GTU

CHEM 113

GENERAL

CHEMISTRY

LABORATORY-I

BOOKLET

2016 FALL

2

PREFACE

Üniversitelerde Fen bölümleri, tıp ve mühendislik öğrencileri ilk yıllarında Kimya

Laboratuvarı dersi ile karşılaşmaktadır. Kimya Laboratuvarı dersi ile ilgili bu kitapçığın

başında öncelikle kimya laboratuvarı için son derece önemli olan kimya laboratuvarında

uyulacak kurallar ve laboratuvarda alınması gereken güvenlik önlemleri açıklanmıştır.

Laboratuvar güvenliği ile ilgili detaylı bilgiler, verilen web bağlantısındaki kaynakta yer

almaktadır. Kimya Laboratuvarında kullanılan malzemelerin tanıtıldığı bölümden sonra,

Genel Kimya Ders konuları ile bağlantılı seçilen deneylerin föyleri yer almaktadır.

Laboratuvara ilk geldiğiniz gün, kimya laboratuvarında uyulacak kurallar ve laboratuvarda

alınması gereken güvenlik önlemleri konularında bilgilendirilerek, yangın tüpleri, güvenlik

duşu, göz yıkama muslukları gibi kaza anında kullanılacak malzeme ve sistemlerin yerleri ve

nasıl çalıştıklarını öğreneceksiniz.

Kimya laboratuvarları, her an istenmeyen kazaların yaşanabileceği yerler olması nedeni ile

herşeyden önce güvenlik konularındaki bilgilerin öğrenilmesi çok önemlidir. İlk hafta

güvenlik kuralları ve önlemleri bölümünü ve her deney haftası deney föyünü mutlaka okuyup

öğrenerek geliniz.

Hepinize sağlıklı, başarılı ve kazasız bir eğitim, öğretim dönemi dilerim.

Saygılarımla

Prof. Dr. Ayşe Gül GÜREK

Kimya Bölüm Başkan V.

3

CONTENT

PREFACE .................................................................................................................................. 2

CONTENT ................................................................................................................................. 3

LABORATORY RULES FOR CHEM 113 GENERAL CHEMISTRY LAB. I COURSE ...... 4

LABORATORY EQUIPMENTS ............................................................................................ 11

EXPERIMENT 1: IDENTIFICATION OF PURE SUBSTANCES BY THEIR PROPERTIES

.................................................................................................................................................. 16

EXPERIMENT 2: IDENTIFICATION OF SUBSTANCES BY THEIR

REACTIONS………………………………………………………………………………...20

EXPERIMENT 3: STOCHIOMETRIC CALCULATIONS ................................................... 26

EXPERIMENT 4: ACID - BASE TITRATION ..................................................................... 29

EXPERIMENT 5: DETERMINATION OF A SALT SOLUBILITY ..................................... 36

EXPERIMENT 6: DETERMINATION QUANTITY OF Fe2+

WITH KMNO4 ..................... 40

EXPERIMENT 7: BUFFERS, BUFFER CAPACITY, AND BUFFERING ZONE .............. 44

EXPERIMENT 8: POLYPROTIC ACIDS: DETERMINATION OF pKA VALUES USING

pH TITRATION CURVES ...................................................................................................... 53

4

LABORATORY RULES FOR CHEM 113 GENERAL CHEMISTRY LAB.

I COURSE 1. The students must attend the General Chemistry Laboratory course in exact day, time and

laboratory, which are announced on the course programs’ of the departments.

2. Attendance compulsion is %80 for the laboratory semester. If a student does not attend for at

least 2 experiments, a VF will be issued.

3. During the lab course, the students have to wear their lab coat, lab goggles and latex gloves.

The students must bring their own lab coat, lab goggles and latex gloves to the lab. Be aware that

failing to wear lab coat, lab goggles and latex gloves in the laboratory will result in immediate

expulsion from the laboratory, failing the experiment. This rule is strictly enforced.

4. If the student is late more than 10 minutes to the lab course, she/he will not to be allowed to

attend the lab and the experiment.

5. During the lab course, it is strictly forbidden to use mobile phones. The students must not

leave the lab and experiment set without permission of the lab assistant.

6. In order to provide life safety and to ensure the success of experiment in a safe manner, hand

jokes among the students are strictly forbidden. Also, touching to chemicals without gloves, sniffing

and tasting them are dangerous.

7. In each experiment, required chemicals and materials will be supplied to the students by lab

assistants. The students will not borrow the chemicals and materials from the other students. At the

end of the each experiment, all the glass or metal materials will be cleaned by the students. Also, the

experiment set will be tidy and clean.

8. Before coming to the lab, the students are expected to read the experiment of the week in this

booklet thoroughly and to know about the procedure of the regarding experiment and theoretical

information.

9. At the start of each laboratory, the students will be given a prelab quiz. The object of this quiz

is to assess whether they are well prepared for the experiment.

10. The total grade of an experiment will be the sum of the report grade (%50) and prelab quiz

grade (%50).

11. At the beginning of the semester, the students will be informed about the safety rules of

laboratory. After that, the students will be required to pass the written examination about the safety

rules in two attempts. If the student does not get 60 out of 100 from the written examination in the first

attempt, which will be held at the2nd

week of semester, the student will have the second right for the

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examination as a make-up exam in the 3rd

week of semester. Failing to get the score over 60 in both

exams will result in failing the General Chemistry Laboratory course.

12. The final grade of General Chemistry Laboratory course will be based on the sum of the

laboratory safety rules exam (%20) and the average grade of the all experiments for the semester

(%80).

13. Students who do not obey the rules and fulfill the requirements of the lab will be considered as

not attended to the lab. (look at entries 1, 3, 4, 11) and no reports for the mentioned lab session will be

accepted.

14. Make-up experiments will be available only for those with health reports. The health reports

must be handed over within 5 days to the lab assistant.

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LABORATORY SAFETY RULES AND CAUTIONS

1. Lab coats must be worn at all times when in the laboratory! All students have to wear

lab coat and students without lab coat will not be allowed in the lab.

2. Safety glasses must be worn at all times when in the laboratory! All students have to

wear safety glasses in the lab and students without safety glasses will not be allowed in the

lab. Contact lenses are not allowed in the lab because acid and organic chemical vapors could

get in between eye and lenses, lenses can be glued on the eye in the case of an accident and

can be diffıcult to remove.

3. Dress appropriately! Wide dresses, sandal type shoes shouldn’t be worn in the lab and

long hair should be tied.

4. No Foods or Drinks! Food, drink and chewing gum are not allowed in the lab.

5. Know the location of exits and safety equipment! All students should know where the

fire extinguisher, first aid cabinet and shower are placed in the lab.

6. Learn the speediest exit from the lab. in the case of a fire.

7. Use the shower when your dresses or hair catches fire.

8. Do not run and do not make jokes in the chemistry lab.

9. Benches should not be used to sit on or to leave bags or personal things.

10. Smoke and vapor released during chemical reactions should not be exposed directly.

11. Do not attempt unauthorized experiments! It is not allowed to work in chemistry labs.

without attendance of assistants or instructors.

12. Do not use Bunsen burners next to flammable chemicals (e.g. ethers)

13. Read the labels carefully on the bottle before use of any chemicals.

14. Read the experimental procedure before coming to the lab. Students who come to lab

having no knowledge about experiment could create risks for themselves and other students.

15. In the case of any accident (glass cut, acid/base burn, fainting etc.), know where to get

help fast and immediately inform your assistant or instructor.

16. Do not orientate the test tube toward yourself and your friend. Reaction carried out in the

test tube could be dangerous.

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17. Water should not be added on concentrated acids. Acids should be added to the water

slowly and by stirring.

18. It is forbidden to smell, to taste the chemicals or to pull solutions by mouth when using

pipet. Immediately clean up any chemicals that you spill. If necessary, obtain advice on the

cleanup procedure from the lab assistant.

19. Chemicals (solid, liquid or solution) must not be dumped into the sink. Waste bottles in

the lab should be used. (Learn where the waste bottles in the lab are.)

20. Use the ‘’ broken glass labeled container in the lab for the broken glass pieces.

21. Matchstick, litmus paper must not be disposed to the sink.

22. Mercury vapor is invisible and toxic. Mercury in the broken thermometer is very

dangerous. You must inform your assistant in the case of thermometer breakage.

