Download MEDICAL CHEMISTRY STUDY GUIDE

DANYLO HALYTSKY LVIV NATIONAL MEDICAL UNIVERSITY
DEPARTMENT OF GENERAL, BIOINORGANIC, PHYSICAL AND COLLOIDAL
CHEMISTRY
MEDICAL CHEMISTRY
STUDY GUIDE
st
for the 1 year students of dentistry faculty
L’VIV – 2015
Medical chemistry study guide for the 1st year students of dentistry
faculty
Методичні вказівки з медичної хімії для студентів
стоматологічного факультету
Методичні вказівки уклали: доценти Кленіна О.В., Роман О.М.,
Огурцов В.В., асистент Маршалок О.І.
За загальною редакцією: доцента Кленіної О.В.
2
THEMATIC SCHEDULE
of practice and laboratory studies on medicinal chemistry
Module 1. The Basics of Medical Chemistry
№
1.
2.
3.
4.
5.
6.
7.
8
9
10
11
12
Number
of hours
The ways of expression concentrations of solutions. Preparation
3
the solution with known concentration
Colligate properties of solutions. Experimental determination of
3
the molecular mass of a solute, osmotic concentration of solutions
with the kriometry method
The equilibrium and processes with coordination compounds.
3
Preparation and properties of complex and inner complex
compounds. Complexonometry
Bio-elements in medicine and dentistry. Their chemical properties
3
and biological role
Acid-base
equilibrium.
Calculation
and
experimental
3
determination of the рН of biological liquids. Protolytical
processes in living organisms
Properties of buffer solutions and their role in biological systems.
3
Preparation of buffer solutions. Determination of the buffer
capacity
3
The basic principles of the volumetric analysis. Acid-base titration.
Precipitation and dissolution reactions
3
Thermodynamical and kinetical regularities of biochemical
processes passage
Measuring the electrical driving force of electrochemical elements
3
and electrodes potentials
Adsorptive processes and ions exchange in bio-systems.
3
Chromatography
Preparation, purification and properties of colloidal solutions
3
Electrolytic coagulation of colloids. Physical chemistry of bio3
polymers solutions
The topics
3
Safety Rules
The chemistry laboratory is not a dangerous place to work as long as all
necessary precautions are taken seriously. In the following paragraphs, those
important precautions are described. Everyone who works and performs
experiments in a laboratory must follow these safety rules at all times. Students who
do not obey the safety rules will not be allowed to enter and do any type of work in
the laboratory and they will be counted as absent. It is the student’s responsibility to
read carefully all the safety rules before the first meeting of the lab.
Eye Protection: Because the eyes are particularly susceptible to permanent
damage by corrosive chemicals as well as flying objects, safety goggles must be
worn at all times in the laboratory. Prescription glasses are not recommended since
they do not provide a proper side protection. No sunglasses are allowed in the
laboratory. Contact lenses have potential hazard because the chemical vapours
dissolve in the liquids covering the eye and concentrate behind the lenses. If you
have to wear contact lenses consult with your instructor. If possible try to wear a
prescription glasses under your safety goggles. In case of any accident that a
chemical splashes near your eyes, immediately wash your eyes with lots of water
and inform your instructor. Especially, when heating a test tube do not point its
mouth to anyone. 4
Always assume that you are the only safe worker in the lab. Work defensively.
Never assume that everyone else as safe as you are. Be alert for other’s mistakes.
Cuts and Burns: Remember you will be working in a chemistry laboratory and
many of the equipment you will be using are made of glass and it is breakable.
When inserting glass tubing or thermometers into stoppers, lubricate both the tubing
and the hole in the stopper with water. Handle tubing with a piece of towel and push
it with a twisting motion. Be very careful when using mercury thermometer. It can
be broken easily and may result with a mercury contamination. Mercury vapor is an
extremely toxic chemical.
When you heat a piece of glass it gets hot very quickly and unfortunately hot
glass look just like a cold one. Handle them with a tong. Do not use any cracked or
broken glass equipment. It may ruin an experiment and worse, it may cause serious
injury. Place it in a waste glass container. Do not throw them into the wastepaper
container or regular waste container.
Poisonous Chemicals: All of the chemicals have some degree of health hazard.
Never taste any chemicals in the laboratory unless specifically directed to do so.
Avoid breathing toxic vapors. When working with volatile chemicals and strong
acids and bases use ventilating hoods. If you are asked to taste the odor of a
substance does it by wafting a bit of the vapor toward your nose. Do not stick your
nose in and inhale vapor directly from the test tube. Always wash your hands before
leaving the laboratory.
Eating and drinking any type of food are prohibited in the laboratory at all times.
4
Smoking is not allowed. Anyone who refuses to do so will be forced to leave the
laboratory.
Clothing and Footwear: Everyone must wear a lab coat during the lab and no
shorts and sandals are allowed. Students who come to lab without proper clotting
and shoes will be asked to go back for change. If they do not come on time it will be
counted as an absence. Long hair should be securely tied back to avoid the risk of
setting it on fire. If large amounts of chemicals are spilled on your body,
immediately remove the contaminated clothing and use the safety shower if
available. Make sure to inform your instructor about the problem. Do not leave your
coats and back packs on the bench. No headphones and Walkman are allowed in the
lab because they interfere with your ability to hear what is going on in the Lab. 5
Fire: In case of fire or an accident, inform your instructor at once. Note the
location of fire extinguishers and, if available, safety showers and safety blankets as
soon as you enter the laboratory so that you may use them if needed. Never perform
an unauthorized experiment in the laboratory. Never assume that it is not necessary
to inform your instructor for small accidents. Notify him/her no matter how slight it
is.
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Topic 1
The ways of expression concentrations of solutions.
Preparation the solution with known concentration
1. Objectives
Knowledge of the modern theory of solutions is necessary because most of the
important biological processes in living organisms occurs in solutions. Large
number of drugs used in the form of liquid dosage forms. In addition, the modern
theory of electrolytes is a scientific basis for the study of electrolytic balance of the
human body and elucidate the mechanism of electrolyte homeostasis
2. Learning Targets
– know the factors which depend on solubility of substances;
– know the role of solutions in life;
– know the ways of expressing concentration of solutions;
– to be able to solve situational problems;
– to be able to prepare solutions of given concentration
3. Self Study Section
3.1. Syllabus Content
Role of solutions in the organisms life. Classification of solutions. Mechanism of
dissolution processes. Thermodynamic approach to the process of the dissolution.
The solubility of the substances.
The solubility of gases in liquids. The dependence of the solubility of gases on
the pressure (Henry-Dalton’s law), nature of the gas and solvent, temperature. Effect
of electrolytes on the solubility of gases (Sechenov’s law). Solubility of gases in the
blood. Decompression sickness. The solubility of liquids and solids in liquids. The
dependence of solubility on temperature and the nature of the solute and solvent.
Nernst law of distribution and its importance in the phenomenon of the permeability
of biological membranes.
The values that characterize the quantitative composition of solutions.
Preparation of solutions of a given quantitative composition.
3.2. Overview
Solution is a homogeneous mixture of two or more substances, consisting of
ions or molecules. This term is usually used to describe homogeneous mixtures of
two or more liquids or of a liquid and one or more solids. Solutions may exist as
gases, liquids, or solids. Nonreactive gases can mix in all proportions to give a
gaseous solution. Liquid solutions are the most common types of solutions found in
the chemistry lab. Many inorganic compounds are soluble in water or other
suitable solvents. Solid may form a solid solution with another solid. Solid
solutions of metals are referred to as alloys. Types of solutions:
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1) gaseous solutions (For example, air is a solution of oxygen, nitrogen, and
smaller amounts of other gases)
2) gases also dissolve in many liquids (For example, a solution of ammonia,
NH3, in water)
3) gases may also dissolve in solids (For example, hydrogen is soluble in
platinum)
4) homogeneous mixture of two or more liquids (For example, ethanol
dissolves in water). Fluids that dissolve in each other in all proportions are
said to be miscible fluids. If two fluids do not mix, they are said to be
immiscible.
5) homogeneous mixture of a liquid and one or more solids (For example,
sodium chloride, NaCl, dissolves in water)
6) solid solutions (For example, sterling silver (solution of copper in silver)
and different alloys: brass is an alloy composed of copper and zinc; bronze
is an alloy of copper and tin; pewter is an alloy of zinc and tin).
The solute (A) is the dissolved substance. In the case of a solution of a gas or
solid in a liquid, it is the gas or solid. Otherwise, it is the component of lesser
amount.
The solvent (B) is the dissolving medium. Generally it is the component of
greater amount. However, if the solution consists of a solid in liquid, the solid is
always called the solute and the liquid the solvent.
The concentration of a solute is the amount of solute dissolved in a given
quantity of solvent or solution. The quantity of solvent or solution can be expressed
in terms of volume or in terms of mass or molar amount. Thus, there are several
ways of expressing the concentration of a solution:
The mass percentage of solute is the percentage by mass of solute contained in
100 grams of solution:
mass of solute
Mass percentage of solute =
×100%
mass of solution
m(A)
m(A)
CP =
⋅100% =
⋅100%
m(S)
m(A) + m(B)
The molar concentration (molarity) of a solution is the moles of solute that are
in 1 liter of solution:
moles of solute
Molarity =
[mol/L] or M
volume of solution ,
m(A)
m(A)
n(A)
CM =
n( A) =
CM =
V(S) ,
M(A) →
M(A) ⋅ V(S)
The molar concentration of equivalent (normality) is the quantity of molequivalents of solute in 1 L of solution:
7
Normality =
CN =
Eq(A)
V(S) ,
moles of solute equivalent
[mol-eq/l] or N
volume of solution
,
Eq( A) =
m(A)
m(A)
CN =
M
M eq (A) →
eq (A) ⋅ V(S)
The molality of a solution is the moles of solute in 1 kilogram of solvent:
moles of solute
Molality =
[mol/kg] or m
mass of solvent
,
n(A)
m(A)
m(A)
Cm =
n( A) =
Cm =
m(B) ,
M(A) →
M(A) ⋅ m(B)
The titr of a solution is the grams of solute in 1 ml of solution:
m( A)
mass of solute
T=
=
[g/ml]
volume of solution
V (S ) ,
The mole fraction of a substance A in a solution is the moles of component
substance divided by the total moles of solution (that is, moles of solute plus
solvent):
n( A)
n( A)
moles of substance A
χ (A) =
=
=
total moles of solution n( sol ) n( A) + n( B )
3.3. References
1. V.O. Kalibabchuk, V.I. Halynska, L.I. Hryshchenko et al. Medical
Chemistry. – AUS MEDICINE Publishing. – 2010. – 224 p.
2. Raymond Chang. Chemistry (6th Edition). – WCB/McGraw-Hill. – 1998. – 995
p.
3. Rodney J. Sime Physical Chemistry. Methods. Techniques. Experiments. –
Saunders College Publishing. – 1990. – 806 p.
4. John McMurry, Robert C. Fay. Chemistry (3rd Edition). – Prentice Hall. – 2001.
– 1067 p.
5. David E. Goldberg. Fundamentals of Chemistry (2nd Edition). – WCB/McGrawHill. – 1998. – 561 p.
6. Theodore L. Brown, H.Eugene LeMay, Bruce E. Bursten. Chemistry. The
Central Science. – Prentice Hall. – 2000. – 1017 p.
7. John Olmsted III, Gregory M. Williams. Chemistry. The Molecular
Science. – Mosby. – 1994. – 977 p.
8. Steven S. Zumdahl. Chemistry (4th Edition). – Houghton Mifflin Company. –
1997. – 1031 p.
9. Gary L. Miessler, Donald A. Tarr. Inorganic Chemistry. – Prentice Hall. – 1991.
– 625 p.
8
3.4. Self Assessment Exercises
а) Review Questions
1. How changes the Gibbs energy at dissolving? What is its value for saturated,
unsaturated, oversaturated solution? Role enthalpy and entropy in the process of
dissolution.
2. What factors affect the solubility of gases? Definition of Henry’s, Sechenov’s
laws.
3. Explain different solubility of gases (O2, N2, CO2) in water, plasma, whole blood
based on the law of Sechenov.
4. What factors determine the unlimited and the limited solubility of two liquids?
b) Types of Numerical Problems and Their Solving Strategies
Numerical Problem 1. Calculate the molarity and normality of 49 % H3PO4
solution (d = 1.88 g/mL).
For solution of such type of numerical problems the
Given:
Cp(H3PO4)= 49% following equations could be used:
d = 1,88 g/mL
10 ⋅ C p ⋅ d
10 ⋅ C p ⋅ d
CM =
; CN =
Calculate:
M
E
СМ – ?
Calculate the molarity and normality of 49% H3PO4 solution
СN – ?
using data from the task:
10 ⋅ 49 ⋅1,88
10 ⋅ 49 ⋅1,88
= 9,4 М; C N =
= 19,95 N
CM =
98
32,7
Answer: molar concentration of 49 % Н3РО4 solution is 9,4 mol/mL, and its
normality is 19,95 mol-eq/mL.
Numerical Problem 2. What mass (in grams) of 36 % and 20 % solutions of
hydrochloric acid should be used to prepare 100 g of 26 % HCl solution?
Given:
For the solution of this type of problems we can use
Cp1 = 36 % HCl
the formula of ‘cross rule’ or ‘mixing rule’:
Cp2 = 20 % HCl
Cp1
Cp3 – Cp2
36
26 – 20 = 6
Cp3 = 26 % HCl
Cp3
26
m(26% HCl sol.) = 100 g
Cp2
Cp1 – Cp3
20
36 – 26 = 10
Calculate:
After reduction: 6:10 = 3:5.
m(36 % sol.). – ?
So, the mass ratio between initial solutions is: 3:5
m(20 % sol.) – ?
(of 36 % solution and 20 % solution respectively).
Calculate the masses of solutions:
m(36 % sol.). = 100⋅3:8 = 37,5 g; m(20 % sol.) = 100⋅5:8 = 62,5 g
Answer: 37,5 g of 36 % and 62,5 g of 20 % HCl solutions should be used to prepare
100 g of 26% solution.
Numerical Problem 3. What should be the volume (in mL) of 40 % HCl solution
(with density 1.2 g/mL) which is necessary for preparation 100 mL of 0.15 M
9
hydrochloric acid solution?
Calculate how many grams of HCl are contained in
Given:
100 mL of 0.15 M HCl:
CP1(HCl) = 40%
d1(sol.) = 1.2 g/mL
m(HCl)
→
С M 2 ( HCl ) =
V2(sol.) = 100 mL = 0.1 L
M(HCl) ⋅ V2 (sol.)
CM2(HCl) = 0.15 mol/L
m(HCl) = CM2(HCl) · M(HCl) · V2(sol.)
Сalculate:
Molar
mass
of hydrochloric acid is:
V1(sol.)-?
M(HCl) = 1+35.5 = 36.5 g/mol
The mass of hydrochloric acid in the 2nd solution is:
m(HCl) = 0.15 mol/L · 36.5 g/mol · 0.1 L = 0.55 g
Calculate how many grams of 40 % solution contain 0.55 g of HCl:
From the formula for the mass percentage of HCl in the 1st solution the mass of the
1st solution is:
m (HCl) ⋅100%
m (HCl)
C P1( HCl ) = 1
⋅100% → m1 ( sol ) = 1
C P (HCl)
m1 (sol.)
1
0.55 g ⋅100%
m1 ( sol ) =
= 1.38 g
40%
Finally, calculate the volume of 1.38 g of 40 % solution:
m (sol.)
1.38 g
V1 ( sol ) = 1
=
= 1.15 mL
d1 (sol.) 1.2 g/mL
Numerical Problem 4. 200 mL of 35% solution (density = 1.15 g/mL) and 100
mL of 25% solution (density = 1.2 g/mL) were mixed. Calculate the mass
percentage of the solute in the prepared solution.
Given:
3rd solution was obtained by mixing of 1st and 2nd solutions that is
V1 = 200 mL
why the mass of 3rd solution equals the sum of 1st and 2nd
CP1 = 35%
solutions as well as the mass of solute in 3rd solution equals the
d1 = 1.15 g/mL sum of masses of solutes in 1st and 2nd solutions:
V2 = 100 mL
m 3 ( A)
m1 ( A) + m 2 ( A)
С P3 =
⋅100% =
⋅100%
CP2 = 25%
m 3 (sol.)
m1 (sol.) + m 2 (sol.)
d2 = 1.2 g/mL
Calculate the mass of the 1st solution:
Calculate:
m1(sol.) = V1 · d1 = 200mL · 1.15 g/mL = 230 g
CP3 - ?
Calculate the mass of the 2nd solution:
m2(sol.) = V2 · d2 = 100mL · 1.2 g/mL = 120 g
Calculate the mass of the solute in the 1st solution:
С P ⋅ m1 (sol.) 35% ⋅ 230 g
m ( A)
1
C P1 = 1
⋅100% → m1 ( A) =
=
= 80.5 g
m1 (sol.)
100%
100%
Calculate the mass of the solute in the 2nd solution:
10
С P ⋅ m 2 (sol.) 25% ⋅120 g
2
=
= 30 g
⋅100% → m 2 ( A) =
100%
100%
m 2 (sol.)
Calculate the mass percentage of solute in the 3rd solution:
m1 ( A) + m 2 ( A)
80.5 + 30
110.5
C P3 =
⋅100% =
⋅100% =
⋅100% = 31.57 %
m1 (sol.) + m 2 (sol.)
230 + 120
350
C P2 =
m 2 ( A)
Numerical Problem 5. How many grams of crystalline soda Na2CO3⋅10H2O should
be used to prepare 250 sm3 of 0,1 N soda Na2CO3?
Given:
The numerical problem could be solved using the
V = 250 sm3 = 0,25 dm3
following equations:
CN = 0,1 mol-eq/dm3
m = CN⋅E⋅V; E = M/2;
Сalculate:
M (Na2CO3⋅10H2O) = 106 + 180 = 286 g/mol
m–?
Е = 286/2 =143 g/mol
m = 0,1⋅ 143 ⋅ 0,25 = 3,575 g
Answer: 3,575 g of crystalline soda should be dissolved in measuring flask 250 sm3
by volume.
c) Problems to Solve
1. How would you prepare 150 g of an aqueous solution containing 6.8 % by mass
of sodium acetate, CH3COONa?
2. What is the molality of a solution containing 5.5 g of sodium hydroxide, NaOH,
dissolved in 65.5 g of water?
3. How many milliliters of concentrated sulfuric acid, which is 98 mass % H2SO4
and has a density of 1.842 g/ml are required to prepare 500 ml of a 0.175 M
solution of H2SO4?
4. What volume of water is required to prepare 250 ml of 25 % H3PO4 (density
1.659 g/ml)?
5. What should be the volume (in ml) of 38 % HCl solution (with density 1,19
g/ml) necessary for preparation of 400 ml of 0.125 M hydrochloric acid
solution?
4. Laboratory Activities and Experiments Section
The content and methods of practice and laboratory studies
4.1. Practical Skills and Suggested Learning Activities:
– prepare a titrated solution of hydrochloric acid;
– prepare 0.1N solution of sodium carbonate using the Na2CO3·10H2O as a solid
substance;
– prepare isotonic solution of sodium chloride.
4.2. Experimental Guidelines
4.2.1. Preparation of the titrated solution of hydrochloric acid.
Determine the density of initial solution by hydrometer. From the table to find
11
the mass percentage of substance in the initial solution and calculate how much of
this solution should be taken to prepare V cm3 solution of given concentration. Add
into the volumetric flask same amount of distilled water and measure by the
graduated cylinder calculated volume of initial solution. Then flask is filled with a
solution exactly to a graduation mark on the neck of the flask.
4.2.2. Preparation of 0.1N solution of sodium carbonate using the
Na2CO3·10H2O as a solid substance.
Calculate the mass of solid needed for the preparation of V cm3 solution given
concentration. Weigh the required amount of substance. Insert funnel in volumetric
flask V cm3 and transfer sample of matter. Completely dissolved substance and fill
the volumetric flask with water till graduation mark on the neck of it. The last
portion of water should be added dropwise with a pipette. The fluid level is
determined by the lower level of the meniscus. Tightly close the flask and mix the
solution. Calculate the titer of the solution.
4.2.3. Preparation of sodium chloride isotonic solution.
Isotonic solution of NaCl - a 0.9% solution of sodium chloride. To prepare 500
cm3 of this solution must weigh 0,9⋅5 = 4,5 g NaCl, transfer to a volumetric flask of
500 cm3 and fill it with distilled water to the mark on the neck of the flask.
5. Conclusions and Interpretations. Lesson Summary
Topic 2
Colligate properties of solutions. Experimental determination of
the molecular mass of a solute, osmotic concentration of
solutions with the cryometric method
1. Objectives
Knowledge of the colligative properties of dilute solutions give the possibility to
understand the mechanisms of osmotic pressure and the importance of osmosis for
biological systems, especially for the human body.
2. Learning Targets:
– know colligative properties of solutions;
– to be able to solve situational problems;
– to be able to prepare hypo-, hyper- and isotonic solutions and know their
application in medicine;
– to be able to determine some of the parameters of matter by cryometry method;
– to be able to explain such phenomena as plasmolysis, hemolysis, turgor.
3. Self Study Section
3.1. Syllabus Content
Colligative properties of diluted nonelectrolytes solutions. Lowering of the
vapor pressure of the solvent above the solution. Raoult’s law. Ideal solutions.
Depression of the freezing point of a solution and boiling point elevation of a
12
solution. Osmosis and osmoticpressure. Vant’ Hoff’s law. Colligative properties of
diluted nonelectrolytes solutions. Isotonic coefficien. Hypo-, hyper- and isotonic
solutions.
Cryometry, ebuliometry, osmometry, and their use in biomedical research. The
role of osmosis in biological systems. Osmotic pressure of blood plasma. Haller
equation. Oncotic pressure. Plasmolysis and hemolysis.
3.2. Overview
Colligative properties of solutions are properties of solutions that depend on
the number of solute and solvent molecules (or ions), but not on the nature of the
solute or of the solvent as well as on the nature of the forces between them.
1. Vapor pressure lowering is a colligative property equal to the vapor
pressure of the pure solvent minus the vapor pressure of the solution:
P0>P1, ∆P = P0 – P1
I Raoult’s law states that the depression of vapor pressure of the solution of
non-volatile non-electrolyte over the vapor pressure of the solvent is equal to the
mole fraction of the solute:
P0 − P1
= N ( A) ,
P0
∆P
n(A)
=
P0
n(A) + n(B)
where P0 is the vapor pressure of the pure solvent, P1 is the vapor pressure of the
solvent in the solution, N(A) is the mole fraction of solute.
2. The boiling point elevation for solutions: The boiling point of a solution is
greater than that of the solvent by ∆Tb:
Tb(S) > Tb(B), Tb(S) = Tb(B) + ∆Tb
The elevation of the boiling point is proportional to the number of moles of
solute in a given amount of solvent:
∆Tb = Kb Cm
where Cm is the molality of the solute, [mol/kg], and Kb (or E) is the boiling point
elevation constant.
3. The freezing point depression of solutions: the freezing point of a solution
is lower than that of pure solvent by ∆Tf:
Tf(S) < Tf(B), Tf(S) = Tf(B) – ∆Tf
The relationship between the depression of the freezing point and the number
of molecules of solute in 1 kg of solvent is similar to that for the boiling point
elevation:
∆Tf = Kf Cm,
where Cm is the molality of the solute, [mol/kg], and Kf (or K) is the freezing point
depression constant.
10. Osmotic pressure
Osmosis is the transport of solvent molecules through a semi-permeable
membrane from a solution of higher solvent concentration to a solution of lower
solvent concentration.
The pressure that is just sufficient to stop osmosis is called the osmotic
13
pressure. The osmotic pressure is related to the concentration of the solution, and
the absolute temperature (Vant’ Hoff’s law):
π = CМ R T,
where R = 8.314 J/mol·K – gas constant per mole, T – Kelvin temperature.
The osmotic pressure is related to the number of moles of solute by an equation
that closely resembles the ideal gas equation:
πV = nRT, n( A) = m(A) → π = m(A) ⋅ R ⋅ T
M(A) ⋅ V (S)
M(A)
where V(S) is the volume of the solution, n is the number of moles of solute.
For solutions of electrolytes: π = i·CM·R·T,
where i is isotonic coefficient
By electrolytic dissociation total number of particles in the electrolyte solution
increases in i times. Therefore, for the electrolytes in the right part of the formula 1,
2, 3 and 4 is used isotonic coefficient і.
3.3. References
1. V.O. Kalibabchuk, V.I. Halynska, L.I. Hryshchenko et al. Medical
Chemistry. – AUS MEDICINE Publishing. – 2010. – 224 p.
2. Raymond Chang. Chemistry (6th Edition). – WCB/McGraw-Hill. – 1998. – 995
p.
3. Rodney J. Sime Physical Chemistry. Methods. Techniques. Experiments. –
Saunders College Publishing. – 1990. – 806 p.
4. John McMurry, Robert C. Fay. Chemistry (3rd Edition). – Prentice Hall. – 2001.
– 1067 p.
5. David E. Goldberg. Fundamentals of Chemistry (2nd Edition). – WCB/McGrawHill. – 1998. – 561 p.
6. Theodore L. Brown, H.Eugene LeMay, Bruce E. Bursten. Chemistry. The
Central Science. – Prentice Hall. – 2000. – 1017 p.
7. John Olmsted III, Gregory M. Williams. Chemistry. The Molecular
Science. – Mosby. – 1994. – 977 p.
8. Steven S. Zumdahl. Chemistry (4th Edition). – Houghton Mifflin Company. –
1997. – 1031 p.
9. Gary L. Miessler, Donald A. Tarr. Inorganic Chemistry. – Prentice Hall. – 1991.
– 625 p.
3.4. Self Assessment Exercises
а) Review Questions.
1. What is the difference between diffusion and osmosis? What factors affect the
value of osmotic pressure?
2. What solutions are called iso-, hyper- and hypotonic? Will vary in the same
temperature osmotic pressure of 0.1% glucose, 0.1% protein solution and 0.1%
solution of NaCl?
3. Explain why the vapor pressure above the solution is lower than on pure solvent?
14
4. Show graphically solution must boil at a higher and freeze at a lower temperature
than the solvent. Formulate Raoult’s laws and write down their mathematical
expression.
5. How to calculate the osmotic pressure of the solution, if is known lowering of the
freezing point?
6. What is the relationship between the isotonic coefficient and the degree of
electrolytic dissociation?
b) Types of Numerical Problems and Their Solving Strategies
Numerical problem 1. What is the boiling point of 20% salicic acid solution
(C7H6O3) in ethanol, if the pure solvent has the boiling point of 78,5 °C?
Boiling point elevation constant of ethanol Kb(C2H5OH) = 1.22
°C·kg/mol.
Given:
Calculate the molality of the 20% salicic acid
Cp(C7H6O3) = 20%
solution:
Tb(C2H5OH) = 78.5 oC
m(С7 H 6O3 )
Cm =
Kb(C2H5OH) = 1.22 oC·kg/mol
M(C 7 H 6O3 ) ⋅ m(C 2 H 5OH)
Calculate:
The molar mass of salicic acid:
Tb(S) - ?
M(C7H6O3) = 7·12 + 6·1 + 3·16 = 138 g/mol
The mass of salicic acid in solution:
С (С H O ) ⋅ m(sol.) 20 % ⋅100 g
m(С7 H 6 O3 ) = P 7 6 3
=
= 20 g
100%
100 %
The mass of the solvent m(C2H5OH) = 100 – 20 = 80 g. Then the molality of the
20% salicic acid solution is
20 g
Cm =
= 1.81 mol/kg
138 g / mol ⋅ 0.08 kg
Calculate the elevation of the boiling point:
∆Tb = Kb(C2H5OH) · Cm = 1.22 oC·kg/mol · 1.81 mol/kg = 2.21 oC
Calculate the boiling point of 20% salicic acid solution is:
Tb(sol.) = ∆Tb + Tb(C2H5OH) = 78.5 oC + 2.21 oC = 80.71 oC
Numerical problem 2. What is the molar mass of organic substance, if solution
which contains 0,83 g of it in 20 sm3 of water has the freezing point of
–0,7 0C.
Given:
Molar mass of the substance could be calculated according
m = 0,83 g
to the II Raoult’s law:
L = 20 sm3
m(A)
then
∆Tf = Kf(H2O) · Cm(A) and С m (A) =
Tfr(s) = –0,7 0С
M(A)
⋅
m(H
O)
2
Kf (H2O) = 1,86 0С
K (H O) ⋅ m(A) ⋅1000 .
f 2
Calculate:
M(A) =
∆T ⋅ m(H O)
M–?
2
f
15
Calculate the depression of the freezing point:
Tf(S) = Tf(H2O) – ∆Tf, → ∆Tf = Tf(H2O) – Tf(S) =
00 – (–0,70) = 0,7 0C
Calculate the molar mass of the substance:
186
, ⋅ 0,83 ⋅ 1000
M=
= 110,27 (g/mol).
20 ⋅ 0,7
Answer: the molar mass of the substance 110,27 g/mol.
Numerical problem 2. Calculate the isotonic coefficient і and the degree of
electrolytic dissociation α for saline (0,9 % NaCl), if it freezes at t = –
0,55 0C.
Given:
Calculate the experimentally determined freezing point
Cp = 0,9 %
depression ∆Тexpr:
t = –0,55 0С
∆Тexpr= 0 – (–0,55) = 0,55 0С
і–?
Calculate the theoretical freezing point depression :
α–?
K ⋅ m ⋅ 1000 186
, ⋅ 0,9 ⋅ 1000
∆Тtheor= К⋅С =
=
= 0,285 0С
58,5 ⋅ (100 − 0,9)
M ⋅L
Calculate the isotonic coefficient і:
і = ∆Тexpr/ ∆Тtheor = 0,55 : 0,285 = 1,93.
Calculate the degree of electrolytic dissociation α:
i − 1 1,93 − 1
α=
=
= 0,93 or 93 %.
n −1
2 −1
Answer: the isotonic coefficient і for saline is equval 1,93, the degree of electrolytic
dissociation α= 0,93 or 93 %.
c) Problems to Solve
1. Calculate the molar mass of a non-electrolyte, if the solution, which contains 45 g
of this substance in 500 mL of H2O (Tf = 0 oC) has the freezing point of – 0.93
o
C. Freezing point depression constant of the water Kf(H2O) = 1.86°OC·kg/mol.
Answer: 180 g/mol.
2. Osmotic pressure of blood plasma is 780 kPa at 370C . What mass of sucrose
should be taken to prepare 0.25 L of solution isotonic to serum?
Answer: 25,65 g.
3. Osmotic pressure of the liquid in some protoplasts is 5 atm. What is the molar
concentration of the aqueous solution of sucrose if it isotonic in relation to liquid
in these cells at 30 oC.
Answer: 0,2 mol/l
4. Calculate the osmotic pressure of a saline in which mass percentage of NaCl
=0,9%.The density of the solution is equal to1g/ml, M (NaCl)= 58,5g/mol, R =
8,314, i = 1,9.
16
5. Calculate the osmotic pressure of a solution at 20 oC. This solution contains 0,1
mol of glucoseC6H12O6 in 200 ml, R = 8,314.
6. What is the boiling point of 3.6% salicic acid solution (C7H6O3) in ethanol, if the
pure solvent has the boiling point of 78,5 °C? Boiling point elevation constant of
ethanol Kb(C2H5OH) = 1.22 °C·kg/mol.
Answer: 78,83 0С.
7. Calculate the osmotic pressure of a solution that contains 18,4 g of glycerine
C3H5(OH)3 in 1 L. The temperature of the solution is 170С. R=8,314.
4. Laboratory Activities and Experiments Section
The content and methods of practice and laboratory studies
4.1. Practical Skills and Suggested Learning Activities:
– determine the molecular mass of the substance (nonelectrolytes) by cryoscopyc
method;
– determine the isotonic coefficient and degree of ionization of NaCl hypertonic
solution;
– determine the isotonic coefficient, degree of ionization and the osmotic pressure
of the electrolyte solution.
4.2. Experimental Guidelines
4.2.1. Determination of the relative molecular mass by depression of freezing point
(cryoscopyc method).
