water yields H3O+ and an anion, A–. You can use a balanced

equation to write the equilibrium-constant expression for a reaction.

The equilibrium constant (KC):

Recall that the concentration of water is constant in dilute solutions. This constant can be

combined with the KC for the acid to give the dissociation constant (Ka) for the acid.

The acid dissociation constant, Ka: is the ratio of the concentration of the dissociated form

of an acid to the concentration of the undissociated form (HA). The dissociated form

includes both the H3O+ and the anion, A-.

The acid dissociation constant (Ka) reflects the fraction of an acid that is ionized. For this

reason, dissociation constants are sometimes called ionization constants.

For example:

K = [H3O+][CH3COO-]

[H2O]

28


Dissociation constants (Ka) values for some monoprotic acids at 25 oC

[Acid strength decreases down the table (smaller Ka values)]

pKa is another expression for the acid dissociation constant

pKa = -log Ka

A low pKa corresponds to a high Ka.

stronger acid has larger Ka or smaller pKa.

weaker acid has smaller Ka or larger pKa.

Chapter 1 29


The relation between Ka and pKa for some acids

Dissociation constants of polyprotic acids

Some acids have more than one dissociation constant because they have more than one

ionizable hydrogen.

For example:

Oxalic acid is a diprotic acid. It loses two hydrogens, one at a time. Therefore, it has

two dissociation constants.

Phosphoric acid is a triprotic acid. It loses three hydrogens, one at a time.

Therefore, it has three dissociation constants.

Dissociation constants (Ka) values for Oxalic acid and Phosphoric acid at 25 oC

Observe what happens to the Ka with each ionization.

The Ka decreases from first ionization to second.

It decreases again from second ionization to third.

Base dissociation constant (Kb)

Dissolve a base (B) in water yields H3O+ and a cation, BH+

30


4

4

The base dissociation constant (Kb) is the ratio of the concentration of the conjugate acid

(BH+) times the concentration of the hydroxide ion (OH-) to the concentration of the base

(B).

For example:

When equilibrium is established, only about 1% of the ammonia is present as NH +. This ion is

the conjugate acid of NH3. The concentrations of NH + and OH− are low and equal.

The equilibrium-constant expression for the dissociation of ammonia in water is as follows:

Recall that the concentration of water is constant in dilute solutions. This constant can be

combined with the KC for ammonia to give a base dissociation constant (Kb) for ammonia.

Dissociation constants (Kb) values for some weak bases at 25 oC

Ionization equilibrium Kb at 25 oC

Similarly, pKb is another expression for the base dissociation constant

pKb = -log Kb

Chapter 1 31


A low pKb corresponds to a high Kb.

stronger base has larger Kb or smaller pKb.

weaker base has smaller Kb or larger pKb.

Acid and base dissociation constants are the measure of the strengths of acids and bases.

The larger the Ka/Kb value the stronger the acid or base.

The values of Ka indicate the relative strength of the acids. Strong acids have very

large Ka, while weak acids have small Ka‘s (Ka << 1).

For example: Nitrous acid (HNO2) has a Ka of 4.4 × 10−4, but acetic acid (CH3COOH) has a Ka

of 1.8 × 10−5.

This means that nitrous acid is more ionized in solution than acetic acid. Nitrous acid is a

stronger acid than acetic acid.

The values of Kb indicate the relative strength of the bases. The smaller the value of

Kb, the weaker the base. Weak bases have small Kb‘s (Kb < 1).

The magnitude of Kb indicates the ability of a weak base to compete with the very

strong base OH– for hydrogen ions. Because bases such as ammonia are weak relative

to the hydroxide ion, the Kb for such a base is usually small. The Kb for ammonia is

1.8 × 10−5.

Concentration Versus Strength

Sometimes people confuse the concepts of concentration and strength.

The words concentrated and dilute indicate how much of an acid or base is dissolved

in solution. These terms refer to the number of moles of the acid or base in a given

volume.

The words strong and weak refer to the extent of ionization or dissociation of an

acid or base.

The table below shows four possible combinations of concentration and strength for acids.

32


The gastric juice in your stomach is a dilute solution of HCl.

The relatively small number of HCl molecules in a given volume of gastric juice are

all dissociated into ions.

Even when concentrated hydrochloric acid is diluted with water, it is still a strong

acid.

Conversely, acetic acid is a weak acid because it ionizes only slightly in solution.

Vinegar is a dilute solution of acetic acid.

Even at a high concentration, acetic acid is still a weak acid.

The same concepts apply to bases.

A solution of ammonia can be either dilute or concentrated. However, in any solution of

ammonia, the relative amount of ionization will be small.

Thus, ammonia is a weak base at any concentration.

Likewise, sodium hydroxide is a strong base at any concentration.

Calculating dissociation constants, Ka & Kb

To calculate the acid dissociation constant (Ka) of a weak acid, you need to know the

initial molar concentration of the acid and the [H+] (or alternatively, the pH) of the

solution at equilibrium.

You can use these data to find the equilibrium concentrations of the acid and the

ions.

These values are then substituted into the expression for Ka.

Example 8:

In a 0.1M solution of acetic acid, [H3O+] = 1.34 × 10−3M. Calculate the Ka of this acid.

Solution:

CH3COOH(aq) + H2O(l) ⇄ H3O+(aq) + CH3COO-(aq)

Each molecule of CH3COOH that ionizes gives an H+ ion and a CH3COO– ion.

Determine the equilibrium concentrations of each component:

Chapter 1 33


a

[H3O+] = [CH3COO−] = 1.34 × 10−3M

[CH3COOH ] = (0.1 – 0.00134)M = 0.0987M

Concentration [CH3COOH] [ H3O+] [CH3COO−]

Initial 0.1000 M 0 0

Change −1.34 × 10−3

M

1.34 × 10−3

M

1.34 × 10−3

M

Equilibrium 0.0987 M 1.34 × 10−3

M

1.34 × 10−3

M

K = [H3O+][CH3COO-]

[H2O]

Percent ionization

Percent ionization is the fraction of acid molecules that dissociate compared with the initial

concentration of the acid.

Percent ionization (%ionization) =

��

��

��

� ��%

For a monoprotic acid HA:

Percent ionization (%ionization) =

[H3O+]

� ��% [HA]0

Where, [HA]0 = initial concentration of the acid

34


For the previous example:

Percent ionization of strong and weak acids

[H3O+]

%ionization of CH3COOH acid = [CH3COOH]0

x 100%

1.34 × 10-3 M =

�.� � x 100%

= 1.34%

Percent ionization of weak acid increases as the dilution increases.

The more we dilute the acid solution, the higher the fraction of the acid that will ionize,

which increases the degree of ionization.

For example, if [CH3COOH]0 = 0.0100 M and [H3O+] = 4.24 x 10-4 M

4.24 x 10-4

Percent ionization 0.0100

x 100 4.24%

At [CH3COOH]0 = 0.00100 M, and [H3O+] = 1.34 x 10-4 M

1.34 x 10

-4

Percent ionization 0.00100

x 100 13.4%

Ka, for a weak acid, HA, at 25 °C is 2.2 x 10-4.a) Calculate [H3O

+] of a 0.20 M solution of HA.

b) Calculate the percent ionization of HA.

Chapter 1 35


Autoionization (Self-ionization) of water

In pure water (no solute) water molecules behave as both an acid and base!!

It is called amphoteric meaning it will act as either an acid or a base depending on whether

a base or acid is present in solution.

In pure water, the water molecules spontaneously react with each other as shown:

Water ionizes to produce both H3O

+ and OH-, thus it has both acid and base properties

(amphoteric).

A water molecule that gains a hydrogen ion becomes a hydronium ion (H3O+).

A water molecule that loses a hydrogen ion becomes a hydroxide ion (OH−).

This equilibrium lies very much to the left. i.e. mostly water.

This reaction is called the autoionization (self-ionization) of water.

The ion product constant of water, Kw

H2O(l) + H2O(l) ⇌ H3O+(aq) + OH-(aq)

The equilibrium-constant expression for this system is:

[H O ][OH

]

Kc 3

2

[H2O]

The concentration of ions is extremely small (equilibrium lies to far left), so the

concentration of H2O remains essentially constant. This gives:

Kc[H2O]2 = constant = [H3O+][OH-]

36


The equilibrium value for the ion product [H3O+][OH-] is called the ion-product constant for

water, which is written Kw.

Kw = [H3O+][OH-]

Because we often write H3O+ as H+, the ion-product constant expression for water can be

written:

Kw = [H+][OH-]

At 25 oC, the value of Kw is 1.0 x 10-14 M2. which means there are such small amounts of ions

in pure water that water will not conduct electricity.

Kw = [H+][OH-] = 1.0 x 10-14 M2 at 25 oC

The ion-product constant (Kw)

is the product of the molar concentrations of H+ and OH- ions at a particular temperature.

For pure water at 25oC: [H+] = [OH-] = 1.0 x 10-7 M

If you add acid or base to water they are no longer equal but the Kw expression still holds. A

change in [H3O+] causes an inverse change in [OH-]. One goes up other goes down to keep

product equal to Kw.

By dissolving substances in water, you can alter the concentrations of H+(aq) and OH-(aq).

In a neutral solution, the concentrations of H+(aq) and OH-(aq) are equal, as they

are in pure water.

[H+] = [OH-] = 1.0 x 10-7 M

In an acidic solution, the concentration of H+(aq) is greater than that of OH-(aq).

[H+] > 1.0 x 10-7 M > [OH-]

In a basic solution, the concentration of OH-(aq) is greater than that of H+(aq).

[H+] < 1.0 x 10-7 M < [OH-]

Chapter 1 37


Comparison between the concentrations of H+(aq) and OH-(aq) in acidic, basic and neutral solutions

Example 9:

A research chemist adds a measured amount of HCl gas to pure water at 25 oC and obtains a

solution with [H3O+] = 3.0 x 10-4 M. Calculate [OH-]. Is the solution neutral, acidic or basic?

Solution:

Kw = 1.0 x 10-14 M2 = [H3O+][OH-]

[OH-] = Kw/[H3O+] = 1.0 x 10-14 M2/3.0 x 10-4 M = 3.3 x 10-11 M

[H3O+] > [OH-]; the solution is acidic.

If the [H+] in a solution is 1.0 × 10−5M, is the solution acidic, basic, or neutral?

What is the [OH−] of this solution?

The pH and the pH scale

Although you can quantitatively describe the acidity of a solution by its [H+], it is often more

convenient to give acidity in terms of pH.

The pH of a solution is defined as the negative logarithm of the molar hydrogen-ion

concentration.

pH = -log[H+]

For a solution in which the hydrogen-ion concentration is 1.0 x 10-3 M, the pH is:

pH = -log(1.0 x 10-3 M) = 3.00

What is the pH of a solution with a hydrogen-ion concentration of 4.2 × 10−10M?

The pH of an unknown solution is 6.35. What is the hydrogen-ion concentration of

the solution?

38


pH scale

pH gives values that ranges from 0 to 14 that is called pH scale. The pH scale is a way of

expressing the strength of acids and bases.

Note that as [H3O+] increase, the pH-value will decrease and vice versa.

In a neutral solution, whose hydrogen-ion concentration is 1.0 x 10-7 M, the pH =

7.00.

For acidic solutions, the hydrogen-ion concentration is greater than 1.0 x 10-7 M, so

the pH < 7.00.

For basic solutions the hydrogen-ion concentration is smaller than 1.0 x 10-7 M, so

the pH > 7.00.

The pH scale

Similarly, pOH = -log[OH-]

For water, Kw = [H3O+][OH-] = 1.0 x 10-14

-log(Kw) = -log [H3O+] + (-log[OH-])

pKw = pH + pOH = 14.00

At 25oC, pOH = 14 – pH

Measuring the pH

The pH of a solution can be measured electronically using a pH meter. It can also be

measured visually using pH paper, which is embedded with indicators that change color

based on the pH of a solution.

Chapter 1 39


Calculating the pH

pH of strong acids and bases

Strong acids like HCl, HClO4 and H2SO4 ionize completely in aqueous solution:

HCl(aq) + H2O H3O+(aq) + Cl-(aq)

HClO4(aq) + H2O H3O+(aq) + ClO4

-(aq)

H2SO4(aq) + H2O 2H3O+(aq) + SO4

2-(aq)

In solutions of strong monoprotic acids HA, such as HCl and HClO4,

[H3O+] = [HA]0

In solutions of strong diprotic acids H2A, such as H2SO4,

[H3O+] = 2 x [HA]0

Where, [HA]0 is the initial concentration of the acid

For example,

in 0.10 M HCl, [H3O+] = [HCl]0 = 0.10 M, and pH = -log(0.10) = 1.00,

in 0.050 M H2SO4, [H3O+] = 2 x [H2SO4]0 = 2 x 0.050 M = 0.10, and pH = -log(0.1) = 1.00

Like strong acids, strong bases also ionize completely in aqueous solution:

NaOH(aq) Na+(aq) + OH-(aq)

Ba(OH)2(aq) Ba2+(aq) + 2OH-(aq)

In a base solution such as 0.050 M NaOH,

[OH-] = [NaOH]0 = 0.050 M;

40


Where, [NaOH]0 is the initial concentration of the NaOH

pOH = -log(0.050) = 1.30; pH = 14.00 - 1.30 = 12.70

In a base solution such as 0.050 M Ba(OH)2,

[OH-] = 2 x [Ba(OH)2]0 = 2 x 0.050 M = 0.10 M;

Where, [Ba(OH)2]0 is the initial concentration of the Ba(OH)2

pOH = -log(0.10) = 1.00; pH = 14.00 - 1.00 = 13.00

A sample of orange juice has a hydrogen-ion concentration of 2.9 x 10-4 M. What

is the pH?

The pH of human arterial blood is 7.40. What is the hydrogen-ion concentration?

An ammonia solution has a hydroxide-ion concentration of 1.9 x 10-3 M. What is

the pH of the solution?

Calculate the pH, and pOH in solution prepared by dissolving 10.0g of Ba(OH)2

per liter.

pH of weak acids and bases

In weak acid solutions, [H3O+] < [HA]0;

[H3O+] and pH can be calculated from the initial concentration of the acid and its Ka value.

Example 10:

What is the pH of a 0.100 M acetic acid, CH3COOH, with Ka = 1.8 x 10-5?

Solution:

CH3COOH(aq) + H2 O H3O+(aq) + CH3COO-(aq)

Concentration

Initial

Change

Equilibrium

0.100 M

- x M

(0.100 - x M)

0.00

+x M

x M

0.00

+x M

x M

[H O

][CH CO ] x2

K 3 3 2 1.8 x 10-5

a [CH COOH]

3 (0.100 - x)

Chapter 1 41


a 0

3

a

[H O

][CH CO ] x2

Ka 3 3 2 [CH CO H]

(0.100 - x) 1.8 x 10-5

3 2

K [HA] (1.8 x 10-5 0.100) x 0.100,

and

(0.10 - x) ~ 0.10. This

makes x2

~ (0.100 - x)

x2

0.100 1.8 x 10-5 ;

x2 (0.100)(1.8 x 10-5 ) 1.8 x 10-6 ; and x 1.34 x 10-3 ;

[H O ] x 1.34 x 10-3 M; pH - log(1.34 x 10-3 ) 2.873

Example 11:

The pH of a 0.100 M monoprotic acid HA is 1.28, calculate the Ka for the acid.

Solution:

pH = 1.28

[H+] = 10-1.28

[H+] = 0.05248 M

At equilibrium:

[H+] = [A-] = 0.05248 M

[HA] = 0.100 M – 0.05248 M = 0.04752 M

K = [H+][A-]

[HA] =

(0.05248 )(0.05248 ) = 5.8 x 10-2

(0.04752 )

Similarly, In a weak base, [OH-] < [Base]0

Where, [Base]0 is the initial concentration of the base

[OH-] and pH can be calculated from the initial concentration of the base and its Kb value.

Example 12:

What is the pH of a 0.100 M ammonia, NH3, with Kb = 1.8 x 10-5?

1.8 x 10-6

42


b 0

4

4

Solution:

Concentration

NH3(aq) + H2O(l) NH4

+(aq) + OH-(aq)

Initial 0.100 M 0.00 0.00

Change - x M +x M +x M

Equilibrium (0.100 - x M) x M x M

Kb [NH

][OH

- ]

4 [NH3 ]

x2

(0.100 x) 1.8 x 10

5

Kb [NH

][OH

- ]

4 [NH3 ]

x2

(0.100 x) 1.8 x 10

5

K [B] (1.8 x 10-5 0.100) x 0.100, and

(0.100 - x) ~ 0.100, which

makes x2

~ (0.100 - x)

x2

0.100 1.8 x 10

-5 ;

x2 (0.100)(1.8 x 10

-5 ) 1.8 x 10

-6 , which yields x 1.34 x 10

-3 ;

[OH- ] 1.34 x 10

-3 M pOH 2.873, and pH 11.127

What is the pH of a 0.122 M monoprotic acid whose Ka is 5.7 x 10-4?

Nitrous acid, HNO2, has Ka = 4.0 x 10-4 at 25 oC. Calculate the pH and percent

ionization of HNO2 in 0.10 M solution of the acid.

If the pH of 0.40 M NH3 at 25 oC is 11.427, calculate the Kb.

Calculate the concentrations of H2SO4, H3O

+, HSO -, and SO42-, in 0.10 M H2SO4

solution. What is the pH of the solution? (H2SO4 is a strong acid and HSO - has Ka =

1.2 x 10-2)

Hydrolysis of salts

A salt is one of the products of a neutralization reaction. A salt consists of an anion from an

acid and a cation from a base.

All salts are strong electrolytes – this means that they are fully ionized in dilute

aqueous solution.

If the salts are formed from either weak acids or bases, then their ions may react

with water. These interactions between salts and water are called hydrolysis.

Chapter 1 43


In salt hydrolysis, the cations or anions of a dissociated salt remove hydrogen ions

from, or donate hydrogen ions to, water. Salts that produce acidic solutions have

positive ions that release hydrogen ions to water. Salts that produce basic solutions

have negative ions that attract hydrogen ions from water and produce OH- ions.

Types of salts and the acid-base properties of their solutions

Salts of strong acid-strong base reactions.

Such as NaCl, NaNO3, KBr; solutions are neutral.

Salts of weak acid-strong base reactions.

Such as NaF, NaNO2, CH3COONa; solutions are basic.

Salts of strong acid-weak base reactions.

Such as NH4Cl, NH4NO3; solutions of these salts are acidic.

Salts of weak acid-weak base reactions.

Such as CH3COONH4, NH4CN, NH4NO2; solutions of these salts can be acidic, basic, or

neutral, which depends on the relative strength of the acid and the base.

Salts of strong acid-strong base reactions

For example: NaCl

NaCl(s) + H2O(l) Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq)

The concentrations of H+(aq) and OH-(aq) in NaCl solution are the same as in pure water →

solution is neutral (pH = 7).

Salts of weak acid-strong base reactions

For example: CH3COONa

In solution, the salt is completely ionized.

CH3COONa(aq) → CH3COO−(aq) + Na+(aq)

The CH3COO− ion is a Brønsted-Lowry base, which means it is a proton acceptor. It reacts

with water to form acetic acid and hydroxide ions. At equilibrium, the reactants are

favored.

Kb of CH3 [CH3COOH][OH-]

[CH3COO-] = 5.6 x 10

-10 COO- =

44


4

]

This process is called hydrolysis because a hydrogen ion is split off a water molecule.

In the solution of CH3COONa, the hydroxide-ion concentration is greater than the hydrogen-

ion concentration, [OH-] > [H+]. Thus, the solution becomes basic.

Salts of strong acid-weak base reactions

For example: NH4Cl

It is completely ionized in solution.

NH4Cl(aq) → NH4+(aq) + Cl−(aq)

The NH + ion is a Bronsted-Lowry acid and donate a hydrogen ion to a water molecule. The

products are ammonia molecules and hydronium ions. The reactants are favored at

equilibrium.

K of NH + = [NH3][H3O+]

-10

a 4 [NH +

= 5.6 x 10 4

This process is another example of hydrolysis. At equilibrium the [H3O+] is greater than the

[OH–]. Thus, a solution of ammonium chloride is acidic.

Salts of weak acid-weak base reactions.

Salts produced by reactions of weak acids and weak bases can be neutral, acidic, or basic,

depending on the relative magnitude of the Ka of the weak acid and the Kb of the weak base.

If Ka ~ Kb, salts are neutral; example: CH3COONH4

If Ka > Kb, salts are acidic; example: NH4NO2

If Ka < Kb, salts are basic; example: NH4CN

Predicting Acid-Base Property of Salts

To determine if a salt will form an acidic or basic solution, remember the following rules:

When the negative ion is from a weak acid then the salt is basic by hydrolysis.

When the positive ion is from a weak base then the salt is acidic by hydrolysis.

If the salt is formed from a strong acid and strong base then it is neutral.

Chapter 1 45


4 4

4 2 4 b

If the salt is formed from a weak acid and weak base then its hydrolysis is

determined by the relative Ka and Kb values:

a) Consider a solution containing CH3COONH4

CH3COONH4(aq) NH4+(aq) + CH3COO-(aq)

NH4+(aq) + H2O H3O

+(aq) + NH3(aq); Ka = 5.6 x 10-10

CH3COO-(aq) + H2O CH3COOH(aq) + OH-(aq); Kb = 5.6 x 10-10

Ka = Kb = 5.6 x 10-10 CH3COONH4 is neutral.

b) Consider a solution containing (NH4)2SO4

(NH4)2SO4(aq) 2NH +(aq) + SO 2-(aq)

NH4+(aq) + H2O H3O

+(aq) + NH3(aq); Ka = 5.6 x 10-10

SO 2-(aq) + H O HSO -(aq) + OH-(aq); K = 8.3 x 10-13

Ka > Kb (NH4)2SO4 is acidic.

c) Consider a solution containing NH4CN

NH4CN(aq) NH4+(aq) + CN-(aq)

NH4+(aq) + H2O H3O

+(aq) + NH3(aq); Ka = 5.6 x 10-10

CN-(aq) + H2O HCN(aq) + OH-(aq); Kb = 1.6 x 10-5

Kb > Ka NH4CN is basic.

Common ion effect

When a compound containing an ion in common with an already dissolved substance is added

to a solution at equilibrium, the equilibrium shifts to the left. This phenomenon is known as

the common ion effect.

The presence of a common ion suppresses the ionization of a weak acid or a weak base.

Example 13:

consider 1.0 L of 0.10 M solution of CH3COOH

Add 0.050 mol of CH3COONa

46


a

a

Effect:

Effect of common ion on equilibrium calculations

Without the common ion:

K 1.8x10 5

[CH COO ][H

] x

2 x

2 5

Ka 3

[CH3COOH]

0.10 M x

0.10 M 1.8x10

[H ] 1.34x10

3

pH log(1.34x103

) 2.87

With the common ion

K 1.8x10 5

[CH COO ][H ] Ka

3 [CH3COOH]

(x)(0.050 M x)

0.10 M x

(x)(0.050 M)

0.10 M

1.8x105

Chapter 1 47


[H ] 3.6 x 10

5

pH log( 3.6 x 105

) 4.44

2.87 4.44

So, when 0.050 mol of CH3COO- (common ion) is added (from CH3COONa) to 1.0 L of 0.10 M

solution of CH3COOH, [H+] decreases from 1.34 x 10-3 to 3.6 x 10-5 and the pH increases from

2.87 to 4.44

Discuss the effect of the addition of 0.5 mole of NH4Cl to 1.0 L of 0.10 M aqueous

NH3 solution upon the pH value of the solution.

Buffer solutions

Our bodies operate under strict conditions of temperature, concentration, and pH. How do

our bodies maintain pH in our bloodstream when we consume a variety of foods at different

pH‘s?

Our bodies contain solutions of weak acids, containing both acids and conjugate bases, to

help neutralize incoming acids and bases.

A buffer is a solution that contains both a weak acid and its conjugate base, or a weak base

and its conjugate acid.

A buffer will resist a change in pH if small amounts of an acid or a base are added.

Buffers are what helps our body maintain the proper pH in our bloodstream when we

consume a variety foods at different pH‘s.

Buffers can be acidic or basic:

Acidic buffer: consisting of a mixture of the weak acid and its salt.

For example: a mixture of CH3COOH and CH3COONa

Basic buffer: consisting of a mixture of the weak base and its salt.

For example: a mixture of NH4OH and NH4Cl

We only concerned with acid buffers.

Buffer action

Is the ability of a buffer solutions to resist change in pH upon addition of an acid or a base.

A buffer consisting of a mixture of the weak acid(CH3COOH) and its salt (CH3COONa) will

undergo the following changes on the addition of acid or base:

48


(aq)

The salt dissociates completely since it is ionic:

CH3COONa(aq) → Na+(aq) + CH3COO-(aq)

The acid is partially dissociated:

CH3COOH(aq) ⇌ H+(aq) + CH3COO-(aq)

Dissociation of CH3COOH and CH3COOH

This results in an equilibrium mixture containing large concentrations of the undissociated

acid, CH3COOH and its conjugate base, CH3COO- .

The very large concentration of the base shifts the acid equilibrium left, so the

concentration of H+ ions is very small.

If acid (H+) is added to the solution

An increase in H+(aq) concentration would rapidly lower the pH of water, but in the buffer it

simply combines with the CH3COO-(aq) ions and shifts the acid dissociation back to the left.

CH3COO-(aq) + H+(aq) → CH3COOH(aq)


← shift to left

The pH will not change significantly because the CH3COOH formed is a weak acid.

Chapter 1 49


Change of pH in water and in buffer solution.

If base is added (OH-) to the solution

OH-(aq) ions react with the H+(aq) ions present:

H+(aq) + OH-(aq) → H2O(l)

Some of the acid dissociates, returning the [H+(aq) ] to near its original concentration:


So the pH will not change significantly.

Buffer capacity

Shifting the buffer equilibrium

Buffer capacity is the amount of protons or hydroxide ions that can be absorbed without a

significant change in pH. It depends on the amount of acid and its conjugate base from

which the buffer is made.

For example:

A buffer consisting of a mixture of 1M of an acid HA and 1M of its salt NaA.

Another buffer consisting of a mixture of 0.1M of an acid HA and 0.1M of its salt

NaA.

Both two buffer solutions will have same pH, but the first one will have greater buffer

capacity.

50


3

Note that:

- The buffering power is greatest when pH = pKa , i.e. when the acid and the salt are at the

same concentration

- A compound can buffer the pH of a solution when:

Its concentration is sufficient. Most buffers work best at concentrations between 0.1

M and 10 M.

The pH of the solution is close (within about one pH unit) of the acid/ conjugate

base pKa.

Calculating the pH of buffer solutions

Since there is an acid dissociation taking place the pH of a buffer solution must depend on

the Ka of the acid (HA) present and the equilibrium concentrations of the conjugate acid-

base pair (HA/A-).