23. Materials like hot test tube, crucible, and beaker must not hold by hand. Theyshould be

cautiously held by tube tongs or left to cool on an asbestos wire.

24. Please use the amounts of chemicals given in the procedure. Use of excess amounts can

make difficult to control the reactions or cause the side reactions.

25. The chemicals left behind the experiment should not be returned to the stock bottles,

should be discarded into the waste bottle.

26. Always keep your working area, the balance and its environment clean. Work clean and

tidy in the lab.

27. Don’t change the locations of the stock chemicals during the experiment.

28. At the end of the experiment, hand in all the materials you used to your assistant as

cleaned.

29. Before leaving the lab, make sure that gas and water taps are closed (turned off).

30. Wash your hands before leaving the lab.

31. Learn the information given in the link below and Table 1.

http://www.gtu.edu.tr/Files/kimyaBolumu/documents/LabGuvenlik.pdf

Note: This web page is important for Lab. Security Examination

General laboratuvary safety exam. covers the information given in the link above (slides ; 1-

32, 40-47 , 52 ) and Table 1.

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The following cases, must be notified firstly to the assistant or lecturer.

BURN: expose the burned area to the tap water (5-10

min.), apply first aid.

CUT / INJURY: Wash with water

and apply first aid.

FAINTING: Provide fresh air. Lay down and put the

head lower than the body.

FIRE: (Notify the assistant

immediately) Put the bunsen burner

off. Use shower in case of hair and

clothes caches fire. Use the fire

extinguisher when necessary.

BLEEDING: compress on the wound, keep the wound

above the heart level and get medical help.

CHEMICAL SPILL: clean in a

manner appropriate to the chemical.

Aqueous solutions can be removed

with water. Information your assistant.

ACID BURNS: Use NaHCO3 solution

BASE BURNS: Use Boric acid or Acetic acid solution

CHEMICALS SPILLED IN THE

EYE: The Eye is washed immediately

with plenty of water for at least 15

minutes (use the eye-wash shower

rooms) Get medical help.

9

Laboratory Security Pictograms According To the United Nation’s Globally

Harmonized System of Classification and Labeling of Chemicals (GHS)

Pictogram Usage

E (Explosive):

Spark, heat, flame, pressure, or friction can cause explosion (R1-

R3). Keep away from spark, flame, heat, or pressure Protective

dress should be worn and a safe distance should be kept.

F (Flammable):

F+ (Extremely Flammable): Flammable and flashing (R10-

R12). Flashing point is below 35 °C. Keep away from open fires,

sources of heat and sparks.

O (oxidazing):

Oxidizing substances can ignite flammable and combustible

material or worsen existing fire and thus make fire fighting more

difficult. Caution! Keep away from flammable, combustible and

spontaneously combustible materials. (R7-R9). Protective dress

should be worn and a safe distance should be kept.

G (Gas):

Compressed, liquidified, refrigerated, or dissolved gas. May

cause its container to explode if heated or dropped.

C (Corrosive):

Causes severe eye and skin irritation upon contact. Prolonged

contact can causes severe tissue damage. May be harmful if

inhaled. Do not breathe vapors; avoid contact with skin and eyes.

(R34, R35).

10

T (Toxic):

T+ (Very Toxic-):

Toxic (R23-R-25) and very poisonous (R26-R28). Very

hazardous to health when inhaled, swallowed or when they come

in contact with the skin. May even lead to death. Avoid contact

with the human body and immediately contact a physician in

case of contact.

Xi (Irritant):

Xn (Sensitizing):

Causes severe eye and skin irritation upon contact. May be

harmful if inhaled. Do not breathe vapors; avoid contact with

skin and eyes. (R20-R22i R36-R38)

H (Healt effect):

Sign for carcinogens, mutagens, teratogens, respiratory

sensitizers and substances with target organ toxicity. Avoid

contact with body/skin. Avoid inhalation. (R40, R45-R47).

N (Toxic to environment):

Hazards to the aquatic environment material. Must not be

disposed or released to the environment.

All the chemical substances are identified according to the risk

of hazard. The numbers and/or letters on the pictogram identify

the hazard degree of the substance.

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LABORATORY EQUIPMENTS

12

13

14

15

16

EXPERIMENT 1: IDENTIFICATION OF PURE SUBSTANCES BY

THEIR PROPERTIES

1.1. Purpose

Identification of the substances by utilizing the physical and chemical properties of them.

1.2. Theory

Matter is anything that occupies and displays the properties of mass and inertia. All

matter could be classified as pure substances and mixtures.

Pure substances:

The composition and properties of an element or a compound are uniform throughout a given

sample and from one sample to another. Element and compounds are called substances. Pure

substances have characteristic chemical and physical properties.

1) Elements: An element is a substance made from only one type of atom, that cannot be

broken down into simpler components by any non-nuclear chemical reaction (C, Na, Mg,

N2).

2) Compounds: Chemical compounds are substances comprising atoms of two or more

elements joined together. (H2O, AgCl, CaCO3).

Mixtures:

1) Heterogeneous mixtures: The composition and physical properties vary from one part of

the mixture to another. (oil + water, sand + water).

2) Homogeneous mixtures: The mixtures is uniform in composition and properties

throughout is said to be homogeneous mixtures (sugar + water, salt + water)

A compound can be identified with the following characteristics:

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Figure 1.1. Classification of matter

Physical properties

A physical property is one that a sample of matter displays without changing its composition.

Melting point, boiling point, density, solubility, physical state, colour, and crystal structure

can be given as an example of physical properties. In physical changes, appearance of the

substance changes, but the composition of it doesn’t changes.

For example;

- Melting of ice

- Dissolution of salt in water

- Melting of cupper at high temperatures

- Melting of candle are physical changes.

Chemical properties

In a chemical changes the composition of matters changes and new kids of matters occur.

According to this chemical properties is the ability of a substance to a variation in the

composition under certain conditions.

Chemical reactions are understood from the following observations:

-Color change

- Precipitation formation

- Gas release

- Heat and sound formation

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For example;

- Burning of sugar,

- Making cheese from milk

- Combustion of wood are chemical changes.

1.3 Materials and Chemicals

Materials Chemicals

Glass stick

Test tubes

Bunsen burner

Wooden tongs

Litmus paper (blue

and red)

Tube holder.

Salt (NaCl)

Sugar (C6H12O6)

Okzalic asid (H2C2O4)

Sand (SiO2)

Hypo (Na2S2O3)

Lime (CaO)

Cupper sulfate (CuSO4)

1.4. Experimental Procedure

Tests given below is performed for all of these substances:

1. Heat change

2. Solubility in water

3. Acidity and basicity of the water solutions

1) Put a small amount of substance (spatula-tip) is into a clean and DRY test tube and heat on

bunsen burner.

* Observe the changes in the substance carefully and write down the type of the changes.

2) Put a small amount of substance (spatula-tip) is into a clean and DRY test tube and fill half

of the test tube with pure water. Stir with glass stick and observe.

Is the solution homogeneous, heterogeneous, is there any suspension formation? Write down

your observations.

If the solution is not homogeneous, heat gently and note if there is a change or not.

Keep the solutions to use in part 3.

3) Put a drop of the solutions obtained in part 2 on a litmus paper using glass stick. Note if

there is a color change.

If RED litmus turns into BLUE → compound is BASE

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If BLUE litmus paper turns into RED → compound is ACID

If there is no change in the color → compound is NEUTRAL

4) Ask your assistant for your unknown substance. Perform the testes done above to identify

the substance. (Unknown substances: All, except sand, cupper sulfate and sugar)

*All the test tubes used should be clean. Any contamination may cause you to find wrong

compound.

Cautions:

Wear protector glasses to protect your eyes.

Don’t smell released gasses.

1.5. Results and discussion

Heating Type of

change

Solubility in

water

Acidity/Basicity

Salt (NaCl)

Hypo (Na2S2O3)

Oxalic asid (H2C2O4)

Cupper (II) sulfate (CuSO4)

Sugar (C6H12O6)

Lime (CaO)

Sand (SiO2)

Unknown No:

Formula:

1.6. Questions

1. What is chemical and physical change? Explain and give an example for each.

2. What are the observations to identify chemical change?

3. How do we identify whether a compound is acidic or basic?

4. Is heat always cause same change? Think about your observations for the answer

20

EXPERIMENT 2: IDENTIFICATION OF SUBSTANCES BY THEIR

REACTIONS

2.1. Purpose

Investigation of the reactions given by the compounds

2.2. Theory

Chemical reaction is change of one or more pure substances into new substances as a result of

electron mobility causing bond breakage /formation under specific conditions. There are

many types of reactions and can be classified in different ways.

