POZNAŃ UNIVERSITY OF MEDICAL SCIENCES DEPARTMENT … · POZNAŃ UNIVERSITY OF MEDICAL SCIENCES...

POZNAŃ UNIVERSITY OF MEDICAL SCIENCES

DEPARTMENT OF PHYSICAL PHARMACY AND PHARMACOKINETICS

PHYSICAL CHEMISTRY IN PHARMACYLaboratory Practical Course for PharmD Students

Franciszek Główka, Marta Karaźniewicz-Łada, Katarzyna Kosicka MichałRomański, Dorota Danielak, Andrzej Czyrski and Matylda Resztak

Edited by Franciszek Główka

Poznań 2014

1

PREFACE

The first edition of Physical Chemistry in Pharmacy; Laboratory Practical Course is designed for

students of second year of Doctor of Pharmacy Program (Pharm.D.). Physical Chemistry for

Pharmacists, widely known as a Physical Pharmacy, concerns with the use of physical and chemical

principles to many disciplines of pharmacy. Laboratory course for Physical Pharmacy is integral part

curriculum of Pharm.D. Program including also lectures and seminars. The schedule of the

laboratory course is prepared to give the opportunity to students to better understand not always easy

material of Physical Pharmacy. The laboratory course includes the following issues: properties of

colloids, thermodynamics, electrochemistry, kinetics, pharmacokinetics, phase equilibria, surface

phenomena and quantum mechanics. Each experimental part was preceded by theory. More details

readers will find in the chapter Experiments. The labs were designed for manual as well as computer

performance. The English edition was partially based on the Polish guide “Ćwiczenia laboratoryjne z

chemii fizycznej dla studentów farmacji i analityki medycznej” by G. Uchman and T Hermann. In

meantime most laboratories have been greatly improved as well as new were introduced. The

modern tensiometer was applied for performance of the experiment Critical micelle concentration

prediction for Tween 20. To strengthen the pharmaceutical aspects of the laboratory course in some

of the experiments: electrochemistry, kinetics, surface phenomena were introduced medicinal

substances. Authors believe that the course of Physical Pharmacy will be helpful in understanding

subjects such as pharmaceutical technology, pharmacology, pharmacokinetics, pharmaceutical

chemistry, biopharmacy, tightly connected with pharmacist profession.

Authors expect also some remarks from teachers and students, who will perform the experiments, to

improve next editions of the guide.

Prof. Franciszek Główka, Ph.D.

2

Experiments

Electric Properties of Colloids Marta Karaźniewicz-Łada

1a Separation of chosen proteins by gel electrophoresis

1b Determination of the charge of colloidal particles

2

Thermodynamics of Phase Transitions Franciszek Główka

Determination of change of enthalpy of vaporisation of a volatile liquid.

3

Phase Equilibria Andrzej Czyrski, Matylda Resztak

Temperature-composition diagram for the systems consisting of water and phenol

4

Thermodynamics of Mixtures. Colligative Properties Franciszek Główka

Determination of molecular mass and van’t Hoff factor i using method of determination of osmotic pressure

5

Distribution of Solutes Between Immiscible Solvents Marta Karaźniewicz- Łada

Spectrophotometric determination of specific partition coefficient o/w for benzoic acid

Electrochemistry Michał Romański, Franciszek Główka

6aDetermination of pKa of acetylsalicylic acid by potentiometric titration

6b Determination of the solubility product of the chosen calcium salt

7

Kinetics, Franciszek Główka, Michał Romański

Determination of the rate constant and the thermodynamic parameters for the hydrolysis of acetylsalicylic acid

8

Pharmacokinetics Franciszek Główka

Determination of pharmacokinetic parameters of salicytates

9

Quantum mechanics Michał Romański, Dorota Danielak,

Application of molecular modeling to determination of physicochemical properties of medicinal substances

10a

Properties of colloids. Viscosity Marta Karaźniewicz- Łada

Determination of the isoelectrical point of gelatin

10b Determination of the emulsion type

3

11

Surface tension. Surfactants Franciszek Główka, Marta Karaźniewicz-Łada

Critical micelle concentration (CMC) prediction for Tween 20

12

Surface phenomena. Adsorption isotherms Katarzyna Kosicka, Franciszek Główka

Adsorption of acetaminophen on activated charcoal

4

ELECTRIC PROPERTIES OF COLLOIDS

Experiment 1a. Separation of chosen proteins by gel electrophoresis

The aim of the experiment is an electrophoretic separation of human serum proteins on agarose gel and determination of the molecular weight of α1 - and α2 – globulins

Required knowledge: definition and application of electrophoresis, electrophoretic mobility, types of electrophoresis

Introduction

Electrophoresis is a technique widely used for separation and characterization ofproteins, DNA and other charged molecules. Under the influence of an electric field,positively charged molecules will migrate to the cathode and negatively charged moleculeswill migrate to the anode. There are a number of different types of electrophoresis, but allinvolve generating an electric field between to points and placing a matrix of some sort in-between through which the macromolecules must travel. The speed at which they passthrough this matrix in the presence of the electric field is called their electrophoreticmobility. Electrophoretic mobility is a result of a constant drift speed, s, reached by an ionwhen the driving force zeE (where ze is the net charge and E is the field strength) is matchedby the frictional force fs. The drift speed is then:

f

zeEs = (9.1)

Therefore, the mobility of a macromolecule in an electric field depends on its net charge,size and shape.

In gel electrophoresis a porous gel may act as a sieve by retarding the movement oflarge macromolecules while allowing smaller molecules to migrate freely. The pulling forceof the electrical voltage and the opposing force of the gel matrix result in different migrationrates for the proteins and their separation into fractions. After separation components of themixture are visualized by staining with e.g. amidoblack, coomassie stain, cooper or silversalts, ethidium bromide and then, they are identify by the use of many methods. Gelelectrophoresis is applied for separation of molecules with small as well as large molecularweight, such as amino acids, small peptides, proteins, nucleotides, RNA and DNA.Electrophoretic separation is performed on agarose and polyacrylamide plates. Agarose is apolysaccharide extracted from seaweed. It is typically used at concentrations of 0.5 to 2%.Agarose gels are extremely easy to prepare. Agarose powder is mixed with buffer solution,melted by heating and after cooling to 500C it is poured on glass plates. Polyacrylamide is across-linked polymer of acrylamide. Polymerization of acrylamide is initiated by addingammonium persulfate and N,N’-tetramethylenediamine. Polyacrylamide gels have a rathersmall range of separation but very high resolving power. They are used for separatingfragments of less than about 500 bp. Fragments of DNA or RNA differing in length by asingle base pair are easily resolved. However, acrylamides used for preparation ofpolyacrylamide gels are neurotoxic.

5

All amino acids contain ionizable groups that cause the amino acids, in solution, to act ascharged polyelectrolytes that can migrate in an electric field. Depending on pH of theelectrolyte, amino acid molecule is positively or negatively charged. The amino acids with anet positive charge will migrate toward the negative electrode. Those with a negative netcharge will move toward the positive electrode.

H2N - R - COO - H3N + - R-COO - H3N + - R - COOH

Electrophoresis of proteins is applied to comparative study on different protein formulationsas well as in clinical practise in plasma protein analysis. Gel electrophoresis enables toseparate of serum proteins into 5 fractions: albumin, α1, α2, β and γ – globulins.Concentration of each fraction can be measured by the use of densitometer.

Proper values of protein fractions in human serum:

- albumin 53-66 %

- α1 – globulins 2 – 5.5 %

- α2 – globulins 6 – 12 %

- β – globulins 8-15 %

- γ – globulins 11 -21 %

Changes observed in protein fraction pattern can be caused by inflammatory response of the body or many chronic diseases.

Experiment

Materials:An electrophoresis chamber and power supply, blotting – paper (“Whatman” number 3), plates with agarose, templates for sample application, a glass plate, 1% solution of the human serum proteins: albumin, α – and β - globulin, γ – globulin, dilution of the human serum, solution of the egg proteins, barbital buffer of pH 8.6, fixative solution, staining solution (amidoblack), destaining solution

ProcedureA. Performance of the electrophoretic separation

1. Remove cover from the chamber and pour 1 l of buffer into each part of chamber.2. Take the gel out of package not touching its surface and place it on a sheet of paper

or blotter.3. Use blotter to dry area where samples will be placed. Remove wet blotter

immediately.4. Place the sample template on the gel. Move finger longwise the template to press it to

the gel. Template should firmly adhere to the gel.5. Apply 5 μl of the sample to each slit and leave it for 5 min. since the application of

the last sample. 6. Remove excess of the serum with blotter.7. Take the sample template off.8. Place agarose plate on the chamber plate, samples should be on the cathodic side (-).9. Place two pieces of blotter on the both edges of the plate and immerse their

extremities in the buffer. Cover the agarose plate with the glass plate. Wet the blotter with the buffer.

6

10. Close chamber with cover.11. Perform electrophoresis for 40-50 min. at 200 kV.12. After migration remove plate and immerse it in the fixative solution for 15 min. in

vertical position.13. Dry the gel under hot air stream or in the oven at the temperature up to 800C until it is

completely dry.14. Immerse the plate in the staining solution for 10 min., then destain it in 2 -3

successive baths of destaining solution.15. Rinse the gel with distilled water and dry it with hot air up to 800C.

B. Determination of protein fraction mobility and percentage content of fractions in serum 1. Insert a stained agarose plate into scanner. 2. Open the program Epson Scan.3. When the program is initiated, click on the button Preview to start prescanning.4. After appearance of a picture of the plate, rotate the image right and mark an area of

scan.5. Click on the button Scan, define location of the file, its name and image format type

(jpg). Close the program.6. Start the program Image.j.7. Open the file with scanned image.8. With pressed left button of the mouse mark an area of protein separation on the plate

from the point, where the samples had been applied, to the point, where the stain had moved during electrophoresis.

9. Click on the option Analyze in the Toolbar, select Gels and then Select First Lane. Confirm that the lanes are really horizontal.

10. Select Gels in the Analyze menu and then Plot lanes.11. Select an icon Multipoint selection in the Toolbar and mark tops of the peaks.

12. Click on the option Analyze in the Toolbar and select Set measurements. No option in the list should be ticked. Then, choose Measure from Analyze menu and rewrite the values for protein and stain distances from the second column (values given in piksels convert to meters).

13. Select an icon Lines in the Toolbar and limit the edges of the peaks.

14. Then, select an icon Wand and indicate the peaks.

7

15. Select Gels in the Analyze menu and then Label peaks.16. Rewrite the value of percentage content for each protein fraction.

Calculate mobility (μ) of each fraction using the following equation:

µ = Ut

Ll

where:l – distance of protein migration [m]L – distance of dye migration [m]t – time of electrophoresis [s]U – applied voltage [V]

C. Determination of molecular weight (Mw) Calculate parameters of the calibration curve using mobility of the protein fraction versuslogarithm of molecular weight for the known molecular weight of standards:Albumin – 69 kD (kilodalton)β-globulin – 120 kDγ-globulin – 156 kD

Plot a graph: μ = f(logMw) using the calculated parameters.Use the standard curve to determine the weight of α1 and α2 – globulin.

Experiment 1b. Determination of the charge of colloidal particles

The aim of the exercise is to produce a colloid of ferrum hydroxide and to determine its charge using electrophoresis in a glass U-pipe.

Required knowledge: a structure of the colloidal micelle, properties of colloids, Smoluchowski’s equation

IntroductionThe charge of colloidal particles is the result of electrostatic adsorption of ions to the surface.In colloids, two regions of charge can be distinguished. First, there is a fairly immobile layer of ions that adhere tightly to the surface of the colloidal particle. This inner sub-layer is called Stern layer. The radius of the sphere that captures this rigid layer is called the radius of shear and is the mayor factor determining the mobility of the particles. The electric potential at the radius of shear relative to its value in the distant bulk medium is called zeta potential or the electrokinetic potential. It is also denoted as ζ-potential. Second, the charged unit attracts an oppositely charged atmosphere of mobile ions. The outer part of the screening layer is usually called the diffuse layer.The inner shell of charge and the outer ionic atmosphere is called electrical double layer.

8

http://en.wikipedia.org/wiki/Diffuse_layer

Structure of the colloid:{m Fe(OH)3 n Fe3+; (3n-x) Cl -} + x Cl –

Electrokinetic potential can be determined from Smoluchowski’s equation:

E

vk

=

where:k – factor connected with particle shapeη – viscosity of the dispersing mediumv – molecule velocity in an electric fieldE – potential of an electric fieldε – dielectric constant of the dispersing medium

Electrokinetic potential is an important parameter characterizing colloidal dispersion. When the absolute value of zeta potential is above 50 mV the dispersions are very stable due to mutual electrostatic repulsion and when the zeta potential is close to zero the coagulation (formation of larger assemblies of particles) is very fast and this causes a fast sedimentation.

Experiment

Materials: beaker of 250 ml, glass pipette of 10 ml, 2% FeCl3, 0.1 M HCl

Procedure1. Pour 100 ml distilled water into beaker and heat it until it boils. Add 10 ml of 2%

FeCl3 solution using glass pipette. Cool the colloid in cold water.2. Prepare 10 ml of water acidified by few drops of 0.1 M HCl.3. Pour 50 ml of the colloid to the container with a tap locked. Pour acidified water into

one of the arm of U-pipe. Open the tap slowly and wait until the level of the electrolyte raises to the marked height. Place electrodes and observe the movement ofthe colloid.

Fe(OH)

3

Fe3+

Fe3+

Fe3+

Fe3+

ξ

positively chargedStern layer

negatively charged diffuse layer

Cl-Cl-

Cl-

Cl-

Cl-

pote

ntia

l

distance from the micelle nucleus

φ

Fig.1. Changes of electric potential in the colloidal particle

9

4. Apply voltage of 200 V to the electrodes and observe the movement of the colloid.5. Turn off an electric field after 15 minutes and replace electrodes from the electrolyte.

Determine the charge of the colloid.

References:1. Atkins P., Julio de Paula: Elements of Physical Chemistry. Oxford University Press

Inc., New York 2005.2. Uchman G., Hermann T.W.: Ćwiczenia laboratoryjne z chemii fizycznej dla

studentów farmacji i analityki medycznej. Akademia Medyczna , Poznań 2002.3. Jeppsson J.O., Laurell C.B., Franzen B.: Agarose Gel Electrophoresis. Clin. Chem.

1979, 25, 629-638.

10

Report 1

Name and surname:....................................................... Date:................................

Experiment 1a

A. Separation of chosen proteins by gel electrophoresis

Aim of the experiment

……………………………………………………………………………….……………..….

…………………………………………………………………………………………………

Results

1. Percentage content of the protein fractions in plasma

Fraction Obtained value [%] Proper value [%]

Albumin

α1 – globulins

α2 – globulins

β – globulins

γ – globulins

2. Migration distances of protein fractions

Fraction Distance [pixels] Distance [m]

Albumin

α1 – globulins

α2 – globulins

β – globulins

γ – globulins

amidoblack

3. Electrophoretic mobility () vs. molecular weight (Mw)

Fraction Log(Mw) Electrophoretic

mobility [m2/V·s]Albumin

α1 – globulins

α2 – globulins

β – globulins

γ – globulins

11

4. Parameters of the calibration curve =f(logMw)a =b = r = Equation of the calibration curve: = …………………………………… Theoretical points for the graph =f(logMw):x1 = …………… y1 = ……………… x2 = …………… y2 = ………………

5. Molecular weight of 1- and 2-globulins calculated from the calibration curve:

Mw(1) =

Mw(2) =

Conclusion……………………………………………………………………………

…………………………………………………………………………………………

…………………………………………………………………………………………

…………………………………………………………………………………………

Experiment 1bB. Determination of the charge of colloidal particles

Purpose…………………………………………………………………………………

…………………………………………………………………………………………

…………………………………………………………………………………………

Conclusion…………………………………………………………………………………………

…………………………………………………………………………………………

…………………………………………………………………………………………

…………………………………………………………………………………………

…………………………………………………………………………………………

…………………………………………………………………………………………

12

Thermodynamics of Phase Transitions

Experiment 2. Determination of change of enthalpy of vaporization of a volatile liquid.

AimThe aim of the class is practical knowledge of thermodynamic functions. Using of Clausius Clapeyron’s equation to calculate enthalpy (vapH) of a volatile liquid.

Required knowledgeThermodynamic systems; definitions of energy, work (W) and heat (Q); expansion work andreversible, isothermal expansion work, thermodynamic functions, internal energy (U) and enthalpy (H), entropy (S), Gibs’ free energy (G), thermodynamics of transitions, phase equilibria of pure substances, Clausius-Clapeyron equation.

IntroductionSome physical processes like boiling, freezing, sublimation or the conversion of graphite to diamond are all of phase transitions or changes of phase without changes of physical composition. Many phase changes are common everyday phenomena and their description isimportant part of physical chemistry. They occur whenever a liquid changes into solid or liquid into solid, like in case of freezing of water or melting of ice or a liquid changes into a vapour, as in the vaporization of water in lungs. Translational energy of motion (kinetic energy) is not distributed evenly among molecules. Some of the molecules have more energy and hence higher velocities than others at any moment. When a liquid is placed in an evacuated container at constant temperature, the molecules with the highest energies break away from the surface of the liquid and pass into the gaseous state and some of the molecules subsequently returns to the liquid state or condense. When the rate of condensation equals of the rate of vaporization at a definite temperature, the vapour becomes saturated and a dynamic equilibrium is established. The pressure of the saturated vapour above the liquid is then known as the equilibrium vapour pressure. As the temperature of the volatile liquid is elevated, more molecules approach the velocity necessary for escape and pass into gaseous state. As result, the vapour pressure increases with rinsing temperature, as shown in Figure 1. At any point on one of the curves represents a condition in which the liquid and the vapour exist together in equilibrium. As observed in the diagram, if the temperature of any liquids is increased while the pressure is held constant or if the pressure is decreased while temperature is held constant, all the liquid will pass into vapour state.Gibbs energy, G=H-TS, of a substance is centre stage of deliberations on thermodynamics oftransitions. Thermodynamics provides us with a way of predicting the location of the phase boundaries. Suppose two phases are in equilibrium at a given pressure and temperature. Then, if we change the pressure, we must adjust the temperature to a different value to ensure that the two phases remain in equilibrium. In other words, there must be a relation between the changes in pressure, p, that we exert and the change in temperature T, we must make to ensure that two phases remain in equilibrium. The relation between the changes of vapour pressure and changes of absolute temperature of a liquid is expressed by Clausius-Clapeyron equation.

13

Clausius-Clapeyron equation: Heat of vaporization.The relationship between the vapor pressure and the absolute temperature of liquid is expressed by the Clausius - Clapeyron equation:

2vap

RT

H

dT

plnd or dTx

RT

HΔdlnp

2

vap or for larger changes Tx RT

HΔlnp

2

vap ;

the equation shows that as the temperature of a liquid is raised (T>0), its vapour pressure increases (an increase in the logarithm of p, lnp>0, implies that p increases).

The equation can be rearranged to form to obtain the explicit expression for the ln vapour at

any temperature:

'

vap,

T

1

T

1

R

HΔlnplnp

Derivation:

The CC equation is rearranged to the form: dTx RT

HΔdlnp

2

vap and integrate both sides. If

the vapour pressure is p at temperature T and p’ at temperature T’, we can write the integration as:

dT RT

HΔdlnp

lnp'

lnp

T

T 2

vap x

The integral on the left evaluates to lnp’-lnp, which simplifies to ln(p’/p). To evaluate the integral right, we suppose that the enthalpy of vaporization is constant over the temperature range of interest, so together with R it can be taken outside the integral:

'

'

2

111'ln

TTR

HdT

TR

H

p

p vapT

T

vap

To obtain this result we have used the standard integral

tconsxx

dxtan

12

'

vap,

T

1

T

1

R

HΔlnplnp - the equation lets us calculate the vapour pressure at one

temperature provided we know it at another temperature. The equation tells us that, for givenchange in temperature, the larger the enthalpy of vaporization, the greater the change in vapour pressure. The vapour pressure of water for instance responds more sharply to changesof temperature than that of benzene doesThe last equation can be rearranged into linear form:

TBApln or

T

1BAlnp

where A and B are constants of the linear regression equation. The value of A depends on the units adopted for pEnthalpy of vaporisation can be calculated from regression coefficient B (slope), because

RBHR

HB vap

vap

References

14

1. Atkins P., de Paula J.: Elements of Physical Chemistry (4thEdition) W.H. Freeman andCompany, Oxford Iniversity Press Inc, 2005.

2. Martin A. Physical Pharmacy. Physical Chemical Principles in The PharmaceuticalSciences (4thEdition). Lippincott Williams & Wilkins, Philadelphia 2006.

3. Uchman G., Hermann T.W.: Ćwiczenia laboratoryjne z chemii fizycznej dla studentówfarmacji i analityki medycznej. Akademia Medyczna , Poznań 2002.

Experimental partEquipmentDept. Physical Pharmacy and Pharmacokinetics computer program, barometer, calculation software

Software manual

Turn on the computer and open icon “parowa”. Press the key F5, to display information: Determination of change of enthalpy of vaporisation of a volatile liquid. Computer simulation.

Displayed information: “If you are prepared to perform the laboratory press the key T”. Yourdecision?.

Write down letter T, to accept it press Enter. You will see a table with names of some volatile liquids.

Information displayed on monitor: Choose the volatile liquid by writing its number. For example: write down 10, and press Enter. You receive information: Number 10 represents ethyl acetate with the melting point –83.5 oC and the boiling point 77.1 oC, thesubstance was chosen to perform the experiment). Enter.

Pay attention ( pink colour): The method requires that the boiling point of a substance is not overtaken. Recommendation. Write down the number of temperatures in centigrades (oC)at which the vapour pressure will be measured. (take 3 - optimal number of temperatures). Press Enter.

Read and write down the value of atmosphere pressure in Pa (Pascals, 1 Pa= Nm-2) from a barometer (! Barometer is placed in the second class room. 1hPa=100 Pa). For instance: 99000 Pa. Enter. On the monitor will be displayed scheme of instrument for the measurement of vapour pressure and question: (Fill? yes/no). Write down Y (yes) and press Enter.

Write value of a temperature in centigrades, ie. 20 oC, and press Enter.Write down the number of measurements of vapour pressure: 3. and press Enter.When value of pressure will be stable (you will see windows of the cursor), note it and close

the upper valve. Close it ? yes/no. Press Y (yes).

When value of pressure will be stable (you will see windows of the cursor), note it and open the upper valve. Open it ? Y/N (yes/no). Press Y and press Enter}. In the lower part of monitor will be displayed information in blue colour. Write down value of difference between pressure in Pa, when valve was open and close. Repeat it for temperatures 40 and 60 o C or others, chosen by student remembering about information given above, at point 5.

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Report 2

Name and surname:............................................................... Date:.................................

Determination of change of enthalpy of vaporization of a volatile liquid.


……………………………………………………………………………….……………..….

…………………………………………………………………………………………………

Results

Temperature: 20 oC (293K)

Pressure [Pa] (close valve) Pressure [Pa] (open valve) Difference of pressure p [Pa]

p1 = p1’= p=

p2 = p2’= p=

p3 = p3’ = p=

Mean difference pmean =

Temperature 40 oC (313K)

p1 = p1’ = p=

P2 = P2’ = p=

P3 = P3’ = p=

Mean Difference pmean=

16

Temperature 60 oC (333K)

p1 = ……………………… p1’ = p=

p2 = p2’= p=

p3 = p3’ = p=

Mean difference pmean=

The least square method will be used to calculate linear regression parameters A (point of intercept with 0Y) and B (slope) of Clausius-Clapeyron equation:

TR

Hpp vap 1

lnln ,

For the purpose become to:a) Calculate value 1/T, where T [K]=273+T[oC]. It represents value of x in the linear

equation.b) Calculate mean value of difference of pressure (pmean) for each temperature.c) Calculate value of natural logarithm of mean pressure of vapor (lnpmean). It represents y in

linear regression.d) Calculate value of product 1/Tlnpmean. (It represents product x y).e) . Calculate value of (1/T2). (It repesents x2).f) Calculate value (lnp)2. It repesents y2).g) Calculate sum of particular parameters: x; y; xy; x2; y2

h) The equations to calculate slope (a), intercept (b) and correlation coefficient (r) are:

n

T

T

n

pTp

Ta2

2

1

)1

(

ln1

ln1

; nT

an

pb

1

ln

n

pp

n

T

T

n

pTp

Tr

22

2

2 ln)(ln

1

)1

(

ln1

ln1

i) After analysis of the instruction you can start to calculate above parameters, for this purpose press discretional key.

j) Fill the displayed table. In case when you want to write very small values, for example: 0.00001164845 you can write 1.164845e-5. Expression e-5 is equal to 110-5.

17

x y x y x2 y2

Temp. [oC]

T [K] = 273+T[oC]

1/T[K] pmean [Pa].

lnpmean 1/Tlnpmean (1/T)2 (lnp)2

16 Calculate of Hvap with error.

Slope a=

S

:equation from calculated becan a slope oferror standard The

a. slope oferror standard represents S whereR;S and

Ra- ;

a

aa

H

HthenR

Ha

vap

vapvap

17 The plot lnp=f(1/T) should be drawn on a semilogarithmic paper sheet .

Summary:

vapH ( vapH )= ................................................[ ]

18

Semi logarithmic paper

19

PHASE EQUILIBRIA.

Experiment 3. Temperature-composition diagram for the systems consisting of water and

phenol

The purpose of the experiment is to examine the phase equilibrium for the systems consisting ofwater and phenol. The several mixtures of water and phenol will be prepared and the criticaltemperature will be evaluated then the temperature-composition diagram will be plotted and theupper critical temperature of the system estimated.

Required knowledge: The phase rule, phase diagrams, Raoult’s law, Henry’s law, thedistillation of mixtures, miscible, partially miscible and immiscible liquids, temperature-composition diagrams, lever rule

Introduction

The phase rule. A phase of a substance is a form of matter that is uniform throughout in chemicalcomposition and physical state. Therefore we speak of solid, liquid and gas phases of a substance,and of its various solid phases (allotropes). The number of phases in a system is denoted P. Anexamples of a single phase are a gas, gas mixture or two totally miscible liquids (P = 1). A slurry ofice and water as well as an alloy of two immiscible metals are a two-phase system (P = 2). By acomponent we mean a chemically independent constituent of a system. The number of components,C, in a system is the minimum number of independent species needed to identify the composition ofall the phases present in the system. For example, pure water is a one-component system (C = 1), forthe reason that we need only the species H2O to specify its composition. A mixture of ethanol andwater is a two-component system (C = 2); to specify its composition the species of H2O and C2H5OHare needed.