The experimental determination of freezing point depression is made by
Beckman’s method. The apparatus used in this method consists of the following
three parts:
a) A freezing point tube having a side tube for introduction the solute, and
fitted with a stirrer and a Beckman termometer.
b) A wider tube or air jacket in which the freezing tube is fitted, so living an
air space between the two tubes. This insures gentle and slow cooling of the liquid
and avoids super cooling.
c) A cooling bath provided with a stirrer, the temperature of which is not more
than 5 °C below the freezing point of the pure solvent contained in the freezing
point tube.
A known weight (about 15-20 grams) of the pure solvent is placed in the freezing
tube. A Beckmann thermometer is suspended inside the tube so that its bulb is immersed in
the solvent. The tube is fitted inside the air jacket and this assembly is immersed in a
suitable cooling bath. When the temperature of the pure solvent has fallen to about 0.5 °C
below its freezing point the solvent is stirred vigorously to induce crystallization. When the
solvent begins to freeze, the temperature rises sharply to the true freezing point due to the
release of the latent heat of fusion. The steady temperature attained is noted. The
freezing point tube is then removed and warmed slightly to melt the crystals, and the
process is repeated until two concordant results are obtained.
A weighted pallet of solute is then introduced through the side tube, and allowed
17
to dissolve in the solvent. The freezing point of the solution is determined in the same
way. The procedure is then repeated with several successive additions of the solute.
The difference in the two freezing points represents the depression of freezing point
(∆T).
The molecular weight of the solute can be determined by substituting the values
of Kf, ∆Tf, the mass of solvent m(B) and the weight of solute m(A) in the formula:
K
M(A) =
f
⋅ m(A) ⋅ 1000
∆T ⋅ m(B)
f
4.2.2. Determination of the isotonic coefficient and degree of ionization of NaCl
hypertonic solution.
This experiment is carried out similarly to the described in task 1, only at first is
necessary to take distilled water, and then NaCl hypertonic solution. Calculate:
∆Тexp = Т0 – Т
∆Тtheor = K ⋅ m ⋅1000 .
L⋅M
Calculate і = ∆Тexp/∆Тtheor .
i −1
, where n – number of molecules of electrolyte that dissociate
Calculate α =
n −1
to forms ions.
5. Conclusions and Interpretations. Lesson Summary
Topic 3
The equilibrium and processes with coordination compounds.
Preparation and properties of complex and inner
complex compounds. Complexonometry
1. Objectives
Complex compounds are important in chemistry and biology, because most of
the metals are part of tissues of a living organism and they are in the form of
chelates.
Chelate compounds used for excretion of salts of toxic metals and radioactive
nuclides. Tetacinum, unitiol, ferracen – compounds with chelated structure are used
in medical practice.
2. Learning Targets:
– study the structure of molecules and chemical properties of the complex
compounds :
– write down formulas of coordination compounds, give the name of the complex;
– write down equations of complexation reactions.
18
3. Self Study Section
3.1. Syllabus Content
Complex formation reactions. Werner coordination theory and modern
understanding of the structure of complex compounds. The concept about
complexing agent (central ion). Nature, coordination number, hybridization of
central atom orbitals. The concept about ligands. Denticity of ligands. The inner
and external sphere of the coordination compounds. Geometry of the complex ion.
The nature of the chemical bond in complex compounds. Classification of
compounds according to the charge on the inner sphere and the nature of ligands.
Chelate compounds.
Concept about metal-ligand homeostasis. Violation homeostasis and the
application of chelate compounds in medicine as antidotes to remove toxic metal
ions from the organism.
3.2. Overview
A complex (or coordination compound) is a compound consisting of complex
ion and other ion of opposite:
[Cu(NH3)4]SO4 ⇄ SO42– + . [Cu(NH3)4]2+
Components of coordination compound according to Werner’s coordination
theory:
K3[Fe(CN)6]
Fe+3 – central metal atom;
CN– – ligand;
6 – coordination number;
[Fe(CN)6]3– – complex ion or internal coordination sphere;
K+ – external coordination sphere.
A complex ion is a metal atom or cation with ligands attacked to it through
coordinate covalent bonds.
Ligands are the Lewis bases attached to the metal atom in a complex. So may be
neutral molecule (such as H2O or NH3) or anions (such as CN– or Cl–).
The coordination number of a metal atom in a complex is the total number of
bonds the metal atom forms with ligands. It equals the number of ligands in the
internal coordination sphere if each ligand is attached to the central atom by only
one coordinate covalent bond.
Reactions which lead to formation of coordination compounds are named
complex formation reaction
HgI2 + 2KI = К2[HgI4]
The quantitative characteristic of complex ion stability in solutions is the
constant of its dissociation. The dissociation constant Kd of a complex ion is the
equilibrium constant for the dissociation of the complex ion on the metal ion and the
ligands. Thus, the dissociation constant of [Ag(NH3)2]+ is
[Ag(NH3)2]+ ⇄ Ag+ + 2 NH3
19
[Ag+ ] ⋅ [NH3 ]2
[[Ag(NH3 ) 2 ]+ ]
The greater value of dissociation constant the more stable complex ion. For
example, from comparing of dissociation constants for the following complex ions
of Ag [Ag(NН3)2]+ and [Ag(CN2]–:
[Ag + ] ⋅ [ NH 3 ]2
Кd([Ag(NН3)2]+)=
= 6⋅10–8
+
[Ag( NH 3 ) 2 ]
Kd =
Кd([Ag(CN)2]–) =
[Ag + ] ⋅ [CN − ]2
−
= 1⋅10–21
[Ag(CN ) 2 ]
[Ag(NH3)2]+ dissociates better, so it is less stable.
3.3. References
1. V.O. Kalibabchuk, V.I. Halynska, L.I. Hryshchenko et al. Medical
Chemistry. – AUS MEDICINE Publishing. – 2010. – 224 p.
2. Raymond Chang. Chemistry (6th Edition). – WCB/McGraw-Hill. – 1998. – 995
p.
3. Rodney J. Sime Physical Chemistry. Methods. Techniques. Experiments. –
Saunders College Publishing. – 1990. – 806 p.
4. John McMurry, Robert C. Fay. Chemistry (3rd Edition). – Prentice Hall. – 2001.
– 1067 p.
5. David E. Goldberg. Fundamentals of Chemistry (2nd Edition). – WCB/McGrawHill. – 1998. – 561 p.
6. Theodore L. Brown, H.Eugene LeMay, Bruce E. Bursten. Chemistry. The
Central Science. – Prentice Hall. – 2000. – 1017 p.
7. John Olmsted III, Gregory M. Williams. Chemistry. The Molecular
Science. – Mosby. – 1994. – 977 p.
8. Steven S. Zumdahl. Chemistry (4th Edition). – Houghton Mifflin Company. –
1997. – 1031 p.
9. Gary L. Miessler, Donald A. Tarr. Inorganic Chemistry. – Prentice Hall. – 1991.
– 625 p.
3.4. Self Assessment Exercises
а) Review Questions.
1. Summary of Werner’s theory of coordination compounds. Give the name the
parts of complex compounds: Na3[Ag(S2O3)2], [Ag(NH3)2]Cl.
2. Determine the charges of the complex ions formed by atoms Palladium(ІІ),
Platinum(II), Iron(II), Nickel(II): [PdCl3(NH3)], [PdCl(NH3)2H2O],
[PtNO2(NH3)3], [Fe(CN)5NH3], [Ni(CN)4], [Fe(CNS)6]. Add external
coordination sphere and name the coordination compounds.
3. Write down formulas of the following complex: а) potassium dicyano (І)
argentate; b)hexaaminenickel(ІІ) chloride; c) tetraaminedicarbonatecromium
(ІІІ) sulfate; magnesium trifluorohydroxo (ІІ) beryllate.
20
4. The basic methods of complex compounds preparation. Which reactions can
be used for
obtaining: K4[Fe(CN)6], K2[HgI4], [Cu(NH3)4]SO4,
Na3[Al(OH)6], K[Ag(CN)2], [Ag(NH3)2]Cl. Give the name of this complex
compounds.
5. Give the formula of: a) monodentate ligands; b) bidentate ligands.
b) Types of Numerical Problems and Their Solving Strategies.
1. Algorithms for solving typical problems
Numerical problem 1. Calculate the concentration of silver cations in 0,1 М
solution of potassium tetra-cyanocadmitate (II) K2[Cd(CN)4], dicyanoargentate, which in addition contains 6,5 g/l of KCN.
1. Find the dissociation constant for complex ion [Cd(CN)4]2– in the Table:
Кd=
[Cd 2 + ] ⋅ [CN − ]4
2−
= 7,8 ⋅10 −18
[Cd (CN ) 4 ]
2. Under the secondary dissociation complex ion gives the central metal ion and the
ligands ions or molecules according to the following equation:
[Cd(CN)4]2– ⇄ Cd2+ + 4CN–
In the presence of the excess of CN– ions, which formed in consequence of the
KCN dissociation, the equilibrium is so displaced to the left, that quantity of CN–
ions, which formed by dissociation of the complex ion, we can ignore. Then, the
concentration of CN– ions equals to the concentration of KCN.
The equilibrium concentration of [Cd(CN)4]2– ion equals to the general
concentration of complex salt: [Cd(CN)4]2– = 0,1 mol/l.
3. Calculate the molar concentration of CN– ions:
[CN − ] =
T
6,5 g / l
=
= 0.1mol / l
M(KCN) 65 g / mol
[CN–] = 6,5/65 = 0,1 моль/дм3 (дорівнює загальній концентрації KCN);
4. Calculate the [Cd2+] ions concentration:
[Cd ] = 7,8 ⋅10
(0,1)
2+
−18
⋅ 0,1
4
= 7,8 ⋅ 10–15 mol/dm3.
c) Problems to Solve
1. Calculate the concentration of silver cations in 0,05 М solution of potassium
dicyanoargentate, which also contains 0.01 mol/l of KCN. The general stability
constant of complex ion [Ag(CN)2]– is 1·10-21.
Answer: 5⋅10–19 mol/l.
2. Determine the general hardness of water when for the titration of 100 ml of water
in the presence of indicator chromogene black were applied 8,2 ml of 0.05 N
solution of Trilon B.
Answer: 4,1 mol-eq/l.
21
3. Haw many grams of ethylenediaminetetraacetic acid (trilon B) is needed for the
preparation of 0,5 l 0,05 N solution?
Answer: 0,4653 g.
4. Laboratory Activities and Experiments Section
The content and methods of practice and laboratory studies.
4.1. List of practical tasks for laboratory work:
– preparation of compounds with a complex cation;
– preparation of compounds with a complex anion;
– properties of coordination compounds;
– determination of hardness of water from different sources.
4.2.Instructions for laboratory work
4.2.1. Preparation of compounds with a complex cation.
To a solution of the copper sulfate (CuSO4) in a test-tube to add identical
volume of a solution of the ammonium hydroxide (NH4OH). To a precipitate, which
was received, to add surplus of a solution of the ammonium hydroxide. What is
observed? Write equations of reactions.
4.2.2. Preparation of compounds with a complex anion.
To a solution of the mercury(II) nitrate (Hg(NO3)2) in a test-tube to add identical
volume of a solution of the potassium iodide (KI). To a precipitate, which was
received, to add surplus of a solution of the potassium iodide. What is observed?
Write equations of reactions.
4.2.3. Coordination compounds in the reactions of interchanging.
To a solution of the copper sulfate (CuSO4) in a test-tube to add identical
volume of a solution of the potassium hexacyanoferrate(II) (K4[Fe(CN)6]). What is
observed? Write equations of reactions.
4.2.4. Determination of general hardness of water
For the determination of general hardness of water to the conical flask (250 ml
by volume) pour 50 ml of water, add 50 ml of distilled water, 5 ml of ammonium
buffer solution and the solution of indicator chromogene black by drops till the
solution color turn to dark-red. Then, the content of the flask is titrated by 0.05 N
solution of Trilon B till it colours to greenish-blue.
The general hardness of water (number of Ca2+ and Mg2+ ions) is:
H(H2O) = C N ( s ) ⋅ V ( s ) ⋅ 1000 [mol-eq/l],
V (H 2 O)
where CN – normality of the Trilon B solution, mol-eq/l; V(s) – volume of the Trilon
B solution, consumed for titration, ml; V(H2O) – volume of water, which underwent
the analysis, ml.
5. Conclusions and Interpretations. Lesson Summary
22
Topic 4
Biogenic elements in medicine and dentistry. Their chemical
properties and biological role
1. Objectives
Most of the 70 chemical elements found in the human body. They have an
important biological role in the life of the organism. Six elements - C, H, O, N, P, S
- are part of proteins, nucleic acids, hormones, etc. S-elements Na, K, Mg and Ca
are in the intracellular and extracellular fluids, and Ca is also a part of the bone
tissue. Eight d-elements are a group of nutrients minerals and others have an
important role to be a part of the structure of enzymes active centers. Microelements
positively influence on immune system and duration of human life.
These elements as a simple substances have the great importance for the national
economy. Thus, iron is the basis of steel (manufacture of iron and steel of various
grades), cobalt, nickel, copper, manganese are components to produce special
alloys.
2. Learning Targets:
– know the chemical properties of the macro- and microelements, based on their
position in the periodic system;
– know the biological role of macro-, microelements and use of their compounds
in medical practice;
– to carry out reactions that characterize the acid-base and redox properties of
bioelements;
– to carry out qualitative reactions of identification of Na+, K+, Mg2+, Ca2+, CO32–
PO43–, NO3–, NO2– ions.
3. Self Study Section
3.1. Syllabus Content
General information about nutrients. Qualitative and quantitative content of
nutrients in the body. Macronutrients, micronutrients and impurity elements. The
concept of Vernadsky's doctrine about biosphere and the role of living matter (living
organisms). Relationship between the contents of biogenic elements in the human
body and its contents in the environment. Problems of biosphere pollution and
purification because of toxic chemicals.
Electronic structure and electronegativity of s-, p- and d-elements. Chemical
properties of s-, p- and d-elements and their compounds (reactions without changing
of oxidation charge, oxidation and reduction reactions, complex formation
reactions). The relationship between the location of s-and p-elements in the periodic
table and their content in the body. Use of medicine. Toxic effects of compounds.
Reactions of identification of СО32–, SO42–, NO2–, S2O32– , MnO4–, Fe3+, Cu2+,
+
Ag ions.
23
3.2. Overview
About 16 elements are used in formation of chemical compounds from which
living organisms are made. These 16 elements and a few others, which occur in a
particular organism, are called bioelements. Bioelement is any chemical element
that is found in the molecules and compounds that make up a living organism. Some
of the more prominent representatives are called macronutrients, whereas those
appearing only at the level of parts per million or less are referred to as
micronutrients. These nutrients perform various functions, including the building
of bones and cell structures, regulating the body's pH, carrying charge, and driving
chemical reactions.
The main six elements are: C, H, O, N, P, and S, and they're called primary
bioelements. These elements are present as constituents of biomolecules, in
inorganic matrix substances, and in water. Minerals are rarely present in large
amounts. The above 6 bioelements plus Ca, K, Na, Cl, Mg and Fe make up 99.9%
of the biomass. The remaining elements occur mainly as trace elements, which are
needed only in catalytic quantities. While the light metals are usually present as
mobile cations, the heavy metals are generally fixed as stable components of organic
complexes.
Main macroelements: Oxygen, Carbon, Nitrogen, Phosphorus, Sulfur,
Hydrogen are the part of albumens (proteins), fats, nucleic acids, and also hormones
and enzymes. Their mass in an organism is about 100 and more grammes per 70 kg
of living mass.
Carbon (18%) is a macronutrient. The element carbon is perhaps the one of the
most important elements for life. It is ideal to build big biological molecules. Its
central role is due to the fact that it has four bonding sites that allow for the building
of long, complex chains of molecules. Moreover, carbon bonds can be formed and
broken with a modest amount of energy, allowing for the dynamic organic chemistry
that goes on in our cells.
Phosphorus (1%) is a macronutrient. It is one of the most abundant minerals in
the human body:
Phosphorus is a key component of nucleic acids and many other biologically
important molecules such as DNA and RNA;
It is required for the healthy formation of bones and teeth (it is a component of
hydroxyapatite), and is necessary for our bodies to process many of the foods that
we eat;
It is a part of the body's energy storage system. Phosphorus is found in the
molecule ATP, which provides energy in cells for driving chemical reactions, and in
creatine phosphate, for the energy derived from glycolysis and the citric acid cycle.
Hydrogen is present in all biomolecules linked to C, N, O and S. Removal of H
is equivalent to oxidation; when the H is combined with O by operation of the
electron transport chain of respiring cells, ATP is generated. In most biological
reactions. H participates as the ion H+ + e-; the coenzymes NADH and NADPH are
carriers of H+ + 2e- (equivalent to a hydride ion).
24
Nitrogen (3%) is a macronutrient. It is found in many organic molecules,
including the amino acids that make up proteins, and the nucleic acids that make up
DNA. It plays an important role in digestion of food and growth; almost 80% of the
air we breathe is made up of nitrogen; nitrogen is one of the 3 main elements that
make plant life possible.
Sodium, potassium, and chloride ions are all required in the human diet. These
three minerals are called blood electrolytes because their ions can conduct electrical
currents. Sodium is found primarily in the extracellular fluids, and potassium is
found predominantly within the cell. Both of these elements are needed to maintain
a proper fluid balance inside and outside of the cell. Because these three minerals
are found in most foods, deficiency is rare. The major dietary source of sodium and
chloride is table salt (40% sodium and 60% chloride). Physicians still recommend
that the intake of sodium be restricted to 1–2 g daily. The recommended intake of
chloride is approximately 1.7–5.1 g daily. A high intake of table salt, sodium
chloride, is one factor that may cause high blood pressure, hypertension, in
susceptible individuals. There has been considerable emphasis on “low-salt” diets as
a means of avoiding hypertension. However, it appears that sodium is not the only
culprit. It
Sodium, Potassium, Magnesium, Calcium are found in biological liquids (thus,
К in intracellular and Na - in extracellular). Besides, Mg and Ca are included in the
composition of bone tissue. They participate in the processes of excitation and
inhibition of the central nervous system, and also stimulate some metabolic
processes.
Potassium (0.25%) is the major intracellular cation (meaning it carries a positive
charge when dissolved in water). Dietary intake of potassium is about 1.9–5.6 g/day.
It is found in citrus fruits, bananas, and tomatoes. Potassium deficiency is rare, but
loss of potassium in severe diarrhea, suchas can occur in cholera, and the excretion
of potassium by a person suffering from diabetes mellitus can lead to a debilitating
deficiency. However, potassium deficiency is seen most commonly in individuals
who are taking diuretics.
Potassium helps regulate the heartbeat and is vital for electrical signaling in
nerves.
Sodium (0.15%) is another electrolyte that is vital for electrical signaling in
nerves. It also regulates the amount of water in the body.
Calcium (1.5%) is the most common mineral in the human body. Nearly all of it
is a component of bones and teeth. It is found in a crystalline calcium phosphate
mineral known as hydroxyapatite, [Ca10(PO4)6(OH)2]. In addition, calcium is
required for normal blood clotting (protein regulation) and muscle function (such as
muscle contraction).
In fact, the body will actually pull calcium from bones (causing problems like
osteoporosis) if there's not enough of the element in a person's diet. The RDA for
calcium is 1200 mg/day for adults between nineteen and twenty-four years of age
and 800 mg/day for adults over age twenty-five. Milk, cheese, canned salmon, and
25
dark green leafy vegetables are all rich sources of dietary calcium.
Magnesium (0.05%) plays an important role in the structure of the skeleton and
muscles. Magnesium ions are important in normal muscle function, nerve
conductance, and bone development.
They also are necessary in more than 300 essential cellular metabolic reactions.
They are required for the reactions in the liver that convert glycogen to glucose.
Many enzymes that are involved in the catabolic breakdown of glucose require
magnesium ions as cofactors.
A typical adult contains about 25 g of magnesium, and the recommended daily
intake is about 300 mg/day. Magnesium is plentiful in leafy green vegetables,
legumes, cereal grains, and lean meats.
Microelements are contained in an organism within the limits of 10-2-10-6 %
mass. They are included in a significant number of enzymes (metaloenzymes), some
vitamins (B12) and hormones (insulin). They are involved in the processes of
hematopoiesis, reproduction, growth, and metabolism. Their biological functions
are closely related to the processes of complexing between bioligands and metal ion
due to free atomic orbitals.
From the large group of d-elements eight major elements may be referred to as
bio-elements, concentration of which in an organism is significant and their specific
physiology role is well-proven. They are d-elements: Fe, Cu, Zn, Mn, Co, Ni, Cr,
Mo.These elements enter in the complement of plenty of enzymes (more than 200),
and also some vitamins.
How was already marked, the biological role of chemical elements is concerned,
above all things, by the structure of electronic shells of their atoms. From the bioelements listed above only Zinc has completed up to 18 electronic shells, is
characterized by the permanent value of the oxidation state +2 and that is why it
does not belong to the transitional elements.
Chemical properties of both elements and their compounds depend on the
structure of their atoms. These properties appear in ability of d-elements to take part
in the varied chemical reactions: acid-basic, protolytical, oxidizing-reduction,
coordination compounds formation and others like that.
The biological role of chemical elements is defined the structure of their atoms.
d-elements of the fourth period of the periodic system have the electronic
configuration [Ar]ns2(n–1)d1–10. All these elements, with the exception of zinc, have
incomplete d-orbitals and show different valence and degrees of oxidation.
Therefore they are called transition elements.
Oxides, hydroxides and salts of d-elements show their acid-base properties in the
reactions with water, acids and bases. Redox properties associated with the ability to
loss or gain electrons.
Iron Fe (0.006%) is a key element in the metabolism of almost all living
organisms.
It is required mineral for heme-containing proteins and is an element that is
absolutely essential for normal physiological functioning.
26
It is found in the oxygen transport and storage proteins, hemoglobin and
myoglobin, and is also a component of the cytochromes that participate in the
respiratory electron transport chain.
The total content of iron in an organism is within the limits of 4-5 gs; much of it
is in a liver (500-600 mg), muscles (400-450 mg), marrow (250-300 mg), however
60-70 % from its general mass is contained in red corpuscles and nervous cages.
The requirement for iron is 20-30 mg per day. Iron can be absorbed by the body
only in its ferrous, Fe2+ oxidation state. The iron in meat is absorbed more
efficiently than that from most other foods. Deficiency of iron leads to irondeficiency anemia, a condition in which the amount of hemoglobin in red blood
cells is abnormally low.
Copper Cu (0.0001%) is essential for all life, but only in small quantities. The
RDA for copper in normal healthy adults is 0.9 mg/day. General symptoms of not
getting enough copper in your diet include anemia (a condition in which your blood
can’t supply enough oxygen to your body), arthritis (painful swelling of the joints),
and many other medical problems. Seafood, vegetables (broccoli), dried beans,
soybeans, pease, whole-wheat products, nuts (almond), garlic, and meats such as
liver are excellent sources of copper ions. Copper is carried mostly in the
bloodstream on a plasma protein called ceruloplasmin. Though when copper is first
absorbed in the gut it is transported to the liver bound to albumin. An inherited
condition called Wilson's disease causes the body to retain copper, as it is not
excreted by the liver into the bile. This disease, if untreated, can lead to brain and
liver damage.
Copper is important as an electron donor in various biological reactions.
Without enough copper, iron won't work properly in the body.
Copper is the key component of a variety of enzymes, including the copper
centers of cytochrome-C-oxidase, the Cu-Zn containing enzyme superoxide
dismutase, and is the central metal in the oxygen carrying pigment hemocyanin. It is
also required by some of the enzymes that are responsible for the synthesis of
connective tissue proteins;
Copper is a major component of the oxygen carrying part of blood cells. The
respiratory electron transport chain contains an enzyme, cytochrome oxidase, that
contains both heme groups and copper ions;
Copper protects our cells from being damaged by certain chemicals in our body;
It helps your body produce chemicals that regulate blood pressure, pulse;
Copper, along with vitamine C, is important for keeping blood vessel and skin
elastic and flexible.
This important element is also required by the brain to form chemicals that keep
us awake and alert. Copper also helps your body produce chemicals that regulate
blood pressure, pulse, and healing.
Zinc Zn (0.0032%) is an essential trace element for all forms of life. Several
proteins contain structures called "zinc fingers" help to regulate genes. Zinc is
involved in different reactions in the body and controls how every single part of our
27
bodies is made and works. Zinc is needed for the growth and repair of tissues
throughout our bodies. It is one of the most important elements to a healthy immune
system.
Manganese Mn (0.000017%) is essential for certain enzymes, in particular those
that protect mitochondria — the place where usable energy is generated inside cells
— from dangerous oxidants.
Cobalt Co (0.0000021%) is a component of vitamin B12, which is important in
protein formation and DNA regulation.
Chromium Cr (0.0000024%) helps regulate sugar levels by interacting with
insulin, but the exact mechanism is still not completely understood.
Molybdenum Mo (0.000013%) is essential to virtually all life forms. In humans,
it is important for transforming sulfur into a usable form, it is required by various
enzymes. In nitrogen-fixing bacteria, it is important for transforming nitrogen into a
usable form.
Fluorine F (0.0037%) is found in teeth and bones. Outside of preventing tooth
decay, it does not appear to have any importance to bodily health.
Fluoride aids in the prevention of dental caries (cavities). BUT: an excess of
fluoride is, in fact, toxic, but at the level found in fluoridated water supplies,
approximately 1 part per million, no toxic effects are observed. Fluoride works by
displacing hydroxide in calcium hydroxyapatite to give a crystalline mineral in teeth
known as fluorapatite, [Ca3(PO4)2 _ CaF2], that is far more resistant to the acid
produced by oral bacteria than is hydroxyapatite itself.
Iodine I (0.000016%) is required for making of thyroid hormones, which
regulate metabolic rate and other cellular functions. The thyroid gland extracts
iodine from nutrients and incorporates it into various hormones.
Iodine deficiency, which can lead to goiter and brain damage, is an important
health problem throughout much of the world. An enlargement of the thyroid gland
is an abnormality that results from an effort to compensate for low iodine intake.
Goiter can be prevented if iodine is included in the diet. Seafood is one of the best
sources of iodine. In areas where seafood is not available, dietary iodine is easily
obtained in the form of iodized salt, found in most grocery stores.
Selenium Se (0.000019%) is essential for certain enzymes, including several
antioxidants. Unlike animals, plants do not appear to require selenium for survival,
but they do absorb it, so there are several cases of selenium poisoning from eating
plants grown in selenium-rich soils.
3.3. References
1. V.O. Kalibabchuk, V.I. Halynska, L.I. Hryshchenko et al. Medical
Chemistry. – AUS MEDICINE Publishing. – 2010. – 224 p.
2. Raymond Chang. Chemistry (6th Edition). – WCB/McGraw-Hill. – 1998. – 995
p.
3. Rodney J. Sime Physical Chemistry. Methods. Techniques. Experiments. –
Saunders College Publishing. – 1990. – 806 p.
28
4. John McMurry, Robert C. Fay. Chemistry (3rd Edition). – Prentice Hall. – 2001.
– 1067 p.
5. David E. Goldberg. Fundamentals of Chemistry (2nd Edition). – WCB/McGrawHill. – 1998. – 561 p.
6. Theodore L. Brown, H.Eugene LeMay, Bruce E. Bursten. Chemistry. The
Central Science. – Prentice Hall. – 2000. – 1017 p.
7. John Olmsted III, Gregory M. Williams. Chemistry. The Molecular
Science. – Mosby. – 1994. – 977 p.
8. Steven S. Zumdahl. Chemistry (4th Edition). – Houghton Mifflin Company. –
1997. – 1031 p.
9. Gary L. Miessler, Donald A. Tarr. Inorganic Chemistry. – Prentice Hall. – 1991.
– 625 p.
3.4. Self Assessment Exercises
a) Review Questions.
1. What chemical elements are called bioelements and what classifications of
chemical elements that are members of living organisms are known?
2. Biological role of s-elements and their daily needs.
3. What biometals are parts of chlorophyll and what is its function in plant growth?
4. List mineral buffer system of blood and specify their composition.
5. What substances are called crown ethers and what is their function in the body
tissues?
6. Write down the electronic formulas of the following atoms and ions: Fe, Mn, Mo,
Fe3+, Mn2+, Mo6+. What is the valency of the atoms and what are the coordination
numbers for the ions?
7. What properties of d-elements are called acid-base properties? Write the
equations of hydrolysis of the following salts: FeCl3, CuSO4 та Zn(NO3)2.
8. What are the chemical properties indicate biological role of microelements
manganese, copper, chromium and zinc?
9. Describe the biological role of microelements copper, zinc, manganese and
molybdenum and specify their containing in the body and daily requirement.
10.What are the most important medicines from a family of d-elements specify their
chemical composition, the conditions under which they are assigned to treat.
11.What is the structure of vitamin B12 molecule and what is its biological role?
b) Types of Numerical Problems and Their Solving Strategies.
Numerical problem 1. In animal bones contained 2.12% phosphorus, 7.56%
calcium and 1.51% magnesium. Find the mass percentage of these
elements in the ash of the bones, which is 27% of their mass.
Given:
According to the problem, with 100 g of bone obtain 27
Cp(P) = 2,12 %
g of ash.
Cp(Ca)=7,56 %
So, mass percentage of the elements will be:
Cp(Mg)=1,51 %
Cp(P) = 2,12 : 27 = 0,0785 or 7,85 %
Cp(ash) = 27 %
Cp(Ca) = 7,56 : 27 = 0,28 or 28 %
29
Cp(elements) – ?
Cp(Mg) = 1,51 : 27 = 0,056 or 5,6 %
Answer: mass percentage of the elements:
Р – 7,85 %; Са – 28 %; Mg – 5,6 %.
Numerical problem 2. Human blood contains 60% plasma and 40% blood cells.
Calculate the mass percentage of water in the blood, if its mass percentage in plasma
is 92%, mass percentage in cells - 64%.
Given:
Denote mass percentage of waner in a blood –
Cp (plasma) = 60 %
Cp(Н2О)blood.
Cp (cells) = 40 %
100 g of blood consist 60 g of plasma and 40 gof blood
Cp (Н2О)plasma =91 % cells. Mass of water in 100 g of blood equals
Cp (Н2О)cells= 64 %
100⋅Cp(Н2О)blood and this mass contains water of plasma
Cp (Н2О) blood – ?
60⋅0,91 and water of cells 40⋅0,64.
So, 100 ⋅ Cp(Н2О)blood = 60 ⋅ 0,91 + 40 ⋅ 0,64
and Cp(Н2О)blood = 0,802
Answer: Cp (Н2О) blood = 80,2 %
Numerical problem 3. 1.5 g of commercial zinc was affected by the excess of
hydrochloric acid and 0.448 l of hydrogen were escaped. Calculate the
mass percentage of pure zinc in commercial.
Given:
Zn + 2 HCl → H2 ↑ + ZnCl2
Calculate the mass of pure zinc in commercial. According
m (Zn) = 1,5 g
V = 0,448 l
to the equation:
Cp (Zn) – ?
from 65 g of Zn we obtain 22.4 l of H2
from x – 0.448 l
x=
65 g ⋅ 0.448l
= 1.3 g
22.4l
Calculate the mass percentage of pure zinc in commercial:
m(pure Zn)
C ( Zn) =
⋅100% =
P
m(com. Zn)
1.3g ⋅100%
= 86.67%
1.5g
Answer: Cp (Zn) = 86,7 %
Problems to Solve
1. 108 g of iron bromide (II) were oxidized by potassium permanganate in the
sulfuric acid medium. Determine the mass of iron salt, which was obtained as the
result of the reaction.
Answer: 100g.
2. For the purpose of steel wire covering with copper it was passed through the
solution of copper sulfate (II). Determine the mass of copper sulfate solution with
mass percentage of salt 25%, which is needed for the chalcosis of steel wire if
11.2 tones of iron joined to the reaction with copper sulfate.
30
Answer: 128 tones.
3. The content of magnesium in plasma and cellular elements of the blood are
respectively 1.33 and 2.125 mmol/kg. This blood consists 58% of plasma and 42%
of the cells. Find magnesium content in a blood (mmol/kg).
Answer: 1,66 mmol/kg.
4. Laboratory Activities and Experiments Section
4.1. Practical Skills and Suggested Learning Activities:
– hydrolysis of salts ;
– reactions of identification of Na+ and Са+2 ions ;
– color of the flame of Na+, K+, Ca2+, Li2+ ions;
– reactions of identification of РО43– and S2– ions ;
– preparation and properties of copper and zinc hydroxides;
– reduction properties of the ions of d-elements in the lowest and highest oxidation
states;
– preparation and properties of coordination compounds;
– reactions of identification of Fe(II) and Fe(III) ions;
– reaction of identification of chromate-ion;
– reaction of identification of Zn2+ ions.