Chemical reactions may be grouped in four main groups:

1. Coupling reactions: A compound forms by combination of the two or more elements.

A + B → AB

CO2 + H2O→ H2CO3

2. Decomposition reactions: The reaction is the opposite of the coupling reaction and

degradation of a compound of two or more simple substances.

AB → A + B

NaCl → Na+ + Cl

-

3. Single displacement reaction: A reaction that an element is displaced to the other element

in a compound.

A + BC → AC + B

Mg + 2HCl → MgCl2 + H2

4. Double substitution reactions: A reaction that displacement takes place between plus or

minus charged ions in the reaction. Such reactions can take place only in solution and may

end with,

a) Precipitation formation

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b) Gas release

Positive charge exchange

↓ ͞ ͞ ͞ ↓

AB + CD → AD + CB

KCl(aq) + AgNO3(aq) → KNO3(aq) + AgCl(k)

K2CO3(aq) + HCl (aq) → KCl(aq) + H2O(s) + CO2(g)

Precipitation formation is defined by the solubility rules given below.

General Rules for the solubility of ionic compounds in water.

Water Soluble Compounds

Na+, K

+ and NH4

+ salts

NO3 -, C2H3O2

- and ClO3

- salts

Cl -, (Br

-) and (I

-) salts, (except Ag

+, Pb

+2 and Hg2

+2 compounds)

SO4 -2

salts (except Ba+2

, Pb+2

and Sr+2

compounds)

CO3 -2

and PO4 -3

salts (except Na+, K

+ and NH4

+ compounds)

OH- and O

-2 salts (except Na

+, K

+, NH4

+ and Ba

+2 compounds)

S-2

salts ( except Na+, K

+, NH 4

+ and Ba

+2 compounds)

Compounds which are not soluble in water:

If a compound is not soluble in water, can be made soluble by the methods below.

1) pH of the solution can be adjusted.

2) Oxidation- reduction reactions can be used.

3) Water- soluble complex of the compound can be formed.

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2.3 Materials and Chemicals

Materials Chemicals

Test tubes (6-8)

Bunsen burner

Wooden tongs

Tube holder.

Water bottle

Baget

Sodium carbonate (Na2CO3)

Magnesium oxide (MgO)

Sodium sulfate (Na2SO4)

Magnesium sulfate (MgSO4)

Lead nitrate(Pb(NO3)2)

Barium chloride (BaCl2)

Barium nitrate solution

(BaNO3)2(aq)

Hydrochloric acid solution

HCI(aq)

Sulfuric acid solution H2SO4(aq),

Nitric acid solution HNO3(aq)

2.4. Experimental Procedure

List of the compounds are given below:

Cupper sulfate (CuSO4)

Ammonium chloride (NH4Cl)

Sodium carbonate (Na2CO3)

Sodium sulfate (Na2SO4)

Magnesium oxide (MgO)

Magnesium sulfate (MgSO4)

Lead nitrate (Pb(NO3)2)

Barium chloride (BaCl2)

Tests given below is performed for all of these substances:

1) Put a small amount of substance (spatula-tip) is into a clean and DRY test tube and heat on

bunsen burner.

* Observe the changes in the substance carefully and write down the type of the changes.

2) a) Find the solubility of each substances in water by using solubility rules and save in your

report as soluble or insoluble

b) Prepare the solutions of the compounds soluble in water. Put a very small amount of the

substance in a test tube and fill half of the tube with water. Dissolve the compound by stirring

with glass stick.

c) Divide the homogeneous solutions obtained into 3 parts. Add

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* 2-3 drops of Ba(NO3)2 (Barium nitrate) solutions, into first test tube

* 2-3 drops of HCl (Hydrochloric acid) solutions, into second test tube

* 2-3 drops of H2SO4 (Sulfuric acid) solutions, into third test tube

Note any change. If there is a change (as precipitation, color change, gas release, suspension

formation) write down the reaction for each substance.

3) a) If the substance is not soluble in water. Dissolve it in HNO3 (Nitric acid) solution. Note

if there is a change or not.

b) repeat the test in 2-c

4) Ask your assistant for your unknown substance. Perform the testes above to identify the

substance. (Unknown substances: All, except cupper sulfate)

(NOTE: All the test tubes used should be clean. Any contamination may cause you to find

wrong compound)

Cautions:

Wear protector glasses to protect your eyes.

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2.5. Results and discussion

Heating Solubility

H2O/HNO3

Ba(NO3)

addition

HCl

addition

H2SO4

addition

Sodium

carbonate

(Na2CO3)

Sodium sulfate

(Na2SO4)

Lead nitrate

(Pb(NO3)2)

Cupper sulfate

(CuSO4)

Ammonium

chloride(NH4Cl)

Magnezium

oxide(MgO)

Barium

chloride(BaCl2)

Unknown No: Formula:

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2.6. Questions

1. Write down the type of reactions below.

a) Fe3+

+ 3OH−

→ Fe(OH)3________________

b) NaClO3→ NaCl + 3/2 O2 ________________

c) Zn+ 2HCl → ZnCl2 + H2 ________________

d) 2NaOH + CaCl2 → Ca(OH)2 + 2NaCl ________________

2. Give the compound name for the following ;

a) Dissolves NaCl but not the AgCl ________________

b) Dissolves Na2CO3 but not the BaSO4’ı ________________

c) Dissolves both ZnS and Ag2S ________________

d) Dissolves both CaO and Sr(OH)2 ________________

3. Give the compound name for the following ;

a) Forms precipitate both with NaCl and FeCl2 ________________

b) Forms precipitate both with KCl and K2CO3 ile ________________

c) Forms precipitate with AgNO3 but not with Zn(NO3)2 ________________

4. Complete the possible ones in the reactions below

a) CaCO3 (k) + HNO3 (sulu) →

b) CuBr2 (aq) + K2S (aq) →

c) Sr (s) + O2 (g) →

d) KCl (aq) + AgNO3 (aq) →

e) BaSO4 (s) + NaCl(aq) →

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EXPERIMENT 3: STOCHIOMETRIC CALCULATIONS

3.1. Purpose

Determination of the reaction stoichiometry, calculation of the theoretical and actual yield.

3.2. Theory

Many information like mol ratios and physical states of the reactants and products can be

obtained from a balanced reaction equation. For example;

2Cu(NO3)2 (s) → 2Cu (aq) + 4NO3-(aq)

Ba(NO3)2 (aq) + K2SO4 (aq) → BaSO4(s) + 2KNO3 (aq)

In all the reactions, reactant compounds combine in specific ratios. If one of the reactants is

used in excess, excess amount remains in the reaction medium without reacting. Completely

used reactant in a chemical reaction is called limiting component and this component

determines the amount of products.

Amount of a product calculated from the balanced equation of a chemical reaction is called as

theoretical yield. Many reactions are completed, in such cases the amount obtained is always

less than theoretical yield. Amount obtained at the end of a reaction is called real yield.

Percentage yield of the reaction is calculated using the following formula:

Real Yield

% yield = x 100

Theoretical yield

In this experiment the following reaction is going to be performed and percentage yield will

be calculated.

FeCl3(aq) + 3K2C2O4.H2O (aq) → K3Fe(C2O4)3.3H2O(s) + 3KCl(aq)

(Iron(III) chloride + potassium okzalate → potassium ferritriokzalate trihidrate + potassium

27

chloride).

3.3 Materials and Chemicals

Materials Chemicals

100 ml beaker

100 ml erlen mayer

Funnel

filter paper

glass rod

bunsen burner

balance

desiccator

FeCl3(aq)

K2C2O4.H2O (aq)

3.4. Experimental Procedure

Synthesis of K3Fe(C2O4)3.3H2O salt

1. Weigh 4 gr. K2C2O4.H2O into a 100 ml. beaker and add 8 ml distilled water.

2. Heat gently (not be boiled) to dissolve the compound.

3. Add 4 ml FeCl3 solution into the hot solution by using a measuring cylinder . Color of the

solution will turn to green. (1 ml solution contains 0,4 g FeCl3)

4. Solubility of the substance to be synthesized increases by temperature. Therefore, leave the

beaker in ice bath for a while to precipitate the compound.