The number of intensive variables that can be changed independently without disturbing thenumber of phases in equilibrium is known as the variance and is denoted F. In a single-component,single-phase system (C = 1, P = 1), the pressure and temperature may be changed independentlywithout changing the number of phases, therefore F = 2. Such system is called bivariant, or that ithas two degrees of freedom. However if a liquid andits vapor are in equilibrium (two phases) in a single-component system (C = 1, P = 2), the temperature (orpressure) can be changed at will but the change intemperature (or pressure) have to be parallel with thechange in pressure (or temperature) to preserve thenumber of phases in equilibrium. Thus the variance ofthe system has fallen to 1 (F = 1).

The phase rule, deduced by J.W. Gibbs,demonstrates the relation between the degree offreedom, F, the number of components, C, and thenumber of phases at equilibrium, P, for a system ofany composition: F = C – P + 2

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Vaporwater

CA

T3 T

2 T

1

Temperature

Vap

or p

ress

ure

P1

P2

Fig. 3.1. Phase diagram for water at moderate pressure

Systems containing one component

Phase diagram for a one component system, such as pure water is illustrated in Figure 3.1. Thecurve OA in the pressure-temperature diagram is known as the vapor pressure curve. Its upper limitis at the critical temperature, 374 C , and its lower end terminates at 0.0098 C, called triple point.Along the vapor pressure curve, vapor and liquid coexist in equilibrium. Curve OB is thesublimation curve, and here the vapor and solid are in equilibrium. Curve OC is the melting pointcurve, at which liquid and solid exist together in equilibrium. Since the slope of OC curve isnegative the freezing point of water decreases with increasing external pressure. If the temperature isheld constant at T1, where water is in the gaseous state above the critical temperature, no matter howmuch the pressure is raised, the system remains as a gas. At a temperature T2 below the criticaltemperature, water vapor is converted into liquid water by an increase of pressure because thecompression brings the molecules within the range of the attractive intermolecular forces. At atemperature below the triple point, T3, an increase of pressure on water in the vapor state convertsthe vapor first to ice and then at higher pressure into liquid water. At the triple point, all three phasesare in equilibrium (F = 0, T = 0.0098 C, p = 0.61 kPa). In any of the three regions in which puresolid, liquid or vapor exist and P = 1, the phase rule gives F = 2 (F = 1 – 1 + 2 = 2). Therefore wemust fix two conditions (temperature and pressure) to describe the system completely. Along anythree of the curves where two phases exist in equilibrium, F = 1. Hence only one condition need begiven to define the system.

Systems containing two components – partially miscible liquids

It is known from experience that ethyl alcohol and water are miscible in all proportions, whereaswater and mercury are completely immiscible. Between these two extremes lies a whole range ofsystems that exhibit partial miscibility. Partially miscible liquids do not mix in all proportions at alltemperatures. Examples for such systems are phenol and water or hexane and nitrobenzene.Temperature-composition diagram for the system consisting of two partially miscible liquids A andB is shown on Figure 3.2. The curve demonstrates the limits of temperature and concentration withintwo liquid phases exist in equilibrium. The region outside this curve contains systems having but oneliquid phase. Suppose a small amount of liquidB is added to a sample of liquid A attemperature T1. It dissolves completely, and thebinary system remains a single phase. Whenmore B is added, a stage comes (I) at which nomore dissolves. The sample now consists of twophases in equilibrium with each other (P = 2),the most abundant one consisting of A saturatedwith B, the minor one a trace of B with A. Thecompositions of the A- and B-rich phase aredenoted by the points I and II on the diagram,consequently. The relative abundances of thetwo phases are given by the lever rule. Whenmore B is added, A dissolves in it slightly. Thecomposition of the two phases in equilibriumremain I and II because P = 2 implies that F’ =0, and hence that the compositions of the phasesare invariant at a fixed temperature andpressure. However the amount of one phaseincreases at the expense of the other. Finally astage is reached when so much B is added that it

T1

Tem

pera

ture

, T

Mole fraction of B, XB

P = 1A B

Tuc

P = 2

Fig. 3.2. The temperature-composition diagram for two partially miscible liquids A and B

I II

21

can dissolve all the A, and the system reverts to a single phase. The composition of the two phases atequilibrium varies with the temperature; for phenol and water, raising the temperature increases theirmiscibility. One can construct the entire phase diagram by repeating the observations at differenttemperatures and drawing the envelope of the two-phase region.

Critical solution temperatures

The upper critical solution temperature, Tuc, is the highest temperature at which phase separationis observed. Above Tuc the two partially miscible liquids are fully miscible. This temperature occursfor the reason that thermal motion overcomes a potential energy advantage in molecules of one typebeing close together. In case of the phenol-water system, the upper critical solution temperature is66.8C (Fig. 2). All combinations of phenol and water above this temperature give one-phase liquidsystem.

There are systems that show a lower critical solution temperature, Tlc, below which they mix inall proportions and above which they form two phases. An example is water and triethylamine. Suchliquids at low temperatures the two components are more miscible because they form a weakcomplex; at higher temperatures the complexes break up and the two components are less miscible.

Some systems have both upper and lower critical solution temperatures. They exist because, afterthe weak complexes have been broken up, leading to partial miscibility, the thermal motion at highertemperatures homogenizes the mixtures again as in the case of ordinary partially miscible liquids.An example of such behavior is the system consisting of nicotine and water.

References

4. Atkins P., de Paula J.: Physical chemistry. W.H. Freeman and Company, New York 2002.

5. Sinko P.J. (Ed): Martin’s physical pharmacy and pharmaceutical sciences. Lippincott Williams& Wilkins, Philadelphia 2006.

6. Hermann T.W. (Ed): Farmacja fizyczna. PZWL, Warszawa 1999.

7. Uchman G., Hermann T.W.: Ćwiczenia laboratoryjne z chemii fizycznej dla studentów farmacji ianalityki medycznej. Akademia Medyczna , Poznań 2002.

Experimental

The virtual software is used during the experiment. In order to construct the phase diagram onemust analyze at least five compositions of phenol and water. Phenol concentration in the systemsshould be between 10 and 90 %. There are examples of five mixture compositions in the tablebelow.

Composition # 1 2 3 4 5 6 7

Mass of phenol [g] 5 5 5 5 5 5 5

Mass of water [mL] 3 5 7 10 15 20 27

The following equations are needed to calculate the concentration, ΔCp, and solution, Δsolution,errors:

%1002

waterphenol

phenol

waterphenol

phenol

mm

msolution

mm

mCp

22

Δsolution = Δmphenol + Δmwater,

Δmphenol = 0.1g, accuracy of phenol measurement

Δmwater = 0.05 mL –accuracy of water measurement

Each composition is virtually heated up and cooled down so the temperature of phasehomogenization and phase separation may be observed and written down in the lab report. One mustcalculate the mean temperature of 4 measurements and the standard deviation, SD. Then draw thetemperature-composition diagram for phenol and water and read the upper critical solutiontemperature, Tuc. Utilize obtained diagram in order to solve the numerical problem presented below.

A mixture of .... g of phenol and .... g of water was prepared at the temperature of .... ºC.Calculate the mass of water-rich phase (the upper one), and the mass of phenol-rich phase (thelower one) and the content of phenol in each phases.

The following equations and hints may be helpful in the calculations:

The mass of the mixture, mu

mu = mf + mw

Using the diagram find:

the concentration of phenol in water-rich phase, Cp1

The concentration of phenol In phenol-rich phase, Cp2

Calculate a mass of water-rich phase (the upper one), mww, using the following equation:

12

2 %100

pp

fupww

CC

mmCm

mww – a mass of water-rich phase

Cp1 – concentration of phenol in water-rich phase

Cp2 – concentration of phenol in phenol-rich phase

mf – mass of phenol

mu – mass of the mixture

In order to estimate:

o a mass of phenol-rich phase, mwf

o mass of phenol in water-rich phase, mfw

o mass of phenol in phenol-rich phase, mff

utilize the formulas listed below:

%1001 ww

fwp

m

mC

mu = mww + mwf

mf = mfw + mff

Write down all the results in the table and compare with those given by the calculation software.

23

Software manual

1. The title of the experiment: “Determination of the critical solution temperature for the mixture ofphenol and water”

2. If you are ready, press „y”

3. Procedure:

The purpose of the experiment is to prepare a few systems of phenol in water, and observe thephase transformation while the system is heated or cooled. For each composition the temperature ofphase transformation is estimated. Two-phases and one-phase systems are formed when the mixtureis cooled down and heated up, subsequently.

Would you like to see examples of phase transformations (y/n)?

4. The example of phase transformation during heating

a) muddy two-phase solution

b) transformation into one-phase solution

c) clear one-phase solution

The example of phase transformation during the temperature lowering

a) clear one-phase solution

b) transformation into two-phase solution

c) muddy two-phase solution

5. Type the number of prepared systems.

The concentration of phenol in water should be in the range of 10-90%.

6. The composition of the first solution.

Mass of phenol [g]

Phenol measurement accuracy [g]

Volume of water [ml]

Water measurement accuracy [ml]

The concentration of the solution [%]

Concentration error [%]

Cp = (mphenol/(mphenol + mwater)) · 100%

mphenol – mass of phenol, mwater – mass of water

%1002

waterphenol

phenol

waterphenol

phenol

mm

msolution

mm

mCp

Δsolution = Δmphenol + Δmwater,

Δmphenol, Δmwater – accuracy of phenol and water measurements

24

7. How many measurements of phase transition do you want to perform?

If you want to heat, write „h” and press ENTER

If you want to cool, write “c” and press ENTER.

From what temperature do you want to observe the phase transformation?

Press any key, when you notice the transformation.

8. Calculate a mean value of the observed phase transition temperature.

Calculate SD (standard deviation) for the measurements.

9. Repeat steps from 6 to 8 for next compositions.

10. The temperature-composition diagram for phenol and water appears on the screen

Choose the option:

o – calculation

w – diagram

k – exit

11. A mixture of ….. g of phenol and ..... g of water was prepared at the temperature of …ºC.Calculate the mass of water-rich phase (the upper one), mww, and the mass of phenol-rich phase(the lower one), mwf, and the content of phenol in each phases, mff, mfw.

Draw the temperature-composition diagram for phenol and water. Read the phenol content ineach phase.

1. Calculate the following values:

a) The mass of the mixture - mu

b) Read from the diagram the phase transformation temperatures, Cp1 and Cp2. c) calculate a mass of water-rich phase,mww, using following equation:

12

2 %100

pp

fupww

CC

mmCm

where

mww – a mass of water-rich phase

Cp1 – concentration of phenol in water-rich phase

Cp2 – concentration of phenol in phenol-rich phase

mf – mass of phenol

mu – mass of mixture

25

d) calculate a mass of phenol-rich phase, mwf

e) calculate a mass of phenol in each phase, mff and mfw

f) write your results into a lab report

Concentration of phenol in water-rich phase (Cp1)

Concentration of phenol in phenol-rich phase (Cp2)

A mass of water-rich phase

12

2 %100

pp

fupww

CC

mmCm

mu= mf + mw

Calculate a mass of phenol in water-rich phase using following equation:

%1001 ww

fwp

m

mC

Calculate a mass of phenol-rich phase using following equation:

mu = mww + mwf

Calculate a mass of phenol in phenol-rich phase from the following equation:

mf = mfw + mff

12. Compare your results with these calculated by software

A solution consisted of ….. g of phenol and ….. g of water at …..0C contains ….. g of water-rich phase (….. % consisting ….. g of phenol) and ….. g of phenol-rich phase (….. %consisting ….. g of phenol).

26

Report 3

Name and surname:.................................................................... Date:....................................

Temperature-composition diagram for the systems consisting of water and phenol

Aim of the experiment ……………………………………………………………………………….……………..….…..………………………………………………………………………………………………......……………………………………………………………………………………………….………………………………………………

Results

1. Mass of phenol and water (mf, mw), concentration of mixtures Cp, and concentration errors, ΔCp

# mf [g] mw [mL] Cp [%] ΔCp

1

2

3

4

5

6

7

2. Temperatures of phase separation, T [C], for each composition

# T1 T2 T3 T4 Tav SD

1

2

3

4

5

6

7

27

3. Using the average temperatures of phase separation draw the temperature-composition diagram on the graph paper. Estimate each point with the concentration error, ΔCp, and standard deviation, SD, of the phase separation temperature.

4. The upper critical solution temperature (read from the diagram), Tuc, of phenol-water mixture.

Tuc = ………….

5. The numerical problem.

Subject Your resultsResults calculated by the

software

Concentration of phenol inwater-rich phase [%]

Concentration of phenol inphenol-rich phase [%]

A mass of water-rich phase[g]

A mass of phenol in water-rich phase [g]

A mass of phenol-rich phase[g]

A mass of phenol in phenol-rich phase [g]

6. Summary

28

THERMODYNAMICS OF MIXTURES. COLLIGATIVE PROPERTIES

Experiment 4. Determination of molecular mass and van’t Hoff factor i using method of

determination of osmotic pressure

Aim Determination of molecular mass of unknown nonionizable substance. Calculation of van’t Hoff factor i and degree of dissociation α of NaCl and Na2SO4 diluted solutions.

Required knowledgeColligative properties, phenomenon of osmosis, thermodynamic treatment of osmosis, chemical potential and van’t Hoff equation

Introduction

Colligative properties

When a non-volatile solute is combined with a volatile solvent, solely the solvent provides the vapour above the solution. We can therefore expect a solute to modify the physical properties of the solution. The solute reduces the escaping tendency of the solvent and, on the basis of Raoult’s law the vapour pressure of a solution containing a non-volatile solute is lowered proportional to the relative number of the solute molecules, rather than the weigh concentration. Apart from lowering the vapour pressure of the solvent, a non-volatile solute has three main effects:

1-it raises the boiling point of a solution 2- it lowers the freezing point3-it gives rise to an osmotic pressure

Because these properties all stem from changes in the disorder of the solvent (ΔS>0) and the increase in disorder is independent of the identity of the species, all of them depend only on the number of solute particles present, not their chemical identity. For this reason the properties are called colligative properties (from Greek: collected together). Thus, a 0.01 mol kg -1 aqueous solution of any non-electrolyte should have the same boiling point, freezing point, and osmotic pressure.

Lowering of the freezing point

Fig.4.1. Depression of the freezing point of the water by solute

29

If a non-volatile solute is dissolved in a liquid solvent at triple point, a vapor pressure of the liquid solvent is lowered below that of the pure solid solvent. The temperature has to decrease to establish new equilibrium state between the liquid and the solid. For that reason, the freezing point of the solution is always lower than that of pure solvent. The lowering o f the freezing point Tf is proportional to the molal concentration m of the solute.It is expressed by equation:

Tf = Kf m (1) where m=w/M ( w- mass of solute, M-molecular mass of solute)

or

21

2ff Mw

1000wKΔT (2)

Tf – depression of freezing point, Kf – molal depression constant or cryoscopic constant of the solvent Table 1.

Table 1. Cryoscopic constants

Solvent Cryoscopic constant Kf (K kg mol-1)

Acetic acid 3.90Benzene 5.12Camphor 40.0Carbon disulfide 3.8Naphtalene 6.94Phenol 7.27Tetrachloromethane 30.0Water 1.86

The equation (2) is useful form of the (1), to calculate the freezing point in term molality in mol/kg; m=w/M=ni-number of moles, but molality is expressed in mol/kg (see below)

Molality. The molality of a solution mi is defined as the amount of a constituent ni divided by the

mass of the solvent msolvent (not the mass of the solution):solventm

in

im

The SI-unit for molality is mol/kg.

Molar concentration. The molar concentration ci is defined as the amount ci of a constituent ni divided by the volume of the mixture V:

Vin

ic

The SI-unit is mol/m3. However, more commonly the unit mol/L is used.

30

To understand better lowering freezing point following example is given to solve.

Example: Calculate freezing point of a solution consisted of 1,8 g of glucose C6H12O6 and 250 g of water? Molar mass of glucose is 180 g/mol. Cryoscopic constant Kf of water in the solution is approximately equal 1.86 K kg mol-1.Solution:

21Mw2

1000w

fKfΔT

180 x 250

1.81000x x1.86fΔT

Tf = 0.074 oC

The freezing point of the aqueous solution of glucose is -0.074 oC.

Molecular mass determination from depression of the freezing point

Molecular mass of non-volatile solutes can be determined from depression of the freezing point of the solvent by rearranging of equation (2) to obtain, if :

1

wfΔT1000w

KM 2f2

Where w2 represents amounts of grams of solute (i.e. glucose, urea) dissolved in w1 grams of solventdiluted in w2 grams of solvent. If we have cryoscopic data molecular mass can be calculated.

Example. Using cryoscope 0.2 oC lowering freezing point of unknown substance was received for solution consisting of 6 g substance (w2) and 1000 g of solvent. Calculate molecular mass of the unknown substance.

1

2f2 wfΔT

w1000KM = 55.8g/mole

1000 x 0.21000x61.86xM2

To receive more accurate molecular mass the measurements of depressing of freezing point should be also more precise because if we will take 0.2 oC instead of 0.186 0.2 the molecular mass will be 55.8 instead 60 g/mol.

Osmosis

The phenomenon of osmosis is the passage of a pure solvent into a separated from it by a semipermeable membrane. A semipermeable membrane is a membrane that is permeable to the solvent but not the solute. The osmotic pressure, (uppercase pi), is the pressure that must be applied to the solution to stop the inward flow of solvent.

31

Fig. 4.2. Pressure opposing the passage of solvent into the solution arises from the hydrostatic pressure of the column of solution that the osmosis itself produces. The osmotic pressure of a solution is proportional top the concentration of solute.

Fig. 4.3. Influence of an osmotic pressure on red blood cells

Thermodynamic explanation of osmotic pressure - van’t Hoff equation

van’t Hoff equation, (bears an uncanny resemblance to the expression for the pressure of a perfect gas):

RTV

nΠ RTnΠV B

B

32

because BV

nB , the molar concentration of the solute, a simpler form of this equation is

= [B] RT

This equation applies only to solutions that are sufficiently dilute to behave as ideal - dilute solutions.

Thermodynamic explanation of osmotic pressure. Derivation of the van’t Hoff equation

The thermodynamic treatment of osmosis makes use of the fact that, at equilibrium, the chemical potential of the solvent A (A) is the same on each side of the membrane (Fig. 4.4).

The starting relation is therefore

A (pure solvent at pressure p) = A (solvent in the solution at pressure p + )

Fig. 4.4. Thermodynamic base of osmotic pressure.

The pure solvent is at a pressure, p and the solution is at a pressure p+ on account of additional pressure, , that has to be exerted on the solution to establish equilibrium. We shall write the chemical potential of the pure solvent at the pressure p as (p)*μA .The chemical potential of the solvent in the solution is lowered by the solute, but it is raised on

account of the greater pressure p+ acting on the solution. We denote this chemical potential by A

(xA, p+).

Our task is to find the extra pressure needed to balance the lowering of chemical potential caused by the solute. The condition for equilibrium is

1) Π)p;A(xAμ(p)*Aμ

we must take the effect of the solute into account by using chemical potential equation:

ARTlnx*AμAμ - chemical potential of a solvent A present in solution at mole fraction xA;

*Aμ is the chemical potential o pure solvent A

Chemical potential * of solvent in solution at pressure p+

Chemical potential

33

2) ARTlnxΠ)(p*AμΠ)p,A(xAμ

The effect of pressure on an (assumed incompressible) liquid is given by equation:

ΔGA=VA Δp dp)VG( Πpp A

(where VA is molar volume of pure solvent A), but now expressed in terms of the chemical potential and the partial molar volume of the solvent :

3) Πpp A

*A

*A dpV(p)μΠ)(pμ

At this point we identify the difference in pressure p as .

)ΠAVp)Π(pAVdpΠppA(V

When the last three equations:

1) Π)p ;A(xAμ(p)*Aμ

2) ARTlnxΠ)(p*AμΠ)p,A(xAμ

3) Πp

p dpAV(p)*AμΠ)(p*

Aμ

are combined, we get:

(1-2) ARTlnxΠ)(p*Aμ(p)*

Aμ

3) Πp

p dpAV(p)*AμΠ)(p*

Aμ

: ARTlnxΠp

p dpAV(p)*Aμ(p)*

Aμ

assumption: the VA= (molar volume of solvent) in small range of pressure is constants, then

ARTlnxΠAV(p)*Aμ(p)*

Aμ and therefore

-RTlnxA= VA

The mole fraction of the solvent A, expressed as xA is equal 1-xB, where xB is the mole fraction of solute molecules. In dilute solution, ln(1-xB) is approximately equal to –xB,

If x 1, then the terms involving x raised to power greater than 1 are much smaller than x, so ln(1-x)=-x. For example, ln(1-0.050) = ln0.95 = -0.051, which is close to - 0.05.

so this equation becomes RTxB= VA

34

When the solution is dilute, xB=nB/n= nB/nA.:RTnB/nA= VA

Moreover, because nAVA=V, the total volume of the solution, this equation becomes

RTVBn

Π ΠV BRTn

Because nB/V=[B], the molar concentration of the solute, a simpler form of this equation is

= [B] RT

Derive of osmotic pressure SI units of molecular mass

= [B]RT

Pa2m

N3m

m N3m

JK Kmol

J 3m

molΠ

Molecular mass calculation

The above equation is used to calculate molecular mass of tested non-ionisable substance:

= [B]RT, this equation applies only to solutions that are sufficiently dilute to behave as ideal –dilute solutions

Where nB/V=[B], (the molar concentration of the solute B) and nB=mB/MB or generally n=m/M , then

RT V1

MmRT

V

nBRTΠ

])/(

//

33[

⇒RTV M

mΠ

KmmJ

KmolJgmolg

V ΠTR m

M

Where:M - molecular mass M - mass of non-ionisable substance used in the experimentR - gas constantT - temperature in KelvinsΠ - pressure in Pascals V - volume of solvent (≈solution)

van’t Hoff factor i

theor

obserΠ

Π=i

Where:Πobser – measured pressure in Pa

35

Πtheor - calculated from equation =BRT (B-concentration of analite in mol/m3)

a degree of dissociation α

1-n1 -i

=α

Where: i- van’t Hoff factor , and n is number of ions per molecule

References

1. Atkins P., de Paula J.: Physical chemistry. Ed.W.H. Freeman and Company, New York 2005.

2. Martin A.: Physical Pharmacy. Ed. Lippincott Williams & Wilkins, Philadelphia 2004.

3. Uchman G., Hermann T.W.: Ćwiczenia laboratoryjne z chemii fizycznej dla studentów farmacjii analityki medycznej. Akademia Medyczna, Poznań 2002.

Experimental part

Apparatus: Osmometer “Krioskop 800cl”

Solutions: 0.1 mol/l glucose; 0.9%NaCl and 0.1 mol/l Na2SO4 and tested nonionizable substance. Solution of tested substance has been prepared by dissolution of 6 g the substance in 1000g H2O.

Laboratory procedure of measurement of osmotic pressure (mOsm/kg ) of solutions of 0.1 mol/l glucose; 0.9%NaCl and 0.1 mol/l Na2SO4 and tested non-ionisable substance.

1. Turn on the osmometer2. Information”- - c0” displayed on the screen of osmometer means that apparatus is ready to do

a measurement of osmotic pressure.3. Before carry out of a measurement, clean the termistor (temperature sensitive resistor) and

dry it using soft cotton - make it very carefully.4. Osmometer „Krioskop 800 cl” is fully automated, digital measuring device for determination

of osmotic pressure in easy and precise way. The lowering of freezing temperature of 0.1 M solution glucose, 0,9% NaCl, 0,1 M Na2SO4 ie. ) In the method using

5. Calibration of osmometer. For the purpose 100 μl of deionized water has to be measured to plastic vial and it place in the head of apparatus Fig.5. Then the head of apparatus with vial insert in hole of the Krioskop. Lowering of the temperature to -7oC and then grow up 0oC is observed. The information “0000” displayed on the screen means that water possesses osmotic pressure 0 mOsm/kg and osmometer is ready to measurements. Water in vial is frozen therefore be careful in operation of the apparatus. Wait some minutes to allow an ice to thaw.

36

Fig. 4.5. Sample cell for measuring of freezing point in osmometer.

When osmotic pressure of the following solutions: 0.1 M glucose, 0,9% NaCl, 0,1 M Na2SO4 or unknown substance is measured lowering temperature is observed. The lowering of the freezing point of the solutions if compare with their solvent (water, p.a.) is proportional to the molal concentration (m) of the solute. Changes of temperature of solution during determination of osmotic pressure are showed in Fig. 4.6., the lowest value represents the freezing point of the solution. Decreasing of temperature of the solution in the sample cell up to -7oC, it means several degrees below of freezing point, cause that supecooled liquid is obtained. Stirring of thewire initiates crystallization process in the solution. The temperature of the sample increases to its freezing point and heat of crystallization (Fig. 4.6.).

With Move of agitator initialize crystallization of the solution. Fig.5

Fig 4.6. Changes of temperature of solution during determination of osmotic pressure using osmometer.

37

6. Measurements of the osmotic pressure prepared solutions in the same way like water. Each measurements repeat 3 times. Results write in Table 1.

7. Calculate of van’t Hoff factor i using of equation:

theor

obserΠ

Π=i

and a degree of dissociation α of 0.9%NaCl and 0.1 mol/l Na2SO4 from equation:

1-n1 -i

=α

where n is number of ions per molecule; n = 2 – for NaCl and n = 3 for Na2SO4.

38

Report 4

Name and surname:................................................................... Date:....................................

Determination of molecular mass and van’t Hoff factor i using method of determination of osmotic pressure

Aim of the experiment ……………………………………………………………………………….

……………..….

…………………………………………………………………………………………………

Result of measurements of osmotic pressure.