4.2. Experimental Guidelines
І. Acid-Base Properties
4.2.1. Preparation and properties of copper and zinc hydroxides.
To 2-3 ml of solutions of copper and zinc salts add slowly solution of an
alkali. Note the color of the precipitates which are forming. Divide the precipitates
to two test-tubes. To the first one add the excess of an alkali solution, to the second
one – some of a mineral acid solution. What are the colors of the precipitates and
solutions. Write the equations of the reactions.
II. Reduction-Oxidation Reactions
4.2.2. Reduction properties of the ions of d-elements in the lowest oxidation states.
To 1-2 ml of the solution of Fe2+ salt add 0.5 ml of diluted sulfuric acid and
potassium permanganate KMnO4 solution. What is observed? Write the equations of
the reactions.
4.2.3. Oxidation properties of the ions of d-elements in the highest oxidation states.
Oxidizing properties of potassium permanganate KMnO4 depending of the pH of
solution.
In tree test-tubes add 2-3 ml of potassium permanganate KMnO4. To the first
test-tube add some sulfuric acid, to the second one – some water, to the second one
– some alkali solution. Add 2-3 ml of sodium sulfite Na2SO3 to every test-tube.
What is observed? Write the equations of the reactions.
ІІІ. Identification of the ions
31
4.2.4. Tests on iron (II) and iron (III) ions.
а) reaction with potassium hexacyano(III)ferrate.
To 5-8 drops of iron (II) sulfate FeSO4 add 2-3 drops of potassium
hexaciano(III) ferrate solution. What is the color of precipitate which forms? Write
the equations of the reactions, name the complex compound.
б) reaction with potassium hexacyano(II)ferrate.
To 5-6 drops of iron (III) chloride FeCl3 add 2-3 drops of potassium
hexaciano(II) ferrate solution. What is the color of precipitate which forms? Write
the equations of the reactions, name the complex compound.
4.2.5. Tests on chromate-ion.
To 5-6 drops of barium salt solution add the same volume of potassium
chromate K2CrO4. What is the color of precipitate? Does the precipitate dissolves in
hydrochloric and acetate acids? Write the equations of the reactions.
4.2.6. Test on Са2+ ion.
To 2-3 ml of a calcium salt solution in two test-tubes add the identical volumes
of ammonium oxalate (NH4)2C2O4 solution to the first test-tube and ammonium
carbonate (NH4)2CO3 solution – to the second one. What are the colors of the
precipitates? Write the equations of the reactions.
4.2.7. Flame painting with the salts of Na+, K+, Ca2+, Li+ ions.
Moisten a platinum wire which was cleared with hydrochloric acid and hardened
in lithium salt solution and then bring it into the lower part of the flame. Repeat the
experiment with the solutions of another metals salts. Note the colour of the flame
with these metals ions.
4.2.8. Test on ortho-phosphate ion.
To 1–2 ml a solution of phosphoric acid (H3PO4) in a test-tube add the solution
of ammonium molybdenate ((NH4)2MoO4) and heat up. What is observed? Write
the equation of the reaction and name the product.
4.2.9. Test on sulfide- ion.
To 2–3 ml of a sodium sulfide Na2S solution in two test-tube add the identical
volumes of lead acetate Pb(CH3COO)2 solution to the first test-tube, and cadmium
nitrate Cd(NO3)2 solution – to the second one. What are the colors of the
precipitates? Write the equations of the reactions.
5. Conclusions and Interpretations. Lesson Summary
32
Topic 5
Acid-base equilibrium. Calculation and experimental
determination of the рН of biological liquids. Protolytical
processes in living organisms
1. Objectives
Most chemical reactions are reversible and occur in the direction of equilibrium.
Equilibrium processes are of great importance in chemistry and biology.
Equilibrium processes also include hydrolysis - the interaction of substances with
water molecules, which are based on important metabolic processes - the hydrolysis
of polysaccharides, fats and proteins that occur in the tissues of a living organism
and buffer action of some protein and salt systems.
An important characteristic of electrolyte solutions particularly those that are
used as blood substitutes is their ion force because it has equal ionic force of
plasma.
Equilibrium processes include hydrolysis - the interaction of substances with
water molecules, which are based on important metabolic processes - the hydrolysis
of polysaccharides, fats and proteins that occur in the tissues of a living organism
and buffer action of some protein and salt systems.
2. Learning Targets:
– to know the theory of weak and strong electrolytes;
– to know the pH for fluids of the human body in norm and pathology;
– to know about the role of hydrolysis in biochemical processes;
– to be able to experimentally determine the pH value;
– to be able to solve situational problems on this topic.
3. Self Study Section
3.1. Syllabus Content
Electrolyte solutions. The degree and the dissociation constant of weak
electrolytes. Properties of solutions of strong electrolytes. Activity and activity
coefficient. Ionic force of solution. Water and electrolyte balance - a necessary
condition for homeostasis. Dissociation of water. Ionic product of water. pH. The
pH values for different liquids of the human body in normal and pathological
conditions.
Theories of acids and bases. Types of protolytic reactions: neutralization,
hydrolysis and ionization. Hydrolysis of salts. The degree of a hydrolysis, its
dependence on concentration and temperature. Constant of a hydrolysis. The role of
hydrolysis in biochemical processes.
3.2. Overview
When electrolytes are dissolved in water, they undergo dissociation and produce
ions. The phenomenon of the production of ions in solution is called dissociation or
ionisation:
33
CtxAny ⇄ xCt+ + yAn–
Strong electrolytes are almost completely dissociated into ions in solution,
degree of their dissociation α ~ 100%. Strong electrolytes are present only as ions in
solution; the acid or base molecule does not exist in aqueous solution. The
concentration of ions largely dominates the concentration of the undissociated
molecules of the electrolyte and hence the concentration of the undissociated
molecules is negligible.
Weak electrolytes are only feebly ionised in solution, they are present partly as
ions and partly as undissociated molecules which are in dynamic equilibrium with
each other. Weak electrolytes are found to contain lesser concentration of ions and
an appreciable concentration of undissociated molecules of the electrolyte.
The constant of electrolytical dissociation (ionization constant):
y
[Сt + ] x ⋅ [ An − ] , pKd = – lg Kd
K =
d
[Сt An ]
x y
Percent of dissociation (degree of dissociation or degree of ionization):
n
α = ⋅100% , according to the Ostwald’s law of dilution α = K d
N
C
M
pH refers to hydrogen ion concentration and is applied to aqueous (water-based)
solutions. pH is defined as the negative of the logarithm of the molar concentration
of hydrogen-ion:
pH = –log10[H+]
For strong acids (HCl, HNO3, H2SO4, HBr, HI, HClO4, HClO3, HBrO3,
H2SeO4):
pH = – log [H+], [H+] = CM(acid)·α·n
For strong bases (LiOH, NaOH, KOH, RbOH, CsOH, Ca(OH)2, Ba(OH)2):
pH = 14 – pOH, pOH = – lg [OH–]
For weak acids:
pH = – log [H+],
[ H + ] = K (acid ) ⋅ С (acid )
d
M
For weak bases:
pH = 14 – pOH, pOH =– lg [OH–],
[OH − ] = K (base) ⋅ С (base)
d
M
Hydrolysis is a chemical process in which a certain molecule is split into two
parts by the addition of a molecule of water. One fragment of the parent molecule
gains a hydrogen ion (H+) from the additional water molecule. The other group
collects the remaining hydroxyl group (OH−). The hydrolysis occurs when 1) an
acidic or basic salt or 2) a weak acid or weak base forms.
The quantitative characteristics of a hydrolysis (the percent of hydrolysis h and
the hydrolysis constant Kh) for different types of salts could be calculated
34
according to following equations:
– salts which are formed by a strong base and a weak acid hydrolyze by anion
(рН>7):
K H 2O
Kh
, h=
, [OH–] = K ⋅ [salt ]
Kh =
h
K acid
[salt ]
– salts which are formed by a weak base and a strong acid hydrolyze by cation
(рН<7)
K H 2O
Kh
, [H+] = K ⋅ [salt ]
h
K base
[salt ]
– salts which are formed by a weak base and a weak acid hydrolyze by cation and
Kh =
anion. Kh =
K H 2O
K acid ⋅ K base
,
,
h=
h=
Kh ,
[OH–] =
K H 2O ⋅ K acid
K base
3.3. References
1. V.O. Kalibabchuk, V.I. Halynska, L.I. Hryshchenko et al. Medical
Chemistry. – AUS MEDICINE Publishing. – 2010. – 224 p.
2. Raymond Chang. Chemistry (6th Edition). – WCB/McGraw-Hill. – 1998. – 995
p.
3. Rodney J. Sime Physical Chemistry. Methods. Techniques. Experiments. –
Saunders College Publishing. – 1990. – 806 p.
4. John McMurry, Robert C. Fay. Chemistry (3rd Edition). – Prentice Hall. – 2001.
– 1067 p.
5. David E. Goldberg. Fundamentals of Chemistry (2nd Edition). –
WCB/McGraw-Hill. – 1998. – 561 p.
6. Theodore L. Brown, H.Eugene LeMay, Bruce E. Bursten. Chemistry. The
Central Science. – Prentice Hall. – 2000. – 1017 p.
7. John Olmsted III, Gregory M. Williams. Chemistry. The Molecular
Science. – Mosby. – 1994. – 977 p.
8. Steven S. Zumdahl. Chemistry (4th Edition). – Houghton Mifflin Company. –
1997. – 1031 p.
3.4. Self Assessment Exercises
а) Review Questions
1. Give the definition of the dissociation process. What compounds are called
electrolytes?
2. Give the mathematical equation of Ostwald’s dilution law and define all the
symbols.
3. Define the term ionic product of water. How it is affected when an acid or a base
is added to water?
4. Define pH. What is the value of pH of a neutral, acidic and basic solution?
35
5. What are the pH values of some biological liquids (blood, gastric juice, urine)?
6. What is acidosis and alkalosis? What are the pH values of the blood plasma at
acidosis and alkalosis?
7. What factors affect the hydrolysis equilibrium shifting? Give the examples.
8. Write formulas for calculation of the constants, degree of hydrolysis and pH of
solutions of different types of salts.
9. The role of hydrolysis in biochemical processes.
b) Types of Numerical Problems and Their Solving Strategies.
Numerical problem 1. Calculate percent of dissociation α and concentration of
[H+] ions in 0,3 М НСООН solution if its ionization constant is 2,1⋅10–4.
Given:
1. Calculate α using the following equation:
См(НСООН) = 0,3 М
Kд
2,1 ⋅ 10 −4
α=
=
= 2,64⋅10–2 or 2,64 %.
К = 2,1⋅10–4
C
0
,
3
M
Calculate:
α – ? [H+] – ?
2. Calculate the concentration of [H+] ions using the following equation:
[H+]= C⋅α = 0,3⋅2,64⋅10–2 = 7,9⋅10–3 mol/dm3
Numerical problem 2. Calculate рН of gastric juice, if its acidity is caused by the
presence of HCl 1,5 % by mass (d = 1 g/sm3).
Given:
1. Calculate the molar concentration of 1.5% HCl solution
Сp(HCl) = 1,5 %
using the following equation:
d = 1 g/sm3
10 ⋅ C p ⋅ d 10 ⋅ 1,5 ⋅ 1
= 0,4109 mol/dm3.
CM =
=
Calculate:
M ( HCl )
36,5
рН – ?
+
2. Calculate the concentration of [H ] ions according to the equation:
[H+] = Cмαn = 0,4109 ⋅ 1 ⋅ 1 = 0,4109 mol/dm3
3. Calculate рН of solution:
рН = –log[H+] = –log0,4109 = 0,3862 ≈ 0,4.
Numerical problem 3. pН value of arterial blood equal 7,36. Calculate molar
concentration of Hydrogen-ions in the blood.
Given:
Calculate the concentration of [H+] ions from the following equation:
рН= 7,36
pH = –lg[H+]; 7,36 = –lg[H+]; –7,36 = lg[H+]; lg[H+] = 8,64 ;
+
[Н ] – ?
[H+] = 4,36⋅10–8 mol/l.
Numerical problem 4. Calculate the hydrolysis constant, percent of hydrolysis
and pH of 0.1 M solution of ammonium chloride NH4Cl. The dissociation constant
Kd(NH4OH) = 1.77·10–5.
Calculations
Given:
Ammonium chloride is a salt formed by a weak base
36
СМ(NH4OH)=0.1 М
Кd(NH4OH)= 1.77⋅10–5
Кh – ?
α–?
pH – ?
and a strong acid, therefore:
Kh =
KH O
− 14
2 = 10
= 0.565 ⋅10− 9
Kbase 1.77 ⋅10− 5
Equations of NH4Cl hydrolysis:
NH4Cl + HOH ⇄ NH4OH + HCl;
NH4+ + HOH ⇄ NH4OH + H+; pH < 7.
h=
Kh
[ salt ]
0.565 ⋅10− 9
= 0.565 ⋅10− 8 = 0.752 ⋅10− 4
0 .1
=
[ H + ] = K h ⋅[ salt ] = 0.565 ⋅10 − 9 ⋅ 0.1 = 0.565 ⋅10 − 10 = 0.752 ⋅10
pH = –lg [H+] = –lg 0.752·10–5 = –(lg 0.752 + lg 10–5) =
–(–0.12 – 5) = 5.12
Numerical problem 5. Calculate pH of 0.1 M solution of KCN. The
dissociation constant for Kd(HCN) = 7.2·10–10.
Calculations
Given:
Potassium cyanide undergoes hydrolysis according to
СМ(HCN)=0.1 М
the equations:
Кd(HCN)= 7.2⋅10–10
KCN + HOH ⇄ HCN + KOH
pH – ?
CN– + HOH ⇄ HCN + OH–
It is salt formed by strong base and weak acid,
therefore:
KH O
2
K =
h K
acid
and
[OH − ] = K ⋅ [ salt ]
h
[OH − ] =
therefore
K H O ⋅ [ salt ]
2
K
acid
Calculate the [OH–] concentration in 0.1 M potassium
cyanide solution:
[OH − ] =
10− 14 ⋅ 0.1
= 0.0139 ⋅10− 4 = 0.12 ⋅10− 2
7.2 ⋅ 10− 10
Calculate the pOH of 0.1 M potassium cyanide solution:
pOH = –lg [OH–] = –lg 0.12·10–2 = –(lg 0.12 + lg 10–2) =
= –(–0.92 – 2) = 2.92
Calculate the pH of 0.1 M potassium cyanide solution:
pH = 14 – pOH = 14 – 2.92 = 11.08
c) Problems to Solve
1. Calculate α and [H+] of 0,1 М hypochloric acid solution HClO.
Answer: 7⋅10–4; 7⋅10–5 mol/l.
37
2. Degree of dissociation of 0,1 М solution of СН3СООН equal 1,32⋅10–2.
Calculate dissociation constant and pK.
Answer: 1,7⋅10–5; 4,75.
3. The acidity of gastric juice mostly caused by hydrochloric acid. Mass percentage
of HCl approximately 1 %. Calculate concentration of [H+] in mol/l.
Answer: 274 mmol/l.
4. Write the molecular and ionic equations of the hydrolysis of the following salts:
KCN, Li3PO4, Cr2(SO4)3, CuCl2, CH3COONH4. Indicate the pH.
5. Calculate the pH of the solution, constant of hydrolysis and percent of hydrolysis
of ammonia bromide NH4Br in its 0.01 M solution.
6. Calculate the pH of 0.36 M CH3COONa solution. Kd(CH3COOH) = 1.75·10–5.
7. Calculate the pH of 0.42 M NH4Cl solution. Kd(NH4OH) = 1.8·10–5.
8. Calculate the pH of an aqueous solution of 0.1 M ammonium formate
HCOONH4 assuming its complete dissociation. Kd(HCOOH) = 1.8·10–4,
Kd(NH4OH) = 1.8·10–5.
9. Calculate pH, hydrolysis constant and percent of hydrolysis of 0.1 M solution of
sodium carbonate. The dissociation constant for H2CO3 is 4.5·10–11.
Answer: рН = 11,7; Кh.= 2.2⋅10–4; h = 4.7 %.
4. Laboratory Activities and Experiments Section
The content and methods of practice and laboratory studies.
4.1. Practical Skills and Suggested Learning Activities:
– reactions of formation of feebly ionised compounds;
– shifting the equilibrium in a solution of ammonia;
– reactions with the formation of the precipitate;
– reactions of formation of feebly ionised compounds;
– shifting the equilibrium in a solution of ammonia;
– conditions for precipitation;
– conditions of precipitate dissolution resulting from chemical ;
– effect of the nature of salt on the reaction medium;
– effect of temperature on the degree of hydrolysis;
– effect of dilution on the hydrolysis of salts;
4.2. Experimental Guidelines
4.2. 1. Reactions of formation of feebly ionised compounds:
1) place 3-4 ml of NaOH solution into the clean test-tube. Add 3 drops of
phenophthaleine indicator to the test-tube and record the indicator color in the
solution. Then add the sulfuric acid to discoloration of solution in the test-tube;
2) place 2-3 ml of CuSO4 solution into the clean test-tube. Add NaOh solution
tillprecipitate will forms. Then add the sulfuric acid to dissolve the precipitate.
Write the equations of the reactions.
4.2.2. Shifting the equilibrium in a solution of ammonia.
Place into the clean test-tube ammonium hydroxide (NH4OH) solution. Add 3
38
drops of phenophthaleine indicator to the test-tube and record the indicator color in
the solution. Colored solutions divide into 4 tubes. Add some crystalline
ammonium acetate to the first test-tube, diluted solution of HCl to the second, the
third test-tube heat to boiling, and a fourth left for comparison. Observe and explain
the effect of adding CH3COONH4, HCl and heating to shift the equilibrium in the
system ammonia-water.
4.2.3. Effect of the nature of salt on the reaction medium.
Into four clean test-tubes add approximately 2 ml of 0.1 M sodium carbonate
(Na2CO3), 0.1 M zinc sulfate (ZnSO4), 0.1 M sodium chloride (NaCl) and 0.1 M
ammonium acetate (CH3COONH4) solutions. Add one drop of phenolphthalein
indicator and methyl orange indicator solutions to each test-tube. Observe and
record any changes in color. Interpret your observations and write equations of
reactions.
Table
Solution of salt
Color of indicator
Universal indicator Methyl
Phenolphthalein
paper
orange
The reaction
medium
рН
of solution
Na2CO3
ZnSO4
NaCl
CH3COONH4
Are all salts hydrolyzing? Write hydrolysis equation in molecular and ionic
form.
4.2.4. Effect of temperature on the degree of hydrolysis.
Into two clean test-tubes add 2–3 ml of sodium acetate (CH3COONa) solutions
and add 2–3 drops of phenolphthalein indicator. The first test-tube to heat till
boiling. Compare the color of cold and hot solutions. Cool the tube under running
water and observe changing color of the solution. Write equationf of reaction.
4.2.5. Effect of dilution on the hydrolysis of salts.
Into clean test-tube with 1-2 cm3 solution antimony (III) chloride adds 2-3 cm3
of water. Explain the reason for formation of the precipitate. Write the reaction
equation.
4.2.6. A complete hydrolysis.
To 3 cm3 aluminum chloride solution or aluminum sulphate add 3 cm3 of sodium
carbonate (test-tube number 1). Test tube close with a stopper vent tube, which dip
into the tube of a limy water (test-tube number 2). Observe the formation of
precipitates in both test-tubes: an amorphous in the test-tube number 1 and
crystalline in test-tube number 2. Separate precipitate from the the test-tube number
1 and divide it into two test-tubes and check its solubility in acids and alkalis. Write
all the reactions.
5. Conclusions and Interpretations. Lesson Summary
39
Topic 6
Properties of buffer solutions and their role in biological
systems. Preparation of buffer solutions. Determination of the
buffer capacity
1. Objectives
Constant value of pH of biological liquds is provided by several physiological
mechanisms, as well as buffer systems, the main ones being protein, hemoglobin,
phosphate, and carbonate. Changing the pH of biological liquds indicates the
pathological processes in a human body. Therefore, the study of buffer systems,
mechanisms of action will contribute to a better knowledge of the biological
processes.
2. Learning Targets:
– to know the concept of buffer systems, their composition, types and mechanism
of their action;
– to know the formula for calculation of the pH of buffer systems;
– to know the definition of buffer capacity and be able to carry out calculations of
value;
– to be able to carry out calculations related with the preparation of buffer
solutions;
– to be able to prepare buffer solutions and determine the pH value.
3. Self Study Section
3.1. Syllabus Content
Buffer solutions and their classification. Henderson-Hasselbach equation.
Mechanism of buffer action.
Buffer capacity. Buffer systems of the blood. Bicarbonate (hydrogencarbonate)
buffer, phosphate buffer. Protein buffer system. The concept of acid-base condition
of blood.
3.2. Overview
A buffer is a solution characterized by the ability to resist changes in pH when
limited amounts of acid or base are added to it. Biological fluids, such as blood, are
usually buffer solutions because the control of pH is vital to their proper
functioning. To the biological buffer systems, which help to support the value of рН
of bioliquids in certain limits, belong:
phosphoric: NaH2PO4 + Na2HPO4
hydrogen-carbonate: NaHCO3 + CO2
proteinic (albuminous): Prt(NH2)COOH
haemoglobinous: HHb + KHb
The change of рН of biological liquids testifies the proceeding of pathological
processes that can be the diagnostic sign of some diseases.
Buffers contain either a weak acid and its anion or a weak base and its cations.
40
Blood, for example, contains H2CO3 and HCO3–, as well as other conjugate acidbase pairs.
рН of buffer solution depends on Кd of a base (or an acid) and the ratio between
base (or acid) and salt concentrations. This dependence is expressed by HendersonHasselbach equation:
[acid ]
for any acidic buffer solution pH = pKd(acid) – lg
[ salt ]
[base]
for any basic buffer solution pH = 14 – pKd(base) + lg
[ salt ]
Buffer solutions also are characterized by buffer capacity. The buffer capacity
is defined as the number of moles of strong acid (or base) necessary to change the
pH of 1 litter of the buffer solution by one unit. It is calculated by the equation:
C ⋅ V1 ⋅ 1000
BC =
[mmol/ml–1],
V2 ⋅ ∆pH
where C – concentration of an acid or base (mol/l), V1 – volume of an acid or base
(ml), V2 – volume of buffer solution (ml).
The buffer capacity depends on the concentration of components of the buffer
system and also on their ratio (optimal is 1:1). The buffer capacity can be calculated
using the following equation:
3.3. References
1. V.O. Kalibabchuk, V.I. Halynska, L.I. Hryshchenko et al. Medical
Chemistry. – AUS MEDICINE Publishing. – 2010. – 224 p.
2. Raymond Chang. Chemistry (6th Edition). – WCB/McGraw-Hill. – 1998. – 995
p.
3. Rodney J. Sime Physical Chemistry. Methods. Techniques. Experiments. –
Saunders College Publishing. – 1990. – 806 p.
4. John McMurry, Robert C. Fay. Chemistry (3rd Edition). – Prentice Hall. – 2001.
– 1067 p.
5. David E. Goldberg. Fundamentals of Chemistry (2nd Edition). –
WCB/McGraw-Hill. – 1998. – 561 p.
6. Theodore L. Brown, H.Eugene LeMay, Bruce E. Bursten. Chemistry. The
Central Science. – Prentice Hall. – 2000. – 1017 p.
7. John Olmsted III, Gregory M. Williams. Chemistry. The Molecular
Science. – Mosby. – 1994. – 977 p.
8. Steven S. Zumdahl. Chemistry (4th Edition). – Houghton Mifflin Company. –
1997. – 1031 p.
9. Gary L. Miessler, Donald A. Tarr. Inorganic Chemistry. – Prentice Hall. –
1991. – 625 p.
3.4. Self Assessment Exercises
а) Review Questions.
41
1. Definition of a buffer system. Classification of of a buffer systems? What are
the most important examples of buffer systems and write down their
composition.
2. Explain the mechanism of action of phosphate buffer system.
3. What factors influence on the pH buffer systems?
4. Explain influence of dilution and addition of strong acid and alkali solutions
on the pH value of the buffer.
5. Buffer blood system, their composition, mechanism of action and biological
significance.
6. What is buffer capacity? What is the factors of influence of buffer capacity ?
Write the equation for the calculation of buffer capacity by acid and alkali.
7. What buffer system of blood have the greatest buffer capacity?
b) Types of Numerical Problems and Their Solving Strategies.
Numerical problem 1. Calculate рН of buffer solution, which was prepared by
mixing 40 cm3 of 0.1 М Н2СО3 solution and 60 cm3 of 0.1 М NaHCO3
solution. The dissociation constant Kd(Н2СО3) = 4.4⋅10–7.
Calculations
Given:
1. Calculate the concentration of H+ ions in acidic
3
С(Н2СО3) = 0.1 mol/dm
buffer solution:
С(NaHCO3)= 0.1 mol/dm3 [ H + ] = K (acid ) ⋅ С (acid ) ⋅V (acid ) = 4.4 ⋅10− 7 ⋅ 0.1 mol / l ⋅ 0.4 l =
d
С ( salt ) ⋅V ( salt )
0.1 mol / l ⋅ 0.6 l
V(Н2СО3) = 40 cm3
3
−
V(NaHCO3) = 60 cm
= 2.93 ⋅10 7 mol/dm3
К1(Н2СО3) = 4.4⋅10–7
2. Calculate pH of buffer solution:
Calculate:
pH = –log[H+]= –log(2.93⋅10–7) – (log2.93 + log10–7)
рН (buffer sol.) – ?
= –(0.466 – 7) = 6.5.
3. We can use Henderson-Hasselbach equation for acidic buffer for solving of this
type of problem:
рН = рК – lg
Cacid Vacid
;
Csalt Vsalt
if buffer is basic, then:
рН = 14 – рК + lg Cbase Vbase .
Csalt Vsalt
Numerical problem 2. Calculate the volumes of 0.1 M solutions of acetic acid
CH3COOH and sodium acetate CH3COONa which should be used to
prepare 200 cm3 of buffer solution with pH 5.24. The dissociation
constant of acetic acid Kd(CH3COOH) = 1.75·10–5.
Calculations
Given:
1. Calculate the concentration of hydrogen ions in
42
C(CH3COOH)= 0.1 mol/dm3
C(CH3COONa)=0.1 mol/dm3
V(buffer sol.) = 200 cm3
рН = 5.24
К(CH3COOH )=1.75⋅10–5
V(salt) – ?
V(acid) –?
acidic buffer solution:
[H+] = antilog (–5.24) = 10–5.24 = 5.76·10–6 mol/dm3
2. The concentration of hydrogen ions in acidic
buffer solution is:
С (acid ) ⋅ V (acid )
[H+] = Kd(acid)·
C ( salt ) ⋅ V ( salt )
V (acid )
V ( salt )
If we denote the volume of acetic acid as x, then the volume of salt is
x
V(salt) = (200–x), then:
[H+] = Kd·
200 − x
3. Solve the equation using data from the task:
x
x
5.76·10–6 = 1.75·10–5 ·
,
= 0.329,
200 − x
200 − x
x = 0.329·(200–x), 1.329·x = 65.8, x = 49.5 ml
Thus, the volume of acetic acid is V(CH3COOH) = 49.5 ml and the volume of
sodium acetate V(CH3COONa) = 200 – 49.5 = 150.5 ml.
4. In the case of basic buffer solution we have to calculate concentration of [H+],
then usiing the expression of the ionic product constant of water we can
calculate the concentration of [OH-]:
[OH–] = 10–14 / [H+] .
If we denote the volume of base as x, then the volume of salt is V–x. then:
[OH–] = К⋅ Cbase Vbase ; if Сbase = Сsalt, then
Csalt Vsalt
x
.
[OH–] = К⋅ Vbase = K ⋅
Vsalt
Vbuff − x
Hence we find the volumes of the base and salt needed to prepare basic buffer.
[H+] = Kd·
If C(acid)=C(salt), then:
Numerical problem 3. Calculate the buffer capacity of blood on acid and base if to
100 cm3 of blood:
1) 36 cm3 of 0.05 М solution of HCl should be added to change its pH
from 7.36 to 7.00;
2) 14 cm3 of 0.1 M solution of NaOH should be added to change its pH
from 7.4 to 9.4.
Calculations
Given: 1) V(blood) = 100 cm3
2) V(blood) = 100 cm3
3
V(HCl) = 36 cm
V(NaOH) = 14 cm3
3
СМ(HCl) = 0.05 mol/dm
СМ(NaOH) = 0.1 mol/dm3
∆рН = 7,36 – 7,00 = 0,36
∆рН = 9,36 – 7,36 = 2
Вa – ?
Вb – ?
43
1) The buffer capacity of blood on acid could be calculated according to the
equation:
0.05 ⋅ 36
C (acid ) ⋅ V (acid )
Bb =
=
= 5⋅10–2 mol/dm3
V (blood ) ⋅ ∆pH
100 ⋅ 0.36
2) The buffer capacity of blood on base could be calculated according to the
equation:
0.1⋅14
C (base) ⋅ V (base) ⋅ 1000
Bb =
=
= 7⋅10–3 mol/dm3
V (blood ) ⋅ ∆pH
100 ⋅ 2
c) Problems to Solve
1. Calculate the volumes of 0.2 M solution of NaH2PO4 and 0.1 M solution of
Na2HPO4 which should be used to prepare 200 cm3 of buffer solution with pH
7.38.
Answer: 50 сm3 and 150 сm3.
2. Calculate pH of buffer solution which was prepared by mixing 150 cm3 of 0.02
molarity solution of ammonium hydroxide NH4OH and 200 cm3 of 0.015
molarity solution of ammonia chloride NH4Cl (Kb(NH4OH) = 1.8 ×10-5).
Answer: рН=9,26.
3. Calculate the buffer capacity of blood on acid and on base if to 100 cm3 of :
1) 10 cm3 of 0.2 M solution of HCl should be added to change its pH from 7.4
to 3.4;
2) 0.8 cm3 of 0.1 M solution of NaOH should be added to change its pH from
7.4 to 9.4.
Answer: Вa = 5 mmol/dm3; Вb = 0,4 5 mmol/dm3.
4. Laboratory Activities and Experiments Section
The content and methods of practice and laboratory studies
4.1. Practical Skills and Suggested Learning Activities:
– preparation 50 (100) cm3 buffer solution with a given pH;
– influence of dilution and adding of the small amounts of acid or alkali on pH
buffer solution;
– determination of buffer capacity of blood serum.
4.2. Experimental Guidelines
4.2.1. Prepare 50 or 100 cm3 buffer solution with a pH equal – 3.9; 4.28; 6.4; 7.4;
8.7.
а) using the data in the table to determine the optimal component composition
buffer mixture indicated pH.
Table
Name of the buffer
system
Weak electrolyte
Salt
Dissociation constant
of acid (base)
рН range
Acetate
carbonate
CH3COOH
H2CO3
CH3COONa
NaHCO3
1,8⋅10–5
4,3⋅10–7
3.7-5.6
5.5-7.4
44
Phosphate
Ammonium
NaH2PO4
NH4OH
Na2HPO4
NH4Cl
6,3⋅10–8
1,8⋅10–5
6.2-8.2
8.4-10.3
a) transform the pH value of buffer solution in the concentration of hydrogen
ions, as, рН = 4,28:
[H+]= alg(–pH) = alg(–4.28) = 5.25·10–5 (mol/dm3).
b) using the equation: [H+] = К Cacid Vacid ; calculate volumes of the acid (х), and
Csalt Vsalt
salt (50 – х) to prepare for example, 50 cm3 of buffer solition. According to the
table, for pH = 4.28 is optimal acetate buffer mixture, so use for the calculation
value of Кc(СН3СООН)= 1,8·10-5. Volume of acid (х) calculates by the formula:
x
50 ⋅ [ H + ]
50 ⋅ 5.25 ⋅ 10−5
[H+] = K ⋅
;
x=
=
= 37.2
+
50 − x
K d + [ H ] (1,8 + 5.25) ⋅ 10− 5
х = 37.2 cm3 СН3СООН; 50 – 37.2 = 12.8 cm3 CH3COONa
c) Mix in a simple flask 37.2 mL of 0.1 M solution of CH3COOH and 12.8 mL
of 0.1 M solution of CH3COONa.
d) Using the universal indicator to check up a рН of the prepared buffer
solution.