The equation for the acid equilibrium is:

HA(aq) + H2O(l) ⇌ H3O+(aq) + A-(aq)

acid conjugate base

Ka

A H

HA

O

H O K

HAa A

log H

O logK

log HA

A

A pH pKa log

HA

This equation is known as Henderson-Hasselbalch equation and allows determination of pH in

acidic buffer systems. This is often written as:

pH = pKa + log [salt]

[acid]

[salt]/[acid] ratio is known as the molar ratio of salt to acid. Sometimes it is desired to know

this ratio in order to prepare a buffer of a definite pH.

If [salt] = [acid], [salt]/[acid] = 1 pH = pKa (greatest buffering power)

3

3 a

Chapter 1 51


Example 14:

A buffer solution is made by adding 0.3 mol of acetic acid and 0.3 mol of sodium acetate to

enough water to make 1L of solution.

a) Determine the pH of the buffer solution. The Ka for acetic acid is 1.8 x 10-5

b) Calculate the pH of the buffer solution after 0.02 mol of NaOH is added

Solution:

a) pH = pKa + log [A-] / [HA]

A- = sodium acetate, HA = acetic acid

pH = - log (1.8 x 10-5) + log (0.3/0.3)

pH = 4.74

b) 1 litre of buffer contains 0.3 moles of sodium acetate and 0.3 moles of acetic acid.

Sodium hydroxide is a strong base. The acetic acid will react with the base added to try to

maintain the pH. What the acid loses in concentration, the salt (sodium acetate) will gain.

CH3COOH + OH- → CH3COO- + H2O

pH = pKa + log [A-] / [HA]

pH = - log (1.8 x 10-5) + log (0.32/0.28)

pH = 4.74 + 0.058

pH = 4.8

Calculate the pH of a solution containing:

a) 0.01 mol/L benzoic acid and 0.04 mol/L sodium benzoate. Ka for benzoic acid

is 6.3 x 10-5.

b) 0.100 mol/L acetic acid and 0.200 mol/L sodium acetate. pKa for acetic acid is

4.74

Calculate the pH of a buffer made by dissolving 18.5 g of propanoic acid,

C2H5COOH and 12.0 g of sodium propanoate, C2H5COONa, in water and then

making up the volume to 250 cm3. pKa for propanoic acid = 4.87

52


3 4 4

3

3

The role of a natural buffer in the control of blood pH

To remain healthy human blood has to be maintained at a constant pH 7.4 (7.35-7.45).

If the blood becomes too acidic and the pH drops ( as in the medical condition acidosis), we

have to breathe rapidly to expel more carbon dioxide.

The main mechanism for maintaining pH is the buffering action of several acid/base pairs

e.g bicarbonate buffer system [H2CO3(aq)/HCO -(aq)] and H2PO -(aq)/HPO 2-(aq), together with the buffering action of plasma proteins and haemoglobin.

Bicarbonate buffer system

The bicarbonate buffer system [H2CO3(aq)/HCO -(aq)] is the main buffer system in our blood.

Dissolved CO2, produced during cellular respiration, is equilibrated through carbonic acid

into bicarbonate ions prior to exhalation at the lungs. The intermediates, carbonic acid and

water, are often omitted since they are short lived in the reaction.

The bicarbonate buffer system in our bloodstream is shown in the following figure:

The bicarbonate buffer system can be denoted by showing the acid and its conjugate base

like this: H2CO3/HCO -.

The bicarbonate system helps our bodies maintain its optimal physiological pH:

The ability of an organism to maintain its internal environment by adjusting such factors as

pH, temperature, and solute concentration is called homeostasis.

In normal breathing, CO2 is removed from the bloodstream and blood pH is maintained.

A person who hypoventilates may fail to remove enough CO2 due to shallow breathing

causing CO2 to build up in the bloodstream. A buildup of CO2 in the bloodstream makes the

blood more acidic, a condition that is known as respiratory acidosis.

Individuals suffering this condition must be treated to raise blood pH back to normal.

A bicarbonate solution can be administrated intravenously. This will drive the equilibrium to

the left, when the excess bicarbonate present reacts with the excess acid.

Chapter 1 53


A person who hyperventilates will exhale too much CO2 from the lungs. This will draw H3O+

from the bloodstream, making the blood more basic. This condition is known as respiratory

alkalosis.

This condition necessitates getting more CO2 back into the bloodstream, which can be done

by having the person breathe into a paper bag. This will enrich CO2 in the bloodstream,

shifting the equilibrium back to the right, thereby producing more H3O+.

Solubility product

What is the relationship between the solubility product constant and the solubility of a

compound?

Most ionic compounds containing alkali metals are soluble in water. For example, more than

35 g of NaCl will dissolve in only 100 g of water.

By contrast, some ionic compounds are insoluble in water. For example, compounds that

contain phosphate, sulfite, or carbonate ions tend not to dissolve in water.

The following table provides some general rules for the solubility of ionic compounds in

water.

54


Most insoluble ionic compounds will actually dissolve to some extent in water. These

compounds are said to be slightly soluble in water.

When the ―insoluble‖ compound silver chloride is mixed with water, a very small amount of

silver chloride dissolves in the water.

AgCl(s) ⇌ Ag+(aq) + Cl–(aq)

An equilibrium is established between the solid and the dissolved ions in the saturated

solution.

You can write an equilibrium-constant expression for this process.

To compare the solubility of salts, it is useful to have a constant that reflects only the

concentrations of the dissolved ions.

This constant is called the solubility product constant (Ksp):

Is the product of the concentrations of the ions in a saturated solution of slightly soluble salt

each raised to a power equal to the coefficient of the ion in the equilibrium state.

AaBb(s) ⇌ aA+(aq) + bB–(aq)

Ksp = [A+]a [B-]b

The smaller the numerical value of the solubility product constant, the lower the solubility

of the compound.

Chapter 1 55


Molar solubility (mol/L) is the number of moles of solute dissolved in 1 L of a saturated

solution.

Solubility (g/L) is the number of grams of solute dissolved in 1 L of a saturated solution.

Which of the following compounds is least soluble at 25oC?

The following table lists the Ksp values for some ionic compounds that are slightly soluble in

water.

Example 15:

What is the concentration of lead ions and chromate ions in a saturated solution of lead(II)

chromate at 25oC? (Ksp = 1.8 10–14).

Solution:

PbCrO4(s) ⇌ Pb2+(aq) + CrO42–(aq)

56


4

4

Ksp = [Pb2+] [CrO42–] = 1.8 10–14

At equilibrium, [Pb2+] = [CrO42–]

Ksp = [Pb2+] [Pb2+] = [Pb2+]2 = 1.8 10–14

[Pb2+] = 1.3 10–7 M, also [CrO42–] = 1.3 10–7 M

Example 16:

The solubility of lead(II) chromate (PbCrO4) is 4.5 x 105 g/L. Calculate the solubility product

(Ksp) of lead(II) chromate.

Solution:

Solubility(mol/L) =s 4.5 x 105 g

L x

mol

323.2 g 1.4 x 107 M

PbCrO4(s) ⇌ Pb2+(aq) + CrO 2–(aq)

Ksp s

2 [Pb

2 ][CrO

2 ]

Ksp [1.4 x10

7 M ][1.4 x10

7 M ] 2.0 x10

14

Calculate the solubility of SnS in g/L at 25 °C. Ksp = 1.0 x 10-26

The solubility of BaCO3 is 5.1 x 10-5 M at 25 oC. Calculate the solubility product

(Ksp) constant.

How can you predict whether precipitation will occur when two solutions are mixed?

A precipitate will form if the product of the concentrations of two ions in the mixture is

greater than the Ksp value for the compound formed from the ions.

Example 17:

What concentration of Ag is required to precipitate only AgBr in a solution that contains both

Br- and Cl- at a concentration of 0.02 M? Ksp for AgBr = 7.7 x 10-13, and for AgCl = 1.6 x 10-10

Solution:

AgBr(s) ⇌ Ag+(aq) + Br-(aq)

Chapter 1 57


4

Ksp = [Ag+][Br-]

[Ag+] = Ksp/[Br-] = 7.7 x 10-13/0.02 = 3.9 x 10-11 M

AgCl(s) ⇌ Ag+(aq) + Cl-(aq)

Ksp = [Ag+][Cl-]

[Ag+] = Ksp/[Cl-] = 1.6 x 10-10 /0.02 = 8.0 x 10-9 M

3.9 x 10-11 M < [Ag+] < 8.0 x 10-9 M

Does a precipitate form when 1 L of 0.034 M Na2SO4 is mixed with 1 L of 0.0012

M Ba(NO3)2? The Ksp for BaSO4 is 1.1 10–10.

The common ion effect and solubility

For example:

In a saturated solution of lead(II) chromate, an equilibrium is established between the solid

PbCrO4 and its ions in solution.

PbCrO4(s) ⇌ Pb2+(aq) + CrO 2–(aq)

What would happen if you added some Pb(NO3)2 to this solution?

Pb(NO3)2(s) → Pb2+(aq) + 2NO3–(aq)

Lead(II) nitrite is soluble in water, so adding Pb(NO3)2 causes the concentration of Pb2+ ion to

increase. The addition of Pb2+ ions is a stress on the equilibrium. Applying Le Châtelier‘s

principle, the stress can be relieved if the reaction shifts to the left, resulting in decrease

the solubility of PbCrO4.

In this example, the Pb2+ ion is a common ion. The lowering of the solubility of an ionic

compound as a result of the addition of a common ion is called the common ion effect.

Chapter 2 57

Objectives

Apply the titrimetric in determination of concentrations of solutions.

Identify different methods of precipitation reactions.

Apply Solubility product concept in precipitation of substances.

Explain fractional precipitation.

Describe the complex formation reactions.

Measure the concentrations in human fluids using the different methods of titrimetric titration.

Solve the problems of titrimetric analyses.

Design and set up an experiment.

Collect and analyze data.


Apply appropriate techniques to arrive at a solution.


Test the correctness of their solution.

Titrimetry

Is a term which includes a group of analytical methods based on determining the quantity of

a reagent of known concentration that is required to react completely with the analyte.

Titrimetric methods

There are three main types of titrimetry: volumetric titrimetry, gravimetric titrimetry, and

coulometrtic titrimetry. The benefits of these methods are that they are rapid, accurate,

convenient, and readily available.

Volumetric titrimetry

A type of titrimetry in which the volume of a standard reagent that is needed to

react completely with the analyte is being measured.

Chapter 2

Titrimetric analysis

58

Gravimetric titrimetry

Is like volumetric titrimetry, but the mass is measured instead of the volume.

Coulometric titrimetry

Is where the reagent is a constant direct electrical current of known magnitude that

consumes the analyte; the time required to complete the electrochemical reaction is

measured.

Types of volumetric titrimetry

There are four main types of volumetric titrimetry:

Acid-Base titrations.

Redox titrations.

Precipitation titrations.

Complex formation titrations.

Titration

A process in which a standard solution is added to a solution of an

analyte until the reaction between the analyte and the reagent is

believed to be complete. The quantity of analyte is calculated from

the amount of titrant added.

N.B. Substance to be analysed is known as the analyte, the solution added to the analyte is

known as the titrant and usually delivered from a buret.

Requirements for titration

Known reaction stoichiometry.

Rapid reaction.

No side reactions.

Large change in some solution property at equivalence point.

Coincidence of equivalence and end points.

Quantitative reaction.

Standards in volumetric titrimetry

Standard solution

A solution of a known concentration and play a key role in titrimetric methods. Overall

accuracy of titrimetric analysis limited by accuracy of the concentration of standard solution

used in analysis.

Chapter 2 59

Desirable Properties of Standard Solutions:

Sufficiently stable.

React rapidly with analyte.

React completely with analyte.

Undergo selective reaction with analyte.

Primary standard

A highly purified compound that serves as a reference material in all volumetric and mass

titrimetric methods. The accuracy depends on the properties of a compound and the

important properties are:

High purity (should be 99.9% pure) .

Atmospheric stability.

Absence of hydrate water.

Readily available at a modest cost.

Reasonable solution in the titration medium.

Reasonably large molar mass.

Compounds that meet or even approach these criteria are few, and only a few primary

standards are available.

Some primary standards compounds

60

4 2 2

4

4

Standardization

Required when a primary standard titrant is not available. In this case, the titrant is

prepared with approximately the desired concentration, and the concentration of the titrant

solution is determined by using it to titrate a carefully measured quantity of a primary

standard.

after standardization, the titrant is known as secondary standard solution

Secondary standard solution

titrant that is standardized against a primary standard solution.

Equivalence point

The point in which the quantity of added titrant is the exact amount necessary for

stoichiometric reaction with the analyte.

End point

Equivalence point is the ideal theoretical result we seek in a titration. What we actually

measure is the end point, which is marked by a sudden change in the physical property of

the solution: Change in color, pH, voltage, current, absorbance of light, presence/absence

ppt.

5 HOOC-COOH + 2 MnO - + 6 H+ → 10 CO + 2 Mn2+ + 8 H O

Analyte Titrant

(Colorless) (Purple) (Colorless)

Equivalence point occurs when 2 moles of MnO - is added to 5 moles of Oxalic acid. After

equivalence point occurs, excess MnO - turns solution purple (end point)

Titration error

difference between the equivalence point and the end point, corrected by a blank titration:

Repeat procedure without analyte.

Determine amount of titrant needed to observe change.

Subtract blank volume from the volume observed in the titration.

Chapter 2 61

4

4

4 2 2

Titration error should be very small and by choosing a physical property whose change is

easily observed one can minimize the titration error so that the end point is very close to

the equivalence point

Back titration

This is a process that is sometimes useful if better/easier to detect endpoint, in which an

excess of the standard titrant is added, which completely react all the analyte. Titrate

excess standard titrant by back titration with a second standard titrant to determine how

much is left. Difference is related to amount of analyte.

For example: Back titration of oxalic acid by MnO -

5 HOOC-COOH + 2 MnO - + 6 H+ → 10 CO + 2 Mn2+ + 8 H O

Analyte Titrant

(Colorless) (Purple) (Colorless)

Add enough MnO4- so all oxalic acid is converted to product

Titrate Fe2+ to determine the amount of MnO - that did not react with oxalic acid

Differences is related to amount of analyte (oxalic acid).

End point Detection

Methods for determining the end point are:

detecting a sudden change in the voltage

or current between a pair of electrodes.

observing an indicator color change.

monitoring the absorption of light.

Indicators

Compounds are added to analyte solution in order to give an observable physical change

(end point) at or near the equivalence point. Typically appearance or disappearance of

color, change in color, ppt formed, etc.

Acid-Base titrations

A salt is formed between the reaction of an acid and a base. Usually, a neutral salt is formed

when a strong acid and a strong base is neutralized in the reaction:

H+ + OH- → H2O

62

When weak acids or bases react, the relative strength of the conjugated acid-base pair in

the salt determines the pH of its solutions.

The salt, or its solution, so formed can be acidic, neutral or basic.

A salt formed between a strong acid and a strong base is a neutral salt (pH=7), for

example NaCl.

A salt formed between a strong acid and a weak base is an acid salt (pH<7), for

example NH4Cl.

A salt formed between a weak acid and a strong base is a basic salt (pH˃7), for example

CH3COONa

Titration curves

A titration curve is a graph showing how concentrations of analyte and titrant vary during a

titration. From it we can:

understand the chemistry that occurs during a titration.

learn how experimental control can be exerted to influence the quality of an

analytical titration.

Two types of titration curves routinely encountered in titrimetric methods; they are

sigmoidal curve and linear segment curve.

Chapter 2 63

Acid-Base titration curves

A plot of pH versus the amount of titrant added. Typically the titrant is a strong

(completely) dissociated acid or base. Such curves are useful for determining end points and

dissociation constants of weak acids or bases.

When acid base titration is at the equivalence point, the acid has neutralized the base

leaving only a salt and water. The pH of the equivalence point depends on type of salt.

You need to be able to recognize each and then choose a suitable indicator for that

titration.

When you choose an indicator, you must pick one so that the transition point of the indicator

matches the equivalence point of the titration. The transition point refers to when an

indicator changes color. pH meter can be used to monitor pH during the titration.

There are three types of titration curves:

1. Titration of a strong acid with a strong base

Assume strong acid and base completely dissociate

Any amount of OH- added will consume a stoichiometric amount of H+. Reaction

assumed to go to completion:

Analyte Titrant

64

Three regions of the titration curve:

Before the equivalence point

The pH starts out low, reflecting the high [H3O+] of the strong acid and increases

gradually as acid is neutralized by the added base. The pH is determined by excess

H+ in the solution

At the equivalence point

OH- is just sufficient to react with all H+ to make H2O. Suddenly the pH rises steeply,

this occurs in the immediate vicinity of the equivalence point. For this type of

titration the pH is 7.0 at the equivalence point.

After the equivalence point

Beyond this steep portion, the pH increases slowly as more base is added. pH is

determined by excess OH- in the solution.

2. Titration of a weak acid with a strong base

(HPr = Propionic acid)

Four Regions to Titration Curve

Before any added base

just weak acid (HA) in water and pH is determined by Ka

With addition of strong base

Adding OH- creates a mixture of HA and A- (Buffer).

Chapter 2 65

pH is determined by Henderson Hasselbach equation:

[ A ]

pH pKa log [HA]

At equivalence point

Exactly enough OH- to consume HA and all HA is converted into A- (Weak base).

The solution only contains A- and pH is determined by Kb

pH will always be < 7.00 for titration of a weak acid because acid is converted into

conjugate base at the equivalence point.


excess strong base is added to A-, OH- is a much stronger base than A- and pH is

determined by excess of OH-.

The major differences between a strong acid-strong base titration curve and a weak

acid-strong base titration curve

The initial pH is higher.

A gradually rising portion of the curve, called the buffer region, appears before the

steep rise to the equivalence point.

The pH at the equivalence point is greater than 7.00.

The steep rise interval is less pronounced.

3. Titration of a weak base with a strong acid

Simply the Reverse of the Titration of a Weak Acid with a Strong Acid. Again, Titration

Reaction Goes to Completion:

66

The major differences between a weak acid-strong base titration curve and a weak base-

strong acid titration curve

The initial pH is above 7.00.

A gradually decreasing portion of the curve, called the buffer region, appears before

a steep fall to the equivalence point.

The pH at the equivalence point is less than 7.00.

Thereafter, the pH decreases slowly as excess strong acid is added.

Titrations of polyprotic acids and bases

Principals for monoprotic systems apply to polyprotic systems.

Multiple equivalence points and buffer regions with multiple inflection points in titration

curve.

Chapter 2 67

Features of the titration of a polyprotic acid with a strong base

The loss of each mole of H+ shows up as separate equivalence point (but only if the

two pKas are separated by more than 3 pK units).

The pH at the midpoint of the buffer region is equal to the pKa of that acid species.

The same volume of added base is required to remove each mole of H+.

Acid-Base Indicators

An interesting group of weak acids and bases derived from organic dyes. Such compounds

have at least one conjugate acid-base species that is highly colored, their titration results in

a change in both pH and color. This change in color can serve as useful means for

determining the end point of a titration, provided that it occurs at the equivalence point.

Ka

H In - HIn

68

Some common indicators

1. Litmus

Litmus is a weak acid. It has a seriously complicated molecule which we will simplify to HLit.

The "H" is the proton which can be given away to something else. The "Lit" is the rest of the

weak acid molecule.

unionized litmus ionized litmus

(Red) (blue)

What happened when:

Adding H+ ions

Adding OH- ion

Chapter 2 69

If the concentrations of HLit and Lit - are equal:

At some point during the movement of the position of equilibrium, the concentrations of the

two colours will become equal. The colour you see will be a mixture of the two.

(Red) (blue) (brown)

2. Phenolphthalein

is a chemical compound with the formula C20H14O4 (often written as "HIn" or "phph"). Often

used in titrations, it turns colorless in acidic solutions and pink in basic solutions.

70

3. Methyl orange

Methyl orange is an intensely colored compound used in an acid-base titration to allow end

point detection.. It changes from red (at pH 3.1) to orange-yellow (at pH 4.4). pH-

related color changes result from changes in the way electrons are confined in a molecule

when hydrogen ions are attached or detached.

Methyl orange in acidic solution

Methyl orange in basic solution

The pH range of indictors

Indictors does not change colour sharply at one particular pH, they change over a narrow

range of pH, this range is termed the color change interval. It is expressed as a pH range.

Chapter 2 71

indctors pKind pH range

litmus 6.5 5-8

Methyl orange 3.7 3.1-4.4

phenolphthalein 9.3 8.3-10.0

Choosing an indicator

Indicators are chosen, such that they change colors at the range of the pH of interest. Want

Indicator that changes color in the vicinity of the equivalence point and corresponding pH.

The closer the two match, the more accurate determining the end point will be.

In general we seek an indicator whose transition range (±1pH unit from the indicator pKa)

overlaps the steepest part of the titration curve as closely as possible.

1. For titration of strong base with strong acid

Both of ph.ph and M.O are useful

72

2. For titration of strong acid with weak base

M.O is useful and ph.ph is useless

3. For titration of weak acid with strong base

ph.ph is useful and M.O is useless

Chapter 2 73

3 3

3 2 2

4. For titration of Na2CO3 with HCl

CO 2- + H+ ↔ HCO -

HCO - + H+ ↔ CO + H O

Ph.ph useful for detect first endpoint

M.O is useful for detect the second end point

Redox titrations

Oxidation – Reduction reactions

Why does cut fruit turn brown?

Some fruits, including apples, turn brown when you cut them.

What do you think is happening on the surface of the fruit that

causes it to turn brown?

Oxygen in air reacts with chemicals on the surface of the cut fruit. The oxygen oxidizes the

chemicals in the fruit, causing a oxidation-reduction reaction and therefore the color

change.

One class of chemical reactions is oxidation-reduction reactions, in which electrons are

transferred from one reacting species to another. Oxidation-reduction reactions are also

known as redox reactions. This class of reactions include: formation of a compound from its

elements, all combustion reactions, reactions that generate electricity and that produce

cellular energy.

74

For example, potassium metal reacts violently with water to

produce hydrogen gas (which ignites) and potassium hydroxide.

Many reactions in which color changes occur are redox reactions. For example:

Oxidation and Reduction

Oxidation

A species is oxidized when it loses one or more electrons, and it is called a reducing agent.

2Mg + O2 → 2MgO (notice the magnesium is losing electrons)

Oxidizing agent (Oxidant)

A substance which oxidizes somebody else. It is reduced in the process. Such as: Iodine I2,

potassium dichromate K2Cr2O7, potassium permanganate KMnO4.

Reduction

A species is reduced when it gains one or more electrons, and it is called an oxidizing agent.

MgO + H2 → Mg + H2O (notice the Mg2+ in MgO is gaining electrons)

Reducing agent (Reductant)

A substance which reduces somebody else. It is oxidized in the process. Such as Sodium

thiosulphate Na2S2O3, oxalic acid HOOC-COOH.

Chapter 2 75

Oxidation – Reduction (Redox) reactions

Oxidation and reduction always occur together, never in isolation. One does not occur

without the other , if something gains electrons, something else had to lose them.

Oxidation number

Oxidation number (O.N.) is also known as oxidation state. It is defined as the charge the

atom would have in a molecule (or anionic compound) if electrons were not shared but were

transferred completely.

76

An increase in oxidation number of an atom signifies oxidation and a decrease in oxidation

number of an atom signifies reduction.

For a binary ionic compound, the O.N. is equivalent to the ionic charge. For covalent

compounds or polyatomic ions, the O.N. is less obvious and can be determined by a given set

of rules.

Rules for assigning an oxidation number

General rules

1. For an atom in its elemental form: O.N. = 0

O.N. of Na, Be, K, Pb, H2, O2, P4 = 0

2. For a monatomic ion: O.N. = ion charge.

O.N. of Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2

3. The sum of O.N. values for all the atoms in a molecule or formula unit of a compound

equals to zero.

4. The sum of O.N. values for all the atoms in a polyatomic ion equals to the ion‘s charge.

Rules for Specific Atoms or Periodic Table Groups

1. For group 1A: O.N. = +1 in all compounds.

2. For group 2A: O.N. = +2 in all compounds.

Chapter 2 77

4

3. For hydrogen: O.N. = +1 in combination with nonmetals O.N. = -1 in combination with

metals and boron.

4. For fluorine: O.N. = -1 in all compounds.

5. For oxygen: O.N. = -1 in peroxides O.N. = -2 in all other compounds (except with F).

6. For group 7A: O.N. = -1 in combination with metals, nonmetals (except O), and other

halogens lower in the group.

Example 18:

Determine the oxidation number (O.N.) of each element in the following compounds:

a) CaO b) KNO3 c) NaHSO4 d) CaCO3 e) N2 f) H2O

Solution:

Simply apply the rules for assigning an oxidation number as described earlier

a) Ca+2O-2

b) K+1N+5O-2

O.N. of N = 0-[+1+3(2-)] = +5

c) Na+1H+1S+6O-2

O.N. of S = 0-[+1+1+4(2-)] = +6

d) Ca+2C+4O-23

O.N. of C = 0-[+2+3(2-)] = +4

e) N20

f) H+12O

-2

Example 19:

Identify the oxidizing agent and reducing agent in each of the following reaction:

a) 2H2(g) + O2(g) → 2H2O(g)

b) Cu(s) + 4HNO3(aq) → Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)

Solution:

Assign oxidation numbers and compare.

3

78

4 2 2

3 6 2 2

Oxidation is represented by an increase in oxidation number.

Reduction is represented by a decrease in oxidation number

a) 2H02(g) + O0

2(g) → 2H+12O

-2(g)

- O2 was reduced (O.N. of O: 0 → -2); O2 is the oxidizing agent.

- H2 was oxidized (O.N. of H: 0 → +1); H2 is the reducing agent.

b) Cu0(s) + 4H+1N+5O-2 (aq) → Cu+2(N+5O-2 ) (aq) + 2N-4O-2 (g) + 2H+1 O-2(l) 3 3 2 2 2

- Cu was oxidized (O.N. of Cu: 0 → +2); Cu is the reducing agent.

- HNO3 was reduced (O.N. of N: +5 → +4); HNO3 is the oxidizing agent.

Identify the oxidizing and reducing agents in each of the following reaction:

a) 8H+(aq) + 6Cl-(aq) + Sn(s) + 4NO -(aq) → SnCl 2-(aq) + 4NO (g) + 4H O(l)

b) 2MnO -(aq) + 10Cl-(aq) + 16H+(aq) → 5Cl (g) + 2Mn2+(aq) + 8H O(l)

Balancing redox equations

Two methods can be used for balancing redox equations:

1. Oxidation number change method

2. Half-reaction method

These two methods are based on the fact that the total number of electrons gained in

reduction must equal the total number of electrons lost in oxidation.

Method 1: Oxidation number change method

In the oxidation number change method, you balance a redox equation by comparing the

increases and decreases in oxidation numbers.

1. Assign oxidation numbers to all atoms in the equation.

2. From the changes in O.N., identify the oxidized and reduced species.

3. Compute the number of electrons lost in the oxidation and gained in the reduction from

the O.N. changes.

4. Multiply one or both of these numbers by appropriate factors to make the electrons lost

equal the electrons gained, and use the factors as balancing coefficients.