Name of formulation

Osmotic pressure [mOsm/kg H2O]

Osmotic pressurein [Pa]; conversion factor: x 2270 Pa (1mOsm/kg =2270Pa)

Theoretical osmotic pressure[Pa] =BRT *

van’t Hoff factor i

theor.

0bser

Π

.Π=i

degree of dissociation

1-n

1-i=α

0.1 M/l Na2SO4

1.2.3.

Mean=0.9 % NaCl (... M/l NaCl

1.2.3.

Mean=

0.1 M/l glucose

1.

X2.3.Mean=

Tested subst. 6.0g/1000 g H2O

1.

X2.3.Mean=

7. Calculate molecular mass of analyzed substance with SD (standard deviation)

MSD= …………………………………………………………

8. Summary.

39

DISTRIBUTION OF SOLUTES BETWEEN IMMISCIBLE SOLVENTS

Experiment 5. Spectrophotometric determination of specific partition coefficient o/w for

benzoic acid

Aim

The aim of the experiment is to determine the partition coefficient o/w for benzoic acid onthe basis of measurements of the total concentration of benzoic acid in water phase in equilibriumwithout the need for oil phase analysis.

Required knowledge: definition of partition coefficient, Garrett – Woods equation forpartition coefficient determination, significance of the partition coefficient knowledge in reference tocharacteristics of drugs.

Introduction

According to Nernst partition law the ratio of concentrations of a compound in the twophases of a mixture of two immiscible solvents, e.g. oil (co) and water (cw), at equilibrium is constantand is called partition coefficient

w

o

C

C=K (4.1)

The value of constant K does not depend on concentration but depends on temperature. Thisequation is valid if the substance does not change its form in any of the two phases. If such moleculeundergoes association or dissociation then this equation still describes the equilibrium betweenconcentrations in both phases, but only for the same form (monomers, undissociated molecules).Concentrations of all remaining forms must be calculated by taking into account all the otherequilibria. The partition coefficient is usually given in the form of its logarithm to base ten (logK).

When an ionizable compound is equilibrated in a two-phase system at a pH at which it ispartially ionized, its concentration in the organic phase can be determined by distribution coefficient.The distribution coefficient (D) is defined as the ratio of the concentration of un-ionized compoundin the lipid phase (co) to the concentration of all species in the aqueous phase (cw) at given pH.

ionized-unw

ionized

ionized-uno

c+c

c=logD

w

log (4.2)

The distribution coefficient is pH dependent, therefore the pH at which the log D was measuredshould be specified. Of particular interest is the log D at pH = 7.4 (the physiological pH of bloodserum). For un-ionizable compounds, log P = log D at any pH.

Benzoic acid in the form of sodium salt is often used for emulsion preservation at aconcentration of 0.84 g/l. Its antibacterial properties depend on the concentration of undissociatedform in water phase, because bacterial growth in oil phase is not observed. The most effectivepreservation concentration of the substance in oil-water system can be calculated on the basis ofdissociation constant, partition coefficient, volumes of the both phases, minimal antibacterialconcentration of the undissociated form of the substance in water phase and water phase pH.Benzoic acid undergoes dissociation depending on pH of water solutions (pKa = 4.2). Common

40

http://en.wikipedia.org/wiki/Dissociation_(chemistry)

http://en.wikipedia.org/wiki/Equilibrium

http://en.wikipedia.org/wiki/Solvent

http://en.wikipedia.org/wiki/Immiscible

http://en.wikipedia.org/wiki/Chemical_compound

http://en.wikipedia.org/wiki/Concentration

forms for benzoic acid in oil and water phases are undissociated and unassociated molecules.Partition coefficient o/w for benzoic acid is defined as the ratio of acid concentration in oil ([HA]o) toconcentration of undissociated form in water ([HA]w):

w

o

w [HA]

c=

[HA]

[HA]=K (4.3)

Total concentration of benzoic acid in water is the sum of concentrations undissociated [HA]w anddissociated [A-]w form.

c = [HA]w + [A-]w (4.4)

Therefore, the measured partition coefficient of benzoic acid is in fact an apparent partitioncoefficient K’ (or distribution coefficient), defined as:

w

o-

o

c

c=

A+HA

[HA]=K'

ww ][][(4.5)

The value of partition coefficient depends on the partition of undissociated form of benzoic acidbetween oil and water as well as on dissociation constant (Ka), described below:

w

-+

a HA]

AH=K

[

]][[(4.6)

Using this equation we obtain the following formula describing partition coefficient:

])[(][

][+

a

+o

H+KHA

H[HA]= K'

w

where K

HA]=[HA] o

w

[

][

]+

a

+

H+K

[H K=K' (4.7)

][

]+

a

+

w

o

H+K

[H K=

c

c(4.8)

][

]+

a

w+

o H+K

c[H K=c (4.9)

Total concentration of benzoic acid in water phase before partition between oil and water equals:

c = co + cw

][

])[(]+

a

w+

aw+

H+K

cH+K+cK[H=c (4.10)

From the above equation we obtain the final formula known as Garrett – Woods equation:

c

K+H

c

1+K=

c

H+K a+

w

+a ][

][(4.11)

The presented equation is used to determine the actual partition coefficient o/w for benzoic acid onthe basis of water phase pH measurements and total concentration of the substance in water phase.

41

Fig. 5.1. Diagram presenting the relation ])[][

105 +5

w

+a Hf(10=

A

H+K for benzoic acid in oil-water system, where

Aw is absorbance of benzoic acid in water phase at =229 nm, at a definite pH.

Intercept for the presenting equation equals:

c

K=b a

Using the value of Ka = 6.5 · 10-5, we could calculate c from the following equation:

b

K=c a

Partition coefficient K we obtain using the value of parameter a:

c

1+K=a

K = a · c – 1

The partition coefficient is taken into account in many chemical processes, such as extractionor chromatography. Moreover, it is an important factor in the formulation of topical ointments, dyesand many other consumer products. In the context of pharmacokinetics, the partition coefficientplays a major role in drug absorption, distribution, metabolism and excretion. Hydrophobic drugswith high partition coefficients are preferentially distributed to hydrophobic compartments such aslipid bilayers of cells while hydrophilic drugs (low partition coefficients) preferentially are found inhydrophilic compartments such as blood serum. In the context of pharmacodynamics, thehydrophobic effect is the major driving force for the binding of drugs to their receptor targets.However, hydrophobic drugs tend to be more toxic because they are retained longer, have a wider

42

http://en.wikipedia.org/wiki/Receptor_(biochemistry)

http://en.wikipedia.org/wiki/Hydrophobic_effect

http://en.wikipedia.org/wiki/Pharmacodynamics

http://en.wikipedia.org/wiki/Blood_plasma

http://en.wikipedia.org/wiki/Lipid_bilayers

http://en.wikipedia.org/wiki/Pharmacokinetics

distribution within the body, are less selective in their binding to proteins, and finally are oftenextensively metabolized.

Experimental

Principle: To determine partition coefficient of benzoic acid a shake flask method is used. Thesample is shaken to achieve equilibrium between all interacting components of the system.Concentration of benzoic acid dissolved in water phase is determined by spectrometer.

Materials: pH meter, glass electrode, 5 Erlenmeyer flasks of 50 ml with glass stoppers, 5 volumetricflasks of 100 ml, volumetric glass pipette of 1, 2, 5 and 10 ml, 5 plastic vials, benzoic acid solution,0.05 M NaOH, 0.05 M HCl, oil

Procedure:

1. Transfer 0.4, 0.6, 0.8, 1.0 and 1.2 ml of NaOH solution into flasks and fill with water to the volume of 2 ml.

2. Add 8 ml of benzoic acid solution and 10 ml of oil. Shake the mixtures for 4 min. Wait for 20 min until the both phases are separated.

3. Transfer about 5 ml of the lower water phase into plastic vials and centrifuge for 10 min.

4. Transfer about 1 ml of the solutions following centrifugation into volumetric flasks, add 1 ml of HCl and fill with water to 100 ml.

5. Measure the absorbance values of the solutions at =229 nm in reference to water, using quartz cuvette.

6. Measure the pH of the solutions in plastic tubes.

7. From pH measurements calculate hydrogen ions concentrations and values of the following ratio: Ka + [H+]/Aw (Ka = 6.5 10-5, Aw – absorbance of water phase)

8. Draw the graph for the function: Ka + [H+]/Aw = f([H+]) according to Garrett – Woods’ equation:

A

K]H[

A

1K

A

]H[K a

w

a

where:

43

Ka – dissociation constant

cw – concentration of the substance in water phase

K – partition coefficient

9. Use the least squares method to determine the parameters of the function

Ka + [H+]/Aw = f([H+])

10. Calculate partition coefficient of benzoic acid.

References:

4. Scherrer R.A., Howard S.M.: Use of distribution coefficients in quantitative structure-activity relationships. J. Med. Chem. 1977, 20, 53-58.

5. Uchman G., Hermann T.W.: Ćwiczenia laboratoryjne z chemii fizycznej dla studentów farmacji i analityki medycznej. Akademia Medyczna , Poznań 2002.

44

Report 5

Name and surname:............................................................... Date:....................................

Spectrophotometric determination of specific partition coefficient o/w for benzoic acid

Aim of the experiment:

…………………………………………………………………………………………………………

……………………………………………………………………………………………………..

Nr Absorbance

(Aw)

pH [H+] (mol/l)

w

a

A

HK ][ ++

12345

Parameters of the Garrett-Wood equation:

A

K]H[

A

1K

A

]H[K a

w

a

a = ……………….

b = ……………….

r = ………………

Partition coefficient K = ……………………..

Conclusion:

45

Electrochemistry

Experiment 6a. Determination of pKa of acetylsalicylic acid by potentiometric titration

The aim of the experiment is to determine the acid dissociation exponent, pKa, foracetylsalicylic acid by potentiometric titration.

Required knowledge: the Arrhenius and Brønsted-Lowry theory of acids and bases,acid/base ionization constant, the Henderson-Hasselbalch equation, potentiometric titration, methodsfor determination of pKa, Debeye-Hückel theory.

Introduction

Most of drugs are weak acids or weak bases. If the drug substance is a weak electrolyte, thepH of the environment significantly affects its solubility in water and penetration across biologicalmembranes that underlays absorption, distribution and excretion of drugs from the body. For thatreason the knowledge of the acid base equilibria has a practical meaning in such disciplines ofpharmacy as pharmaceutical technology and pharmacokinetics. According to the law of mass action,the equilibrium of the partial ionization (deprotonation) of a weak acid:

HA + H2O H3O+ + A

is described by the equilibrium constant: OHHA

AOHK

2

3

(1)

When deprotonation of a weak acid proceeds in a dilute solution the concentration of water may beconsidered constant, thus the product K [H2O] is also a constant quantity called the acidityconstant, acid ionization constant, or dissociation constant, Ka

1:

HA

AOHK 3

a

(2) (because Ka = K [H2O])

From eqn (2) the hydronium ion concentration in the solution of a weak acid can be determined:

A

HAKOH a3 (3)

Rearranging of eqn (3) leads to the expression describing the pH of the dilute solution of a weakacid:

HA

AlogpKpH a

(4)

where pKa = logKa. Based on eqn (4) it can be stated that the pKa value of a weak acid HAcorresponds to the pH of the solution at which the concentrations of the protonated and deprotonatedmolecules of a weak acid are equal to each other. Moreover, eqn (4) shows that the degree ofdissociation of a weak acid increases with pH. If a salt of the weak acid with a strong base is addedto the solution of that weak acid, the buffer solution is obtained. Due to dissociation of theintroduced salt, concentration of A ion increases in the solution and, according to Le Chatelier'sprinciple, equilibrium of the weak acid dissociation shifts toward the reactants. Consequently, it canbe assumed that the dissociation of the weak acid practically does not occur, hence [HA] isapproximately equal to the formal (total) molar concentration of the acid ([HA] = [acid]), while [A]

1 with respect to electrolytes a term of ‘dissociation’ is less appropriate because deprotonation of an acid and protonation of a base does not include a simple fragmentation into ions but is a more complex process is which the solvent moleculesparticipate. However, the term ‘dissociation’ is still commonly applied in the literature, so is in this Laboratory Guide.

46

corresponds to the formal molar concentration of the salt which is a strong electrolyte ([A] = [salt]).Application of the above relationships to formula (4) leads to the Henderson-Hasselbalch equationfor an acid buffer, that is the buffer composed of a weak acid HA and a salt that supplies itsconjugate base A:

acid

saltlogpKpH a (5)

Similar consideration can be applied to the dissociation (protonation) of a weak base:B + H2O BH+ + OH

From the equilibrium constant of the above reaction:

OHB

OHBHK

2

(6)

the following equation for the basicity constant, also called the base ionization constant or basedissociation constant, in dilute solutions (then [H2O] = const) can be obtained:

B

OHBHK b

(7) (as Kb = K [H2O])

Using the above equation, the concentration of hydroxide ions in the solution of a weak base can bederived:

BH

BKOH b (8)

Taking the logarithms of the both sides of eqn (8) and including the ionic product of water ([H3O+][OH] = Kw) leads to the equation describing the pH of the dilute solution of a weak base:

BH

BlogpKpKpH bw (9)

where pKb = logKb. Taking into account the relationship between the value of pKb of a weak base Band the pKa value of its conjugate acid BH+ (pKa + pKb = pKw), eqn (9) rearranges to:

BH

BlogpKpH a (10)

The above equation shows that the pH of a solution of a weak base B in which concentrations of theundissociated and dissociated molecules are equal, corresponds to the pKa of the conjugate acid BH+.Moreover, the degree of dissociation of a weak base decreases with pH of the solution. If theequilibrium of protonation of a weak base is shifted toward the reactants by addition of a salt of thebase with a strong acid (the source of BH+ ions), then [B] will be practically equal to the formal(total) concentration of the weak base ([B] = [base]), and [BH+] will reflect the formal saltconcentration (([BH+] = [salt]). Eventually, the Henderson-Hasselbalch equation for a basebuffer, that is the buffer composed of a weak base B and a salt that supplies its conjugate acidBH+, is obtained:

salt

baselogpKpH a (11)

According to the Debeye-Hückel theory, in solutions of electrolytes that are not very dilutethe interactions between the ions are of essential importance. Properties of these solutions do notdepend on the molar concentrations of the ions but on their activities, also known as effectiveconcentrations. For this reason, a more accurate determination of the pH of solutions of electrolytesoften requires replacement of the molar concentration of the ions present in the Henderson-Hasselbalch equations by their activities. Then, eqn (5), applied to an acid buffer, takes the followingform:

47

Aa γlog

acid

saltlogpKpH (12)

where γ±A is the mean activity coefficient of A ion that can be calculated from the ionic strength ofthe solution. In turn, eqn (11), applied to a base buffer, yields the following form:

BHa logγ

salt

baselogpKpH (13)

Determination of pKa of a weak acid by potentiometric titrationIn this method the pH of a solution is measured while titrating a weak acid with a strong

base, usually NaOH or KOH, until the stoichiometric point (SP), at which the acid is neutralized in100%, is exceeded2. The pKa value is found as the pH at which 50% of the weak acid has beendeprotonated, because then [HA] = [A], so according to eqn (4), pH = pKa. Taking into account the degree of dissociation of a weak acid (α), the concentration of theprotonated ([HA]) and deprotonated molecules ([A]) of the acid can be expressed as follows:

[HA] = c(1 α) [A] = cα

Then, eqn (4) yields the form:

α1

αlogpK

α)c(1

cαlogpKpH aa

(14)

Eventually, using the properties of logarithms, it can be written:

α

α1logpKpH a

(15)

The graph of the relationship (15) yields a straight line in the form of y = ax + b (Fig. 1), with theslope a = 1 and the intercept b equal to the searched pKa value of the weak acid (substituting α =

0.5 to the equation α

α1logpKpH a

gives aa pK0,5

0,51logpKpH

). Total dissociation of

a weak acid being titrated with a strong basecomprises its spontaneous dissociation (HA +H2O H3O +A) and its neutralization (HA +OH H2O + A). The degree of thespontaneous dissociation is reflected by a ratioof [H3O+]/c (c total concentration of theweak acid), and the degree of theneutralization yields a ratio of the volume ofthe base added at given moment of the titration(Vi) to the volume of the base at thestoichiometric point (VSP), that is Vi/VSP. TheVSP is most often determined by the Hahn’smethod, assuming that it is equal to the volumeof the base at the titration end point. Inpractice, a degree of the total dissociation of aweak acid (e.g. acetic acid, pKa 4.8) is oftenassumed to be equal to a degree of itsneutralization, as justified by the low degree of the spontaneous deprotonation (α ≈ 0.04) that evenfurther decreases when concentration of the A anions in the solution increases due to the titrationwith the strong base. However, at the beginning step of the titration of stronger acids (e.g.acetylsalicylic acid, pKa 3.5) a degree of the spontaneous deprotonation has a significant contribution

2 for historical reasons, the stoichiometric point (SP) is also widely called the equivalence point (EP).

Fig. 6.1. A graph of the pH of a weak acid solution being titrated with a strong base plotted against the logarithm of (1 – α)/α.

α

α1log

48

to an overall degree of the dissociation and cannot be ignored, otherwise the plot of the graph

α

α1logfpH would be non-linear.

In addition to the above algebraic method, the pKa of a weak acid may be determined by asimple but less accurate graphical method, also known as the tangent method. In this method, thepKa is directly found on the pH curve, that is the plot of the pH of the analyte against the volume ofthe titrant, pH = f(Vtitrant). First, the stoichiometric point (VSP) of the weak acid is located (in practiceit is the titration end point) where the degree of the neutralization is 100 % (Fig. 2). Then, the recordfor the pH at half-way to the stoichiometric point (½VSP) is found, when enough base has been addedto neutralize half the weak acid. If the degree of dissociation of the acid is equal to the degree of itsneutralization, the found pH value gives the pKa (α = 0.5 pH = pKa). The drawback of thegraphical method lies in the limited accuracy of determination of the VSP on the pH curve, especiallywhen the change in pH near thestoichiometric point is not abrupt. In orderto increase the accuracy of location of theVSP, the method of the first or the secondderivative of pH with respect to the titrantvolume (V) is applied, in which ΔpH/ΔVand Δ(ΔpH/ΔV)/ΔV, respectively, isplotted on the Y-axis instead of the pH.The additional disadvantage of thegraphical method, especially in case ofstronger acids (e.g. salicylic acid, pKa 3.0),is assuming the equality of the dissociationdegree to the degree of neutralization ofthe acid. In reality, after addition of thetitrant volume equal to ½VSP into thesolution of the weak acid, the degree of thetotal dissociation of this acid is greaterthan the degree of its neutralization (α >0.5), and the stronger the acid, the greaterthe difference.

Importance of ionization of weak electrolytes in pharmacyDepending on the pH, drugs which are weak acids (e.g., acetylsalicylic acid, warfarin,

phenytoin, penicillin G, phenobarbital) may exist in the form of anions or neutral molecules, whilethose which belong to weak bases (e.g., morphine, codeine, procainamide, imipramine, quinidine) inthe form of neutral molecules or cations. Unionized and ionized species of drugs differ in a numberof physicochemical properties that affect their behavior in vitro (solubility, stability, ease offormulation) and in vivo (absorption, distribution, binding to receptors, excretion, and evenmetabolism). In the context of the behavior of pharmacologically active weak electrolytes in theorganism, of the great importance is the degree of their ionization in the pH range 1 8, encounteredin the human physiological fluids (stomach: 1 3; intestine: 6 8; rectum 7.6 8; blood: 7.4; urine:5.5 6.5; milk from the mammary gland: 6.4). The two most important differences betweenunionized and ionized molecules of drugs are solubility in water and permeation across biologicalmembranes. The ionized forms of drugs are highly soluble in water, whereas the solubility of theneutral molecules is often poor. In turn, most of the neutral molecules of drugs are capable to crossbiological membranes by passive diffusion, in opposition to the ionized forms. The practicalimportance of ionization of drugs is illustrated by the following examples:

Fig. 6.2. The pH curve for the titration of a weak acid (the analyte) with a strong base (the titrant), and presentation of determination of the pKa by the graphical method.

49

in the stomach, absorption of drugs having a character of weak acid is better than those which areweak bases, while in the intestine the weak bases are better absorbed than the weak acids;

when pH of the urine is being increased, the renal excretion rate of weak acids increases, whilethe rate of excretion of weak bases decreases;

in pharmaceutical technology, salts of weak electrolytes are preferred over the free acids andbases because of greater ease of their formulation, higher solubility as well as rate of dissolutionin water, and also better stability (longer shelf life).

Hence, knowledge of the degree of ionization of the therapeutic substances which are weak acidsand bases enables to anticipate their behavior in vitro and in vivo. This degree depends on the pH ofthe environment and the pKa (in case of the weak base the pKa refers to the conjugate acid) and canbe calculated from the Henderson-Hasselbalch equation.

References:

1. Hermann T.W. (Red.): Chemia fizyczna. Wydawnictwo Lekarskie PZWL, Warszawa 2007.

2. Szczepaniak W.: Metody instrumentalne w analizie chemicznej. Wydawnictwo naukowe PWN,Warszawa 2007.

3. Atkins P., de Paula J.: Elements of Physical Chemistry, 5th Ed., Oxford University Press Inc.,Oxford 2009.

4. Pandit N.K.: Introduction to the Pharmaceutical Sciences. Lippincott Williams & Wilkins,Philadelphia 2007.

5. Amiji M.M., Sandman B.J. (Eds): Applied Physical Pharmacy. The McGraw-Hill Companies,Inc., New York, 2003.

6. Some I.T. et al.: Improved kinetic parameter estimation in pH-profile data treatment. Int. J.Pharm. 2000, 198, 29-49.

7. Uchman G., Hermann T.W.: Ćwiczenia laboratoryjne z chemii fizycznej dla studentów farmacji ianalityki medycznej. Akademia Medyczna, Poznań 2002.

50

Experimental

Acetylsalicylic acid (ASA), the active ingredient of a popular Aspirin, isformed by esterification of a phenolic group of salicylic acid with aceticacid. Due to the presence of a free carboxyl group, the drug is a weak acid.During the exercise, pKa of ASA will be determined by potentiometrictitration method using 0.05 M NaOH. Half-life for hydrolysis of the estermoiety of ASA in aqueous solutions at room temperature is greater than 16

h at pH range 1 – 10, and becomes significantly shorter at pH above 11. For this reason,potentiometric titration of the ASA solution should be carried so that the pH = 11 is not exceeded.

Equipment and materials: pH-meter, combination electrode (couple of glass electrode and silver–silver-chloride electrode), magnetic stirrer, magnetic bar, automatic burette, 100 mL beaker (high),50 mL beaker, 50 ml volumetric flask, 1 mL graduated pipette, 1 M solution of ASA in methanol,0.05 M NaOH.

Procedure

1. Turn on the automatic burette. If necessary, remove the air bubble from the tip of the burette bystarting dispensing the titrant. Handling of the equipment is explained by the assistant.

2. Pour approximately 45 mL of deionized water into a 50 mL volumetric flask, then add 0.5 mL of1 M methanolic solution of ASA and mix the contents. Thereafter, fill up the solution withwater, close the flask with the cap and mix again. Concentration of ASA in the resulting solutionis 0.01 M

3. Pour the entire volume of the prepared solution of ASA (50 ml) into a high 100 mLbeaker andplace the combination electrode and the magnetic bar inside the solution. Turn on the magneticstirrer, paying attention that the rotating magnetic bar has no contact with the combinationelectrode

4. Record the initial pH of the ASA solution (it should be about 3). Thereafter, start thepotentiometric titration of the solution by dosing 0.05 M NaOH in portions of 0.5 ml, using theautomatic burette. After addition of each portion of the titrant, record its total volume (V i) andthe stabilized readings of the pH of the solution, read on the monitor of the pH-meter.

In order to perform the calculations described in the following sections use the ready-to-usetemplate prepared in Microsoft Excel.

5. Plot the pH curve, i.e. a graph of pH of the solution against the added volume of the NaOHsolution (Vi). In the drawn graph find the volume of the titrant at the stoichiometric point VSP (inpractice the volume at the titration end point is determined and assumed to be equal to VSP).Thereafter, find the pKa of ASA graphically, according to the Fig. 2.

6. Plot the course of the first and the second derivative of the pH with respect to the titrant volumeand find the VSP therein. An example of calculations of the derivatives is presented in the tablebelow.

No.VNaOH

[mL]pH

First derivative Second derivativeVav NaOH

(x)pH/V

(y)Vav NaOH

(x)(pH/V )/V

(y)1. 0 2.75

25.02

5.00

14.0

05.075.282.2

COOH

OC

CH3

O

51

2. 0.5 2.82 5.02

75.025.0

0

25.075.0

14.014.0

75.02

0.15.0

14.0

5.00.1

82.289.2

3. 1,0 2.89 0.12

25.175.0

04.0

75.025.1

14.016.0

25.12

5.10.1

16.0

0.15.1

89.297.2

4. 1.5 2.97 5.12

75.125.1

0

25.175.1

16.016.0

75.12

0.25.1

16.0

5.10.2

97.205.3

5. 2.0 3.05

7. Calculate the accurate value of the VSP by the Hahn’s method, assuming it is equal to the titrationend point. For that purpose, find the biggest increase of the pH (pHmax) and the next two largestpH increases, that is pH1 and pH2, where pH1 > pH2. The increase pHmax lies betweenpH1 and pH2. From these values, calculate the correction factor a (ml). The factor a allows tocorrect one of the two added titrant volume Vi, between which the pHmax has been recorded, sothat it corresponds to the VSP. The correction factor a is calculated according to the followingformula:

a = V qa

where V is the volume of one portion of the titrant added (0.5 ml), and qa is the symmetry factordefined by the formula:

1

2a pH 2

ΔpHq

The most accurate results are obtained by the Hahn’s method when pH1 and pH2 are notsignificantly different from each other and simultaneously they are both considerably smaller thanpHmax, which is expressed by the two conditions:

qa 0.25 (preferentially qa 0.5) and pHmax 2.5 pH1

but the former condition is more important. If qa < 0.25 it is recommended to repeat the calculationsusing every second point of the experimental data.In order to calculate the VSP, the correction factor a is added to the titrant volume Vi recorded justbefore occurrence of the pHmax (if pH1 has been recorded before pHmax) or subtracted from thetitrant volume Vi recorded just after occurrence of the pHmax (if pH1 has been recorded afterpHmax):

aVVmax1 pH/ΔpHSP (when pH1 precedes pHmax)

oraVV

1max pH/ΔpHSP (when pH1 follows pHmax)

8. Calculate the total degree of dissociation of ASA (α) in the solution after the addition of eachtitrant volume Vi smaller than VSP this degree is the sum of the degree of the spontaneousdissociation of ASA and the degree of its neutralization by the NaOH:

SP

i

i

i3i V

V

c

OHα

where [H3O+] denotes the concentration of hydronium ion and c i denotes the formal (total) molarconcentration of ASA in the titrated solution (the sum of the concentrations of undissociated anddissociated molecules) after addition of the titrant volume V i. The values of [H3O+] and ci arecalculated with the following formulas:

52

ipHi3 10OH

io

ooi VV

Vcc

where pHi stands for the pH of the solution after addition of the titrant volume V i, co initialconcentration of ASA in the solution (0.01 M), Vo initial volume of the ASA solution (50 mL).9. Using the Microsoft Excel, plot a graph of the pH of the solution of ASA against the logarithm

of (1 α)/α and define the line equation and its correlation coefficient (r). The value of theintercept (b) corresponds to the searched value of pKa of ASA in an aqueous solution at a giventemperature.