5. Filter the content of beaker by using funnel and weighed filter paper. Discard the

decantate. Wash the crystals remained on the filter paper using a little amount of water.

6. Take the filter paper carefully from the funnel and place on a watch glass. (Don’t forget to

write your name on the watch glass.)

7. Weigh the synthesized compound next week and use in the calculations.

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3.5. Results and discussion

FeCl3 solution volume :

Filter paper mass:

Filter paper + product mass

Product mass

FeCl3 mass:

FeCl3 mole:

K2C2O4.H2O mass :

K2C2O4.H2O mole:

Excess reactant mass and mole:

Product mole (obtained):

P roduct mole ( theoretical):

% Yield:

3.6. Questions

1. In the precipitation reactions, obtained product mass is always less than the expected one.

Explain, why?

2. Reaction of 1.5 gr. CaCl2 and 3.4 gr. AgNO3 gives 2.5 gr. AgCl. Write the equation for this

reaction and calculate the % yield. (CaCl2: 111 gr/mol, AgNO3: 170 gr/mol, AgCl: 144

gr/mol)

3. 3 ml, 0.1 M Pb (NO3)2 reacts with 12 ml 0.1 M K2CrO4. Calculate the amount of the

precipitate (PbCrO4) and the concentrations of the ions remained in the solution. (PbCrO4:

323 g/mol)

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EXPERIMENT 4: ACID - BASE TITRATION

4.1. Purpose

Preparation of acid and base solutions, learning of neutralization reaction.

4.2.Theory

We encounter the concept of acids and bases in our daily life. For example, acid rain

phenomenon is known as a current environmental issue. Acids and bases are one of the most

important topics in chemistry since a majority of the chemical reactions are the acid and base

reaction.

Despite the wide variety of definitions of acids and bases, Arrhenius, Bronsted-Lowry and

Lewis definitions are used today.

Arrhenius Acid-Base Definition:

Acid: Acid turns blue litmus paper red; is active with some metals (active metals) exposing

hydrogen gas; has sour taste and forms hydrogen ions in the aqueous solution.

HCl H+ Cl-+

H2SO4 2H+ SO4-2

+

HNO3 H+ NO3-

+

H2O

H2O

H2O

Base: Base turns red litmus paper blue and neutralizes acid. It forms OH- ions in aqueous

solution;. Bases form water and carbonated compounds reacting with CO2. They don’t react

with many of metal (except amphoteric metals).

NaOH Na+ OH-+

Ca(OH)2 Ca+2 2OH-+

H2O

H2O

Al(OH)3 Al+3 3OH-+

H2O

Lowry-Bronsted Acid-Base Definition:

30

Acids are substances which give proton, bases are substances which take proton. According to

scientists conjugate base is formed when acid gives proton and similarly conjugate acid is

formed when base takes proton.

Asit BazProton

Baz AsitProton

+

+

NH3 H2O NH4+ OH-+ +

Baz (1) Asit (2) Baz (2)Asit (1)

Konjüge Asit Konjüge Baz

Lewis Acid-Base Definition:

Acid is a substance which takes an electron or accepts electron pair. Base is a substance which

gives electron or carries an electron pair.

B: A BA+

A = Lewis Asiti

B: = Lewis Bazi

N:

H

H

H

+ B

F

F

F

N

H

H

H

B

F

F

F

Lewis Bazi Lewis Asiti

If acid gives only one proton, it is named as monoprotic, mono functional, mono basic, mono

equivalent. According to ionization capability in their aqueous solutions, acids and bases are

classfied as ;

a. Strong

b. Mild

c. Very mild

31

a) Strong Acids and Bases: Acids or bases which are 100% ionizable in aqueous solution is

called "strong acid or base". It is accepted that H3O+ ions completely originated from acid in

strong acid solutions and OH- ions completely originated from bases in strong base solutions.

HCl H2O H3O+ Cl-+ +

NaOH H2ONa+ OH-+

b) Weak Acids and Bases (1): Weak acids or bases cannot ionized 100% in their aqueous

solutions. Ionization of acids and bases is a chemical equilibrium reaction either with acidity

constant, Ka or basicity constant, Kb.

CH3COOH H2O H3O+ CH3COO-+ +

NH3 NH4+ OH-+H2O+

Ka

Kb

Ka = 1,8. 10-5

Kb = 1,8. 10-5

c) Weak Acids and Bases (2): Weak acids or bases can partially ionized in their aqueous

solution. Their Ka and Kb constants are very low whereas pKa and pKb are very high.

H3C C OH

H

H

Ka = 1,3. 10-16 H2N pKb = 9,13

Volumetric Analysis:

Analysis of the synthesized or natural samples is one of the fundamental subjects of the

chemistry. In chemical analysis, identification of the analytes in a sample called ‘’qualitative

analysis’’ and determination the amounts of the analytes in a sample called ‘’qualitative

analysis’’. There are two basic methods for quantitative analysis of the chemical samples in

which no instruments are used. These methods are “gravimetric analysis” and “volumetric

analysis”. In gravimetric analysis, amount of a known (identified) analyte is determined by

measurement of mass. A water insoluble compound of the analyte is prepared and mass of

the compound is used to calculate the amount of this analyte. Volumetric analysis is based on

the measurement of the volume of a solution prepared in exact concentration of a substance

which gives a complete reaction with the analyte. By using the measured volume, the

concentration or mass of the substance can be calculated.

32

Figure 4.1. Titration setup

Titration, also known as titrimetry, is a common laboratory method of quantitative chemical

analysis that is used to determine the unknown concentration of an identified analyte.

Since volume measurements play a key role in titration, it is also known as volumetric

analysis.

In titrations, known concentration and volume of titrant reacts with a solution of analyte to

determine the concentration . Standard solutions are used as titrant to determine the unknown

concentrations of substances. Standard solution is a solution containing a precisely

known concentration of an element or a substance. Chemical substances generally contain

impurities, or therefore solution concentration can not be known precisely. Titrant solutions

are standardized by titration of primary standard (pure, high molecular weight,..) substances.

Volume of titrant used to reach the equivalent point is called titration volume.

The equivalence point, or stoichiometric point, of a chemical reaction is the point at which

chemically equivalent quantities of acid and base have been mixed. In other words, the moles

of acid are equivalent to the moles of base. It can be found by means of an indicator, most

often phenolphthalein. In a reaction, the equivalence of the reactants as well as products is

conserved. The endpoint refers to the point at which the indicator changes colour in a

colourimetric titration. Indicators are substances which show whether the substance tested is

33

basic or acidic. pH indicator is a chemical detector for protons in acid-base titrations.

The reagent of known concentration and volume used in titrations.

An acid–base reaction is a chemical reaction that occurs between an acid and a base. It is

shown as;

H+(aq) OH-(aq) H2O(s)+ Ksu= 10-14

Reaction is balanced. If the reacting acid and base are both too strong, pH would be 7 at the

time that the titration was completed.

· Equivalence factor: The number of hydrogen and hydroxide ions that can be substituted for

the acids and bases (e).

· Equivalent weight: The ratio of the molecular weight to equivalence factor (E).

· Amount of equivalent gram: The ratio of the grams of material to equivalent weight (EA).

· Normality: Normality is a measure of concentration equal to the gram equivalent weight per

liter of solution (N).

Volume of solution: liter (V).

Amount of dissolved matter: gram (m).

Molecular weight: g/mol (MA)

These terms using;

According to principle of equivalence; Na . Va = Nb . Vb

Students will find the amount of acid to neutralize 25 mL of 1 M sodium hydroxide. Acid and

base will react to form sodium chloride which can be crystallized further.

Neutralization reaction ;

HCl + NaOH NaCl + H2O

34

4.3 Materials and Chemicals

Materials Chemicals

Erlenmeyer flask (250 mL)

Burette fastener

Burette (50 mL or 100 mL)

Volumetric flask (25 mL or larger)

HCl solution

1 M NaOH solution

Methyl orange (indicator)

4.4. Experimental Procedure

Figure 4.2. Experimental set up for pH titration

In this experiment, sodium hydroxide forms sodium chloride-soluble salt with

hydrochloric acid neutralization. This solution is concentrated and crystallized in the

crystallization vessel.