5. Finally check to be sure that the equation is balanced both for atoms and charge.

Chapter 2 79

Example 20:

Use the oxidation number change method to balance the following equation.

Fe2O3(s) + CO(g) → Fe(s) + CO2(g)

Solution:

Step 1. Assign oxidation numbers to all the atoms in the equation.

Step 2. Identify oxidized and reduced species.

– CO was oxidized (O.N. of C: +2 → +4)

– Fe2O3 was reduced (O.N. of Fe: +3 → 0)

Step 3. Compute e- lost and e- gained.

– In the oxidation: 2e- were lost from C

– In the reduction: 3e- was gained by Fe

Step 4. Multiply by factors to make e- lost equal to e- gained, and use the factors as

coefficients

C lost 2e- and Fe gained 3e-, so the 2e- lost by C should be multiplied by 3 and the 3e-

gained by Fe should be multiplied by 2. Put the coefficient 3 before CO and CO2 and the

coefficient 2 before Fe. i.e. the oxidation number increase should be multiplied by 3 and the

decrease by 2.

Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)

Step 5: Finally, make sure the equation is balanced for both atoms and charge.

Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)

Balance the following redox equations by using the oxidation number change method.

a) K2Cr2O7(aq) + H2O(l) + S(s) → KOH(aq) + Cr2O3(s) + SO2(g)

b) Al(s) + H2SO4(aq) → Al2(SO4)3(aq) + H2(g)

c) PbS(s) + O2(g) → PbO(s) + SO2(g)

80

Method 2: Half-reaction method

A half-reaction

Is an equation showing just the oxidation or just the reduction that takes place in a redox

reaction.

In the half-reaction method, you write and balance the oxidation and reduction half-

reactions separately before combining them into a balanced redox equation.

The procedure is different, but the outcome is the same as with the oxidation-number-

change method.

1. Divide the skeleton reaction into two half-reactions, each of which contains the oxidized

and reduced forms of one of the species.

2. Balance the atoms and charges in each half-reaction.

– Atoms are balanced in order: atoms other than O and H, then O, then H. In acidic

solutions, H2O and H+ can be used to balance oxygen and hydrogen as needed. In basic

solution, H2O and OH– are used to balance these species.

– Charge is balanced by adding electrons:

• To the left in reduction half-reactions.

• To the right in oxidation half-reactions.

3. If necessary, multiply one or both half-reactions by an integer to make the number of e-

gained equal to the number of e- lost.

4. Add the balanced half-reactions, and include states of matter.

5. Check that the atoms and charges are balanced.

Chapter 2 81

Example 21:

Use the half-reaction method to balance the following equations:

a) ClO3-(aq) + I-(aq) → I2(s) + Cl-(aq) [acidic solution]

b) Fe(OH)2(s) + Pb(OH)3-(aq) → Fe(OH)3(s) + Pb(s) [basic solution]

Solution:

a) ClO3-(aq) + I-(aq) → I2(s) + Cl-(aq)

Step 1. Divide the reaction into half-reactions

ClO3-(aq) → Cl-(aq) I-(aq) → I2(s)

Step 2. Balance atoms and charges in each half-reaction

– Atoms other than O and H

ClO3-(aq) → Cl-(aq) Cl is balanced

2I-(aq) → I2(s) I now balanced

– Balance O atoms by adding H2O molecules

ClO3-(aq) → Cl-(aq) + 3H2O(l) add 3H2O

2I-(aq) → I2(s) no change

– Balance H atoms by adding H+ ions (acidic solution)

ClO3-(aq) + 6H+(aq) → Cl-(aq) + 3H2O(l) add 6H+

2I-(aq) → I2(s) no change

– Balance charge by adding electrons

ClO3-(aq) + 6H+(aq) + 6e- → Cl-(aq) + 3H2O(l) add 6e-

2I-(aq) → I2(s) + 2e- add 2e-

Step 3. Multiply each half-reaction by an integer to equalize number of electrons

ClO3-(aq) + 6H+(aq) + 6e- → Cl-(aq) + 3H2O(l) x 1

2I-(aq) → I2(s) + 2e- x 3

Step 4. Add the half-reactions together

ClO3-(aq) + 6H+(aq) + 6e- → Cl-(aq) + 3H2O(l)

6I-(aq) → 3I2(s) + 6e-

ClO3-(aq) + 6H+(aq) + 6I-(aq) → Cl-(aq) + 3H2O(l) + 3I2(s)

82

Step 5. Check that atoms and charges balance

– Reactants (Cl, 3O, 6H, 6I, -1) → products (Cl, 3O, 6H, 6I, -1)

• ClO3- is the oxidizing agent

• I- is the reducing agent

b) Fe(OH)2(s) + Pb(OH)3-(aq) → Fe(OH)3(s) + Pb(s)

The only difference in balancing a redox equation that takes place in basic solution is in step

4.

• At this point, we add one OH- ion to both sides of the equation for every H+ ion present

• The H+ ions on one side are combined with the added OH- ions to form H2O, and OH- ions appear on the other side of the equation

Step 1. Divide the reaction into half-reactions

Pb(OH)3-(aq) → Pb(s) Fe(OH)2(s) → Fe(OH)3(s)

Step 2. Balance atoms and charges in each half-reaction

– Atoms other than O and H

Pb(OH)3-(aq) → Pb(s) Pb is balanced

Fe(OH)2(s) → Fe(OH)3(s) Fe is balanced

– Balance O atoms by adding H2O molecules

Pb(OH)3-(aq) → Pb(s) + 3H2O add 3H2O

Fe(OH)2(s) + H2O → Fe(OH)3(s) add H2O

– Balance H atoms by adding H+ ions

Pb(OH)3-(aq) + 3H+ → Pb(s) + 3H2O add 3H+

Fe(OH)2(s) + H2O → Fe(OH)3(s) + H+ add H+

– Balance charge by adding electrons

Pb(OH)3-(aq) + 3H+ + 2e- → Pb(s) + 3H2O add 2e-

Fe(OH)2(s) + H2O → Fe(OH)3(s) + H+ + e- add e-

Step 3. Multiply each half-reaction by an integer to equalize number of electrons

Pb(OH)3-(aq) + 3H+ + 2e- → Pb(s) + 3H2O x 1

Fe(OH)2(s) + H2O → Fe(OH)3(s) + H+ + e- x 2

Chapter 2 83

3

3 2

3 2 3

3 2 2 3

3 2 2 3

3 4

6 4

Step 4. Add the half-reactions together

Pb(OH) -(aq) + 3H+ + 2e- → Pb(s) + 3H O

2Fe(OH)2(s) + 2H2O → 2Fe(OH)3(s) + 2H+ + 2e-

Pb(OH) -(aq) + H+(aq) + 2Fe(OH) (s) → Pb(s) + H O(l) + 2Fe(OH) (s)

Step 4(basic). Add OH-. Here, we add 1 OH-

Pb(OH) -(aq) + H+(aq) + OH- + 2Fe(OH) (s) → Pb(s) + H O(l) + 2Fe(OH) (s) + OH-

Pb(OH) -(aq) + 2Fe(OH) (s) → Pb(s) + 2Fe(OH) (s) + OH-(aq)

Step 5. Check that atoms and charges balance

– Reactants (Pb, 7O, 7H, 2Fe, -1) → products (Pb, 7O, 7H, 2Fe, -1)

• Pb(OH) - is the oxidizing agent

• Fe(OH)2 is the reducing agent

Use the half-reaction method to balance the following equations and then identify the

oxidizing and reducing agents:

a) Mn2+(aq) + BiO -(aq) → MnO -(aq) + Bi3+(aq) [acidic]

b) Fe(CN)63-(aq) + Re(s) → Fe(CN) 4-(aq) + ReO -(aq) [basic]

Redox titration

The concentrations of redox-active species can be determined by redox titrations. In a

redox titration, a measured sample of the unknown is titrated against a standard solution of

a substance that will oxidize or reduce the unknown.

Oxidation and reduction reactions always occur simultaneously. One can not take place in

isolation from the other. During a redox reaction the oxidizing agent itself undergoes

reduction while the reducing agent undergoes its oxidation. Thus reduction reaction is

represented as:

Oxidant + ne- = Reductant

Oxidation of a substance leads to increase in its oxidation number while Reduction of a

substance lead to decrease in the oxidation number. The relative tendency to accept or lose

electrons by any reagent is measured in terms of their standard reduction potential values.

Relative values are only obtained by comparison with the normal hydrogen electrode

84

red

red

potential (SEP) when the hydrogen gas at 1 atm. pressure is in contact with a piece of

platinum and with hydrogen ion concentration at 1M. The standard electrode potential

enable us to predict which ions will oxidize or reduce other ions.

The electrode potential which is established when an inert or unattackable electrode is

immersed in a solution containing both the oxidant and the reductant is given by the

expression:

ET = E0 + RT/nF ln [oxd]/[Red]

Where ET is the observed potential of the redox electrode at temperature T , E0 is the

standard reduction potential, n the number of electrons gained by the oxidant in being

converted to the reductant.

The Electrochemical series enlists a number of systems according to decreasing standard

reduction potentials at 25 0C. The most powerful oxidizing agents lie at the top of

electrochemical series (High positive E0 values) and the most powerful reducing agents are

present at the bottom (High negative E0 values).

In redox titrations the concentration of the substances or ions involved in the reaction

continuously keeps changing in the course of the titration. Hence the redox potential of the

solution must also change. By plotting the redox potential corresponding to different points

in the titration, a titration curve similar to the curve obtained in an acid-base method. The

titration curve in redox reactions can be drawn by plotting the potential of half cell against

the volume of the titrant.

Chapter 2 85

Standard electrode potentials at 25 oC

86

4 2 red

Some of the commonly used oxidizing and reducing agents in the redox titrations

Oxidizing agents

i. KMnO4 in presence of dil H2SO4

MnO – + 8H+ + 5e- → Mn2+ + 4H O E° = +1.52V

ii. K2Cr2O7 in dil. H2SO4

a moderately strong oxidizing agent; oxidizing ability depends strongly on pH, decreasing

rapidly as solution becomes more neutral

Cr2O72– + 14H+ + 6e– → 2Cr3+ + 7H2O E°red = +1.33V

iii. Iodine solution

I2 + 2e- → 2I– E°red = +0.54V

Reducing agents

i. Mohr’s salt FeSO4.(NH4)2SO4.6H2O

Fe2+ → Fe3+ + e- E°red = +0.77V

ii. Oxalic acid H2C2O4.2H2O

C O 2- → 2CO + 2e- E° = +0.77V 2 4 2 red

iii. Sodium thiosulphate Na2S2O3.5H2O

2S2O32- → S4O6

2- + 2e- E°red = +0.08V

There is no universal oxidizing agent which can be titrated against every reducing agent and

vice-versa. Hence, the choice of an oxidizing agent to be used against a particular reducing

agent depends upon the reaction conditions and standard reduction potential of the

oxidizing agent.

pH dependence of oxidizing behaviour

It is important to note that for many oxidants the pH of the medium is of great importance

and hence their oxidizing strength may vary depending on the medium in which its reaction

is studied. For example potassium permanganate is oxidizing agent in all three mediums,

acid, alkaline and neutral. However it is strongest in acidic medium.

a. Strongly alkaline medium

MnO4– + e- → MnO4

2- E°red = +0.56V

Permanganate ion Manganate ion

Chapter 2 87

4 2 2 red

4 2 red

b. Neutral medium

MnO – + 2H O + 3e- → MnO ↓ + 4OH- E° = +1.23V

Manganese dioxide ppt.

c. Acidic medium

MnO – + 8H+ + 5e- → Mn2+ + 4H O E° = +1.51V Manganous ion

Redox Indicators

Indicator which undergoe a definite color change at a specific electrode potential. The

requirement for fast and reversible color change means that the oxidation- reduction

equilibrium for an indicator redox system needs to be established very quickly. Therefore,

only a few classes of organic redox systems can be used for indicator purposes.

Type of Redox Indicators

There are two common types of redox indicators: metal-organic complexes (such as

phenanthroline) and true organic redox systems (such as Methylene blue). Almost all redox

indicators with true organic redox systems involve a proton as a participant in their

electrochemical reaction. Therefore, sometimes redox indicators are also divided into two

general groups: independent or dependent on pH.

Self Indicators

Many a times the titrant itself may be so strongly coloured that after the

equivalence point, a single drop of the titrant produces an intense colour in the

reaction mixture. e.g. potassium permanganate. Such Indicators are called self

indicators. Self indicators generally are strongly coloured as a result of charge

transfer transitions in them.

Internal indicators

Such indicators are added into the reaction mixtures, these indicators always have

reduction potential values lower than the analyte system so that they react with the

titrant only when whole of the analyte has been consumed, producing a readily

detectable color change.

External indicators

In case a suitable redox indicator is not available for a given system, an indicator

may be employed which will indicate the completion of reaction by physically or

chemically reacting with the analyte (not through redox reaction). This reaction

between indicator and the analyte may sometimes be an irreversible one and in

some cases may even lead to precipitation. In those case indicators are not added

to the reaction mixture on the whole, rather used externally on a grooved tile. Such

indicators are called external indicators.

88

A redox indicator should be produces a sudden change in the electrode potential in

the vicinity of the equivalence point during a redox titration. This is possible when

the indicator itself is redox active i.e., capable of undergoing oxidation or reduction

process which is a reversible one. The oxidized and reduced form of the indicator

should have a contrast difference in the colours.

Inoxd + ne- = Inred

At potential E, the ratio of the concentration of two forms is given by the Nernst equation:

ET = E0 + RT/nF ln [Inoxd]/[IRed]

Potentiometric Methods

The most accurate method to judge the completion of redox titration is however

potentiometric method which deals with the measurement of e.m.f. between a reference

electrode and the indicator (redox) electrode during the stages of redox titration. This

method infact tells us the equivalence point of the reaction and not the end point.

Titrations by using potassium permanganate

Potassium permanganate is a powerful oxidizing agent because of the high positive charge on

manganese, and is used in a wide variety of chemical analyses of reducing agents such as the

determination of iron in iron ore or nitrites in aquarium water.

Potassium permanganate is not a primary standard. It is not easily obtained in perfectly pure

form which is completely free from manganese dioxide. Also the aqueous solutions are not

stable for long as the ordinary distilled water usually contains reducing substances (traces of

organic matter, etc.) which react with permanganate resulting in precipitation of MnO2. This

reaction is catalyzed by light. Presence of manganese dioxide is totally undesirable as it

catalyzes the auto decomposition of the permanganate ion on standing:

4MnO4– + 2H2O → 4MnO2 + 3O2 + 4OH¯

Permanganate solution decompose in presence of manganous ions also:

2MnO4– + 3 Mn2+ + 2H2O → 5MnO2 + 4H+

The reaction is quite slow in acidic medium but attains a fast rate in neutral solutions. Thus

permanganate solution can not be considered as a primary standard solution by simply

making the calculations based on the amount of permanganate weighed and dissolved. The

permanganate solutions once prepared are either left for a day or two or the freshly

prepared solutions are boiled for an hour. In both the cases the resulting solution is filtered

through sintered glass crucibles. The clear solutions so obtained are then standardized by

titrating with a primary standard solution.

Permanganate ion in the acidic medium is a very strong oxidizing agent

MnO4– + 8H+ + 5e- → Mn2+ + 4H2O E°red = +1.51V

permanganate Manganous ion

Chapter 2 89

4

4 2 2

The acidity is introduced by addition of dil. H2SO4 only and no other acid. Conc. H2SO4 ,

HNO3 acid (both conc. and dilute) can‘t be used as these are oxidizing agent and may

interfere in the reaction of permanganate depending on the reductant‘s strength.

Also HCl acid can‘t be used where, permanganate can oxidize chloride to chlorine, which

can be a source of positive errors as permanganate is consumed in this reaction. (E°red

Cl2/Cl-)= +1.36V)

2MnO – + 16H+ + 10Cl -(aq) → 2Mn2+ + 5Cl + 8 H O(l)

Acetic acid is too weak an acid to provide the desired acidity in the solution.

Permanganate acts as self indicator. Since MnO – is intense purple while Mn2+ is colourless,

the reaction mixture at equivalence point is colourless and even a single drop of the

permanganate would impart sufficient pink colour to the solution acting as self indicator.

This also leads to precaution that while titrating a reducing agent with KMnO4 , constant

stirring of reaction mixture is must otherwise local reduction in acidity may lead to

precipitation of MnO2 according to this reaction.

Precipitation titrations (precipitimetry)

Based on the determination of some compounds or ions through the formation of insoluble

salt (precipitate) by using precipitating agents. Analyte is titrated with a standard solution

of a precipitating agent in accordance with defined reaction stoichiometry. The reaction

should be rapid and the precipitate must be practically insoluble (KSP = 10-10 or lower).

The solubility product (KSP) constant

the product of the ion concentrations in a saturated solution raised to the power of their

coefficients in the equilibrium.

AB(s) AB(aq) A- + B+

Slightly soluble salt Saturated solution Ions

Generally for AnBm salt: KSP = [Am-]n [Bn+]m

e.g. For AgCl, Ksp = [Ag+] [Cl-],

For Ag2CrO4, Ksp = [Ag+]2[CrO42-],

For Pb3(PO4)2 , Ksp = [Pb2+]3[PO43-]2

90

The solubility product (KSP) constant has only one value for a given solid at a given

temperature. Only temperature changes can change the value of KSP. The higher the value of

KSP, the higher the solubility.

Factors affecting solubility

Temperature usually slightly increases the solubility (except PbCl2).

Common ion depress the solubility (enhances precipitation). The lowering of the

solubility of an ionic compound as a result of the addition of a common ion is called

the common ion effect.

Complex formation usually increases the solubility.

AgCl + 2 Cl- ↔ [AgCl3]

2-

AgCl + 2 CN- ↔ [Ag(CN)2]- + Cl-

AgCl + 2 NH3 ↔ [Ag(NH3)2]+ + Cl-

Solvent inorganic salts more soluble in water than organic solvents.

pH salts of weak acids dissolve at low pH (high H+ conc.).

Diverse ion increase solubility; due to interionic attraction which stabilize the ions

of the precipitate in their ionic form.

Argentometric titrations (Argentometry)

In order for a titrimetric method to be viable, the titration reaction must be complete and

should be rapid. There are many precipitation reactions that can satisfy the first

requirement, but far fewer that satisfy the second. Precipitation reactions of silver salts are

usually quite rapid, and so argentometric titrations, which use AgNO3 as the precipitating

agent, are the most common precipitation titrations. Argentometric titrations can be used

to analyze samples for the presence of a number of anions that form precipitates with Ag+;

e.g. halides (Cl-, Br-, I-) and SCN-.

End point detection for argentometric titration

There are three classical methods based on color indicators can be used for end point

detection in titration with silver nitrate:

Mohr’s method

formation of colored precipitate at the end point.

Volhard method

formation of a soluble, colored complex at the end point.

Fajans method

adsorption of a colored indicator on the precipitate at the end point.

Chapter 2 91

Mohr’s method (formation of a secondary colored precipitate)

This method is used for determination of halides (Cl-, Br-, I-) in neutral or slightly alkaline

medium (pH = 6.5-9) using potassium chromate as indicator.

NaCl + AgNO3 = AgCl ↓ + NaNO3, KSP (AgCl) =1.82 x 10-10

Analyte Titrant White ppt

The indicator combines with excess Ag+ at the equivalence point.

2AgNO3 + K2CrO4 = Ag2CrO4 ↓ + 2KNO3, KSP (Ag2CrO4) =1.2 x 10-12

Titrant Indicator Brick red ppt

Before the equivalence point, the solubility of the Ag-halide precipitate must be less than

the Ag-indicator, otherwise the latter would precipitate out during the titration.

Silver ion concentration at equivalence point:

Chromate ion concentration require to initiate formation of Ag2CrO4:

Concentration of pot. Chromate is critical?

To prevent precipitation of silver chromate until complete precipitation of AgCl as

Ksp of Ag2CrO4 is less than Ksp of AgCl.

Not acidic medium?

To prevent formation of Ag2Cr2O7 (soluble). Cr2O72- become lower, more Ag+ to be

added to reach end point, which cause error.

2 CrO42 + 2 H3O

+ → Cr2O72 + H2O

Not strong alkaline medium?

To prevent Formation of AgOH (silver hydroxide) then Ag2O (silver oxide, black ppt)

which may be precipitated before Ag2CrO4 and interfere with end point detection.

2Ag+ + 2OH → 2AgOH(s) → Ag2O↓+ H2O

Black ppt

If ammonium salts present, pH shouldn’t exceed 8?

Free NH3 will be produced and dissolve AgCl forming Ag[NH3]2Cl complex.

92

Volhard’s method

This is a good method for the analysis of Ag+ in acidic medium (HNO3 acid) with thiocyanate

(SCN-) as a titrante and ferric alum (Fe3+) as indicator. This method can used for

determination of SCN- (direct titration). We can extend the applicability of this method to

halides (Cl−, Br−, I−) through the procedure known as back-titration.

Determination of SCN- (Direct titration)

AgNO3 + NH4SCN = AgSCN↓ + NH4NO3

white ppt

Fe3+ + SCN- = [Fe (SCN)]2+

red, complex ion

Determination of halides (Back titration)

Excess AgNO3 is added to precipitate out all of the halide ion. The AgCl precipitate is

filtered, and the excess Ag+ is back-titrated with SCN- producing a white precipitate of

AgSCN. Once the Ag+ is consumed, the excess SCN- reacts with the Fe3+ ion (the indicator)

producing a red [Fe(SCN)]2+ complex. Thus, the appearance of the red color at the end

point.

NaCl + AgNO3(known excess) = AgCl↓ + NaNO3

analyte white ppt

AgNO3 + NH4SCN = AgSCN↓ + NH4NO3

titrant white ppt

Fe3+ + SCN- = [Fe (SCN)]2+

indicator red, complex ion

Total amount of Ag+ is known, so amount consumed by Cl- can be calculated. Subtract excess

[Ag+] from total [Ag+] used to precipitate Cl-.

Notes:

The titration is usually done in acidic medium

- Favorable to formation of the red complex ion.

- To dissolve the silver salts of carbonate, oxalate, and phosphate (remove

interferences).

- To prevent precipitation of Fe(III) hydroxide, Fe(OH)3 then Fe(III)-hydrated

oxide, Fe2O3.X H2O.

- Not HCl or H2SO4? Due to formation of AgCl ppt or slightly dissociate Ag2SO4

- Not basic medium? Due to formation of AgOH then Ag2O ( black ppt).

-

Chapter 2 93

If applied to chloride

The AgCl precipitate must be separated from SCN- to prevent the reaction of SCN-

with it:

AgCl + SCN- = AgSCN + Cl-

Since AgSCN is less soluble than AgCl (Ksp of AgSCN < Ksp of AgCl), equilibrium will shift to

the right causing a negative error for the chloride analysis. To eliminate this error, AgCl

must be filtered or add nitrobenzene before titrating with SCN-; nitrobenzene will form an

oily layer on the surface of the AgCl precipitate, thus preventing its reaction with SCN-.

Fajans Method

This method uses an adsorption indicator. Adsorption indicators are organic compounds that

tend to be adsorbed onto the surface of the solid precipitate (colloidal particles) in a

precipitation titration.

Adsorption indicators work best when:

They do not precipitate out silver ion when the indicators are at low concentration.

They bind to the precipitate only when excess silver ion is present to produce color.

Example of Adsorption Indicators

Fluorescein (H2In)

A polycyclic compound that ionizes in solution to yield yellow-green fluoresceinate ions.

H2In (Fluorescein) ↔ 2H+ + ln2- (Fluoresceinate) yellow green

Fluoresceinate adsorbs to silver ions on the surface of a precipitate when excess silver ion is

present, producing a reddish-colored surface. Only the ionized fluoresceinate produces the

red color.

AgCl(s) + Ag+ + In2- = AgCl(s)Ag+ln2-

red

94

Titration of NaCl with AgNO3

During the titration, colloids are formed.

Ag+ + Cl = AgCl↓

titrant analyte white ppt


The surface of the precipitant particles will be negatively charged due to the

adsorption of excess Cl on the surface of the particles. A diffuse positive counter-

ion layer will surround the particles. The primary adsorption layer is negatively

charged and the anionic indicator is repelled.

At the equivalence point

There is no longer an excess of analyte Cl-, and the surface of the colloidal particles

are largely neutral.


There will be an excess of titrant Ag+, some of these will adsorb to the AgCl

particles. Now, negatively charged fluorescein can penetrate the counter ion layer

and adsorb onto the AgCl lattice due to its affinity to Ag+. The fluorescein

concentration is too small to form an AgIn precipitate in solution. The formed

complex has a red color.

Conditions that favor the formation of colloids:

Fast addition of reagents.

Add dextrin to retard coagulation.

pH control is important to generate the dye anion in solution.

Avoid light, since Ag-indicator complex is light sensitive and decomposes after a

while.

Chapter 2 95

Applications of precipitation titrations

Precipitation titration curves

It is a plot of the precipitated ion concentration against volume of titrant added. Plots of

argentometric titration curves are normally Sigmoidal curves (S shape) consisting of p

function of titrant (pAg) versus volume of titrant (AgNO3 solution) added.

Example: Titration of iodide with AgNO3

I- + Ag+ → AgI↓

↓AgCl ppt

AgCl ppt ↓


AgCl Precipitate surface is

initially negatively charged

due to excess Cl-

After equivalence point, have excess Ag+

so surface is now positively charged and

then the negative adsorption indicator

will bind and change color.

96

AgI(S) ↔ I- + Ag+ , KSP = [Ag+][I-] = 8.3 x 10-17

Since KSP is so small, each addition of Ag+ reacts completely with I-

At equivalence point, sudden increase in Ag+ concentration. All I- has been consumed.

Equivalence point is the steepest point of the curve (point of maximum slope, inflection

point).

Effect of KSP on titration curve

Lowest solubility gives steepest change at equivalence point. The smaller Ksp for a titration

reaction, the more pronounced will be the change in concentration near the equivalence

point. The sharpness of the end point and the magnitude of the inflection:directly related to

the initial conc. & inversely related to KSP.

Chapter 2 97

Observations about argentometric titrations

High reagent concentrations give sharper, more dramatic equivalence point changes

in pAg and better end points.

The smaller the Ksp, the more complete the precipitation reaction and the sharper

the equivalence region changes.

Both Ksp and the reagent concentrations affect the choice and use of an endpoint

indicator.

Fractional precipitation

When a mixture of two ions are titrated, the less soluble precipitate will be precipitated

first. When the equivalence point of the less soluble precipitate is obtained, the

concentration of titrant suddenly increases and then the more soluble salt begins to

precipitate. Therefore, we should observe two breaks in the titration curve, one

corresponding to each equivalence point.