10. Prepare a graph of the pH of the ASA solution against log(1 α)/α on a semilogarithmic paperand determine herein the pKa, bearing in mind that it corresponds to the pH of the solution atwhich α = 0.5.

11. Compare the determined value of the pKa of ASA to its tabular value. Assess the suitability ofthe algebraic method and the graphical (tangent) method for determination of the pKa.

53

Report 6a

Name and surname:................................................................ Date:....................................

Determination of pKa of acetylsalicylic acid by potentiometric titration

Aim of the experiment: ................................................................................................................

....................................................................................................................................................... ..........

.............................................................................................................................................

Results

1. Titration of 0.01 M ASA solution with 0.05 M NaOH

No.VNaOH [mL]

pH pHFirst derivative Second derivative

α

α1Vav NaOH [ml] pH/V Vav NaOH [ml] (pH/V)/V

12345678910111213141516171819202122232425

26

54

2. Determination of the titrant volume at the stoichiometric point

Graphical method (Tangent method): VSP = ................... [………….]

The first derivative method: VSP = ....................... [.................]

The second derivative method: VSP = .................... [................]

The Hahn’s method:

pHmax = ....................................

pH1 = ....................................

pH2 = ....................................

qa = ................... / (2 ...................) = .............................

a = ...................... ........................ = ............................... [.........]

VSP = .................. ..... .................... = .............................. [.........]

3. Parameters of the line

α

α1logfpH

Slope (a) = ................................

Intercept (b) = ..............................

Correlation coefficient (r) = ....................................

Final form of the equation

α

α1logfpH :

......................................................................................................................................................

Determined value of pKa of ASA = ............... Tabular value of pKa of ASA = .................

4. Enclosures:

a) the pH curve, pH = f(VNaOH), with graphical determination of the pKa

b) the graph of the first derivative of the pH with respect to VNaOH with determination of the VSP

c) the graph of the second derivative of the pH with respect to VNaOH with determination of the VSP

d) the graph

α

α1logfpH on a semilogarithmic paper with indication of the pKa

5. Conclusions (overleaf)

55

Experiment 6b. Determination of the solubility product of the chosen calcium salt

The aim of the exercise is to determine the solubility of calcium hydrogen phosphate(CaHPO4) in water by the potentiometric method using a calcium ion-selective electrode and tocalculate the solubility product of CaHPO4 (Ksp) and its negative exponent (pKsp).

Required knowledge: solubility, the solubility product, rate of dissolution, zero-current cellpotential, types of half-cells, Nernst equation, structure of ion-selective half-cells (electrodes).

Introduction

In pharmaceutical sciences the knowledge of the issues related to dissolution of solids inliquids is essential for a proper design of medicinal substances as well as preparation, storage anduse of liquid dosage forms. The dissolution process determines the possibility of preparation of adrug solution in a particular solvent and the interactions to occur in so-called pharmaceutical phase(in vitro), e.g. precipitation of a drug in the presence of the other one. It also affects the behavior ofdrugs in vivo (in so-called pharmacokinetic phase), especially their absorption followingextravascular administration. It is important to be aware that all medicines, regardless of their dosageform and route of administration, get into the place of their action in the body (receptors, enzymes,transport proteins, etc.) only after the complete dissolution, i.e. as the solution in the physiologicalfluids. Taking into account that the extracellular fluid (including blood) as well as the intracellularfluid are aqueous media, and the cell membranes predominantly have a lipophilic character,medicinal substances must be at least somewhat soluble in both water and lipids in order to reachtheir biological targets. Otherwise, they cannot exert their pharmacological action, according to theLatin sentence ‘Corpora non agunt nisi soluta’ (The bodies do not act if not dissolved).

The basic parameters that characterize the process of dissolution of solids in liquids aresolubility, a dissolution rate and additionally, in case of sparingly soluble inorganic and organicsalts, a solubility product. It should be noted that both the solubility and the solubility product referto the heterogeneous equilibrium between a sediment and a solute, established in the saturatedsolution at a given temperature. In contrast, the dissolution rate is a kinetic parameter that indicateshow quickly this heterogeneous equilibrium state is obtained.

Solubility (S) of a substance in the specific solvent, usually in water, is defined as its molarconcentration in the saturated solution obtained at a given temperature (molar solubility). Moreover,the solubility is also expressed by the number of grams of a solute that forms a saturated solution in100 g of the solvent at a given temperature.

The heterogeneous equilibrium between the molecules of a sparingly soluble salt AnBm thatform the sediment and its ions in the saturated aqueous solution is described by the equation:

AnBm (s) nAm+ (aq) + mBn (aq)

When writing the above equation we assume that the sparingly soluble salt AnBm is a strongelectrolyte of which the dissolved molecules are dissociated in 100%, hence only ions Am+ and Bn,not undissociated molecules (AnBm (aq)), are present in the solution. According to the law of massaction, the equilibrium of the dissolution process obtained in the saturated solution, can be describedby thermodynamic equilibrium constant:

mn

nm

BA

m

B

n

A

a

aaK

(1)

where mAa and nB

a denote the activity of the solvated ions Am+(aq) and Bn

(aq), and mBnAa stands for

the activity of the solid salt contained in the sediment. At a fixed temperature (effect of pressure on

56

the equilibrium constant of processes occurring in condensed systems is negligible) the K and mBnAa

are constant, hence eqn (1) can be written as:

mB

n

ABA nmmnaaaK (2)

The product mnBAaK is a constant quantity called the thermodynamic solubility product of the

salt AnBm (mnBAspK ) or, according to the current nomenclature, the thermodynamic solubility

constant (mnBAsK ):

mB

n

ABAsp nmmn

aaK (3)

Taking into account that an ion activity is a product of a molar concentration and an activitycoefficient (a = c γ), the thermodynamic solubility product can be written as:

mB

mnnA

nmBAsp nm

mnγBγAK

(4)

If a sparingly soluble salt is dissolved in pure water, ionic strength of the saturated solution, which isa function of the concentrations of all ions present in the solution, will have a low value, hence theactivity coefficients of the ions will be approximately equal to 1. Consequently, eqn (4) will be aproduct of the molar concentrations of the ions in the saturated solution, each raised to the powerequal to the stoichiometric coefficient. The expression obtained in this way is termed the solubilityproduct or, according to the current nomenclature, the solubility constant:

mnnmBAsp BAK

mn

(5)

When the concentration of both the ions in the saturated solution is expressed as a function of themolar solubility of the salt AnBm (S), that is [Am+] = nS and [Bn] = mS, a general equation thatrelates the solubility product to the molar solubility of the sparingly soluble salt in pure water maybe given:

n)(mmnmn

BAsp SmnmSnSKmn

(6)

From the above equation, the solubility product may be calculated provided the solubility of a givensparingly soluble salt is known. For example, the solubility product of CaHPO4 will be calculatedaccording to the formula:

2CaHPOsp SSSK

4

The magnitude of the thermodynamic solubility product as well as the solubility product of asparingly soluble salt depends on the type of the solvent, temperature, and the pH of the solution if itaffects the salt dissociation equilibrium.

Factors affecting solubility:a) Physical and chemical properties of a solute, such as a dipole moment, the presence of atoms and

functional groups that are donors (O, S, N) or acceptors for hydrogen bonds (OH, SH, NH2), thenumber of the carbon atoms; and physicochemical properties of the solvent (dipole moment, thepresence of donors or acceptors for hydrogen bonds, dielectric constant).

The general principle is 'similia similibus solvantur’ (Latin: ‘like dissolves like’): polar substances (such as salts, alcohols, phenols, acids) dissolve better in polar solvents (e.g.,

methanol, acetonitrile or water). Here the attraction forces such as dipole dipole, ion dipoleand hydrogen bonds plays the main role in the interaction between the molecules of the soluteand the solvent. In case of salts having the structure of ionic crystal, such as NaCl, a veryimportant factor affecting the solubility is the ability of the polar solvent to weaken the forcesof electrostatic attraction between the oppositely charged ions in the crystal lattice of thesolute. This ability is expressed by the solvent dielectric constant. For example, the solubilityof NaCl in water that has a high dielectric constant (78.5) is as much as 24 times greater than

57

in methanol (dielectric constant 31.5), although methanol, like water, has a polar structure andcapability of forming hydrogen bonds .

nonpolar substances (such as hydrocarbons) dissolve better in nonpolar solvents (e.g., carbontetrachloride). The van der Waals forces (dipole induced dipole, induced dipole induceddipole) are responsible for the interaction between the molecules of the solute and the solvent.

b) Physical form (amorphous, crystalline) and polymorph of the solute: Amorphous forms are generally more soluble than the crystalline forms because in the crystal

lattice the stronger attractions exist between the species (ions or molecules). For example, theamorphous form of insulin is more soluble as well as faster dissolves in water compared tocrystalline insulin. For this reason, aqueous suspensions containing the amorphous form ofinsulin are intermediate-acting formulations, while the suspensions of crystalline insulin areslowly but long-acting formulations.

Plenty of drugs (e.g., barbiturates, sulfonamides, steroids) have polymorphs that differ insolubility in water. For example, only the polymorph B of chloramphenicol palmitate(antibiotic) accounts for absorption of the drug from a gastrointestinal tract as the polymorphA is practically insoluble in water.

c) TemperatureThe increase in temperature causes an approximately exponential rise or (rarely) reduction ofsolubility of substances, depending on the value of the molar enthalpy (heat) of solution at thesaturation equilibrium (Hsoln). If the dissolution is associated with absorption of heat from the

surroundings by the system of the solute – the solvent (Hsoln > 0), then according to Le Chatelier's

principle an increase in temperature leads to an increase in the solubility of the substance (e.g.,KNO3, KBr). If heat is released to the surroundings (Hsoln < 0), the solubility decreases with

temperature (e.g., CaHPO4, CaO, methylcellulose). Quantitative relationship between the solubility(S) and the absolute temperature (T) is described by the equation:

21

12soln

1

2

TT

TT

R

Δ

S

Sln

H= (7)

where S1 and S2 denote solubility of a solid at temperature T1 and T2, respectively, and Hsoln is the

molar enthalpy of solution at the saturation.

d) pH of the solution (if the solute is a weak electrolyte)The process of dissociation of a weak electrolyte substantially increases its solubility in a particularsolvent. For this reason, the solubility of weak acids is greater in alkaline medium (the greater thedifference pH – pKa, the greater the solubility), and the one of weak bases – in acidic solutions (thegreater the difference pKa – pH, the greater the solubility). A clinical example of the aboverelationships is the increase in the solubility of warfarin (an oral anticoagulant with a character of aweak acid) in a gastrointestinal tract in the presence of an antacid Mg(OH)2, which leads to theincreased absorption of the drug and undesirable strengthening of its action.

e) Common-ion effect (if the solute is a sparingly soluble salt)When a freely soluble electrolyte (usually water-soluble acid, hydroxide or salt) is introduced into asolution of a sparingly soluble salt having a common ion (cation or anion), the solubility of thesparingly soluble salt becomes smaller. For example, the solubility of AgCl is reduced in thepresence of AgNO3 (common cation) or NaCl (common anion). The effect of the common ion stemsfrom the constancy of the thermodynamic solubility product at a given temperature: an increasedactivity of the common ion in the solution, following introduction of the additional electrolyte, mustentail a reduction in the activity of the second ion that originates only from dissolution of thesparingly soluble salt, to maintain the solubility product unchanged. Consequently, some portion of

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the ions must precipitate from the saturated solution which means the solubility of the sparinglysoluble salt has been decreased.

f) Diverse ion effect (= salt effect) (if the solute is a sparingly soluble salt)Introduction of the electrolyte into a solution of a sparingly soluble salt that have no ion in commonresults in increased solubility of the sparingly soluble salt. For example, the solubility of CaCO3

increases in the presence of NaNO3. This effect stems from the increase in the ionic strength of thesolution, which then entails a reduction of the activity coefficients of the ions. As thethermodynamic solubility product, described by eqn (4), must remain constant, the concentrations ofthe ions in the solution must increase, which is achieved by increasing the solubility of the sparinglysoluble salt.

g) The presence of solubilizersSolubilization consists in increasing the solubility of a substance that is inherently poorly soluble ina specific solvent by adding the adequately selected compounds (solubilizers). In pharmaceuticaltechnology a phenomenon of solubilization is mainly applied to increase the solubility of medicinalsubstances in aqueous solutions administered orally or parenterally. The solubilizers used inpharmacy are a group of the diverse compounds that vary in structure and mechanism of action (e.g.,glycerol, salicylic acid, surfactants of type Span and Tween, urea).

The rate of dissolution of a solid in the liquid at constant temperature is described by the Noyes-Whitney equation, although it should be noted that the similar equation has been earlier developedby the Polish scholar, Professor Joseph Boguski:

ccVh

DA

dt

dcs (8)

where:dc/dt rate of dissolution of the solidD – the diffusion coefficient of the solute in the solvent

A – total area of the solid contacting with the solution

V – volume of the liquid phase (solution)h – the thickness of the diffusion layer, i.e. a thin layer of the solvent formed on the surface of thesolid particle, where the concentration of the solute infinitely fast reaches a value characteristic forthe saturated solution (cs). From the diffusion layer the dissolved molecules (or the ions) diffuse intothe solution bulk, leaving a free space therein to which another molecule can pass from theundissolved particles of the solid.cs – the concentration of the substance in the saturated solution in a particular solvent at a giventemperature (solubility)c – the concentration of the solute in the solution at time t.Under specific conditions, in which D, V and h are constant, and assuming that cs >> c, the followingequation is obtained:

sKAcdt

dc (9)

where Vh

DK . Based on the Noyes -Whitney equation it is readily inferred that the rate of

dissolution of a solid in the liquid may be increased by: fragmentation of the solid (reduction of the solid particles size causes an inversely proportional

increase in the effective surface area A of the solid being in contact with the solvent)

59

mixing of the solution (it decreases the thickness h of the diffusion layer and, additionally, makesthe dissolved molecules of the solute are faster removed outside the diffusion layer)

all factors that increase the solubility of the substance cs, including replacement of the crystallineform by the amorphous one (e.g. insulin).

In case of solid dosage forms (e.g. tablets) the dissolution rate of medicinal substances depends alsoon a variety of technological parameters, such as the type and the amount of excipients used or theforce exerted on the tablet by the punch during the tablet formulation. It should be noted that, incontrast to the dissolution rate, the solubility of a substance depends on neither size of the solidparticles nor mixing of the solution. These manipulations only accelerate achievement ofsaturation of the solution, but have no effect on the concentration of the saturated solution.

The importance of solubility and the solubility product in pharmaceutical sciencesSolubility and the solubility product are of practical importance for a pharmacist during thepreparation of medicines in the form of multi-component solutions such as oral mixtures (mixturae),eye drops, intravenous injections and infusions, or mixtures for parenteral nutrition, that oftencontain ions capable of forming sparingly soluble salts. A precipitation of the solute will occur whenthe product of the actual concentrations of the ions present in the solution, calculated according toeqn (5), exceeds the solubility product of the sparingly soluble salt at a given temperature (insolutions having higher ionic strength the ion activities and the thermodynamic solubility productshould be used). Hence, the knowledge of the solubility product enables avoiding suchpharmaceutical incompatibilities, which is extremely important as they cause the loss of thepharmacological activity of the active ingredients and, in case of the solutions administeredintravenously, also threaten the patient’s life. One of the most known pharmaceutical incompatibilityis a precipitation of calcium hydrogen phosphate, CaHPO4, when preparing the multielectrolytemixtures for parenteral nutrition. To prevent precipitation of this sparingly soluble salt, the productof the concentrations of Ca2+ and HPO4

2 ions in the whole mixture must be smaller than thesolubility product of CaHPO4 (1.0 10-7, temp. 25 oC). As the inorganic phosphate concentrates thatare used to prepare the nutritional mixtures contain HPO4

2 as well as H2PO4 ions (pH about 6.4), in

pharmaceutical practice a simplified rule is applied that the product of the Ca2+ concentration and thetotal concentration of the phosphate ions cannot exceed the threshold of 72 mM2 (7.2 10-5 M2).This value is higher than the solubility product of CaHPO4, which results from the much highersolubility of Ca(H2PO4)2 than CaHPO4, and the stabilizing influence of the amino acids(complexation of Ca2+) and the magnesium ions (the diverse ion effect), which both are contained inthe nutritional mixtures. Besides providing the suitable amount of the Ca2+ and HPO4

2 ions in thenutritional mixture, a pharmacist must also remember about the appropriate order of mixing of thecomponents. For example, he should first introduce the phosphate concentrate into the nutritionalbag, then all the neutral components, including the solvent, and just finally CaCl2. Also of greatimportance, especially in the summer months, is to protect the prepared nutritional mixture againsttemperature increase during its storage as well as administration to the patient, because the solubilityof CaHPO4 decreases with temperature. Nowadays, instead of the inorganic salts of calcium andphosphorus, their organic compounds are often used, such as calcium gluconate or calciumglubionate, and sodium glycerol phosphate or glucose-1-phosphate, which greatly reduce the risk ofCaHPO4 precipitating.

Types of half-cells. Determination of the solubility product by potentiometric titrationCalculation of the solubility product usually consists in determining the concentrations of the ions ofa sparingly soluble salt in its saturated solution. Assuming that in this solution all the molecules arecompletely dissociated, the concentration of each ion can be calculated from the solubility of the saltdetermined by potentiometric, conductometric or chromatography method.

60

The potentiometric methods are based on measurements of the zero-current cell potential (formerlytermed the ‘electromotive force’), that is a difference in potentials of two half-cells. Each half-cell iscomposed of a suitable solution wherein the electrode made of the specific metal (formerly also ofgraphite) is immersed. In the laboratory jargon of the electrochemists the term ‘electrode‘ is oftenused interchangeably with the term ‘half-cell’. Hence, a half-cell having a higher potential istraditionally called the cathode and the half-cell having a lower potential – the anode. For example,in the Daniell galvanic cell the half-cell Zn2+/Zn, with a potential of 0.763 V, is the anode, while thehalf-cell Cu2+/Cu, with a potential of 0.337 V, is the cathode.The potential of the half-cell type I is described by the Nernst equation:

nMo aln

nF

RTEE (10)

where:Eo [V] – the standard half-cell potential, i.e. the potential of the half-cell in which the electrode madeof a given metal is immersed into a solution of the ions of the same metal having the activity of 1 M.Eo value is usually given for standard conditions (298.15 K, 1 bar), although it may also refer to anyarbitrary selected temperature and pressure. To emphasize this fact the symbol Eo is sometimes usedinstead.n – the number of electrons participating in the half-reaction at the metal electrode: (Mn+ + ne M)F Faraday’s constant, the value of electric charge per one mole of electrons (96500 C/mol)

nMa [M] – the activity of the metal ions in the solution (in case of dilute solutions with low ionic

strength the molar concentration may be used instead of the activity).For determination of the solubility product of sparingly soluble salts, ion-selective half-cells,

traditionally called the ‘ion-selective electrodes’ (ISE), are often applied. Just as the name implies,the potential of these half-cells depends highly selectively on the activity of one specific ion presentin the solution (the primary ion), and is relatively independent of the presence of the other ions(interfering ions). The most important element in the structure of all ISE, which is not met in theother types of half-cells, is a membrane that separates the interior of the half-cell from the analyzedsolution. For this reason, ISE are also called the membrane half-cells. The principle of the selectiveaction of ISE is the fact that the membrane exhibits ionic permeability mainly for these ions whichform its electroactive material. As a result, when the half-cell is placed in the tested solution, apreferential passage of the analyzed ion (primary ion) occurs from the solution into the membraneand vice versa. This generates a difference of the potential across the membrane, known as themembrane potential, which is a function of the activity of the primary ion in the solution. Here itshould be noted that, contrary to the other half-cells, no redox reactions occur in ISE because onlyions, not electrons, are involved in the electrode processes!

From a formal point of view, the potential of ISE depends on the activity of a primary ionand interfering ions present in the solution. In practice, however, ISE have so high selectivity thatthe effect of the interfering ions on their potential can be ignored. Then, the equation for the half-cellpotential takes the form of the Nernst equation:

io aln

nF

RTEE (11)

and in case of dilute solutions (with low ionic strength):

io cln

nF

RTEE (12)

In addition to selectivity, the advantages of ISE are simplicity and rapidness of the measurements,great possibility of their miniaturization (for example, needle-sized ISE are used in the studies ofhuman body electrolytes) and automation of the determinations. Therefore, introduction of ISE intothe quantitative analysis of ions has contributed to the intensive development of potentiometricmethods in medical, environmental and industrial applications.

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References:




4. Amiji M.M., Sandman B.J. (Eds): Applied Physical Pharmacy. The McGraw-Hill Companies,Inc., New York, 2003.

5. Helms R.A., Quan D.J.: Textbook of Therapeutics: Drug and Disease Management. LippincottWilliams & Wilkins, Philadelphia 2006.

6. Mirtallo J. et al.: Task force for the revision of safe practices for parenteral nutrition. JPEN 2004,28, S39-S70.

7. Szmal Z., Lipiec T.: Chemia analityczna z elementami analizy instrumentalnej. WydawnictwoNaukowe PWN, Warszawa 1996.

8. Danek A.: Chemia fizyczna. Wyd. V popr. i uzup., Państwowy Zakład Wydawnictw Lekarskich,Warszawa 1982.

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Experimental

A sparingly soluble salt of CaHPO4 precipitates as a white flocculent sediment from solutionscontaining Ca2+ and HPO4

2 ions at pH close to neutral (at low pH, CaHPO4 does not precipitate dueto formation of a more soluble salt of Ca(H2PO4)2). In a saturated solution of CaHPO4 theequilibrium of dissolution process is established, as described by the equation:

CaHPO4 (s) Ca2+ (aq) + HPO4

2 (aq)

During the exercise the solubility of CaHPO4 will be determined by measuring the calcium ionconcentration in the saturated solution of this salt by potentiometric method using the calcium ion-selective half-cell (electrode).

Equipment and materials: potentiometer, calcium ion-selective electrode, silver–silver-chlorideelectrode, seven 100 mL beakers, 50 mL beaker, 1 mL and 5 mL graduated pipette, six 50 mLvolumetric flasks, funnel, baguette, Whatman filter paper, 0.1 M CaCl2, 0.1 M Na2HPO4.

Procedure

1. Precipitate calcium hydrogen phosphate, CaHPO4, in a 50 mL beaker by mixing 10 mL of 0.1 MCaCl2 solution and 10 mL of 0.1 M Na2HPO4 solution. Filter the obtained precipitate on aWhatman filter paper and repeatedly rinse it with deionized water using at least 100 mL of thewater. Thereafter, prepare a saturated solution of CaHPO4 in deionized water. For that purpose,flush the precipitate from the filter paper with approximately 50 ml of deionized water into a 100mL beaker, stir it vigorously and allow to stand for 1 hour to obtain saturation of the solution.

In the meantime, follow the instructions described in Sections 2 6.

2. In a 50 mL volumetric flasks, prepare the standard solutions of CaCl2 in deionized water:a. standard solutions of CaCl2 of 1 10-2, 4 10-3, 1 10-3 and 4 10-4 M are prepared by

appropriate diluting the 0.1 M stock solution of CaCl2

b. standard solutions of CaCl2 of 1 10-4, and 2.5 10-5 M are prepared usingthe 1 10-3 M standard solution of CaCl2, obtained earlier in the point a. Before startingperforming the solutions, consult your calculations with the assistant.

3. Measure the zero-current cell potential E (V) for each of the prepared standard solutions ofCaCl2 using3:- calcium ion-selective electrode (the indicator electrode) playing the role of the cathode- silver–silver-chloride electrode (the reference electrode) playing the role of the anode.

For that purpose, pour about 40 mL of the standard solution of CaCl2 into a 100 mL beaker, immersethe two electrodes into the solution and after the E readings have been stabilized (about 1 minute)record its value. Caution: If after 1 min the changes of the E are still large, it is most probablycaused by the air bubble located on the electrode membrane. In that case, re-immersing of theelectrodes is recommended. During the measurements the calcium ion-selective electrode must beimmersed not more than up to the level of the black rubber ring. After each measurement of the E,both the electrodes must be accurately rinsed with deionized water and dry with the blotting paper.

In order to perform the calculations described in the following sections use the ready-to-usetemplate prepared in Microsoft Excel. Assume that due to the low ionic strength of the studiedsolutions, the activity of the ions may be replaced by their molar concentration (activity coefficientsof the ions are equal to 1).

3 In very accurate measurements of the E of a cell with the usage of solutions that differ in ionic strength, it is recommended to add the so-called ionic strength buffer, for example 1 M KNO3, to all the studied solutions. However, in case of the measurements carried out during the exercise it is of no practical significance.

63

4. Using the Microsoft Excel determine the parameters of the calibration curve E = f(ln[Ca2+]) inthe form of E = b + a ln[Ca2+]. The b parameter corresponds to the theoretical value of thestandard E of the cell, i.e. when the concentration of Ca2+ ions in the solution is equal to 1 M (asln1 = 0).

