’ students MEDICAL CHEMISTRY Handbook for the medicine and odontology faculties

(1)

Lithuanian University of Health Sciences

Department of Biochemistry

Laima Ivanovienė, Ramunė Morkūnienė, Julius Liobikas,

Dalia Marija Kopustinskienė, Svaja Dubickienė, Artūras Kašauskas

MEDICAL CHEMISTRY

Handbook for the medicine and odontology faculties’

students

The teaching book

Kaunas, 2014

(2)

Aprobavo Lietuvos sveikatos mokslų universiteto Medicinos akademijos

Medicinos fakulteto taryba ir LSMU leidybos komisija 2014-02-28. Protokolas Nr 3/14

Recenzentai:

prof.dr. Hiliaras Rodovičius doc.dr. Zita Naučienė

Autoriai:

© Artūras Kašauskas, 2014

(3)

EXAMPLE OF THE LABORATORY WORK REPORT

Name Family name

Group N Year Faculty

Laboratory work N ….. Title ……… ……… Aim……… ……… ……….. ……… Principle of the lab-work

……… ……… ……… ……… ………

Equipment and materials………. ……… ……… ……… ……… ……… ……… ………

(12)

Results and calculations

(13)

(14)

1. Preparation of solutions. Percent concentration

Background. Solution is a homogenous mixture containing two or more components. One of them, solvent, generally is present in the largest amount. All the remaining components are called solute and are equally distributed in the solvent. The physical state of solvent does not change. If solute is an ionic compound, it dissociates into ions when dissolved in the polar solvent (water).

The amount of solute per amount unit of solution (mass or volume unit) is called concentration, abbreviated as [C]:

[C] = amount of solute/amount unit of solution

Depending on the units used to express the amount of solute and solvent there are several ways to indicate concentration. Thus, there is percentage concentration, molar concentration, molar concentration of equivalent and a few others.

A widely used type of concentration is the percentage concentration. The term percent literally means number of parts in the total of one hundred parts. Consequently, the percentage concentration means number of solute parts in one hundred parts of solution. There are several types of the percentage concentration:

mass / mass – indicates number of solute mass units per 100 solution mass units and is denoted in parenthesis as (mass/mass), (m/m), (w/w). For example, 5 % NaCl (mass/mass) means that there are 5 grams of NaCl in 100 g of the solution;

volume / volume – indicates number of solute volume units per 100 solution volume units and is denoted in parenthesis as (vol./vol.) or (v/v). 10 % (vol./vol.) ethanol means that there are 10 ml of pure ethanol in 100 ml of the solution;

mass / volume – indicates the number of solute grams per 100 ml of solution. 3 % NaCl (mass/vol.) means 3 grams of NaCl in 100 ml of the solution.

If the percentage concentration is shown without an indication in parenthesis, generally it means grams of solute per 100 g of solution. In clinical trials sometimes the obsolete milligram percentage concentration mg% is used, which indicates milligrams of solute in 100 grams of solution.

Experimental part:

Prepare 300 g of CaCl2 solution in water of the following concentration:

A 2% B 3% C 5% D 7% E 9% F 10%

(15)

1.Calculate the amount of salt and water needed.

Example: Prepare 300 g of 20 % CaCl2 solution in water.

20 % solution will contain 20 g CaCl2 in 100 g of solution; x g should be in 300 g of solution; then x = 20g  300g/100g = 60 g CaCl2

300g – 60g = 240g H2O Thus, 300 g of the solution will be made of 60 g of CaCl2, and the rest 240 g will be water. 2. Weigh CaCl2 on a piece of paper and transfer it to the flask.

3. Since the density of water is approximately1g/cm3= 1g/mL, the weight of water will be equal to its volume. Use a graduated cylinder to measure the needed amount of water.

4. Pour the water into the flask with the salt and dissolve the salt by mixing.

5. Transfer the obtained solution into the cylinder and measure its density (d) with the aerometer. Find the real % concentration of this solution from the density tables. Indicate how much it differs from the theoretical value. Explain the possible error.

General questions

1. What systems are called solutions? 2. What is a solvent and what is a solute?

3. Describe the structure of water molecule in short. 4. What is concentration of solution?

5. What types of concentration do you know?

6. What is a percentage concentration? What types of percentage concentration are there? 7. What does it mean: mass/mass, volume/volume, mass/volume concentration?

8. How to determine the density of solution and what density units are there? 9. How can aerometer be used to determine the density of solution?

(16)

Name Family name

Group N Year Faculty Laboratory work N 1 Title ……… ……… Aim……… ……… ……….. ……… Principle of the lab-work

……… ……… ……… ……… ………

(17)

(18)

2. Preparation of molar solutions. Molar concentration

Background. Molar concentration - the method of expressing concentration that indicates how many moles of solute are present per unit volume of solution. The molarity of solution is the number of moles of solute per liter of solution. The abbreviation for molarity is M.

Molarity = number of moles of solute/ number of liters of solution,

Or using abbreviations: M = mol / L

For example, if we have 4 moles of NaCl in 2 liters of solution, the molar concentration will be

4 mol / 2 L = 2 M (= mol/L). Number of moles n = m/M (m- mass of substance, M- molar mas of substance) To remind you, one mole of a compound is numerically equal to the sum of atomic masses of all elements making up that particular compound.

A concentration lower than 1M can be expressed using prefixes which mean: milli- = 10-3; 1 mmol = 1 10-3 mol; 1mM = 1 10-3 M (mol/L) micro- = 10-6; 1 mol = 1 10-6 mol; 1M = 1 10-6 M (mol/L) nano- = 10-9; 1 nmol = 1 10-9 mol; 1nM = 1 10-9 M (mol/L) pico- = 10-12; 1 pmol = 1 10-12 mol; 1 pM = 1 10-12 M (mol/L) Here are some examples of calculations using molar concentration:

Example 1.