Titration of a mixture of Cl- and I- with AgNO3

(Two Stage Titration Curve)

Ag+ + Cl → AgCl(S)

KSP(AgCl) = 1.8 x 10-10

Ag+ + I → AgI(S)

KSP(AgI) = 8.3 x 10-17

KSP(AgI) << KSP(AgCl)

Product with the smaller Ksp Precipitates. Because we add AgNO3 gradually from the burret,

the concentration of Ag+ starts low and increase gradually, therefore, all Agl is precipitated

firstly while Cl- ions remains in solution. Then, as [Ag+] increases, AgCl starts to precipitate.

98

Titration of (a) Mixture of I- & Cl-, (b) I-

Complex formation titrations

In this titration, the titrant is a complexing agent and forms water-soluble complex with the

analyte which is a metal ion. The titrant is often a chelating agent.

The formation of complexes

Complex compounds are those formed as a result of a coordinate bond formation between a

donor group (ligand) and acceptor group (metal), to produce a complex which has different

properties from those of the free metal ion

Complex: metal + ligand

Metals

They are Lewis acids that has one or more of empty orbitals and accept electron pairs from

donating ligands (electron acceptors, e.g. Mg2+, Ca2+, Al3+,…).

Ligands

They are Lewis bases that has at least one pair of unshared electrons available for bond

formation. They are neutral or negatively charged species (molecules or ions) containing

lone pairs electron that can be donated to metal ion (electron donors).

Ligands classified as uni/monodentate, bidentate or multidentate according to the number

of bonding sites (lone pair electrons) that they have. For example: H2O, NH3, Cl-, Br-, I- ……

Chapter 2 99

Unidentate ligand

A ligand that has a single donor group, such as NH3, H2O

Bidentate ligand

A ligand that has two groups available for coordinate bonding, such as ethylenediamine

H2NCH2CH2NH2

Multidentate ligand

A ligand that can form several bonds to single metal ion, e.g. ethylenediaminetetraacetic

acid (EDTA)

Most metal ions form coordination compounds with electron-pair donors (ligands). The

number of coordinate bonds that a cation tends to form with electron donors is its

coordination number (C.N.). Typical values for coordination numbers are 2, 4, and 6. For

example, [Cu(NH3)4]2+ (C.N. = 4), [CuCl4]

2- (C.N. = 4), [Cr(NH3)6]3+ (C.N. = 6)

The species formed as a result of coordination can be electrically positive, neutral, or

negative. For example, copper(II), which has a coordination number of 4, forms a cationic

ammine complex. Cu(NH3)42+; a neutral complex with glycine, Cu(NH2CH2COO)2 and an

anionic complex with chloride ion, CuCI42-.

Chelates

A chelate is a complex that involves ligands with two or more bonding sites forming a ring

structure complex. Reaction called chelation reaction. Chelate complexes are more stable

than non-chelate complex.

The Cu complex of glycine is an example for chelation:

100

Chelating agent

A substance whose molecules can form several bonds to single metal ion. In other words, a

chelating agent is a multidentate ligand.

Examples of chelating agents

Chelating agents can be used for titration of metals to form complex ions (complexometric

titration).

Ethylenediamenetetraacetic acid (EDTA)

Chelating agent has six potential sites for bonding a metal ion (the four carboxyl groups and

the two amino groups providing six lone pairs electrons). Thus, EDTA is a hexadentate

ligand.

O O

HO C C C H2 H2

N C C N

C OH

HO C C C C OH

O O

EDTA is fully protonated (H4Y) at pH < 3, half protonated (H2y2-) between pH 3-10, and fully

deprotonated (Y4-) at pH > 10

Chapter 2 101

EDTA is commercially available as H4Y and the dihydrate of the sodium salt, Na2H2Y.2H2O

EDTA is found to be sparingly soluble in water (0.2% w/v) whereas its corresponding

disodium salt is almost 50 times more soluble than the parent compound (solubility 10%

w/v). Therefore, it is the disodium salt of EDTA (H2Y2-) which is normally used in

complexometric titrations.

EDTA is a valuable titrant and is the most widely used complexometric titrant, because:

react more completely with metal ion.

usually react in a single step.

all metal-EDTA complexes have a 1:1 stoichiometry.

they are very stable complexes.

provide sharper end-points.

Acidic properties of EDTA: EDTA (H4Y) has four carboxylic acid groups

H4Y H+ + H3Y

-

H3Y- H+ + H2Y2-

H2Y2- H+ + HY3-

HY3- H+ + Y4-

K1 = 1.02 x 10-2

K2 = 2.14 x 10-3

K3 = 6.92 x 10-7

K4 = 5.50 x 10-11

Structure of EDTA (H4Y) and its dissociation products

102

Note that the fully protonated species H4Y exists as the double zwitterion with the amine

nitrogens and two of the carboxylic acid groups protonated. The first two protons dissociate

from the carboxyl groups, while the last two come from the amine groups.

Complexes of EDTA with metal ions

The reagent combines with metal ions in a 1:1 ratio regardless of the charge on the cation.

Metal-EDTA complex forms a cage-like structure around the metal ion.

Structure of a metal/EDTA complex.

Note that EDTA behaves here as a hexadentate ligand in that six donor atoms are involved in

bonding the metal cation. We usually express the equilibrium for the formation of complex

ion in terms of the Y4- form (all four protons dissociated).

Ag+ + Y4- AgY3-

Al3+ + Y4- AlY-

Chapter 2 103

Where, KMY is the equilibrium constant and called formation constant for EDTA complex

Note that, the higher the formation constant (KMY) the more complete and spontaneous the

reaction would be, the more it would be suitable for titrimetric analysis.

metal ion indicators for EDTA Titrations

The metal ion indicator is a dye which is capable of acting as a chelating agent to give a

dye-metal complex. The latter is different in color from the dye itself and also has a lower

formatiom constant (Kf) than the EDTA-metal complex.

The color of the solution, therefore, remains that of the dye complex until the end point,

when an equivalent amount of EDTA has been added. As soon as there is the slightest excess

of EDTA, the indicator is displaced and the metal-indicator complex decomposes to produce

free indicator; this is accomplished by a change in color.

Mn+ + Y4- MY(n-4)+

[MY(n-4)+] KMY=

[Mn+] [Y4-]

104

Common metal ion indicators

Eriochrome Black T

Structure and molecular model of Eriochrome Black T

Chapter 2 105

is a typical metal ion indicator that is used in the titration of several common cations. Its

behavior as a weak acid is described by the equations:

It has blue color in the pH range (7-11) in the uncomplexed form. The metal complexes of

Eriochrome Black T are generally red,, therefore, it is necessary to adjust the pH to 7 or

above so that the blue form of the species, HIn2-, predominates in the absence of a metal

ion.

Until the equivalence point in a titration, the indicator complexes the excess metal ion so

that the solution is red. With the first slight excess of EDTA, the solution turns blue as the

free uncomplexed indicator accumulated.

Complexometric titration curves

Titration curve is usually a plot of pM = - log [M] as a function of the volume of titrant

added.

Most often, in complexometric titrations the ligand is the titrant and the metal ion the

analyte, although occasionally the reverse is true.

As titrants, multidentate ligands, particularly those having four or six donor groups (such as

EDTA) have two advantages over their unidentate counterparts:

First, they generally react more completely with cations and thus provide sharper

end points.

Second, they ordinarily react with metal ions in a single step process (1:1 ratio),

whereas complex formation with unidentate ligands usually involves two or more

intermediate species.

Like a strong acid/strong base titration, there are three points on the titration curve of a

metal with EDTA: before, at, and after the equivalence point.

Generic titration curve

106

EDTA titration curves for 50.0 ml of 0.005 M Ca2+ (KCaY = 1.75×1010) and Mg2+ (KMgY = 1.72×108) at pH 10

Note that because of the larger formation constant, the reaction of Ca2+ with EDTA is more

complete, and a larger change occurs in the equivalence-point region. The shaded areas

show the transition range for the indicator Eriochrome Black T.

Effect of pH on complex formation

EDTA has four protonated states, and since the actual complexing species is Y4-, complexes

will form more efficiently and stable in alkaline solution. However, some complexes of

divalent and trivalent metals, e.g. copper, lead, nickel, cobalt are stable down to pH 3.

Therefore, solutions are usually buffered at a pH at which the complex is most stable and at

which the color change of the indicator is most distinct.

Influence of pH on the titration of 0.0100 M Ca2+ with 0.0100 M EDTA

Chapter 2 107

Note that the end point becomes less sharp as the pH decreases because the complex

formation reaction is less complete under these circumstances.

Minimum pH needed for satisfactory titration of various cations with EDTA

EDTA titration techniques

Direct titration

A suitable buffer solution and indicator are added to the metal ion solution and the solution

titrated with standard EDTA solution until the indicator just changes color. A blank titration

may be performed, omitting the analyte as a check on the presence of traces of metallic

impurities in the reagents.

108

Back titration

Back-titration procedures are used when no suitable indicator is available. when the

reaction between analyte and EDTA is slow, or when the analyte forms precipitates at the

pH required for its titration.

The sample is heated with measured excess of EDTA to form the soluble complex with the

analyte (metal), and then the excess EDTA is back-titrated with Mg2+ or Zn2+ of known

concentration using a suitable indicator.

Displacement titration

For metals without a good indicator ion, the analyte can be treated with excess Mg(EDTA)2-.

The analyte displaces Mg2+, and then Mg2+ can be titrated with standard EDTA

Indirect titration

Anions can be analyzed by precipitation with excess metal ion and then titration of the

metal in the dissolved precipitate with EDTA.

Applications of EDTA

1. Water hardness determination

EDTA titrations are routinely used to determine water hardness in lab. In this reaction, the

EDTA ligand will react with the minerals present in the water (mainly calcium and

magnesium carbonates, sulfates, etc.).

Hard water is water that has a high mineral content. Hard water usually consists of calcium

(Ca2+), magnesium (Mg2+) ions, and possibbly other dissolved compounds such as bicarbonates

and sulfates.

The mineral ions along with other metal ions such as Fe3+ and Pb2+ can be removed from

hard water by the addition of EDTA.

2. Food industry

EDTA is used as a stabilizing agent in the food industry. EDTA deactivates enzymes (that

contain metal) responsible for food spoilage by removing the metal ions from them and

forming stable chelates with them.

3. Medical

As an anticoagulant for stored blood in blood banks; it prevents coagulant by sequestering

the calcium ions required for clotting.

As an antidote for lead poisoning, calcium disodium EDTA exchanges its chelated calcium for

lead, and the resulting lead chelate is rapidly excreted in the urine.

Chapter 3 109

Objectives

Discuss the principle of gravimetric analysis.

Identify technique of gravimetric analysis.

Calculate the gravimetric factor in gravimetric analysis.

Solve the problems of gravimetric analyses.

Estimate the quantity of precipitates of the analytes from solutions.


The principle of gravimetric analysis

Gravimetric analysis is a method of the quantitative chemical analysis which is based on

exact measurement of weight of defined substance or its components allocated in

chemically pure compound or in the form of corresponding compounds (precisely known

constant structure).

Gravimetric methods

There are two major types of gravimetric methods:

Precipitation gravimetry: is a gravimetric method in which the signal is the mass of a

precipitate. In this method the analyte is converted to a sparingly soluble precipitate. This

precipitate is then filtered, washed free of impurities, and converted to a product of known

composition by suitable heat treatment, and the product is weighed.

Volatilization gravimetry: is a gravimetric method in which the loss of a volatile species

gives rise to the signal. In this the analyte or its decomposition products are volatilized at a

suitable temperature. The volatile product is then collected and weighed, or, alternatively,

the mass of the product is determined indirectly from the loss in mass of the sample.

Direct method of volatilization gravimetry. A defined volatile component absorb a specific

absorber and on increase of the weight of the last calculate the weight of a volatile defined

component.

Chapter 3

Gravimetric analysis

110

For example: NaHCO3 in antacid tablets

NaHCO3(aq) + H2SO4(aq) → CO2(g) + H2O(l) + NaHSO4(aq)

↓ absorption tube

2NaOH + CO2 → Na2CO3 + H2O

Indirect method volatilization gravimetry. In indirect methods define weight of the rest of

substance after full removal of a defined volatile component.

BaCl2∙2H2O → BaCl2 + 2H2O↑

Technique of precipitation gravimetry

How to Perform a Successful Gravimetric Analysis?

What steps are needed?

After appropriate dissolution of the sample, the following steps should be followed for a

successful gravimetric procedure:

Preparation of the solution

Precipitation

Digestion

Filtration

Washing

Drying or igniting

Weighing

Calculation

1. Preparation of the Solution:

This may involve several steps including adjustment of the pH of the solution in order for the

precipitate to occur quantitatively and get a precipitate of desired properties, removing

Chapter 3 111

interferences, adjusting the volume of the sample to suit the amount of precipitating agent

to be added.

2. Precipitation:

This requires addition of a precipitating agent to the sample solution to form precipitate.

How do precipitates form?

Precipitates form in two ways, by nucleation and by particle growth. The particle size of a

freshly formed precipitate is determined by which way is faster.

In nucleation, Upon addition of the first drops of the precipitating agent, supersaturation

occurs, and then nucleation starts to occur where every few ions, atoms, or molecules

(perhaps as few as four or five) come together to form a nucleus. At this point, addition of

extra precipitating agent will either form new nuclei or will build up on existing nuclei to

give a precipitate. Further precipitation then involves a competition between additional

nucleation and growth on existing nuclei (particle growth). If nucleation predominates, a

precipitate containing a large number of small particles results; if growth predominates, a

smaller number of larger particles is produced.

Requirements to precipitates and precipitating agents

Requirements to precipitates:

The ideal precipitating reagent would react with the analyte to give a precipitate that is:

1. Of sufficiently low solubility (Ksp10-8) so that no significant loss of the solid occurs during filtration and washing.

2. The precipitate should be whenever possible largely crystals for easily filtered and washed

free of contaminants.

3. Unreactive with constituents of the atmosphere

4. Of known chemical composition after it is dried or, if necessary, ignited

5. The precipitate should turn easily enough in weighed form.

Requirements to precipitating agents (precipitants):

It is desirable, that a precipitant was volatile compound.

A precipitant should be react specifically, and selectively with the analyte to precipitate a

defined ion in the presence of others ions.

*Specific precipitating agent (rare): react only with a single chemical species.

Dimethylglyoxime → Ni2+

112

*Selective precipitating agent (common): react with a limited number of species.

AgNO3 → Cl-, Br-, I- & SCN-

Organic precipitants:

Organic precipitants are chelating agents. They form insoluble metal chelates.

For examples:

Dimethylgloxime (DMG): Specific precipitant for Ni2+ in NH3

Dimethylgloxime (DMG) Ni(DMG)2

(Organic precipitant) (Precipitated form)

8-Hydroxyquinoline (oxine): Useful precipitant for Al3+ and Mg2+

8-Hydroxyquinoline(oxine) Mg(Oxinate)2


Chapter 3 113

Oxalic acid: Useful precipitant for Ca2+

O

C OH

C OH

O

O

C

+ Ca2+

C

O

O

Ca + 2H+

O

Oxalic acid Ca(Oxalate)


Cupron (α-Benzoinoxime): Organic precipitant for Cu2+ in NH3

OH NOH

CH C

Cupferron (Ammonium nitrosophenylhydroxylamine): Organic precipitant for Fe3+

NO

N O NH4+

Advantage of organic precipitants consists in the following:

Solubility of precipitate with organic precipitants is less.

Precipitates with organic reagents are crystal.

Precipitates with organic reagents are purer as on their surface impurity are less

adsorbed.

Organic precipitant have higher selectivity and specificity.

The gravimetric factor at definition with organic reagents on much less so, accuracy

of definition increases.

Particle Size and Filterability of Precipitates

Precipitates made up of large particles are generally desirable in gravimetric work because

large particles are easy to filter and wash free of impurities. In addition, such precipitates

are usually purer than are precipitates made up of fine particles.

114

What Factors Determine Particle Size?

The particle size of solids formed by precipitation varies enormously. At one extreme are

colloidal suspension, whose tiny particles are invisible to the naked eye (10-7 to 10-4 cm in

diameter). Colloidal particles show no tendency to settle from solution, nor are they easily

filtered. At the other extreme are particles with dimensions on the order of tenths of

millimeter or greater. The temporary dispersion of such particles in the liquid phase is called

a crystalline suspension. The particles of a crystalline suspension tend to settle

spontaneously and are readily filtered.

The particle size of a precipitate is influenced by experimental variables as precipitate

solubility, temperature, reactant concentrations, and the rate at which reactants are mixed.

The particle size is related to a single property of the system called its relative

supersaturation, where

Relative supersaturation = (Q – S) / S

In this equation, Q is the concentration of the solute at any instant and S is its equilibrium

solubility.

When (Q – S)/ S is large, the precipitate tends to be colloidal.

When (Q – S) / S is small, a crystalline solid is more likely.

Controlling Particle Size

Experimental variables that minimize supersaturation and thus lead to crystalline

precipitates include:

1. elevated temperatures to increase the solubility of the precipitate (S in Equation).

2. dilute solutions to minimize concentration (Q in Equation).

3. slow addition of the precipitating agent with good stirring to keep Q as low as possible.

The last two measures also minimize the concentration of the solute (Q) at any given

instant.

4. Larger particles can also be obtained by pH control, provided the solubility of the

precipitate depends on pH.

Colloidal Precipitates

Colloidal suspensions are stable because all the particles present are either positively or

negatively charged. This charge results from cations or anions that are bound to the surface

of the particles. The process by which ions are retained on the surface of a solid is known as

adsorption. colloidal precipitate tends to adsorb its own ions present in excess, forming

what is called a primary ion layer which attracts ions from solution forming a secondary or a

counter ion layer. Individual particles repel each other keeping the colloidal properties of

Chapter 3 115

the precipitate. We can readily demonstrate that colloidal particles are charged by

observing their migration when placed in an electrical field.

Representation of silver chloride colloidal particle and adsorptive layers when Cl- is in excess.

116

Coagulation (or agglomeration) of Colloids

Converting a colloid suspension into a filterable solid. Coagulation can be hastened by

heating, stirring, and adding an electrolyte to the medium in order to shield the charges on

colloidal particles and force agglomeration.

Peptization of Colloids

Peptization refers to the process by which a coagulated colloid reverts to its original

dispersed state. When a coagulated colloid is washed, some of the electrolyte responsible

for its coagulation is leached from the internal liquid in contact with the solid particles.

Removal of this electrolyte has the effect of increasing the volume of the counter-ion layer.

The repulsive forces responsible for the original colloidal state are then reestablished, and

particles detach themselves from the coagulated mass. The washings become cloudy as the

freshly dispersed particles pass through the filter. Peptization is prevented by washing ppt

with a solution of a volatile electrolyte. For example, AgCl ppt washed with a dil. HNO3.

Crystalline Precipitates

Crystalline precipitates are generally more easily filtered and purified than coagulated

colloids. In addition, the size of individual crystalline particles, and thus their filterability,

can be controlled to a degree.

The particle size of crystalline solids can often be improved significantly by minimizing Q,

maximizing S, or both in Equation. Minimization of Q is generally accomplished by using

dilute solution and adding the precipitating from hot solution or by adjusting the pH of the

precipitation medium.

Digestion of crystalline precipitates (without stirring) for some time after formation

frequently yields a purer, more filterable product. The improvement in filterability results

from the dissolution and recrystallization.

Contamination of precipitate

Contamination during precipitation may be due to coprecipitation & true precipitation

I. Coprecipitation

a process in which normally soluble compounds are carried out of solution during precipitate

formation.

There are five types of coprecipitation:

1. surface adsorption, 2. Mixed-crystal formation, 3. Occlusion, 4. Inclusion, 5.

Mechanical entrapment

Surface adsorption and mixed crystal formation are equilibrium processes, whereas occlusion

and mechanical entrapment arise from the kinetics of crystal growth.

Chapter 3 117

1. Surface adsorption

Adsorption is a common source of coprecipitation that is likely to cause significant

contamination of precipitates with large specific surface areas, that is coagulated colloids. a

normally soluble compound is carried out of solution on the surface of a coagulated colloid.

Coagulation of a colloid does not significantly decrease the amount of adsorption because

the coagulated solid still contains large internal surface areas that remain exposed to the

solvent. The coprecipitated contaminant on the coagulated colloid consists of the lattice ion

originally adsorbed on the surface before coagulation and the counter ion of opposite charge

held in the film of solution immediately adjacent to the particle. The net effect of surface

adsorption is therefore the carrying down of an otherwise soluble compound as a surface

contaminant.

For example, in chloride analysis primary adsorbed ion: Ag+ counter-ion layer: NO3- or other

anions → AgNO3 (normally soluble) is coprecipitated with the AgCl.

Minimizing Adsorbed Impurities on Colloids

The purity of many coagulated colloids is improved by:

i) Digestion

During this process, water is expelled from the solid to give a denser mass that has a smaller

specific surface area for adsorption.

ii) Washing with volatile electrolyte solution

Washing a coagulate colloid with a solution containing a volatile electrolyte may also be

helpful because any nonvolatile electrolyte added earlier to cause coagulation is displace by

the volatile species. Washing generally does not remove much of the primarily adsorbed ions

because the attraction between these ions and the surface of the solid is too strong.

Exchange occurs, however between existing counter ions and ions in the wash liquid.

iii) Reprecipitation

A drastic but effective way to minimize the effects of adsorption is reprecipitation, or

double precipitation. Here, the filtered solid is redissolved and reprecipitated. The first

precipitate ordinarily carries down only a fraction of the contaminant present in the original

solvent. Thus, the solution containing the redissolved precipitate has a significantly lower

contaminant concentration than the original, and even less adsorption occurs during the

second precipitation. Reprecipitation adds substantially to the time required for an analysis.

2. Mixed-Crystal Formation (Isomorphous replacement)

In mixed-crystal formation, one of the ions in the crystal lattice of a solid is replaced by an

ion of another element. For this exchange to occur, it is necessary that the two ions have

the same charge and that their sizes differ by no more than about 5%. Furthermore, the two

salts must belong to the same crystal class. For example, MgKPO4, in MgNH4PO4, SrSO4 in

BaSO4, and MnS in CdS.

118

The extent of mixed-crystal contamination increases as the ratio of contaminant to analyte

concentration increases. Mixed-crystal formation is troublesome because little can be done

about it.

Separation of the interfering ion may have to be carried out before the final precipitation

step. Alternatively, a different precipitating agent may be used.

3. Occlusion

When a crystal is growing rapidly during precipitate formation, foreign ions in the counter-

ion layer may become trapped, or occluded, within the growing crystal.

4. Inclusion

If the precipitation medium contains ions of the same charge and size as one forming the

crystal structure of the precipitate, this extraneous ion can replace an ion from the

precipitate in the crystal structure.

For example, in the precipitation of NH4MgPO4 in presence of K+

, ammonium leaves the

crystal magnesium ammonium phosphate and is replaced by K+ since both have the same

charge and size. However, the FW of NH4+ is 18 while that of K+ is 39. In this case a positive

error occurs as the weight of precipitate will be larger when K+ replaces NH4+. In other

situations we may get a negative error when the FW of the included species is less than the

original replaced species.

Chapter 3 119

NH4MgPO4 in presence of K+

5. Mechanical entrapment

Occurs when crystals lie close together during growth. Here, several crystals grow together

and in so doing trap a portion of the solution in a tiny pocket.

Occlusion, inclusion and mechanical entrapment are at a minimum when the rate of

precipitate formation is low, that is, under conditions of low supersaturation. Digestion is

often remarkably helpful in reducing these types of copreipitation. The rapid solution and

reprecipitation that goes on at the elevated temperature of digestion opens up the pockets

and allows the impurities to escape into the solution.

II. True precipitation

a process in which impurity is precipitated with the precipitate.

There are two types of True precipitation:

1. Simultaneous precipitation

In this process the analyte and impurity are precipitated at the same time. Where, they

have the same Ksp. For example, during precipitation of Fe3+ as Fe(OH)3, Al3+ and Cr3+ ions

will precipitate also if present.

To minimize Simultaneous precipitation, separation of the interfering ion may have to be

carried out before the final precipitation step. Alternatively, a specific or selective

precipitating agent may be used.

2. Post precipitation

In cases where there are ions other than analyte ions which form precipitates with the

precipitating agent but at much slower rate than analyte, and if the precipitate of the

analyte is left for a long time in contact with the mother liquor without filtration then the

other ions start forming a precipitate over the original precipitate leading to positive error.

For example, precipitation of Cu2+ as sulfide in presence of Zn2+ ions. CuS is formed first but

if not directly filtered, ZnS starts to precipitate on the top of it. The same is observed in the

120

precipitation of Ca2+ as oxalate in presence of Mg2+. Rapid filtration can avoid post

precipitation.

Precipitation from homogeneous solution

Precipitation from homogeneous solution is a technique in which a precipitating agent is

generated in a solution of the analyte by a slow chemical reaction. Local reagent excesses

do not occur because the precipitating agent appears gradually and homogeneously

throughout the solution and reacts immediately with the analyte. As a result, the relative

supersaturation is kept low during the entire precipitation. In general, homogeneously

formed precipitates, both colloidal and crystalline, are better suited for analysis than a solid

formed by direct addition of a precipitating agent.

Reactions for homogeneous generation of selected precipitating agents:

3. Digestion of the Precipitate:

Heating the precipitate within the mother liquor (or solution from which it precipitated) for

a certain period of time to 30 min to 1 hour in order for the particles to be digested.

Digestion involves dissolution of small particles and reprecipitation on larger ones resulting

in particle growth and better precipitate characteristics. This process helps to produce

larger crystals that are more easily filtered from solution and called Ostwald ripening. An

important advantage of digestion is observed for colloidal precipitates where large amounts

of adsorbed ions cover the huge area of the precipitate. Digestion forces the small colloidal

particles to agglomerate which decreases their surface area and thus adsorption.

Ostwald ripening of AgCl precipitate

Chapter 3 121

4. Filtering the precipitate:

The process in which the precipitate is isolated from the solution by use filter Papers

(weight of ashes 0,00003 – 0,00007 g), ashless filter Papers (do not remain residue during

ignition) or filtering crucibles.

Filtering crucibles

Proper procedure for filtering solids using filter paper.

The filter paper circle in (a) is folded in half (b), and folded in half again (c). The filter paper is parted (d),

and a small corner is torn off (e). The filter paper is opened up into a cone and placed in the funnel (f). Note

that the torn corner is placed to the outside.

122

2 4

Procedure for filtering through a filtering crucible.

The trap is used to prevent water from a water aspirator from backwashing into the suction flask.

5. Washing the precipitate:

It is crucial to wash the precipitate very well in order to remove all adsorbed species which

will add to weight of precipitate.

Properties of washing liquid:

Dissolve impurities but not precipitate.

Do not react with the precipitate.

Volatile, for easily to removed during drying or ignition.

Choice of a washing liquid:

Crystalline precipitates with low solubility are rinsed by water.

Colloidal precipitates are rinsed by solutions of volatile electrolytes (such as dilute

nitric acid, ammonium nitrate, or dilute acetic acid) to avoid of peptization of a

precipitate.

Precipitates with high solubility are rinsed by solutions of electrolytes which contain

the same ion with a precipitate to decrease solubility of it.