5. Determine the characteristics of the calcium ion-selective electrode. To do this, calculate thestandard potential of the calcium ion-selective electrode (Eo

Ca) and the slope of the electrode (S),which are both constant under the given conditions of the measurements.

Hint: The E of the applied cell is equal to the difference between the potential of the calcium ion-selective electrode ECa (cathode) and the potential of the reference electrode (anode) that is constantunder the given conditions (EAgCl/Ag = 0.22 V):E = ECa EAgCl/Ag

When in the above equation the ECa is expressed according to the Nernst equation for the calciumion-selective electrode, it turns out that the b parameter of the calibration curve E = f(ln[Ca2+]) is equal to the difference between the standard potential of the calcium ion-selectiveelectrode (Eo

Ca) and the reference electrode potential (EAgCl/Ag):b = Eo

Ca EAgCl/Ag; hence it can be calculated that EoCa = b + EAgCl/Ag.

In turn, the slope a of the calibration curve E = f(ln[Ca2+]) corresponds to the experimental value ofRT/2F (n = 2) for the calcium ion-selective electrode, and characterizes its slope (S) under the givenconditions of the performed measurements.

6. Based on the parameters determined in Section 5, write the Nernst equation for the calcium ion-selective electrode that specifically applies to the experimental conditions, in the form of E =f(ln[Ca2+]) and E = f (log[Ca2+])

7. Write the equation of dissolution of CaHPO4 and based on it do the following:a. give the relationship between the molar solubility of CaHPO4 and the molar concentration of

Ca2+ ions in the saturated solution of this saltb. give the equation for the solubility product of CaHPO4

c. express the equation for the solubility product of CaHPO4 in terms of the solubility of thissalt, according to eqn (6).

8. Measure the E (V) of the cell after immersing the electrodes into the clear saturated solution ofCaHPO4 located above the precipitate of this salt.

9. Calculate the following values:a. concentration of Ca2+ ions (M) in the saturated solution of CaHPO4, which is equal to the

solubility of CaHPO4 in water: a

bE2 eCa

, where E is the zero-current cell potential

measured for the saturated solution of CaHPO4, while a and b are the parameters of thecalibration curve E = f(ln[Ca2+]) determined in Section 4.

b. solubility product of CaHPO4 (Ksp) and its negative logarithm (pKsp).

10. Compare the experimentally determined value of pKsp of CaHPO4 to the tabular one given for 25°C (7.00). Explain the possible difference.

11. What importance has the knowledge of the solubility product of slightly soluble salts, such asCaHPO4, in a hospital pharmacist practice? What are the ways of avoiding the precipitation of thissalt?

64

Report 6b

Name and surname:.................................................................. Date:................................

Determination of the solubility product of the chosen calcium salt

Aim of the experiment:.................................................................................................................

...................................................................................................................................................... ...........

............................................................................................................................................

.......................................................................................................................................................

Results

1. Measurements of E of the cell using the standard solutions of CaCl2

No. Conc. of CaCl2 [M] ln[Ca2+] E [V]1.2.3.4.5.6.

2. Parameters of the calibration curve E = f(ln[Ca2+])

Slope (a) = ...........................................

Intercept (b) = ......................................

Correlation coefficient (r) = ............................................

Final equation of the line E = f(ln[Ca2+]):

.................................................................................................................................

3. Characteristics of the calcium ion-selective half-cell (electrode)

Slope of the electrode (S) = ................................

Standard potential (EoCa) = ...........................

Nernst equation for the calcium ion-selective electrode:

a) in the form of ECa = f(ln[Ca2+]): .........................................................................................

b) in the form of ECa = f(log[Ca2+]): ......................................................................................

4. Analysis of the saturated solution of CaHPO4

Equation of dissolution of CaHPO4 in water:

……………………………………………………………………

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Relationship between the molar solubility of CaHPO4 (cs) and the concentration of Ca2+ ([Ca2+]) in

the saturated solution of CaHPO4: cs = ………………….............…………….

Equation for the solubility product of CaHPO4: Ksp = .........................................................

Equation for the solubility product of CaHPO4 as a function of the molar solubility of this salt (cs):

Ksp = ..................................................................

E of the cell for the saturated solution of CaHPO4 (V) = ...................................................

Concentration of Ca2+ in the saturated solution of CaHPO4 (M) = .........................................

Solubility of CaHPO4 in water (M) = ....................................................................

Solubility product of CaHPO4 in water: Ksp = .......................................................

Determined value of the exponent of the solubility product of CaHPO4: pKsp = ...................

Tabular value of the pKsp = ...........................

5. Enclosures:

The graph E = f(ln[Ca2+]) on a semilogarithmic paper

6. Conclusions

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KINETICS

Experiment 7. Determination of the rate constant and the thermodynamic parameters for the hydrolysis of acetylsalicylic acid

The aim of the exercise is to determine the shelf life of acetylsalicylic acid in the buffersolution at pH 5.0 and 20 °C using the accelerated aging testing.

Required knowledge: rate of a chemical reaction, molecularity and order of a reaction, zero-order, first-order, and second-order reaction, pseudofirst-order reaction, effect of temperature on thereaction rate, the Arrhenius equation, accelerated aging testing, shelf life.

Introduction

Basic concepts of chemical kinetics. Kinetics of a pseudofirst-order reactionThe application of chemical kinetics in pharmaceutical sciences is mainly related to

evaluation of stability of drugs in pharmaceutical formulations, starting from the moment of theirproduction, then over a time of their storage at the pharmacy or the patients’ home, untiladministration to the patient. A particular question of the kinetic research in pharmacy is todetermine the chemical stability of active contents in solutions prepared ex tempore, i.e. bydissolving a solid dosage form immediately prior to administration to the patient, which is importantespecially in relation to drugs administered as long-term infusions, e.g. imipenem – a carbapenemantibiotic.

A basic concept of a chemical kinetics is the rate of a chemical reaction (instantaneous andaverage rate). If a chemical reaction involves the conversion of one molecule of the reactant into onemolecule of the product (A B), the instantaneous and the average rate (velocity) of the reactionare expressed by the formulas:

dt

dcν (instantaneous rate) (1);

t

cν

(average rate) (2)

where the minus sign refers to the changes in the reactant concentration, and the plus sign to thechanges in the product concentration. The unit of the reaction rate is (concentration unit)/(time unit),e.g. M/s. The average rate of a reaction can be directly computed from formula (2), but calculationof the instantaneous rate is more complicated it requires the knowledge of the equation for thereagent concentration as a function of time and calculation of the value of a derivative dc/dt for theparticular time t. The much easier way to designate the instantaneous rate is the graphical method, inwhich the graph of the reagent (reactant or product) concentration against time is used. Theinstantaneous rate is equal to the negative (for the reactant) or positive (for the product) tangent (tan)of the angle between the positive time-axis and the tangent line to the curve c = f(t) at the specifictime t at which the reaction rate is of interest. The average rate is determined in the analogous wayexcept for the secant line is drawn that intersects the graph c = f(t) at the two time points betweenwhich the average rate will be calculated, instead of the tangent line that only touches the graph atonly one point (see Fig. 7.1).

67

The mechanism of a chemical reaction comprises one (rarely) or more elementary reactions(steps) in which the collisions between chemical species (atoms, molecules, ions or radicals) occur.The number of chemical entities that collide with one another in an elementary reaction is called themolecularity of the reaction. It can only take on a value of one or two, rarely three, since theprobability of a simultaneous collision of the greater number of molecules is practically equal tozero. It should be noted that in a multistage reaction, of which mechanism involves a lot ofelementary reactions, the term of molecularity applies only to each individual elementary reaction.Hence, the molecularity cannot not be identified with the number of molecules found in thestoichiometric equation of the overall reaction, as this equation is only a summary notation of thereaction that does not at all reflect its mechanism. For example, the synthesis of hydrogen chloridethat proceeds according to the following stoichiometric equation:

H2 + Cl2 2HClis not a bimolecular reaction (the molecularity is not equal to 2) because the individual elementarysteps are as follows:

1) Cl2 + hν Cl + Cl2) H2 2H3) H + Cl2 HCl + Cl4) Cl + H2 HCl + H (etc.)

Accordingly, the first two steps (elementary reactions) of this complex reaction are unimolecular,while the third and the forth ones are bimolecular, but giving the molecularity of the overall reactionis meaningless. In contrast, the synthesis of hydrogen iodide, described by the equation analogous tothe synthesis of hydrogen chloride, that is:

Fig. 7.1. An example of determination of the instantaneous rate of a reaction (at 5 and 15 min) andthe average rate (between 5 and 15 min) with the graphical method on the basis of the changes in thereactant concentration. The value of the tan of the acute angle in the triangle formed by the two axesof the graph and the tangent () or the secant (----) is determined trigonometrically. The value of theaverage rate of the reaction can also be calculated from the formula νav = ∆c/∆t = (c1 c2)/(t2 t1).

68

H2 + I2 2HIis a bimolecular reaction because its mechanism involves only this one elementary reaction.

Under given conditions of pressure and temperature, for each chemical reaction, regardless ofits complexity, the rate law (rate equation) can be experimentally determined that takes a generalform:

= k f(concentration of the reactants, products, and catalysts) (3)

wherein is the reaction rate (velocity), and k is the reaction rate constant that depends ontemperature (for condensed systems the effect of pressure on the k is negligible). In contrast to thereaction rate, the unit of the reaction rate constant depends on the reaction order (n) and is given bythe formula:

(concentration unit)1n/(time unit)

For example, the unit of the first-order reaction rate constant is s1. The rate of most chemicalreactions depends only on the concentrations of the reactants. Then, a rate of the reaction thatproceeds according to a general equation:

aA + bB + .... cC + dD + .... (4)

is described by the following rate law:

21 nn BAkν (5)

The exponents n1 and n2 define how the concentration of the reactants affect the reaction rate. Theyconstitute the order of the reaction with respect to the given reactant. Thus, in accordance with therate law (5), reaction (4) is n1-th order in the reactant A and n2-th order in the reactant B. The sum ofall the exponents to which the reactants concentrations in the rate law are raised (in this case n1 + n2)is the overall reaction order. It is very important to note that the reaction rate law and,consequently, the reaction order can be established only experimentally and it is not generallypossible to infer them from the stoichiometric equation for the reaction. The only exception is theelementary reaction because its equation strictly reflects its mechanism, therefore the order of such areaction is always equal to its molecularity. Interestingly, reactions having a complex mechanismmay be characterized by fractional orders, e.g. one-half, three-halves, or even can have indefiniteorder. Most often, however, the overall order of chemical reactions (n), even those involving severalelementary steps, has a value of small integers, such as 0, 1, and 2 (the reaction is then zero-, first-,and second-order, respectively). One example is the mentioned synthesis of hydrogen chloridewhich, though composed of many elementary reactions, proceeds as a second-order reaction. This isdue to the fact that the rate of chemical reactions is usually limited by its slowest elementaryreaction, called the rate-determining step. Additionally, the reaction order may vary depending onthe initial concentrations of the reactants. For example, according to eqn (5), the reaction (4) is n-thorder overall, provided that the concentrations of both the reactants A and B do not differsignificantly. However, if one of the reactant, for example B, is present in so large excess that itsconcentration practically does not change in the course of the reaction ([B ] = const), the reactionwill be n1-th order overall, more precisely pseudo-n1-order:

121 nnn Ak'constAkνconstB (6)

where k' is the pseudo-n1-th-order reaction rate constant that equals 2nconstk . Bimolecularhydrolysis reactions of drugs are often pseudofirst-order due to the large excess of water and, in caseof the acid or alkaline hydrolysis, the presence of the buffers which maintain the constantconcentration of H3O+ or OH, respectively. Derivation of the integrated rate law for theconcentration of a reactant at specific time in the course of a bimolecular pseudofirst-order reactionis identical as in an usual first-order reaction:

69

A + H2O products

tk'AlnAln

tk'AlnAln

0tk'AlnAln

dtk'A

Ad

dtk'A

Ad

dtk'A

Ad

Ak'dt

Adν

o

o

o

t

0

A

A

t

0

A

A

1

o

o

where k 'is the pseudofirst-order reaction rate constant expressed in s1, and [A] is the initialconcentration of the reactant at time t = 0. A graph of ln[A] plotted against time yields a straight linewith the slope a = k’ and intercept b = ln[A]o (Fig. 2). To designate the pseudofirst-order reactionrate constant, one must determine the reactant concentrations at different times following the start ofthe reaction, then calculate the slope of the line ln[A] = f(t) by the least squares method, and change the sign of the slope from negative to positive(k’ = a). The linear form of eqn (7) can be converted to the exponential form using the properties oflogarithms:

tk'o

tk'Aln

tk'Alno

eAA

eeA

eAtk'AlnAlno

o

The graph of [A] against time yields anexponential curve that connects the Y-axis at [A]o

(Fig. 3). The half-life of a pseudofirst-orderreaction (t1/2 or t50), that is the time required for50% of the reactant to convert into the product,can be defined from eqn (7) by substituting [A] =0.5 [A]o

k'

0.693t

k'

ln2

k'

A 0.5

Aln

t

k'A

Aln

k'

AlnAlnttk'AlnAln

1/2

o

o

1/2

o

oo

However, when carrying out the studies of chemical stability of drugs, the more important questionthan t1/2 is the time of decomposition of 10% of the active pharmaceutical ingredient (t90). If the

(7)

(8)

(9)

Fig. 7.2. Plot of a natural logarithm of the reactantconcentration against time in a pseudofirst-orderreaction.

Fig. 7.3. Plot of the reactant concentration againsttime in a pseudofirst-order reaction. Theconcentration of the reactant decays exponentiallywith time.

70

decomposition follows the pseudofirst-order kinetics, the t90 is determined from eqn (7) bysubstituting [A] = 0.9 [A]o:

k'

0.1054t

k'9

10ln

k'

A 0.9

Aln

t

90

o

o

90

Notice that for the first-order kinetics both the t1/2 and t90 are independent of the initial drugconcentration.

The practical application of the time t90 in pharmaceutical sciences amounts to its usage as ashelf life – a criterion for establishment of the expiration date of a drug. According to the generalprinciple, the shelf life of a medicinal product is defined as the time of its storage under theconditions specified by the manufacturer (temperature, humidity, access to light) after which lessthan 10% of the active ingredient is decomposed, leaving 90% of the drug viable. For example, if anew drug in the form of a tablet will be stored ‘on the shelf’, the manufacturer has to specify the t90

for decomposition of the active substance in its solid form at room temperature that is typical for thegiven climate zone (e.g. 20 °C in Poland). The pharmacists must, however, be aware that the timerequired for decomposition of 10% of the active substance is not a sufficient criterion forestablishing the chemical stability of all drugs. More stringent requirements are applied to, amongother, drugs that are highly potent, highly toxic, have a narrow therapeutic window, or decompose totoxic products. Obviously, in addition to the chemical stability, the medicinal product stored over thetime specified by the expiration date, must maintain also the physical and microbiological stability.

Arrhenius equationThe rate of all chemical reactions increases with temperature. From mathematical point of

view, this fact is reflected by the increase in the reaction rate constant in the general rate law (3) thatdescribes the reaction rate. Regardless of the reaction order, the quantitative relationship between thereaction rate constant and the absolute temperature (on the Kelvin scale) is given by the Arrheniusequation. It has been developed by the Swedish scientist on assumption that the effect of temperature(T) on the reaction rate constant (k) is similar to the effect of temperature on the reaction equilibriumconstant described by the van't Hoff equation:

2a

RT

E=

dT

lnk d(van’t Hoff equation: 2

r

RT

H=

dT

lnK d )

constT

1

R

E=lnk

T

dT

R

E=lnk d

dTRT

E=lnk d

a

2a

2a

∫∫

Assuming that the constant of integration const is equal to lnA, the linear Arrhenius equation isobtained:

T

1

R

ElnA=lnk a (11)

wherein:A the pre-exponential factor, also known as the frequency factor (expressed in the same unit asthe reaction rate constant, i.e. M s1 for a zero-order reaction, s1 for a first-order reaction, and M1 s1

for a second-order reaction). As the name suggests, the frequency factor is a measure of the number

(10)

71

of collisions in which the spatial orientation of the molecules is suitable to initiate a chemicalreaction. In unimolecular reactions the value of this factor usually ranges between 1012 and 1015 s1,whereas in bimolecular reactions it amounts from 108 to 1012 M1 s1.Ea the activation energy [J/mol], that is the minimumkinetic energy needed for the colliding molecules toenter the active state known as the transition state, inwhich they are able to give the reaction product. Thehigher the activation energy Ea, that is the higher theenergy barrier which must be ‘overcome’ by thereactants to enter the transition state, the lower thereaction rate constant k. On the other hand, the higherthe activation energy, the greater increase in k followingthe temperature increase, so the stronger effect oftemperature on the reaction rate. The activationenergies of bimolecular reaction are generally smallerthan of unimolecular reaction. In the hydrolysis reactions of most drugs the Ea usually rangesbetween 40 120 kJ/mol.The frequency factor A and the activation energy Ea are collectively called the Arrheniusparameters. In the adequately narrowtemperature range both of theseparameters are considered to beconstant.

The graph of the linear form ofthe Arrhenius equation yields a straightline with the slope a = Ea/R and theintercept b = lnA. To calculate theparameters of this line by the leastsquares method, the reaction rateconstants are determined at severaldifferent temperatures.

The linear form of the Arrheniusequation can be transformed to theexponential form:

T

1

R

E

T

1

R

ElnA

T

1

R

ElnA

a

a

a

a

Aek

eek

ekT

1

R

ElnAlnk

Accelerated aging testingA rate of the decomposition of active contents in the majority of medicinal products stored in

accordance with the manufacturer's recommendations is small. This fact is reflected in the length oftheir shelf lives that are usually equal to 2 5 years. So much time would be needed fordetermination of the t90 under conditions the drug is stored in everyday pharmaceutical practice.Therefore, for practical reasons, in the drug stability studies various types of stress (such as highertemperature, higher humidity, artificial exposure to high intensity light) are applied to accelerate thedecomposition reactions. The obtained reaction rate constants are then extrapolated to ‘normal’conditions. This kind of procedure is known as an accelerated aging testing. The most common

(12)

Fig. 7.4. The Arrhenius plot – natural logarithm of the rateconstant plotted against the reciprocal of the absolutetemperature.

72

approach of this testing is to examine the decomposition of the active compound at adequately hightemperatures, then use the Arrhenius equation to determine the reaction rate constant at the lowertemperature of interest, and based on that result calculate the t90. For example, the accelerated agingtesting of a medicinal product applied in the form of an aqueous solution stored at room temperature,in which the active ingredient undergoes hydrolysis that follows pseudofirst-order kinetics,comprises the following steps: carrying out the drug hydrolysis reactions at several higher temperatures (usually 50 90 °C)

while maintaining the other parameters, such as initial concentration of the active substance ,pH, ionic strength, light intensity, unchanged;

determination of the concentration of the active substance (or the product of its decomposition)in the collected samples of the solution, using a suitable analytical method;

designation of the equation of the line ln[A] = f(t) for the hydrolysis of the substance at eachstudied temperature and calculation of the pseudofirst-order rate constants based on the slopeof the obtained lines (k = a);

determination of the Arrhenius equation in its linear form lnk = f (1/T) on the basis of the rateconstants designated at the several higher temperatures;

calculation of the hydrolysis reaction rate constant at room temperature (20 °C) by substituting1/T = 1/(273 + 20) into the previously designated Arrhenius equation;

calculation of the t90 for the decomposition of the active substance at 20 °C using the formula:t90 293 K = 0.1054/k293 K.

References:



10. Amiji M.M., Sandman B.J. (Eds): Applied Physical Pharmacy. The McGraw-HillCompanies, Inc., New York, 2003.

11. Rainsford K.D. (Ed.): Aspirin and related drugs. Taylor & Francis Inc., New York, 2004.


13. Molski A.: Wprowadzenie do kinetyki chemicznej. Wydawnictwa Naukowo-Techniczne,Warszawa 2001.

73

Experimental

In aqueous solutions, acetylsalicylic acid (ASA) undergoes a non-enzymatic hydrolysis tosalicylic acid (SA) and acetic acid, in accordance with the pseudofirst-order reaction kinetics

The rate of the decomposition of ASA varies considerably with the pH of the solution. One of thepopular formulations of ASA, used as an OTC medicine (over-the-counter a medicine availablewithout a prescription) are effervescent tablets that additionally contain vitamin C (brand names:Aspirin C, Upsarin C). Once they are dissolved in water, a solution having a pH of about 5 isobtained (the solution of alone ASA, without vitamin C, would have pH of approximately 3). Duringthe exercise the shelf life (t90) for hydrolysis of ASA in the aqueous solution at pH 5 and 20 °C willbe determined using the accelerated aging testing. The obtained result will enable to define how longthe solution formed after dissolution of the mentioned effervescent tablet can be kept at roomtemperature so that the degree of thedecomposition of ASA does not exceed theacceptable level of 10%. In the experimentthe changes in concentration of ASA in thestudied solution will be designatedindirectly by measuring the increase inconcentration of SA, the hydrolysisproduct. This is because a selectivedetermination of ASA in the presence ofSA is not feasible using a simplespectrophotometric method as at anywavelength the UV spectrum of ASA isoverlapped by the spectrum of SA (Fig.7.5). Nevertheless, at λ > 290 nm anadequately selective determination of SAin the presence of ASA can be carried out.

Equipment and reagents: thermostated water bath, spectrophotometer, quartz cuvette, 50 mLvolumetric flask with a stopper, seven 5 mL pipettes, 100 mL graduated cylinder, set of graduatedpipettes at volumes 100 μL 5 mL, six 25 mL volumetric flasks, seven glass test tubes, 250 mLbeaker, 1 L beaker, glass rod, 0.2 M solution of ASA in methanol, 0.1 M sodium acetate solution,0.1 M acetic acid solution, 2 mM solution of SA in the acetate buffer at pH 5 and ionic strength of0.1 M, 4 M NaCl solution.

Procedure

1. Turn on the water bath with the thermostat set at 80 oC.

COOH

OC

CH3

O

COOH

OHHO

CCH3

O

+ +H2O

ASA SA

Fig. 7.5. UV spectra of 100 μM solutions of ASA andSA in acetate buffer of pH 5 and ionic strength 0.1 M.

74

2. Prepare the acetate buffer of pH 5 (at 80 °C) and ionic strength of 0.1 M. For that purposetransfer 168 mL of 0.1 M sodium acetate solution, 32 mL of 0.1 M acetic acid solution and 790μL of 4 M NaCl solution into a 250 mL beaker. Stir the solution with a glass rod.

3. In the 25 mL volumetric flasks, prepare the standard solutions of salicylic acid (SA) atconcentration of 10, 20, 50, 100, 200, and 300 μM in the acetate buffer of pH 5 and ionicstrength 0.1 M, using the 2 mM stock solution of SA. Before starting preparing the dilutions,consult your calculations with the assistant.

4. Measure the absorbance of the standard solutions of salicylic acid (ASA) in a quartz cuvette (l = 1cm) at a wavelength = 310 nm using the acetate buffer prepared in Section 2 as a blanksample.

For the calculations described in the following sections use the ready-to-use template preparedin Microsoft Excel.

5. In the Microsoft Excel (Sheet 1) prepare the calibration curve of the absorbance of SA against itsconcentration in the standard solutions, in the form of ASA = a cSA. Based on the slope a of thecalibration curve, calculate the molar absorption coefficient ε of salicylic acid: ε [M1 cm1] = a106.

6. Transfer 49.9 mL of the acetate buffer prepared in Section 2 into a 50 mL volumetric flask. Putthe closed flask into the water bath set at 80 oC. Once the solution has reached the desiredtemperature, add 100 mL of the 0.2 M methanolic solution of acetylsalicylic acid (ASA).Quickly mix the contents of the flask, place it back in the water bath and immediately start thestopwatch. The initial concentration of ASA in the solution is 400 μM.

7. Collect approximately 3 mL samples of the solution from the flask using 5 mL pipettes, at thefollowing time points: 1, 5, 10, 15, 20, 25, and 30 min. Transfer the collected sample directlyinto the test tube placed in the 1 L beaker filled with cold tap water. After cooling the contents ofthe tube (1 min), measure the absorbance of the solution in the quartz cuvette (l = 1 cm) at =310 nm.

8. In the Microsoft Excel (Sheet 2) calculate:

a. Concentration of SA (μM) at the particular time points in the course of the ASA hydrolysisin the acetate buffer at pH 5 and 80 °C (use the equation of the calibration curve prepared inSection 5)

b. Concentration of ASA (μM) at the particular time points in the course of the ASAhydrolysis: cASA = coASA – cSA, where coASA is the initial concentration of ASA in the solution,equal to 400 μM.

9. Plot the changes in ASA concentration against time of the hydrolysis reaction on asemilogarithmic paper. In the Microsoft Excel (Sheet 2) prepare a graph lncASA = f(t) and determine the equation of the line and its correlation coefficient. Based on theslope a of the obtained line, give the pseudofirst-order reaction rate constant (k = −a) for thehydrolysis of ASA in the acetate buffer at pH 5 and 80 °C. Moreover, from the intercept b,calculate the theoretical value of the initial concentration of ASA in the buffer, that is theconcentration at time t = 0 (coASA = eb). Based on the determined values of the k and coASA, givethe exponential equation describing the concentration of ASA as a function of time (cASA = coASA ekt).

10. On a semi-logarithmic paper and in the Microsoft Excel (Sheet 3) prepare a graph of theconcentration of ASA plotted against time of its hydrolysis. For each time point, calculate thetheoretical concentration of ASA using the equation cASA = coASA ekt determined in Section 9,and draw a plot of the theoretical concentrations of ASA against time. Using the graphical

75

method (Fig. 1), determine the instantaneous rate of hydrolysis of ASA at 5 and 25 min, and theaverage rate between 5 and 25 min.

According to the above procedures the pseudofirst-order reaction rate constants have beendetermined for the hydrolysis of ASA in the acetate buffer of pH 5 and ionic strength of 0.1 M, at50 , 60 and 70 oC. The results are shown in the Table below:

Temperature[oC]

Duration of the collection[h]

Percent of the decomposed ASA[%]

Rate constantk

[h1]50 6 64 0.169060 2 49 0.319870 1 49 0.6384

11. The ASA hydrolysis reaction rate constant for 80 °C, determined during the exercise, as well asthe rate constants for the other temperatures, depicted in the Table, will be automaticallyexpressed in s−1 in the Microsoft Excel (Sheet 4). Based on the obtained values of the constantsk, prepare a graph of lnk = f(1/T) on a semilogarithmic paper .