What is the molarity of NaOH solution, if it contains 16.0 grams of NaOH in 2 liters of water solution? Molecular mass of NaOH = 23 + 16 + 1 = 40, thus, 1 mol of NaOH = 40 g

Number of moles of NaOH = grams of NaOH in solution / grams of 1 mol = 16 g / 40 g = 0.4 mol Molarity of this solution = number of moles of solute / volume of solution =

= 0.4 mol / 2 L = 0.2 mol/L= 0.2 M Example 2.

24.5 mL of 1.5 M NaOH is added to the 20.5 mL of 0.85 M NaOH. What is the molarity of the final solution?

It is equal to the amount of NaOH in moles in the initial portions of both solutions divided by the final volume of the mixture (in liters!)

moles of NaOH in the first solution:

in (1 liter) 1000 mL there is -- 1.5 mol NaOH in 24.5 mL -- x mol

x = 24.5 mL  1.5 mol / 1000 mL = 0.0368 mol moles of NaOH in the second solution:

in 1000 mL -- 0.85 mol in 20.5 mL -- x

x = 20.5 mL  0.85 mol / 1000 mL = 0.0174 mol

the total number of moles of NaOH in both solutions = 0.0368 + 0.0174 = 0.0542 mol the final volume of the solution in liters = 24.5 mL + 20.5 mL = 45 mL = 0.045 L the molarity of NaOH in the final solution:

there is 0.0542 mol NaOH in 0.045 L of solution x -- in 1 L

x = 0.0542 mol  1L / 0.045 L = 1.2 M

The following example is calculation used in everyday lab work. Solutions usually are stored as stock-solutions of relatively high concentration (they are more stable), and working solutions are made from them by an appropriate dilution with water or other solvent. Such dilution is based on the fact that the number of moles of solute does not change during dilution:

moles of solute = molar concentration  volume of solution; thus: Mbefore dilution  Vbefore dilution = Mafter dilution Vafter dilution

(19)

Example 3.

What volume of 1M KNO3 has to be diluted with water in order to get 250 mL of 0.2 M KNO3?

Put the numbers given into the equation above: 1 M  x = 0.2 M  0.250 L

then x = 0.2 M  0.250 L / 1 M = 0.05 L = 50 mL

Experimental: Prepare 250 ml of CaCl2 solution in water of following concentration:

A 0.3M B 0.4M C 0.5M D 0.6M E 0.7M F 0.8M G 0.9M

Find the percent concentration (w/w and w/v) of this solution.

Procedure:

1. Calculate the amount of the salt needed.

Example: Prepare 250 ml of 2M CaCl2 solution in water. Calculate its % (weight/volume) concentration. 2.0 M CaCl2 means that for 1 liter (1000 ml) you need 2.0 moles of CaCl2;

then in 250 ml there should be X moles of CaCl2

X = 250  2.0/ 1000 = 0.5 moles CaCl2 is needed for making 250 ml of 2.0M solution.

1 mol of CaCl2 amounts to 40 + 2×35.5 = 111 g;

0.5 moles is X g

X = 0.5  111/ 1 = 55.5 g

So, to make the required solution we need 55.5 g of CaCl2

2. Weigh a salt on a piece of paper, and then transfer it into the flask. 3. Use a graduated cylinder to measure approximately 150-200 ml of water.

4. Pour the water into the flask; dissolve all the salt by stirring with a magnetic stirrer.

5. Transfer the solution of salt back into the cylinder. Adjust the volume of the solution to 250 ml. Now you have the right amount of salt in the required volume of solution. However, the solution is not yet homogenous – you may notice disturbances and flows in the cylinder, indicating zones of various concentrations.

6. Transfer the solution to the flask and mix it. The solution is ready. 7. Calculate % (weight/volume) concentration of this solution:

Example: Solution has 55.5 g of salt in 250 ml X g in 100 ml X = 55.5  100/ 250 = 22.2 % (w/vol) concentration 8. Determine the % (weight/weight) concentration:

Transfer the solution into the graduated cylinder and measure its density with an aerometer. Find the real % (w/w) concentration to which the density of the obtained solution corresponds.

(20)

1. What is the mole?

2. How many elementary units does the mole contain?

3. What kind of concentration is called molar concentration (molarity)? 4. What is the dilution of solution?

(21)

Name Family name

Laboratory work N 2 Title ……… ……… Aim……… ……… ……….. ……… Principle of the lab-work

……… ……… ……… ……… ………

(22)

(23)

3. Volumetric analysis. Acid-base titration

Background. In chemical analysis, it is often necessary to determine the concentration of ions in solution. For this purpose, a technique called titration is often used. It is based on the measurement of the volume of one reactant required to react with a measured mass or volume of another reactant in the chemical reaction. For this, a standard solution of a titrant (a reactant of known concentration) is added drop by drop to the measured volume of the solution of unknown concentration until the reaction is stoichiometrically complete. The task of titration is to determine what volume exactly of the titrant is needed to complete the reaction. The reaction is completed when all the particles of the substance have reacted with the particles of the titrant – this is called reaction endpoint or equivalence point. To make this point visible we use indicators. An acid-base indicator is a weak acid or a weak base. The undissociated form of the indicator has a different colour from its ionic form.

HInd <---> H+ + Ind-

(colour (1) <---> different colour (2))

The colour change occurs over a range of hydrogen ion concentrations. This range is called colour change interval and is expressed as a pH range.

Examples of the most widely used indicators: Methylorange -- red (pH 3.2) <<>> yellow (pH 4.4)

Phenolphtalein -- colourless (pH 8.2) <<>> pink (pH 10.0)

Problem. Determine the amount of NaOH in solution of unknown concentration by titration with 0.1 M HCl.

Procedure

Reaction which will take place NaOH + HCl ----> NaCl + H2O

1 mol of NaOH is neutralized by 1 mol of HCl

1. Fill the burette with 0.1 M HCl (titrant). Burette is a long tube, graduated in mL and tenths of mL, at the bottom it has a stopper which allows dripping of the solution. Mark the initial volume of the solution in the burette.