4.2.2. Study the influence of dilution and addition of small amounts of acid or alkali
on pH buffer solution.
а) measure 1 cm3 of buffer into the test-tube and add 9 cm3 of distilled water.
Measure the pH before and after the dilution of the buffer solution with a universal
indicator.
б) measure into the two test-tubes 10 cm3 of buffer solution. Add 0,5 cm3 of 0.1
М HCl solution in to the first test-tube and 0.5 cm3 of 0.1 М NaOH solution in to
the second test-tube. Determine of the pH of prepared solutions.
4.2.3. Determination of the buffer capacity.
а) determination the buffer capacity of blood serum by acid.
Place 15 cm3 of blood serum in to the flask number 1 and determine the pH
value. Measure 15 cm3 of the solution with an exactly known pH in to the second
flask.
Add 1 drop of corresponding indicator.
Titrate with burette prepared blood serum using 0.1 M solution of HCl to obtain
the same color of the indicator with the control solution.
The calculation of buffer capacity by acid carried out by the formula:
Ba =
Vacid Cacid ,
∆pH ⋅ Vbuff
where Vacid and Cacid – volume and molarity acid solution;
Vbuff – volume of the buffer solution;
∆рН – changeof the pH of blood serum and control solution.
б) determination the buffer capacity of blood serum by base.
45
Determination is carried out similarly as in the previesled experement, but for
the titration is ussed 0.1 M solution of potassium hydroxide.
The calculation of buffer capacity by acid carried out by the formula:
Bbase =
Vbase Cbase
∆pH ⋅ Vbuff
,
where Vbase and Cbase – volume and molarity base solution;
Vbuff – volume of the buffer solution;
∆рН – changeof the pH of blood serum and control solution.
5. Conclusions and Interpretations. Lesson Summary
Topic 7
The basic principles of the volumetric analysis. Acid-base
titration. Precipitation and dissolution reactions
1. Objectives
Titrimetric analysis - a method of quantitative analysis, which is is widely used
in various fields of chemistry, biology and medicine because it is rapid, convenient,
accurate, and readily automated. It is important to know the composition of different
biological liquids of living organisms for understanding the processes that occur in
them. Exact data on their composition allows for a reasoned diagnosis and
treatment. Skills mastery of methods of titrimetric determination has practical
importance for the doctors.
Since heterogeneous equilibrium make a significant contribution to the overall
homeostasis, it is important to study heterogeneous processes on the interfaces
features.
2. Learning Targets:
– to know the theoretical basis of the method of acid-base titration;
– to know the methods of calculations in titration analysis;
– to know the methods of standardization of titrated solutions;
– to choose an indicator for the method of neutralization titration;
– to determine the acidity of gastric juice;
– study the conditions of precipitates formation.
3. Self Study Section
3.1. Syllabus Content
Principles of titrimetric analysis. Titrimetric methods of analysis.
The method of acid-base titration. Law equivalents and its use in quantitative
analysis. Equivalence point. Acidimetry and Alkalimetry. Indicators. Theory of
indicators, their quantitative characteristics. Titration curves. Choice of indicators.
Precipitation and dissolving reactions. Solubility product constant. Precipitates
46
formation conditions. The heterogeneous equilibrium role in general homeostasis of
the organism.
3.2. Overview
The procedure of using a neutralization reaction to determine the amount of acid
or base in a solution is called titration.
This is a quick and accurate method for determining acidic or basic substances
in many samples. This method enable to determine some inorganic and hundred of
organic acids and bases of different types; frequently organic compounds are
titrated in waterless environment. The used titrant is typically a strong acid or base.
The sample species can be either a strong or weak acid or base. The neutralisation
method based on acid-basic reactions (exchange reactions by protons):
Н3О+ + ОН– → 2Н2О,
or
Н+ + ОН– → Н2О.
Titrations according to the applied titrant are
1) acidimetric (titrants are the acids solutions) – uses for determination of strong
and weak bases, salts of strong bases and weak acids and organic compounds;
2) alkalimetric (titrants are solutions of bases) – uses for titration of strong and
weak acids, sour salts, salts of strong acids and weak bases, organic compounds
having acidic disposition (acids, phenols).
In a titration, a solution with a known concentration of base from a burette is
added to a solution of unknown concentration of acid (or a known acid can be added
to an unknown base). A measured volume of an acid or base of known concentration
is reacted with a sample to the equivalence point.
The neutralization reaction is monitored by an acid-base indicator. An acid-base
indicator is a weak acid that has a conjugate base with a different color from that of
the acid. In the laboratories various organic dyes such as phenolphthalein, which is
colorless in acid solution, and pink in basic solution, and methyl orange, which is
red in acid solution and yellow in basic solution are used. Table 1 lists several
indicators together with their useful pH ranges.
Indicators
Methyl violet
Methyl orange
Methyl red
Bromothymol blue
Phenolphthalein
Lithmus
Table. Properties of Some Indicators
Color
Effective pH range
Acid Form
Base Form
0.0–3.0
Yellow
Violet
3.1–4.1
Red
Yellow
4.2–6.2
Red
Yellow
6.0–7.8
Yellow
Blue
8.0–10.0
Colorless
Red
5.0–8.0
Red
Blue
Methods of neutralization are widely used in clinical laboratories to determine
47
the acidity of gastric juice, pancreatic juice, urine and other biological liquids.
These methods are used to determine acidity of various foods in the analysis of
water, waste water and air.
When we use the solubility rules to predict whether or not the precipitate will
form when two solutions are mixed we should strictly speaking say that the
precipitate may form, rather then it will form, because if the solution were very
diluted a precipitate may not form. What concentrations of ions will, in fact, give a
precipitate? We can answer this question by considering the equilibrium between a
solid salt and its ions in saturated solution of the salt in a more quantitative manner.
In general, for an ionic compound with the formula AxBy, the equilibrium in a
saturated solution can be written as:
AxBy (s) ⇄ xAm+(aq) + yBn-(aq).
The solubility product constant is then:
Ksp = ([Am+]x × [Bn-]y)eq.
In other words, the solubility product constant is equal to the product of the
concentrations of the ions involved in the equilibrium, each raised to the power of
its coefficient in the equation for the equilibrium.
If we’ll express the concentration of Am+ ions in the saturated solution as xS
(where S is solubility), and the concentration of Bn- ions in the saturated solution as
yS, then the equation for solubility product constant may be written as:
Ksp = (xS)x ⋅ (yS)y = xx ⋅ yy ⋅ Sx+y.
So the solubility of a slightly soluble salt (in moles/l) may be found out as:
S =
x+ y
K
sp
xxy
y
If [A ] ⋅ [B ] > Ksp, then the solution is over-saturated.
If [Am+]x ⋅ [Bn-]y < Ksp, then the solution is non-saturated.
If [Am+]x ⋅ [Bn-]y = Ksp, then the solution is saturated.
So, in case when [Am+]x ⋅ [Bn-]y ≥ Ksp, the precipitate will form.
3.3. References
1. V.O. Kalibabchuk, V.I. Halynska, L.I. Hryshchenko et al. Medical
Chemistry. – AUS MEDICINE Publishing. – 2010. – 224 p.
2. Raymond Chang. Chemistry (6th Edition). – WCB/McGraw-Hill. – 1998. – 995
p.
3. Rodney J. Sime Physical Chemistry. Methods. Techniques. Experiments. –
Saunders College Publishing. – 1990. – 806 p.
4. John McMurry, Robert C. Fay. Chemistry (3rd Edition). – Prentice Hall. – 2001.
– 1067 p.
5. David E. Goldberg. Fundamentals of Chemistry (2nd Edition). – WCB/McGrawHill. – 1998. – 561 p.
6. Theodore L. Brown, H.Eugene LeMay, Bruce E. Bursten. Chemistry. The
Central Science. – Prentice Hall. – 2000. – 1017 p.
m+ x
n- y
48
7. John Olmsted III, Gregory M. Williams. Chemistry. The Molecular
Science. – Mosby. – 1994. – 977 p.
8. Steven S. Zumdahl. Chemistry (4th Edition). – Houghton Mifflin Company. –
1997. – 1031 p.
9. Gary L. Miessler, Donald A. Tarr. Inorganic Chemistry. – Prentice Hall. – 1991.
– 625 p.
3.4. Self Assessment Exercises
а) Review Questions.
1. What is the process of titration?
2. What solutions are called titrated?
3. What is the equivalence point, the point of neutrality?
4. Requirements to chemical reaction used in titrimetric methods of analysis.
5. What law is used for the calculations in titrymetric analysis?
6. What is titration curves and how they help carry out choice of indicator?
7. What substances can be determined using methods of neutralization?
8. What is the standart solution (titrant) that is used in the method of neutralization.
How to prepare it?
9. What indicators can be used in the titration of hydrochloric, phosphoric acid and
ammonium chloride?
10. What are the types of gastric acidity and in what units it is expressed?
11.What is the solubility measure for feebly soluble compounds. Write down the Ksp
expressions for the following salts: CaF2, Ag2S, Mg3(PO4)2.
b) Types of Numerical Problems and Their Solving Strategies.
Numerical problem 1. How many grams of borax Na2B4O7·10H2O are necessary
to prepare 400 ml of 0.05 N solution?
Calculations
Given:
1. The mass of borax Na2B4O7·10H2O could be
V(sol.) = 400 ml3
calculated from the formula of normality of the
СN = 0.05 mol-eq/ml
solution:
m
Calculate:
→ m = CN·ME·V(sol.)
CN =
M E ⋅V(sol.)
m(Na2B4O7·10H2O) – ?
2. Calculate the molar mass of Na2B4O7·10H2O
equivalent:
M(Na 2 B4O 7 ⋅ 10H 2O) 381.36
MЕ(Na2B4O7·10H2O)=
=190.68 g/mol-eq.
=
2
2
3. Calculate the mass of borax:
m(Na2B4O7·10H2O) = 0.05·0.4·190.68= 3.8136 g.
Numerical problem 2. 0.2664 g of anhydrous soda were dissolved in water to get
100 cm3 of solution. 12.4 cm3 of 0.0502 N НСl solution were used for the
49
titration of 20 cm3 of prepared solution (in the presence of methyl
orange). Determine the mass percentage (in %) of Na2CO3 in soda.
Calculations
m(soda) = 0.2664 g
1. Calculate the concentration of prepared
V(soda sol.) = 100 cm3 = 0.1 dm3 solution according to the equivalent’s law:
V(sample) = 20 cm3
V(HCl) ⋅ HCl) 12.4 ⋅ 0.0502
C N (soda) =
=
=
V(HCl) = 12.4 cm3
V(sample)
20
CN(HCl) = 0.0502 N
= 0.03112 mol-eq/dm3.
Calculate:
2. Calculate the mass of Na2CO3 in 0.1 dm3
Cp(Na2CO3) – ?
of soda solution:
m(Na2CO3) = СN(Na2CO3)·V(soda sol.)·ME(Na2CO3) = 0.03112·0.1·53 = 0.1649 g,
where molar mass of anhydrous soda equals
M (Na 2CO 3 ) 106 = 53 (g/mol-eq),
M E (Na 2CO 3 ) =
=
2
2
3. Calculate the mass percentage of Na2CO3 in the anhydrous soda:
m(Na 2 CO 3 ) ⋅ 100% 0.1649 ⋅ 100
C p ( Na 2 CO 3 ) =
=
= 61.9 %
m(soda)
0.2664
Numerical problem 3. What is the acidity of gastric juice if 2.5 ml of 0.1 N of
alkali were used for the titration of 5 ml of gastric juice.
Given:
General acidity of gastric juice characterizes the
V(alkali) = 13.6 cm3
total content of substances of acid character in
СN(alkali) = 0.0485 N
gastric juice. Acidity of gastric juice is the volume
Calculate:
(in ml) of 0.1 N of alkali solution which is used for
Acidity of gastric juice – ? the titration of 100 ml of filtered gastric juice and is
calculated according to the following equation:
V(0.1N alkali)·М(g.j.)·0.1 = V(alkali)·CN(alkali)·100,
V(0.1 N alkali) = CN (alkali) ⋅ V (alkali) ⋅ 100
V (g.j.) ⋅ 0.1
Calculate acidity of gastric juice:
0.0485 ⋅13.6 ⋅100
V(0.1N alkali) = CN (alkali) ⋅ V (alkali) ⋅ 100 =
= 66 cm3
1
0
⋅
0.1
V (g.j.) ⋅ 0.1
Answer: The acidity of gastric juice equaks 66 clinical units.
Numerical problem 4. Calculate the volume (in ml) of 55% solution of sulfuric
acid (density 1.1 g/ml) which is necessary to prepare 500 ml of 1.5 N
H2SO4 solution.
Given:
1. From the formula of normality of the solution the mass of
Сp(H2SO4) = 55 %
H2SO4 in 1.5 N solution could be calculated:
V2(sol.) = 500 ml
m(H SO )
2 4
C (H SO ) =
d(sol.) = 1.1 g/ml
N 2 4
M (H SO ) ⋅ V (sol.)
СN(H2SO4) = 1.5 N
E 2 4
2
50
m(H2SO4) = CN(H2SO4)·ME(H2SO4)·V2(sol.)
where molar mass of H2SO4 equivalent is:
Calculate:
V1(sol.) – ?
M (H SO ) =
E 2 4
M(H SO ) 98
2 4 =
= 49 g / mol − eq
2
2
Calculate the mass of H2SO4 in 500 ml of 1.5 N solutions:
m(H2SO4) = 1.5 mol-eq/l · 49 g/mol-eq · 0.5 l = 36.75 g
2. From the formula for the mass percentage of H2SO4 in the 1st solution the mass of
the 1st solution could be determined:
CP(H 2 SO4 ) =
m(H SO )
2 4 ⋅ 100%
m (sol.)
1
→
m (H SO ) ⋅ 100% 36.75g ⋅ 100%
4
=
= 66.82 g
C (H SO )
55%
P1 2 4
m1(sol.) = 1 2
3. Calculate the volume of the 1st solution:
m (sol) 66.82 g
V ( sol ) = 1
=
= 60.75 ml
1
d (sol) 1.1 g/ml
1
Numerical Problem 5. The solubility of an electrolyte AB is 0.00714 g/l at the
temperature of 25 OC. Calculate the Ksp quantity if the
molar mass of the electrolyte is 100 g/mol (CaCO3).
Steps to Solution:
Calculate the solubility of the electrolyte in mol/l:
S ( g/l )
0 . 00714
S ( mol/l ) =
=
= 0 . 0000714
M ( g/mol )
100
The dissociation equation for the electrolyte AB is:
AB ⇄ A– + B+
The solubility product constant for the electrolyte AB is:
Ksp = [A–]1 ⋅ [B+]1
In the saturated solution ions concentrations is: [A–] = [B+] = S.
Calculate the solubility product constant for the electrolyte AB:
Ksp = 0.0000714 · 0.0000714 = 5·10–9.
mol/l
Numerical Problem 6. The solubility of an electrolyte A2B is 8.59 mg/l at the
temperature of 25 OC. Calculate the Ksp quantity if the
molar mass of the electrolyte is 58 g/mol (Mg(OH)2).
Steps to Solution:
Calculate the solubility of the electrolyte in mol/l:
S ( mol / l ) =
S (g / l)
8 . 59 ⋅ 10
=
M ( g / mol )
58
−3
= 0 . 0001481
The dissociation equation for the electrolyte A2B is:
A2B ⇄ 2A– + B2+
The solubility product constant for the electrolyte A2B is:
51
mol/l
Ksp = [A–]2 × [B2+]1
In the saturated solution ions concentrations is: [A–] = 2S and [B+] = S.
Calculate the solubility product constant for the electrolyte A2B:
Ksp = (2·0.0001481)2 · 0.0001481 = 1.3·10–11.
c) Problems to Solve
1. Calculate the mass of oxalate acid, which reacts with KOH according to the
equation:
H2C2O4 + 2КОН = K2C2O4 +2H2O,
necessary for preparing 250 cm3 of solution with molar concentration of
equivalent of 0.1 mol-eq/dm3.
Answer: 1.125 g.
2. Determine the mass of sodium carbonate, which should be used to prepare 500
cm3 of 0.1 N soda solution.
Answer: 2.65 g.
3. 1.2046 g of sample of potassium hydroxide were dissolved in water to get 250 ml
of the solution. 14.82 ml of 0.1050 M solution of hydrochloric acid were used for
the titration of 20 ml of the prepared solution. Calculate the mass percentage of
the КОН in the sample.
Answer: 90.42 %.
4. 6.3 g of acetic acid sample was diluted by water to volume of 200 cm3. 10 cm3 of
this solution were titrated with 7.8 cm3 of 0.1 M KOH solution. Determine the
mass percentage of CH3COOH in the sample.
Answer: 14.86 %.
5. 33 cm3 of 1.010 M solution of alkali were used for the titration of 3.204 g of
hydrochloric acid. Determine the mass percentage of acid in solution.
Answer: 37,97 %.
6. To determine the total amount of gastric juice 5 cm3 of it were titrated with 2.8
cm3 0.095 N alkali solution in the presence of phenolphthalein. Determine the
acidity of gastric juice in titrimetric units.
Answer: 53.2.
7. Determine the content of hydrochloric acid and total acidity of gastric juice in
titrimetric units if for the titration of 10 cm3 of gastric juice 3.1 cm3 of 0.098 N
alkali solution were used in the presence of methyl orange and 6.0 cm3 of the
same alkali solution solution were used in the presence of phenolphthalein.
Answer: 30.4; 58.8.
8. Calculate the solubility of silver chloride AgCl in water in g/ml. Ksp(AgCl) =
1.8·10–10.
Answer: 1.9·10–3 g/ml
9. Calculate the solubility product constant for Mg(OH)2, if its solubility equals to
1.4·10–4 mol/l at 180 оС.
52
Answer: Ksp = 1.1·10–11
4. Laboratory Activities and Experiments Section
4.1. Practical Skills and Suggested Learning Activities.
– to choose the indicator in titration of strong acid with alkali;
– to choose the indicator in titration of weak acid with alkali;
– to determine the total acidity of gastric juice.
4.2. Experimental Guidelines:
4.2.1. Titration of a strong acid with a strong base using methyl orange indicator
1. Carefully fill the burette with the 0.1 M KOH to a point above the top
calibration mark.
2. Run sufficient solution through the tip of the burette into a beaker to remove air
bubbles trapped in the tip of the burette.
3. Lower the meniscus of the solution until the meniscus is at a point on the
calibrated portion of the burette.
4. Read the burette to the nearest ± 0.1 ml and record the initial burette reading on
the data sheet.
5. By use a pipette measure 10 ml of the HCl solution and add it in a sample flask.
When draining the pipette, hold the tip of the pipette against the inside surface to
avoid splattering.
6. Add three drops of the solution of methyl orange. Swirl the flask and contents to
mix the solution thoroughly.
7. Place the flask containing 10 ml of HCl solution under the burette. Lower the
burette so that the tip of the burette is inserted a 2 cm or more into the mouth of the
flask. Gently swirl - do not shake! The flask with one hand as the stopcock is
controlled with the other hand.
8. Begin titrate the solution of HCl by slowly adding the 0.1 M KOH. As titration
progresses, approach of the endpoint of the titration is indicated by appearance of
the yellow endpoint color where the titrant first comes in contact with the acidic
solution (red color). Titration is complete when the solution will change red color
on orange. Read to the nearest 0.1 ml and record the final burette reading.
9. Repeat the titration until three results of the volume of 0.1 M KOH added are
obtained that agree to within 0.2 ml.
10. Obtained outcome note in table 1.
11. Calculate the normality of HCl solution.
4.2.2. Titration of a strong acid with a strong base using phenolphthalein indicator.
1. Repeat items 1–5.
2. Add three drops of the solution of phenolphthalein. Swirl the flask and contents
to mix the solution thoroughly.
3. Begin titrate the solution of HCl solution by slowly adding the 0.1 M KOH. As
titration progresses, approach of the endpoint of the titration is indicated by
appearance of the red endpoint color where the titrant first comes in contact with the
53
acidic solution (colorless). Titration is complete when the solution will change
colorless on red. Read to the nearest 0.1 ml and record the final burette reading.
4. Repeat the titration until two results of the volume of 0.1 M KOH added are
obtained that agree to within 0.2 ml.
5. Obtained outcome note in table 1.
6. Calculate the normality of HCl solution.
4.2.3. Titration of a weak acid with a strong base using methyl orange indicator.
1. Repeat items 1–8, using instead of HCl solution the solution of CH3COOH.
2. Repeat the titration until two results of the volume of 0.1 M KOH added are
obtained that agree to within 0.2 ml.
3. Obtained outcome note in table 1.
4. Calculate the normality of CH3COOH solution.
4.2.4. Titration of a weak acid with a strong base using phenolphthalein indicator.
1. Repeat items 2.1–2.3, using instead of HCl solution the solution of CH3COOH.
2. Repeat the titration until two results of the volume of 0.1 M KOH added are
obtained that agree to within 0.2 ml.
3. Obtained outcome note in table 1.
4. Calculate the normality of CH3COOH solution.
Table 1. Outcomes of titration of strong and weak acids with various indicators
Volume of titrant - 0.1 M KOH (ml)
Analyte
with methyl orange
with phenolphthalein
1.
1.
2.
2.
10 ml of HCl solution
3.
3.
CN =
CN =
1.
1.
2.
2.
10 ml of CH3COOH
3.
3.
solution
CN =
CN =
4.2.5. Determination of general acidity of gastric juice.
General acidity of gastric juice characterizes the total content of substances of
acid character in gastric juice. Acidity of gastric juice is expressed as volume of 0.1
N alkali solutions which was used for the titration of 100 ml of filtered gastric juice.
Method of determination. Place some amount of filtered gastric juice in the
flask, add 1-2 drops of phenolphthalein and titrate with the 0.1N alkali solution until
light pink color of the gastric juice that does not disappear for 20-30 seconds. To
calculate the total acidity of gastric juice by the equation:
54
V(0,1 N base) = V (base) ⋅ С N (base) ⋅100 , where
V (base) ⋅ 0,1
V(base) – volume of KOН (cm3), that is used for titration;
V(juise) – volume of gastric juice (cm3) for determination;
CN(base) – normality of the titrated alcali solution.
5. Conclusions and Interpretations. Lesson Summary
Topic 8
Thermodynamical and kinetical regularities of biochemical
processes passage
1. Objectives
Most of the world energy is currently obtained from the combustion of fossil
fuels, which are mainly hydrocarbons. We are all familiar with the idea of energy
and we have a qualitative idea of what we mean by energy from everyday life. We
will be particularly concerned with the energy, usually in the form of heat that can
be obtained from chemical reactions. The quantitative study of the heat changes
associated with chemical reactions is called thermochemistry. Thermochemistry is
part of a subject of much wider scope called thermodynamics. Thermodynamics is
the science of the transformations of energy.
Bioenergetics is based on the basic principles of thermodynamics and describes
the energy transformations in living organisms.
Chemical kinetics is the field of physical chemistry, which studies the rates and
mechanisms of chemical and biochemical reactions. Chemical kinetics, catalysis and
equilibrium laws are of great theoretical and practical importance, since they allows
to select the optimal conditions for the reactions progress.
In general, reaction kinetics is the study of rate of chemical change and the way
in which this rate is influenced by conditions of concentration of reactants, products
and other chemical species which may be present, and the factors such as solvent,
pressure and temperature. Reaction kinetics permits formulation of models for the
intermediate steps through which reactants are converted into other chemical
compounds and is a powerful tool in elucidating the mechanism by which chemical
reactions proceed.
It provides a rational approach to stabilization of drug products and prediction of
shelf-life and optimum storage conditions. Study of the reactions rates is the basis
for the drugs pharmacokinetics studying, clinical diagnosis, biochemistry.
Assimilation characteristics and mechanism of enzymes action as biocatalysts is
important for the metabolism processes understanding, diagnosis and treatment of
certain diseases.
2. Learning Targets:
− to learn the main terms and basic laws of thermochemistry;
55
− to make thermochemical calculations for the foods fuel capacity evaluation;
− to get skills of theoretical calculation and experimental determination of
chemical reactions and processes heat effects;
− to be able to apply the knowledge of the thermodynamics laws for the chemical
processes direction prediction;
− to explain the characteristics of living systems and the basic processes of energy
transformations in them;
− identify all the terms in a kinetical equation. Be familiar with the terminology of
order of a reaction;
− giving the rate-law expression for a specific reaction at a certain temperature,
calculate the initial rate of reaction at the same temperature for any initial
concentrations of reactants. Use changes in concentrations to predict changes in
initial rates;
− perform various calculations with the integrated rate equations for zero-, first-,
and second-order reactions relating rate constants, half-lives, initial
concentrations, and concentrations of reactants remaining at some later time.
− study the reactants concentrations influence on the reaction rate;
− analyse the chemical equilibrium shifting.
3. Self Study Section
3.1. Syllabus Content
The special fields of chemical thermodynamics. Basic terms of chemical
thermodynamics: thermodynamical system (isolated, closed, open, homogeneous,
heterogeneous), the state variables (extensive and intensive), thermodynamical
processes (reversible, irreversible). Living organisms as open thermodynamical
systems. Irreversibility of life processes.
The first law of thermodynamics. Enthalpy. Thermochemical equations.
Standard enthalpies of formation and combustion. Hess's law. Calorimetry
techniques. Biochemical processes energetic characteristics. Thermochemical
calculations for the foods fuel capacity (caloricity) evaluation and making rational
and therapeutic diets.
Spontaneous and non-spontaneous processes. The second law of
thermodynamics. Entropy. Thermodynamic potentials: Gibbs’ free energy,
Helmholtz’ free energy. Termodynamical equilibrium conditions. The criteria for
the spontaneous processes direction.
The basic principles of thermodynamics applying to living organisms. ATP as an
energy source for biochemical reactions. Macroergic compounds.
Chemical kinetics as the basis for the rates and mechanism of biochemical
reactions studying. The reaction rate. Concentration affection the reaction rate. The
law of mass action for the reaction rate. Rate constant. The reaction order. Kinetical
equations for zero-, first- and second-order reactions. Half-life. The reaction
mechanism concept and the reaction molecularity.
The temperature influence the reaction rate. Van't Hoff’s rule.
56
Activation energy. Сollision theory. Arrhenius equation. The concept of the
transition state theory.
The kinetics of complex reactions: parallel, successive, conjugated, chain. The
concept of antioxidants. Free radical reactions in living organisms. Photochemical
reactions, photosynthesis.
Catalysis and catalysts. Features of catalysts. Homogeneous, heterogeneous and
microheterogeneous catalysis. Acid-base catalysis. Autocatalysis. The mechanism of
catalytical action. Promoters and catalytic poisons.
The kinetics of enzymatic reactions. Enzymes as biological catalysts. Enzymes
features: selectivity, efficiency, temperature and reaction medium affections. The
concept of the enzymes action mechanism.
Chemical equilibrium. Equilibrium constant and its expression. Chemical
equilibrium shifting. Le Chatelier principle.
3.2. Overview
Definition of the first law of thermodynamics is: Energy can neither be created
nor destroyed but only changed from one form to another or The energy of a system
that is isolated from its surroundings is constant.
If an amount of heat Q flows into a system from the surroundings, then the
internal energy of the system will increase and the system can do an amount of work
W on the surroundings:
Q = ∆U + W.
Enthalpy: The heat, QP, that flows into the system at constant pressure is equal
to the enthalpy change, ∆H:
Qp = – ∆Hр.
Enthalpy is defined by the expression:
Н = U + pV.
The enthalpy of formation of the most stable form of an element in its standard
state is zero.
From the ∆Hf values for the reactants and products of a reaction, we can
calculate enthalpy ∆H° for reaction. For a chemical reaction the enthalpy change is
given by the equation:
∆Hf = Σ∆Hf(products) – Σ∆Hf(reactants),
where Σ∆Hf(products) is the sum of the enthalpies of the products, and
Σ∆Hf(reactants) is the sum of the enthalpies of the reactants.
When the total enthalpy of the products, Σ∆Hf(products), is greater than the total
enthalpy of the reactants, Σ∆Hf(reactants), the enthalpy change, ∆H, is positive
There is a flow of heat Qp= –∆H from the surroundings to the reaction system. In
other words, the reaction is endothermic.
When the total enthalpy of the products is less than that of the reactans, the
enthalpy change, ∆H, is negative. Thus, Qp= –∆H is also negative, and heat flows
from the reaction system to the surrounding. In other words, the reaction is
exothermic.
57
Hess’s law (the law of constant heat summation): the energy change for any
chemical or physical process is independent of the pathway or number of steps
required to complete the process provided that the final and initial reaction
conditions are the same.
∆H1 = ∆H2 + ∆H3 = ∆H4 + ∆H5 + ∆H6
∆Hformation = –∆Hcombustion
Entropy (S) is a quantity that is a measure of the disorder of the particles (atoms
and molecules) that make up the system and the dispersal of energy associated with
these particles. The disorder in a system depends only on the conditions that
determine the state of the system, such as composition, temperature, and pressure.
The change in entropy therefore depends only on the initial and final states of the
system. Entropy, like enthalpy is a state function.
Definition of the second law of thermodynamics is: In any spontaneous process
the total entropy of a system and its surrounding increases.
For any spontaneous process we may write:
Suniverse = Ssystem + Ssurroundings > 0
In other words, for any spontaneous process the total entropy change must be
positive.
Gibbs free energy:For any change at constant temperature and pressure we have:
∆G = ∆H – T·∆S
For a spontaneous process the change in the Gibbs free energy, ∆G must be
negative. In other words, the Gibbs free energy decreases during a spontaneous
process.
The standard free energy of formation, ∆Gf, is the free energy change for the
formation of 1 mol of compound from its elements in their standard states. The
standard free energies of formation, of the elements in their standard states are taken
to be zero, for the free energies of formation of the substances you need to refer to a
table of ∆Gf values.
There are four possible combinations of ∆H and ∆S:
∆H
–
∆S
+
–
–
+
–
∆G
– at all T
– at low T
+ high T
– at all T
58
Spontaneous Reaction
Yes
Yes
No
No
+ at low T
No
– at high T
Yes
For any reaction nAA + nBB → nXX + nYY:
The enthalpy change for the reaction is the difference between sum of the
standard enthalpies of formation of the products and sum of the enthalpies of
formation of the reactants:
∆H° = Σn·∆H°(products) – Σn·∆H°(reactants) =
= [nX·∆H°(X) + nY·∆H°(Y)] – [nA·∆H°(A) + nB·∆H°(B)]
The standard entropy change for a reaction is easily calculated from the
standard molar entropies, using the expression:
∆S° = Σn·S° (products) – Σn·S°(reactants) =
= [nX·S°(X) + nY·S°(Y)] – [nA·S°(A) + nB·S°(B)]
The standard free energy change for any reaction, ∆G may be found from the
standard free energies of formation, ∆Gf of the reactants and products in just the
same way as a standard enthalpy change is calculated:
∆G = ∆H – T·∆S = Σn·∆Gf (products) – Σn·∆Gf (reactants) =
= [nX·∆G°(X) + nY·∆G°(Y)] – [nA·∆G°(A) + nB·∆G°(B)].
Chemical kinetics studies the rate and the mechanism of chemical and
biochemical processes.
The rate of a chemical reaction is defined as the change in the concentration of
a reactant (or product) in a given time interval.
An expression which relates the rate of a reaction to the concentrations of the
reactants, is called a rate law. In general, for a reaction
aA + bB + cC + • • • → Products
the rate law often has the form:
v = k[A]x[B]y[C]z ...
where x is called the order of the reaction with respect to A, y is the order with
respect to B, z is the order with respect to C, and the sum, x+y+z+ ... , is called the
overall order.
For a homogeneous reaction:
А(g) + В(g) → АВ(g)
the expression of the reaction rate is:
V = k CACB or V = k [A][B],
where V is the rate of the reaction, CA and CB or [A] and [B] are concentrations of
reactants А і В respectively (mol/l), k is rate constant.
For homogeneous reaction 2А + В → 2С kinetic equation is:
V = k [A]2·[B].
In the case of heterogeneous reactions concentration of solid phase is not
included to the rate equation, for example:
2Al(s) + 3Cl2 → 2AlCl3;
V = k [Cl2]3.
Therefore, for heterogeneous reactions the order of the reaction and
+
+
59
molecularity are not identical.