For example, ammonium oxalate (NH4)2C2O4 used as washing liquid for calcium oxalate

CaC2O4 precipitate.

CaC2O4 → Ca2+ + C O 2-

(NH4)2C2O4 → 2NH4

+ + C2O42-

Common ion

Decrease solubility of ppt

Chapter 3 123

6. Drying and ignition of the precipitate:

After filtrationand washing, a gravimetric precipitate is dried at about 120-150 oC in an oven

until its mass becomes constant. Drying removes the solvent and any volatile species carried

down with the precipitate. Some precipitates are also ignited in a muffle furnace at

temperatures ranging from 600-1200 oC to decompose the solid and form a compound of

exactly known composition so that the amount of analyte can be accurately determined.

This new compound is often called the weighing form. The temperature required to produce

a suitable weighing form varies from precipitate to precipitate.

Ignition of some precipitates

Requirements to the weighed form of the precipitate:

Exact conformity of structure to the chemical formula (precipitate form Fe(OH)3

Fe2O3 xH2O, and the weighed form Fe2O3).

Chemical stableness of the weighed form.

The contents of a defined element in the weighed form should be as it is possible

smaller.

Importance of the low contents of defined substance into the weighed form

WWeeiigghheedd ffoorrmm

CCrr22OO33

WWeeiigghheedd ffoorrmm

BBaaCCrrOO44

152 g CCrr22OO33 - 104 g Cr

1 mg CCrr22OO33 - Х mg Cr

Х = 1041 / 152 = 0.7 mg Cr loss

253.3 g BBaaCCrrOO44 - 52 g Cr

1 mg BBaaCCrrOO44 - Х mg Cr

Х = 521 / 253.3 = 0.2 mg Cr loss

124

Selected gravimetric method for inorganic cations based on precipitation

Gravimetric calculations

The point here is to find the weight of analyte from the weight of precipitate. We can use

the concepts discussed in stoichiometric calculations but let us learn something else. Assume

Cl2 is to be precipitated as AgCl, then we can write a stoichiometric factor reading as

follows: one mole of Cl2 gives 2 moles of AgCl. We can define a quantity called the

gravimetric factor (GF) where:

a/b are stoichiometric coefficients and = moles of analyte/moles of precipitate

GF for Cl2 = (1 mol Cl2/2 mol AgCl) X (FW Cl2/FW AgCl)

or,

Gravimetric Factor (GF) (analyte) = (a/b) X (FWanalyte/FWppt)

Weight of analyte (g) = Weight of precipitate (g) X GF

% analyte = [Weight of precipitate (g) X GF /Weight of sample (g)] X

10

% analyte = [Weight of analyte (g)/Weight of sample (g)] X 10

Chapter 3 125

Example 22:

Calculate the gravimetric factor for each analyte of the following:

Analyte Precipitate form

1) Mg (wt = 24 g/mol) Mg2P2O7 (FW = 221.94 g/mol)

2) P (wt = 30.97 g/mol) Ag3PO4 (FW = 711.22 g/mol)

3) Bi2S3 (FW 514.15 g/mol) BaSO4 (FW = 233.40 g/mol)

Solution:

1) 2Mg → Mg2P2O7

2 mol of Mg = 1 mol of Mg2P2O7

GF (analyte) = (moles of analyte/moles of precipitate) X (FWanalyte/FWppt)

GF (Mg) = (moles of Mg/moles of Mg2P2O7) X (FW Mg/FW Mg2P2O7)

GF (Mg) = (2/1) X (24 g/mol /221.94 g/mol) = 0.216

2) P → Ag3PO4

1 mol of P = 1 mol of Ag3PO4

GF (P) = (moles of P/moles of Ag3PO4) X (wt P/FW Ag3PO4)

GF (P) = (1/1) X (30.97 g/mol /711.22 g/mol) = 0.0435

3) Bi2S3 → 3BaSO4

1 mol of Bi2S3 = 3 mol of BaSO4

GF (Bi2S3) = (moles of Bi2S3/moles of BaSO4) X (FW Bi2S3/FW Ag3PO4)

GF (Bi2S3) = (1/3) X (514.15 g/mol /233.40 g/mol) = 0.734

Example 23:

Phosphate in a 0.2711 g sample was precipitated giving 1.1682 g of (NH4)2PO4.12 MoO3 (FW =

1876.5 g/mol). Find percentage P (wt = 30.97 g/mol) and percentage P2O5 (FW = 141.95

g/mol) in the sample.

126

Solution:

P → (NH4)2PO4.12 MoO3

1 mol of P = 1 mol of (NH4)2PO4.12 MoO3

GF (analyte) = (moles of analyte/moles of precipitate) X (FWanalyte/FWppt)

GF (P) = (moles of P/moles of (NH4)2PO4.12 MoO3) X (FW P/FW (NH4)2PO4.12 MoO3)

GF (P) = (1/1) X (30.97 g/mol /1876.5 g/mol) = 0.0165

Weight of analyte (g) = Weight of precipitate (g) X GF

Weight of P (g) = Weight of (NH4)2PO4.12 MoO3 (g) X GF (P)

Weight of P (g) = Weight of 1.1682 g X 0.0165 = 0.0192735 g

% analyte = [Weight of analyte (g)/Weight of sample (g)] X 100

% P = [Weight of P (g)/Weight of sample (g)] X 100

% P = (0.0192735 g/0.2711 g) X 100 = 7.11 %

Apply the same procedure to find the percentage of P2O5

Problems

1. Manganese in a 1.52 g sample was precipitated as Mn3O4 (FW = 228.8 g/mol) weighing

0.126 g. Find percentage Mn2O3 (FW = 157.9 g/mol) and Mn (wt = 54.94 g/mol) in the

sample.

2. What weight of sulfur (FW = 32.064 g/mol) ore which should be taken so that the weight

of BaSO4 (FW = 233.40 g/mol) precipitate will be equal to half of the percentage sulfur in

the sample.

3. A mixture containing only FeCl3 (FW = 162.2 g/mol) and AlCl3 (FW = 133.34 g/mol) weighs

5.95 g. The chlorides are converted to hydroxides and ignited to Fe2O3 (FW = 159.7 g/mol)

and Al2O3 (FW = 101.96 g/mol). The oxide mixture weighs 2.62 g. Calculate the percentage

Fe (wt = 55.85 g/mol) and Al (wt = 26.98 g/mol g/mol) in the sample.

4. Consider a 1.0000 g sample containing 75% potassium sulfate (FW 174.25 g/mol) and 25%

MSO4. The sample is dissolved and the sulfate is precipated as BaSO4 (FW 233.39 g/mol).

If the BaSO4 ppt weighs 1.4900 g, what is the atomic weight of M2+ in MSO4?

Chapter 3 127

5. A mixture of mercurous chloride (FW 472.09 g/mol) and mercurous bromide (FW 560.99

g/mol) weighs 2.00 g. The mixture is quantitatively reduced to mercury metal (wt 200.59

g/mol) which weighs 1.50 g. Calculate the % mercurous chloride and mercurous bromide

in the original mixture.

6. A 10.0 mL solution containing Cl- was treated with excess AgNO3 to precipitate 0.4368 g of AgCl. What was the concentration of Cl- in the unknown? (AgCl = 143.321 g/mol)

7. The calcium in a 200.0 ml sample of a natural water was determined by precipitating the

cation as CaC2O4. The precipitate was filtered, washed, and ignited in a crucible with an

empty mass of 26.6002 g. The mass of the crucible plus CaO (FW 56.077 g/mol) was

26.7134 g. Calculate the concentration of Ca (FW 40.078 g/mol) in the water in units of

grams per 100 ml.

8. An iron ore was analyzed by dissolving a 1.1324 g sample in concentrated HCl. The

resulting solution was diluted with water, and the iron(III) was precipitated as the hydrous

oxide Fe2O3.xH2O by addition of NH3. After filtration and washing, the residue was ignited

at high temperature to give 0.5394 g pure Fe2O3 (FW 159.69 g/mol). Calculate (a) the %

Fe (FW 55.847 g/mol) and (b) % Fe3O4 (FW 231.54 g/mol) in the sample.

9. A 0.2356 g sample containing only NaCl (FW 58.44 g/mol) and BaCl2 (FW 208.23 g/mol)

yielded 0.4637 g of dried AgCl (FW 143.32 g/mol). Calculate the percent of each halogen

compound in the sample.

10. A mixture containing only Al2O3 (FM 101.96) and Fe2O3 (FM 159.69) weighs 2.019 g. When

heated under a stream of H2, Al2O3 is unchanged, but Fe2O3 is converted into metallic Fe

plus H2O (g). If the residue weighs 1.774 g, what is the weight percent of Fe2O3 in the

original mixture?

129

Introduction

The aim of this laboratory course is to provide the student with experimental work which enhances the various topics covered in the lecture portion of the course. Experiments such as complexometric, neutralization, precipitation and redox titrations will be performed. As the course title indicates, the ultimate goal is to determine how much of a particular entity is contained in a sample. Therefore quantitative methods of preparing and handling solutions and reagents will be stressed.

Required materials

1. Safety goggles or safety glasses and a lab apron must be worn at all times.

2. All of your data will be collected in a lab notebook.

Attendance

All students are required to be present at the start of the lab, as this is when the theory and techniques of the laboratory experiment will be explained and demonstrated. The student should attempt to keep pace with the laboratory schedule as extra lab time will be limited. Should you find it necessary, for any reason, to miss a lab. it would be helpful to notify the instructor.

Laboratory grade

The lab portion of the course will contribute 40 pts to your overall grade and will be determined by computing your numerical grade according to the following:

Lab notebook and lab report (includes accuracy) 20 pts

Final practical exam. 20 pts

Before you come to lab, you must enter the title and purpose of the experiment into the laboratory notebook. Any required calculations needed for the experiment need to be completed in your lab notebook before coming to lab. You will given a zero for the experiment if the calculations are not completed before starting the experiment. It will also be helpful to set up any data tables or charts in your lab notebook. This material will be checked during each experiment.

Practical analytical chemistry

130

Your final lab report must be typewritten. All chemical structures are to be drawn with a molecular drawing software.

After you have finished the experiment, you will complete the Final laboratory

report in which you organize all your data and present your results. You will also answer any questions that may be included in the "include in the final laboratory report" section. You should also check your final value with the instructor to find out if it is too high or too low. Then analyze what factors would lead to the error in your value. The Conclusion section should briefly summarize the goal of the experiment and state whether it was achieved or not. It should also state the results of the experiment.

Good writing is essential. Lab notebooks and the final lab reports must be

grammatically correct and comprehensible; otherwise they will not be accepted.

Laboratory reports are due one week from the last scheduled date of the

experiment.

No late lab reports will be accepted!!

Please note that in order to pass the course, you must receive a passing grade in both

the lecture and laboratory parts of the course.

Academic Integrity

It is your responsibility to maintain a high degree of integrity in your work. Cheating of any kind will not be tolerated and will result in a failure in the course! The following are considered cheating: (a) Sharing of results and answers on lab reports, graded assignments, quizzes and exams; (b) Use of unauthorized materials during an exam; (c) Plagiarism, including copying a fellow student‘s lab report or homework.. When in doubt, both parties involved in plagiarism (both the copier and the copyee) will be held responsible for the integrity violation.

131

Basic tools and operations of analytical chemistry

Borosilicate glassware (Pyrex, Kimax): is normally used, because it is thermally stable.

Balances: Modern balances are electronic. They still compare one mass against another since they are calibrated with a known mass. Common balances are sensitive to 0.1 mg.

132

The single pan balance: operates by removing weights equal to the mass of the sample. Small residual imbalances are read optically from the deflection of the beam.

The single-pan balance is as accurate as electronic balances, and almost as fast. But it can‘t be interfaced to a computer to collect and process data. And you have to read a scale instead of a digital number.

Weighing bottles: are used for drying samples. Hygroscopic samples are weighed by difference, keeping the bottle capped except when removing the sample.

133

A weighing dish or boat: is used for direct weighing of samples.

Volumetric flasks: are calibrated to contain an accurate volume.

Volumetric pipettes: accurately deliver a fixed volume. A small volume remains in the tip.

134

Measuring pipettes: are straight-bore pipettes marked at different volumes. They are less accurate than volumetric pipettes.

Syringe pipettes: precisely deliver microliter volumes. They are commonly used to introduce samples into a gas chromatograph.

The syringe pipettes can reproducibly deliver a selected volume. They come in fixed and variable volumes. The plastic tips are disposable.

Single-channel and multichannel digital

displacement pipettes and microwell plates.

135

Burette: A 50-mL burette is marked in 0.1 mL increments. You interpolate to 0.01 mL, good to about ±0.02 mL. Two readings are taken for every volume measurement.

Position the black field just below the meniscus. Avoid parallax error by reading at eye level.

Proper technique for titration: Place the flask on a white background. Place the burette tip in the neck of the flask while your swirl.

136

Desiccators: are used to cool a dried or ignited sample. Cool a red hot vessel before placing in the desiccator.

Do not stopper a hot weighing bottle (creates a partial vacuum on cooling).

Desiccator and desiccator plate.

Drying agents: CaCl2 is commonly used. It needs periodic replacement when wet or caked.

137

Muffle furnace: Used to ignite samples at high temperatures, e.g., to dry ash organic matter.

Drying oven: Used to dry samples before weighing. Usually 110o C used.

A fume hood: is ―dirty‖ since it draws in laboratory air. A laminar-flow hood filters air (0.3 mm HEPA filter) and flows it out into the room. Use it as a workstation for trace analysis.

Laminar-flow workstation.

138

Wash bottles: Use these for quantitative transfer of precipitates and solutions, and for washing

precipitates.

Wash bottles: (a) polyethylene, squeeze type; (b) glass, blow type.

Filtering crucibles: Use for filtering non-gelatinous precipitates.

Filtering crucibles: (a) Gooch crucible; (b) sintered-glass crucible; (c) porcelain filter crucible.

Crucible holders: Mount the filtering crucible in a crucible holder and connect the filtering flask to a water aspirator.

139

Filter papers: are used for isolate the precipitate from the solution.

Properly folded filter paper: This provides a good seal and prevents air bubbles from being drawn in. Suction from the weight of the water in the stem increases the filtration rate. Let the precipitate settle in the beaker before beginning filtration.

Ashless filter papers: They are ignited away after collection of the precipitate. Use for gelatinous precipitates.

Proper technique for transfer of a precipitate: Decant the solution by pouring down the stirring rod. After decanting the mother liquor, add wash water to the precipitate and decant again, repeating 2-3 times. Then wash the precipitate into the filter.

140

Rubber policeman: Use this to scrub the walls of the beaker and collect all the precipitate (by washing).

Heat or ignite the crucible to a constant weight (to 0.3-0.4 mg) before adding the filtered precipitate. Fold the filter paper over the precipitate. Drive off moisture at low heat. Then gradually increase heat till the paper begins to char. After the paper is gone, ignite the precipitate.

Crucible and cover supported on a wire triangle for charring off paper.

Microwave ovens: provide rapid drying. Acid decomposition times are reduced from hours to minutes. Lower blank levels are achieved with reduced amounts of reagents.

Schematic of a microwave system.

141

Kjeldahl flasks: Use these for acid digestions. They are tilted while heating to avoid losses from ―bumping‖.

142

Acid-Base titrations

Experiment (1): Preparation of a NaOH standard solution using direct titration

Purposes

This experiment demonstrates the most common method for obtaining standard solutions for

titrimetric analysis. It involves preparation of a solution that has the approximate

concentration desired (usually within 10%), determination of the concentration by direct

titration against a primary standard, and a test of the accuracy of your determined

concentration by comparison with a known standard. It is important to standardize your

solution carefully, as it will be used in later experiments. You should be able to determine

your NaOH concentration to ± 0.5% of its actual value. You will be graded on your accuracy.

Introduction

Solid sodium hydroxide cannot be massed accurately because it absorbs water and carbon

dioxide from the air. Consequently, it is not possible to make an aqueous solution to a very

specific and accurate concentration.

The solution of NaOH(aq) will titrate against a known amount of an acid to determine the

concentration of NaOH.

The acid used to standardize the concentration of NaOH is potassium hydrogen phthalate

(KHP). It can be massed very accurately. KHP is actually the potassium salt of the phthalate

ion:

phthalate ion (acid)

The phthalate ion is the acid that is titrated by NaOH in the standardization:

KHP

143

Safety issues and chemical hazard information

1. Avoid skin contact with chemicals.

2. Any acid or alkali spilt should be thoroughly washed with tap water.

Physical hazards Health hazards

Hydrochloric acid water-reactive, corrosive toxic

Phenolphthalein none irritant

Potassium hydrogen phthalate none irritant Sodium hydroxide water-reactive, corrosive toxic, irritant

Materials, equipments and apparatus

Potassium hydrogen phthalate (KHP, 204.23 g/mol), phenolphthalein indicator, sodium

hydroxide ~50 % w/v aqueous solution, hydrochloric acid solution (~0.1 M), distilled water,

graduated cylinder, analytical balance, weigh paper, conical flasks (3 x 250 mL), 1 L glass

bottle, weighing bottle, burette (50 mL), pipette (10 mL), stand and burette clamp.

Experimental procedures

Part A: Preparation of a primary standard (KHP).

1. KHP is a primary standard. It is pre-dried and kept in desiccators.

2. Using a clean dry weighing bottle, weigh accurately by difference, triplicate ~0.8 g

samples of KHP, into labeled 250 mL conical flasks.

3. Add 75 mL of distilled water (measure by graduated cylinder) and three drops of

phenolphthalein indicator and swirl gently.

4. Cover the flasks and leave to dissolve, swirling periodically.

Part B: Preparation of approximate solution (0.1M NaOH).

5. Using the 50 % NaOH solution provided, calculate (for 1 L) and prepare your ~0.1 M NaOH solution, by adding the calculated amount of 50 % solution to a 1 L glass bottle (half filled with freshly boiled distilled water ).

6. Once the concentrated NaOH has been added, fill the bottle up to the neck with the distilled water.

Part C: Standardization of NaOH with KHP.

7. Clean your burette, and rinse with a small aliquot of your NaOH solution, before

filling it with the NaOH solution. Make sure the meniscus is on or below the 0 mark,

but do not spend time trying to get it to exactly 0.00.

8. Do a calculation to estimate the approximate volume of base that will be required to

reach the endpoint in this first titration. Check this value with your instructor before

you proceed.

144

9. Begin to titrate your first KHP solution by adding NaOH rapidly until a pink color is

noticed (A piece of white paper under the titration conical flask will aid in observing

the color change).

10. At this point, slow the addition of NaOH so that only a localized pink color is observed

in the stirred solution. Continue slowing the rate of addition to keep the amount of pink color localized. Stop every so often to be sure that all color disappears upon mixing. As the endpoint is approached, a single drop of NaOH may turn the solution pink for a moment.

11. Proceed very slowly with drop by drop additions until the entire solution remains light

pink for at least 60 seconds. If the color fades add one drop more of the NaOH soln. At this point the endpoint has been reached. Record the final volume of the burette.

12. Similarly, titrate your other two flasks of KHP. You may titrate more quickly since you can approximate where the endpoint will be. Be sure to slow down when approaching the endpoint. The correct endpoint is a clear solution with a faint pink tinge. If your solution is hot pink, you have passed the endpoint. Try to do better on your other replicate measurements.

Be careful not to let the fluid level in the burette drop below the 50 mL line. If more than 50 mL will be required for a titration, stop the titration before reaching the 50 mL line, record the exact volume, refill the burette, record the new initial volume, and continue with the titration. Upon completion, the two volumes can be added to obtain the total titrant volume.

Part D: Standardization of HCl acid solution by NaOH solution.

13. To each of three clean labeled 250 mL conical flasks. Add 5, 10 and 15 mL of the HCl acid solution (~0.1 M) and 3 drops of phenolphthalein to flasks #1, #2 and #3, respectively. Titrate the HCl with your standardized NaOH solution in the same way you titrated the KHP.

Results

Burette reading(mL)

KHP

HCl

Initial reading

Final reading

Difference

Mean reading (volume of NaOH)

145

Data analysis

1. Using the weight of KHP in each flask, and the molecular weight of KHP, calculate the

number of moles of KHP in each flask.

2. Calculate the molarity of your NaOH solution using the total volume of NaOH used to

titrate each KHP flask to its endpoint. (Don't forget to convert your volume from mL

to liters.) Report your standardized NaOH concentration as the average of your three

replicates.

3. Using your experimentally determined standardized NaOH concentration (use the

average value), determine the molarity of your HCl acid solution. Report your

determination of the HCl concentration as the average of your three replicates.

Discussion questions

1. What characteristics of KHP make it a good primary standard?

2. Why can't we just dissolve a known mass of NaOH in a known volume of water and

calculate the NaOH concentration? In other words, explain why you standardized your

NaOH solution.

3. Ideally, when performing aqueous acid-base titrations, all distilled water should be

boiled prior to use to remove dissolved carbon dioxide. All solutions should then be

kept tightly closed when not in use; NaOH solutions are especially capable of

adsorbing large

amounts of CO2 from the atmosphere. Why would we ideally want to eliminate CO2

from solutions when performing acid-base titrations?

4. Describe how you would weigh your KHP samples, if you didn‘t know about weighing

by differences. Mentioning any differences between your method and the method we

made you use, explain why we use the technique of weighing by difference.

146


Purposes

To determine the composition of the following mixture by double indicator method:

1. NaOH(aq) and Na2CO3(aq)

2. NaHCO3(aq) and Na2CO3(aq)

Introduction

Consider a mixture of NaOH(aq) and Na2CO3(aq). Reaction between HCl(aq) and Na2CO3(aq)

takes place in two stages:

HCl(aq) + Na2CO3(aq) → NaHCO3(aq) + H2O(l) ………………… (1)

HCl(aq) + NaHCO3(aq) → NaCl(aq) + CO2(g) + H2O(l) ………… (2)

While that between HCl(aq) and NaOH(aq) completes in only one step:

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) ……………………… (3)

Solution mixture of reaction (1) at the equivalence point is alkaline, that of reaction (2) is

acidic and that of reaction (3) is neutral. Thus the whole titration should have three breaks

in the pH curve, corresponding to the above three stages. Reactions (1) and (3) can be

indicated by phenolphthalein and that of reaction (2) can be indicated by methyl orange.

Safety




Phenolphthalein indicator, methyl orange indicator, NaOH solution(~0.2 M), HCl solution

(~0.15 M), a mixture of NaOH(aq) and Na2CO3(aq), a mixture of NaHCO3(aq) and Na2CO3(aq),

distilled water, conical flask (250 mL), burette (50 mL), pipette (10 mL), stand and burette

clamp.


Part A: Standardization of the sodium hydroxide solution

1. The solution of NaOH(aq) that you are given is approximately 0.20 M. You will titrate

it against a known amount of a primary standard acid to determine the concentration

of NaOH. The acid used to standardize the concentration of NaOH is potassium

hydrogen phthalate (KHP). For experimental procedures, see parts A and C of the

previous experiment No. 1

Experiment (2): Acid-base titration using method of double indicators

147


Part B: Comparison of HCl to your standardized NaOH

2. To each of three clean labeled 250 mL conical flasks. Add 5, 10 and 15 mL of the

~0.15 M HCl solution and 3 drops of phenolphthalein indicator to flasks #1, #2 and #3,

respectively.

3. Clean, rinse, and fill a burette with the ~0.2 M NaOH solution that you standardized.

4. Titrate the three HCl flasks with the NaOH solution to the phenolphthalein endpoint

(to the first permanent pink colour).

Part C: Titration of a mixture of NaOH(aq) and Na2CO3(aq) with HCl(aq) using

phenolphthalein indicator followed by methyl orange indicator.

5. Pipette 10 cm3 of the solution mixture into a clean conical flask and add 2 drops of

phenolphthalein indicator.

6. Fill the burette with standardized HCl(aq).

7. Begin to titrate the solution mixture by adding HCl rapidly. When approaching the first

end point, as judged by the rate of disappearance of the pink color of the solution in

the conical flask, proceed very slowly with drop by drop additions until the entire

solution remains colorless for at least 60 seconds.

8. Observe the color change of the reaction mixture when the titration passes the first

end point.

9. Add 3 drops of methyl orange indicator when the reaction mixture becomes colorless.

Look for the second end point and continue titration until the color of the solution

change from yellow to orange-red.

Part D: Titration of a mixture of NaHCO3(aq) and Na2CO3(aq) with HCl(aq) using

phenolphthalein indicator followed by methyl orange indicator.

10. As described above, titrate 10 cm3 of the solution mixture of NaHCO3(aq) and

Na2CO3(aq) with standardized HCl(aq) using phenolphthalein indicator followed by methyl orange indicator.

148


Results

Table A: Titration of a mixture of NaOH(aq) and Na2CO3(aq).

Burette reading(cm3)

Ph. Ph.

M.O.

Initial reading

Final reading

Difference

Mean reading (volume of HCl)

Table B: Titration of a mixture of NaHCO3(aq) and Na2CO3(aq)

Burette reading(cm3)

Ph. Ph.

M.O.

Initial reading

Final reading

Difference


Data analysis

Part A

1. Using the weight of KHP in each flask, and the molecular weight of KHP, calculate the

number of moles of KHP in each flask.

2. Calculate the molarity of your NaOH solution using the total volume of NaOH used to titrate each KHP flask to its endpoint. (Don't forget to convert your volume from mL to liters.) Report your standardized NaOH concentration as the average of your three replicates.

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Part B

3. Using your experimentally determined standardized NaOH concentration (use the average value), determine the molarity of your HCl solution from the volumes of NaOH needed to titrate your flasks of HCl, remembering that you designed it to be ~0.15 M. Report your standardized HCl concentration as the average of your three replicates.

Part C

4. From the methyl orange end point, calculate the number of moles of HCl(aq) added

and hence the number of moles of Na2CO3(aq) in 10 cm3 of the alkaline solution mixture.

5. From the phenolphthalein end point, calculate the number of moles of HCl(aq) added

and hence the total number of moles of NaOH(aq) and Na2CO3(aq) in 10 cm3 of the solution mixture.

6. Calculate the number of mole of NaOH(aq) in 10 cm3 of the solution mixture.

7. Calculate the mass of Na2CO3 and NaOH in 1 L of the solution mixture respectively.

Part D

8. From the methyl orange end point, calculate the number of moles of HCl(aq) added

and hence the number of moles of Na2CO3(aq) in 10 cm3 of the solution mixture.

9. From the phenolphthalein end point, calculate the number of moles of HCl(aq) added

and hence the number of moles of NaHCO3(aq) in 10 cm3 of the solution mixture.

10. Calculate the mass of Na2CO3 and that of NaHCO3 in 1 L of the solution mixture.


1. Suggest other indicators that can be used in place of methyl orange and

phenolphthalein. Explain.

2. Can the same method be applied to determine the concentrations of Na3PO4(aq) and NaH2PO4(aq) in a solution mixture of the two salts? What factors should be considered?