2. Dilute the sample. Take a volumetric flask X with NaOH solution of unknown concentration (provided by the technician), and add distilled water up to the 100 mL mark. Close the flask and mix the contents thoroughly by turning over several times.

3. Pipette exactly 10 mL of diluted NaOH solution into a 100 ml Erlenmeyer flask; add 2 drops of phenolphthalein indicator. The solution colours pink.

4. Put a magnetic bar in the flask and place it on the magnetic stirrer. While opening the stopper of the burette with one hand, slowly add HCl acid into the reaction flask. If titrating by hand swirl flask contents with another hand continuously. Near the endpoint slow the rate of addition to drops; the last few drops should be added at 3 -5 second intervals.

5. Titrate until the colourless endpoint, which indicates that the neutralization reaction is complete. You need to catch the first moment of the colour change otherwise there would be too much titrant added. Record

the final volume of the burette. The difference between the initial and final volumes is the volume of HCl required to neutralize all NaOH in the investigated sample.

6. Repeat the titration once more. If the second measurement is close to the first, go to step 7, if not - titrate for the third time. Find the average of the closest two measuring.

(24)

7. Calculate the concentration of the given NaOH solution from the obtained titration data:

-- reaction is completed when number of HCl (acid) moles added is equal to the number of NaOH (base) moles in the flask:

molesNaOH = molesHCl

moles = Molarity  Volume

Molarity NaOH VolumeNaOH = Molarity HCl VolumeHCl

Molarity NaOH = Molarity HCl VolumeHCl / VolumeNaOH

8. Calculate the amount of NaOH in the flask X in grams.

mass NaOH = Molarity NaOH Total volume NaOH Molar mass NaOH

1. What is the method of volumetric analysis based on?

2. What is the method of volumetric analysis (titration) used for? 3. What solution is called “titrant”?

4. What process is called “titration”?

5. What is the equivalence point and what is its pHin the case of acid-base titration? 6. What substances are acid-base indicators? How to choose a suitable one?

7. What is the colour change interval of the indicator? 8. At what pH does Methylorange change its colour?

9. What colour does Phenolphtalein have when the pH is less than 8.2; higher than 10? 10. What is the “titration curve”?

(25)

Name Family name

……… ……… ……… ……… ………

(26)

(27)

4. Buffer solutions

Theory: Solutions of substances (usually in water) which can resist against changes of pH are called buffer solutions or buffer systems. The buffer solutions are made from a weak acid and its salt with a strong base (as the acetic acid/sodium acetate buffer) or a weak base and its salt with a strong acid (as ammonium hydroxide/ammonium chloride buffer).

Preparations of buffers solutions.

1. Selected weak acid is mixed with a salt containing common ion in the structure. This common ion

acts as a conjugated base to the selected acid. For example, acetic acid (CH3-COOH) and sodium

acetate (CH3-COONa) have acetate (CH3-COO-) as a common ion. When acetic acid solution is

mixed with sodium acetate solution, the obtained buffer is known as acetate buffer.

2. Salts of different acidities can also be used for preparation of buffers. For example, NaH2PO4 (acts

as an acid) and Na2HPO4 (acts as a base), which dissociation results in ions of different acidities:

NaH2PO4  Na+ + H2PO4-

Na2HPO4  2Na+ + HPO42- (base)

H2PO4- ion contains 2H+ ions and acts as a weak acid, HPO42- ion has a single H+ therefore it acts

as a base conjugated to H2PO4-.

pH of buffer solutions.

Let consider a case when buffer comes from an acid HA and a salt of the acid MeA. In water, the acid undergoes ionisation by this equation (for simplicity take dissociation):

HA  H+ +A-

Constant of acidity for such dissociation is described by an equation:

Concentration of protons (H+ ions) can be calculated from re-arranged equation:

If we add salt in the solution of such acid, dissociation of the acid will be depressed by increased amounts

of the ion A-. It means, that concentration of non-dissociated acid [HA] will be equal to concentration of

the acid added at the beginning. Therefore H+ concentration in such acidic buffer can be calculated by the

formula:

We can calculate pH of such a buffer solution using Henderson-Hasselbalch equation:

We can also prepare a buffer solution if we mix different volumes of solutions of an acid and a base. In this case Henderson-Hasselbalch equation is:

In Henderson-Hasselbalch equation, Vsalt is a volume of salt solution

Vacid is a volume of acid solution

[salt] is a concentration of salt solution taken for preparation of a buffer

[acid] is a concentration of acid solution taken for preparation of a buffer

Ka= [H +_]x[A-_] [HA] Ka= [H +_]x[A-_] [HA] H+_{= Ka x} [A-_] [HA] H+_{= Ka x} [A-_] [HA] H+_{= Ka x} [salt] [acid] H+_{= Ka x} [salt] [acid] pH= -lg[H+_{]= lg}[salt] [acid] - lgKa pH= -lg[H+_{]= lg}[salt] [acid] - lgKa pH= -lg[H+_{]= lg} [salt] x Vsalt [acid] x Vacid - lgKa pH= -lg[H+_{]= lg} [salt] x Vsalt [acid] x Vacid - lgKa

(28)

Mechanism of buffer action. H+ concentration and pH of buffer solution do not change (or change very little) when a strong acid or base is added to them. This phenomenon comes from interaction of buffer components with added acid or base. Suppose we add HCl (strong hydrochloric acid) into acetate buffer (composition was considered above):

CH3-COONa + HCl CH3-COOH + NaCl

CH3-COO- + H+  CH3-COOH

As CH3-COOH (acetic acid) undergoes very low dissociation, pH is little changed. A little change of pH

comes from alterations in the ratio between salt and acid amounts.