At given temperature k is a constant characteristic of the reaction. Its value is
independent of the concentrations of the reactants, although it does depend on the
temperature and the nature of reactants. The rate constant k is a measure of the
intrinsic rate of the reaction: it is the rate when the concentrations of all the reactants
are 1 mol/l. Fast reactions have large k values, while slow reactions have law k
values.
The rate constants expressions for zero-, first-, and second-order reactions (k0,
k1, k2 – relatively) are given below:
1
k0 = (С0 – Сτ), k1 = 2,303 lg C0 , k2 = 1 ⋅ C0 − Cτ .
τ
τ
Cτ
τ C0 Cτ
The Van’t Hoff’s and Arrhenius equations show the temperature affecting the
reaction rate and the rate constant:
V t= V 0 γ
∆T
10
E
− a
, k = A ⋅ e RT .
Activation energy Еа may be calculated according to Arrhenius equations
equation:
2,303RT1T2 k 2 .
Ea =
lg
T2 − T1
k1
The fermentative reactions rates may be calculated using Michaelis-Menten
equation:
ν
[S ] ,
V = max
K M + [S ]
where [S] – the substrate concentration;
КМ –Michaelis-Menten constant.
Few chemical reactions proceed in only one direction. Most are reversible, at
least to some extend. At the start of the reversible process, the reaction proceeds
towards the formation of products. As soon as some products molecules are formed,
the reverse process begins to take place and reactants molecules are formed from
product molecules. Chemical equilibrium is achieved when the rates of the forward
and the reverse reactions are equal and the concentrations of the reactants and
products remain constant.
Experiments have shown that for any reaction at equilibrium the expression
involving the concentrations of the products and the reactants at equilibrium has a
characteristic value. We can write a general equation far a reaction as follows:
aA + bB + cC + ... ⇄ pP + qQ + rR + ...
In this equation A, B, C, and so on, are the reactants; and P, Q, R, and so on, are
the products. The letters a, b, c, . . ., p, q, r, . .. represent the number of moles of
each substance involved in the balanced equation for the reaction. For this general
reaction at a particular temperature the equilibrium constant is:
K =(
[ P]p [Q]q [ R]r ...
[ A]a [ B]b [C ]c ...
60
)eq
where [A], [B], [C], ..., [P], [Q], [R], . . . are the concentrations of the reactants
and the products at equilibrium.
To obtain the expression for the equilibrium constant for any reaction, we raise
the equilibrium concentration of each product to the power given by the number of
moles of that product in the balanced equation for the reaction, and we multiply
these. We then multiply the concentrations of each reactant, similarly raised to the
power given by the number of moles of the reactant in the balanced equation.
Finally, we divide the resulting expression for the products by that for the reactants.
There is a general rule that helps us to predict the direction in which an
equilibrium reaction will move when a change in concentration, pressure, volume,
or temperature occurs. The rule, known as Le Chatelier's principle, states that if an
external stress is applied to a system at equilibrium, the system adjusts in such a
way that the stress is partially offset. The word "stress" here means a change in
concentration, pressure, volume, or temperature that removes a system from the
equilibrium state.
3.3. References
10. V.O. Kalibabchuk, V.I. Halynska, L.I. Hryshchenko et al. Medical
Chemistry. – AUS MEDICINE Publishing. – 2010. – 224 p.
11. Raymond Chang. Chemistry (6th Edition). – WCB/McGraw-Hill. – 1998. – 995
p.
12. Rodney J. Sime Physical Chemistry. Methods. Techniques. Experiments. –
Saunders College Publishing. – 1990. – 806 p.
13. John McMurry, Robert C. Fay. Chemistry (3rd Edition). – Prentice Hall. – 2001.
– 1067 p.
14. David E. Goldberg. Fundamentals of Chemistry (2nd Edition). –
WCB/McGraw-Hill. – 1998. – 561 p.
15. Theodore L. Brown, H.Eugene LeMay, Bruce E. Bursten. Chemistry. The
Central Science. – Prentice Hall. – 2000. – 1017 p.
16. John Olmsted III, Gregory M. Williams. Chemistry. The Molecular
Science. – Mosby. – 1994. – 977 p.
17. Steven S. Zumdahl. Chemistry (4th Edition). – Houghton Mifflin Company. –
1997. – 1031 p.
18. Gary L. Miessler, Donald A. Tarr. Inorganic Chemistry. – Prentice Hall. –
1991. – 625 p.
3.4. Self Assessment Exercises
а) Review Questions
1. Define the basic terms of thermodynamics: a system, a phase, a component, a
state variable, a state function. What is the principle of systems classification
into isolated, open and closed ones?
2. Define the system type for the Earth and the cell of a living organism? Give the
characteristics of the system internal energy.
3. State the first law of thermodynamics and give its mathematical expression.
61
4. Define the term of enthalpy, standard enthalpy, enthalpies of formation and
combustion of substances?
5. The Hess’ law, enthalpy diagrams examples.
6. Write a mathematical expression and give several formulations of the second
law of thermodynamics. Is it possible to apply this law to biological systems?
7. Define entropy, Gibbs’ and Helmholtz’ free energies.
8. What is the role of ATP in energy transformations in biological systems? Write
down the thermochemical equation for the ATP hydrolysis reaction.
9. Chemical equilibrium and equilibrium constant. Write down the equilibrium
constant expressions for the following reactions:
а) 2SO2 + O2 ⇄ 2SO3;
b) 3Н2 + N2 ⇄ 2NH3.
10. On what factors does the rate of a reaction depend? Describe the effect of
reactants concentrations on the reaction rate.
11. State and explain the law of mass action.
12. Explain what is meant by the kinetical equation. Write down the kinetical
equations for the following reactions: a) sulfur dioxide oxidation; b) the Haber
synthesis of ammonia; c) nitrogen (V) oxide decomposition.
13. What is meant by the order and the molecularity of a reaction? Give a few
examples of chemical reactions with the same and different values of the
reaction order and molecularity.
14. When the concentration of reactants is 1 mol/l, what is the special term by which
the reaction rate is known? What are the units in which the rate constants of the
first order and the second order reactions are expressed?
15. Define the following terms: effective collision, proper orientation of the
colliding species, activation energy, activated complex. Plot and explain the
potential energy profiles in the reaction progress for exothermic and
endothermic reactions.
16. What do we mean by the mechanism of a reaction and its elementary step?
17. Explain the mechanism and kinetics of parallel, successive, conjugated, and
chain reactions. Which kinds of reactions occur in living organism?
18. Define the following terms: catalysts, promoters and catalytic poisons. Give the
examples. How does a catalyst increase the rate of a reaction?
19. Explain the term auto catalysis and its significance for biochemical processes.
20. Distinguish between homogeneous catalysis and heterogeneous catalysis and
their mechanisms. Illustrate your answer with h examples.
21. Enzymes, their classification. Enzymes and chemical catalysts divergences.
22. Write down and explain Michaelis-Menten equation.
23. Define equilibrium. What is the rule for writing the equilibrium constant
expressions for the reactions? Write the expressions of the equilibrium constants
for homogeneous and heterogeneous reactions.
24. Explain Le Chatelier’s principle. List factors that can shift the position of an
equilibrium. Give the examples.
62
b) Types of Numerical Problems and Their Solving Strategies
Numerical problem 1. What amount of heat will flow from the reaction system
where 36 g of aluminium will burn in the excess of
oxygen? The heat effect of the reaction is –1676 kJ.
Steps to solution:
1. 4 Al + 3 O2 → 2 Al2O3
The thermochemical equation of aluminium oxide formation is:
2Al(s) + 3/2 O2(g) → Al2O3(s) + Q, Q = –∆Hf (1 mole Al2O3) [kJ/mol]
2. The amount` of heat could be calculated from such proportion:
during the burning of 2·27 g of aluminium 1676 kJ of heat flows
36 g of aluminium – Q
36 g ⋅ 1676kJ
Q=
= 1117 kJ
2 ⋅ 27 g
Numerical problem 2. The heat effect of 1 mol blue vitriol CuSO4·5H2O dissolving
is –11.5 kJ and the heat effect of the same amount of copper
sulfate dissolving is 66.1 kJ. Calculate the enthalpy of
hydration for copper sulfate.
Steps to solution:
Aqueous copper sulfate solution could be prepared in 2 ways: by dissolving of
CuSO4 in water as well as by formation of CuSO4·5H2O crystalline hydrate and its
further dissolving:
According to the Hess’s law the enthalpy of hydration for copper sulfate is:
∆H1 = ∆H2 + ∆H3
where ∆H1 = –66.1 kJ – the enthalpy of CuSO4 dissolving, ∆H2 – the enthalpy
of CuSO4 hydration, ∆H3 = 11.5 kJ – the enthalpy of CuSO4·5H2O dissolving.
Calculate the enthalpy of hydration for copper sulfate:
∆H2 = ∆H1 – ∆H3= –66.1 – 11.5 = –77.6 kJ
Numerical problem 3. For the rusting of iron: 4Fe(s) + 3O2(g) → 2Fe2O3(s)
calculate: a) the standard enthalpy change, b) the standard
entropy change, c) the free Gibb’s energy change.
Steps to solution:
1. The enthalpy change for this reaction is:
∆Hf0 = Σn·∆Hf0(products) – Σn·∆Hf0(reactants) =
63
= 2·∆Hf0(Fe2O3, s) – [4·∆Hf0(Fe, s) + 3·∆Hf0(O2, g)]
∆Hf0(Fe2O3, s) = –822.2 kJ/mol, ∆Hf0(O2, g) = 0 kJ/mol,
∆Hf0(Fe, s) = 0 kJ/mol.
2. Using the values, calculate the standard enthalpy change for the reaction:
∆Hf0 = 2·(–822.2 kJ/mol) – 4·0 – 3·0 = –1644.4 kJ
3. For this reaction the standard entropy change is:
∆S0 = Σn·∆S0(products) – Σn·∆S0(reactants) =
= 2·∆S0(Fe2O3) – [4·∆S0(Fe) + 3·∆S0(O2)]
The standard entropy of formation for substances at 25 °C are:
∆S0(Fe, s) = 27.3 J·K–1·mol–1, ·∆S0(O2, g) = 205.0 J·K–1·mol–1,
∆S0(Fe2O3, s) = 87.4 J·K–1·mol–1
4. Using the values, calculate the standard entropy change for the reaction:
∆S0 = 2·(87.4 J·K–1·mol–1) – [4·27.3 J·K–1·mol–1 + 3·205.0 J·K–1·mol–1] =
= –549.4 J·K–1·mol–1.
5. The standard free energy change for the reaction we can calculate from the
equation: ∆G° = ∆H° – T·∆S°
For this reaction ∆Hf° = –1644.4 kJ and ∆S° = –549.4 J·K–1·mol–1 =
–0.549 kJ·K–1·mol–1. The standard free energy change for the reaction is:
∆G° = –1644.4 – 298·(–0.549) = –1480.68 kJ.
6. Alternatively, we can calculate the standard free energy change for this
reaction using the standard Gibbs free energies of formation for substances at
25 °C:
∆G°f(Fe2O3, s) = –740.3 kJ/mol, ∆G°f(O2, g) = 0 kJ/mol,
∆G°f(Fe, s) = 0 kJ/mol.
Using the values, calculate the standard free energy change for the reaction:
∆GO = Σn·∆G°f(products) – Σn·∆G°f(reactants) =
= 2·∆Gf°(Fe2O3, s) – [4·∆Gf°(Fe, s) + 3·∆Gf°(O2, g)] =
= 2·(–740.3)] – [4·0 + 3·0] = –1480.6 kJ
Numerical problem 4. Calculate the standard entropy change for the reaction
CO + 2Н2 ⇄ СН3ОН.
Steps to Solution:
1. Write down the thermochemical equation for the reaction, pointing out the
states of matters of substances. Values of the standard entropies of formation
for substances are given in Appendix А.
СО(g) + 2 Н2(g) ⇄ СН3ОН(l)
 J 
∆S° 
197,5
130,5
126,8

 mol ⋅ K 
2. The equation for the enthalpy change for this reaction is:
∆Sor = Σ ∆Sfo(products) – Σ ∆Sfo(reactants) =
= 1·∆Sfo(CH3OH(l)) – [1·∆Sfo(CO2(g)) + 2 ·∆Sfo(H2(g))]
3. Calculate the enthalpy change for the reaction using the values of the standard
64
entropies of formation for substances:
∆Sor = 126.8 – (197.5 + 2⋅130.5) = –331.7 (J/К).
Answer: ∆Sor ≈ – 0.33 kJ/(mol·К), ∆Sor < 0 – reaction is impossible at standard
conditions.
Numerical problem 5. Calculate the caloricity of 60 g of an egg which contains 12
% by mass of fats, 3.8 % of carbohydrates, and 68.5 % of
proteins.
Steps to Solution:
1. The caloricity of the egg can be calculates from such equation:
∆Qcomb = ∆Qcomb(fats) + ∆Qcomb(carbohydrates) + ∆Qcomb(proteins)
where ∆Qcomb(fats) = Qcomb(fats)·m(fats),
∆Qcomb(carbohydrates) = Qcomb(carbohydrates)·m(carbohydrates),
∆Qcomb(proteins) = Qcomb(proteins)·m(proteins).
2. Calculate masses of fats, carbohydrates and proteins in 60 g of egg:
m(fats) =
m(egg ) ⋅ C p ( fats )
60 g ⋅12%
=
= 7.2 g
100%
100%
m(carbohydrates) ⋅ C p (carbohydrates)
m(carbohydrates) =
100%
m(proteins) =
m( proteins ) ⋅ C p ( proteins )
100%
=
=
60 g ⋅ 3.8%
= 2.28 g
100%
60 g ⋅ 68.5%
= 41.1 g
100%
3. Calculate the caloricity of the egg using values of oxidation heats of products
in physiological conditions Qcomb(fats) = 37.8 kJ/g, Qcomb(carbohydrates) =
19.8 kJ/g, and Qcomb(proteins) = 16.8 kJ/g:
∆Qcomb(fats) = Qcomb(fats)·m(fats) = 37.8 kJ/g · 7.2 g = 272.16 kJ
∆Qcomb(carboh.) = Qcomb(carboh.)·m(carboh.) = 19.8 kJ/g · 2.28 g = 45.144 kJ
∆Qcomb(proteins)=Qcomb(proteins)·m(proteins) = 16.8 kJ/g·41.1 g = 690.48 kJ
∆Qcomb = 272.16 + 45.144 + 690.48 = 1007.78 kJ.
Numerical Problem 6. Calculate the rate of chemical reaction:
Br2 + HCOOH → 2Br– + 2H+ + CO2, if after 3 min.
concentration of Br2 decreased from 0.1 mol/dm3 to 0.04
mol/l.
Steps to Solution:
As the rate of a chemical reaction is defined as the change in the concentration
of a reactant (or product) in a given time interval, so:
–4
U = – C2 − C1 = − 0.04 − 0.1 = 3.3·10 mol/(l·sec).
τ 2 − τ1
3 ⋅ 60
Numerical Problem 7. For the homogeneous chemical reaction:
N2 (g) + 3 H2 (g) → 2 NH3 (g)
a) How will change the rate if the concentration of the reactants increases
twice?
65
Steps to Solution:
The rate of the reaction is:
3
U 1 = k·[N2]1·[H2]1
If the concentrations of reactants increase twice, [N2]2 = 2[N2]1 and [H2]2 =
2[H2]1, the rate of the reaction will be:
3
3
U 2 = k·[N2]2·[H2]2 = k·2[N2]1·(2[H2]1)
The ratio between rates is:
υ2
υ1
=
2[ N 2 ]1 ⋅ ( 2[H 2 ]1 ) 3
[ N 2 ]1[H 2 ]13
=
2 ⋅ 23
= 2 4 = 16 .
1
So, the rate of the reaction will increase 16 times.
b) How will change the rate if the concentration of the reactants decreases 3
times?
Steps to Solution:
The rate of the reaction is:
3
U 1 = k·[N2]1·[H2]1
If the concentrations of reactants decrease 3 times, [N2]2 = 1/3·[N2]1 and [H2]2
= 1/3·[H2]1, the rate of the reaction will be:
3
3
U 2 = k·[N2]2·[H2]2 = k·1/3·[N2]1·(1/3·[H2]1)
The ratio between rates is:
υ2
υ1
=
1 / 3[ N 2 ]1 ⋅ (1 / 3[H 2 ]1 )3
=
1 / 3 ⋅ (1 / 3)3 1 1
1
= ⋅
= .
1
3 27 81
[ N 2 ]1[H 2 ]13
So, the rate of the reaction will decrease 81 times.
c) How will change the rate if the pressure of system decreases in 3 times?
Steps to Solution:
The rate of the reaction is:
3
U 1 = k·P1(N2) ·P1 (H2)
If the concentrations of reactants decrease 3 times, P2(N2) = 1/3·P1(N2) and
P2(H2) = 1/3·P1(H2), the rate of the reaction will be:
3
3
U 2 = k·P2(N2)· P2 (H2) = k·1/3·P1(N2)·(1/3·P2(H2))
The ratio between rates is:
υ2
υ1
=
1 / 3P1 ( N 2 ) ⋅ [1 / 3P1 (H 2 )]3
P1 ( N 2 )[ P1 (H 2 )]3
=
1 / 3 ⋅ (1 / 3)3 1 1
1
= ⋅
=
.
1
3 27 81
So, the rate of the reaction will decrease 81 times.
d) How did the pressure of system change if the rate of the reaction increased
in 16 times?
Steps to Solution:
The rates of the reaction are:
66
3
3
U 1 = k·P1(N2)·P1 (H2) and U 2 = k·P2(N2)· P2 (H2)
The ratio between rates is:
υ2
υ1
=
P1 ( N 2 ) ⋅ [ P2 (H 2 )]3
P1 ( N 2 )[ P1 (H 2 )]3
4
P 
=  2  .
 P1 
4
 P2 
P
  = 16, 2 = 4 16 = 2 , P2 = 2P1.
P
P1
 1
So, the pressure of system increased twice.
e) How will change the rate if the volume of system decreases twice?
Steps to Solution:
The rate of the reaction is:
v1 = k·[N2]1·[H2]13
The decreasing of volume of gas system in 2 times is proportional to increasing
of concentrations of reactants in 2 times: [N2]2 = 2[N2]1 and [H2]2 = 2[H2]1.
Therefore, the rate of the reaction will be:
3
3
U 2 = k·[N2]2·[H2]2 = k·2[N2]1·(2·[H2]1) .
The ratio between rates is:
υ 2 2[ N 2 ]1 ⋅ (2[H 2 ]1 ) 3 2 ⋅ 23
=
=
= 2 4 = 16 .
3
υ1
1
[ N 2 ]1[H 2 ]1
So, the rate of the reaction will increase 16 times.
Numerical Problem 8. How many times the rate of a chemical reaction will change
at the increasing of temperature from 20 to 40 OC, if the
temperature coefficient γ=3?
Steps to Solution:
The rates of almost all chemical reactions increase with increasing temperature
by a factor γ = 2 – 4 for every 10 K rise in temperature:
υ2
υ1
=γ
∆T
10
=γ
40− 20
10
= 32 = 9.
So, the rate of the reaction will increase 9 times.
Numerical Problem 9. What is activation energy of the reaction if its rate at 100 0С
is 10 times greater than at 80 0С?
Steps to Solution:
According to the Arrhenius equation:
2,303RT1 ⋅ T2 k 2
Ea =
lg .
T2 − T1
k1
67
If
k2
k
= 10, then log 2 = 1, therefore:
k1
k1
Ea =
2,303 ⋅ 8,31 ⋅ 353 ⋅ 373
= 125.8 (kJ/mol).
373 − 353
Numerical Problem 10. How many grams of radioactive isotope Bi will remain in
4 hours if its initial mass was 200 mg and the half-time of
decomposition is 2 hours?
Steps to Solution:
Decomposition is the reaction of 1st order, therefore the half-time of
decomposition is:
τ1/ 2 =
0 . 693
k1
From this equation:
k1 =
0 . 693
τ 1/ 2
=
0 . 693
= 0 . 3465
2
The rate constant for the reaction of the 1st order is:
k1 =
2 . 303
τ
lg
C0
Cτ
=
2 . 303
τ
lg
m0
m
Calculate the mass of isotope after 4 hours of decomposition:
m
2 . 303
→ lg m 0 = 0 . 3465 ⋅ 4 = 0 . 6018
lg 0 = 0 . 3465
4
m
m
2 . 303
m0
m
200 mg
0
= 10 0 . 6018 = 3 . 9976 , m =
=
= 50 . 03 mg
m
3.9976
3 . 9976
Answer: after 4 hours 50 mg of radioactive isotope Bi will remain.
c) Problems to Solve
1. Calculate the enthalpy change for the reaction of glucose complete oxidation in a
living organism using the values of standard enthalpies of formation of
substances.
Answer: ∆Hр = –2812.7 kJ
2. The heat of 2360.8 kJ is flowing out when phosphine PH3 burns. Calculate the
phosphine standard enthalpy of formation.
Answer: ∆Hf(PH3) = –5.2 kJ/mol
3. How much heat does the organism loss if it losses 650 g of water through skin?
Answer: 1589 kJ
4. Calculate the entropy change for the reactions:
а) CaO(т) + СO2(г) = СаСО(к);
b) 2С(гр.) + СО2(г) = 2СО(г).
Answer: а) –164.7 J/К; b) 170.3 J/К
5. How will the reaction 2NO + O2 → 2NO2 rate change if the volume in the
system would be 3 times increased?
68
Answer: the rate will increase by a factor of 27
6. The rate of a reaction increases by a factor of 1024 with 50 оС rise in
temperature. Calculate the temperature coefficient γ for this reaction.
Answer: γ = 3.98
7. Calculate the half-life for the radionuclide Radon-220, if the rate constant for the
reaction is 1.26·10–2 sec–1.
Answer: τ1/2 = 55 sec
8. The equilibrium constant for the reaction Н2 + І2 ⇄ 2НІ is 50. Calculate the
equilibrium concentrations of hydrogen and iodine if their initial concentrations
were equal to 1 mol/l.
Answer: [H2]eq = [I2]eq = 0.22 mol/l
4. Laboratory Activities and Experiments Section
4.1. Practical Skills and Suggested Learning Activities
− to solve tasks and exercises on the thermodynamical functions calculations
(enthalpy, entropy, Gibbs’ free energy) under standard conditions;
− to find out the thermodynamical possibility of chemical and biochemical
reactions;
− to calculate the equilibrium constant and its relationship with Gibbs’ energy
studying;
− to calculate the temperature at which a chemical reaction can take place;
− to determine experimentally the enthalpy of neutralization of a strong base with
a strong acid;
− summarize the factors that affect reaction rates;
− for a given reaction, express reaction rate in terms of changes in concentrations
of reactants and products per unit time;
− describe simple one-step reactions in terms of collision theory and transition
state theory;
− describe activation energy and illustrate it graphically for exothermic and
endothermic reactions;
− understand, in terms of the distribution of energies of the reactant molecules,
how the reaction rate depends on temperature;
− be able to use the Arrhenius equation to find the rate constant from collision
frequencies, A, and activation energies, Ea, and to relate rate constants at two
different temperatures. Know how to represent and interpret this equation
graphically;
− understand, in terms of potential energy diagrams, how a reaction rate is altered
by the presence of a catalyst. Give examples of homogeneous and heterogeneous
catalysts.
4.2. Experimental Guidelines
4.2.1. Experimental Determination of the Enthalpy of Neutralization of a
69
Strong Base With a Strong Acid
Weigh a flask and fill it with 100 cm3 of the solution of KOH with the
concentration of 0.5 M. Measure the temperature of the solution reading to the
nearest 0.1 °C and record the initial temperature.
Measure by the graduated cylinder 100 ml of 0.5 M HCl solution. Add the
solution of HCl to the solution of the base, mix it and measure the highest
temperature.
Calculate the heat effect of the reaction:
∆Q = (msCs + mgCg) ⋅ ∆t,
where ms and mg – masses of the solution and the flack, Cs – specific heat
capacity of the solution, Cs = 4.18 J/(g⋅K), Cg – specific heat capacity of glass,
Cg = 0.753 J/(g⋅K).
Calculate the enthalpy of neutralization of the acid in the account of 1 mole of
the hydrogen ions:
M
(J),
m
where M – the molar mass of the acid; m – the mass of the acid in grams.
Calculate the error of the experiment (∆H = -57.2 kJ is the theoretical value of
the enthalpy of neutralization).
∆Ηneutr= – ∆Q
4.2.2. The Reactants Concentrations Affecting the Reaction Rates Studying
The reaction of sulfur formation (turbidity of solution appearance) should be
studied:
Na2S2O3 + H2SO4 → Na2SO4 + H2S2O3,
H2S2O3 → H2O + SO2 + S↓.
Using the following table, set up a series of 10 test-tubes with the quantities of
1 М Na2S2O3, 1 М H2SO4 solutions and water as indicated.
Table 1
Test-tube No.
Reagents
Na2S2O3
H2O
H2SO4
Na2S2O3
mol/l
1
2
1
4
3
2
3
5
solution
concentration,
4
0.1
5
3
2
5
0.2
6
7
4
1
5
0.3
8
9
5
0
5
0.4
10
5
0.5
The time of turbidity occurrence τ, sec.
The reaction rate, v = 1/τ
When you would be ready to time the reactions, mix the solutions of test-tubes
pairs (1 and 2, 3 and 4 and so on). Begin the timing with the stopwatch as soon as
solutions have been mixed. Shake test-tubes to ensure good mixing. Stop the
stopwatch when the first sign of the turbidity appears for each mixed pair. Record
the measurements of time (in sec).
70
Calculate the reaction rate as V = 1/τ. Graph the reaction rate V versus the
reagent Na2S2O3 concentration. Make the conclusions about the reactants
concentrations affecting the reaction rate.
5. Conclusions and Interpretations. Lesson Summary
Topic 9
Measuring the electrical driving force of electrochemical
elements and electrodes potentials
1. Objectives
The mechanisms of the electrode potential, diffusion, membrane, and redox
potentials origin and their magnitude affecting with different factors studying allows
to realize the trends of most biochemical reactions.
Commonly, a biological cell contains 25 times more K+ inside than is on the
outside. The Na+-K+ pump is orientated so that it pumps Na+ out of the cell and K+
into the cell. ATP located on the inside of the pump drives the system. Decades of
observations concerning membranes potentials followed, and bioelectrochemisity
developed as an integral facet of the biomedical sciences.
Bio potentials measurements are the basis of electrocardiography,
electroencephalography and other diagnostic methods. The electromotive forces
measurements allows to determine concentration of physiologically active ions
(Н3О+, К+, Na+, Са2+, Сl–, NO3–etc.) in biological liquids and body tissues.
A very large part of chemistry is concerned, either directly or indirectly, with
determining the concentrations of ions in solution. Any method that can accomplish
such measurements using relatively simple physical techniques is bound to be
widely exploited. Cell potentials are fairly easy to measure, and although the Nernst
equation relates them to ionic activities rather than to concentrations, the difference
between them becomes negligible in solutions where the total ionic concentration is
less than about 10–3 M.
Potentiometry as the method of determining pH of solutions has a few
advantages as compared with another methods: it is more exact and accurate (allows
to measure pH with the accuracy of 0.03 – 0.05 points), allows to measure pH of
multi-component systems and colour solutions. This method is widely used in
biology, medicine, pharmacy. It can be also used for measuring pH of different
objects of environments and biological liquids. The large importance has the
determining of pH for the researches of biochemical and physiological processes.
2. Learning Targets:
− to learn the skills of constructing galvanic cells using different half-cells;
− to measure the electromotive forces of galvanic cells produced by each cell
using pH-meter;
− to learn the electrode potential determining technique;
71
− to learn the skills of the potentiometer (pH-meter) applying for the potentiometry
experiments;
− to learn the techniques of pH potentiometry measurements for biological liquids
(blood plasma, gastric juice, urine etc.) using a glass electrode;
− to perform a potentiometry titration of acids, bases and their mixtures.
3. Self Study Section
3.1. Syllabus Content
The electrochemical phenomena significance for biochemical processes.
Electrodes potentials and their origin mechanisms. Nernst equation. The
standard electrode potential. Half-cells potentials measurement.
Indicator
electrodes and reference electrodes. Silver-silver chloride electrode. Ion-selective
electrodes. Glass electrode.
Galvanic (electrochemical or voltaic) cells. Diffusion potential. Membrane
potential. The biological role of diffusion and membrane potentials.
Redox reactions significance for biochemical processes. Redox potential as a
measure of the half-cell tendency to act as oxidizing or reducing agent. Peters’
equation. A standard redox potential.
The spontaneity and the direction of redox reaction proceeding prediction by
their redox potentials values. Equivalent factors of reduction and oxidizing agents.
Redox potentials role for the biological oxidation mechanism.
Potentiometry. The technique of potentiometry determining of pH and ions
activity in analyte solutions. Indicator electrodes (hydrogen, glass electrode) and
their using in electrometric methods based on electromotive force (EMF)
measurement of a galvanic cell.
The technique of potentiometry titration. The proper choice of the indicator
electrode. The integral and differential curves of potentiometry titration plotting and
the equivalence point determining. The analyte solution concentration determining.
3.2. Overview
The galvanic (electrochemical) cell is a device in which the chemical energy of
redox reaction is transforming into electrical one. The most common galvanic cell is
constructed of two connected half-cells (electrodes). The compensatory method and
multimeter of pH-meter are used for the galvanic cells electromotive forces
experimental determining.
All electrodes may be divided into 4 basic types:
Electrodes of the 1st kind, reversible to cation. Metal plate immersed into a
solution of its salt may be an example of the 1st kind electrodes. The schematical
representation of the 1st kind electrodes:
Men+Me,
2+
2+
for example, Zn Zn, Cu Cu.
Their potential is calculated according to the Nernst equation:
ϕ =ϕ0 +
72
2.303RT
log a n+
Me
nF
or at standard conditions
ϕ =ϕ0 +
0.059
log a n +
Me
n
,
where ϕ0 – standard electrode potential; n – the number of electrons taking part in
the electrode reaction.
Electrodes of the 1st kind are often used as indicator electrodes. Since such
electrodes respond rapidly to the concentration of the analyte ion in a solution, it is
possible to calculate the activity of the ions in a solution. For example, the
concentration of H+ ions may be determined using the hydrogen electrode
(Pt) ½ H2 | H+, whose potential may be defined as:
φ = 0.059 a(H+) or φ = – 0.059 pH.
nd
Electrodes of the 2 kind are constructed of a metal plate covered with a layer
of its insoluble compound (salt, oxide, hydroxide) being in contact with a solution
containing an anion of the salt. The schematical representation of the 2nd kind
electrodes: MeMeA, An–,
The 2nd kind electrodes potential may be calculated according to the Nernst
equation:
ϕ = ϕ0 −
2.303RT
log a n− .
A
nF
The examples of the 2nd kind electrodes are:
− silver-silver chloride electrode Ag, AgCl | KCl;
− calomel electrode Hg, Hg2Cl2 | KCl.
Silver-silver chloride electrode is a silver wire covered with silver chloride and
immersed into the solution of potassium chloride. Under its proceeding the
following reactions occur:
Ag → Ag+ + e− and Ag+ + Cl– → AgCl or Ag + Cl– → AgCl + e−.
The Nernst equation for the silver-silver chloride half-cell may be given as:
2.3 R ⋅ T
ϕ Ag,AgCl|KCl = ϕ o Ag,AgCl|KCl +
lg a( Аg + ) or
n⋅F
K sp (AgCl)
2.303 R ⋅ T
0
ϕ Ag, AgCl|KCl = ϕ
+
log
n⋅ F
Ag, AgCl| KCl
a(Cl− )
The saturated silver-silver chloride electrode is often used as a reference
electrode (φ = 0,021 V) instead of the standard hydrogen electrode, which is
difficult to operate.
Oxidation-reduction (redox) electrodes consist of an inert metal (platinum,
gold, iridium, graphite etc.) immersed into a solution containing oxidized and
reduced forms of the same substance, for example:
PtFe3+, Fe2+.
During the operating of redox half-cell the reactions proceed without the
involving of the inert electrode material. It serves only as a conductor of electrons,
oxidation or reduction products remain in the solution.