150


Experiment (3): Determination of aspirin using back titration

Purposes

This experiment is designed to illustrate techniques used in a typical indirect or back titration. You will use the NaOH you standardized last week to back titrate an aspirin solution and determine the concentration of aspirin in a typical analgesic tablet. You will be graded on your accuracy.

Introduction

Many reactions are slow or present unfavorable equilibria for direct titration. Aspirin is a weak acid that also undergoes slow hydrolysis; i.e., each aspirin molecule reacts with two hydroxide ions. To overcome this problem, a known excess amount of base is added to the sample solution and an HCl titration is carried out to determine the amount of unreacted base. This is subtracted from the initial amount of base to find the amount of base that actually reacted with the aspirin and hence the quantity of aspirin in the analyte.

Aspirin (Acetylsalicylic acid)




3. Ethanol is flammable — keep away from heat, sparks, and open flames.

Physical hazards Health hazards

Hydrochloric acid water-reactive, corrosive toxic

Phenolphthalein none irritant

Ethanol flammable irritant Sodium hydroxide water-reactive, corrosive toxic, irritant Potassium hydrogen phthalate none irritant

151



Ethanol, distilled water, sodium hydroxide (standardized solution), hydrochloric acid (37 wt

%), Potassium hydrogen phthalate (KHP, 204.23 g/mol), phenolphthalein indicator solution,

aspirin tablet, analytical balance, weigh paper, conical flasks (3 x 250 mL), burette (50 mL),

pipette (10 mL), graduated cylinder, water bath, weighing bottle, boiling chips, mortar and

pestle, burette stand and clamp.


Part A: Preparation of approximate acid solution (~ 0.1M HCl)

1. Using the 37 wt % HCl solution provided, calculate the volume of concentrated HCl

you will need to prepare 250 mL of 0.1M HCl (concentrated reagent grade HCl has

a density of 1.188 g/cm3 and is 37 wt %).

2. Put ~100 mL distilled water into a 250 mL glass bottle and measure approximately

the volume of concentrated HCl using a graduated cylinder.

3. Gradually add the acid to the water in your bottle, mixing well (remember, always

add acid to water, not water to acid.).

4. Add more distilled water, mixing thoroughly with each addition, until the total

solution volume is ~ 250 mL. You do not have to measure the quantities accurately

because you are going to standardize this solution in the next steps to determine

its actual concentration.

Part B: Standardization of the sodium hydroxide solution

5. The solution of NaOH(aq) that you are given is approximately 0.10 M. You will

titrate it against a known amount of a primary standard acid to determine the

concentration of NaOH. The acid used to standardize the concentration of NaOH is

potassium hydrogen phthalate (KHP). For experimental procedures, see parts A

and C of the previous experiment No. 1

Part C: Comparison of HCl to your standardized NaOH

6. To each of three clean labeled 250 mL conical flasks. Add 5, 10 and 15 mL of the

~0.1 M HCl solution and 3 drops of phenolphthalein indicator to flasks #1, #2 and

#3, respectively.

7. Clean, rinse, and fill a burette with the ~0.1 M NaOH solution that you

standardized.

8. Titrate the three HCl flasks with the NaOH solution to the phenolphthalein

endpoint (to the first permanent pink colour).

152


Part D: Sample preparation

9. Accurately record the weight of a group of three aspirin tablets so that you can

determine an average tablet weight.

10. Use a mortar and pestle to crush enough tablets to produce ~ 1 g tablet powder.

11. Using a clean dry weighing bottle, weigh accurately, by difference, triplicate ~0.3

g samples of tablet, into labeled 250 mL conical flasks.

12. To each flask, add 20 mL of ethanol (measure by graduated cylinder) and three

drops of phenolphthalein indicator. Swirl gently to dissolve.

Note that: Aspirin is not very soluble in water — the ethanol helps the aspirin dissolve. An

aspirin tablet contains other compounds in addition to aspirin. Some of these are not very

soluble. Your solution will be cloudy due to insoluble components of the tablet.)

Part E: Aspirin titration with base

13. Titrate the first aspirin sample with the NaOH solution that you standardized to

the first permanent cloudy pink colour.

14. The aspirin/NaOH acid-base reaction consumes one mole of hydroxide per mole of

aspirin. The slow aspirin/NaOH hydrolysis reaction also consumes one mole of

hydroxide per mole of aspirin, and so for a complete titration we will need to use

a total of twice the amount of NaOH that you have already used, plus we will add

some excess NaOH to ensure that we really have reacted with all of the aspirin in

your sample. Calculate how much extra NaOH you will need to add, following this

reasoning: The volume of base to add for the hydrolysis reaction is equal to the

volume of base you have already used to titrate to the acid-base endpoint in Step

6 plus an additional 10 mL of excess base. (For example: if you used 26 mL of base

in the previous step, the volume of base you would add now would be 26 + 10 = 36

mL. Thus, you would have added a total of 26 + 26 + 10 = 62 mL of base). Record

the total volume of NaOH added to each flask.

Part F: Heating the reaction to completion

15. Add two or three boiling chips to each flask and heat in a water bath to speed up

the hydrolysis reaction. Avoid boiling, because the sample may decompose. While

heating, swirl the flasks occasionally. After 15 minutes, remove samples from the

water bath and cool for 5 minutes.

16. If the solution is colorless, add a few more drops of phenolphthalein. If it remains

colorless, add 10 mL more of the base and reheat. (Don't forget to add this

additional volume of base to the previously recorded total volume).

Part G: Back titration with acid

17. The only base remaining in each flask will be excess base that has not reacted with

the aspirin. Titrate the excess base in each flask with HCl until the pink color just

disappears. The endpoint is best described as ―cloudy white‖.

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Result

Table A: Direct titrations.

Burette reading(mL)

KHP

HCl

Aspirin

Initial reading

Final reading

Difference

Mean reading (volume of NaOH)

Table B: Back titration of aspirin.

Burette reading(mL)

Excess NaOH

Initial reading

Final reading

Difference


Data analysis

1. Using the weight of KHP in each flask, and the molecular weight of KHP, calculate

the number of moles of KHP in each flask.

2. Calculate the molarity of your NaOH solution using the total volume of NaOH used to titrate each KHP flask to its endpoint. (Don't forget to convert your volume from mL to liters.) Report your standardized NaOH concentration as the average of your three replicates.

3. Using your experimentally determined standardized NaOH concentration (use the average value), determine the molarity of your HCl solution from the volumes of NaOH needed to titrate your flasks of HCl, remembering that you designed it to be ~0.1 M. Report your standardized HCl concentration as the average of your three replicates.

154


4. Using the total volume of NaOH added to each aspirin sample, and the molarity of your NaOH standard solution, calculate the total number of moles of NaOH added to each sample.

5. Using the volume of HCl needed to back-titrate each aspirin flask and the average

HCl concentration determined in step 3, calculate the number of moles of excess NaOH left in each flask after the reaction with aspirin.

6. From the total number of moles of NaOH added to the sample, and the number of moles of excess NaOH left after the reaction was complete, calculate the number of moles of base that reacted with aspirin.

7. Hence determine the number of moles of aspirin in each of your samples.

8. Calculate the mass of aspirin in each of your samples. Using the mass of tablet you

weighed into each sample, calculate the weight percent of aspirin in the tablets.

9. From the mass of the three tablets you weighed, calculate the average mass of a tablet. Use this value, and the weight percent of aspirin you have just calculated, to calculate the mass of aspirin in one of the original tablets.

10. Compare the aspirin weight per tablet that you determined to the weight per

tablet claimed by the manufacturer on the label.


1. Why did you use your burette and not a graduated cylinder to add the excess NaOH?

2. What would be the consequence of competing equilibria, such as from other

ingredients in the tablet, on your results? What if there were several pain relievers in the same tablet? How could the error from these interferences be minimized?

3. Ethanol was used in the solutions to help dissolve the acetylsalicylic acid. Ethanol is slightly acidic, and will react with NaOH. Describe a blank correction experiment; i.e., what experiment you might do to determine how much NaOH reacts with the ethanol in your solution. Once you have determined this, how would you use the results of a blank correction experiment in the data analysis?

155


4 2

4

4

4 2 2

2 4

4

2 4 2

Redox titrations

Purposes

In this experiment we will determine the percent purity of an impure sample of sodium oxalate, Na2C2O4 by titration with standardized permanganate.

Introduction

In this experiment you will perform a redox titration using potassium permanganate as the titrant. Potassium permanganate is a powerful oxidizing agent because of the high positive charge on manganese, and is used in a wide variety of chemical analyses of reducing agents such as the determination of iron in iron ore or nitrites in aquarium water.

Titrations involving permanganate are normally carried out in acidic solutions, and the half reaction for permanganate under these conditions is:

MnO -(aq) + 8H+(aq) + 5e- Mn2+(aq) + 4H O(l)

There is one problem with using permanganate ion in titrations, and that is it is able to autocatalyze its own destruction. This is because as Mn2+ is produced during the titration, the Mn2+ can actually react with additional MnO -, producing solid MnO2 as a result:

3Mn2+(aq) + 2MnO -(aq) + 2H O(l) 5MnO (s) + 4H+(aq)

The unfortunate result of this side reaction is that you overestimate the amount of analyte that is being consumed because you must add more titrant to reach the endpoint. There are several methods that have been developed to overcome this difficulty, and we will use the McBride method where the titration is carried out at an elevated temperature. This speeds up the reaction with oxalate ion, and since the reaction of Mn2+ with MnO - is relatively

slower, the titration error can be minimized.

The oxidation half reaction in the titration occurs when oxalate ion, C O 2-, is oxidized to carbon dioxide:

C O 2-(aq) 2CO (g) + 2e-

The combination of half-reactions yields the overall reaction -

5C2O42-(aq) + 2MnO -(aq) + 16H+(aq) 10CO2(g) + 2Mn2+(aq) + 8H2O(l)

An additional advantage of permanganate titrations is that it serves as its own indicator due to its deep purple color. The titration is stopped at the first persistent light pink color.

In Part I of the procedure you will standardize the unknown concentration of your permanganate solution using a known quantity of primary standard sodium oxalate. In Part II you then use the standardized permanganate to titrate a sample of sodium oxalate with an unknown percent purity.

Experiment (4): Determination of the percent purity of an unknown sample of sodium oxalate

156


Safety

Lab coats, safety glasses and enclosed footwear must be worn at all times in the laboratory.

Concentrated sulfuric acid is dangerous; don't spill or splash any. Always slowly add acid to water, never the other way around.


KMnO4 solution (~ 0.1M), 28 g/L Na2C2O4 solution (to standardize the KMnO4), H2SO4 solution (3M), sodium oxalate with unknown percent purity, distilled water, 10-mL pipette, 50-mL burette, 250-mL conical flask, 250-mL beaker, graduated cylinder, hot plate or Bunsen burner, thermometer, analytical balance, weighing papers, burette stand and clamp.


Part A: Determining the concentration of KMnO4 using primary standard sodium oxalate

1. Prime a clean burette with the permanganate solution by rinsing it with approx. 2-3

mL of KMnO4 into a beaker.

2. Fill the burette with permanganate to approx. 0.0 mL (it doesn‘t have to be exact), and make sure there are no air bubbles in the burette tip by dispensing a small portion into a beaker.

3. Record the initial volume of permanganate.

4. Using a 10-mL pipette, transfer 10 mL of sodium oxalate solution (28 g/L) into the

250-mL conical flask. Add 25 mL of distilled water and 10 mL of 3M H2SO4.

5. Heat the contents of the flask to about 80 oC, and then turn off the Bunsen burner.

6. Now titrate slowly while stirring until a permanent faint pink coloration is attained.

You will see a brown solid forming that quickly disappears, which is completely normal. If the solution changes completely to brown, then you‘ll need to do another sample.

7. Record the volume used in the first titration.

8. Repeat this titration (steps 4-6) two more times to obtain 3 good values for the

volume of KMnO4 required to titrate the sodium oxalate solution.

Part B: Determination of the percent purity of an unknown sample of sodium oxalate

9. Thoroughly clean your conical flask.

10. You have been given approx. 4 g of a sample of sodium oxalate with an unknown percent purity. Accurately weigh out three 1 g of the sample using weighing paper.

157


11. Place the first sample into the 250-mL conical flask, and dissolve the sample in 25 mL of distilled water. This will take some vigorous stirring since sodium oxalate is hard to dissolve. You might have to gently heat.

12. Carefully add 10 mL of 3M H2SO4 to the flask.

13. Heat the contents of the flask to about 80 oC again, and then turn off the Bunsen burner.

14. Titrate with permanganate to a light pink color. Gently swirl the flask while adding titrant.

15. Record the volume used in the first titration.

16. Repeat this titration (steps 3-6) two more times to obtain 3 good values for the

volume of KMnO4 required to titrate the sodium oxalate solution.

Results

Burette reading(mL)

Primary

standard

Na2C2O4

Na2C2O4

with an unknown

percent purity

Initial reading

Final reading

Difference

Mean reading (volume of KMnO4)

Data analysis

Part A: Determining the concentration of KMnO4 using primary standard sodium oxalate

1. Calculate the number of moles of Na2C2O4 pipetted.

2. Calculate the number of moles of KMnO4 that will react with the number of moles of oxalate you calculated in step 1 using the equation:

5C2O42-(aq) + 2MnO4

-(aq) + 16H+(aq) 10CO2(g) + 2Mn2+(aq) + 8H2O(l)

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4

3. Calculate the average molarity of the KMnO4.

Part B: Determination of the percent purity of an unknown sample of sodium oxalate

4. Calculate the number of moles of KMnO4 used.

5. Calculate the number of moles of Na2C2O4 that will react with the number of moles of KMnO4 you calculated in step 1 using the equation:

5C2O42-(aq) + 2MnO -(aq) + 16H+(aq) 10CO2(g) + 2Mn2+(aq) + 8H2O(l)

6. Calculate the number of grams of Na2C2O4 in the sample.

7. percent purity = mass Na2C2O4 X 100%

mass sample


1. Balance the following redox reaction under acidic conditions.

C2O42-(aq) + MnO4

-(aq) CO2(g) + Mn2+(aq)

2. A 5.00 gram sample of impure sodium oxalate required 36.91 mL of 0.100 M KMnO4 to reach the endpoint. What was the percent purity of the sample? You will need the balanced equation from question #1.

159


4

4 2

2 2 4

Experiment (5): Determination of the percentage of iron in a sample

Purposes

To determine the percent of iron in a sample by titration with standardized permanganate.

Introduction

The concentrations of redox-active species can be determined by redox titrations. In a redox titration, a measured sample of the unknown is titrated against a standard solution of a substance that will oxidize or reduce the unknown.

In the present experiment you will take a sample containing iron, add acid to dissolve it [thereby converting all the iron to iron(II)], then use a solution containing permanganate ion, MnO4

-, to oxidize this Fe2+ to Fe3+ ion. The percent of iron in the sample will be calculated from the amount of permanganate needed to oxidize fully all the Fe2+ ions.

A solution of permanganate ion in sulfuric acid efficiently oxidizes Fe2+ to Fe3+

MnO - + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4 H O

The permanganate ion acts as its own indicator, as MnO - is highly colored while Mn2+ is essentially colorless. The product of oxidation, the Fe3+ ion, is itself, slightly colored. To avoid any possible interference with the equivalence point determination a little phosphoric acid, H3PO4, is added so as to complex Fe3+ to a completely, colorless ion.

Safety


Concentrated sulfuric acid is dangerous; don't spill or splash any. Always slowly add acid to water, never the other way around.


KMnO4 solution (~ 0.1M), oxalic acid dihydrate (H C O .2H2O), concentrated sulfuric acid,

H2SO4 solution (3M), 85% H3PO4 acid, an unknown mixture containing iron, distilled water, 10-

mL pipette, 50-mL burette, conical flasks (3 x 250-mL), 250-mL beaker, graduated cylinder,

hot plate or Bunsen burner, thermometer, analytical balance, weighing papers, burette stand

and clamp.


Part A: Standardization of permanganate solution

1. Weigh three clean dry labeled 250-mL conical flasks on an analytical balance. Place

about 0.135 g of oxalic acid dihydrate, H C O .2H O, into each of the three separate 2 2 4 2 flasks and reweigh the flasks containing the acid.

2. Set up a burette with KMnO4 solution (~ 0.1M) to be standardized by titration.

3. Dissolve each acid sample in about 25 mL of distilled water. Again don't mix up the

samples. Take one flask and add 1-2 mL of concentrated sulfuric acid.

160


4

4

4. The solution to which the acid has been added should get quite warm, but, since the titration is to be done at elevated temperatures to prevent side reactions, this is desirable. Heat the solution further to 70 oC; during the titration the solution should be kept between 60 and 80 oC.

5. Read the level in the permanganate burette (the initial reading), and then add the

solution slowly from the burette into the flask with the warmed acid sample with

constant stirring. The equivalence point is the first appearance of a pink color (excess

MnO -) that lasts, with stirring, for 30 seconds. When this is obtained, read the

burette again (the final reading).

6. Take a second flask with oxalic acid in it and add 1-2 mL of concentrated sulfuric acid. Repeat steps 4 and 5 above with this sample. Do a third trial with the third flask in the same manner.

Part B: Determination of iron in a sample

7. Weigh out accurately three samples of your unknown mixture containing iron and put

each sample in a separate, clean 250-mL conical flask. The amount of the sample should be around 1g.

8. Fill a burette with standardized KMnO4 solution.

9. Dilute 3 M H2SO4 with distilled water to prepare 150 mL of 1 M H2SO4 (acid into water,

never the other way). Stir well.

10. Put one-third of this 1 M H2SO4 solution (50 mL) into only one of your conical flasks

and dissolve the iron sample quickly and completely. Take an initial reading of the burette and, with stirring, as speedily as possible, start titrating by slowly adding the permanganate solution to the acidified sample.

11. When the solution turns a light yellow color, add 3 mL of the 85% H3PO4, and continue

immediately.

12. Continue adding the permanganate with stirring. The equivalence point of the

titration is the first appearance of a pink color (excess MnO - that lasts, with stirring, for 30 seconds). When this occurs take the final reading of the burette.

13. Repeat steps (10), (11) and (12) for the second and third samples.

161


Results

Table A

Trials

Trial

1

Trial

2

Trial

3

Mass of flask (g)

Mass of flask + oxalic acid (g)

Mass of oxalic acid (g)

Moles of oxalic acid used

Table B

Trials

Trial

1

Trial

2

Trial

3

Mass of flask (g)

Mass of flask + iron sample (g)

Mass of iron sample (g)

162


4

Table C

Burette reading(mL)

Oxalic acid

The unknown

mixture

containing iron

Initial reading

Final reading

Difference

Mean reading (volume of KMnO4)

Data analysis

Part A: Standardization of permanganate solution

1. From the measured mass of the oxalic acid samples that you used, calculate the

number of moles of oxalic acid in each case. Remember the oxalic acid was weighed out as a dihydrate.

2. Write a balanced oxidation-reduction equation for the reaction of oxalic acid with

potassium permanganate in an acidic solution then, from the indicated molar ratio, calculate how many moles of MnO - must have been used in each of the three titrations. The products are carbon dioxide and manganese(II) ion.

3. With the calculated number of moles and the measured volume of solution used,

calculate three values for the molarity of the permanganate solution. Report an average molarity.

Part B: Determination of iron in a sample

4. With the known molarity of the permanganate solution and the measured volumes used in the titration, calculate the number of moles of permanganate used in each of the trials.

5. Write a balanced oxidation-reduction equation for the reaction of iron(II) with

permanganate in an acidic solution and, from the indicated molar ratio, calculate how many moles of iron were in each of your weighed out samples.

6. Calculate how many grams of iron were in each of your weighed out samples and then

what mass percent of iron was present in each case. Report an average mass percent of iron in the unknown mixture.

% of iron in the sample = mass iron X 100%

mass sample

163


Experiment (6): Iodometric titration of vitamin C

Purposes

In this experiment, you will be acting as the quality control laboratory for a pharmaceutical manufacturer. The product line that you support produces 100-mg Vitamin C supplements. You are to determine the average amount of Vitamin C per tablet in a sample of tablets and report this value and its uncertainty to the product line manager.

Introduction

Vitamin C (ascorbic acid), is a mild reducing agent [it accepts electrons from an electron donor, leaving the oxidation state of the donor at a value less than original (reduced)]. The ascorbic acid itself is oxidized to a higher oxidation state. This class of reactions is known as a reduction/oxidation reaction or simply, a redox reaction. One such redox reaction is the reduction of the aqueous iodine molecule (I2(aq)) with ascorbic acid, as shown below.

(1) KIO3(aq) + 6 H+(aq) + 5 I-(aq) 3 I2(aq) + 3 H2O(l) + K+(aq) generation of I2

(2) C6H8O6(aq) + I2(aq) C6H6O6(aq) + 2 I-(aq) + 2 H+(aq) oxidation of vitamin C Vitamin C

Reaction one generates aqueous iodine, I2(aq). This is then used to oxidize vitamin-C (ascorbic acid, C6H8O6) in reaction two. Both of these reactions require acidic conditions and so dilute hydrochloric acid, HCl (aq), will be added to the reaction mixture. Reaction one also requires a source of dissolved iodide ions, I-(aq). This will be provided by adding solid potassium iodide, KI(s), to the reaction mixture.

The two relevant half reactions for reaction (2) above are:

(Oxidation half reaction for vitamin-C at pH 5)

I2 + 2e- 2 I- (Reduction half reaction for Iodine at pH 5)

A few drops of starch solution will be added to help determine the titration endpoint. When vitamin-C (ascorbic acid) is completely oxidized, the iodine, I2(aq), will begin to build up and will react with the iodide ions, I-(aq), already present to form a highly colored blue I3---- starch complex, indicating the endpoint of our titration.

Each of the iodine atoms is reduced to the I- ion and the ascorbic acid gains two electrons to form dehydroascorbic acid as in the chemical reaction below:

ascorbic acid + I2(aq) + H2O dehydroascorbic acid + 2I- + 2H+

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This reaction has a high K value and goes to completion. Unfortunately, the solubility of I2(solid) in water is not very high. The saturated I2(aq) concentration is only 1.3 mM.

I2(solid) I2(aq) K = 1.3 x 10-3

To increase the solubility of the I2 molecule, we can create a complex between the I2(aq) and the iodide ion (I-) as below:

I2(aq) + I- I3- K = 7 x 102

The I - complex is known as triiodide. Rather than starting with solid I2 and taking the time to dissolve it in solution, triiodide can also be produced by reacting iodate ion (IO iodide (I-) as below:

-) with

IO3- + 8I- + 6H+ 3I3- + 3H2O

Triiodide is reduced by ascorbic acid in the same way that the I2(aq) species reacted:

ascorbic acid + I3- + H2O dehydroascorbic acid + 3I- + 2H+

It is the reaction that we use to measure indirectly the amount of ascorbic acid in the sample. We do have a titration solution and indicator that can measure the amount of I - in the sample. The titration solution is sodium thiosulfate, Na2S2O3, and it reacts with the triiodide species in the 1:2 reaction below:

I3- + 2S2O32- 3I- + S4O6

2-

The indicator used is a starch solution. In the presence of the triiodide, starch and triiodide form a complex that is intensely dark blue in color. In the absence of triiodide, the starch indicator is a milky-white.

Safety

The sulfuric acid used in this experiment will cause injury if in contact with the skin. The proper use of laboratory safety glasses and/or splash goggles and gloves will be expected and strictly monitored.


Starch solution, sodium carbonate (Na2CO3), 0.5 M and 0.3 M sulfuric acid (H2SO4), Potassium Iodide (KI), Potassium Iodate (KIO3), 100 mg vitamin C tablets, sodium thiosulfate pentahydrate (Na2S2O3.5H2O), 100-mL and 500-mL beakers, hot plate, analytical balance, 1-L brown plastic bottle, 1-L white plastic bottle, 500-mL volumetric flask, weigh paper, 250-mL conical flask, burette (50 mL), pipette (10 mL), graduated cylinder, burette stand and clamp.


Part A: Standardization of sodium thiosulfate solution

I. Preparation of starch indicator

1. Clean and rinse a 100-mL beaker with distilled water. Fill the beaker to the 100-mL

mark and place on a hot plate until boiling.

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2. Weigh out around 1 g of soluble starch and slowly add to the beaker of boiling water with stirring. Continue boiling until the solution is clear.

II. Preparation of sodium thiosulfate solution

3. Clean and rinse a 500-mL beaker with distilled water. Fill the beaker to the 500-mL mark and place on a hot plate. Boil the water for 5 minutes to expel dissolved CO2

gas. Allow solution to cool.

4. Weigh out around 0.05 g of Na2CO3 and place into the 500-mL beaker of boiled

water.

5. Weigh out around 8.7 g of Na2S2O3.5H2O and dissolve into the 500-mL beaker of boiled water buffered with Na2CO3.

6. Clean and rinse a 1-L brown plastic bottle with distilled water. Rinse the bottle with

a small amount of the sodium thiosulfate solution. Transfer the remainder of the sodium thiosulfate solution to the brown plastic bottle. Label this bottle ―sodium thiosulfate solution‖ and keep this bottle tightly capped when not in use.

III. Preparation of standard iodate solution

7. Clean and rinse a 500-mL volumetric flask with distilled water. Fill the clean flask

with around 400 mL of distilled water.

8. Accurately weigh out around 1 g of KIO3 using the analytical balance.

9. Dissolve the KIO3 into the distilled water contained in the 500-mL volumetric flask.

After all of the solid has dissolved, fill the volumetric flask to the mark with distilled water.

10. Clean and rinse a 1-L white plastic bottle with distilled water. Rinse the bottle with

a small amount of standard KIO3 solution. Transfer the remainder of the KIO3 solution from the volumetric flask into the while plastic bottle. Label this bottle ―standard KIO3 solution‖.

IV. Standardization of sodium thiosulfate solution

11. Clean and rinse a 50-mL burette with distilled water.

12. Rinse the 50-mL burette with a small amount of the sodium thiosulfate solution. Fill

the burette with the sodium thiosulfate solution.

13. Clean and rinse a 250-mL conical flask with distilled water.

14. Accurately pipette 10.00 mL of standard KIO3 solution into the flask.

15. Weigh out around 0.5 g of KI and place into the conical flask.

16. Add 2 mL of 0.5 M H2SO4 to the conical flask.

17. The solution should start out as a ―reddish‖ solution due to the presence of the triiodide. Titrate the solution with the sodium thiosulfate until the solution has lost most of the reddish color (should be a pale yellow). At this point add 0.5 mL of the starch indicator (may turn smoky-blue or remain yellowish). Carefully add sodium thiosulfate until the solution turns colorless (may be a milky-white). Record this volume as the end point.

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Note: The indicator is not added until just before the end point as the triiodide/starch complex can ―hold on to‖ the triiodide in concentrated solutions and prevent it from reacting with the sodium thiosulfate.

18. Repeat this titration (steps 13 – 17) two more times to obtain 3 good values for the

volume of sodium thiosulfate required to titrate the sample of KIO3.