1. Prepare 25 ml 1 M NaOH solution.

35

2. Mix pour 25 ml sodium hydroxide solution into an erlenmeyer flask and add two

drops of methyl orange.

3. Full the burette with unknown molarity HCl solution end of the 0 point.

4. Add in a small volume of hydrochloric acid solution onto sodium hydroxide solution

and rinse after each addition. Shake after any addition. Continue the addition of acid

until a red solution appears. Please read through the burette volume of added acid.

5. Please read through the burette volume of added acid.

6. Repeat titration (4. step) two more times. And use the average value of these 3

titrations for your calculations.

Cautions:

Wear protective gloves, gowns, glasses and a mask.

Avoid breathing vapors.

4.5. Results and discussion

HCl quantitaties which is needed neutralization.

1.Titration 2. Titration 3. Titration

Used NaOH volume (ml)

Used HCl volume (ml)

Normality of the HCl solution

Normality of the NaOH solution

Milliequivalent gram of used

HCl

Milliequivalent gram of used

NaOH

4.6. Questions

1. Explain acids and bases definitions of Arrhenius, Bronsted-Lowry and Lewis.

2. Explain definitions of Titration, The equivalence point, Indicator, titrant briefly?

3. A 25 ml solution of 0.5 M NaOH is titrated until neutralized into a 50 ml sample of HCl.

What was the concentration of the HCl?

36

EXPERIMENT 5: DETERMINATION OF A SALT SOLUBILITY

5.1. Purpose

Learning the general principles of solubility of salts.

5.2.Theory

Solubility of a salt, generally in a certain amount of a solvent (typically 100 g) and at a given

temperature, is calculated by taking into account how much salt is dissolved in a solution.

Temperature is the main factor which influences the solubility. To increase or to reduce the

temperature of a saturated solution changes the equilibrium and according to Le Chatelier's

Principle, the direction of the equilibrium shifts to the direction of reducing changes.

Increasing of temperature in the endothermic reaction, the equilibrium shifts to the

endothermic direction; unlike decreasing of temperature shifts the equilibrium to the

exothermic direction. Increasing the temperature increases the solubility of the most ionic

compounds. Dissolution of more substances in a saturated solution is an endothermic process

which requires extra energy (heat) and high temperature. It is more suitable to show changes

in solubility versus temperature in a graph.

5.3 Materials and chemicals

Materials Chemicals

Mortar and pestle

Test tube (20 cm)

Two-hole stopper

Thermometer

Copper wire

Beaker (100 mL)

Pasteur pipette

NH4Cl or another salt

Distilled water

37

5.4. Experimental Procedure

Figure 5.1. Experimental set up for solubility of a salt

1. Choose one of the salts.

2. Prepare about 8 g of salt fine powder in a mortar.

3. Take a large test tube (20 cm) covered with an appropriate two-hole stopper.

4. Place a 110°C thermometer in one of the hole. Be careful that the thermometer must

not touch the bottom of the tube.

5. Dip a copper-ring wire in the tube by using the another hole of stopper. Using the tip

that the ring of copper wire should be able to move easily up and down manually.

6. Add 5 gr of the salt which crushed into fine powder and weight it again on analytical

balance.

7. Add 3 ml of distilled water carefully by using a pipette and close the stopper on the

tube.

8. Dip the tube into the boiling water in a 400ml beaker. Solution inside the tube should

remain in the water bath, with a chelator attached spores and mixed by the ring bar

until all the salt dissolved in the tube. Be careful not to break the thermometer in this

process.

38

9. Add 1 ml of water, if the salt in the tube in the boiling water does not dissolve

completely in a few minutes.

10. If necessary add 1 ml of distilled water again and continue until fully dissolved. Do

not add too much water because it is very close to the desired degree of saturation of

the solution. Do not hold the tube in boiling water more than necessary. Thus

evaporative water loss is prevented in this way.

11. Remove the tube from hot water after all of the salt fully resolved.

12. Observe carefully the crystallization during cooling of the tube and determine

temperature when the crystallization starts. This temperature is the saturation

temperature of the solution.

13. Repeat this process several times to obtain an accurate saturation temperature in order

to determine heating, cooling and crystallization temperature.

14. Add 1 ml of distilled water in the tube.

15. Keep the tube in the boiling water until all crystals dissolve again.

16. Remove the tube from hot water when the dissolution is completed.

17. Cool it down by applying continuous stirring.

18. Detect the temperature which crystals start to form. In this case repeat the heating and

cooling process several times to obtain an accurate saturation temperature. Repeat this

process (addition of 1 ml of pure) until obtaining at least 5 or 6 saturation temperature

in a wide temperature range. So a different saturation temperature is obtained by

adding 1 ml of distilled water. If addition of 1 mL of distilled water cause to a big

change in saturation temperature, other experiments can be repeated by adding 0.5 ml

of distilled water instead of 1 mL of distilled water. Conversely, if the addition of 1 ml

of distilled water cause to a little change in saturation temperature, other experiments

can be repeated by adding 2 or 3 ml of distilled water instead of 1 mL of distilled

water.

Cautions:

Wear protective gloves, gowns, glasses and a mask.

Avoid breathing dust / smoke / gas / mist / vapor / spray.

Keep away the explosives from the open flame and avoid releasing to the environment.

.

39

5.5. Results and discussion

Data for used salt:

Experiment 1 Experiment 2 Experiment 3

Amount of the salt (g)

Weight of test tube (g)

Weight of test tube and salt (g)

Weight of distilled water used to solve

the salt (mL)

Total weight of used distilled water (mL)

Crystallization temperature (°C)

5.6. Questions

1. Please predict the solubility of your salt at some different temperatures.

2.

Temperature (oC) / x-axis Solubility (g Pb(NO3)2/100 g water) / y-axis

92.1 115

70.1 91.2

49.6 78.6

31.2 61.4

Draw a line between the points and beyond in both directions past the data points. Use this

line to answer the following questions.

a. What would the solubility be at 35 oC of lead nitrate?

b. What would the solubility be at 25 oC of lead nitrate?

c. What temperature would be necessary to make a solution with solubility of

65.0 g/100 mL?

40

EXPERIMENT 6: DETERMINATION QUANTITY OF Fe2+

WITH

KMNO4

6.1. Purpose

Learning the general principles of induction reduction reactions

6.2.Theory

Oxidation is the loss of electrons or an increase in oxidation state by a molecule, atom, or ion.

Reduction is the gain of electrons or a decrease in oxidation state by a molecule, atom, or ion.

The total increase in the oxidation number must be equal to the total decrease in the oxidation

number.

S0

+ O20

S4+

O22-

Oxidized Reduced

(reducer) (oxidizer)

When the oxidation number of sulfur atom was increased from 0 to +4, S was oxidized. When

the oxidation number of oxygen atom was decreased from 0 to -2, O was reduced.

Reduced agents are responsible for oxidation. Therefore this substance is called oxidizer. The

two processes are interdependent so the oxidized agent is reducing.

Two methods are commonly used for balancing of oxidation-reduction reactions.

1. The oxidation number method

2. Ion-electron method

Redox titrations: Because of the large number of elements can be easily oxidized or reduced

compared to other types of reactions, there are much more titration methods based on redox

reactions.

It is the most beautiful example the abundance and variety of material that can be assigned

directly or indirectly with strong oxidation power and at the same time indicator feature

showing the standard KMnO4 solution. Redox titrations similar to acid-base titration.

Oxidizing and reducing agents react in equivalent amounts with each other.

In equvalent value meg oxidizing substance=meg reducing substance (meg=miliequvalent

gram).

Knowing received or given electron numbers in redox reaction is required to calculate the

equivalent weight of a compound used in a redox titration

Example: KMnO4- Mn

2+ semi reaction

41

1 mol MnO4- takes 5 mol e

-. Mn atom in permanganate ion has 7+ oxidation number and is

reduced to Mn2+

ion by taking 5 e-. 1 mol KMnO4 is 158 g and combines with 5 mol e

-. 1 mol

e- is 31.6 g (1/5 mol). Generally, the equivalent weight of a compound used in a redox

titration is found by the following equation.