The value of the redox system potential is defined according to the NernstPeters equation:
73
ϕ red/ox = ϕ o red/ox +
2.3 RT
a (Ox) ,
lg
nF
a (Re d )
0.059
o
or at standard conditions ϕ
lg
red/ox = ϕ red/ox +
n
a (Ox )
a(Re d )
.
where a(Ox) and a(Red) are activities of the oxidized and reduced forms, φored/ox is
standard electrode potential of a redox system, which is equal to the potential of an
electrode if a(ox)=a(red).
The potential of the redox system depends on the ratio a(Ox) : an increase of
a(Re d )
this ratio increases the potential (intensifies oxidizing action) and a decrease of this
ratio reduces the potential (strengthens reducing action).
The φored/ox value serves as a measure of the oxidizing or reducing ability of a
system: the greater the value φored/ox, the better its oxidizing action.
Ion-Selective electrodes are electrochemical sensors their potentials magnitudes
being affected the certain kind ions activity in a solution.
Glass electrodes are manufactured in huge numbers for both laboratory and field
measurements. They contain a built-in Ag-AgCl reference electrode in contact with
the HCl solution enclosed by the membrane. The glass membrane of a pH glass
electrode consists of a silicate framework containing lithium (or sodium) ions.
When a glass surface is immersed in an aqueous solution then a thin solvated layer
(gel layer) is formed on the glass surface in which the glass structure is softer. This
applies to both the outside and inside of the glass membrane:
Reference
Analyte solution
electrode (external)
Membrane
Inner solution
Reference electrode
(inner)
Ion selective electrode
Н+glass + Ме+solution ⇄ Н+solution + Ме+glass
As the proton concentration in the inner buffer of the electrode is constant, a
stationary condition is established on the inner surface of the glass membrane. In
contrast, if the proton concentration in the measuring solution changes then ion
exchange will occur in the outer solvated layer and cause an alteration in the
potential at the glass membrane. Only when this ion exchange has achieved a stable
condition will the potential of the glass electrode also be constant. This means that
the response time of a glass electrode always depends on the thickness of the
solvated layer
Ion selective electrodes consist of a frame 1, auxiliary electrode 2, immersed
into the inner solution 3, and the membrane 4.
74
Figure 1. Ion selective electrode
Scheme 2. Glass electrode
The potential of glass electrode depends on the activity of hydrogen ions:
RT
.
ϕ glass = ϕ 0 +
lna
+
F
H3O (solution)
Potentiometry as the method of determining pH of solutions has a few
advantages as compared with another methods: it is more exact and accurate (allows
to measure pH with the accuracy of 0.03 – 0.05 points), allows to measure pH of
multi-component systems and colour solutions. This method is widely used in
biology, medicine, pharmacy. It can be also used for measuring pH of different
objects of environments and biological liquids. The large importance has the
determining of pH for the researches of biochemical and physiological processes.
For the pH determining by potentiometry titration the electrochemical cell
should consist of the reference electrode (silver/silver chloride or calomel electrode
usually) and the indicator electrode, which is immersed in the analyte solution:
reference electrode | salt bridge | analyte solution | indicator electrode.
A reference is an electrode that has the half-cell potential known, constant, and
completely insensitive to the composition of the solution under study. In
conjunction with this reference is the indicator (or working) electrode, whose
response depends upon the analyte concentration (φ depends on [H+]).
The following galvanic cells are the most commonly used for the potentiometry
determining of pH.
Hydrogen-silver/silver chloride galvanic cell. In this cell hydrogen electrode
acts as an indicator electrode while silver/silver chloride acts as a reference one.
The schematical representation of the cell is given below:
(–)
(Pt) H2  2H3O+  KCl, AgCl  Ag (+).
At 298 К EMF of hydrogen-silver/silver chloride galvanic cell is:
EMF = ϕ ( +) Ag,AgCl| KCl − ϕ (−) 2H + |H 2 = ϕ o Ag,AgCl|KCl + 0.059pH ,
then
pH =
EMF − ϕ o Ag, AgCl|KCl EMF − 0.22 ,
=
0.059
0.059
(1)
0
where ϕ Ag,
is the potential of saturated silver/silver chloride electrode, which
AgCl |KCl
equals +0.22 V at 298 К.
The saturated calomel electrode may be used instead of the saturated
75
silver/silver chloride electrode (ϕcalomel = +0.25 V), then the рН value may be
calculated by the equation:
рН =
Е − ϕ calom E − 0.25 .
=
0.059
0.059
The hydrogen electrode applying for pH measuring has a few limitations one of
which is the complexity of its manufacturing. It is also not appropriate to measure
the pH of biological liquids because some organic substances, such as proteins, are
able to precipitate on the platinum plate surface, resulting in pH values obtained to
be false.
Glass electrode is often used as an indicator electrode being widely used in
biomedical researches, as its potential depends on activity of hydrogen ions in
solution.
Glass-silver/silver chloride galvanic cell. The glass electrode in this cell acts
as an indicator while silver/silver chloride is a reference electrode. The schematical
representation of the cell is given below:
(–)
Ag AgCl, HCl glass membrane
analyte solution with рНх
KCl, AgCl Ag(+)
The electromotive force of this cell may be calculated by the equation:
(+)
(-) .
EMF = ϕ Ag,
AgCl| KCl − ϕ glass
Taking into account that Nernst equation for glass electrode is:
0
ϕ glass = ϕ glass
− 0.059pH ,
(+)
0
EMF = ϕ Ag,
AgCl|KCl − ϕ glass + 0.059pH
If we assume ϕ(+)Ag,AgCl|KCl – ϕ 0glass = ϕ 0/, then EMF = ϕ 0/ + 0.059 pH, so
EMF − ϕ 0 / .
(2)
рH =
0.059
The standard potential of a glass electrode depends on the grade of glass the
electrode membrane is made of. Moreover it varies with time.
That is why the calibration of the cell including glass electrode is commonly
carried out. The cell calibration includes electromotive force measurements of the
cell with standard buffer solutions of known рН values. Then the calibration curve
should be plotted as the pH value versus the pH of buffer solutions. For the
unknown pH value of a solution determining the EMF of the cell should be firstly
measured. EMF value is then used for determination of the solution рН value by
using plot on which find the EMF value (Ex value on Figure 2), draw horizontal line
to the intersection with calibration line and find рН value which corresponds to the
measured EMF value of the solution (pHx ).
Figure 2. Calibration curve of the cell including glass
electrode and its using for pH determining
76
pH measurements are now probably the most numerous type of quantitative
chemical assay performed, and nearly all of these measurements are determined
potentiometrically (electrochemically) using a pH meter, and a glass indicator and
calomel or silver/silver chloride reference electrode.
In many situations, accurate determination of an ion concentration by direct
measurement of a cell potential is impossible due to the presence of other ions and a
lack of information about activity coefficients. In such cases it is often possible to
determine the ion indirectly by titration with some other ion.
The potentiometry titration is based on the fixing the equivalence point using
the results of measurements of a galvanic cell potential during the titration process.
Near the equivalence point there will be the significant change of the indicator
electrode potential.
The potentiometry titration is the electrochemical method based on the
equivalent volume of a titrant determining by a major change in electromotive force
(EMF) of the proper galvanic cell. For potentiometry titration the electrochemical
cell should consist of the reference electrode (silver/silver chloride or calomel
electrode usually) and the indicator electrode, which is immersed in the analyte
solution. For example, for acid-base titration the indicator electrode potential should
vary under the concentration changing of H+ ions. So for acid-base titration
quingydrone, hydrogen and glass electrodes can be used as indicator ones. So the
indicator electrode potential should vary under the concentration changing of the
ions which take part or are forming in the process of titration while the reference
electrode potential should remain constant.
The potentiometry titration may be carried out using the reactions of
neutralization, reduction-oxidation, precipitation, etc.
The galvanic cell electrodes during the titration should be immersed in the
analyte solution while the titrant being added with a burette in small portions (0.51.0 ml). After each portion adding the EMF should be measured using the
potentiometer (pH meter).
The titration data are used then for the integral curve plotting in the coordinates
of determined EMF values as a function of the titrant volume (Fig. 3).
∆E
∆V
EEV
, ?mV
200
100
2000
0
1000
- 10 0
equivalence point
mV/ml
equivalence point
8 00
.
- 20 0
6 00
- 30 0
4 00
- 40 0
2 00
- 50 0
Volume of titrant, ml
0
2
4
6
8
0
2
77
4
6
Volume of titrant, ml
Figure 3. The curves of potentiometry titration of a solution of sulfuric acid with
an alkali solution. a) Integral curve; b) Differential curve.
Lowering the perpendicular from the middle of the jump (point of equivalence)
it is possible to determine the equivalent volume of a titrant as the intersection with
x-axis.
The equivalence point may be also determined by the differential curve of
titration which is plotted as ∆EMF/∆V – V (Fig. 2).
The perpendicular, dropped from the top of the curve on x-axis indicates the
equivalent volume of a titrant.
The analyte solution concentration may be calculated by the equation:
Cx⋅V = Ct⋅Vt , so:
Cx =
C t ⋅ Vt
,
V
where Сx, Сt are the concentrations of the analyte solution and the titrant,
respectively;
V, Vt are the volumes of the analyte solution and the titrant, respectively.
3.3. References
19. V.O. Kalibabchuk, V.I. Halynska, L.I. Hryshchenko et al. Medical
Chemistry. – AUS MEDICINE Publishing. – 2010. – 224 p.
20. Raymond Chang. Chemistry (6th Edition). – WCB/McGraw-Hill. – 1998. – 995
p.
21. Rodney J. Sime. Physical Chemistry. Methods. Techniques. Experiments. –
Saunders College Publishing. – 1990. – 806 p.
22. John McMurry, Robert C. Fay. Chemistry (3rd Edition). – Prentice Hall. – 2001.
– 1067 p.
23. David E. Goldberg. Fundamentals of Chemistry (2nd Edition). –
WCB/McGraw-Hill. – 1998. – 561 p.
24. Theodore L. Brown, H.Eugene LeMay, Bruce E. Bursten. Chemistry. The
Central Science. – Prentice Hall. – 2000. – 1017 p.
25. John Olmsted III, Gregory M. Williams. Chemistry. The Molecular
Science. – Mosby. – 1994. – 977 p.
26. Steven S. Zumdahl. Chemistry (4th Edition). – Houghton Mifflin Company. –
1997. – 1031 p.
27. Gary L. Miessler, Donald A. Tarr. Inorganic Chemistry. – Prentice Hall. –
1991. – 625 p.
3.4. Self Assessment Exercises
а) Review Questions
1. Explain the mechanism of the electrode potential appearance.
2. Write Nernst equation and mention the factors which determine its magnitude.
78
3. Explain the construction of electrochemical (galvanic) cell and the electromotive
force of a cell. Give Danielle-Jacobi cell as an example.
4. Give a few examples of the 1st and 2nd kind electrodes. Write Nernst equation for
them.
5. What is meant by the standard half-cell potential?
6. What is hydrogen electrode? How is it utilized to measure the potentials of halfcells?
7. What is electrochemical series of metals? The relative strengths of the reducing
and oxidizing agents.
8. Ion-selective electrodes: construction, classification, application.
9. What is meant by the redox electrode? Write Peters equation and explain it.
10. The biological significance of diffusion, membrane and redox potentials.
11. Explain the matter of the potentiometry method of pH determining.
12. Give the examples of indicator electrodes and reference electrodes used for the
potentiometry determining of pH.
13. Construction and uses of glass electrode in the potentiometry measuring of pH.
14. Explain the potentiometry titration technique.
15. Plot the integral and differential curves for the potentiometry titration of a strong
acid with an alkali.
16. Explain the titrant equivalent volume determining using the curves of titration.
b) Types of Numerical Problems and Their Solving Strategies
Numerical problem 1. Calculate the potential of cadmium electrode in 0.01 M
CdSO4 solution at standard conditions, if its standard
potential is –0.40 V.
Steps to Solution:
According to the Nernst equation the electrode potential at standard conditions
is:
ϕ
Cd 2 + |Cd o
=ϕ0
Cd 2 + |Cd
+
0.059
0.059
lg a 2+ = ϕ 0 2+
+
lg([Cd 2+ ] ⋅ α ⋅ n)
Cd
Cd |Cd
n
n
СdSO4 ⇄ Cd2+ + SO42–, Cd2+ + 2e ⇄ Cdo
Two electrons are transferred from Cd2+ ion to cadmium metal in the balanced
equation for this reaction, so n equals 2 for this electrode. One Cd2+ ion is formed at
CdSO4 dissociation, so n equals 1. The standard potential of cadmium electrode is
–0.40 V. Therefore, the electrode potential in 0.01 M CdSO4 solution is:
ϕ Cd 2+ |Cdo = ϕ o Cd 2+ |Cdo
+ 0.059 log C M (Cd 2+ ) = −0.40 + 0.03⋅ log 0.01 = −0.40 + 0.03⋅ (−2) = −0.46 V
2
Numerical problem 2. Calculate the electrode potential of the Cu|CuSO4 electrode
at standard conditions, if concentration of CuSO4 solution is
4 mol/L and percent of CuSO4 dissociation α=0.8.
Steps to Solution
According to the Nernst equation the electrode potential at standard conditions
79
is:
ϕ Cu 2+ | Cu o = ϕ o Cu 2+ | Cu o +
0.059
0.059
log a 2+ = ϕ o Cu 2+ | Cu o +
log([ Cu 2 + ] ⋅ α ⋅ n)
Cu
n
n
СuSO4 → Cu2+ + SO42–, Cu2+ + 2e → Cuo
Two electrons are transferred from Cu2+ ion to copper metal in the balanced
equation for this reaction, so n equals 2 for this electrode. One Cu2+ ion is formed at
CuSO4 dissociation, so n equals 1. The standard potential of copper electrode is
+0.34 V. Therefore, the electrode potential is:
ϕCu 2+ |Cu o = 0.34 + 0.059 log(CM ⋅α ) = 0.34 + 0.03⋅ lg(4 ⋅ 0.8) =
2
= 0.34 + 0.03 ⋅ lg 3.2 = 0.34 + 0.015 = 0.355 V
Numerical problem 3. Calculate the electromotive force of the galvanic cell
consisting of two silver electrodes at standard conditions if
the concentration of Ag+ ions in the electrodes solutions are
10–2 and 10–5 mol/l.
Steps to Solution:
The electromotive force of the galvanic cell is:
EMF = φcathode – φanode
According to the Nernst equation the electrode potential at standard conditions
is:
ϕ =ϕo +
0.059
0.059
log a + = ϕ o +
log[ Ag + ]
Ag
n
n
The standard electrode potential of silver electrode is φo = 0.799 V.
Calculate the electrode potential in the 1st solution:
ϕ1 = 0.799 +
0.059
log10 − 2 = 0.799 + 0.059·(-2) = 0.799 - 0.118 = 0.681 V
1
Calculate the electrode potential in the 2nd solution:
ϕ 2 = 0.799 +
0.059
log10 − 5 = 0.799 + 0.059·(-5) = 0.799 - 0.295 = 0.504 V
1
φ1 > φ2, therefore, the 1st electrode in this galvanic cell is cathode and the 2nd is
anode. Calculate the electromotive force of the galvanic cell:
EMF = 0.681 – 0.504 = 0.177 V.
Numerical problem 4. Calculate the potential of the hydrogen electrode immersed
in solution with pH=5.
Steps to Solution:
According to the Nernst equation the electrode potential at some moment of time
is:
+
φ = φo + R ⋅ T ln [H ] = φo + 0.059 lg[H + ] .
n⋅F
[H 2 ]
n
If pH = 5, then [H+] = 10–5 mol/l. The standard electrode potential of hydrogen
80
electrode φO = 0 V. Therefore, the electrode potential is:
φ = 0 + 0.059·lg10–5 = 0.059·(–5) = - 0.295 V.
Numerical problem 5. Calculate the electromotive force of the hydrogen galvanic
cell at standard conditions if the pH values of the
electrodes solutions are 4 and 10 respectively.
Steps to Solution:
The electromotive force of the galvanic cell is:
EMF = φcathode – φanode
According to the Nernst equation the electrode potential at some moment of time
is:
+
φ = φo + R ⋅ T ln [H ] = φo + 0.059 lg[ H + ] .
n⋅ F
n
[H 2 ]
The standard electrode potential of hydrogen electrode φO = 0 V.
The electrode potential in the 1st solution is (if pH = 4, then [H+] = 10–4 mol/l):
φ1 = 0 + 0.059 log 10 − 4 = 0 + 0.059·(–4) = –0.24 V.
1
The electrode potential in the 2nd solution is (if pH = 10, then [H+] = 10–10 mol/l)
φ2 = 0 + 0.059 log 10− 10 = 0 + 0.059·(–10) = –0.59 V.
1
Calculate the electromotive force:
EMF = –0.24 – (–0.59) = 0.35 V.
Numerical problem 6. The electrode potential of zinc electrode Zn|ZnSO4 (0.2 M)
is –0.79 V. Calculate the percent of dissociation of ZnSO4
in this solution.
Steps to Solution:
According to the Nernst equation the electrode potential at some moment of time
is:
2+
2+
φZn2+|Zno = φoZn2+|Zno + R ⋅ T ln [ Zn ] = φoZn2+|Zno + 0.059 log [ Zn ]
0
n⋅F
n
[ Zn ]
2+
o
1
Zn + 2e → Zn
Two electrons are transferred from Zn2+ ion to zinc metal in the balanced
equation for this reaction, so n is 2 for this electrode. The standard potential of zinc
electrode is –0.76 V. Therefore, the electrode potential is:
0.059
φZn2+/Zno = –0.76 +
log[ Zn 2 + ] ⋅ α
2
Calculate the percent of ZnSO4 dissociation:
0.059
–0.79 = –0.76 +
log 0.2α
2
81
0.03 log 0.2α = –0.03
log 0.2α = –1
0.2α = 10–1
α = 0.1/0.2 = 0.5 or 50 %.
Numerical problem 7. Calculate the value of the quinhydrone electrode potential
if it is immersed in solution with pH = 4.5.
Steps to Solution:
The electrode potential at some moment of time is:
φ = φ0 – 0.059pH.
The standard electrode potential of quinhydrone electrode φ0 = 0.704 V.
Therefore:
φ = 0.704 – 0.059·4.5 = 0.44 V.
Numerical problem 8. Calculate the pH of solution, in which the quinhydrone
electrode potential is 0.20 V at the temperature of 298 K.
Steps to Solution:
The electrode potential at some moment of time is:
φ = φ0 – 059pH.
Therefore, the pH of the solution is:
o
pH = ϕ − ϕ .
0.059
The standard electrode potential of quinhydrone electrode φ0 = 0.704 V.
Calculate the pH of solution:
pH = 0.704 − 0.2 = 8.54.
0.059
Numerical problem 9. Calculate the pH of solution, if the electromotive force of
the galvanic cell consisting of quinone-hydroquinone and
silver-silver chloride electrodes is 0.08 V at the
temperature of 298 K.
Steps to Solution:
Electromotive force of the galvanic cell is:
EMF = φcathode – φanode = φQ/H – φAg,AgCl|KCl = φ0Q/H – 0.059pH – φAg,AgCl|KCl.
The standard electrode potential of silver-silver chloride electrode is 0.222 V.
The standard electrode potential of quinhydrone electrode φ0 = 0.704 V. Therefore,
EMF = 0.704 – 0.222 – 0.059 pH = 0.482 – 0.059 pH
Calculate the pH of the solution is:
0.482 − EMF
0.482 − 0.08
pH =
=
= 6.8.
0.059
0.059
Numerical problem 10. Calculate the electrode potential for the Pt | Fe3+, Fe2+
electrode, if the concentrations of Fe3+ and Fe2+ ions in the
solution are 0.1 and 0.05 mol/l respectively.
82
Steps to Solution:
According to the Nernst equation the electrode potential at some moment of time
is:
3+
3+
0.059
log [Fe ]
φ(Fe3+/Fe2+) = φo(Fe3+/Fe2+) + R ⋅ T ln [Fe ] = φo(Fe3+/Fe2+) +
2
+
n
n ⋅ F [Fe ]
[Fe 2 + ]
Fe3+ + 1e → Fe2+
One electron is transferred from Fe2+ ion to Fe3+ in the balanced equation for this
reaction, so n is 1 for this electrode. The standard electrode potential φo(Fe3+/Fe2+) =
0,77 V.
Therefore, the electrode potential is:
0.059
0.1
φ(Fe3+/Fe2+) = 0.77 +
= 0.77 + 0.059·lg 2 = 0.77 + 0.02 = 0.79 V.
log
1
0.05
Numerical problem 11. Calculate the pH value and hydrogen ions concentration in
a gastric juice into which hydrogen and saturated calomel
electrodes are immersed and the galvanic cell EMF is
measured as 0.32 V at 25 0С. The standard potential value
for the saturated calomel electrode is 0.25 V.
Steps to Solution:
Write down the schematical representation of the galvanic cell and the equation
for its EMF calculation:
(–}
(Pt) Н2, 2Н+| KCl, gastric juice | KCl | KCl, Hg2Cl2 | Hg (+}
EMF = ϕ+– ϕ– = 0.25 + 0.059 рН
Calculate the pH value of gastric juice:
рН = EMF − ϕ calomel = 0.32 − 0.18 = 1.19.
0.059
0.059
Calculate the hydrogen ions [H+] concentration in gastric juice:
pH = –log [Н+] = 1.19; [Н+] = alog(–1.19) = 6.46·10–2 mol/l.
с) Problems to Solve
1. Calculate the electrode potential value for the aluminium half-cell lowered into
0.2М Al2(SO4)3 solution at 25 оС, if its standard potential (ϕо) = –1.663 V, the
activity coefficient of aluminium ions in the solution f (Al+3) = 0.255.
Answer:–1.674 V
2. Give the schematic representation of the galvanic cell consisting of silver and
zink 1st kind half-cells. Calculate the EMF for the cell if both metals activities
are 1 mol/l in respective solutions.
Answer: 1.50 V
3. Give the schematic representation of the galvanic cell consisting of calomel
electrode and redox half-cell PtFe3+, Fe2+. Calculate the EMF for the cell at
standard conditions.
Answer: 0.52 V
83
4. Calculate the pH value of the blood if the EMF of hydrogen-silver/silver
chloride galvanic cell value is 0.65 V at 25 оС. The electrode potential
silver/silver chloride half-cell is 0.22 V.
Answer: рН = 7.29
5. Calculate the pH and pOH values of gastric juice into which hydrogen electrode
is immersed. Another electrode in the galvanic cell is saturated calomel
electrode. The EMF value of this cell is 0.315 V at 25 оС and saturated calomel
electrode potential is 0.247 V.
Answer: рН = 1.15, рОН = 12.85
6. The potentiometry titration of HCl solution with 1 M NaOH solution was carried
out. Determine the mass of HCl in the analyte solution using the following
experimental data:
VNaOH , ml
0
0.50
0.90
0.99
1.0
1.01
1.10
ϕ of hydrogen
398
416
457
516
683
890
929
electrode, mV
Answer: 0.0365 g
4. Laboratory Activities and Experiments Section
4.1. Practical Skills and Suggested Learning Activities
− to measure the electromotive force (EMF) for Weston cell;
− to provide the student with the opportunity to construct a number of voltaic
cells;
− to measure the EMF produced by each cell;
− to determine the 1st kind electrode potentials and redox potentials;
− to determine pH of a biological liquid of human’s organism;
− to determine a strong acid or an alkali solution concentration with the
potentiometry titration method.
4.2. Experimental Guidelines
4.2.1. The electromotive force measurement for Weston cell with multimeter or
pH-meter
A Weston cell is an example of a cell that can be made to definite specifications,
has a definite EMF, is long lived, and produces an EMF that changes little with
temperature. Such cells are used as standard cells in potentiometry circuits to
determine the EMF of another cell.
Weston cell is an H type cell. One electrode consists of Cd amalgam covered
with crystals of CdSO4⋅8/3H2O. Another electrode contains Hg with solid Hg2SO4
and covered with crystals of CdSO4⋅8/3H2O. The whole cell is filled with a
saturated solution of CdSO4.
The cell is represented as follows:
(-) Cd(Hg)CdSO4⋅8/3H2OCdSO4(sat’d)Hg2SO4Hg (+)
84
The cadmium electrode is treated as if it were the anode, the electrode at which
oxidization occurs:
Cd(Hg) + SO42- + 8/3H2O → CdSO4⋅8/3H2O + 2e−.
The mercury electrode is the cathode, at which reduction occurs:
Hg2SO4 + 2e− → 2Hg + SO42-.
The overall reaction:
Cd(Hg) + Hg2SO4 + 8/3H2O → CdSO4⋅8/3H2O + 2Hg.
The EMF of this cell is 1.01807 V at 25 oC.
To connect Weston cell with pH-meter and measure its electromotive force.
Compare it with the known value.
4.2.2. To measure the electromotive force produced by the galvanic cell
Disconnect Weston cell. Construct the galvanic cell of two 1st kind electrodes as
it’s shown on the Figure:
Lower the salt bridge into position so that one end of the U-tube is immersed in
the copper (II) chloride solution and the other end is in the zinc sulfate. Clamp the
salt bridge into position and determine the voltage of the cell with pH-meter.
Compare the EMF of the cell with the standard EMF calculated with the
standard electrode potentials.
(–)
Zn
KCl
Cu
(+)
(–) Zn ZnCl2 KCl CuCl2 Cu(+)
.
Figure 1. A galvanic cell constructed of two 1st kind electrodes
4.2.3. The 1st kind electrode potential determining
Construct the galvanic cell consisting of the 1st kind electrode with the unknown
potential and saturated silver-silver chloride half-cell as a reference electrode with
the known potential (ϕ = +0.22 V). Write the schematical representation of the cell,
for example:
(–)
Mе | Men+ | KCl | KCl, AgCl | Ag(+)
Determine the voltage of the cell with pH-meter.
As EMF = ϕ+ – ϕ– it is possibly to calculate the potential of the 1st kind
electrode ϕ x :
−
85
ϕ x−
= – EMF + ϕsilver -silver chloride
Compare the calculated potential value of the 1st kind electrode with its standard
potential. Make conclusions. Comment on any cases in which the standard potential
does not agree with the experimentally observed value.
4.2.4. The redox electrode potential determining
Construct the galvanic cell consisting of redox electrode with the unknown
potential and saturated silver-silver chloride half-cell as a reference electrode with
the known potential (ϕ = +0.22 V). Write the schematical representation of the cell,
for example:
(–)
Ag | AgCl, KCl | KCl | I2, 2KI | С(graphite)(+) .
Determine the voltage of the cell with pH-meter, calculate the redox potential as:
ϕ
− = EMF + 0.22 .
(I 2 | 2I )
Compare the calculated redox potential value with the standard potential. Make
conclusions. Comment on any cases in which the standard potential does not agree
with the experimentally observed value.
4.2.5. The pH value determining for a biological liquid of human’s organism
1. Compose a galvanic cell consisting of glass indicator electrode and saturated
silver-silver chloride reference electrode. Connect the galvanic cell to pHmeter-millivoltmeter.
2. Press the "POWER" button on the device.
3. Set the temperature of the solution by pressing the "MODE" button to set the
unit of measurement " оС " and, turning the knob "SET TEMPERATURE
MANUALLY".
4. Switch to pH measurement by pressing the "MODE" to set units "pH".
5. Immerse electrodes into the analyte solution, which was poured into a beaker
and carry out the pH measurement.
6. After the measurement electrodes should be rinsed with distilled water and
immersed into distilled water or 0.1 M HCl solution.
4.2.6. Determining the concentration of a strong acid or an alkali solution with
the potentiometry titration method
1. Use a volumetric pipette to transfer 10.00 ml of the hydrochloric acid to a
100 ml beaker and, using a graduated cylinder, add about 20 ml of distilled
water to the acid.
2. Place the beaker on a magnetic stirrer, place a small stirrer bar in the
solution, and set the stirrer in motion—not too fast, otherwise some of the
acid may splash from the beaker.
3. Compose a galvanic cell consisting of glass indicator electrode and silversilver chloride reference electrode:
(–)
Ag | AgCl, KCl | glass membrane | analyte solution | KCl, AgCl | Ag (+)
4. Fill your burette with 0.1 M sodium hydroxide solution.
86
5. Connect the galvanic cell to pH-meter-millivoltmeter. Record the pH of the
solution to two places past the decimal.
6. Position your burette so that you can begin to add sodium hydroxide to the
beaker of hydrochloric acid. Add the sodium hydroxide approximately 0.5
ml at a time until a total of 9.0 ml has been added. Record the pH of the
solution after the addition of each 0.5 ml aliquot. Continue adding the
sodium hydroxide, but now in approximately 0.2 ml aliquots, until a total of
11.0 ml has been added. As before, record the pH and total volume of
sodium hydroxide after each aliquot has been added. Finally, add more
sodium hydroxide in approximately 0.5 ml aliquots until the total volume of
sodium hydroxide added is 18.0 ml. Again the pH of the solution should be
recorded after each addition. The above procedure may be summarized as
follows:
Table 1
The experimental data of potentiometry titration
The volume of titrant EMF (or ϕ), ∆EMF (∆ϕ),
∆V,
∆EMF /∆V,
V, ml
mV
mV
ml
mV/ml
0
1
2
so on
7. Plot a graph of EMF (y-axis) against total volume of sodium hydroxide
added (x-axis). Determine the equivalence point for the titration between
hydrochloric acid and sodium hydroxide. In a titration curve, the equivalence
point is at the mid-point of the steepest part of the curve.
8. Plot also the differential curve of titration as ∆EMF /∆V (y-axis) against total
volume of sodium hydroxide added (x-axis). Determine the equivalence
point for the titration between hydrochloric acid and sodium hydroxide. In a
titration curve, the equivalence point is in the top of the curve.
9. From the volume and concentration of NaOH solution used and the volume
of HCl solution used for titration, calculate the concentration of HCl
solution.
5. Conclusions and Interpretations. Lesson Summary
Topic 10
Adsorptive processes and ions exchange in bio-systems.
Chromatography
1. Objectives
Surface tension has significant impacts on the survival as well as the processes
needed for day to day existence of living beings. . For example: phospholipids are a
87
key component in cell membranes, which act as a protective surface against the
environment. Most household detergents contain sodium dodecyl sulfate which
reduces the surface tension of water. If local water reaches specific concentration of
detergent, water striders will break the surface tension and sink. Surface tension is
also essential for the transfer of energy from wind to water to create waves. Waves
are necessary for rapid oxygen diffusion in lakes and seas. Understanding surface
tension and what affects it is vitally important for researchers and engineers alike
because new solutions for treating water have to maintain water's delicate and
complex balance.
Surfactants have many commercial uses and depending on the use can be called
by many names, including: -wetting agents, emulsifying agents, solubilizing agents,
and detergents or soap. Surfactants are also widely used in pharmaceuticals. They
are commonly added to drug suspensions to hinder caking of medications during
storage, for reconstitution of powdered forms of medication into water at later use,
or as an additive to tablets to aid in the penetration of moisture into the tablet for
ready disintegration upon administration.
Gas adsorption is of practical consequence to engineers and chemists in many
ways. It can provide a convenient, cheap and reusable method for fluid purification
and purification. More significantly, perhaps, the phenomenon of surface adsorption
has been used to modify the rates of product yields of chemical reactions through
heterogeneous catalysis.
There are many environmental applications of adsorption in practice and many
others are being developed. Activated carbons and clays are frequently used for the
removal of organic contaminants, such as phenol and aniline. Moreover, the
adsorption on inexpensive and efficient solid supports has been considered a simple
and economical viable method for the removal of dyes from water and wastewater.
The most frequently used procedures in pharmaceutical processing for solid
dosage formulations are mixing, granulation, and compaction, as well as storage of
finished dosage forms. The effects of adsorption on these procedures have been
utilized widely in the pharmaceutical industry.
Enterosorbents are very effective for adsorbing bacterial enterotoxins and
endotoxins, so they are the common nonspecific treatment for intoxications. Kaolinpectin formulations are popular for symptomatic adsorption therapy.
2. Learning Targets:
– to understand the concept of surface tension, and that liquids tend to minimize
their area;
− to demonstrate that surface tension is a physical property of liquids;
− to evaluate the surface activity of substances on the basis of their structure;
− to analyze the structural features of the surface layer of adsorbed surfactant
molecules, to explain the principles of biological membranes structure;
− to analyze the adsorption equations, to distinguish a monomolecular and
polymolecular adsorption;
88
− to interpret the physical-chemical properties of proteins, which are structural
components of all tissues;
− to interpret the adsorption of substances from solutions on solid surfaces
regularities;
− to explain the physical-chemical foundations of the adsorption therapy methods;
− to distinguish the selective and ion exchange adsorption of electrolytes;
− to interpret the chromatographic analysis methods and their application in
medical and biological researches.