If a strong base is added to acetate buffer, it combines with acidic component of the buffer:

CH3-COOH + NaOH CH3-COONa + H2O

CH3-COOH + OH- CH3-COO-+ H2O

This example demonstrates increasing salt amount in the buffer solution.

Buffer capacity. A buffer solution can keep pH stable only up to a certain amount of acid or base added. After reaching this threshold pH of a buffer solution changes if extra acid/base is added, as it does in the case of regular solutions. Therefore, every buffer solution is characterized by the buffer capacity (B) which indicates how many moles of a strong acid or base should be added to 1 litre of the buffer in order to change its pH by 1 pH unit:

B = C / (pHafter addition of acid or base - pHinitial); where B - buffer capacity, C - acid or base concentration

mol/L

The buffer capacity depends on the nature and concentration of the buffer components as well as on the ratio of these concentrations:

- The buffer capacity increases when the concentration of buffer components increases;

- The capacity for both acid and base is highest in a buffer where ratio of concentrations equals 1. Experimental

Prepare acetate/sodium acetate buffer. Calculate its pH and buffer capacity. Each pair of students has to prepare and analyse only one buffer solution.

Procedure

1. Take 2 flasks and mixing the appropriate volumes of two components -- 0.1 M acetic acid and 0.1 M sodium acetate in each flask prepare 2 identical acetic buffers (one - for determining buffer capacity for acid, and the other - buffer capacity for base). The volumes are indicated in the table.

Vbuffer=VCH3COOH +VCH3COO-Na =10 ml pH pH V (titrant) Buffer capacity N CH3COOH, mL CH3COONa, mL Initial pHi After titration with acid After titration with base

pHacid pHbase Vacid Vbase Bacid Bbase

1 2 3 4 5 6 2 3 4 5 6 7 8 7 6 5 4 3 3.4 3.4 3.4 3.4 3.4 3.4 6.3 6.3 6.3 6.3 6.3 6.3

(29)

2. Calculate the initial pH of the prepared buffer using Henderson-Hasselbalch equation: pH = lg ([salt] / [acid]) - lg Ka;

Ka - ionization constant for acetic acid, Ka= 1.85  10-5, and pKa=-lg (1.85  10-5) = 4.7.

Since the concentrations of components in the buffer are equal, the volumes in mL used in preparation of the buffer can be used instead of the concentration:

pH = lg ( volume of salt / volume of acid ) + 4.7

2. Add 3--4 drops of methyl-orange indicator to one flask. Titrate this flask with 0.1 M HCl until pink-orange colour appears (indicating pH = 3.4 – that is pH after titration with acid). Use the titration volume of HCl (Vacid) to calculate the buffer capacity for acid Bacid:

Bacid = (Vacid Cacid / pHacid Vbuffer)

Vbuffer - volume of buffer in ml,

pHacid - change of pH after titration with acid.

3. Add 3-4 drops of methyl-red indicator to the second flask. Titrate this flask with 0.1 M NaOH until yellow colour starts to appear (pH = 6.3 – the pH after titration with base). Use the titration volume of NaOH (Vbase) to calculate the buffer capacity for base Bbase:

Bbase = (Vbase Cbase / pHbase Vbuffer)

pHbase -- change of pH after titration with base.

(30)

1. What solutions are called “buffers”? 2. How are buffer solutions prepared?

3. What does the pH of buffer solution depend on? 4. What equation is used to calculate the buffer pH?

5. How does acetate buffer maintain solution pH at constant level if a small amount of strong acid is added? Write the equations.

6. How does acetate buffer maintain a solution pH at constant level if a small amount of strong base is added? Write the equations.

7. What is buffer capacity?

8. How can the buffer capacity for an acid be calculated? Write the formula. 9. How can the buffer capacity for a base be calculated? Write the formula. 10. What does the buffer capacity depend on?

(31)

Name Family name

……… ……… ……… ……… ………

(32)

(33)

5. Colloidal solutions

Background. Colloidal solutions are disperse systems where disperse phase particles are 1-100 nm in size. Such solutions are stable over time. The solution where a disperse phase is solid and medium is liquid is called sol. To prepare a colloidal solution, particles of the disperse phase have to be made of the size of colloidal particles. There are two ways to do this: condensation (to make the colloidal size particles from smaller ones) and dispersion (to disperse larger particles into colloidal size particles).

Dispersion methods:

- Colloidal mill (a mechanical way to grind large particles); - Ultrasound;

- Peptization (disaggregates large particles by chemical agents, which increase repulsion of particles).

Condensation methods (produce compounds of relatively low solubility from soluble ones): - Exchange of solvent (by another solvent in which the same substance has lower solubility); -Oxidation (obtains neutral chemical elements (mainly non-metals) from their ionic forms);

- Reduction (obtains neutral elements (mainly metals) from their ionic forms); - Hydrolysis reaction (makes compounds of lower solubility);

-Exchange reactions (makes insoluble compounds).

Exchange reaction: AgNO3(aq) + KCl(aq) → AgCl↓(s) + KNO3(aq).

Micelle when it is excess of AgNO3: {m(AgCl)nAg+(n-x)NO3-}x+xNO3- .

Micelle when it is excess of KCl : {m(AgCl) nCl-(n-x)K+}x-xK+.

Experimental

A. Prepare colloidal Fe(OH)3 solution (sol) by chemical condensation (hydrolysis reaction):

1. Put 1 mL of 2% FeCl3 into a test tube, add 10 mL of distilled water.

2. Mix thoroughly and boil the mixture until brown transparent Fe(OH)3 sol. is formed.

3. Write the hydrolysis reaction and micelle of Fe(OH)3.

Reaction of hydrolysis: FeCl3 + H2O→………..

………... ... Micelle: {m(Fe(OH)3●●●●●. . . .

B. Preparation of colloidal colophony (pine resin) sol by the exchange of solvent

1. Add several drops of 2% colophony/ ethanol solution to 10 mL of distilled water. 2. Mix the mixture. Milky sol should be obtained.