Equvalent weight =𝐹𝑜𝑟𝑚𝑢𝑙𝑒 𝑤𝑒𝑖𝑔ℎ𝑡

𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠 𝑝𝑒𝑟 𝑓𝑜𝑟𝑚𝑢𝑙𝑎

6.3 Materials and Chemicals

Materials Chemicals

Burette

Flask

Beaker

Sulfuric acid

KMnO4

FeSO4

Pure water

6.4. Experimental Procedure

42

Figure 6.1. Experimental setup

Potassium permanganate is oxidizer and this effect changes depending on the acidic or

alkaline environment. Usually KMnO4 titrations are performed in the acidic environment and

in this case action valence of KMnO4 is 5. According to this, 1 liter 1/10 N KMnO4 solution

has 1/50 mol/mol (3,16 g) KMnO4.

1. Solve 3.16 g KMnO4 in 1 L boiling water to prepare 0.1 N KMnO4 solution and boil 5

minute.

Close the top and leave to rest for a night. Filter with glass fiber and keep in the bottle.

2. Take 250 ml clean, dry flask. Receive certain volume of fresh prepared Fe2SO4 solution.

3. Add distile water to be total volume 30 ml to solution. Add 10 ml 6N H2SO4 solution on.

4. Titrate with 0.1 N KMnO4 solution until the pink color seen.

5. Repeat the same process 2. and 3. times , if you have enough time.

6. Note your results in the table.

Used reactions in this experiment:

2KMnO4 + 10 FeSO4 + 8H2SO4 2MnSO4 + 5Fe2(SO4)3 + 8H2O + K2SO4

MnO4- + 5Fe

2+ + 8H

+ Mn

2+ + 5Fe

3+ + 8H2O

Semi reactions;

Mn7+

O4- Mn

2+

Fe2+

Fe3+

In this reaction Fe2+

ion is oxidized to Fe3+

by losing an electron.

Equvalent weight of ferrous will be 55.85/molx1/1=55.85 equvalent gram.

Cautions:

Wear protective gloves, lab coat, glasses and a mask.

Avoid breathing vapors.

Keep away from flammable materials and prevent spreading into the environment.

43

6.5. Results and discussion

Determination Preucation of Quantity of Fe2+

:

1. Experiment 2.Experiment 3.Experiment

Used FeSO4 Volume (ml)

Used MnO4 Volume (ml)

Normality of KMnO4 solution

Normality of FeSO4 solution

Used miliequvalent gram of KMnO4

Miliequvalent gram weight of Fe(II)

Miligram quantities of Fe(II) in the

sample

Equvalent weight of Fe(II) (original

value)

6.6. Questions

1. NH4+

, SO42-

, H2O2, N2O, S2O32-

, CN-, H3PO4, Cr2O7

2-, CrO4

2- Determine the oxidation state

of ions groups and each atoms in the compounds.

2. Balance this reaction which accur in acidic solutions VO43-

+ Fe2+

VO2+

+ Fe3+

3. 5 SO32-

+ 2 MnO4- + 6 H

+ 5 SO4

2- + 2 Mn

2+ + 3 H2O

What are the reduced and oxidized species in this redox reaction?

4. How to decide whether a chemical reaction is redox rection or not?

5. Balance the following redox reaction in basic solution.

a. MnO2(k) + ClO3- MnO4

- + Cl

-

b. Fe(OH)3(k) + OCl- FeO4

2- + Cl

-

c. ClO2 ClO3- + Cl

-

44

EXPERIMENT 7: BUFFERS, BUFFER CAPACITY, AND BUFFERING

ZONE

7.1. Purpose

To learn how to prepare a buffer solution

To determine buffering capacity of a buffer solution

To determine buffering zone of a buffer solution

7.2.Theory

Buffers are aqueous systems that resist changes in pH when small amounts of acid or base are

added. Buffer solutions are composed of a weak acid (the proton donor) and its conjugate

base (the proton acceptor). Buffering results from reaction equilibrium in a solution between

a proton donor (HA) and its conjugate proton acceptor (A-).

The pH of any given buffer can be related to the pKa of the weak acid involved and the

logarithm of the molar ratio of conjugate base to acid by means of the Henderson-

Hasselbalch equation:

Stated more generally,

Whenever H+ or OH

- is added to a buffer, the result is a small change in the ratio of the

relative concentrations of the weak acid and its anion and thus a small change in pH. The

decrease in concentration of one component of the system is balanced exactly by an increase

in the other. The sum of the buffer components does not change, only their ratio.

HA A-

OH- H2O

H+

Buffer capacity generally depends on the concentration of buffer solution. Buffers with

higher concentrations offer higher buffering capacity. On the other hand, pH is dependent not

45

on the absolute concentrations of buffer components, but on their ratio. Buffer capacity is a

term used to describe the ability of a given buffer to resist changes in pH on addition of

acid or base. A buffer capacity of 1 is when 1 mol of acid or alkali is added to 1 liter of buffer

and pH changes by 1 unit.

Buffering Zone: Each conjugate acid-base pair has a characteristic pH zone in which it is an

effective buffer. Since buffering is due to the equilibrium between HA and A-, this

equilibrium is mostly established at pH values equal to pKa ± 1.

Temperature coefficient (∆pKa/C) values: The pKa of a buffer, and hence the pH, changes

slightly with temperature. Although the mathematical relationship of activity and temperature

may be complicated, the actual change of pKa with temperature (∆pKa/C) is approximately

linear. Table 1 presents the pKa and ∆pKa/°C for several selected buffers commonly used in

biological experimentation.

Table 7.1. Some buffers used in biological assays.

Most biochemical experiments are influenced by pH, and it is therefore usually necessary to

control the pH of any biochemical experimental system. A common way of accomplishing

this is to perform experiments in buffered systems. Thus biochemical experimentation usually

calls for buffer preparation. The purpose of this experiment is to illustrate buffer preparation

by having you prepare an acetic acid-sodium acetate buffer of a desired pH.

To prepare a buffer of a desired pH, it is necessary to prepare a solution containing the

appropriate amounts of a weak acid and its conjugate base.

46

One way of preparing buffer is to mix the requisite amounts of the appropriate weak acid and

its conjugate base. Another involves starting with the weak acid and generating its conjugate

base by adding a strong base, or conversely starting with the conjugate base and generating

the weak acid by adding a strong acid. If a pH meter is available, the appropriate mixture can

be obtained by “trial and error” mixing of the solutions available for preparation of the buffer.

However, if one does not have a pH meter or does not wish to use this approach, the

appropriate amounts of the substances to be used can be found in biochemical handbooks or

calculated from the Henderson-Hasselbalch equation. The latter approach will be used in this

experiment.

7.3 Materials and Chemicals

Materials Chemicals

pH Meter

buret,

buret clamp, magnetic stirrer stirring bar

two ring stands utiliy clamp

250 mL beakers

100 mL volumetric flasks

glass pipets

1 M acetic acid solution

1 M sodium acetate solution

1 M sodium hydroxide, NaOH, solution

distilled water

7.4. Experimental Procedure

A titration apparatus is built from the available stand materials together with a magnetic

stirrer, beaker and burette (or pipette) (see Fig.1).

47

Figure 7.1. Experimental set-up for pH titration.

Prepare a 0.1 M acetic acid-sodium acetate buffer having a pH of 5.0. Use Henderson-

Hasselbalch equation to calculate the amounts of acetic acid and acetate ion contained in

100 mL of this buffer.

1) Prepare a 100 mL sample of this buffer by mixing the appropriate amounts of the

1 M acetic acid and 1 M sodium acetate solutions that will be provided.

2) Prepare a second 100 mL sample of the desired buffer by mixing the appropriate

amounts of the acetic acid and 1 M sodium hydroxide solutions that will be

provided. Note: in this case, be sure to start with the right amount of acetic acid. After

preparing your two samples buffer, have your laboratory instructor check the pH

of each.

Keep the buffers you just prepared!

Buffer capacity: The purpose of a buffer is to minimize changes in pH when either acids or

bases are added to a solution. The ability of a buffer to resist pH changes depends both on the

absolute concentration of the buffer components and the ratio of these to each other. At a

given pH, a concentrated buffer will have greater buffering capacity than a diluted buffer. At a

given concentration, the buffering capacity of a buffer will be greatest when the pH of the

buffer equals pKa of the acid used to prepare the buffer.