3. Self Study Section
3.1. Syllabus Content
Surface phenomena and their significance for biology and medicine. Surface
tension of liquids and solutions. A surface tension isotherm. Surfactants and surfaceinactive substances. Surface activity. Traube’s rule. Adsorption on a liquid-gas and
liquid-liquid interfaces. Gibbs’ equation. Surfactants molecules orientation in the
interfacial layer. The structure of biological membranes.
Adsorption at the solid-gas interface. Langmuir equation. Adsorption from
solution at the solid-liquid interface. Physical sorption (or physisorption) and
chemical sorption (or chemisorption). General rules for the solutes, vapours and
gases adsorption. Freundlich equation.
Physico-chemical basis of adsorption therapy (hemosorbtion, plazmosorbtsiya,
limfosorbtsiya, enterosorption, Application therapy). Immunosorbents.
Adsorption of electrolytes: specific (selective) and ion exchange. Fajans-Peneth
precipitation and adsorption rule. Naturally occurring ion exchangers and
synthetically produced organic resins. Adsorption and ion exchange significance or
the vital process in plants and living organisms.
Chromatography. Chromatographic methods of analysis classification based on
the phases states of matter, the technique and the separation mechanism.
Adsorption, ion exchange and distribution chromatography. Chromatography
applications in biology and medicine.
3.2. Overview
The molecular basis for surface tension may be explained by considering the
attractive forces that molecules in a liquid exert on one another. If the liquid is not
acted upon by external forces, a liquid sample forms a sphere, which has the
minimum surface area for a given volume. Nearly spherical drops of water are a
familiar sight, for example, when the external forces are negligible.
To break or change the shape or area of a surface it is necessary to do work. This
work counterbalances the resistance the surface is offering to the change. The work
per unit area is called the surface tension.
The surface tension is the ratio between the excess of free surface Gibbs energy
and the surface area of the interface:
σ =
GS
, [J/m2 or N/m]
S
89
where GS – free surface Gibbs energy; S – the surface area of the interface.
Solutes can have different effects on surface tension depending on their
structure. Surface active substances are absorbed at the interface and decrease the
surface tension of water (σsolution < σwater): salts of carbonic acids R-COONa (C11 < R
< C18) (sodium oleate С17Н33СООNa); sulphur-acids salts R-C6H4-SO3–Na+ (R >
C12).
Surface inactive substances increase the surface tension of water (σsolution >
σwater) and are concentrated in the bulk of solution (non-organic acids, bases, salts)
and some strongly polar organic compounds (glycerine, amino acids etc.).
The Gibbs adsorption equation enables the extent of adsorption at a liquid
surface to be estimated from surface tension data:
2
С ∆σ
C (σ 1 − σ ) , [mol/m ]
A=−
2
⋅
=−
⋅
RT ∆C
RT (C1 − C 2 )
where A – adsorption, mol/m2; C – general concentration of solution, mol/m3;
R=8,314 J/(mol·K) – universal gas constant; ∆σ/∆c – change of surface tension with
changing of substance concentration.
In case of surface active substances surface tension decreases at increasing of
∆σ
< 0 and adsorption is positive A > 0.
solute concentration
∆C
In case of surface inactive substances surface tension increases at increasing of
∆σ
solute concentration
> 0 and adsorption is negative A < 0.
∆C
When a gas or vapour is brought into contact with a solid, part of it is taken up
by the solid. The molecules that disappear from the gas either enter the inside of the
solid, or remain on the outside attached to the surface. The former phenomenon is
termed absorption (or dissolution) and the latter adsorption. When the phenomena
occur simultaneously, the process is termed sorption.
The solid that takes up the gas is called the adsorbent, and the gas or vapour
taken up on the surface is called the adsorbate.
Molecules and atoms can attach themselves onto surfaces in two ways. In
physisorption (physical adsorption), there is a weak van der Waals attraction of the
adsorbate to the surface. The attraction to the surface is weak but long ranged and
the energy released upon accommodation to the surface is of the same order of
magnitude as an enthalpy of condensation (on the order of 20 kJ/mol). During the
process of physisorption, the chemical identity of the adsorbate remains intact, i.e.
no breakage of the covalent structure of the adsorbate takes place. Physisorption, to
be a spontaneous thermodynamic process, must have a negative ∆G. Because
translational degrees of freedom of the gas phase adsorbate are lost upon deposition
onto the substrate ∆S is negative for the process. Since ∆G = ∆H – T∆S, ∆H for
physisorption must be exothermic.
90
In chemisorption (chemical adsorption), the adsorbate sticks to the solid by the
formation of a chemical bond with the surface. This interaction is much stronger
than physisorption, and, in general, chemisorption has more stringent requirements
for the compatibility of adsorbate and surface site than physisorption. The
chemisorption may be stronger than the bonds internal to the free adsorbate which
can result in the dissociation of the adsorbate upon adsorption (dissociative
adsorption). In some cases ∆S for dissociative adsorption can be greater than zero,
which means endothermic chemisorption, although uncommon, is possible.
The adsorption quantity on immovable interfaces may be defined as the ratio of
an adsorbate moles n by the adsorbent surface area S (m2), or by its mass m (kg):
A=
n
, [mol/m 2 ];
S
A=
n
, [mol/kg]
m
Adsorption is usually described through isotherms, that is, functions which
connect the amount of adsorbate on the adsorbent, with its pressure (if gas) or
concentration (if liquid).
Langmuire adsorption theory is based on four assumptions:
1. The surface of the adsorbent is uniform, that is, all the adsorption sites are
equivalent.
2. Adsorbed molecules do not interact.
3. All adsorption occurs through the same mechanism.
4. At the maximum adsorption, only a monolayer is formed: molecules of
adsorbate do not deposit on other, already adsorbed, molecules of adsorbate,
only on the free surface of the adsorbent.
The Langmuir adsorption isotherm for gases adsorbed on solids:
A = A∞
Kp
1 + Kp
where A is the substance amount of adsorbate adsorbed per gram (or kg) of the
adsorbent, mol/g or mol/kg; A∞ is the number of adsorption centres (maximum
boundary adsorption – maximal substance amount of adsorbate per gram (or kg) of
the adsorbent), mol/g or mol/kg; p – pressure of a gas; K – proportionality
adsorption constant, l/mol.
For liquids (adsorbate) adsorbed on solids (adsorbent), the Langmuir isotherm
can be expressed by:
KC
A = A∞
1 + KC
where C – concentration of solution (of adsorbate in liquid), mol/l.
In practice, activated carbon (also called activated charcoal or activated coal) is
used as an adsorbent for the adsorption of mainly organic compounds along with
some larger molecular weight inorganic compounds such as iodine and mercury. It
is a material with an exceptionally high surface area. Just one gram of activated
carbon has a surface area of approximately 500 m². The three main physical carbon
types are granular, powder and extruded (pellet). All three types of activated carbon
91
can have properties tailored to the application. Activated carbon is frequently used
in everyday life, in: industry, food production, medicine, pharmacy, military, etc. In
pharmacy, activated charcoal is considered to be the most effective single agent
available as an emergency decontaminant in the gastrointestinal tract. It is used after
a person swallows or absorbs almost any toxic drug or chemical.
Empirical Freundlich isotherm:
x
x
A = = kp1 / n or A = = kC 1 / n
m
m
where x is the number of moles of adsorbate, m – mass of adsorbent, K and n are
temperature-dependent parameters. K and n depend on the nature of adsorbate and
adsorbent. K is also known as the specific adsorption (if p = 1, than A = K).
In logarithmic form this equation may be written as:
or log x = logk + 1 logC
x
1
log
m
= log k +
n
log p
m
n
Ion exchange is a reversible chemical reaction where an ion from solution is
exchanged for a similarly charged ion attached to an immobile solid particle. Ion
exchange is the process through which ions in solution are transferred to a solid
matrix (ion exchanger, or ion exchange resin) which, in turn releases ions of a
different type but of the same polarity. In other words the ions in solutions are
replaced by different ions originally present in the solid. Since ion exchange occurs
between a solution and the internal surface of a solid it can be viewed as a special
type of sorption process.
Although ion exchange is similar to sorption since a substance is captured by a
solid in both processes, there is a characteristic difference between them: ion
exchange is a stoichiometric process in contrast to sorption. It means that in the ionexchange process, for every ion that is removed, another ion of the same sign is
released into the solution. In contrast, in sorption, no replacement of the solute takes
place. So ion exchange can be seen as a reversible reaction involving chemically
equivalent quantities
Conventional ion exchange resins consist of a cross-linked polymer matrix with
a relatively uniform distribution of ion-active sites throughout the structure. Ion
exchangers are either naturally occurring inorganic zeolites or synthetically
produced organic resins. An organic ion exchange resin is composed of highmolecular-weight polyelectrolytes that can exchange their mobile ions for ions of
similar charge from the surrounding medium. Each resin has a distinct number of
mobile ion sites that set the maximum quantity of exchanges per unit of resin.
The synthetic organic resins are the predominant type used today because their
characteristics can be tailored to specific applications.
Ionisable groups attached to the resin bead determine the functional capability of
the resin. Industrial water treatment resins are classified into four basic categories:
− Strong Acid Cation (SAC);
− Weak Acid Cation (WAC);
92
− Strong Base Anion (SBA);
− Weak Base Anion (WBA).
SAC resins can neutralize strong bases and convert neutral salts into their
corresponding acids. SAC resins derive their functionality from sulfonic acid groups
(HSO3¯ ). When used in demineralization, SAC resins remove nearly all raw water
cations, replacing them with hydrogen ions.
SBA resins can neutralize strong acids and convert neutral salts into their
corresponding bases. These resins are utilized in most softening and full
demineralization applications. SBA resins derive their functionality from quaternary
ammonium functional groups. When in the hydroxide form, SBA resins remove all
commonly encountered anions.
WAC and WBA resins are able to neutralize strong bases and acids,
respectively. These resins are used for dealkalization, partial demineralization, or
(in combination with strong resins) full demineralization.
Weak acid cation exchange resins derive their exchange activity from a
carboxylic group (-COOH). When operated in the hydrogen form, WAC resins
remove cations that are associated with alkalinity, producing carbonic acid.
Weak base resin functionality originates in primary (R-NH2), secondary (RNHR'), or tertiary (R-NR'2) amine groups. WBA resins readily remove sulfuric,
nitric, and hydrochloric acids.
Weak acid cation resins are used primarily for softening and dealkalization of
high-hardness, high-alkalinity waters, frequently in conjunction with SAC sodium
cycle polishing systems. In full demineralization systems, the use of WAC and SAC
resins in combination provides the economy of the more efficient WAC resin along
with the full exchange capabilities of the SAC resin.
3.3. References
28. V.O. Kalibabchuk, V.I. Halynska, L.I. Hryshchenko et al. Medical
Chemistry. – AUS MEDICINE Publishing. – 2010. – 224 p.
29. Raymond Chang. Chemistry (6th Edition). – WCB/McGraw-Hill. – 1998. – 995
p.
30. Rodney J. Sime. Physical Chemistry. Methods. Techniques. Experiments. –
Saunders College Publishing. – 1990. – 806 p.
31. John McMurry, Robert C. Fay. Chemistry (3rd Edition). – Prentice Hall. – 2001.
– 1067 p.
32. David E. Goldberg. Fundamentals of Chemistry (2nd Edition). –
WCB/McGraw-Hill. – 1998. – 561 p.
33. Theodore L. Brown, H.Eugene LeMay, Bruce E. Bursten. Chemistry. The
Central Science. – Prentice Hall. – 2000. – 1017 p.
34. John Olmsted III, Gregory M. Williams. Chemistry. The Molecular
Science. – Mosby. – 1994. – 977 p.
35. Steven S. Zumdahl. Chemistry (4th Edition). – Houghton Mifflin Company. –
1997. – 1031 p.
93
36. Gary L. Miessler, Donald A. Tarr. Inorganic Chemistry. – Prentice Hall. –
1991. – 625 p.
3.4. Self Assessment Exercises
а) Review Questions
1. What is free surface energy? Explain the reasons of its origin.
2. What is surface tension? The surface tension units.
3. Why does surface tension try to minimize surface area?
4. Factors affecting the surface tension.
5. Define surface active substances (surfactants) and surface inactive substances.
Give examples.
6. Illustrate the surface tension and adsorption magnitudes for surfactants with the
surface tension and adsorption isotherms.
7. Explain the properties of surfactants on the basis of their molecules structure.
8. Illustrate the surfactants molecules orientation in the interfacial layers of waterthe air and benzene-water.
9. Define the surface activity, its units. Different variables affection the surface
activity.
10. State and explain Traube’s rule.
11. What is the specific surface area of the adsorbent and how it may be calculated?
12. What experimental data are needed for the surfactants molecules sizes (surface
area and the length) and the monomolecular layer thickness calculating.
13. Explain the structure of biological membranes.
14. Adsorption on a solid-gas interface. Different variables affection the adsorption
magnitude.
15. Langmuir adsorption theory, its basic statements. Langmuir equation and
Langmuir adsorption isotherm. The equation constants defining.
16. Adsorption on a solid-gas interface. Different variables affection the adsorption
magnitude. The rule of polarities aligning. Freundlich equation and adsorption
isotherm. The equation constants defining.
17. Nature of adsorption forces. Define the sorption processes: adsorption,
desorption, physisorption, chemisorption,
physicosorption, capillary
condensation.
18. Polymolecular
adsorption
theory
of
Polany
and
BET-theory.
The equation of an isotherm of polymolecular adsorption, its analysis.
19. Fundamentals of adsorption therapy (haemosorbtion, enterosorption,
Application therapy). Immunosorbents.
20. Adsorption from solutions. Molecular and ionic adsorption. Ion-exchange
adsorption. Its features.
21. State the Fajans-Peneth adsorption rule. Illustrate if for BaSO4 – K2SO4
system.
22. Define the ion exchangers and give their classification
23. Write down the equations of the processes occurring during water
94
demineralization with ion exchange.
24. Explain chromatographic methods of analysis and give their classifications.
b) Types of Numerical Problems and Their Solving Strategies
Numerical problem 1. Calculate the adsorption quantity (in mol/m2) at 10 °С for
the solution which contains 50 mol/L of pelargonic acid
С8Н17СООН. The surface tension of water is 74.22·10–3
J/m2 and the surface tension of the researched solution is
57.0·10–3 J/m2 at this temperature.
Steps to Solution:
1. According to the Gibbs equation adsorption on the solution-gas interface is:
A=−
С ∆σ
C (σ sol − σ H 2 O )
⋅
=−
⋅
RT ∆C
RT (Csol − CH 2 O )
2. Calculate the adsorption quantity of solution:
A=−
33
50 mol / L
(57 ⋅ 10−3 − 74.22 ⋅ 10−3 ) J / m 2
− 17.22 ⋅ 10 −3
⋅
=−
= 7.32 ⋅ 10−6 mol/m2
8.314 J / mol ⋅ K ⋅ 283 K
(50 − 0) mol / L
2352.86
. σ2<σ1,
∆σ
< 0, A > 0, so pelargonic acid is surface active substance.
∆C
Numerical problem 2. Calculate the free surface energy (in kJ) for the pelargonic
acid С8Н17СООН solution at the temperature of 10°С if the
surface tension of this solution is 57.0·10–3 J/m2. The
surface area of the interface is 50 m2.
Steps to Solution:
1. The free surface energy could be calculated from the formula for surface tension:
→ GS = σ ⋅ S
GS
σ=
S
2. Calculate the free surface energy:
GS = 57·10–3 · 50 = 2.85 J = 2.85·10–3 kJ
Numerical problem 3. Calculate the specific surface area of arsenic (III) suldide
As2S3 if its particles average diameter is 1·10–7 m, and its
density is 3.43·103 kg/m3.
Steps to Solution:
1. Calculate the particles average radius:
r=
d 10 −7
=
= 5 ⋅ 10 −8 m.
2
2
2. The specific surface area of the adsorbent is the total area of all its particles
(Sspecific=4πr2) in the unit volume ( V = 4 πr 3 ) or in the unit mass (m). Accepting
3
the spherical shape of the dispersed phase particles, it may be written:
95
Sspecific =
S
4 πr 2
3
=
= [m -1 ]
4 3 r
V
πr
3
or S specific =
S
S
4 πr 2
3 m2 
=
=
=


m Vρ 4 3
rρ  kg 
πr ⋅ ρ
3
3. Calculate the specific surface area:
S specific =
–1
3
5 ⋅ 10
−8
= 6.7 ⋅ 10 7 m
or S specific =
m2 
= 1.75 ⋅ 10 7 

⋅ 3.43 ⋅ 10
 kg 
3
5 ⋅ 10
−8
.
3
Numerical problem 4. When 2.8 g of oxygen is adsorbed with activated charcoal at
68 K 1.36 kJ of heat release. Calculate the adsorption heat
effect (in kJ/mol) of oxygen with charcoal.
Steps to Solution:
The adsorption heat effect is the amount of heat which flows out of or into a
reacting system under the 1 mole substance adsorption with an adsorbent. 1.36 kJ of
heat is given to flow out while 2.8 g of oxygen been adsorbed, then X kJ of heat will
flow out for 1 mole (32 g) oxygen adsorption:
X=
32 ⋅ 1.36
= 15.54 kJ/mol.
2.8
Numerical problem 5. Calculate the adsorption quantity of acetic acid on activated
charcoal, if the equilibrium concentration of the acid is
3.76 mmol/l, К and n constants in Freundlich equation are
2.82 and 1.21, respectively.
Steps to Solution:
Calculate the adsorption quantity according to Freundlich equation:
A = kC 1 / n = 2.82 ⋅ 3.76 1 / 1.21= 2.82 ⋅ 3.760.826 = 2.82 ⋅ 2.986 = 8.42 .
Numerical problem 6. Calculate the equilibrium concentration of acetic acid, if
adsorption quantity equals 0.012 mol/g, К and n constants
in Freundlich equation are 0.25 and 3.1, respectively.
Steps to Solution:
1. According to Freundlich adsorption isotherm equation:
1
1
1
x
A = = К ⋅ Сeqn ;
=
= 0.32 ; 0,012 = 0,25 ⋅ Сeq0.32 ,
m
n 3.1
therefore: Сeq0.32 = 0,048.
2. Calculate the equilibrium concentration of acetic acid:
C= 0.32 0.048 = 8 ⋅10−5 mmol/g .
Alternatively the equilibrium concentration may be calculated by taking
logarithm of this equation:
log Сeq0.32 = log 0,048, 0.32 log Ceq. = log 0.048,
96
log Ceq = log 0.048 = −1.3188 = − 4.1213
0.32
0.32
Сeq. = alog(–4.1213) = alog (–5 + 0.8787) = 7.57⋅10–5 ≈ 8⋅10–5 mmol/g.
Numerical problem 7. To 5 numbered flasks, each containing 100 ml of acetic acid
solutions of various concentrations at 20 °C 3 g of
activated charcoal were added. The acid solutions
concentrations before and after the adsorption were
evaluated by 25 ml each solution titration with 0.1 M
NaOH solution with phenolphthalein. Calculate the
adsorption magnitudes (A) and acetic acid equilibrium
concentrations (Ceq) for each solution according to the
experimental data listed in the Table below:
The volumes of NaOH solution consumed for 25 ml acids
Flask No.
solution titration, ml
before the adsorption
2.75
3.90
5.60
11.50
22.20
1
2
3
4
5
after the adsorption
0.60
1.05
1.83
5.10
12.50
Steps to Solution:
1. The following values should be calculated firstly:
1) Со – the acid concentration before the adsorption (mmol/l ),
2) Сeq – the acid concentration after the adsorption (mol/l). For example, before
the adsorption the acid concentration in solution No 1 may be calculated
using the equation of the equivalents law: СacidVacid = СalkaliValkali;
C0 =
C alkali ⋅ Valkali 0.1 ⋅ 2.75
=
= 0.0111 mol/l = 11.1 mmol/l;
Vacid
25
2. After the adsorption for the same solution:
Ceq =
0.1 ⋅ 0.6
= 0.0024 mol/l = 2.4 mmol/l.
25
3. The initial acid concentrtions and its concentrations after the adsorption for
other solutions may be calculated in the same manner.
4. The adsorption magnitude A= x/m (mmol per 1.0 g of the adsorbent) should be
calculated according to the equation:
x (C0 − Ceq ) ⋅ V
A= =
m
m
where:
х – the change of the adsorptive moles number owing to
adsorption, mmol;
m – the mass of the adsorbent, g;
Со – the initial cid concentration, mol/l.
97
Сeq – the cid concentration after the adsorption, mol/l;
V – the volume of the acid solution in which the adsorption was studied, l.
For solution No 1:
A=
(C 0 − Ceq ) ⋅ V (0.0111 − 0.0024 ) ⋅ 0.1
=
=
m
3
0.29 mmol/g.
5. The same calculations should be done for other solutions.
Numerical problem 8. Define the surface charge sign of phosphorite in the natural
spring water which contain different soluble salts like
Са(НСО3)2.
Steps to Solution:
Water insoluble phosphorite Са3(РО4)2 is a polar adsorbent of trigonal symmetry
structure in the nodes of which Са2+ and РО43– ions are alternating. According to
Fajans-Peneth adsorption rule Са2+ cations from water are attracting
electrostatically to РО43– anions on the phosphorite surface and are finishing
building its crystal lattice. That is why the phosphorite surface will gain the positive
charge.
с) Problems to Solve
1. Calculate the adsorption quantity (in mol/m2) for the 20% solution of СaCl2 at
the temperature of 293 К, if the surface tension of this solution is 80·10–3 J/m2
and the surface tension of water is 73·10–3 J/m2. The density of solution is 1.177
g/ml.
Answer: –2.86·10–6 mol/m2
2. Calculate the adsorption quantity (in mol/m2) at 15 оС for the aqueous solution
which contains 49 g/l of acetone if the surface tension of this solution is 59.4·10–
3
J/m2. The surface tension of water is 73.49·10–3 J/m2 at this temperature. The
density of solution is 0.990 g/ml.
Answer: 5.8·10–6 mol/m2
3. For aqueous solutions with the concentration of carbonic acid 0.005 and 0.01
mol/l the surface tension quantities at 273 К are 65.8·10–3 and 60.05·10–3 J/m2,
respectively. Calculate the adsorption quantity (in mol/m2).
Answer: 3.82·10–6 mol/m2
4. Calculate the specific surface area of kaolin suspension (its density is 1.8·10–3
kg/m3), if its particles are suppose to be of a spherical shape with the average
diameter of 0.5·10–6 m.
Answer: 4.8·103 m2/kg
5. Calculate the equilibrium concentration of acetic acid if it was found that 1 g of
activated charcoal can adsorb 3.76·10–3 moles of the acid. K and n constants in
Freundlich equation are 2.82 and 2.44, respectively.
Answer: 10–7 mol/l
6. Define К and 1/n constant in Freundlich equation for oxalic acid adsorption with
activated charcoal (m = 1 g) at К using the following experimental data:
98
С, mmol/l
A, mmol/g
280
1.26
440
1.58
660
2.00
1260
2.80
2950
4.00
What does Freundlich equation look like?
Answer: log K = –1.2; К = 0.063; tgφ = 1/n = 0.53
7. Calculate the boundary adsorption magnitude for the Langmuir adsorption
isotherm if the solute adsorption magnitude was determined as 6·10–3 mol/m2 at
the equilibrium solute concentration of 0.03 mol/l. The constant K for the
Langmuir equation is 0.8.
Answer: 0.166 mol/m2.
8. Illustrate schematically the selective adsorption phenomena and define the
surface charge signs:
− for the chalk particles immersed into Ca(NO3)2 solution;
− for calcium silicate CaSiO3 particles immersed into soluble glass ( Na2SiO3)
solution.
Answer: In the first case the adsorbent surface would gain a positive charge, in the
second case – a negative charge
4. Laboratory Activities and Experiments Section
4.1. Practical Skills and Suggested Learning Activities
– To demonstrate that surfactants can reduce the surface tension of a liquid;
– To get practical skills of the surface tension experimental determination on the
liquid-gas interface with the maximal bubble pressure method;
– To learn how Gibbs’ equation should be applied for the liquid-gas interfacial
layer adsorption calculations;
– to determine and interpret gas adsorption isotherms;
– to determine the magnitude of a solute adsorption in terms of the Langmuir
adsorption isotherm;
– to study the adsorption ability of activated charcoal;
– to separate colored salts from their mixture.
4.2. Experimental Guidelines
4.2.1. Analysis of alcohol concentration influence on the surface tension of
solutions
Bubble pressure method consists in the following:
1. In vessel 2 distilled water is poured, preliminary the capillary has to be washed
out by the distilled water.
2. Vessel 2 is covered with plug with inserted capillary 1. The end of the capillary
should only become moist by the liquid meniscus.
3. Then vessel 2 is set in the thermostat, and connected with manometer (pressure
gauge) 3 and aspirator 4.
4. The tap is turned on and the water is poured out gradually. Hereby in the vessel
2 rarefaction occurs, and through the liquid the bubble of ear slips. The bubble
has to tear off of the capillary 1 evenly, almost every 10 s.
99
3
1
45
h
0 00 0
5 43 20
0 0 00
0 2 3 45
2
Fig. 1. Scheme of the Rebinder’s Bubble pressure method device for the surface
tension evaluation
5. In vessel 2 distilled water is poured, preliminary the capillary has to be washed
out by the distilled water.
6. Vessel 2 is covered with plug with inserted capillary 1. The end of the capillary
should only become moist by the liquid meniscus.
7. Then vessel 2 is set in the thermostat, and connected with manometer (pressure
gauge) 3 and aspirator 4.
8. The tap is turned on and the water is poured out gradually. Hereby in the vessel
2 rarefaction occurs, and through the liquid the bubble of ear slips. The bubble
has to tear off of the capillary 1 evenly, almost every 10 s.
Pressure P equals the difference between external pressure (atmospheric
pressure) and the pressure in vessel 2 and is proportional to surface tension
σ = Kh, K is constant for given capillary. As there are tabular data for surface
tension of water at different temperatures, so constant K is measured using
water.
9. Calculation of constant K:
K = σH2O/h
Calculation of surface tension for other liquids and solutions:
σ = Kh, or σ = σH2O · hsol/hH2O
10. In the bubble pressure method the surface tension of 4 solutions with different
concentration has to be defined. Obtained experimental data are put in the table
1 below.
11. According to experimental data students draw a diagram of surface tension
isotherm, where ordinate axis is σ (J/m2) and abscissa axis is C (mole/l)
12. According Gibbs equation adsorption of alcohol is calculated. Students draw a
diagram of Gibbs adsorption isotherm: ordinate axis is G (mole/l), abscissa axis
is C (mole/l).
100
Table 1
Experimental data for the surface tension and adsorption evaluation and the
surfactant solutions concentrations affecting their values
Alcohol
Difference between
Surface tension Adsorption A,
concentration
Substance
the levels
mole/m2
С,
σ, J/m2
h, mm
mol/l
1. Water
0.00
2. Alcohol
0.025
3. Alcohol
0.050
4. Alcohol
0.1
5. Alcohol
0.2
4.2.2. The hydrocarbon radical length in a surfactant molecule affecting the
surface activity studying
1. Determine and calculate the capillary constant K as it is described in the
previous experiment (4.2.1).
2. Determine the surface tension of alcohols solutions of the same homologous
series (СН3ОН, С2Н5ОН, С3Н7ОН, С4Н9ОН, С5Н11ОН) with the concentration
of 0.1 mol/l.
3. Plot the correlation of the solutions surface tensions (J/m2) versus the number of
carbon atoms in the hydrocarbon radical.
4. Calculate the surface tension increasing ∆σ with the hydrocarbon radical length
increasing. Make the conclusions about the Traube’s rule experimental
verification.
4.2.3. The acetic acid adsorption on activated charcoal determining
1. Prepare aqueous solutions of acetic acid into numbered flasks using pipettes and
measuring cylinders from 0.4 M solution following the scheme given in the
Table 2. The total volume of each solution is 50 ml. Use flasks fitted with
stoppers.
Table 2.
Scheme for acetic acid dilution
Flask No.
1
2
3
4
Volume of 0.4 M acetic acid solution V1 (ml)
50
25
12.5
6.25
Volume of distilled water V2 (ml)
–
25
37.5
43.75
Total volume (ml)
50
50
50
50
2. Transfer 10 ml of the solution from each flask into numbered titrimetric flask, so
final volume of acetic acid solution in the numbered flasks is VA=40 ml per
flask.
3. Determine the actual concentration of acetic acid in flasks by titration in this
101
way:
a) For titration, modify the volume in each titrimetric flask. Take away defined
volume of the solution, to obtain in each flask the volume as given in the Table
3.
Table 3.
Volumes of the acetic acid solutions used for titration before and after the
adsorption
Titrimetric flask No.
2
3
4
1
Volume V (ml)
10
10
10
10
b) Add 2-3 drops of phenolphthalein and titrate by NaOH.
c) Once the endpoint has been reached, read the burette. The volume of the base
Vio(ml) that was required to reach the endpoint write down to the Table 3.
d) Calculate the actual concentration of acetic acid Cio in the flasks No. 1 – 4,
respectively, and write it down to the Table 3:
V o ⋅ СT
, [mol/l] (1)
Ci o = i
V
4.
5.
6.
7.
8.
where Vio is the volume of the titrant (NaOH), CT is the concentration of the
titrant, V is the volume of the analyte (acetic acid according to Table 2), i=1–4 is
the number of flask.
Using practical balance and glazed paper, weigh 4 portions of activated
charcoal, each portion 1 g. The accuracy of weighing must be 0.01 g.
Put activated charcoal into numbered flasks with stoppers (1 portion per flask).
Plug up the flasks, and shake them. Wait for 20 minutes, the process of
adsorption is in progress. Mix the mixtures for several times by flasks shaking
within this period. (Remark: The process of adsorption is a function of time too.
It is important to put charcoal into flasks at the same time, to provide
adsorption for the same period in each flask).
Filter the mixtures into clean and dry flasks. To avoid disturbing effect of
adsorption of acetic acid into filtering paper, remove away the first portion of
filtration, app. 5 ml.
Determine the final concentration of acetic acid Ci in each of the flasks after
adsorption. From each solution, transfer the asked volume into clean and dry
titrimetric flask, again following Table 2. Calculate the concentration of acetic
acid after adsorption (Ci), using the Eq. 1 and data form Table 3 after
adsorption.
Repeat points 3a-3d, and from the consumed base Vi (ml) determine the
concentration of acetic acid Ci after adsorption. Write them down to the Table 4.
Table 4.
Experimental data for the adsorption
Flask
No.
Vio
(ml)
Cio
(mol/l)
Vi
(ml)
102
Ci
(mol/l)
A=
х
m
mmol/g
1
2
3
4
9. Finishing experiment, wash carefully used flasks, pipettes, etc.
10. Determine of the adsorption A as the substance amount of acetic acid adsorbed
per gram of the charcoal m (mol/g) in individual flask:
A=
o
х (С i − С i ) ⋅ V
=
m
m
(2)
where Cio is initial concentration of acetic acid, mmol/l; C i is equilibrium
concentration of acetic acid, mmol/l; V is volume of acetic acid taken for the
adsorption (the volume of the liquid phase in the mixture charcoal – acetic acid),
ml; m is mass of adsorbent, (1 g).; i=1–4 is the number of flask. Eq. 2 supposes
that V is the same for i=1–4, and also the mass of the charcoal (g). Write down
the obtained values of Ai to the Table 3.
11. Plot the dependence (isotherm) A = X – f(С) (T = const) for the adsorption of
m
acetic acid on activated charcoal. Make a conclusion about the equilibrium
concentration of the acid affecting its adsorption.
A=
X
m
Ci
Fig. 6.1. Freundlich adsorption isotherm
5. Conclusions and Interpretations. Lesson Summary
Topic 11
Preparation, purification and properties of colloidal solutions
1. Objectives
Colloids and colloidal systems are essential to life. Most of the substances, we
come across in our daily life, are colloids. The meals we eat, the clothes we wear,
the wooden furniture we use, the houses we live in, the newspapers we read, are
largely composed of colloids. Following are the interesting and noteworthy
examples of colloids: blue colour of the sky, fog, mist and rain, food articles: milk,
butter, halwa, ice creams, fruit juices, etc., are all colloids in one form or the other.