3. Explain, how and why this sol was formed. C. Stability and coagulation of colloidal solutions.

The coagulation of colloidal solutions usually occurs when an electrolyte solution is added to it. The coagulation means that the particles became larger (more than 100 nm), and these newly formed particles easily sediment.

Prepare 5 test tubes for each electrolyte (one pair of students experiments with one electrolyte), and fill them according to the table:

(34)

Number of the test tube 1 2 3 4 5

Fe(OH)3 sol (prepared by the technician), mL 5 5 5 5 5

H2O, mL 4.5 4 3 2 1 Electrolyte, mL 1) 3 M KCl 2) 0.01 M K2SO4 3) 0.001 M K3[Fe(CN)6] 0.5 0.5 0.5 1 1 1 2 2 2 3 3 3 4 4 4

Leave the test tubes for 30 min, and then observe where the colloidal solution has coagulated. The test tube with the lowest concentration of electrolyte where coagulation was observed has concentration called coagulation threshold.

Coagulation threshold is the lowest electrolyte concentration (mmol/L) which induces coagulation of the sol. Calculate the coagulation thresholds:

) / ( ) / ( L mmol V V V xV L mmol C C water sol e electrolyt e electrolyt e electrolyt thr    

The opposite number of the coagulation threshold is called the coagulation power:

thr C P 1 .

Write the results in the table given below. Discuss the results with other students who have done the experiment with different electrolytes. Make a conclusion of which electrolyte is a better coagulation agent and explain why.

Electrolyte Coagulating ion Coagulation threshold Coagulation power

KCl K2SO4

K3[Fe(CN)6]

1. What are the disperse phase and the dispersion medium? 2. What solutions are called “colloidal solutions”?

3. What systems are called “sol”?

4. Provide some examples of the disperse systems in living organisms.

5. What properties should chemical substances have, so that they can be used to prepare colloidal solutions?

6. What methods of colloid preparation are there? 7. What is the peptization method?

8. What is micelle and what is it made of?

9. What is called coagulation, coagulation threshold and coagulation power? 10. How can electrolytes cause coagulation?

(35)

Name Family name

……… ……… ……… ……… ………

(36)

Conclusion

(37)

6. High molecular mass compounds and their properties.

Swelling and gelatinization

Background. High-molecular mass compounds are organic compounds with molecular mass as big as hundreds of thousands or even millions of Daltons. By structural features, they usually are polymers two origins:

1. Cellular origin high-molecular mass compounds are called biopolymers, e.g. nucleic acids, proteins and polysaccharides.

2. Synthetic/artificial high-molecular mass compounds.

Sizes of particles of high molecular mass compounds are similar to colloidal particles, but they are water-soluble. Therefore solutions of high molecular mass compounds share some common properties with colloidal solutions and with real solutions. They also have very specific properties.

Properties of solutions of biopolymers:

1. Properties that are common with colloidal solutions:

 Size of molecules of biomolecules is similar to the size of colloidal particles.  Solutions of biopolymers have slow rate of diffusion.

 Molecules of biopolymers cannot pass through semi-permeable membrane, what implies that their solutions have low osmotic pressure. The osmotic pressure of those solutions depends only on number of biopolymer molecules.

2. Properties of biopolymer solutions, that are common with those of real solutions:  Do not form micelles, i.e. they are solutions of a single phase (homogeneous systems).  Solutions of biopolymers with linear structure do not show Tindall effect.

 Solutions of biopolymers are stable systems. They do not show sedimentation phenomenon. 3. Specific properties of biopolymer solutions.

 Biopolymers swell before dissolving. Under swelling solvent surrounds the molecule of biopolymer, then it moves into the empty spaces of molecule, so that the volume of the molecule increases.

 Solutions of high-molecular mass compounds can form a jelly. It is a net-like structure formed in the solution when hydrophobic regions of the molecules interact with each other and the hydrophilic regions are highly hydrated, water molecules filling in the gaps between the molecules.

Swelling of high-molecular mass compounds. Swelling of a high-molecular mass compound results in increasing of both mass and volume of a high-molecular mass compound. There are two sorts of swelling: the limited swelling and unlimited swelling where a solution of high-molecular mass compounds is formed. Swelling depends on the properties of the solvent.

The swelling degree (Q) is the main characteristic of swelling. It shows how much the volume of the high molecular mass compound enlarges during the swelling. The swelling degree shows how many

cm3 of the solvent can be absorbed by 1 cm3 of a high-molecular mass compound:

V1 is the volume of a high-molecular mass compound before swelling,

V2 is the volume of a high-molecular mass compound after swelling.

The swelling degree (Q) can be estimated by weighing of a high molecular mass compound before its swelling and after swelling.

The swelling is affected by:

1. Temperature. As swelling is an exothermic process, increasing in temperature results in decreasing of the swelling degree.

V2 – V1

V1

(38)

2. Pressure. Increasing in pressure results in increasing of the swelling degree.

3. Electrolytes and pH of solution. Swelling is little at isoelectric point, because molecules can stick together and entering of water is slow.

4. Degree of dispersion. High dispersion degree favours swelling. 5. Time.

Jellification (gelatinization). Solutions of high-molecular mass compounds can form a jelly. It is a net-like structure formed in the solution when hydrophobic regions of the molecules interact with each other and the hydrophilic regions are highly hydrated, water molecules filling in the gaps between the molecules. In chemistry, the jelly is called gel. The process when solutions of high-molecular mass compounds lose their fluidity is called jellification. It depends on the following factors:

1. Concentration of high-molecular mass compound solutions. Jellification can occur in concentrated solutions of high-molecular mass compounds.

2. Size and shape of molecules of a high-molecular mass compound. Thread-like molecules can easily form gel, but the ball-like ones can hardly do this.