12. Determine the Arrhenius equation in its linear form lnk = f(1/T) for the hydrolysis of ASA in theacetate buffer at pH 5. Give the value of the slope (a), intercept (b) and correlation coefficient(r). Introduce the calculated values into the Microsoft Excel (Sheet 4) that automatically verifiestheir correctness and additionally calculates the standard error of the slope (Sa). From the valueof the slope a, calculate the activation energy (a = Ea/R Ea = aR) in J/mol and kJ/mol, andfrom the value of the Sa, calculate the absolute error of determination of the activation energy(Ea = SaR). Compare the activation energy determined for the hydrolysis of ASA in the acetatebuffer of pH 5 and ionic strength 0.1 M to the typical range of the Ea for hydrolyticdecomposition of drugs. Using the intercept b, calculate the value of the frequency factor in s−1

(b= lnA A = eb).

13. On the basis of the designated Arrhenius equation, calculate the pseudofirst-order reaction rateconstant for the hydrolysis of ASA in the acetate buffer of pH 5 and ionic strength of 0.1 M atroom temperature (20 °C).

14. Calculate the shelf life of ASA (t90) in the acetate buffer of pH 5 and ionic strength of 0.1 M at20 oC. Give the result in hours. What practical meaning has the obtained value of the t90?

76

Report 7

Name and surname: ............................................................. Date: .....................................

Determination of the rate constant and the thermodynamic parameters for the hydrolysis ofacetylsalicylic acid


...................................................................................................................................................... ...........

............................................................................................................................................

Results

1. Calibration curve for SA

No. Concentration of SA [μM] Absorbance ( = 310 nm)1.2.3.4.5.6.

Parameters of the calibration curve for SA:

Slope (a) = ................................

Correlation coefficient (r) = ....................................

Equation of the calibration curve: ..............................................................

Molar absorption coefficient for SA: ε = ........................................... [ ……….........………]

2. Quantitative analysis of SA and ASA in the acetate buffer pH 5 at 80 oC

No.Time

[min]

Absorbance of SA

( = 310 nm)

Concentration of SA

[μM]

Concentration of ASA

[μM]1.2.3.4.5.6.7.

3. Parameters of the line lncASA = f(t)

SlopeInterceptCorrelation coefficientEquation of the line lncASA = f(t)Determined rate constant for the pseudofirst-order hydrolysis reaction of ASA in the acetate

77

buffer of pH 5, ionic strength 0.1 M and temperature 80 oC:

k = …………........……. [……..….....]Determined value of the initial concentration of ASA in the acetate buffer (at t = 0):

coASA = …....………..……. [……….......]Exponential equation of the curve cASA = f(t):

……………………………………………………………………

4. Determination of the instantaneous and average rate of the ASA hydrolysis by the graphicalmethod

Instantaneous rate at ............. min

ν = …................………........ / ……..........……….. = …………………...... [..........................]

(the rate calculated by the computer: .................................................. [............................]Instantaneous rate at ............. min

ν = …................………........ / ……..........……….. = …………………...... [..........................]

(the rate calculated by the computer: .................................................. [............................]Average rate between ............ and ............. min

ν = …................………........ / ……..........……….. = …………………...... [..........................]

The rate calculated from the equation νav = (c1 – c2)/(t2 – t1):

ν = …................………....... / ……..........……….. = …………………...... [..........................]

5. Determination of the Arrhenius equation for the hydrolysis reaction of ASA in the acetate bufferof pH 5 and ionic strength 0.1 M

No. Temperature [oC] Temperature [K] 1/T [1/K] (x) k [s1] (y)1.2.3.4.

Linear form of the Arrhenius equationSlope (a) = ...................................

Standard error of the slope (Sa) = .................................

Intercept (b) = ..................................

Correlation coefficient (r) = ...........................................

Final form of the equation: ………………………………………....................................The Arrhenius parameters for the hydrolysis of ASA

Activation energy: Ea = …...……..….. ……………. = …………………… […………....]

78

Error of the determination of the Ea: ∆Ea = …….........… ........……. = ………… [..............]

Frequency factor: A = e………………..............… = …………...…………………… [....................]Exponential form of the Arrhenius equation

...........................................................................................................................

6. Determination of the stability of ASA in the acetate buffer of pH 5 and ionic strength 0.1 M atroom temperature (20 oC)

lnk293 K = …...................... (1/….......….) + ……….........…… = ………………………….

k293 K = …........................................ [.............................]

t90 293 K = …..........................................[s] = .................................... [h]

7. Enclosures:

e) Graph of cASA = f(t) on a semilogarithmic paper with determination of the instantaneous and

average rate of the ASA hydrolysis by the graphical method;

f) Graph of lncASA = f(t) on a semilogarithmic paper;

g) The Arrhenius plot of lnk = f(1/T) on a semilogarithmic paper.

8. Conclusions (overleaf)

79

PHARMACOKINETICS

Experiment 8. Determination of pharmacokinetic parameters of salicytates

AimCalculation of the elimination rate constant and the biological half life of salicylates from urinary extrection data after oral administration of acetylsalicylic acid in tablets.

Required knowledgeLinear pharmacokinetics, compartment, one compartment model analysis, first order knetics, area under curve (AUC), elimination rate constant, biological half-life (t0.5).

Introduction

The pharmacokinetic phase covers the relationship between drug input, which comprises such adjustable factors dose, dosage form, frequency, and route of administration, and the concentration achieved with time.The pharmacodynamic phase covers the relationship between concentration and both desired and adverse effects produced with time. In simple terms pharmacokinetics my be viewed as what the body does to the drug, and pharmacodynamics as what the drug does to the body.

Elimination drugs with urine. Drugs are administered different routes, mainly intravenously, orally, intramuscularly, subcutaneously or rectally and eventually are eliminated from the body. The drugs may be eliminated chemically changed as a metabolites (enzymatic route) or as a transformers ( non-enzymatic, ie. pH depends transformation: treosulfan), or they may be eliminated intact. Drugs and especially their water soluble their metabolites are eliminated mainly the kidney route with urine. Some drugs are eliminated by excretion in the bile, some undergo recirculation (piroxicam). The main factors which influence on elimination of drug with urine are following: solubility - drug or metabolite must be water soluble, weakly bound to the blood proteins, acidity of urine affected also by diet and disorders. After oral administration cumulative amounts of the drug or metabolite in urine are showed on Fig. 8.1.

80

Fig. 8.1. Numeric plot of cumulative amount drug excreted in urine versus time described by given equation (1) in one compartment model after intravenous administration.

Linear equation (rearranged form of Eq.1), to calculate elimination rate constant ke represents following equation with corresponding graph (Fig. 8.2):

Fig. 8.2. Semilog amount of drug remaining to be eliminated. Slope represents elimination rate constant

(1) )e - (1X=X tek-UU ∞

(2)t klnXXXln eUUU

81

Fig. 8.3. Cumulative amount drug excreted in urine versus time described by given equation (3) in one compartment model after oral administration.

Linear equation (rearranged form of Eq.3), to calculate elimination rate constant ke afteroral administration, in one compartment model, when fast absorption has been observe represents following equation:

From slope (-a=ke) elimination rate constant is calculated.

The elimination half life for the first order kinetics is calculated from equation:

e0.5 k

0.693t

Elimination Half Life (t0,5) of the drug is the time (in hours, hrs) necessary to eliminate

the drug concentration in blood, plasma or serum to one half after equilibrium is

reached, at terminal phase of the drug elimination. The following factors can influence

on t0,5: dose size, age, protein binding , diseases - especially renal and liver, intersubject

variation, pH of urine, multidrug therapy, specific feature of the drug, what kind of

pharmacokinetics the drug represents (linear or nonlinear).

References

1. Ritschel W.A., Kearns G.L. Handbook of basic pharmacokinetics… including clinical application APhA, Washington, D.C. 2004.2. Shargel L. Wu-Pong S., Yu A.B.C. Applied biopharmaceutics & pharmacokinetics. McGraw Hill 2005.3. Tozer T.N., Rowland M.: Introduction to pharmacokinetics and pharmacodynamics.

)( ][ 3).e ak-.e e(k ek-ak

1 + 1X =X tek -tak -

0DU

(4)t ek - ek -akak •X

ln =)X -ln(XU

U∞

∞

82

The Quantitative basis of drug therapy. Lippincott Williams & Wilkins, 2006.

Experimental

Procedure of taking Polopiryna tablets with acetylsalicylic acid by students

Following oral administration, to student with a tolerance to salicylates, of two “Polopiryna” tablets containing 300 mg acetylsalicylic acid (ASA) each volumes of urine were collected, and measured. Collecting time was also marked and 10 ml of each urine sample were brought to the class.

Detailed procedure for collecting of urine samplesFor a urinary excretion study based upon single dose administration, the following prerequisites must be met:1. to obtain a sufficient number of urine samples during the first hours after

administration, 400 ml of water is taken after overnight fasting one hour before the experiment, and 200 ml of water is taken at the time of administration of the drug product

2. immediately before administration of the drug, the bladder is emptied completely, a sample of urine is kept as a blank

3. for each urine sample obtained, the exact time should be noted and the volume of urine excreted should be measured

4. a small sample (10-15 ml) of the urine is kept in the freezer until analysis is performed

5. urine must be collected until practically all of the unchanged drug is excreted (24 h) 6. it is essential that all urine samples be collected. If only urine sample is lost the

entire experiment is invalidated.

Absorbances of the salicylates after derivatization with Trinder reagent were measured aaccording to procedure given at section Determination of concentration of salicylates in urine samples. The data were used to calculate the elimination rate constant and the biological half life of salicylates from human body.

ApparatusSpectrophotometer Specol (Germany)

Solutions1. Trinder reagent - 40g HgCl2 was dissolved in 850 ml water, subsequently 120 ml of 1M/l HCl and 40 g Fe(NO3)3 was added 2. Stock solution of salcylate. Salicylate sodium 0.5 g was diluted in water and filled up to 250 ml in flask. Final concentration of salicylates was 2 mg/ml.

Calibration curve preparation.

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Volumes of 2 ml urine (without salicylates) were transferred for four tubes and subsequently I- 0.20; II - 0,50; III - 0,80 and IV - 1.0 ml of salicylates stock solutionwere added. At the end water in volumes 7.8 ml (I); 7,5 ml (II); 7.2 (III) and 7.0 ml (IV) was added to I-IV tubes to receive final volume 10 ml. Then the solutions were shaken. Then to 1 ml of each (I-IV) volumes 5ml of Trinder reagent were added andabsorptions were measured at wavelength max= 540 nm.

Blank sample was prepared using 2 ml urine (free of salicylates) and 8 ml water. After shaking, to volume 1 ml of the solution, 5 ml of Trinder reagent was added.

Prepare graph representing relationship of absorbance as a function of salicylate concentrations; A=f (c). Calculate the calibrate curve using last square method.

Table 8.1. Calibration curve for salicylates in urine

Solution[mL]

0 I II III IV

Blank urine 2.0 2.0 2.0 2.0 2.0

Natriumsalicylate[2mg/ml]

0 0.2 0.5 0.8 1.0

Waterp.a.

8.0 7.8 7.5 7.2 7.0

Salicylateconcentration

[mg/ml]0 ……… ……….. .......……. …….....

Absorbance(AU)

………… ………... ………… ………… ………….

Calibration curve equation: y=axA= …….c

Determination of concentration of salicylates in urine samples

2ml urine volume with salicylates was diluted 8 ml was, after shaking 1ml was taken to new tube and volume 5 ml of Trinder reagent was added. The absorption was measured as previously explained.

From the absorbances of salicylates in urine the quantity of acetylsalicylic acid werecalculated using equation:

1.12519VNa

AΣU i

it

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A-absorbancea-slopeN- dilution of the sample (n=5)1.12519 – coefficient received from ratio of molecular mass of acetylsalicylic acid (180.154 g/mol) and sodium salicylate (160. 1 g/mol). Prepare graph representing relationship of amounts of ASA as a function of time;

U=f(t). Calculate amounts of ASA at affinity (U).

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Report 8

Name and surname:.......................................................................... Date:....................................

Determination of pharmacokinetic parameters of salicytatesAim of the experiment ……………………………………………………………………………….……………..….

…………………………………………………………………………………………………………………………………………………

………………………………………………

Results from salicylates urinary excretion

No. Time of collectingof urine [h]

Time elapsedfrom “time 0”

Volume of urine[ml]

Absorbance[AU]

Amounts [mg]

1.12519VNa

AΣX i

iU t

tUU XX

[mg]

)Xln(XtUU

1……………… …..……… …..……… …..……… …..……… …..……… …..………

2……………… …..……… …..……… …..……… …..……… …..……… …..………

3……………… …..……… …..……… …..……… …..……… …..……… …..………

4……………… …..……… …..……… …..……… …..……… …..……… …..………

5……………… …..……… …..……… …..……… …..……… …..……… …..………

6……………… …..……… …..……… …..……… …..……… …..………

7……………… …..……… …..……… …..……… …..……… …..……… …..………

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….. ……………… …..……… …..……… …..……… …..……… …..……… …..………

87

Prepare graph representing relationship ln(XU-XUt)=f(t) on semi-logarithmic paper.

tk)kk

kln(X)Xln(X e

ea

autUU

From the slope a calculated last square method determine the elimination rate constant. The biological half life (t0.5) calculate from equation

e0.5 k

0.693t

Summary. Compare t0,5 obtained from your data with that presented in literature..

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Quantum mechanics

Experiment 9. Application of molecular modeling to determination of physicochemical properties of medicinal substances

Aim of the experiment: 1. Getting acquainted with the methods of molecular modeling using the Marvin software:

geometric and physicochemical parameters used in the QSAR analysis will be calculatedfor the two chosen molecules and the impact of the drug structure on its physicochemicalproperties will be assessed.

2. Analysis of the 1H NMR spectrum of a simple organic compound.

Required knowledge: basic concepts of molecular spectroscopy (electromagneticradiation and its characteristics frequency, wavelength, wavenumber, types of energy ofmolecules, quantization of energy, the Bohr frequency condition; polarization andpolarizability, dipole moment; molar refractivity), nuclear magnetic resonance (basicprinciples and rules of analysis of 1H NMR spectra), methods of molecular modeling (abinitio, first principle, semi-empirical, empirical), drug discovery and optimization, theLipinski rule, SAR and QSAR.

Introduction

Molecular modelingDiscovery of new synthetic drugs requires long-term and expensive studies conducted

first in vitro, and then in vivo in animals (preclinical studies) and finally in humans (clinicaltrials). The objective of the in vitro experiments is to determine physicochemical properties ofa number of chemical compounds and to make some preliminary estimation of their biologicalactivity, thus select one or several most promising compounds that will be directed to theanimal experiments. The usage of computational chemistry for molecular modeling in amodern drug design, known as in silico studies or computer-aided drug design (CADD), playsmore and more increasing role. Its goal is to replace a part of the in vitro studies, therebyspeed up the selection of the best candidates for new drugs. The CADD is particularlyvaluable in the analysis of completely new compounds, i.e. those neither existing in nature norsynthesized, because it enables to screen a large group of chemicals, with reducing the cost ofthe research.

A principle of all methods of molecular modeling, regardless of their complexity, is tofind the structure of a molecule that have the lowest possible energy (global minimumenergy). The total energy of the molecule comprises translational, vibrational, rotational andelectronic energy. Among them, only the translational energy is not quantized, so it can varycontinuously (the average translational energy of the molecule increases with temperaturebeing equal to 3/2 kT, where k is Boltzmann’s constant). The other three types of the energyare quantized, which means they can only take on strictly defined values (the discrete levels).

Methods of molecular modeling may be divided into:

1. Methods using quantum mechanics (quantum mechanical methods) – they are based ondefining the wavefunction which is the solution of the Schrödinger equation:

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a) Ab initio methods (Latin ‘from the beginning’) they describe the energy of amolecule using a many-electron wavefunction that takes into account the contribution ofall the electrons of the molecule. Since an analytical solution of the Schrödingerequation is possible only for a one-electron system (e.g. dihydrogen cation H2

+), thewavefunction is only obtained numerically using a number of approximations, includingthe Born-Oppenheimer approximation (neglect of motion of atomic nuclei) and the one-electron approximation (assumption that each electron moves independently of theothers). The most common approach to a practical realization of the aboveapproximations is the self-consistent field method (SCF), also called the Hartree-Fockmethod. An unquestionable advantage of the ab initio methods is their ability todetermine a vast variety of structural, spectroscopic, and thermodynamic properties ofmolecules, to describe the environmental effects such as solvation, or even to predict thepaths of chemical reactions. Additionally, the calculation results obtained with thesemethods are usually very close to the experimental values. On the other hand, adisadvantage of the ab initio methods is that they require extreme computing power,which limits their application to the analysis of systems containing up to about a dozenatoms and makes they are associated with a high cost of purchase of sophisticatedcomputer systems.

b) First principle methods, often called the density functional theory methods (DFT) –they are based on replacing a complex many-electron wavefunction by a concept ofelectron density. With this simplification, the efficiency of the DFT calculations isgreater than the ab initio methods. As a result, molecules consisted of about 100 atomscan be analyzed with only a little loss of accuracy of the calculations in comparison tothe experimental data.

c) Semi-empirical methods they generally use the same principles as the ab initio orDFT methods but only the valence electrons are included in quantum calculations.Contribution of the atomic core electrons to the energy of the molecule is defined by theappropriate constant values or completely disregarded. Such an approach is partlyjustified by the fact that the valence electrons have the highest energy among the others,therefore they govern the chemical properties of the molecule. A major advantage of thesemi-empirical methods is their speed and the ability to analyze more complex systems,such as active sites and enzymes that number even up to 10,000 atoms. On the otherhand, the simplifications applied in these methods cause the obtained results are quiteoften incompatible with the experimental data. For this reason, the semi-empiricalmethods must be used with great caution. The two most accurate semi-empiricalmethods, and hence the most commonly used, are AM1 and PM3.

2. Methods using molecular mechanics (classical, Newtonian mechanics):

Empirical methods, also called the force fields. In these methods, molecules are treated asrigid balls (atoms) linked by springs that imitate chemical bonds. In other words, the moleculeis a set of harmonic oscillators. The existence of electrons is neglected due to their extremelylow mass when compared to atomic nuclei. The empirical methods assumes that a totalpotential energy of the molecule is a sum of all the potentials present in the molecule, namely,the energies related to the changes of the bond lengths, flat angles and torsion (dihedral)angles, the energies of electrostatic, hydrogen, and dipole – dipole interactions, etc.Application of the force fields relies upon the assumption that particular groups of atoms havesimilar properties in different chemical compounds. Accordingly, creation of the optimalmolecule structure consists in adding the ‘bricks’ (chemical moieties), of which propertieshave been earlier defined by analysis of some other chemical compounds (hence the term

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empirical). A special type of the empirical methods is molecular dynamics that enables tosimulate the time evolution of a system, such as oscillation and rotation of the chemicalbonds. However, due to treating the molecule as a complex harmonic oscillator, the empiricalmethods, in contrast to the quantum mechanical methods, cannot predict any chemicalreaction. Nevertheless, their advantage is a simplicity, a short time of the calculations, and theability to analyze very complex systems, even composed of hundreds of thousands of atoms,e.g. viruses.

Drug designFor the drug designing it is practical to divide the structure of a molecule into two

main elements:

Pharmacophore the part of the molecule that accounts for its interaction with a biologicaltarget (e.g., protein, receptor, enzyme, DNA). The pharmacophore provides the biologicalactivity of the drug (the ligand). It is usually made up by several chemical moieties (e.g., anaromatic ring, a hydrophobic alkyl group, a hydrophilic functional group being a donor or anacceptor for H+ ions, etc.) that can develop different types of interactions with the receptor,such as ion ion, ion dipole, dipole dipole induced, induced dipole induced dipole orhydrophobic interaction. The interaction of the drug with an active center of the receptorrequires a steric, electrostatic, and hydrophobic complementarity. The interaction of the drugwith the receptor via the pharmacophore can activate the receptor (then the drug is called theagonist) or only block the access to the receptor against the natural agonists (e.g. access ofnoradrenaline to β-adrenergic receptors) then the drug is called the antagonist.

Vector groups the fragments of the molecule responsible for pharmacokinetics and toxicityof the drug. They include:▪ carrier groups that affect ionization and lipophilicity of the drug and, consequently, its

absorption, distribution and excretion;

▪ vulnerable groups that are susceptible to enzymes action, thus affect the drug metabolism.

For many past years, new drugs have been discovered by studies of biological activityof substances found in nature or randomly synthesized in the laboratory. Currently, the mostimportant approach in the drug discovery is the so-called rational drug design that comprisesthe following steps:

a) Identification of a molecular target (e.g., receptor, protein, enzyme) involved in themechanism of the disease development or alleviation of its symptoms.

b) Establishment of a pharmacophore structure that provides acceptable affinity andselectivity to the molecular target. It can be done by:

▪ studying the three-dimensional structure of the active site of the receptor or the enzyme(Receptor-Based Design, RBD);

▪ analyzing the structure of at least several ligands that are already known to bind thereceptor (Ligand-Based Design, LBD), in case of the three-dimensional structure of thereceptor is not known.

c) Designing the so-called lead compound by molecular modeling methods or selection of thelead compound from a wide pool of the already synthesized substances. The leadcompound must contain the pharmacophore established at the previous steps, thus exhibitssome biological activity and constitutes a starting point for designing the final structure ofthe future drug.

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d) Optimization of the lead compound structure by modification of the vector groups in orderto:

▪ improve the pharmacokinetic parameters of the compound (absorption, distribution,metabolism and excretion)

▪ increase the selectivity of the biological action of the compound

▪ reduce the compound toxicity.

e) Synthesis and in vitro testing of analogs of the lead compound. The analogs are chemicalcompounds having identical or very similar pharmacophore as the lead compound butother vector groups. The analogs can form a homologous series, so differ from each otherby the repeating fragment, e.g. a methylene group (CH2).

f) Selection of the analog that possesses the best properties and will be further examined invivo in the animals.

At the step of optimization of the lead compound the Lipinski rule as well as SARand QSAR methods are of crucial importance.

The Lipinski rule is an empirical rule used to predict a drug-likeness, that is whethera given compound being a candidate for the drug (e.g. the lead compound) may exhibitdesirable pharmacokinetic properties in terms of its absorption, distribution, metabolism andexcretion. The rule has been developed by making the statistical analysis of physicochemicalproperties of more than two thousand drugs possessing different pharmacophores andmechanisms of action, which were subjects of phase II clinical trials. The Lipinski rule statesthat poor absorption or permeation of the compound through biological membranes is likely ifit has the following physicochemical properties:

▪ molar mass > 500

▪ logarithm of the n-octanol/water partition coefficient (log P) > 5

▪ number of hydrogen bond donors (sum of OH and NH groups) > 5

▪ number of acceptors for hydrogen bonds (sum of O and N atoms) > 10.

The possibility of determining the above parameters by a molecular modeling makesthe Lipinski rule the most commonly used criterion for selection of compounds with drug-likeproperties, especially those which are supposed to be administered orally. A lot of the studiesconfirmed the validity of the Lipinski rule for the majority of medicinal substances. Theexceptions from this rule among the oral drugs are substrates of membrane transporters:antibiotics, antifungals, vitamins and cardiac glycosides.

The SAR method (structure-activity relationship) relies on determination of therelationship between a structure of the structurally related compounds and their biologicalactivity. This relationship may concern physicochemical properties (solubility, stability),pharmacokinetic properties (absorption, distribution, metabolism, excretion) andpharmacodynamic properties (receptor – drug interactions). The relationship between thestructure and the activity of the substances can be determined in a qualitative way only(simple SAR) or quantitative one (quantitative SAR, that is QSAR). The most importantadvantage of both these methods, especially QSAR, in the context of a rational drug design istheir ability to predict the biological activity of new substances at the steps of the leadcompound optimization and selection of its analogs.

QSAR method relies on four basic principles:

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1. Properties of molecules can be described by measurable physicochemical parameterscalled the descriptors.

2. The biological activity of a compound is a measurable quantity, usually expressed as theminimum concentration of the compound inducing a specific biological effect.

3. Compounds that form the analyzed series possess the same mechanism of action.

4. The relationship between the descriptors and the measurable biological activity can beexpressed as the mathematical equation in the form of biological activity = f(chemicalstructure).

The aim of the QSAR is to obtain a linear equation describing the biological activity of agiven series of structurally related compounds as a function of the descriptors. Thesedescriptors can be determined experimentally, but more often they are calculated by themolecular modeling methods. For this reason, the QSAR is currently one of the main toolsused in a rational drug design.

Among more than 1500 descriptors currently used in the QSAR method the followinggroups are distinguished:

1. Molecular descriptors molecular weight, length and width of the molecule.

2. Constitutional descriptors they characterize the structure of the compound molecule, e.g.,the number of atoms, multiple bonds, rings, the value of atomic and ionic radii.

3. Substituent descriptors describe the effect of a substituent on the properties of the wholemolecule, for example Hammett constant describes the influence of the substituent on theacid dissociation of benzoic acid analogs.

4. Geometric and topological descriptors characterize the interatomic distances, the distancefrom the center of the mass, the number of chemical bonds, the value of angles between thebonds.

5. Conformational descriptors describe, for example, the energies of the conformations ofthe molecule or the number of carbon atoms with sp3 hybridization which enable freerotation of the molecule fragments.

6. Steric descriptors are a measure of the size and shape of the whole molecule or itsparticular functional groups:

▪ Molar refractivity (R) characterizes the volume of the whole molecule (unit:

m3/mol). According to the Lorenz-Lorentz equation:

ρ

M

2n

1n2

2

R (1)

where: n refractive index of the substance; M molar mass; ρ density of the substance.Ratio M/ρ corresponds to the molar volume of the substance, so the larger the volume of themolecule, the higher the molar refractivity R.

▪ Taft constant describes a steric effect of the individual substituent.