1) After measuring the pH of your two buffer samples, combine them to give

approximately 200 mL of your acetic acid-sodium acetate buffer.

2) Transfer 20 mL of this combined buffer to a clean 100 mL volumetric flask and make

up to volume by adding water. Mix well. You now have two buffer solutions that

differ five-fold in their concentrations.

3) Place a 25 mL sample of the undiluted buffer in a clean beaker. Measure the pH. Add

1 mL of 0.1 N NaOH. Mix well and measure the pH. Continue adding the NaOH in 1

mL increments and measuring the pH until the pH is approximately 1 pH unit above the

pKa for acetic acid. At this point start adding the NaOH in 0.5 mL increments and

48

measuring the pH until the pH is approximately 2 pH units above the pKa for acetic acid

(pKa for acetic acid = 4.75).

4) Place a 25 mL sample of the diluted buffer in a clean beaker. Measure the pH. Is it the

same as the undiluted buffer? Add 0.5 mL of 0.1 N NaOH. Mix well and measure the

pH. Continue adding the NaOH in 0.5 mL increments and measuring the pH until the

pH is approximately 2 pH units above the pKa for acetic acid.

5) Place a 25 mL sample of the undiluted buffer in a clean beaker. Measure the pH. Add

1 mL of 0.1 N HCl. Mix well and measure the pH. Continue adding the HCl in 1 mL

increments and measuring the pH until the pH is approximately 1 pH unit below the pKa

for acetic acid. At this point, start adding the HCl in 0.5 mL increments and measuring

the pH until the pH is approximately 2 pH units below the pKa for acetic acid.

6) Finally, place a sample of the diluted buffer in a clean beaker. Measure the pH. Add

0.5 mL of 0.1 N HCl. Mix well and measure the pH. Continue adding the HCl in 0.5 mL

increments and measuring the pH until the pH is approximately 2 pH units below the

pKa for acetic acid.

Cautions:

Acids and bases have a corrosive effect. Only small quantities and low concentrations are

used In this experiment, therefore only a slight hazard exists.

Wear protective glasses and a laboratory coat.

7.5. Results and discussion

Data:

Table 1. pH values of the buffers you prepared

Buffer pH

1

2

Table 2. Titration of undiluted buffer

Volume of NaOH (0.1 N), mL pH

Table 3. Titration of diluted buffer

Volume of NaOH (0.1 N), mL pH

49

Table 4. Titration of undiluted buffer

Volume of HCl (0.1 N), mL pH

Table 5. Titration of diluted buffer

Volume of HCl (0.1 N), mL pH

ADDITIONAL KNOWLEDGE

Molarity and Ionic Strength of a buffer:

In some of the biochemical literature, buffers are identified in terms of molarities, in others

they are identified in terms ionic strengths. Accordingly, one needs to be familiar with these

concepts. More important, however, one needs to be aware of the importance of the concept

of ionic strength.

The molarity of a buffer is defined as the sum of the molarities of the Brönsted acid and

conjugate contained in the buffer. For example, in an acetic acid-sodium acetate buffer

containing 0.045 moles of acetic acid and 0.055 moles of sodium acetate per liter the molarity

would be 0.100.

The concept of ionic strength differs from molarity in two ways. First, it considers only the

ionic species present in solution. Secondly, it considers both the concentrations and the

charges of these ionic species. Mathematically the ionic strength (µ) of a solution is defined as

follows:

𝜇 =1

2 ∑𝑐𝑖𝑍𝑖

2

Where ci is the concentration of ion “i” and Zi is the charge of this ion. While µ is used in this

experiment and much of the biochemical literature to denote ionic strength, frequently one

will encounter the use of I for this quantity in the biochemical literature.

50

SAMPLE CALCULATION

The ionic strength of 0.1 M NaCl can be calculated as follows:

𝜇 =1

2([𝑁𝑎+](1)2 + [𝐶𝑙−](−1)2)

𝜇 =1

2[(0.1)(1) + (0.1)(1)]

μ =1

2(0.1 + 0.1) = 0.1

In this case the ionic strength is numerically equal to the concentration of the solution. This

will be true for all solutions of electrolytes composed of two univalent ions.

The ionic strength of 0.1 M (NH4)2SO4 can be calculated as follows:

𝜇 =1

2([𝑁𝐻4](1)2 + [𝑆𝑂4](−2)2)

𝜇 =1

2[(0.2)(1) + (0.1)(4)]

μ =1

2(0.2 + 0.4) = 0.3

In this case the ionic strength is numerically larger than the concentration of the solution. This

will be true for all solutions of electrolytes containing multivalent ions.

The concept of ionic strength is important in systems that involve charged substances since

the attractive and repulsive forces between these substances will be decreased as the ionic

strength of the solution increases. For example, the ionization constant for acetic acid

increases as the ionic strength of the solution increases. Of more obvious importance in

biochemistry, are the interactions between charged solutes and the charged ion exchange

matrices that are used in ion exchange chromatography, the intramolecular reactions between

charged groups in proteins, and the intermolecular reactions proteins and charged substances.

All of these interactions will be affected by changes in ionic strength.

SAMPLE CALCULATION

Part I: Combining 1 M acetic acid and 1 M sodium acetate and water to obtain a buffer at a

specified pH

We will make use of two equations. This first is the following rearrangement of the

Henderson-Hasselbach equation:

𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝐼: 10(𝑝𝐻−𝑝𝐾𝑎) =[𝐴−]

[𝐻𝐴]=

(𝐴−)

(𝐻𝐴)

51

Equation I gives us the required ratio of concentrations (or, equivalently, of amounts) of

conjugate acid and conjugate base (in terms of the pKa of the acid involved and the pH

desired). The second equation we need is the following:

𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝐼𝐼: [𝐻𝐴] + [𝐴−] = 0.1 𝑀

Equation II simply states that the combined concentrations of conjugate acid (acetic acid) and

conjugate base (acetate) must be 0.1 M (this is specified as part of the laboratory write-up).

Notice that we have two equations in two unknowns, [HA] and [A-]. Now we simply solve for

each unknown:

From Equation I

[𝐴−] = [𝐻𝐴] × 10(𝑝𝐻−𝑝𝐾𝑎)

Substituting for [A-] in Equation II,

[𝐻𝐴] + [𝐻𝐴](10(𝑝𝐻−𝑝𝐾𝑎)) = [𝐻𝐴] × (1 + 10(𝑝𝐻−𝑝𝐾𝑎)) = 0.1𝑀

Rearranging, to solve for [HA],

[𝐻𝐴] =0.1𝑀

1 + 10(𝑝𝐻−𝑝𝐾𝑎)

And, of course, we have:

[𝐴−] = 0.1 𝑀 − [𝐻𝐴]

Now, knowing (or being told) that the pKa of acetic acid is 4.76 and being asked to prepare a

buffer of a particular pH (as an example, pH 4.06), we can easily figure what volumes of 1 M

acetic acid and 1 M acetate we must use.

[𝐻𝐴] =0.1𝑀

1 + 10(−0.7)=

0.1𝑀

1 + 0.20=

0.1𝑀

1.20= 0.083𝑀

[𝐴−] = 0.1 𝑀 − 0.083 𝑀 = 0.017 𝑀

These are concentrations in the final solution of 100 mL. It is simple, however, to calculate

what volumes of the 1 M stock solutions must be used to give these final concentrations (with

formulation N1.V1=N2.V2).

Part II. Making a buffer at the same pH using 1 M acetic acid, 1 M NaOH, and water

Once we realize that all of the acetate in this buffer solution will come from acetic acid itself

(it is generated by adding NaOH to acetic acid), the calculations are essentially the same as

performed in Part I. After NaOH and acetic acid and water have been combined in yet-to-be-

determined proportions, we must satisfy the same two equations:

𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝐼: 10(𝑝𝐻−𝑝𝐾𝑎) =[𝐴−]

[𝐻𝐴]=

(𝐴−)

(𝐻𝐴)

and

52

𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝐼𝐼: [𝐻𝐴] + [𝐴−] = 0.1 𝑀

Thus, in the final solution, the ratio of base to acid must be given by Equation I and the

concentration of acetic acid and acetate must sum to 0.1 M (as shown in Equation II). Since

the acetate will come from whatever volume of acetic acid we chose to use, we must select 10

mL of 1 M acetic acid. (10 mL diluted to 100 mL will give a concentration of 0.1 M).