3. Temperature. Low temperatures favour to jellification.

4. Presence of electrolytes and pH. Anions are the most important for the jellification. According to the efficiency of the effect, the anions make a line:

SO42- >citrate>CH3-COO- >Cl- >NO3-->Br- >I- >SCN-

Jellification is promoted by highly hydrated ions. Starting with Cl- the anions diminish the jellification

(gelatinization). These ions are adsorbed on the surface of the macromolecules. They give charge to polymers and prevent macromolecules from jellification.

Lab. Procedure

Effect of solvent on swelling.

1. Take 4 test tubes. Put 2ml of water in the tubes N1 and N2. Put 2 ml of benzene into the tubes N3 and N4.

2. Add one piece of agar (polysaccharide) into the tube N1 and another piece of the same size into the tube N3.

3. Add one piece of synthetic rubber into the tube N2 and another piece of the same size into the tube N4. 4. After 20 min., compare the sizes of the agar and rubber. Make a conclusion about the effect of solvents on the swelling.

Effect of electrolytes on jellification (gelatinization)

1. Take 6 test tubes and put 1.5 ml of 1 M solutions of the electrolytes as indicated in the table:

Number of test tube 1 2 3 4 5 6

Electrolyte K2SO4 CH3COOK KCl KI KSCN H2O

Beginning of gelatinization t1

End of gelatinization t2

Time required for complete gelatinization t2 - t1

2. Add 1.5 ml of 6% hot gelatine solution into each test tube and mix thoroughly. 3. Place the test tubes into the hot water bath (50 - 60º C) for 10 minutes.

4. Remove the test tubes from the bath and place them in cold water. Mark the time t1.

5. Periodically check the fluidity of the solution by inclining the test tubes. If you observe that solution is no longer fluid, mark the end time of gelatinization.

(39)

1. What substances are called the high molecular mass compounds (HMMC)? Give some examples.

2. How are solutions of HMMC similar to colloidal solutions? 3. How are solutions of HMMC similar to real solutions?

4. What properties are specific to high molecular mass compounds? 5. What process is called swelling?

6. What effect on swelling does a solvent have? 7. What does the swelling degree show?

8. What process is called jellification?

9. How does a gel form and what can affect its formation? 10. What effect do electrolytes have on jellification?

(40)

Name Family name

……… ……… ……… ……… ………

(41)

(42)

7. Coordination compound based determination of water hardness

Background. Some amino-polycarbonic acids and their salts can form stable coordination compounds with most of the cations. These compounds are called complexones (chelating agents). An example of such complexone is EDTA (ethylendiaminetetraacetic acid):

Most often the disodium salt of EDTA is used, which has a trivial name of Trilon B. This compound reacts with the divalent ions (such as Ca2+, Mg2+, Ba2+) forming a rather stable colourless coordination compound. One molecule of Trilon B binds one ion:

Na2[H2EDTA] + Ca2+ <---> Na2[CaEDTA] + 2H + [H2EDTA] 2- + Ca2+ <---> [CaEDTA] 2- + 2H +

The term “total water hardness” means a total amount of Ca2+ and Mg2+ ions dissolved in water. Due to the ability

to bind these ions and to form a coordination compound Trilon B is used for the quantitative determination of these ions measuring the water hardness. Since the compound of Trilon B with Ca2+ and Mg2+ ions is colourless, an indicator is needed to pinpoint the equivalence point. The most suitable for this purpose is Eriochrom Black T (for

convenience abbreviated as H3Ind). Depending on pH this indicator has different colours: pH 6 – red;

pH 7-11 – blue; pH 12 – yellow-orange H3Ind ↔ H2Ind- + H+; pH = 6 red H2Ind- ↔ HInd2- + H+; pH = 7-11 blue HInd2- ↔ Ind3- + H+; pH = 12 yellow-orange

(43)

Eriochrom Black T forms a red-violet coordination compound with Ca2+ and Mg2+ (at pH>7), which is less stable than the one formed by Trilon B. While titrated with Trilon B, all Ca2+ and Mg2+ ions from the solution are coordinated by it, then this red-violet compound decomposes and more stable compound of Ca2+ and Mg2+ with Trilon B forms. The free anions of Eriochrom Black T make solution blue-coloured at the equivalence point:

[MgInd] - + [H2EDTA]2- ---> [MgEDTA]2- + HInd2- + H+

Red-violet colourless colourless blue

Water hardness values: < 0.75 mmol/L – soft water

1.5 – 2.7 mmol/L – medium soft/hard water 2.7 – 5.35 mmol/L – hard water

> 5.35 mmol/L – very hard water

Examples: rainwater – 0.05 mmol/ L; sea water – 65 mmol/ L.

Experimental Procedure

1. Dilute the given solution X (solution of MgCl2 whose hardness has to be determined) to the 100 ml in

a volumetric flask (add distilled water up to 100 ml)

2. Put 10 mL of this diluted solution X into the flask, add 2 mL of the buffer solution (1 M

NH4Cl/NH4OH) and 4 drops of the indicator –Eriochrom Black T. Mix well.

3. Titrate the mixture with the solution of 0.05 M Trilon B (mix well during titration) until the red colour starts changing into violet (blue).

4. The titration is over when after one single drop of Trilon B the solution becomes blue (You need to catch the first moment of the change in the initial colour). Therefore the last portion of Trilon B should be added drop-by-drop. Record the volume (mL) of Trilon B used for titration.

5. Repeat the titration three times, and take the mean volume of the three titrations.

6. Calculate the total water hardness. One mole of Trilon B can bind one mole of Mg2+ ions. Use the equation:

HT = (MB VB / Vwater )  1000 mmol/L,

where HT -- total water hardness in mmol/L;

MB – molarity of the Trilon B solution;

VB – volume of the Trilon B used for titration;

Vwater – volume of water used for titration.

(44)

1. What compounds are called coordination compounds? Give some examples. 2. What compounds are called complexones (chelating agents)? Give some examples. 3. Draw the formulas of EDTA and disodium salt of EDTA.