7. Electrostatic descriptors they include, among others, a total electronegativity of theatoms, a molar polarizability, an electric dipole moment, charges and valence of the atoms,the number of donors and acceptors for hydrogen bonds, energy of the highest occupiedmolecular orbital (HOMO), energy of the lowest unoccupied molecular orbital (LUMO):

93

▪ Dipole moment it characterizes a polarity of molecules that do not possess a center ofsymmetry. An electric dipole consists of two equal-sized charges of opposite sign (+qand q) spaced by a distance r. The electric dipole moment is a vector quantity, whichmeans it has its magnitude as well as direction (pointed from the negative to the positivecharge). The magnitude of the electric dipole moment of a single polarized bond isexpressed by the equation:

μ = q r (2)

The total electric dipole moment of a polyatomic molecule is a vector sum of the dipolemoments of all the bonds. The higher value of the dipole moment, the higher the polarity ofthe molecule. The SI unit of a dipole moment is the coulomb meter (C m), and the traditionalnon-SI unit is the debye (1 D = 3.33 1030 C m). An advantage of using a dipole momentexpressed in debye is the fact that its value is then close to the difference in electronegativitiesof the atoms forming a given chemical bond. Consequently, the approximate value of thedipole moment of the given bond can be easily estimated from the Pauling's electronegativityscale. Moreover, the debye is convenient because dipole moments of majority of moleculesare of order of 1 D.

▪ Polarizability (α) is a measure of the ability of a molecule to undergo the phenomenonof polarization, that is adopting an orientation parallel to the direction of an appliedexternal electric field. There are two types of polarizability (and also polarization):

Orientation polarizability it is attributed only to polar molecules that have apermanent dipole moment. The orientation polarizability (αorient) is directlyproportional to a square of the permanent dipole moment and inverselyproportional to the absolute temperature since the thermal motion of the moleculesdisturb their arrangement parallel to the electric field lines: αorient = μ2/(3kT).

Induced polarizability relates to all molecules, both polar and nonpolar ones. Itoccurs due to the fact that an external electric field, for instance the one generatedby a nearby ion or polar molecule, distorts the electron shells and bring about adisplacement of the atomic nuclei in the given molecule. Thus, a temporary induceddipole moment arises that is proportional to the induced polarizability (αind).

The polarizabilities α are often reported as polarizability volumes α‘ (α‘ = α/4πεo, where εo –electric permittivity of a vacuum) that have the dimension of volume, often Å3 (1 Å(angstrem) = 1010 m). Noteworthy, the polarizability (microscopic quantity) of a givenperfectly homogeneous and isotropic substance is connected with its relative permittivity(macroscopic quantity), as expressed by:

o

indorientA

r

r

ε 3

ααN

ρ

M

2ε

1ε

(3)

where: εr – relative permittivity (dielectric constant) of the substance. The right-sideexpression in the above equation defines the so-called molar polarization (P).

The relation analogous to eqn (3) but without the contribution from the orientationpolarizability is called the Clausius – Mossotti equation:

o

indA

r

r

ε 3

αN

ρ

M

2ε

1ε

(4)

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where εr is now measured when the orientation polarizability does not exist, either because themolecules are nonpolar or because of the high frequency of the applied field (as explainedbelow).

Total polarizability of a polar substance is a sum of the orientation polarizability and theinduced polarizability. The alone induced polarizability of polar substances can be determinedby measuring their relative permittivity in a variable (not constant!) electric field of whichfrequency exceeds 1012 Hz. Then, the polar molecules cannot reorientate fast enough to followthe changes in the direction of the electric field and, as a result, the orientation polarizabilityis lost. Interestingly, the similar principle is applied in microwave ovens, where polarmolecules of water that cannot keep up with the fast changes in the electric field frequency,release much energy to the surroundings as heat.

8. Thermodynamic descriptors e.g., molar enthalpy of formation, molar enthalpy ofsolvation, logP, lipophilicity substituent constant (π), and logD:

▪ LogP logarithm of the n-octanol/water partition coefficient, that is a ratio ofconcentrations of the undissociated molecules of the substance in n-octanol and water atpartitioning equilibrium. It is a measure of lipophilicity of the whole molecule.

(water)

octanol)(n

c

cloglogP (5)

▪ Lipophilicity substituent constant (π) specifies the effect of a given substituent onthe lipophilicity of the molecule:

H

XHX P

PloglogPlogPπ (6)

where: PX n-octanol/water partition coefficient of the compound containing the substituentX; PH n-octanol/water coefficient of the reference compound containing a hydrogen atominstead of the substituent X.

▪ LogD logarithm of the n-octanol/water distribution coefficient, that is a ratio of totalconcentrations of the molecules (undissociated and dissociated ones) of the substance inn-octanol and water at the partitioning equilibrium:

(water)

octanol)(n

c

cloglogD (7)

LogD is a measure of distribution of molecules of a weak electrolyte (acid or base) betweenthe nonpolar (n-octanol) and the polar phase (water) at given pH. The dissociated moleculesof the electrolyte (ions) accumulate only in the aqueous phase, while the undissociatedmolecules distribute between the two phases according to the value of the logP. On the otherhand, a ratio of the concentrations of the dissociated to the undissociated molecules dependson a character of the weak electrolyte (acid or base), the value of the dissociation constant andthe pH of the aqueous phase. For example, at high pH, a weak acid exists mainly in adissociated form and accumulates in the aqueous phase, while the predominant form of aweak base are the undissociated molecules that generally have a higher affinity to n-octanol.Consequently, logD values of weak acids decrease with increasing pH of the aqueous phase,while the logD of the weak bases increases. Consequently, the knowledge of the logD can beused in practice to evaluate the extent of absorption of weak electrolytes from various parts ofa gastrointestinal tract and to predict in which compartment of the body they will accumulate.

95

References:

1. Hermann T.W. (Red.): Chemia fizyczna. Wydawnictwo Lekarskie PZWL, Warszawa2007.

2. Pandit NK: Introduction to the Pharmaceutical Sciences. Lippincott Williams & Wilkins,Philadelphia 2007.

3. Atkins P., de Paula J.: Physical Chemistry, 8th Ed., Oxford University Press Inc., Oxford2006.

4. Atkins P., de Paula J.: Elements of Physical Chemistry, 5th Ed., Oxford University PressInc., Oxford 2009.

5. Lipinski C.A. et al.: Experimental and computational approaches to estimate solubilityand permeability in drug discovery and development settings. Adv. Drug Deliv. Rev. 1997, 3,3-25.

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Experimental

Equipment: MarvinSuite 6.2 software (ChemAxon, Budapest, Hungary).

Procedure

A. Analysis of the structure and properties of the therapeutic compounds

1. Open the MarvinSketch of which icon is placed on the desktop.

2. Consult the assistant in order to choose the two molecules that will be analyzed during theexercise.

3. Using the tools of the MarvinSketch draw the structure of the molecule:

a. Single atoms are added by a single click in the program window. The type of the atomcan be changed using the periodic table or the particular icons placed on the right sideof the window.

b. Chemical bonds are drawn by clicking on the atom and then dragging the cursor to the

other or by using the tool Bond located on the toolbar at the left top of thewindow. Multiple bonds are added by clicking on the already drawn single bond. Thisoption also enables to draw the chemical bonds in the appropriate spatial orientation.This is very important in the case of drugs having a chiral center since the individualstereoisomers often possess different pharmacokinetic and pharmacodynamicproperties. Hence, when drawing molecules of such drugs the configuration of thestereogenic center should be taken into account.

c. Carbon chains may be introduced using the tool Chain .

d. Incorrectly drawn atoms and bonds can be deleted by using the eraser located at theleft top of the window.

e. When drawing more complex compounds, the predefined structures (templates) fromthe software database can be used. Their icons are placed on the toolbar at the bottomof the window.

4. When the whole molecule is drawn, select Structure > Clean 2D. The structure of themolecule will be ordered.

5. A final three-dimensional image of the molecule is obtained from the menu Clean 3D. Tooptimize the molecule structure (find the structure with the lowest energy), from the menuStructure > Clean 3D > Cleaning Method, select Fast Build that finds the lowestenergy conformer, and then Gradient Optimize. For optimization of the moleculestructure the MarvinSketch uses the force fields theory, namely the DREIDING method.

6. The resulting molecule can be visualized in various ways:

a. Bezene ring can be represented by the aromatic or the Kekulé structure (Structure >Aromatic Form > Convert to Aromatic Form/Convert to Kekule Form).

b. From the menu View > Structure Display many ways of the molecule visualizationcan be chosen. The molecule can be rotated freely by selecting View > Mouse Mode >Rotate in 3D.

97

c. The option Structure > Add > Add Explicit Hydrogens adds the ‘hidden’ hydrogenatoms.

7. Draw the molecule with the optimized structure into your report. The drawing should belarge and clear.

8. From the menu View, choose Open MarvinSpace. This program enables to measure thebond lengths and the angles between the atoms, including the torsion (dihedral) angles.

a. Length of the bonds is displayed by selecting the tool and clicking on the twoadjacent atoms. The bond length is expressed in Å (angstrem).

b. Flat angles are shown after choosing and then clicking on the three adjacentconsecutive atoms.

c. Torsion angles are displayed by selecting and clicking on the four adjacentconsecutive atoms.

9. Calculations of the physicochemical parameters of the molecule are performed in theMarvinSketch program using the tools from the Calculations menu. The results of thecalculations should be presented in the report.

a. Elemental Analysis displays the approximate molar mass and the elementalcomposition of the compound.

b. Protonation calculates the pKa of the drug. For most of molecules, more than onepKa value will be displayed, depending on a number of the functional groups capableof dissociating. In that case, each value of the pKa should be assigned to theappropriate group.

c. Partitioning calculates the logP and logD. The logD values should be calculated for

pH 1.5 and 7.4.

d. Charge this module estimates the properties of molecules related to their electriccharges.

i. Charge calculates the charges on the individual atoms (the values are expressedin e, which is the charge carried by a single proton or electron, called theelementary charge; 1 e = 1.6 1019 C).

ii. Polarizability calculates the polarizability volume (α’) of the molecule,expressed in Å3.

iii. Dipole Moment Calculation displays the electric dipole moment and its valueexpressed in D (debye). The dipole moment vector should be drawn into thestructural formula of the molecule in the report.

e. Geometry this module calculates the geometric parameters of the molecule.

i. Polar Surface Area calculates the area formed by the polar atoms.

98

ii. Molecular Surface Area calculates the total surface of the molecule, expressedin A2. The result can be displayed as the van der Waals surface or the solventaccessible surface. The default value of the solvent molecule radius, that is 1.4 Å,is attributed to water. The results displayed in the window are listed as follows:ASA the total solvent accessible surface, ASA/ASA+ the solvent accessiblesurface formed by the negatively/positively charged atoms, ASA_H/ASA_P thesolvent accessible surface formed by the hydrophobic/polar atoms.

f. Other

i. H Bond Donor/Acceptor displays the number of donors or acceptors forhydrogen bonds in the molecule (‘donor/acceptor count’ option should bechosen).

ii. Refractivity calculates the molar refractivity in m3/mol; the displayed value is aresult of multiplication of the real refractivity by 106.

10. On the basis of the calculated parameters assess whether the molecule satisfies the criteriaof the Lipinski rule (‘the rule of five’). Draw the appropriate conclusions.

11. Using the online databases (e.g. drugbank.ca) find the experimental value of the pKa and

logP and compare them to the values calculated by the Marvin program.

12. All the above procedures should be repeated for the second of the chosen molecules.

13. With the help of the assistant, discuss the similarities and differences in the structure ofboth the analyzed drugs and deduce how the structural change affects thephysicochemical properties. The conclusions should be placed in the report.

B. Analysis of 1 H NMR spectrum of the chosen simple molecule

1. With the help of the assistant, choose a simple molecule of an organic compound to beanalyzed.

2. On the basis of your knowledge predict what signals will be present in the 1H NMRspectrum of the chosen compound.

3. Draw the molecule in the MarvinSketch and optimize its structure as described in the pointA-5.

4. From the menu Calculations > NMR select HNMR Prediction. The 1H NMR spectrumof the compound will be displayed in the new window. After clicking on the peak presentin the spectrum, the hydrogen atoms that generate the given signal are highlighted. Discussthe results with the assistant.

5. Draw the structural formula of the compound and its 1H NMR spectrum into your reportand describe which hydrogen atoms generate the particular signal.

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Report 9

Name and surname..................................................................... Date: ...................................

Application of molecular modeling to determination of physicochemical properties ofmedicinal substances

Aim of the experiment: ................................................................................................................

.......................................................................................................................................................

.......................................................................................................................................................

.......................................................................................................................................................

Results

A. Analysis of the structure and properties of the therapeutic compoundsMolecule 11. Drug name: …………………………………………………………………….2. Drawing of the molecule with assigning of:- length of 5 bonds of different type- value of 5 flat angles of different type- value of 2 torsion (dihedral) angles- value of the positive and negative charge on a total of 8 chosen atoms- vector of the electric dipole moment

3.

A

n

a

l

y

s

i

s

of the physicochemical properties of the molecule:

100

Molar mass [g/mol]:……………….. Is it greater than 500 g/mol? YES NO

Number of donors for hydrogen bonds:…… Is it greater than 5? YES NO

Number of acceptors for hydrogen bonds:… Is it greater than 10? YES NO

LogP:……………. Is it greater than 5? YES NO

Does the compound satisfy the Lipinski rule? YES NO

Van der Waals molecular surface: …………………...................... [...........]

Water accessible molecular surface: …………........... [...........]

Molar refractivity: ……………........………….. [..................]

Polarizability volume: …….....……… [...........]

Dipole moment: .....................[D] = ....................................................[Cm]

LogDpH=1.5 :………………………… LogDpH=7.4: ……………………………..

Character of the substance: Nonelectrolyte Electrolyte (weak acid weak base )

For the weak acid: pKa = ………… or For the weak base: pKa (of the conjugate acid) = ………(pKb = 14 pKa = ………………)

Assessment of the extent of absorption from a gastrointestinal tract:

Absorption from a stomach (pH 1.5) into bloodcirculation (pH 7.4)

Absorption from an intestine (pH 7.0)into blood circulation (pH 7.4)

............................

...........................

stomach

blood

101

101

c

c

............................

...........................

intestine

blood

101

101

c

c

Conclusion regarding the extent of absorption of the analyzed substance from a stomach andan intestine: ……………………………………………………………………………………………………………………………………………………………………………………………………

4. The selected experimental data found in the online database drugbank.ca:

pKa: ………………….. LogP:……………………………………………….

101

apKypH

apKxpH

y

x

101

101

c

c

Formula for a weak acid

ypHapK

xpHapK

y

x

101

101

c

c

Formula for a weak base

http://www.drugbank.ca/

Molecule 2

1. Drug name: …………………………………………………………………….2. Drawing of the molecule with assigning of:- length of 5 bonds of different type- value of 5 flat angles of different type- value of 2 torsion (dihedral) angles- value of the positive and negative charge on a total of 8 chosen atoms- vector of the dipole moment

3.

A

n

a

l

y

s

i

s

of the physicochemical properties of the molecule:

Molar mass [g/mol]:……………….. Is it greater than 500 g/mol? YES NO

Number of donors for hydrogen bonds:…… Is it greater than 5? YES NO

Number of acceptors for hydrogen bonds:… Is it greater than 10? YES NO

LogP:……………. Is it greater than 5? YES NO

Does the substance satisfy the Lipinski rule? YES NO

Van der Waals molecular surface: …………………...................... [..........]

Water accessible molecular surface: …………............. [..........]

Molar refractivity: ……………........………….. [..................]

102

Polarizability volume: …….....……… [...........]

Dipole moment: .....................[D] = ...............................................[Cm]

LogDpH=1.5 :………………………… LogDpH=7.4: ……………………………..

Character of the substance: Nonelectrolyte Electrolyte (weak acid weak base )

For the weak acid: pKa = ………… or For the weak base: pKa (of the conjugate acid) = ………(pKb = 14 pKa = ………………)

Assessment of extent of absorption from a gastrointestinal tract:

Absorption from a stomach (pH 1.5) into bloodcirculation (pH 7.4)

Absorption from an intestine (pH 7.0)into blood circulation (pH 7.4)

............................

...........................

stomach

blood

101

101

c

c

............................

...........................

intestine

blood

101

101

c

c

Conclusion regarding the extent of absorption of the analyzed substance from a stomach and an intestine: ………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………

4. The selected experimental data found in the online database drugbank.ca:

pKa: ………………….. LogP:……………………………………………….

Conclusions:

103

apKypH

apKxpH

y

x

101

101

c

c

Formula for a weak acid

ypHapK

xpHapK

y

x

101

101

c

c

Formula for a weak base

http://www.drugbank.ca/

B. Analysis of the 1 H NMR spectrum of the chosen molecule1. Structural formula:

2. 1H NMR spectrum:

Conclusions:…………………………………………………………………………………………………………………………………………………………………………………………………..…………………………………………………………………………………………………………………………………………………………………………………………………..…………………………………………………………………………………...........................................................................................................................................................................................................................................................................................................................

104

Properties of colloids. Viscosity

Experiment 10a. Determination of the isoelectrical point of gelatin

The aim of the experiment is measurement of viscosity of gelatin solutions at different pH inorder to determine the pH at which the viscosity possesses the lowest value

Required knowledge: isoelectrical point, viscosity, methods for viscosity determination

Introduction

The isoelectric point (pI) is the pH at which a particular molecule or surface carries no netelectrical charge. Amphoteric molecules called zwitterions contain both positive and negativecharges depending on the functional groups present in the molecule. They are affected by pH of theirsurrounding environment and can become more positively or negatively charged due to the loss orgain of protons (H+). A molecule's pI can affect its solubility at a certain pH. Such molecules haveminimum solubility in water or salt solutions at the pH which corresponds to their pI and are oftenseen to precipitate out of solution. Moreover, pI can affect other physical properties of molecule,such as viscosity, conductivity and osmotic pressure. Biological amphoteric molecules such asproteins contain both acidic and basic functional groups. Amino acids which make up proteins maybe positive, negative, neutral or polar in nature, and together give a protein its overall charge. At apH below their pI, proteins carry a net positive charge. Above their pI they carry a net negativecharge.

Gelatin is a heterogeneous mixture of water-soluble proteins of high molecular weight. On a dryweight basis, gelatin consists of 98 to 99% protein. The molecular weight of these large proteinstructures typically ranges between 20,000 and 250,000, with some aggregates weighing in themillions. Gelatin is amphoteric, meaning that it is neither acidic nor alkali, but possesses bothproperties depending on the nature of the solution. The pH at which gelatin’s charge in solution isneutral is known as the isoelectric point of gelatin. Its value ranges between 4.0 and 9.4, with acidprocessed gelatins having higher isoelectric points than alkali processed gelatins. The pI of proteinscould be determined by measurements of viscosity of protein solutions at different pH. Viscosity is ameasure of the resistance of a fluid to deformation under shear stress. It is determined with varioustypes of viscometer, typically at standard state.

References:

1. Hermann T.W.: Farmacja fizyczna. Wydawnictwo Lekarskie PZWL, Warszawa 1999.


Experimental

Materials: pH-meter, glass electrode, Ostwald’s viscometer, stop-watch, 3 volumetric pipettes of 20ml, a volumetric pipette of 2 ml, 6 Erlenmeyer flasks of 100 ml, 6 beakers of 50 ml, 0.5 % solutionof gelatin in water, acetic acid of 1 mol/l, 1 mol/l solution of NaOH in water.

105

http://en.wikipedia.org/wiki/Proteins

http://en.wikipedia.org/wiki/PH

http://en.wikipedia.org/wiki/Functional_groups

http://en.wikipedia.org/wiki/Amphoteric

http://en.wikipedia.org/wiki/Solution

http://en.wikipedia.org/wiki/Precipitate

http://en.wikipedia.org/wiki/Solubility

http://en.wikipedia.org/wiki/Protons#In_Physics_and_biochemistry

http://en.wikipedia.org/wiki/Functional_groups

http://en.wikipedia.org/wiki/Zwitterion

http://en.wikipedia.org/wiki/Amphoteric

http://en.wikipedia.org/wiki/Electric_charge

http://en.wikipedia.org/wiki/Molecule

http://en.wikipedia.org/wiki/PH

Procedure

1. Prepare 6 buffer solutions in Erlenmeyer flasks at following compositions:

Solution 1 2 3 4 5 6

Acetic acid (1 mol/L) [mL]

NaOH (1 mol/L) [mL]

30

0.2

30

1.5

30

10

30

20

30

25

30

28

2. Pour 20 ml of the acetic buffer into the beaker and add 20 ml of gelatin solution. Mix the bothsolutions together.

3. Pour 20 ml of water into the Ostwald’s viscometer. Draw water over the upper calibrationmark of the viscometer capillary using pipette-pump. Measure twice time necessary for thewater flow from the higher mark to the lower mark on the thin arm of viscometer. Repeat themeasurement if the difference between time values 0.3 s.

4. Determine time necessary for the flow of the gelatin solutions mixed with acetic buffer betweenmarks on the viscometer. Try not to foam the solution during drawing it into the arm. Aftermeasurement pour the solution back to the beaker !

5. Measure the pH of the studied solutions.

6. Determine the values of the relative viscosity of the gelatin solutions using mean time of thesolution flow versus mean time of the water flow.

7. Draw the graph for the function: = f(pH).

8. Use the least squares method to determine the parameters of the both parts of the curve

= f(pH).

9. Calculate the isoelectrical point for gelatin using the following equation:

21

12

a-a

b-b = pH

where:

a1, a2, b1, b2 – parameters of the calibration curves: 1) y = a1x + b1

2) y = a2x + b2

106

Experiment 10b. Determination of the emulsion type

The aim of the experiment is to determine the emulsion type of solutions prepared usingcommon emulsifiers and of two drugs: Lorinden, Sulfarinol

Required knowledge: definition and characteristics of emulsion, methods for emulsion typedetermination, definition, characteristics and types of emulsifier agents, HLB

Introduction

An emulsion is a mixture of two immiscible (unblendable) substances. One substance (thedispersed phase) is dispersed in the other (the continuous phase). Emulsions are part of a moregeneral class of two-phase systems of matter called colloids. Although the terms colloid andemulsion are sometimes used interchangeably, emulsion tends to imply that both the dispersed andthe continuous phase are liquid. In a water-in-oil emulsion oil surrounds droplets of water. In an oil-in-water emulsion water surrounds droplets of oil. Emulsion type can be determined by the use ofone of the following methods:

1. Measure of emulsion conductivity. Electric conductivity of o/w emulsion is greater thanconductivity of w/o emulsion.

2. Dissolution. O/w emulsion dissolves in water and w/o emulsion – in oil.

3. Staining, e.g. using Sudan IV which dissolves in oil. Observation under the microscope whichphase has been stained.

Energy input through shaking, stirring, homogenizers, or spray processes are needed to form anemulsion in the process of emulsification. Emulsions tend to have a cloudy appearance, because themany phase interfaces (the boundary between the phases) scatter light that passes through theemulsion. Emulsions are unstable and thus do not form spontaneously. Over time, emulsions tend torevert to the stable state of oil separated from water. Surface active substances (surfactants) canincrease the kinetic stability of emulsions greatly so that, once formed, the emulsion does not changesignificantly over years of storage. Homemade oil and vinegar salad dressing is an example of anunstable emulsion that will quickly separate unless shaken continuously. This phenomenon is calledcoalescence, and happens when small droplets recombine to form bigger ones. Fluid emulsions canalso suffer from creaming, the migration of one of the substances to the top of the emulsion underthe influence of buoyancy or centripetal force when a centrifuge is used. There are three types ofemulsion instability: flocculation, where the particles form clumps; creaming, where the particlesconcentrate towards the surface (or bottom, depending on the relative density of the two phases) ofthe mixture while staying separated; and breaking and coalescence where the particles coalesce andform a layer of liquid.

An emulsifier (also known as an emulgent) is a substance which stabilizes an emulsion,frequently a surfactant. Examples of food emulsifiers are egg yolk (where the main emulsifyingchemical is lecithin), honey and mustard. Detergents are another class of surfactant, whichchemically interact with both oil and water, thus stabilising the interface between oil or waterdroplets in suspension. This principle is exploited in soap to remove grease for the purpose ofcleaning. A wide variety of emulsifiers are used in pharmacy to prepare emulsions such as creamsand lotions. Whether an emulsion turns into a water-in-oil emulsion or an oil-in-water emulsiondepends on the volume fraction of both phases and on the type of emulsifier. Generally, the Bancroft

107

http://en.wikipedia.org/wiki/Bancroft_rule

http://en.wikipedia.org/wiki/Lotion

http://en.wikipedia.org/wiki/Cream_(pharmaceutical)

http://en.wikipedia.org/wiki/Pharmacy

http://en.wikipedia.org/wiki/Cleaning

http://en.wikipedia.org/wiki/Yellow_grease

http://en.wikipedia.org/wiki/Soap

http://en.wikipedia.org/wiki/Water

http://en.wikipedia.org/wiki/Cooking_oil

http://en.wikipedia.org/wiki/Detergent

http://en.wikipedia.org/wiki/Mustard_seed

http://en.wikipedia.org/wiki/Lecithin

http://en.wikipedia.org/wiki/Egg_yolk

http://en.wikipedia.org/wiki/Surfactant

http://en.wikipedia.org/wiki/Coalescence

http://en.wikipedia.org/wiki/Creaming

http://en.wikipedia.org/wiki/Flocculation

http://en.wikipedia.org/wiki/Centrifuge

http://en.wikipedia.org/wiki/Centripetal_force

http://en.wikipedia.org/wiki/Buoyancy

http://en.wikipedia.org/wiki/Creaming_(chemistry)

http://en.wikipedia.org/wiki/Coalescence_(meteorology)

http://en.wikipedia.org/wiki/Salad_dressing

http://en.wikipedia.org/wiki/Vinegar

http://en.wikipedia.org/wiki/Cooking_oil


http://en.wikipedia.org/wiki/Scattering

http://en.wikipedia.org/wiki/Phase_boundary

http://en.wikipedia.org/wiki/Liquid

http://en.wikipedia.org/wiki/Colloid

http://en.wikipedia.org/wiki/Phase_(matter)

http://en.wikipedia.org/wiki/Dispersion

http://en.wikipedia.org/wiki/Phase_(matter)

http://en.wikipedia.org/wiki/Immiscible

rule applies: emulsifiers and emulsifying particles tend to promote dispersion of the phase in whichthey do not dissolve very well; for example, proteins dissolve better in water than in oil and so tendto form oil-in-water emulsions (that is they promote the dispersion of oil droplets throughout acontinuous phase of water).