Part B: Analysis of vitamin C

I. React triiodide with ascorbic acid

19. Clean and rinse a 250-mL conical flask with distilled water.

20. Place 30 mL of 0.3 M H2SO4 into the clean conical flask.

21. Add a vitamin C tablet to the conical flask and dissolve in the sulfuric acid. You can

use a clean glass stirring rod to help break up the solid tablets. Some solid binding material may not dissolve.

22. Accurately pipette 25.00 mL of standard KIO3 solution into the conical flask.

23. Weigh out around 1 g of KI and place into the conical flask.

24. Gently swirl the conical flask for 1 minute to insure complete formation of the

triiodide complex and complete reaction between the triiodide and the vitamin C.

II. Titrate the solution with sodium thiosulfate

25. Titrate the solution with sodium thiosulfate solution in the same way you standardize

the sodium thiosulfate solution

26. Repeat this titration (steps 19 – 25) two more times to obtain 3 good values for the

volume of sodium thiosulfate required to titrate the remaining triiodide.

Results

Burette reading(mL)

- I3

- Remaining I3

(after complete

reaction with vitamin C)

Initial reading

Final reading

Difference

Mean reading (volume of Na2S2O3)

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Data analysis

1. Using the formula weight of KIO3, calculate the molarity of your standard IO -

solution.

2. Using the known stoichiometry of the sodium thiosulfate/IO3 - reaction, calculate the molarity of your sodium thiosulfate solution. Calculate the average of the concentration.

3. Using the known stoichiometry of the sodium thiosulfate/IO3

- and ascorbic acid/IO -

reactions, calculate the average number of moles of ascorbic acid contained in each vitamin C tablet.

4. Using the formula weight of vitamin C (C6H8O6, FW = 176.13), calculate the

averagenumber of milligrams of vitamin C contained in each tablet.

5. Compare the mass in milligrams of vitamin C per tablet that you determined to the

mass per tablet claimed by the manufacturer on the label information from the bottle of vitamin C tablets.

Discussion question

Why is sulfuric acid used in the Iodometric titration of vitamin C?

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Precipitation titrations

Purposes

To determine the chloride ion concentration of a solution by titration with silver nitrate using Mohr‘s method.

Introduction

The goal of any titrimetric method is to determine the number of moles of titrant needed to reach the equivalence point of the titration reaction:

aA + tT → product(s)

where a and t are the stoichiometric coefficients in the reaction between titrant and analyte. In a titrimetric analysis, solution of titrant is added until the equivalence point of the titration reaction is reached. At the equivalence point, neither analyte nor titrant is present in excess. By definition, the equivalence point is the point during the titration at which the following relationship is true:

nA/а = nT/t

where nA is the number of moles of analyte originally present in the sample solution, and nT

is the number of moles of titrant that must be added to the sample to reach the equivalence point.

From this last expression, we see that, in order to determine the number of moles of analyte originally present in the sample solution, we must (a) know the stoichiometry of the reaction, and (b) determine how many moles of titrant are needed to reach the endpoint. In order to meet this last requirement, we must somehow keep track of the quantity of titrant solution that is added to the sample solution.

Argentometric titrations

In order for a titrimetric method to be viable, the titration reaction (1) must be complete and (2) should be rapid. There are many precipitation reactions that can satisfy the first requirement, but far fewer that satisfy the second. Precipitation reactions of silver salts are usually quite rapid, and so argentometric titrations, which use AgNO3 as the titrant, are the most common precipitation titrations. Argentometric titrations can be used to analyze samples for the presence of a number of anions that form precipitates with Ag+; such as halides: Cl-, Br-, I-

Endpoint detection for argentometric titrations

Another requirement of titrimetric analysis is that there must be some method of determining when the titration reaction has reached its equivalence point. There are three common methods of endpoint detection for argentometric titrations using a chemical indicator:

1. The chromate ion, CrO42− (the Mohr method);

2. The ferric ion, Fe3+ (the Volhard method);

3. Adsorption indicators such as fluorescein (the Fajans method).

In this experiment, we will be using the chromate (CrO42−), a Mohr indicator, to detect the

endpoint chemically.

Experiment (7): Determination of concentration of chloride in seawater using Mohr’s

method

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Mohr‘s method determines the Cl- ion concentration of a solution by titration with AgNO3. As the AgNO3 solution is slowly added, a precipitate of AgCl forms.

Ag+(aq) + Cl–(aq) → AgCl(s)

The end point of the titration occurs when all the Cl- ions are precipitated. Then additional Ag+ ions react with the chromate(CrO4

2−) ions of the indicator, potassium chromate (K2CrO4), to form a red-brown precipitate of silver chromate (Ag2CrO4).

2Ag+(aq) + CrO 2-(aq) → Ag CrO (s) titrant indicator red-brown precipitate

The concentration of titrant rises sharply near the equivalence point, and the solubility of

Ag2CrO4 is exceeded. The appearance of red precipitate marks the endpoint. The

applicability of the Mohr method is limited compared to either of the other chemical

indicator methods: it can be used to analyze for Cl− or Br− anions.

This method can be used to determine the chloride ion concentration of water samples from many sources such as seawater, stream water, river water and estuary water. Seawater is used as the example here.

The Mohr titration should be carried out under conditions of pH 6.5 – 9. If the solutions are acidic, the gravimetric method or Volhard‘s method should be used. At higher pH silver ions may be removed by precipitation with hydroxide ions, and at low pH chromate ions may be removed by an acid-base reaction to form hydrogen chromate ions or dichromate ions, affecting the accuracy of the end point.

The Mohr titration is sensitive to the presence of both chloride and bromide ions in solution and will not be too accurate when there is a significant concentration of bromide present as well as the chloride. However, in most cases, such as seawater, the bromide concentration will be negligible. For this reason, the method can also be used to determine either the total concentration of chloride and bromide in solution, or the concentration of bromide when the chloride concentration is known to be negligible.


1. Lab coats, safety glasses and enclosed footwear must be worn at all times in the

laboratory.

2. The chromate solution needs to be prepared and used with care as chromate is a known carcinogen.

3. Silver nitrate solution causes staining of skin and fabric (chemical burns).

4. Aqueous silver nitrate is photosensitive and should not be exposed to light. It should be stored in darkened storage bottle.

5. Any spills should be rinsed with water immediately.


Silver nitrate (AgNO3), Potassium chromate indicator solution (~1 M), primary standard NaCl salt, a sample of seawater, distilled water, 50-mL burette,10-mL pipette, 100- and 250-mL volumetric flasks, 250-mL conical flasks,10-mL and 100-mL measuring cylinders, 500-mL brown bottle, analytical balance, weighing papers, drying oven, desiccators, burette stand and clamp.

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Part A: Preparation of the solutions

I. Sample preparation

1. If the seawater contains traces of solid matter such as sand or seaweed, it must be filtered before use.

II. Preparation of silver nitrate solution (~0.1 M)

2. Clean and rinse a 250-mL volumetric flask with distilled water. Fill the clean flask with around 150 mL of distilled water.

3. Dry 5 g of AgNO3 for 2 hours at 100°C and allow to cool.

4. Weigh out around 4.25 g of solid AgNO3 using the analytical balance.

5. Dissolve the AgNO3 into the distilled water contained in the 250-mL volumetric flask, fill the volumetric flask to the mark with distilled water.

6. Clean and rinse a 500-mL brown bottle with distilled water. Rinse the bottle with a

small amount of AgNO3 solution. Transfer the remainder of the AgNO3 solution from the volumetric flask into the brown bottle.

III. Preparation of Potassium chromate indicator solution (~1 M)

7. Dissolve 1 g of K2CrO4 in 20 mL distilled water.

Part B: Determination of chloride ion concentration in a sample of seawater

8. You have been given ~ 0.5 g of a primary standard NaCl salt. Dry the pure NaCl salt

for at least one hour at 110 °C. Cool in a desiccator.

9. Accurately weigh by difference three 0.1 g samples of the pure NaCl (to the nearest 0.1 mg) into three separate 250-mL conical flasks. Add approximately 100 mL of distilled water to each of the three flasks to dissolve the solid and 1 mL of chromate indicator.

10. Prime a clean burette with the AgNO3 solution by rinsing it with approx. 2-3 mL of AgNO3 solution into a beaker.

11. Fill the burette with AgNO3 solution to approx. 0.0 mL (it doesn‘t have to be exact), and make sure there are no air bubbles in the burette tip by dispensing a small portion into a beaker.

12. Titrate the first flask with silver nitrate solution. Although the silver chloride that forms is a white precipitate, the chromate indicator initially gives cloudy solution a faint lemon-yellow color (figure 1).

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Figure 1: Before the addition of any silver nitrate the chromate indicator gives

clear solution a lemon-yellow color.

13. The endpoint of the titration is identified as the first appearance of a red-brown

color of silver chromate (figure 2).

Figure 2

Left flask: before the titration endpoint, addition of Ag+ ions leads to formation of silver chloride precipitate, making the solution cloudy. The chromate indicator gives a faint lemon- yellow color.

Centre flask: at the endpoint, all the Cl− ions have precipitated. The slightest excess of Ag+ precipitates with the chromate indicator giving a slight red-brown color.

Right flask: If addition of Ag+ is continued past the endpoint, further silver chromate precipitate is formed and a stronger red-brown color results.

NB: The titration should be stopped when the first trace of red-brown color is observed. Using an incompletely titrated reference flask for comparison is a helpful way to identify the first appearance of red-brown color.

14. Repeat the titration (steps 10 and 11) for the other two flasks.

15. Dilute seawater by pipetting a 20 mL sample into a 100-mL volumetric flask and making it up to the mark with distilled water.

16. Pipette a 10 mL aliquot of diluted seawater into a 250-mL conical flask and add about 50 mL distilled water and 1 mL of chromate indicator.

17. Titrate the diluted seawater sample with silver nitrate solution in the same manner as the primary standard NaCl.

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18. Repeat the titration with further aliquots of diluted seawater until concordant results (titres agreeing within 0.1 mL) are obtained.

Results

Burette reading(mL)

Primary standard

NaCl salt

Cl- ions in the

diluted seawater

sample

Initial reading

Final reading

Difference

Mean reading (volume of AgNO3)

Data analysis

Part A: Determining the concentration of AgNO3 using primary standard NaCl

1. Calculate the number of moles of NaCl used.

2. Calculate the number of moles of AgNO3 that will react with the number of moles of

NaCl you calculated in step 1 using the equation:


3. Calculate the average molarity of the AgNO3.

Part B: Determination of chloride in the seawater sample.

4. Determine the average volume of silver nitrate used from your concordant titres.

5. Calculate the number of moles of silver nitrate used.

6. Use the following reaction equation to calculate the number of moles of chloride

ions that will react with the number of moles of AgNO3 you calculated in step 5:


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7. Calculate the concentration in mol L−1 of chloride ions in the diluted seawater.

8. Calculate the concentration in mol L−1 of chloride ions in the original undiluted seawater.

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Experiment (8): Determination of chloride in cheese using Volhard’s method

Purposes

To determine the chloride in cheese by argentometric titrations using Volhard‘s method.

Introduction

This method uses a back titration with potassium thiocyanate to determine the concentration of chloride ions in a solution. Before the titration an excess volume of a silver nitrate solution is added to the solution containing chloride ions, forming a precipitate of silver chloride. The term ‗excess‗ is used as the moles of silver nitrate added are known to exceed the moles of sodium chloride present in the sample so that all the chloride ions present will react.


The indicator Fe3+ (ferric ion) is then added and the solution is titrated with the potassium thiocyanate solution. The titrate remains pale yellow as the excess (unreacted) silver ions react with the thiocyanate ions to form a silver thiocyanate precipitate.

Ag+(aq) + SCN–(aq) → AgSCN(s)

Once all the silver ions have reacted, the slightest excess of thiocyanate reacts with Fe3+ to form a dark red [FeSCN]2+ complex ion.

Fe3+(aq) + SCN–(aq) → [FeSCN]2+(aq)

The concentration of chloride ions is determined by subtracting the titration findings of the moles of silver ions that reacted with the thiocyanate from the total moles of silver nitrate added to the solution.

This method is used when the pH of the solution, after the sample has been prepared, is acidic. If the pH is neutral or basic, Mohr‘s method or the gravimetric method should be used. The method is illustrated below by using the procedure to determine the concentration of chloride (from sodium chloride) in cheese.


Lab coats, safety glasses and enclosed footwear must be worn at all times in the

laboratory.

Silver nitrate solution causes staining of skin and fabric (chemical burns). Any spills

should be rinsed with water immediately.

Concentrated nitric acid is very corrosive: take great care using the 6 M solution.


Concentrated nitric acid (6 M), silver nitrate (AgNO3), primary standard NaCl salt, potassium thiocyanate (KCN). Potassium permanganate solution (5%). Ferric ammonium sulfate salt (NH4Fe(SO4)2.12H2O), a sample of cheese, distilled water, 50-mL burette,10-mL pipette, 500- mL volumetric flask,10-mL and 100-mL measuring cylinders, (3 x 250-mL) and a 500-mL conical flasks, 500-mL brown bottle, Bunsen burner, analytical balance, weighing papers, drying oven, desiccator, burette stand and clamp.

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Part A: Preparation of the solutions

I. Preparation of silver nitrate solution (~0.1 M)







II. potassium thiocyanate solution (0.1 M)

6. Accurately weigh 2.43 g of solid KSCN and dissolve it in 250 mL of distilled water in a

volumetric flask.

III. Ferric ammonium sulfate indicator solution (saturated)

7. Add 8 g of NH4Fe(SO4)2.12H2O to 20 mL of distilled water and add a few drops of concentrated nitric acid.

Part B: Standardization of silver nitrate solution

8. Prime a clean burette with the 0.1 M KCN solution by rinsing it with approx. 2-3 mL

of KCN solution (0.1 M) into a beaker.

9. Fill the burette with KCN solution (0.1 M) to approx. 0.0 mL (it doesn‘t have to be exact), and make sure there are no air bubbles in the burette tip by dispensing a small portion into a beaker.

10. You have been given ~ 0.5 g of a primary standard NaCl salt. Dry the pure NaCl salt for at least one hour at 110 °C. Cool in a desiccator.

11. Accurately weigh by difference three 0.1 g samples of the pure NaCl (to the nearest

0.1 mg) into three separate 250-mL conical flasks. Add approximately 100 mL of distilled water to each of the three flasks (to dissolve the solid).

12. Add 8 mL of concentrated nitric acid to the NaCl solution in the first flask, then

add 20 mL of silver nitrate solution and heat the solution in a fumehood.

13. Keep the solution hot and away from direct sunlight until the AgCl ppt has

coagulated to yield a clear filtrate.

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14. Cool the solution and filter it in a 250-mL conical flask. Wash the solid residue with a few mL of distilled water.

15. Add 1 mL of saturated ferric ammonium sulfate solution as indicator.

16. Titrate the unreacted silver ions with the 0.1 M potassium thiocyanate solution.

The end point is the first appearance of a dark red colour due to the ferric thiocyanate complex (figure 1).

17. Repeat the titration (steps 12-16) for the other two flasks.

(Figure 1)

Left flask: before the titration endpoint, addition of SCN− ions leads to formation of silver

thiocyanate precipitate, making the solution cloudy. Here the solution also takes a faint yellow color due to the color of the cheese extract.

Centre flask: at the endpoint all the free silver ions have been precipitated by SCN−. The slightest

excess of SCN− forms a dark red colored complex with the Fe3+ ions from the ferric ammonium sulfate indicator, giving the solution a slight orange/red coloration.

Right flask: If addition of SCN− is continued past the endpoint, further ferric thiocyanate complex is

formed and a stronger dark red color results.

NB: The titration should be stopped when the first trace of dark red color is observed. Using an

incompletely titrated reference flask for comparison is a helpful way to identify the first appearance of red coloration.

Part C: Determination of chloride in the cheese sample.

I. Sample preparation

The salt sodium chloride is added during the manufacture of cheddar cheese. In this method, the cheese is ‗digested‘ to release this salt to obtain the concentration of chloride ions. To carry out this digestion, the cheese is reacted with nitric acid and potassium permanganate. The chloride ions are then ‗free‘ to form a precipitate with the added silver ions.

18. Cut or grate the cheese into fine pieces and accurately weigh about 6 g into a 500 mL conical flask.

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19. Precisely add 50 mL of standardized silver nitrate solution, 20 mL of concentrated nitric acid, 100 mL of distilled water and heat the solution to boiling in a fumehood.

20. As the solution boils add 5 mL of 5% potassium permanganate solution. This

addition will cause a very smelly reaction so done in the fumehood. Keep boiling until the purple colour disappears, then add another 5 mL of potassium permanganate solution. Continue this process until 30 mL of potassium permanganate solution has been added and the cheese particles are completely digested (or as close as possible). To find out when digestion is complete, remove the flask from heat and allow it to stand for a few moments. Undigested cheese particles will float upon the surface of the clear liquid, while the white precipitate of silver chloride will sink to the bottom. If there is still too much undigested cheese, the boiling and addition of 5 mL of potassium permanganate should be continued, checking each time until there is a satisfactory level of digestion.

21. Cool the solution and filter it. Wash the solid residue with a few mL of distilled

water.

22. Make the filtrate up to 500 mL in a volumetric flask.

II. Titration

23. Use a volumetric cylinder to measure 100 mL of the cheese extract solution (be

as precise as possible) and pour it into a 250-mL conical flask.

24. Add 1 mL of saturated ferric ammonium sulfate solution as indicator.

25. Titrate the unreacted silver ions with the 0.1 M potassium thiocyanate solution.

The end point is the first appearance of a dark red colour due to the ferric thiocyanate complex (figure 1).

26. Repeat the titration with 100 mL samples of the cheese extract solution until

you obtain concordant results (titres agreeing within 0.1 mL).

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Results

Burette reading(mL)

primary standard

NaCl salt

chloride in the

cheese sample

Initial reading

Final reading

Difference

Mean reading (volume of KCN)

Data analysis

I. Standardization of silver nitrate solution

1. Determine the average volume of potassium thiocyanate used from your

concordant titres.

2. Calculate the number of moles of potassium thiocyanate used.

3. Use the equation of the reaction between silver ions and thiocyanate ions.


to calculate the number of moles of unreacted silver nitrate.

4. Calculate the total number of moles of silver nitrate in the 20 mL of solution

that was added to the NaCl solution.

5. Calculate the number of moles of silver nitrate that reacted with the NaCl salt

by subtracting the number of moles of unreacted silver nitrate (the excess) from the total number of moles of silver nitrate added to the NaCl salt.


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II. Determination of chloride in the cheese sample.

7. Determine the average volume of potassium thiocyanate used from your

concordant titres.

8. Calculate the number of moles of potassium thiocyanate used.

9. Use the equation of the reaction between silver ions and thiocyanate ions.


to calculate the number of moles of unreacted silver nitrate in 100 mL of cheese

extract, and multiply the figure by five to determine the total number of moles of unreacted silver nitrate (the excess) in the 500 mL volumetric flask.

10. Calculate the number of moles of silver nitrate in the 50 mL of solution that was

added during the sample preparation to the cheese.

11. Calculate the total number of moles of silver nitrate that reacted with the salt

from the cheese by subtracting the number of moles of unreacted silver nitrate (the excess) from the total number of moles of silver nitrate added to the cheese.

12. Use the equation of the reaction between the silver ions and the chloride ions to

calculate the number of moles of sodium chloride in the sample of cheese.


13. Calculate the concentration of sodium chloride in the cheese as grams of salt

per 100 g cheese (% NaCl salt).

Additional note

For greatest accuracy it is a good idea to standardize your thiocyanate solution by titrating several samples against your standardized silver nitrate solution (once again using ferric ammonium sulfate indicator). The concentration of SCN– determined by this titration should then be used in all calculations.

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Experiment (9): Titrimetric analysis of chloride using Fajans method

Purposes

The purpose of this experiment is to analyze chloride in a water-soluble solid by titration with silver nitrate using Fajans method.

Introduction

In this experiment, we will be using dichlorofluorescein, a Fajans (adsorption) indicator, to detect the endpoint chemically.

Adsorption indicators function in an entirely different manner than the chemical indicators described thus far, and they can be used in many precipitation titrations, not just argentometric methods. Let‘s imagine that we wish to analyte Cl− in a sample solution by titrating with Ag+; the titration reaction would be


Silver chloride forms colloidal particles. Before the equivalence point, the surface of the precipitant particles will be negatively charged due to the adsorption of excess Cl− to the surface of the particles. A diffuse positive counter-ion layer will surround the particles. When the equivalence point is reached, there is no longer an excess of analyte Cl− , and the surface of the colloidal particles are largely neutral. After the equivalence point, there will be an excess of titrant Ag+, some of these will adsorb to the solid AgCl particles, which will now be surrounded by a diffuse negative counterion layer. The next figure illustrates this concept.

Adsorption indicators are dyes, such as dichlorofluorescein (shown below), that usually exist as anions in the titration solution.

Dichlorofluoroscein

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The doubly charged dichlorofluoroscein anion is attracted into the counter-ion layer immediately following the equivalence point, when the surface charge of the particles changes from negative to positive. The closer proximity of the dye to the particles changes the color of the molecule, providing a visual indication of the titration endpoint. In the case of dichlorofluorescein, the indicator changes to a pinkish color.


1. Lab coats, safety glasses and enclosed footwear must be worn at all times in the

laboratory.

2. Silver nitrate solution causes staining of skin and fabric (chemical burns).

3. Aqueous silver nitrate is photosensitive and should not be exposed to light. It should be stored in darkened storage bottle.

4. Any spills should be rinsed with water immediately.


Silver nitrate (AgNO3), primary standard NaCl salt, a sample of chloride, dextrin (2%), dichlorofluorescein (0.1%), distilled water, 50-mL burette,10-mL pipette, 100- and 250-mL volumetric flasks, conical flasks (6 x 250-mL), 10-mL and 100-mL measuring cylinders, 500-mL brown bottle, analytical balance, weighing papers, drying oven, desiccator, burette stand and clamp.


Part A: Preparation of silver nitrate solution (~0.1 M)







Part B: Determination of chloride ion in the chloride sample

6. You have been given ~ 0.5 g of a sample of chloride and ~ 0.5 g of a primary standard NaCl salt. Dry the sample and the pure NaCl salt for at least one hour at 110 °C. Cool in a desiccator.

7. Prime a clean burette with the AgNO3 solution by rinsing it with approx. 2-3 mL of AgNO3 solution into a beaker.

182

8. Fill the burette with AgNO3 solution to approx. 0.0 mL (it doesn‘t have to be exact), and make sure there are no air bubbles in the burette tip by dispensing a small portion into a beaker.

9. Accurately weigh by difference three 0.1 g samples of the pure NaCl (to the nearest

0.1 mg) into three separate 250-mL conical flasks. Weigh by difference three 0.1 g

samples of the unknown sample into another three 250-mL conical flasks. Add

approximately 100 mL of distilled water to each of the six flasks to dissolve the solid.

10. Add 5 mL of 2% dextrin solution and 10 drops of 0.1% dichlorofluorescein solution to

the first flask. The purpose of the dextrin solution is to stabilize the colloidal suspension of AgCl(s) formed once the titration begins, and the dichlorofluorescein is the adsorption indicator.

11. Titrate with AgNO3 solution at a fairly rapid rate, using continuous swirling. When the rate of fading of the pink color becomes slow, reduce the rate of addition to a dropwise rate, swirl constantly, and continuously observe the color of the suspension to detect the endpoint, which is a change from a light peach color to a darker pink

color.

12. Repeat the titration (steps 10 and 11) for the other five flasks.

13. After completing your titrations, dispose of the AgCl(s) and titrating solutions by

pouring them into the waste bottles. Also return any unused titrant solution to the

instructor. Anything that contained AgNO3 should be rinsed thoroughly before being put away.

Results

Burette reading(mL)

Primary standard

NaCl salt

Cl- ions in chloride

sample

Initial reading

Final reading

Difference

Mean reading (volume of AgNO3)

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Data analysis

Part A: Determining the concentration of AgNO3 using primary standard NaCl

1. Calculate the number of moles of NaCl used.

2. Calculate the number of moles of AgNO3 that will react with the number of moles of NaCl you calculated in step 1 using the equation:



Part B: Determination of chloride in the sample.

4. Determine the average volume of silver nitrate used from your concordant titres.

5. Calculate the number of moles of silver nitrate used.

6. Use the following reaction equation to calculate the number of moles of chloride ions

that will react with the number of moles of AgNO3 you calculated in step 5:


7. Calculate the number of grams of chloride in the sample.

% of chloride in the sample = mass chloride X 100% mass sample

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Experiment (10): Complexometric titration of Zn(II) with EDTA

Purposes

To determine the Zn(II) ion concentration of a solution by complexometric titration with EDTA.

Introduction

This titration is known as a complexometric or chelometric titration because the titrant, a

ligand, reacts with the analyte, a metal ion, to form a complex, more specifically a chelate

in this case. A chelate is a ligand that has two or more sites that bind to the central ion.

EDTA [ethylenediaminetetraacteic acid, C10H16N2O8, (HOOCCH2)2N-CH2CH2-N(CH2COOH)2,

molecular weight = 292.24 g/mol, often symbolized by H4Y] is an excellent chelating agent. It

forms very strong 1:1 complexes with almost every divalent and trivalent metal ion

depending on solution conditions. Ignoring charges for the moment,

EDTA + M MEDTA

Although it is an equilibrium, the reaction lies very far to the right. The equilibrium

formational constants, Kf , are on the order of 108-1025 depending on the metal and other

conditions.

EDTA itself is a tetraprotic 4-acid; it has 4 ionizable protons with pKa's = 1.99, 2.67, 6.16,

10.26. In its fully ionized form, Y4- , the four acetate groups and the lone pairs on the two

nitrogens make it a hexidentate ligand that wraps itself very tightly around a metal ion.

Usually, titrations are performed in basic solution, roughly pH 8-11.

The fully protonated form, H4Y, is only sparingly soluble in water, so the standard form of

EDTA used analytically is usually the disodium salt Na2H2Y.2H2O (372.24 g/mol), which is much more soluble and available in primary standard purity, except for a small (about 0.3%) amount of adsorbed water.

Safety



EDTA (disodium salt), MgCl2 salt, sodium hydroxide (NaOH), Mg-EDTA solution, Potassium

cyanide (KCN), ammonia/ammonium chloride buffer solution (pH 10), zinc unknown

solution, EBT indicator solution, CaCO3, distilled water, concentrated HCl acid, 100-mL, 250-

mL and 1-L beakers, stir bar and magnetic stirrer, analytical balance, drying oven, 250-mL

and 1-L plastic bottles, 250-mL and 500-mL volumetric flask, weigh papers, conical flask (3 x

250-mL), wash bottle, burette (50 mL), pipette (10 mL), graduated cylinder, hot plate,

Buchner funnel and suction filtration, burette stand and clamp.