So, we measure out 10 mL of 1 M acetic acid. Now the only remaining issue is to add an

appropriate volume of 1 M NaOH to generate enough acetate so that the ratio given by

Equation I is satisfied. But this is just the same ratio as in Part I, above. Hence, we add the

same volume of 1 M NaOH as the volume of 1 M acetate used in Part I. Adding this volume

of 1 M NaOH will generate just the right amount of actate and leave just enough acetic acid

behind to give the ratio needed according to equation I.

7.6. Questions

1. Make a plot of pH (ordinate) vs. millimoles of base or acid added to the buffer (abscissa).

Draw a smooth curve (best fit) through the points associated with each of your buffers

(undiluted and diluted).

2. Which buffer is the better buffer? At what pH did your buffers show the greatest

resistance to changes in pH?

3. What is the molarity of the buffer you prepared in the first part of this experiment? What is

its ionic strength?

4. Outline how you could have prepared an acetic acid-sodium acetate buffer of the same pH

but whose ionic strength is 0.1.

5. How many mL of 1 M KH2PO4 and 1 M K2HPO4 would you need to prepare a liter of 0.1

M phosphate buffer of pH 7? The pKa’s for phosphoric acid are 2.1, 7.2, and 11.8.

6. Show by balanced equations how each of the following buffers act when additions of HCl

or NaOH are made: An acetic acid-sodium acetate buffer, a Tris hydrochloride-Tris buffer, a

phosphate buffer of pH 7?

53

EXPERIMENT 8: POLYPROTIC ACIDS: DETERMINATION OF pKA

VALUES USING pH TITRATION CURVES

8.1. Purpose

To perform a titration with an acid.

To determine the strength of an acid.

To calculate the pKa value of an acid

8.2.Theory

pH Titration Curves 'Idealized':To date the equivalence point of an acid base reaction has

been determined using an indicator. In this experiment we are going to monitor the changes in

pH that occurs during the titration of a weak polyprotic acid with a strong base. At the

equivalence point one should expect to see a dramatic change in pH as the solution goes from

acidic to strongly basic. Depicted on Figure 8.1. is an idealized pH titration curve for a weak

diprotic acid. The first thing that you should notice is that there are two regions where we see

a significant pH change. These, if you wish, correspond to two separate titrations. Titration 1

is the reaction of the first proton with the base (in this case sodium hydroxide).

H2X(aq) + NaOH(aq) = NaHX(aq) + H2O (l)

The second titration corresponding to the reaction of the second proton with sodium hydroxide

NaHX(aq) + NaOH(aq) = Na2X(aq) + H2O(l)

54

Figure 8.1. Titration curve of weak polyprotic acid

So, in essence, titrations of a weak polyprotic acid with a strong mono protic base are a

combination of a number of titrations depending on the number of acidic protons on the

polyprotic acid. The overall reaction is the sum of the two titration's

H2X(aq) + 2 NaOH = Na2X(aq) + 2 H2O(l)

In determining the quantity of the acid or the molarity of the acid, we are normally just

interested in the final equivalence point. In a pH titration plot, this is determined by finding

the point of inflection on the final area where we see a significant rise in pH (This can be

approximated by determining the midpoint.) However, this plot contains some other

interesting features. First off, if we look at the area corresponding to the first titration, it

should come as no surprise that its equivalence point corresponds to the addition of exactly

1/2 the volume of NaOH required to reach the final equivalence point. The real neat point

comes at the 1/2 way point of each titration. Let us focus on the Titration 1. At the 1/2 way

point, the concentration of H2X(aq) remaining in the solution is equal to 1/2 the initial

concentration of H2X! The concentration of NaHX(aq) produced is also numerically equal to

1/2 the initial concentration of H2X! Let's focus for a moment on the acid equilibrium

associated with the acid that we are dealing with in titration 1,

H2X(aq) + H2O(l) = HX- + H3O

+

Ka = [H3O+][HX

-]/[H2X]

or written in another way

[H3O+] = Ka{[H2X]/[HX

-]}

using the concentrations that we know for H2X and HX- (=NaHX) at the 1/2 way point we get

[H3O+] = Ka{1/2[H2X]initial/[1/2H2X]initial}

[H3O+] = Ka

55

From the graph we can determine the pH at this point and since pH=-log10[H3O+], we can

determine [H3O+] at this point and thus obtain the Ka for this equilibrium. Since this is a

polyprotic acid, this corresponds to Ka1. Guess what you can determine from the pH at the

midpoint of the second titration. This information can be used to help identify the acid in

question since Ka for a large number of polyprotic acids are known. The first acid that you

will be following today is citric acid which is an acid that falls into the idealized category.

You should see three areas where the pH undergoes significant changes and should be able to

determine the three Ka values for citric acid and compare the result to the three known values

given below.

H3C6H5O7(aq) + H2O(l) ↔ H2C6H7O7 - + H3O

+ Ka1 = 7.4x10

-3 at 25

oC

H2C6H5O7 - + H2O(l) ↔ HC6H6O7

2- + H3O

+ Ka2 = 1.7x10

-5 at 25

oC

HC6H5O7 2-

+ H2O(l) ↔ C6H5O7 3-

+ H3O+ Ka3 = 4.0x10

-7 at 25

oC

pH Titration Curves 'The Real World':

In reality, many polyprotic acids only show one discernible equivalence point! The vast

majority of the time, this corresponds to final equilibrium. If this is the case then, all the other

equivalence points can be determined by knowing what type of polyprotic acid one is dealing

with, i.e., diprotic or triprotic (Fig. 8.2). For a triprotic acid, the other two equivalence points

should correspond to 1/3 and 2/3 of the volume of the base required to reach the final one and

thus one can still determine the Ka values. The second acid that you will be looking at in this

lab is an amino acid, glycine, whose Ka values are given in Fig. 8.2.

56

Figure. 8.2. Conjugate acid-base pairs consist of a proton donor and a proton acceptor. Some compounds,

such as acetic acid and ammonium ion, are monoprotic; they can give up only one proton. Others are diprotic

(H2CO3 (carbonic acid) and glycine) or triprotic (H3PO4 (phosphoric acid)). The dissociation reactions for each

pair are shown where they occur along a pH gradient. The equilibrium or dissociation constant (Ka) and its

negative logarithm, the pKa, are shown for each reaction.

8.3 Materials and Chemicals

Materials Chemicals

buret,

buret clamp,

magnetic stirrer,

stirring bar

two ring stands,

utility clamp

pH Meter

250 mL beakers

0.02 M sodium hydroxide, NaOH, solution

0.02 M citric acid solution

0.02 M glycine solution

distilled water

8.4. Experimental Procedure

A titration apparatus is built from the available stand materials together with a magnetic

stirrer, beaker and burette (Fig.2).

57

Figure 8.3. Experimental set-up for the determination of the acid strength of citric acide.

Your TA will also demonstrate the best set up for this experiment.

1. Using a graduated cylinder, place ~ 20 mL of the ~0.02 M citric acid into a small beaker. If

necessary add distilled water such that the tip of the pH probe is covered.

2. Fill your buret with the ~0.02 M NaOH solution. Record the exact molarity of this

solution. Record the initial buret reading. Remember that this corresponds to 0.00 mL of

NaOH added.

3. Record the initial pH of the Citric acid.

4. Carefully add the NaOH recording the volume of NaOH required to effect a pH change of

0.2. Continue this process until the pH reaches 12.

5. Repeat steps one through four using the ~0.02 M glycine.

8.5. Results and discussion

Data:

Table 1. Your titration data: the volume of NaOH vs. pH

Volume of NaOH (0.02 M),

mL

pH (citric acid) pH (gylicine)

0.00 ---

---- ----

58

Note: There should be separate tables for citric acid and glycine titration data

8.6. Questions

1. Plot a graph of 'pH' versus 'Volume of NaOH" added and from this graph determine:

a) The Ka values for citric acid.

b) The exact concentration of the citric acid.

2. By making same things that is given above.

a) Determine the Ka values for glycine.

b) Find the exact concentration of the glycine.