4. What is the trivial name of disodium salt of EDTA? 5. What does the term “total water hardness” mean?

6. Why is the indicator Eriochrom Black T used when total water hardness is determined? 7. Write down the equation of the reaction between Eriochrom Black T, Trilon B and metal ions.

Explain why the colour of the solution changes at the equivalence point. 8. Write down the formula which is used to calculate the total water hardness.

(45)

Name Family name

……… ……… ……… ……… ………

(46)

(47)

8. Chemical kinetics

Experimental. A. Rate of chemical reaction determined by change of reaction product concentration. This is the reaction of sodium thiosulphate decomposition:

Na2S2O3 + H2SO4 ---> Na2SO4 + S + SO2 + H2O

One of the products is insoluble in water – it is elementary sulphur S. The increased turbidity of the reaction mixture provides a pretty good way to evaluate the reaction rate. The increased turbidity means that more sulphur was formed, and the sooner this process occurs the higher is the rate of the reaction. Procedure

1. Take a small beaker and fill it according to the table (draw it in your notebook) as N 1. As

it is obvious from the table’s second column, the following beakers will have increasing sodium thiosulphate concentrations, as the total volume in each of them will be the same: Number 0.25 M Na2S2O3; mL H2O, mL Time, s (seconds) Rate of reaction, 1/s N 1 N 2 N 3 2 4 6 4 2 0

2. Add 2 mL of 0.25 M H2SO4, stir carefully with a glass rod. Start time countdown from the

moment of adding the acid. Immediately put this flask on the piece of squared paper. While slowly stirring the contents of the beaker, observe when the lines cannot be seen through the layer of reaction

mixture. Record the time elapsed from the moment of H2SO4 addition. Calculate the reaction rate.

3. Wash the beaker carefully and repeat the same procedure with the other concentrations of sodium thiosulphate N 2 and N 3.

4. Draw the graph of the dependence of the time needed for the reaction (in seconds) on the volume of substrate added (in mL), (mL should be on the X axis, and seconds -on the Y axis).

B. Effect of catalyst on the reaction rate

Catalyst is the substance which changes the reaction rate when present in the reaction mixture, but remains unaltered itself. The reaction rate changes because catalyst decreases the activation energy of that particular reaction. This is the reaction of hydrogen peroxide decomposition:

2 H2O2 ---> 2 H2O + 2 [O], then

2[O] ---> O2

The catalyst for this reaction is MnO2. In living organisms this reaction is catalysed by the enzyme

catalase. Decomposition of hydrogen peroxide at the room temperature occurs too slowly for the bubbles

of formed oxygen to be visible. When MnO2 is added, the reaction rate increases significantly, and this

can be noticed by vigorous bubbling. Procedure

Put 1 mL of 3% H2O2 solution into the test tube. There are no bubbles. Add a little of MnO2 powder and

cover the test tube with your finger while the reaction proceeds. At that time get a wooden splint burning and blow the flame out. Stick a glowing wooden splint inside the test tube. Observe what happens and explain the result.

(48)

C. Equilibrium shift.

A chemical reaction consists of direct and reverse reactions, going on in opposite directions. The term “Equilibrium of chemical reaction” means that equilibrium between the direct and reverse reaction has been reached, and the rate of direct reaction is equal to the rate of reverse reaction. Equilibrium of the chemical reaction can be disturbed by variety of factors among which is the change of:

1) Concentration of reactants or reaction products; 2) Temperature;

3) Pressure;

4) Volume of the system.

In living organisms the main factor affecting equilibrium of chemical reaction is the change of substrate or product concentration, since temperature, pressure and volume are kept constant. The effect of various factors on the equilibrium of chemical reaction is described by Le Chatelier’s principle:

If a system in equilibrium has been disturbed by the outside factors (as concentration, temperature or pressure) then equilibrium of this system shifts in the direction which tends to decrease the effect of outside factors.

I.

CoCl2 dissociates when dissolved in water:

CoCl2 ↔ Co2+ + 2Cl-

Then the ions formed make a coordination compound where Co (II) has additional bonds with water: Co2+ + 2 Cl- + 6 H2O ↔ [Co (H2O)6]Cl 2 ↔ 2 Cl- + [Co(H2O)6]2+ (red colour).

When HCl is added, the red-coloured coordination compound transforms into another blue-coloured: (Red colour) [Co (H2O)6]2+ + 2 Cl- + 2 Cl- ↔ 6 H2O + [CoCl4]2- (blue colour)

Keq = [[CoCl4] 2-] [H2O]6 /[ [Co(H2O)6] 2+]  [Cl-]4

When HCl is added, the equilibrium of the reaction shifts towards formation of [CoCl4]2-. If extra water is

added the equilibrium shifts in the opposite direction. Procedure

1. Dissolve a crystal of CoCl2 in a test tube with few drops of water.

2. Add several drops of concentrated HCl until the solution becomes blue. 3. Dilute this solution with water until reappearance of less intense red colour.

4. By heating and cooling the test tube find what is the effect of temperature on this equilibrium.

II.

FeCl3 reacts with NH4SCN and red solution of e iron rhodanide forms:

FeCl3 + 3NH4SCN ↔ Fe(SCN)3+ 3NH4Cl.

This reaction is a reversible one. The shift of equilibrium could be seen by observing the intensity of red

colour in the solution. The colour intensity depends on the concentration of iron rhodanide – Fe(SCN)3.

When the concentration of Fe(SCN)3 in the solution is high, the red colour is intense. When the

concentration of Fe(SCN)3 is low the solution is faded red. It is possible to shift the equilibrium by

(49)

Procedure

1. Pour into a beaker 20 mL of water and add 3 drops of the saturated solution of FeCl3 and 3 drops

of the saturated solution of NH4SCN.