Complexometric titrations

185



Part A: Preparation of solutions

I. Preparation of EDTA solution (~0.01 M)

This solution must be prepared at least one day ahead of time, a week is preferable, to

ensure that the solute is completely dissolved. EDTA solutions are prepared at an

approximate molarity, and then standardized against a solution of a primary standard such

as CaCO3.

1. Dissolve about 3.8 g of the dihydrate of the disodium salt of EDTA (Na2H2Y.2H2O) and 0.1 g MgCl2 in approximately 1 L of distilled water in a large beaker using a magnetic stirrer. A small amount of sodium hydroxide can be added if there is any difficulty in

dissolving the EDTA. Try not to exceed 3.8 g of the disodium salt because much more

than this dissolves only with difficulty.

2. Before use, the EDTA solution should be filtered using a Buchner funnel and suction

filtration.

3. Store the solution in a clean, labeled 1-L plastic bottle that has been rinsed with

distilled. Never store reagent solutions in volumetric flasks.

II. Preparation of ammonia/ammonium chloride buffer stock solution, pH 10.

Each titration will require the addition of pH 10 ammonia buffer. The buffer should only be added immediately before you titrate an individual sample.

4. Dissolve 64.0 g of ammonium chloride in 600 mL of concentrated ammonia (14.8 M,

28% NH3). 5. Slowly and carefully add 400 mL distilled water with stirring. This should be

sufficient for over 120 titrations.

III. Preparation of calcium standard solution.

A CaCO3 solution is prepared as a primary standard for Ca and used to standardize the ~0.01 M EDTA titrant you prepared.

6. Tap out approximately 1 g of predried analytical-reagent-grade CaCO3 in a weigh boat. Accurately weigh (to within 0.1 mg) approximately a 0.25-g sample by

difference into a 250-mL beaker. Note: Never transfer chemicals inside an analytical

balance.

7. Add about 25 mL distilled water and then slowly add concentrated HCl dropwise with

periodic stirring until the sample dissolves completely. Then add 2 drops more. Keep

the beaker covered during the entire dissolution process. Mild heating will speed the

dissolution. Do not boil; this will spatter the calcium solution and lead to losses.

8. Transfer the solution quantitatively into a 250-mL volumetric flask. Rinse the beaker

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thoroughly with distilled water, and carefully dilute to the mark with an eye dropper

or with careful use of your wash bottle. Mix thoroughly.

9. Store the solution in a clean, labeled 250-mL plastic bottle that has been rinsed with

distilled.

Because this Ca2+ standard solution is used to standardize the EDTA titrant, it must be

prepared very carefully so that you know its exact molarity. Therefore, an exactly known (to

± 0.1 mg) mass of CaCO3 must be weighed out, dissolved completely, and transferred quantitatively into the 250-mL volumetric flask. This is critical.

Part B: Standardization of the EDTA solution

10. Attach your 50-mL burette to a ring stand, preferably using one with a white

ceramic base, and a burette clamp. If the only ring stands available have black

bases, cover the base with a completely white sheet of paper before you titrate a

sample.

11. Open the burette valve and drain it completely into a "waste" beaker. Squirt down

the insides with distilled water a couple of times. Squirt down the insides of the

burette a couple of times with a mL or two of the EDTA solution with a dropper to

rinse any remaining distilled water out of the burette.

12. Now close the burette valve and over-fill the burette with your standard EDTA

solution. Check to see if any air bubbles are trapped in the tip of the burette. If so,

open and close the valve quickly as though you were "squirting" reagent from the

burette into the waste beaker until the bubbles have cleared from the tip. Carefully

bring the reagent level to somewhere between the 0- and 1-mL marks. Do not try to

bring the level exactly to the 0.00-mL mark. This is a waste of time. Rinse the

burette tip off with a squirt of distilled water, let it drain, and then touch the tip to

the side of the waste beaker to remove excess water.

13. Pipet 25-mL aliquots of your standard Ca2+ solution into each of three 250-mL

conical flasks. Each aliquot will thus contain one-tenth of the total CaCO3 that was weighed out to prepare the standard solution.

14. Take each sample to completion before starting the next sample. Read the initial

volume on the burette at least twice. Add 7-8 mL of pH 10 buffer from, 15 mL of

distilled water, and 3 drops of Eriochrome Black T indicator, immediately prior to

titrating a sample. The solution should be a pale pink. Do not add more indicator to

make the solution darker as this can cause problems with the endpoint. Titrate the

solution immediately with EDTA against a white background until the light pink

solution turns a light sky blue. Read the final volume at least twice.

15. Repeat the titration for the other two flasks.

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Titrations must be performed swiftly (but carefully) because the NH3 will evaporate

to some degree and thus the pH of the solution will change. In general, the faster the

titrations are performed the better the results will be, as long as the endpoint is not

overshot due to excessive hast.

It is advantageous to perform a trial titration to locate the approximate endpoint and

to observe the color change. In succeeding titrations, titrate very rapidly to within

about 1 or 2 mL of the endpoint, and then titrate very carefully, a drop or half-drop

at a time, to the endpoint. Near the endpoint, periodically squirt the sides of the

flask and the burette tip and swirl the flask to ensure all the titrant has gotten into

the solution in the flask.

The endpoint color change is rather subtle, and sometimes it is slow, so you need to

be careful at the end. If you are having trouble with the endpoint color change, see

notes at the end of the report for the preparation of "before" and "after" flasks.

Calculate the molarity of the EDTA solution from the volume of EDTA used in the

titration of each aliquot. The values (M EDTA and titration volumes) should all agree

very closely. If not, titrate additional aliquots until better agreement is reached.

Part C: Analysis of the zinc unknown solution

16. Carefully dilute your Zn unknown sample in the 250-mL volumetric flask to the mark

with distilled water. Mix thoroughly.

17. Pipet 25.00-mL aliquots into each of three 250-mL conical flasks. Add 15 mL of

distilled water, 9-10 mL of pH 10 buffer, and 3 drops of Eriochrome Black T

immediately prior to titrating a sample.

18. Titrate with standardized EDTA until the pink solution turns light blue.

19. Repeat the titration for the other two flasks.

20. Calculate the milligrams of zinc in the total sample. Remember that each aliquot

represents one- tenth of the total sample volume - a 25-mL aliquot titrated out of

250 mL total volume.

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Results

Burette reading(mL)

Ca2+ standard

solution

Zinc unknown

solution

Initial reading

Final reading

Difference

Mean reading (volume of EDTA)

Data analysis

1. The molarity of the Ca2+ standard solution (M Ca) is calculated in normal fashion using

the molar mass of calcium carbonate (CaCO3) weighed out and the total volume in liters of the standard solution prepared.

2. Calculate the molarity of the EDTA from the volume of EDTA used in the titration of

each aliquot of the Ca2+ standard solution and the known 1:1 stoichiometry

between Ca and EDTA in the reaction. If the reaction has 1:1 stoichiometry, then

mmolEDTA = mmolCa

3. The mmol of each constituent is obtained by multiplying the molarity of each of the

two solutions times the volume in mL of each solution used to reach the endpoint:

MEDTA x VEDTA = MCa x VCa

4. The volume of the Ca standard solution originally taken was 25 mL and the volume of

EDTA used is the volume used to reach the endpoint, Vendpoint = VED TA , in mL.

Therefore,

MEDTA = (MCa x VCa)/VEDTA = 25 x MCa/ VEDTA

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5. The mmol of zinc determined in an individual titration uses the same 1:1 reaction

stoichiometry as for calcium above. Substituting molarity times the volume of EDTA

used in each titration of the Zn unknown produces:

mmolZn = mmolEDTA = MEDTA x VEDTA = MEDTA x Vendpoint

6. The mass of Zn obtained in a single titration, in mg, is equal to the number of mmol

of Zn times its molar mass:

mg Zn = mmol Zn x molar mass of Zn = mmol Zn x 65.38 mg/mmol

And the total mass of Zn in the original 250-mL sample is therefore 10 times this

amount.

Hazardous waste disposal

Empty all the Ca and Zn solutions that were titrated into the proper Hazardous Waste

Bottle for this experiment.

When you are completely done with the experiment, including having received your

grade, mix any remaining EDTA titrant, Ca standard stock solution, and Zn unknown

solution together in a large beaker. Pour down the drain with copious amounts of cold

tap water flowing. The first two solutions are slightly basic and slightly acidic,

respectively. When mixed, they will be near neutral. In addition, EDTA, Ca, and Zn

are not toxic and in very low concentrations, so disposal directly down the drain is

permitted and environmentally safe.

Notes

Eriochrome Black T Indicator. The color change of Eriochrome black T at the endpoint

is rather subtle. It is not an abrupt change from deep red to a dark blue; but rather it

is from a light red (or pink) to a pale blue. At least one trial titration is

recommended.

If you have trouble distinguishing the endpoint, a "before" and an "after" flask are

recommended. Prepare two 250-mL flasks in a similar manner as were the samples -

except do not add the 25 mL of Ca solution. Instead, add a total of about 80-90 mL of

distilled water to approximate the volume of the sample aliquot (25 mL), the volume

of EDTA titrant that would have been titrated into the flask, and the 15 mL of

distilled water. Add the indicator and the ammonia buffer. To one flask (the "before"

the endpoint flask) add a few drops of the Ca solution; to the other flask (the "after"

flask) add a small amount of EDTA solution to get just past the color change at the

190


endpoint. Stopper the flasks and keep them nearby for comparison of the colors.

Titrate against a white background for better discrimination of colors.

Sometimes the Eriochrome black T solution goes bad because of air oxidation. If the

endpoints seem very indistinct or slow to you, try a fresh bottle of indicator.

Alternatively, try adding a small amount of solid Eriochrome black T mixture (1 g

indicator ground with 100 g NaCl). A small amount on the end of a spatula is

sufficient.

191


Experiment (11): Determination of calcium in milk

Purposes

In this experiment, you are to determine the average amount of calcium in a milk sample.

Introduction

Calcium an important mineral for the body, it is an important component of a healthy diet and a mineral necessary for life. Calcium is a mineral that people need to build and maintain strong bones and teeth. It is also very important for other physical functions, such as muscle control and blood circulation.

If we do not have enough calcium in our diets to keep our bodies functioning, calcium is removed from where it is stored in our bones. Over time, this causes our bones to grow weaker and may lead to osteoporosis — a disorder in which bones become very fragile.

Recommended daily allowance of calcium:

Calcium needs vary with age. The Food and Nutrition Board (FNB) of the institute of medicine of the national academies provides guidelines on the amount of calcium needed each day.

Recommended daily allowance in milligrams (mg)

Milk and calcium

Milk is a heterogeneous mixture of proteins, sugar, fat, vitamins and minerals. milk and milk products are some of the natural sources of calcium. Cow‘s milk has good bioavailability of calcium (about 30 to 35%).

Milk is an excellent source of dietary calcium for those whose bodies tolerate it because it has a high concentration of calcium and the calcium in milk is excellently absorbed.

It is estimated that without milk and milk products in the diet, less than half of the calcium requirements would be met.

In this experiment, the determination of calcium in milk is based on a complexometric titration of calcium with an aqueous solution of the disodium salt of EDTA at high pH value (12).

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The common form of EDTA is disodium salt Na2H2EDTA. It is colorless and can be weighed and dissolve in water to form a stable solution.

At high pH (> 10) the remaining protons leave EDTA forming EDTA4- anion:

Complexometric titration is a type of titration based on complex formation between the analyte and titrant. EDTA is capable of forming chelate complexes with many cations in which the cation is bound in a ring structure. The ring results from the formation of a salt- like bond between the cation and the carboxyl groups together with a coordinate bond through the lone pair of electrons of the nitrogen atom.

How to determine calcium in the presence of magnesium?

The method for determining Ca2+ concentration in the presence of Mg2+ relies on the fact that the pH of the solution is sufficiently high (The pH will be approximately 12.5 due to the addition of concentrated NaOH solution) to ensure that all magnesium ions precipitate as magnesium hydroxide before the indicator is added. In this condition, magnesium ions are precipitated as hydroxide and do not interfere with the determination of calcium.

193


The Solochrome dark blue indicator is a suitable indicator in this case. The dye itself has a blue color. This blue dye also forms a complex with the calcium ions changing color from blue to pink/red, but the dye–metal ion complex is less stable than the EDTA–metal ion

complex. As a result, when the calcium ion–dye complex is titrated with EDTA the Ca2+ ions react to form a stronger complex with the EDTA changing the dye color to blue (the color of free indicator).

Ca-Indicator + EDTA4- → Ca-EDTA2- + Indicator

Safety



EDTA (disodium salt), sodium hydroxide solution (8M), sodium hydroxide pellets, , a milk

sample, Solochrome dark blue indicator solution, distilled water, 250-mL beaker, analytical

balance, drying oven, 1-L white plastic bottle, 500-mL volumetric flask, weigh paper, 250-

mL conical flask, burette (50 mL), pipette (10 mL), graduated cylinder, burette stand and

clamp.


Part A: Preparation of EDTA solution (~0.01 M)

1. The solid Na2EDTA has been dried for at least one hour in a drying oven set to at least

110°C.

2. Using an analytical balance, transfer about 2.0 grams (weighed to within 0.0001 gram)

of dry solid Na2EDTA to a 500-mL volumetric flask.

3. Fill the flask about halfway with distilled water and swirl the mixture to dissolve the

solid. It may not all dissolve at first since it is not very water soluble. The solubility

can be enhanced by adding one or two pellets of NaOH to the solution.

4. Dilute to the 500 ml mark. The solution will be ~0.01 M.

5. Clean and rinse a 1-L white plastic bottle with distilled water. Rinse the bottle with a

small amount of Na2EDTA solution. Transfer the remainder of the Na2EDTA solution

from the volumetric flask into the while plastic bottle. Label this bottle ―Na2EDTA

solution‖.

194



6. The solution of Na2EDTA that you are given is approximately 0.01 M. You will titrate it

against a known amount of a primary standard CaCO3 solution to determine the

concentration of Na2EDTA. For experimental procedures, see part B of the previous experiment No. 10

Part C: Determination of calcium in milk

7. Fill a burette with standardized Na2EDTA solution.

8. Combine 10 mL of a milk sample, 40 mL distilled water, and 4 mL of 8M sodium

hydroxide solution into a 250-mL conical flask and allow the solution to stand for

about 5 minutes with occasional swirling.

9. A small of magnesium hydroxide may precipitate during this time. Do not add the

indicator until you have given this precipitate a chance to form.

10. Then add 6 drops of the Solochrome dark blue indicator solution, pink/red color is

formed.

11. After that start to titrate using your standardized Na2EDTA solution. The endpoint

color change is from pink/red to blue. The reaction is slower near the end point, and

the titrant must be added slowly and the solution stirred thoroughly.

12. Repeat this titration (steps 6–9) two more times to obtain 3 good values for the

volume of EDTA required to titrate the Ca2+ ions in milk sample.

Results

Burette reading(mL)

Ca2+ ions in milk sample

Initial reading

Final reading

Difference

Mean reading (volume of Na2EDTA)

195


Data analysis

1. Determine the exact molarity of standardize Na2EDTA solution from your mass of

Na2EDTA and total volume of solution. The molar mass of Na2EDTA is 372.24 g mol-1

2. Using the volume of Na2EDTA titrant and its molarity, calculate the moles of Na2EDTA

used. This is the same as the number of moles of Ca+2 present in the milk sample,

since the titrant and the metal ion(s) react in a one-to-one stoichiometry.

3. Using the moles of Ca+2 above and the molar mass of CaCO3 calculate the number of

mg of CaCO3 in the water sample. This, of course, assumes that all of the Na2EDTA

reacted with Ca+2, which is the standard reporting method for water hardness.

4. Take the density of the milk sample to be 1.00 g/cm3 and calculate the concentration

of Ca2+ in your milk sample in ppm which is equivalent to mg Ca2+/kg (or mg per liter)

of milk. Report the mean value.

196


Experiment (12): Determination of water hardness

Purposes

To determine the water hardness of a water sample.

Introduction

Many metal cations form complexes in solution with substances containing a pair of unshared

electrons. A ligand is a molecule (or ion) which possesses at least one position at which it

can attach itself to a metal ion. Ammonia is a common example. It can attach itself to the

water insoluble AgCl and convert it to the water soluble form [Ag(NH3)]2+. The NH3 molecule

has a single point of coordination which is through its nitrogen atom. For this reason it is

referred to as a unidentate ligand.

Other ligands can attach themselves to many different places: bidentate, tridentae,

tetradentate, and hexadentate ligands are not uncommon. When the number of attachment

points increases, the ligand effectively wraps itself around and "claws" itself to the ion.

When this happens, the complexing agent is called a chelate (pronounced key-late), which is

derived from the Greek word for claw or hoof, representing the characteristics of the metal-

ligand complex. A very widely used complexing reagent for this type of titration reaction is

ethylenediaminetetraacetic acid (EDTA). For solubility reasons the disodium salt will be used

in this experiment. EDTA complexes with the ions contributing to water hardness in a one-

to-one stoichiometry. Even though a ligand may attach itself to the metal ion many different

places within the complex it does not affect the overall stoichiometry, which is what we

really need to know to carry out the calculations. In the case of EDTA it is not known with

certainty whether it attaches itself to the hard water metal ions at either four or six

positions but it does not really matter because regardless of the number of points of

attachment the stoichiometry is known to be one-to-one between the metal ion and the

EDTA. The procedure described below fulfills all of the requirements for a volumetric

titration and is widely used for the routine determination of water hardness.

A water supply is considered hard when the amount of Ca+2, Mg+2, and/or Fe+3 ions becomes

too high for its intended use. Soft water does not contain any significant amounts of these

ions. People with medical intolerance to hard water may need to have a water treatment

cartridge installed on their water supply to remove the hard water ions. We have all

experienced the problems of hard water when using soaps. When you take a shower and use

197


a bar of soap, you will notice a scum or "ring" around the bathtub or the walls of the shower,

which is a direct result of the formation of the insoluble calcium (or other hard water metal

ion) salts of the fatty acids that occurred when the soap solution came in contact with the

hard water ions. Using sodium stearate as a typical soap molecule, it dissolves in water to

form sodium and stearate ions. When hard water is used, the calcium ion reacts with the

stearate ion and forms the slightly soluble compound calcium stearate, which has a solubility

product constant (Ksp) associated with it so that an equilibrium condition is established. The

equation for the dissolving of calcium stearate in water is:

Ca(C17H35CO2)2(s) + H2O(l) Ca(C17H35CO2)2(aq) Ca2+(aq) + 2C17H35CO2-(aq)

slightly soluble

Obviously, this soap scum would cause many problems if soap is used to wash your hair,

clean your clothes, or wash dishes because the scum could deposit itself and remain on the

object being cleaned. The Fe+3 salts in particular are what contribute to laundered clothes

looking dingy or off color if soap is used. For applications such as these, soap is replaced by

a detergent, which is more expensive than soaps so their use has been restricted to

applications where the problems of the soap scum warrant the additional expense.

Detergents became very popular because they do not have the typical "scum" problems

associated with soaps. The Ca+2, Mg+2, and/ or Fe+3 salts of detergents are water soluble and

are then efficiently removed from the cleaning system by merely rinsing with water.

For reasons such as these, it becomes necessary to readily determine just how hard a water

supply is and whether it may need conditioning to reduce the hardness. It is not necessary to

measure each ion contributing to the water hardness separately. Instead, all hard water ions

are determined collectively and reported as an "apparent hardness" assuming that all of the

hardness is derived from CaCO3. This is done more for convenience than any other reason. It

really does not matter whether the hard water ions are Ca+2, Mg+2, and/or Fe+3 because they

all contribute to the problem. The only significant point is that any or all of them exist in

solution and therefore make the water hard. To keep the reporting units consistent from one

location to another, a uniform standard has been adopted by mutually agreeing to report

water hardness as parts per million (ppm) of CaCO3 which is equivalent to mg CaCO3 per kg

of water.

198


In the titration we are going to perform, the indicator is itself also a chelating agent. For

Ca+2, Mg+2, and/or Fe+3 analysis, Eriochrome Black T (EBT), which has 3 ionizable protons

(represented as H3In), is used as the indicator. The color of the indicator complexed with

Mg+2 in solution is red. As EDTA is added the free metal cation is complexed first because the

Mg-EDTA complex is more stable than the Mg(ln)-. After the free metal cation is used up, the

metal is displaced from the indicator and complexed to the EDTA which causes a color

change from red to blue (which is the color of uncomplexed indicator). The first time

through this may be a little tricky. As the titration progresses, the color will change from red

to a reddish-purple, then finally to blue. You want to continue until you see the distinct blue

color.

EBT cannot be used to indicate the titration of Ca+2 alone because it forms too weak a

complex with the indicator and therefore does not give a sharp endpoint. Therefore a very

small amount of Mg+2 is added as Mg-EDTA for a sharper endpoint. This small amount has an

insignificant effect on the final result.

The total water hardness determination is done at pH of around 10, using an ammonia-

ammonium chloride buffer. This is done because if the pH is too high the hard water metal

ion hydroxide(s) may precipitate, causing the reaction with EDTA to be slow. Also, EBT is a

weak acid and its color can depend on the pH, because the different ionized species have

different colors. This change in color could complicate or cloud the endpoint detection.

Safety



EDTA (disodium salt), sodium hydroxide (NaOH), Mg-EDTA solution, Potassium cyanide (KCN),

ammonia/ammonium chloride buffer solution (pH 10), a water sample, EBT indicator

solution, ferrous sulphate (FeSO4·7 H2O), distilled water, brita water filter, 100-mL and 500-

mL beakers, stir bar and magnetic stirrer, analytical balance, drying oven, 1-L white plastic

bottle, 100- and 500-mL volumetric flask, weigh paper, 250-mL conical flask, burette (50

mL), pipette (10 mL), graduated cylinder, burette stand and clamp.

199



Part A: Preparation of EDTA solution (~0.01 M)

1. The solid Na2EDTA has been dried for at least one hour in a drying oven set to at least

110°C.

2. Using an analytical balance, transfer about 2.0 grams (weighed to within 0.0001 gram)

of dry solid Na2EDTA to a 500-mL volumetric flask.

3. Fill the flask about halfway with distilled water and swirl the mixture to dissolve the

solid. It may not all dissolve at first since it is not very water soluble. The solubility

can be enhanced by adding one or two pellets of NaOH to the solution.

4. Dilute to the 500 ml mark. The solution will be 0.01 M.

5. Clean and rinse a 1-L white plastic bottle with distilled water. Rinse the bottle with a

small amount of Na2EDTA solution. Transfer the remainder of the Na2EDTA solution

from the volumetric flask into the while plastic bottle. Label this bottle ―Na2EDTA

solution‖.


6. The solution of Na2EDTA that you are given is approximately 0.01 M. You will titrate it

against a known amount of a primary standard CaCO3 solution to determine the

concentration of Na2EDTA. For experimental procedures, see part B of the previous

experiment No. 10

Part C: Determination of water hardness of a water sample.

7. Obtain a water sample from your instructor. Using 100-mL volumetric flask, transfer

exactly 100 ml of the sample into a 250-ml conical flask, add 2 ml of the

ammonia/ammonium chloride buffer solution (pH 10), 0.5 ml of the Mg-EDTA solution,

and five drops of EBT indicator solution. Do not use too much indicator or you could

make the endpoint detection more difficult. The volumes of buffer and Mg-EDTA are

not critical and will not be used in any calculations so they can be added from a

graduated cylinder. The buffer should be added before the indicator, so that any

small amounts of Fe+3 present will not react with the indicator.

8. This step is to be done only if your instructor indicates it is necessary. If the sample

contains high levels of Fe+3 a violet endpoint color may be seen instead of a blue one.

The interference can be eliminated by adding a few crystals of KCN.

Caution: KCN is poisonous and must not come in contact with an acid since it will

release the deadly gas HCN. Add this only after the alkaline buffer is added. After the

titration, add about 1 g (qualitative measure) of FeSO4·7 H2O to convert CN- to

200


[Fe(CN)6]-4 before disposing of the solution.

9. Add a stir bar to the flask and place it on a sheet of white paper on top of a magnetic

stirrer. The white paper will help show the endpoint color change more clearly.

Titrate the sample using your standardized Na2EDTA solution. The endpoint color

change is from wine red to blue. The first time through, you may not be familiar with

the change and accidentally overshoot the endpoint. If this happens, you will need to

repeat the analysis. The reaction is slower near the end point, and the titrant must be

added slowly and the solution stirred thoroughly.

10. If your water sample was fairly soft and you used less than about 20 ml of the EDTA

titrant, adjust the sample size to require at least 20 ml and repeat the analysis. When

you are done you should have three "good runs" using an acceptable volume of titrant.

11. Now pass your water sample (enough for three more runs) through the brita water

filter.

12. Perform the titration again (three more runs) on the filtered sample.

Results

Burette reading(mL)

Water sample

Filtered water

Initial reading

Final reading

Difference

Mean reading (volume of Na2EDTA)

Data analysis

1. Determine the exact molarity of standardize Na2EDTA solution from your mass of

Na2EDTA and total volume of solution. The molar mass of Na2EDTA is 372.24 g mol-1

2. Using the volume of Na2EDTA titrant and its molarity, calculate the moles of Na2EDTA

used. This is the same as the number of moles of Ca+2 present since the titrant and

the metal ion(s) react in a one-to-one stoichiometry.

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3. Using the moles of Ca+2 above and the molar mass of CaCO3 calculate the number of

mg of CaCO3 in the water sample. This, of course, assumes that all of the Na2EDTA

reacted with Ca+2, which is the standard reporting method for water hardness.

4. Take the density of the water sample to be 1.00 g/cm3 and calculate the water

hardness in ppm of CaCO3 which is equivalent to mg CaCO3/kg (or mg per liter) of

water. Report the mean value.

5. Calculate the water hardness of the filtered water in ppm as the mean value.

6. Include in the report how brita and other water filtration units work. An important

component of these filtration systems is ion-exchange resin. What is it, are there

different types? How do they work? Rationalize your data with this background

knowledge in mind.


1. When Mg-EDTA is added to a sample to compensate for a lack of Mg+2, why is no blank correction necessary?

2. Report the minimum pH for an effective titration of the following metal ions with EDTA.

i. Ca+2 ii. Fe+2 iii. Ga+3

Why do different metal ions have different minimum pH requirements for titration?

3. Write the equation for the titration of Ca+2 with EDTA.

4. Write the equations for the endpoint color change of Eriochrome Black T (EBT) with

Mg+2.

5. Calculate the water hardness as ppm CaCO3 of a 50 ml water sample that required 32. 50 ml of 0.01 M EDTA to reach the end point of the titration.

Bok Coordinator

Mostafa Fathallah

General Directorate of Technical Education for Health

Gender and Social Norms

vi

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Analytical Chemistry, P K Dasgupta, G D Christian, K A Schug, John Wiley & Sons Ltd, Chichester, United Kingdom, 2014.

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حقىق الىشش والحأليف لىصاسة الصحة والسكان ويحزس تيعه

References and Recommended Readings

Analytical Chemistry...solutions. 3 2 General concept of acids and bases. 3 4 3 Strengths of acids...

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