The Hydrophilic-Lipophilic Balance (HLB) of a surfactant is a measure of the degree to which itis hydrophilic or lipophilic, determined by calculating values for the different regions of themolecule. Surfactants with a low HLB are more lipid loving and thus tend to make a water in oilemulsion while those with a high HLB are more hydrophilic and tend to make an oil in wateremulsion.

References:

1. Hermann T.W.: Farmacja fizyczna. Wydawnictwo Lekarskie PZWL, Warszawa 1999.


Experimental

Materials: mortar, glass pipettes of 1 ml, glass plates, Sulfarinol, Lorinden, oil, 0.1 % NaCl, Tween20, Span 60.

Procedure:

A) Preparation of emulsions

1. Mix 1 ml of NaCl solution with 1 drop of Tween 20 in a mortar. Add 1 ml of oil and stirvigorously until emulsion is formed.

2. Mix 1 ml of oil with a pinch of Span 60. Add 1 ml of NaCl and stir vigorously until emulsionis formed.

3. Determine the type of the prepared emulsions.

B) Determination of the emulsion type in common drugs: Add 1 drop of water or oil to 1 drop of thedrug (Lorinden, Sulfarinol) on a glass plate and observe if the fluids mix together. Decide aboutthe emulsion type.

108

http://en.wikipedia.org/wiki/Lipophilic

http://en.wikipedia.org/wiki/Hydrophilic


http://en.wikipedia.org/wiki/Bancroft_rule

Report 10

Name and surname:........................................................ Date:....................................

Experiment 10aDetermination of the isoelectric point of gelatin

Aim of the experiment:……………………………………………………………………………………………. ……………………………………………………………………………………………

Results:

Solution Time of flow [s]

Mean

value [s]

watert

t pH

Water

1.

2.

3.

4.

5.

6.

I.

II.

I.

II.

I.

II.

I.

II.

I.

II.

I.

II.

I.

II.

-

Parameters of the function: pHf :

For first part of the graph:

109

a1 =

b1 =

r1 =

Theoretical points:

x1= x2 =

y1 = y2 =

For second part of the graph:

a2 =

b2 =

r2 =

Theoretical points:

x3= x4 =

y3 = y4 =

Isoelectric point:

21

12

a-a

b-b pI =

Conclusions:

Experiment 10bDetermination of the emulsion type

110


…………………………………………………………………………………………………………

…………………………………………………………………………………………………………

…………………………………………………………………

Method used for determination of the emulsion type:……………………………............

……………………………………………………………………………………………

Type of emulsion:

with Tween –………………………………………………………………………………….

with Span –……………………………………………………………………………………

Lorinden –…………………………………………………………………………………….

Sulfarinol -…………………………………………………………………………………..

Conclusions

111

SURFACE TENSION. SURFACTANTS

Experiment 11. Critical micelle concentration (CMC) prediction for Tween 20

The aim of the experiment is determination of the critical micelle concentration of surfactantTween 20 using tensiometric method.

Required knowledge: definition of surface tension, methods for surface tension determination,definition of critical micelle concentration

Introduction

The occurrence of most of the substance in the state of aggregation depends on the prevailing thermodynamic conditions, namely temperature and pressure, e.g. water pressure of 1 atm (normal pressure) at a temperature below 0 ° C is a solid at temperatures from 0 to 100 ° C liquid, and above 100 ° C becomes a gas.Liquid is one of the states of matter, in the traditional division into: solid, liquid and gas. When comparing the intermolecular interactions in the liquid are much higher than for gases, and smaller than in the solid state. It is difficult to change the volume of liquid, so call them, they are very difficult to incompressible or compressible. In contrast, liquids tend to adopt shapes, which corresponds to the smallest possible area. For this reason, the liquid tends to form a sphere, because as a solid sphere geometry has the lowest surface to volume ratio. The result is that the largest number of molecules of liquid is in the interior, where they are surrounded by affecting neighboring molecule. There are, of course, forces, like the force of gravity, which causes the opposition to the creation of the ideal of a sphere, hence we observe flattened shapes oceans or other bodies of water.

Fluid properties result from the behavior of the molecules:• they have full freedom of movement throughout the volume occupied by the liquid, as it is in the case of gas• there are interactions between them which, however, within the volume of liquid cancel each other.• intermolecular interactions do not stand themselves at liquid phase with another the result that thereis a phenomenon called surface tension.

Fig. 11.1. Comparison of the unequal forces acting on molecule at the surface of a liquid, as compared with molecular forces in the bulk of the liquid. The surface tension at the liquid - gas as a result of the lack of balance the forces of attraction.

Surface active agents –Surfactants

112

Vapour pressure

Resulting force

Surfactants are composed of a polar head group that is hydrophilic and a nonpolar tail group that is hydrophobic. The head groups can be or non-ionic, anionic, cationic, zwitterionic (neutral molecule with a positive and a negative electrical charge) (Fig. 11.2)

Fig. 11.2. Structure of surfactants.

Table 11.1. Common surfactant properties

Surfactant StructureCMC(mM)

ΔG(kJ/mol)

Sodium Dodecyl Sulfate (SDS) 8.2 -22.00

Sodium Octyl Sulfate (SOS) -- -14.71

Cetyl TrimethylammoniumBromide (CTAB)

0.89−0.93

-30.46

CMC – critical micelle concentration, G – Gibb’s free energy of micellization

Surfactants in human physiology has a key role. Occurs on the inner side of the alveoli and reduce the surface tension tending to reduce the volume of bubbles. This reduces the resistance of the elasticwork of breathing occurring in the lungs. Surfactant produce pneumocytes type II in the form of a complex and a carrier protein dipalmitynolecitine - apoprotein.An amphiphilic molecule can arrange itself at the surface of the water such that the polar part interacts with the water and the non-polar part is held above the surface. The presence of these molecules on the surface disrupts the cohesive energy at the surface and thus lowers the surface tension. The proportion of molecules present at the surface or as micelles in the bulk of the liquid depends onthe concentration of the amphiphile. At low concentrations surfactants will favour arrangement on the surface. As the surface becomes crowded with surfactant more molecules will arrange into micelles. At someconcentration the surface becomes completely loaded with surfactant and any further additions must arrange as micelles. This concentration is called the Critical Micelle Concentration (CMC).

113

http://en.wikipedia.org/wiki/File:Sodium_laurylsulfonate_V.1.svg

http://en.wikipedia.org/wiki/File:SOS2.jpg

http://en.wikipedia.org/wiki/File:Cetrimonium_bromide.png

Fig. 11.3. Micelle structures

The Krafft temperature named also Krafft point or critical micelle temperature, represents the minimum temperature at which surfactants form micelles. Below the Krafft temperature, there is no value for the CMC, i.e., micelles cannot form. The Krafft point can be determined by locating the abrupt change in slope of a graph of the logarithm of the solubility against T or 1/T.

Gibbs free energy and enthalpy of micellization

The Gibbs free energy of micellization can be approximated as:

CMCRTGmicelle ln

where Gmicelle is the molar Gibbs energy of micellization, R is the universal gas constant, T is the absolute temperature, and CMC is the critical micelle concentration.

The enthalpy of micelle formation reflects the contributions of interactions between micelle chainswithin the micelles and between the polar head groups and the surrounding medium. The enthalpycan be positive (endothermic) or negative (exothermic). Formation of non-ionic micelles is usuallyendothermic process with H of the order of 10 kJ per mole of surfactant. The enthopy changeduring micellization is positive with a value of about + 140 J K-1mol-1at room temperature, eventhough the molecules are clustering together. Positive enthropy change means that hydrophobicinteractions are important in the micellization process.

Determination of CMC

The CMC is detected by noting a pronounced change in physical properties of the solution, such asmolar conductivity. Fig. 11.4. shows, that there is a transition region corresponding to a range ofconcentrations around the CMC, where physical properties vary with the concentration.

114

Fig. 11.4. The typical variation of some physical properties of an aqueous solution of sodiumdodecylsulfate close to the critical micelle concentration (CMC).

A graph of surface tension vs concentration of surfactant added to a solvent is presented in Fig. 11.5.At certain concentrations surfactant particles form monolayer on the surface and cause decrease ofsurface tension. When surface becomes fully loaded with surfactant molecules, no further change insurface tension is observed.

Fig. 11.5. Changes of surface tension as a function of concentration of surfactant.

There are several methods of surface tension measurements:

1. Capillary rise method

2. Stalagmometer method – drop weight method

3. Wilhelmy plate or ring method

4. Maximum bulk pressure method.

115

Wilhelmy plate method

In this method a thin plate (often made of platinum) is used to measure equ ilibrium surface tension at air-liquid interface. The plate is oriented perpendicularly to the interface and the force acting on the plate due to its wetting is measured by a tensiometer. If the liquid wets completely the plate, the force F needed to detach the plate from the liquid surfaceequals:

)(2)( 0 dlgmmF x

where:F – force needed to detach the plate from the liquidmx – weight of the plate hanging freelym0 – weight of the plate detached from the surfacel – length of the plated – thickness of the plate - surface tension

The surface tension measured by the tensiometer isexpressed by:

)(2 dl

F

Fig. 11.6. Tensiometer.

References

1. P. W. Atkins: Chemia Fizyczna, Wydawanictwo Naukowe PWN, Warszawa 20032. P. Atkins, J. de Paula: Atkin’s Physical Chemistry. 8thedition. Oxford University Press 20063. T.W. Hermann (red.): Chemia Fizyczna, Wydawnictwo Lekarskie PZWL, Warszawa 2007

Experiment

Materials:Tensiometer TD1C Lauda, automatic pipette, 8 volumetric flasks of 50 ml, beaker of 50 ml, 2% Tween 20.

Experimental procedure:

1. Prepare standard solution of 0.001; 0.002; 0.004; 0.008; 0.01; 0.015; 0.04; 0.08; 0.1; and 0.20% of Tween 20 in water using flasks of 50 ml.

For calculation use the following formula: C1·V1 = C2·V2

2. Determine surface tension of the solutions using tensiometer.

Preparation for measurements a) Switch on the tensiometer TD1C at external power supply.

116

b) Take off the plastic tube carefully which is clamped between sample table and transducerhousing to protect the measuring system and the hook.

c) Attach the plate to the hook of the measuring systemd) In Menu select Experiment new. After confirmation select Plate and then unstable. e) After completing the taring procedure the message “Taring successful” appears. f) After confirmation attach the 500 mg calibration weight to the hook in addition to the

plate. Select Calibration and after confirmation select unstable. When the calibration procedure is completed the message “Calibration successful” appears.

g) After confirmation remove the calibration weight.

Performing the plate measurements a) Measure the surface tension of water and the prepared solutions placed in the beakers

(three times for each solution).b) Calculate an average and standard deviation of your valuesc) Plot a graph using surface tension of the Tween solutions versus the values of their

log C. Calculate the parameters for both parts of the curves.d) Determine the critical micellar concentration (CMC) for Tween 20 using the

following equation:

ln 21

12

a-a

b-b=CMC

where:equation of the first curve: y = a1x + b1 (1)equation of the second curve: y = a2x + b2 (2)

117

Report 11

Name and surname:.......................................................... Date:....................................

Critical micelle concentration (CMC) prediction for Tween 20

Aim of the experiment ……………………………………………………………………………….

……………..….……………………………………………………………………………………

Results

1. Surface tension of the prepared Tween solutions:

Nr Concentration(%) lnC

Addedvolume of2% Tween

Surface tension(mN/m)

Meanvalue

(mN/m)SD

water

1

2

3

4

5

6

7

8

9

10

-

0.001

0.002

0.004

0.008

0.01

0.015

0.04

0.08

0.1

0.2

-

2. Slope, intercept and correlation coefficient of the function = f(lnC)

Curve Slope y-Intercept r

1

2

3. ln CMC = .........................

118

CMC = .........................

4. Enclosures

a) The graph = f(lnC)

5. Summary:

119

SURFACE FENOMENA. ADSORPTION ISOTHERMS

Experiment 12. Adsorption of acetaminophen on activated charcoal

The aim of the experiment is to analyze the adsorption process of acetaminophen on activatedcharcoal. Spectrophotometry is used to determine the concentration of paracetamol in solution beforeand after adsorption process on activated charcoal. The obtained results will serve to verify if theFreundlich isotherm can be used for quantitative description of the adsorption of acetaminophen onactivated charcoal.

Required knowledgeAdsorption ad solid interfaces, physical and chemical adsorption, adsorption isotherms (Langmuir,Freundlich and BET), hysteresis phenomena, adsorbents.

IntroductionThe attachment of particles to a surface is called adsorption. The substance that adsorb is called theadsorbate. The adsorbate attaches to the surface of the solid substance that is called adsorbent. Thereverse of adsorption is desorption. Adsorption of material at solid surfaces can take place fromeither adjacent liquid or gas phase.Molecules and atoms can attach to surfaces in two ways. Considering the nature of thoseinteractions, the adsorption process is classified as physical adsorption (physisorption) andchemical adsorption (chemisorption). It is important to note that there are several differencesbetween physical and chemical adsorption: The physisorption is associated with van der Waals interactions or hydrogen bonds that are weak

but have a long range. In chemisorption, the adsorbate is sticked to the adsorbent by forming achemical bond (usually a covalent one). Therefore, the chemical adsorption is considered to bethe specific process.

The energy released as the chemisorption occurs is much greater when compared to physicaladsorption.

The physical adsorption is an exothermic process. Therefore, when the temperature increases thephysisorption decreases (the desorption takes place). In opposite, increase in temperature resultsin increasing chemisorption.

The physical adsorption is reversible (desorption can take place). In the case of chemicaladsorption, it is assumed that it is irreversible (it is very difficult to break the chemical bonds).

In chemical adsorption, due to specific chemical bonds, only monolayer coverage of theadsorbent is possible. Monolayer formation is rare in physisorption where formation of severallayers of adsorbate molecules on adsorbent surface is more common.

Adsorption isotherms are the equations that describe mathematically the equilibrium state betweenan adsorbate and a solid surface (adsorbent) at constant temperature. They represent the amount ofthe adsorbate on the adsorbent as a function of its pressure (if gas) or concentration (if liquid) atconstant temperature. The quantity adsorbed is nearly always normalized by the mass of theadsorbent to allow comparison of different materials. Those equations can be derived theoretically(assuming the specific model of adsorption process) or empirically (considering experimental data).The Langmuir isothermIrving Langmuir presented the simplest adsorption isotherm. It is a semi-empirical isotherm derivedfrom a proposed kinetic mechanism. It is based on following assumptions: Adsorption cannot proceed beyond monolayer coverage.

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The surface of the adsorbent is uniform, that is, all the adsorption sites are equivalent. There are no interactions between adsorbed molecules.

The Langmuir isotherm describes the variation of fractional coverage ϴ with pressure at constanttemperature (derivation of Langmuir isotherm is presented in Appendix):

(1)

where:ϴ - fractional coverage (the ratio of a number of adsorption sites occupied and a number of

adsorption sites available on the surface)K - constant value; the ratio of adsorption rate constant and desorption rate constantp - gas pressure

Replacing ϴ by a/am where a is the mass of gas adsorbed per gram of adsorbent at pressure p and atconstant temperature and am is the mass of gas that 1 gram of adsorbent can adsorb whenmonolayer is complete gives the formula:

(2)

The plot reflecting Langmuir isotherm:

Fig. 12.1. The Langmuir isotherm.

It is characteristic that in the initial part of the graph (at low pressure) the directly proportionalincrease in adsorption with increasing pressure takes place. Then, with the increasing pressure ofadsorbate (if gas), all the sites available for adsorption at the surface are occupied. Thus, despite thefurther increase in pressure, the adsorption does not change because the entire surface of theadsorbent is occupied (and the Langmuir model assumed only the monolayer coverage).

By inverting the equation (2) becomes:

(3)

A plot 1/a against 1/p yields a straight line and the am and K values can be calculated from the slopeand Y-intercept:

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Fig. 12.2. Linear form of the Langmuir isotherm.

The BET isothermOften molecules do form multilayers, that is, some are adsorbed on already adsorbed molecules andthen the Langmuir isotherm is not valid. Stephen Brunauer, Paul Emmett, and Edward Tellerdeveloped a model isotherm that takes that possibility into account. They considered that the initialadsorbed layer can act as a substrate for further adsorption. The BET isotherm serves as the basis foran important analysis technique for the measurement of the specific surface area of a material.

(4)

where:ϴ - fractional coverage (the ratio of a number of adsorption sites occupied and a number of

adsorption sites available on the surface)C - a constant connected with the enthalpies of desorption from monolayer and vaporization

of the liquid adsorbateZ - the ratio of the gas pressure (p) and the saturated vapour pressure (p*)

Fig. 12.3. The BET isotherm.

The Freundlich isothermFreundlich gave an empirical expression representing the variation of adsorption of a quantity ofsubstance adsorbed by unit mass of solid adsorbent with pressure (if adsorbate is a gas) orconcentration (if the adsorbate is liquid) at constant temperature. In Freundlich model there is noassumption on monolayer coverage only, so it can describe multilayer adsorption process.

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The Freundlich isotherm equation is:

(5)

where:a - the mass of a substance adsorbed per gram of adsorbent k - the constant connected to adsorption energy; it is the adsorption value when the

adsorbate concentration equals 1 mol/dm3 n - the constant that reflects the interactions between adsorbate moleculesp - gas pressure

The Freundlich adsorption isotherm is a curve that reflects adsorption increase with increasingpressure:

Fig. 12.4. The Freundlich isotherm.

Constants k and n can be evaluated from linear form of Freundlich equation. Therefore, the decimallogarithms should be taken from equation (5). That gives:

(6)

The plot of loga against logp results a straight line where 1/n is a slope and logk is an Y-intercept.

Fig. 12.5. The linear form of the Freundlich isotherm.

In the case of the adsorption from the solutions, the Freundlich isotherm becomes (c is aconcentration of adsorbate in the solution):

(7)

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The adsorption phenomenon is widely used in laboratory practice and in the industry. It is the basisof processes of purification, separation, analysis and isolation of compounds present in mixtures.Moreover, it plays a major role in adsorption chromatography. Several compunds used in treatmentbecause of their strong abilities to adsorption. Medical charcoal (activated charcoal) is used as an“universal antidote” in reducing the effects of poisoning by the oral route. Activated charcoal anddiosmectite are widely used in the treatment of diarrhea (they adsorb water, bacterial toxins andpatogens such as bacteria or rotaviruses). From the other hand, adsorption may result in druginteractions when the molecules of one drug (eg. antihypertensive) are adsorbed on the surface of co-administered drug (eg. medical charcoal, antacid inorganic salts). It may cause the reducedeffectiveness of pharmacotherapy.

REFERENCES1. Atkins P., de Paula J.: Physical pharmacy. Oxford University Press, Oxford 2006.2. Hermann T.W.: Farmacja fizyczna. Wydawnictwo Lekarskie PZWL, Warszawa 1999.3. Hermann T.W.: Chemia fizyczna. Wydawnictwo Lekarskie PZWL, Warszawa 2007.4. Atwood D., Florence A.T.: Physical pharmacy. Pharmaceutical Press, London 2008.

EQUIPMENT AND MATERIALS Spectrophotometer UV-Vis Thermo-shaker Pipettes 0.1 M HCl Acetaminophen stock solution in 0.1 M HCl of concentration 12.5 mg/mL Acetaminophen standard solutions in 0.1 M HCl of concentrations: 12.5 µg/mL; 10.0 µg/mL;

7.5 µg/mL; 5.0 µg/mL; 2.5 µg/mL; 1.25 µg/mL Activated charcoal

EXPERIMENTALThe task is to determine the acetaminophen concentration in solution (applying spectrophotometry)before and after adsorption (after 45-minute incubation of the drug with activated charcoal).1. Prepare a series of 6 samples (in plastic tubes with the cap) that contain:

Sample Mass of activatedcharcoal [mg]

Volume [mL] of acetaminophensolution 12.5 mg/mL

Volume of 0.1 MHCl [mL]

1 100 8.0 ------2 100 6.0 2.03 100 4.0 4.04 100 2.7 5.35 100 0.8 7.2

blank 100 ------- 8.0

2. Tightly close the tubes, vortex thoroughly and place in thermoshaker. Let the samples incubatefor 45 minutes at the temperature 37˚C. The speed of shaking should be stated at 700 cycles perminute.

3. During the incubation of samples with activated charcoal, prepare a calibration curve ofabsorbance against the concentration of acetaminophen in solution:

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a. Measure the absorbance of the acetaminophen standard solutions (12.5 µg/mL; 10.0µg/mL; 7.5 µg/mL; 5.0 µg/mL; 2.5 µg/mL; 1.25 µg/mL) at λmax = 243 nm. Calibrate thespectrophotometer using 0.1 M HCl solution as a blank sample.

b. Plot the calibration curve A = f (c) and determine the equation of a straight line (consultwith the teacher the significance of the Y-intercept).

4. When the incubation is finished, remove the samples from the thermoshaker and centrifuge thesamples for 10 minutes at 3500 rpm.

5. Prepare the appropriate dilutions of samples 1 – 4 (in volumetric flasks of 50 mL). For thispurpose, measure out (using the pipette) the following volumes of the supernatant:Sample 1 (diluted 1000x) – 0.05 mL (50 µL)Sample 2 (diluted 1000x) – 0.05 mL (50 µL)Sample 3 (diluted 1000x) – 0.05 mL (50 µL)Sample 4 (diluted 100x) – 0.50 mL (500 µL) and fill up to 50 mL with 0.1 M HCl solution.

6. Measure the absorbance of the obtained solutions (after diluting) at λmax = 243 nm. Theabsorbance of the sample 5 should be measured directly (without diluting) by transferring to acuvette about 3 mL of the supernatant. The spectrophotometer should be calibrated using thesupernatant from the blank sample (without acetaminophen) transferred directly to the cuvette.

Draw attention to not shake the sample and check if the obtained solution is clear (withoutparticles of activated carbon) when pipetting the supernatant. If you notice any pollution inthe supernatant, the sample should be centrifuged again and the clear supernatant shouldbe collected.

7. Calculate the concentration of acetaminophen in samples after incubation with carbon using theequation of the calibration curve (the dilutions should be taken into account):

where:Ceq – acetaminophen concentration at the equilibrium state (after incubation) [mg/mL]A – absorbanceb – the Y-intercept of the calibration curvea – the slope of the calibration curveR – the sample dilution

8. Calculate the amount of acetaminophen (in mg):a. That was in the sample before incubation with activated charcoalb. That remained in the solution after incubation

9. Calculate the amount of adsorption a (how many mg of acetaminophen was adsorbed per 1 g ofactivated charcoal).

10. Plot the amount of adsorption a against acetaminophen concentration ceq (take into account theacetaminophen concentrations at the equilibrium state). Discuss the graph with the teacher.

11. Evaluate the linear Freundlich isotherm equation, calculate the values of k and n. Discuss withthe teacher if Freundlich model is suitable for describing the process of acetaminophenadsorption on activated charcoal.

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12. Analyze the amount of acetaminophen adsorbed on activated charcoal in the environment of thestomach (0.1 mol / l HCl) and at the temperature of human body. What clinical implications maybe related to co-administered acetaminophen and medical charcoal? Can the phenomenon ofacetaminophen adsorption on activated charcoal be used in therapy?

13. Analyze the adsorption process in different temperature than 37˚C (considering the resultsobtained from the teacher). Which type of adsorption (physical or chemical) is the binding ofacetaminophen onto the activated charcoal?

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Report 12

Name and surname:................................................... Date:....................................

Adsorption of acetaminophen on activated charcoal Aim of the experiment ……………………………………………………………………………….

……………..….………………………………………………………………………………………

Results

1. Calibration curve A = f(Cst). Absorbance vs acetaminophen concentration in the solution.

Sample Acetaminophen concentration

[µg/mL]

Absorbance (λmax = 243

nm)1 12.52 10.03 7.54 5.05 2.56 1.25

Calibration curve parameters slope

aY-intercept

bY-intercept deviation

Sb

Is the Y-intercept significant?

YES / NOCorrelation coefficient

r

The final calibration curve equation (considering the Y-intercept significance):

…………………………………………………………………..

2. The calculation of initial amount of acetaminophen in the samples that were further incubated with activated charcoal.

Incubation time: ……………………………………………………..

Sample Initial acetaminophen concentrationCin [mg/mL]

Initial acetaminophen amountXin [mg]

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Temperature of incubation: …………………………………………….

Shaking speed: ………………………………………………

3. The calculation of acetaminophen concentration in the samples at the equilibrium state (after incubation).

Sample Absorbance(λmax=243 nm)

dilution Acetaminophen concentration at theequilibrium state - Ceq [mg/mL]

1 10002 10003 10004 1005 ---

4. The calculation of the mass of acetaminophen adsorbed on activated charcoal.

Sample Mass ofactivatedcharcoal

Initial amount ofacetaminophenin the sample

Xin

amount ofacetaminophen inthe sample at theequilibrium state

Xeq

amount ofacetaminophen

adsorbed on 100mg of activated

charcoal Xa = Xin -Xeq

a

[mg] [mg] [mg] [mg][mg/1g ofcharcoal]

1 1002 1003 1004 1005 100

5. The Freundlich isotherm equation (at the temperature 37˚C).

Sample Ceq [mg/mL] logCeqa [mg/1g ofcharcoal]

loga

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Linear equation of Freundlich isotherm: ……………………………………………………………………

Constant k: ………………………………………………………….

Constant n: …………………………………………………………..

6. The Freundlich isotherm equation (at the temperature 70˚C).

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Sample C70 [mg/mL] logC70a70 [mg/1g of

charcoal]loga70

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Linear equation of Freundlich isotherm: ……………………………………………………………………

Constant k70: ………………………………………………………….

Constant n70: …………………………………………………………..

7. Enclosures: Graph of the calibration curve: A = f(Cst) Graph of the Freundlich isotherm: a = f(Ceq) Graph of the linear form of the Freundlich isotherm: loga = f(logCeq)

Summary and conclusions:

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