2. Divide the solution between four test tubes. 3. Add 2 drops of the FeCl3 into the first test tube.

4. Add 2 drops of the NH4SCN into the second test tube.

5. Add some crystals of the NH4C l into the third test tube.

6. Stir the test tubes and compare the colour intensity. 7. Describe and explain the observed differences.

(50)

Calculation of Michaelis-Menten constant and maximal reaction rate

Michaelis-Menten equation describes the relationship between reaction rate and substrate concentration in enzyme-catalysed reaction. [S] K [S] V V m max 0 _  

Vmax is the maximum velocity of the reaction – when all enzyme molecules are fully active. Km –

Michaelis-Menten constant – is the concentration of substrate at Vmax/2.

Task: calculate Michaelis-Menten constant and maximal reaction rate for the given enzyme-catalysed reactions.

Procedure:

1. Calculate reaction rates (v) with different substrate concentrations.

time reaction formed product of amount v mol/min.

2. Plot 1/[S] on x axis and 1/v – on y axis.

3. Find the Km and Vmax values for the given reactions from Lineweaver-Burk plot.

Version A. Assay conditions were following:

1 mg of the enzyme was incubated with the indicated concentration of substrate [S] for 5 minutes, after which the amount of formed product was determined and is presented in the table:

Substrate concentration (μM) 0.63 0.83 1.25 3.3

Amount of product formed per 5 min (μmol) 29 36.2 47.2 76.9

Version B. Assay conditions were following:

1 mg of the enzyme was incubated with indicated concentration of substrate [S] for 5 minutes, after which the amount of formed product was determined and is presented in the table:

Amount of product formed per 5 min (μmol) 37 43.5 53 62

Version C. Assay conditions were following:

1 mg of the enzyme was incubated with indicated concentration of substrate [S] for 5 minutes, after which the amount of formed product was determined and is presented in the table:

(51)

Calculation of Km and Vmax values Lineweaver-Burk Plot (1) -2.0 -1.5 -1.0 -0.5 0.5 1.0 1.5 2.0 -0.1 0.1 0.2 0.3 0.4 0.5 1/[S] 1/v Lineweaver-Burk Plot (2) -2.0 -1.5 -1.0 -0.5 0.5 1.0 1.5 2.0 -0.1 0.1 0.2 0.3 0.4 0.5 1/[S] 1/v Lineweaver-Burk Plot (3) -3.0 -2.5 -2.0 -1.5 -1.0 -0.5 0.5 1.0 1.5 2.0 2.5 3.0 -0.1 0.1 0.2 0.3 0.4 0.5 1/[S] 1/v

(52)

1. What does chemical kinetics study?

2. How is the rate of chemical reaction expressed?

3. What factors can make an impact on the rate of chemical reaction? 4. How does the reaction rate depend on the concentrations of reactants? 5. How is it possible to determine the reaction rate of this reaction? Na2S2O3 + H2SO4---> Na2SO4 + S+ SO2 + H2O

6. What does a catalyst do when it is present in a reaction mixture? How does it work? 7. Which catalyst can catalyze the reaction of hydrogen peroxide decomposition? 8. What does “equilibrium of chemical reaction” mean?

9. What factors can influence the equilibrium of chemical reaction? 10. Describe “Le Chatelier’s principle”.

11. How and why can CoCl2 solution change its colour?

12. What factor determines the colour intensity of iron rhodonide solution? 13. What does the Michaelis-Menten equation describe? Draw this equation. 14. What does „ Lineweaver-Burk plot“ mean? Draw this plot.

15. How could Michaelis-Menten constant and maximal reaction rate for the given enzyme-catalysed reaction be calculated?

(53)

Name Family name

……… ……… ……… ……… ………

(54)

(55)

9. Chemical properties of carbonyl compounds

Esters and lipids

Carbonyl compounds are organic compounds containing carbonyl (oxo) group in a molecule. The compounds fall into two big groups – aldehydes and ketones. In aldehydes, the carbonyl group has both hydrogen atom and hydrocarbon radical attached. In ketones, the group is bound with two hydrocarbon radicals. According to hydrocarbon radical present, aldehydes are grouped into aliphatic ones (a), alicyclic ones (b), and aromatic ones (c):

Ketones are also grouped into aliphatic (a), alicyclic, and aromatic.

Aldehydes and ketones with up to 4 C atoms in molecules are volatile liquids of specific odour. They are soluble in water and in organic solvents. Solubility in water decreases with an increase of a number of C atoms in a chain. Aldehydes containing 8-10 C atoms in a chain have odour of flowers and are used in perfumery.

Chemical properties of carbonyl compounds

Properties of carbonyl compounds are determined by chemical properties of both the carbonyl group and hydrocarbon radical.

1. Reduction producing alcohols (both aldehydes and ketones):

2. Oxidation producing acids (only aldehydes):

H₃C C O H C O H C O H a b c

ethanal cyclohexanecarbaldehyde benzaldehyde

H₃C C CH₃ O a propanone (acetone) H₃C C CH₃ O propanone (acetone) H₃C C CH₃ OH H H₂ H-C-H H-C-OH = = O _O

’ students MEDICAL CHEMISTRY Handbook for the medicine and odontology faculties

Lithuanian University of Health Sciences

Department of Biochemistry

Laima Ivanovienė, Ramunė Morkūnienė, Julius Liobikas,

Dalia Marija Kopustinskienė, Svaja Dubickienė, Artūras Kašauskas

MEDICAL CHEMISTRY

Handbook for the medicine and odontology faculties’

students

The teaching book

Kaunas, 2014

Contents

EXAMPLE OF THE LABORATORY WORK REPORT

1. Preparation of solutions. Percent concentration

2. Preparation of molar solutions. Molar concentration

3. Volumetric analysis. Acid-base titration

4. Buffer solutions

5. Colloidal solutions

6. High molecular mass compounds and their properties.

Swelling and gelatinization

7. Coordination compound based determination of water hardness

8. Chemical kinetics

Calculation of Michaelis-Menten constant and maximal reaction rate

9.

Chemical properties of carbonyl compounds

Esters and lipids