103
Colloids function in every body cell, in the blood, and in all body fluids, especially
the intercellular fluids. Blood is a colloidal solution of an albuminoidal substance.
The styptic action of alum and ferric chloride solution is due to coagulation of blood
forming a clot which stops further bleeding. All life processes take place in a
colloidal system, and that is true both of the normal fluids and secretions of the
organism, and of the bacterial toxins, as well as, in large measure, of the reactions,
which confer immunity.
Atmospheric soils are mainly colloidal in nature. On account of colloidal nature,
soils adsorb moisture and nourishing materials.
Dispersion refers to the spatial distribution of organisms, and is a fundamental
component of a species' ecology and life history. Patterns of dispersion influence
other aspects of a species' behavior and ecology. Dispersion patterns themselves are
affected by the distribution of resources (including sunlight, nutrients, prey species,
etc.) in the environment, as well as through the reciprocal influence of species'
behavioral characteristics.
2. Learning Targets:
– to get practical skills in lyophobic sols preparation;
– to learn their properties;
– to learn the structure of micelle;
– to study capillary analysis bases and its application for the determining the
charge sign of colloidal drugs;
– to learn the theory of light dispersion and Rayleigh’s equation;
− electrical properties of dispersions, introduction to electrokinetic phenomena;
− to learn the micelle structure in isoelectric point;
− to get practical skills in zeta-potential magnitude measurements and to know
how to use it for lyophobic sols stability characteristics;
– to learn electrophoresis applications in medical and biological researches.
3. Self Study Section
3.1. Syllabus Content
The living organism as a disperse systems combination. Classification of
disperse systems according to the aggregative state, interphase interaction,
dispersion. Lyophilic and lyophobic dispersions. A structure of micelle. Structure of
a double electric layer (DEL). The overall performance and history of development
the ideas about DEL structure. Electrokinetial potential of a colloidal particle.
Purification of colloidal solutions. Dialysis, electro-dialysis, ultrafiltration,
compensatory dialysis. Haemodialysis and “artificial kidney” device.
Molecular - kinetic properties of dispersions. Thermal molecular motion and
Brownian motion, diffusion, and osmotic pressure. Optical properties of
dispersions.
Electrical properties of dispersions. Introduction to electrokinetic phenomena.
Electrophoresis and electroosmosis. Helmholz-Smoluchowsky equation. The
104
electrophoresis uses in clinical and laboratory practice researches
3.2. Overview
Colloidal solutions are intermediate between true solutions and suspensions. The
size of the colloidal particles range from 10-9 to 10-7 m. A colloidal system consists
of two phases – the dispersed phase and the dispersion medium. Colloidal systems
are classified in three ways depending upon physical states of the dispersed phase
and dispersion medium; nature of interaction between the dispersed phase and
dispersion medium, and nature of particles of dispersed phase. The colloidal
systems show interesting optical, mechanical and electrical properties.
Colloidal sols can be formed by dispersion methods (e.g. by mechanical
subdivision of larger particles or by dissolution in the case of lyophilic sols) or by
condensation methods (from supersaturated solutions or supercooled vapours, or as
the product of chemical reactions) or by a combination of these two (e.g. in an
electrical discharge).
Dispersion can be done mechanically, in a colloid mill that grinds the substance
into small, equal particles. Another method is with an electric arc. Metal electrodes
are used, at a current of 5-10A and voltage of 30-40V.
When a condensation method is applied, molecules (or ions) are deposited on
nuclei, which may be of the same chemical species as the colloid (homogeneous
nucleation) or different (heterogeneous nucleation).
Peptization may be defined as the process of converting a precipitate into
colloidal sol by shaking it with dispersion medium in the presence a small amount of
electrolyte. The electrolyte used for this purpose is called peptizing agent. This
method is applied, generally, to convert a freshly prepared precipitate into a
colloidal sol. During peptization, the precipitate adsorbs one of the ions of the
electrolyte on its surface. This causes the development of positive or negative
charge on precipitates, which ultimately break up into smaller particles of the size of
a colloid.
According to the micellar theory of colloids solution structure, sol consists of
micelles and intermicellar liquid. A micelle is a colloidal-sized particle formed by
the association of molecules, each of which has a hydrophobic end and a
hydrophilic end.
There are 3 main parts in the micelle structure: nucleus, adsorptive and diffuse
layers. A nucleating agent is a material either added to or present in the system,
which induces either homogeneous or heterogeneous nucleation.
Colloidal particles always carry an electric charge. The nature of this charge is
the same on all the particles in a given colloidal solution and may be either positive
or negative. The charge on the sol particles is due to preferential (selective)
adsorption of ions from solution and/or due to electrical double layer formation. The
sol particles acquire positive or negative charge by selective adsorption of cations or
anions. When two or more ions are present in the dispersion medium, preferential
adsorption of the ion common to the colloidal particle usually takes place according
105
to the Fajans-Peneth adsorption rule. Having acquired a positive or a negative
charge by selective adsorption on the surface of a colloidal particle, this layer
attracts counter ions from the medium forming adsorptive and diffusive layers of
counterions. According to modern views, the first layer of counterions is firmly held
and is termed the adsorptive layer while the second layer is mobile which is termed
the diffusive layer. Since separation of charge is a seat of potential, the charges of
opposite signs on the adsorptive and diffusive parts of the double layer results in a
difference in potential between these layers. This potential difference between the
adsorptive layer and the diffusive layer of opposite charges is called the
electrokinetic potential, or zeta potential.
The magnitude of zeta potential is crucial in determining the stability of a
colloidal suspension. When all the particles have a large negative or large positive
charge they will repel each other, and so the suspension will be stable. A sol is
supposed to be stable when its zeta-potential magnitude is in the range of 30–90
mV.
If the zeta potential is low the tendency for flocculation is increased. Another
important consideration when discussing zeta potentials is pH; in fact, quoting a
zeta potential without an accompanying pH is almost meaningless. This is due to the
fact that, for suspensions of most materials, a plot of zeta potential versus pH
exhibits an isoelectric point, a particular value of solution pH where the net charge
on the particles is zero. At this point the suspension is highly unstable, and
flocculation is at its most likely.
An important consequence of the existence of electrical charges on the surface
of particles is that they interact with an applied electric field. These effects are
collectively defined as electrokinetic effects. There are four distinct effects
depending on the way in which the motion is induced. These are:
Electrophoresis: the movement of a charged particle relative to the liquid it is
suspended in under the influence of an applied electric field.
Electroosmosis: the movement of a liquid relative to a stationary charged
surface under the influence of an electric field.
Streaming potential: the electric field generated when a liquid is forced to flow
past a stationary charged surface.
Sedimentation potential: the electric field generated when charged particles
sediment.
The Helmholts-Smoluchowski equation for zeta-potential calculation by
electrophoresis method:
ξ=
Slη
τEε 0ε
where S – linear motion on a sol boundary, m; l – distance between the
electrodes in the device for electrophoresis, m; η – viscosity of the dispersion
medium, Pa·s; τ – time of electrophoresis, sec.; E – electric field voltage V, ε –
dielectric constant of medium (=81 for water), ε0 = 8.85·10–12 Fh/m – dielectric
constant of vacuum.
106
Systems containing colloid particles are in some properties different from
systems which are homogeneous or composed of macroscopic phases. When the
beam of light is passing through a dispersed system the following phenomena may
be observed: transmission, absorption, refraction, reflection and scattering of light.
The prevailing of any of these processes depends on the ratio of the wavelengths of
the light and the sizes of the suspended particles.
The molecular and ionic solutions are optically transparent as the beam of light
is just transmitting through them. Any medium is able to absorb selectively the light
waves with the certain wavelengths.
The optical properties of colloids depend on the size and structure of suspended
particles and the dispersed phase concentration. The diffraction of light will be
mainly observed for colloids as their particles sizes are equal about half of the initial
light wavelength (λ/2). The diffraction courses the opalescence of colloids – the
lusterless luminarity of soles usually with a touch of blue.
If light passes through a system containing colloid particles, part of the light is
scattered and consequently the ray passage through the environment can be
observed (the Tyndall effect).
Rayleigh equation shows the scattered light intensity affection with different
factors:
2
n 2 − n02 υV particle
I scat = 24π 3 I 0 21
n1 + 2n02
λ4
where Io and Iscat are the intensities of the initial and scattered light beam,
respectively; υ is the number of particles of the volume Vparticle in unit volume; λ is
the wavelength of the incident light, n0 is the refraction index or the dispersion
medium, and n1 is refraction index of the dispersed phase.
From this equation analysis it is obvious that:
− in order for scattered light to originate, the diffraction index of the dispersion
ratio and that of the dispersion environment have to be different;
− light scattering is proportional to the number of particles in the system;
− the scattered light is proportional to the quadrate of the particles volume;
− the scattering is inversely proportional to the fourth power of the wavelength of
the incident light.
3.3. References
37. V.O. Kalibabchuk, V.I. Halynska, L.I. Hryshchenko et al. Medical
Chemistry. – AUS MEDICINE Publishing. – 2010. – 224 p.
38. Raymond Chang. Chemistry (6th Edition). – WCB/McGraw-Hill. – 1998. – 995
p.
39. Rodney J. Sime. Physical Chemistry. Methods. Techniques. Experiments. –
Saunders College Publishing. – 1990. – 806 p.
40. John McMurry, Robert C. Fay. Chemistry (3rd Edition). – Prentice Hall. – 2001.
– 1067 p.
41. David E. Goldberg. Fundamentals of Chemistry (2nd Edition). –
107
WCB/McGraw-Hill. – 1998. – 561 p.
42. Theodore L. Brown, H.Eugene LeMay, Bruce E. Bursten. Chemistry. The
Central Science. – Prentice Hall. – 2000. – 1017 p.
43. John Olmsted III, Gregory M. Williams. Chemistry. The Molecular
Science. – Mosby. – 1994. – 977 p.
44. Steven S. Zumdahl. Chemistry (4th Edition). – Houghton Mifflin Company. –
1997. – 1031 p.
45. Gary L. Miessler, Donald A. Tarr. Inorganic Chemistry. – Prentice Hall. –
1991. – 625 p.
3.4. Self Assessment Exercises
а) Review Questions
1. Define the disperse systems and give their classifications according to different
criteria.
2. Show the divergences between colloids and true solutions.
3. Name and explain the main methods of sols preparation. The difference between
the peptization method and the dispersion and condensation methods.
4. Name the parts of the micelle structure. State the Fajans-Peneth adsorption rule.
Define the potential-defining ions, nucleus, adsorptive and diffuse layers of a
micelle.
5. Consider the examples of AgCl, As2S3, PbS lyophobic sols preparation with
positively and negatively charged granules. Write down the schemes of their
micelles structure.
6. The prevailing of light scattering or absorption and reflection depending on the
sizes of the suspended particles. The Tyndall effect.
7. Write down the Rayleigh’s equation and make its analysis. Which variables do
affect the scattered light intensity?
8. Research techniques of colloid systems, based on the phenomenon of dispersion
of light. Nephelometry. Turbidimetry. Ultramicroscopy. Electronic microscopy.
9. Name the main methods of colloidal solutions purification. Their medicinal
applications. The principle of the "artificial kidney" device operation.
10. Give the definitions of electrophoresis and electroosmosis.
11. Draw the schematical representation of the device for electrophoresis and
electroosmosis.
12. Define the electrokinetical (zeta-) potential and explain the mechanism of its
appearance. Explain its role in determining the stability of colloids.
13. Draw the micelle structure for Fe(OH)3 sol prepared in the presence of Fe(NO3)3
as a nucleating agent. Show the location of zeta-potential occurrence. Draw the
structure of the same micelle in isoelectric state.
14. Application of electrophoresis in medical and biological researches.
b) Types of Numerical Problems and Their Solving Strategies
Numerical problem 1. 12 ml of 0.01 М of FeCl3 solution were added to 10 ml of
0.05 N K4[Fe(CN)6] solution to obtain sol of Berlin blue.
108
Write down the formula of micelle of this sol. What is the
charge of the granule?
Steps to Solution:
1. Calculate the normal concentration of 0.01 M FeCl3 solution:
СN = СМ⋅3 = 0.01⋅3 = 0.03 mol-eq/l.
2. Write down the equation of Berlin blue formation:
4 FeCl3 + 3 K4[Fe(CN)6] → Fe4([Fe(CN)6])3↓ + 12 KCl.
3. Calculate which compound is in excess and is stabilizer:
Number of mmol-eq (FeCl3) = 12⋅0.02 = 0.24 mmol-eq;
Number of mmol-eq (K4[Fe(CN)6]) = 10⋅0.05 = 0.5 mmol-eq.
4. In solution there is the excess of K4[Fe(CN)6], which is the stabilizer. Therefore,
hydrophobic Fe4([Fe(CN)6])3 particles will adsorb [Fe(CN)6]4– ions (FajansPeneth adsorption rule), which determine the negative charge of the granule. К+
ions will be counterions.
5. The scheme of structure of Berlin blue micelle in this case is:
{[(mFe4([Fe(CN)6])3) ⋅ n[Fe(CN)6]4– ]4n– ⋅ 4(n – x) K+}4x– ⋅ 4x K+
(mFe4([Fe(CN)6])3) – aggregate;
[(mFe4([Fe(CN)6])3) ⋅ n [Fe(CN)6]4– ]4n– – nucleus;
{[(mFe4([Fe(CN)6])3) ⋅ n[Fe(CN)6]4– ]4n– ⋅ 4(n – x) K+}4x– – granule;
n[Fe(CN)6]4– – potential determining ions;
4(n – x) K+ – adsorptive layer;
4xK+ – diffuse layer.
Numerical problem 2. What is the minimum volume of 0.1 % AgNO3 solution
which should be added to 50 ml of 0.15 % KI solution for
preparing of sol with positively charged granules?
Steps to Solution:
The reaction of AgI formation is:
AgNO3 + KI → AgI + KNO3
Sol with positively charged granules will form if AgNO3 is the stabilizer of the
sol. The scheme of micelle is:
{[(mAgI)nAg+]n+(n–x)NO3–}x+xNO3–
In this case the AgNO3 solution must be in excess.
According to the equivalent law:
CM(KI)·V(KI) = CM(AgNO3)·V(AgNO3)
The volume of AgNO3 solution is:
V ( AgNO 3 ) =
С M (KI) ⋅ V ( KI)
C M (AgNO 3 )
Calculate the molarity of 0.15% KI solution:
109
С M ( KI) =
m(KI)
0.15 g
=
= 0.009 mol/l .
M (KI) ⋅ V 166 g/mol ⋅ 0.1 l
Calculate the molarity of 0.1% AgNO3 solution:
CM (AgNO3 ) =
m(AgNO3 )
0.1 g
=
= 0.006 mol/l .
M (AgNO3 ) ⋅ V 170 g/mol ⋅ 0.1 l
Calculate the equivalent volume of AgNO3 solution:
0.009 mol/l ⋅ 50 ml
V =
= 75 ml .
0.006 mol/l
Answer: the minimum volume of AgNO3 solution is 76 ml.
Numerical problem 3. Calculate the diameter of an aerosol particles, if 120 of such
particles were found in the volume of 4·10–11 m3. The mass
concentration of sol is 1·10–4 kg/m3, and the dispersed
phase density is 2.2·103 kg/m3.
Steps to Solution:
The quantity of particles in the unit of volume equals:
n
ν= .
V
The volume of 1 particle is equal to the ratio between the mass of 1 particle and
the density of dispersed phase:
m
C
С ⋅V
V0 = =
=
d ν ⋅d n⋅d
For the spherical particles the volume of 1 particle is:
4
V0 = πr 3
3
Then,
С ⋅ V0 4
= πr
n⋅d
3
→ r = 3 3 ⋅ C ⋅ V0
4π ⋅ n ⋅ ρ
Calculate the radius of the spherical particles:
r =3
3 ⋅1 ⋅10− 4 ⋅ 4 ⋅10−11
4 ⋅ 3.14 ⋅120 ⋅ 2.2 ⋅103
= 1.535 10-7 m
Calculate the diameter of the spherical particles:
D = 2r = 1.535⋅10–7 ⋅ 2 = 3.07⋅10–7 m.
Answer: the diameter of sol particles is 3.07⋅10–7 m.
Numerical problem 4. Calculate the value of ξ-potential (in mV) for sol of
collargolum using the following electrophoresis data:
linear motion on a sol boundary S=10 mm, distance
between the electrodes l=20 cm, electric field voltage
E=300 V, time of electrophoresis τ=5 min, viscosity of
the medium η=10–3 Pa·s, dielectric constant of medium
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ε=81 and dielectric constant of vacuum ε0=8.85·10–12
Fh/m.
Steps to Solution:
According to the Helmholts-Smoluchowski equation zeta-potential is:
ξ=
Slη
10 ⋅ 10 −3 m ⋅ 20 ⋅ 10 −2 m ⋅ 10 −3 Pa ⋅ s
=
= 0.031 V = 31 mV .
τEε 0ε 5 ⋅ 60s ⋅ 300V ⋅ 81 ⋅ 8.85 ⋅ 10−12 Fh/m
Answer: zeta-potential of sol particles is 31 мV.
c) Problems to Solve
1. For preparing of silver chloride sol 85 ml of 0.005 М silver nitrate solution were
added to 15 ml of 0.025 М potassium chloride solution. Write down the formula
of obtained sol micelle. What is the charge of its granule?
Answer: positive
2. What volumes of 0.029 % NaCl solution (the density is 1 g/cm3) and 0.001 N
AgNO3 solutions should be mixed to prepare uncharged particles of AgCl sol?
Draw the structure of the micelle in isoelectric state.
Answer: V(NaCl) : V(AgNO3) = 1 : 4.96
3. At ultramicroscopy investigation of gold hydrosol in volume of V0=1.6⋅10–11 m3
70 particles (n) were found. Sol concentration С=7⋅10–6 kg/m3, gold density
d=19.3⋅103 kg/m3. Determine the radius of dispersed phase particles (in m)
assuming their shape as spherical.
Answer: 2.705⋅10–8 m
4. AgI sol was obtained at slowly adding 15 ml of 0.2 % AgNO3 solution (the
density is 1 g/ml) to 20 ml of 0.01 М KI solution. Write down the formula of
micelle of obtained sol. Determine the direction of its particles movement in the
applied electric field.
5. What volume (in ml) of 0.005 М ZnCl2 solution should be added to 20 ml of
0.015 М (NH4)2S solution to prepare ZnS sol with positively charged granules?
Write down the formula of micelle of obtained sol.
Answer: 60 ml
6. Calculate zeta-potential of protein particles if under the electrophoresis its
particles have been moved at a distance of 22.5 mm during 20 min at the applied
voltage of 200 V. Distance between the electrodes is 15 cm. Viscosity of the
solution η = 10–3 Pа⋅sec and constants ε = 81 і ε0 = 8.85⋅10–12 Fh/m.
Answer: 19.6 mV
4. Laboratory Activities and Experiments Section
4.1. Practical Skills and Suggested Learning Activities
– to prepare lyophobic sols with physical condensation method (a solvent
replacement);
– to prepare lyophobic sols with chemical condensation method under the double
exchange reactions, hydrolysis and redox reactions;
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– to prepare sols with the methods of direct and indirect peptization;
– to determine the sign of particles charge by the capillary analysis method;
– to determine the charge sign of a sol granules and zeta potential magnitude for
the particles of a drug.
4.2. Experimental Guidelines
4.2.1. Preparation of sols by physical condensation (by replacing solvent)
а) Preparation of colophony sol
Pour 5 ml of distilled water into the test tube and add the solution of colophony
in ethyl alcohol by drops while stirring. Record your observation. Pay attention to
the opalescence - a characteristic feature of colloidal solutions. What method does a
sol form with?
b) Preparation of sulfur sol
Pour 5 ml of distilled water into the test tube and add 0.5 ml of saturated
solution of sulfur in ethyl alcohol while stirring. What is observed?
4.2.2. Preparation of sols by method of chemical condensation
а) Preparation of silver iodide sol by double-exchange reaction
To 100 ml of 0.002 M KI solution add while stirring 1 ml of 0.01 M of AgNO3
solution. Note the color of sol and write the micelle structure (in the excess of KI).
What is the charge of the granule?
Write the structure of AgI micelle, obtained at the excess of AgNO3.
b) Preparation of Berlin blue sols with different charged granules
In two test tubes prepare Berlin blue sols from 0.005 M FeCl3solution and 0.005
M K4[Fe(CN)6] solution in the following ratios:
1) 3 ml of FeCl3 and 1 ml of K4[Fe(CN)6];
2) 3 ml of K4[Fe(CN)6] and 1 ml of FeCl3.
Write the equations of reactions and the structures of the micelles for both sols.
What factor is affecting the charge of the granules?
c) Preparation of iron hydroxide (III) sol by the reaction of hydrolysis
50 ml of distilled water should be heated to boiling in the flask. Add 5% FeCl3
solution into the boiling water by drops. What is observed? Write the equations of
reactions and the structure of obtained sol micelle.
d) Preparation of copper hexacyano(II)ferrate sol by double-exchange reaction
To 5 ml of 1% CuSO4 solution add a few drops of 0.01% K4[Fe(CN)6] solution.
Write the equation of the reaction and the structure of the obtained sol micelle.
Write the micelle structure of the sol obtained in the K4[Fe(CN)6] excess.
e) Preparation of silicic acid sol
To 5 ml of 5% Na2SiO3 solution add 0.5 ml of 0.1 M HCl solution while stirring.
What is the structure of the obtained sol micelle?
f) Preparation of metallic silver sol by reduction reaction
112
To 10 ml of 0.001 M AgNO3 solution add 2-3 drops of 1% K2CO3 solution and
1 ml of freshly prepared solution of tannin (aldehyde), and heat. Yellowish-brown
sol of metallic silver is formed:
AgNO3 + K2CO3 = AgOK + KNO3 + CO2
stabilizer
2AgOK + R–COH + H2O = 2Ag↓ + R–COOH + 2KOH
Write the structure of obtained sol micelle and explain the method of its
preparation.
g) Preparation of manganese dioxide sol by redox reaction
To 25 ml of 0.1 M KMnO4 solution in the flask add 10% H2O2 solution from the
burette by small portions until the solution droplet placed on filter paper with a glass
stick would not give pink spot. Write the structure of sol micelle.
h) Preparation of lead sulfide sol
To 3 ml of 1% Pb(CH3COO)2 solution add 3 ml of 5% Na2S solution by drops.
Record your observations, the equation of reaction and the structure of the obtained
sol micelle. What is the structure of the micelle of PbS sol, obtained at the excess of
Pb(CH3COO)2?
4.2.3. Preparation of sols by peptization
а) Preparation of Fe(OH)3
Prepare the precipitate of Fe(OH)3 by the reaction of FeCl3 with NH4OH. For
this purpose, 1 ml of saturated FeCl3 solution should be diluted with water to 20 ml.
To the obtained solution add 5% NH4OH solution slowly till bleaching of the liquid
above the precipitate. Write the equation of the reaction. Note the color of the
precipitate.
Pour away the liquid above the precipitate and rinse several times with distilled
water (decantation). Washed precipitate divide into 3 test tubes and add:
into І – 10 ml of distilled water;
into ІІ – 10 ml of 2 % FeCl3 solution;
into ІІІ – 10 ml of 0.02 М HCl solution.
Record your observations after 10 minutes. Explain what happened with the
precipitate in the second and third tubes? What are the types and mechanism of
peptization? Write the structure of the prepared sols micelles.
b) Preparation of Berlin blue sol.
To 5 ml of 2% FeCl3 solution add 1 ml of a saturated K4[Fe(CN)6] solution. The
precipitate should be filtered and washed with distilled water. After washing of the
precipitate on the filter with 0.1 M solution of oxalate acid (Н2С2O4) a blue sol of
Berlin blue is filtered. Write the equation of the reaction and the structure of the sol
micelle, considering that precipitate is peptized by oxalate acid due to С2O42– ions
adsorption.
4.2.4. Determination of the charge sign of colloidal particles of drugs
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In colored sols the sign of particles charge can be determined by method of
capillary analysis when capillaries surface of filter paper is used as charged surface.
It is based on the fact that cellulose capillary walls of filter paper are negatively
charged (because of dissociation of cellulose hydroxyl groups), and water that
moistens the paper – positively. Place the drop of investigated sol on the piece of
filter paper. After the drop absorption sol with positively charged particles is
adsorbed on the paper and gives stain colored at the center and colorless on the
edges; sol with negatively charged particles does not absorbe by the paper and
forms an evenly colored stain.
4.2.5. Determination of the sign and value of zeta-potential of a medicine
(hydrophobic sol directed by the teacher: protargol or collargolum) by
macroelectrophoresis method
1) Into the knee A of the device for electrophoresis (see Figure 1), which is fixed in
tripod, pour in intermediate fluid for a given sol to level ~ 3-5 mm above the
upper edge K. Tap should be opened to let the air come out through it.
2) Turn off the tap K and place the electrodes E into both knees, but do not clamp
the stoppers to let the fluid leak out.
3) Pour a given sol through the funnel В up to the top.
4) Use wire to clean the bottom end of the tube under the funnel В to remove any
air bubbles.
5) Open the tap K slightly to let sol from the funnel leak slowly to the bottom of Ushaped tube and form a clear boundary between sol and intermediate fluid,
which is lightweight than sol, in both knees.
6) Add some more sol into the funnel and withdraw separation boundary between
sol and intermediate fluid in front of the digital divisions on the scale.
7) Turn off the tap K and record the positions of sol boundaries in both knees
respectively to scales (they may not coincide).
8) Turn on the current. Write the pointer of voltmeter will show the voltage
between electrodes, note the start time of electrophoresis.
9) Record the voltage value E by direct indications of voltmeter or calculate it, if
the scale of the device has implicit graduation. Note, what is the voltage (300 V,
450 V or other) supplied by rectifier on the electrophoresis and where the
pointer of the voltmeter stopped.
S
Figure 1. Scheme of the device for electrophoresis:А intermediate fluid; S - sol; В - funnel; Е and Е - electrodes; К - tap.
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10) Stop the electrophoresis, turn off the current in 5-10 minutes (when the
boundary of sol in the knee lower to 5-7 mm - Scale Ш). Record the end time
of electrophoresis and determine the duration of current flow τ.
11) By the difference between the positions of sol boundaries determine the
displacement of sol particles (S). Note the direction (to which of the electrodes
sol boundary shifted during the electrophoresis). Make a conclusion about the
sign of the charge on the granules of investigated colloidal solution.
12) After switching off the current measure the distance l between the electrodes
using thread and ruler (by the course of the current in the U-shaped tube).
13) Pour away the solution from the U-shaped tube, rinse it with water.
14) Briefly describe the procedure of the work done and calculate the results.
15) Convert all measured values into SI system, substitute them in the formula of
Helmholtz-Smoluchowski equation to calculate ζ-potential:
η ⋅l ⋅ S
ζ=
τ ⋅ D ⋅ E ⋅ε .
where besides the already mentioned values l (m), S (m), τ (s), E (V) take the
viscosity η = 10-3 Pa⋅s and D = 81 – dielectric constant for water, and absolute
dielectric permittivity constant for vacuum ε = 8.85⋅10-12 F/m.
16) Recalculate the obtained value of ζ in volts to millivolts (mV).
17) Make a conclusion about the sign of the granules charge of investigated sol
and indicate whether it is stable, comparing the obtained value of zetapotential with the critical value (30 mV).
5. Conclusions and Interpretations. Lesson Summary
Topic 12
Electrolytic coagulation of colloids. Physical chemistry of
biopolymers solutions
1. Objectives
The long-term colloidal stability of dispersion will be of great importance in a
number of industries such as pharmaceutical, ceramic, paints and pigments. The
term “stability” can have different connotations to different applications. When
applied to colloids, a stable colloidal system is one in which the particles resist
flocculation or aggregation and exhibits a long shelf-life. This will depend upon the
balance of the repulsive and attractive forces that exist between particles as they
approach one another. If all the particles have a mutual repulsion then the dispersion
will remain stable. However, if the particles have little or no repulsive force then
some instability mechanism will eventually take place e.g. flocculation, aggregation
etc.
Coagulation processes are widely used in the industry, for example in the
electrical precipitation of smoke. They are also applied for the purification of
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drinking water: the water obtained from natural sources often contains suspended
impurities. Alum is added to such water to coagulate the suspended impurities and
make water fit for drinking purposes.
Most of the medicines are colloidal in nature. For example, argyrol is a silver sol
used as an eye lotion. Colloidal antimony is used in curing kalaazar. Colloidal gold
is used for intramuscular injection. Milk of magnesia, an emulsion, is used for
stomach disorders. Colloidal medicines are more effective because they have large
surface area and are therefore easily assimilated. So the colloidal protection in drugs
manufacturing is of a great importance.
Many polymers occur in nature, such as silk, cellulose, natural rubber, and
proteins. In addition, a large number of polymers have been synthesized in the
laboratory, leading to such commercially important products as plastics, synthetic
fibers, and synthetic rubber.
The variation in the form of macromolecules is largely responsible for molecular
diversity. Much of the variation that occurs both within an organism and among
organisms can ultimately be traced to differences in macromolecules. There are four
basic kinds of biological macromolecules. They are carbohydrates, lipids, proteins
and nucleic acids. These polymers are composed of different monomers and serve
different functions:
− сarbohydrates – composed of sugar monomers and necessary for energy storage;
− lipids – include fats, phospholipids and steroids. Lipids help to store energy,
cushion and protect organs, insulate the body and form cell membranes;
− proteins – composed of amino acid monomers and have a wide variety of
functions including molecular transport and muscle movement;
− nucleic acids – include DNA and RNA that are so important in genes and life
processes. Nucleic acids contain instructions for protein synthesis In fact,
messenger RNA is what makes possible proteins, peptides, and enzymes.
Natural and synthetic polymers are widely used in medicine and pharmacy.
Many biomaterials, especially heart valve replacements and blood vessels, are made
of polymers like Dacron, Teflon and polyurethane. Polymers are used for
orthopedic and reconstructive surgery, and for tissue engineering.
They are used for drug delivery and as excipients in pharmaceutical tablets and
hard gelatin capsules:
− fillers ( starches);
− disintegrants (starches, crosslinked polyvinyl pyrrolidone, sodium starch
glycolate, sodium carboxymethyl cellulose);
− binders (polyvinylpyrrolidone, cellulose derivatives, polyethylene glycol);
− lubricants (polyethylene glycol;
− antiadherents (starches);
− film coatings (cellulose derivates, methylmethacrylate copolymers, methacrylic
acid copolymers, polyvinyl acetate phthalate, polyethylene glycols).
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2. Learning Targets:
– to get practical skills in critical concentration of sols coagulation experimental
determining;
– to learn how is it possible to calculate the coagulating ability of electrolytes with
different valences ions;
– to know the protective action evaluation of polymers or hydrophilic colloids
towards hydrophobic sols and to determine experimentally the protective
number;
– to learn the main methods of polymers preparation;
– to learn the structure of natural and synthesized polymers;
– to get practical skills in the degree of swelling measuring;
– to learn the electrolyted presence and narute affection the degree of polymers
and living tissues swelling;
– to learn the main properties of polyelectrolytes;
– to get practical skills in the experimental determining of polyelectrolytes
isoelectrical point;
– to learn the technique of polyelectrolytes isoelectrical point determining by their
solutions viscosity change and their precipitation.
3. Self Study Section
3.1. Syllabus Content
Kinetic (sedimentation) and aggregative stability of disperse systems. The
reasons of colloidal stability. Coagulation. The mechanism of electrolytes
coagulating action. Coagulation threshold or critical concentration of coagulation.
Schulze-Hardy rule. Mutual coagulation of sols. Coagulation proceedings for the
potable water and wastewater purification.
Macromolecular compounds as the basis of living organisms. Globular and
fibrillar structure of proteins. Macromolecular solutions features and their
similarities and differences with true and colloidal solutions.
Swelling and dissolution of polymers. The mechanism of swelling. Swelling
affecting with pH, temperature and electrolytes nature. The role of swelling in the
organism physiology. Gels creation in polymers solutions. The mechanism of gels
formation. The influence of pH, temperature and electrolytes presence on the gels
formation rate. Thixotropy. Syneresis. Diffusion in gels. Salting out effect of
biopolymers. Coacervation and phase separation and its role in biological systems.
Anomalous viscosity of polymers solutions. The viscosity of the blood. Donnan
membrane equilibrium.
Isoelectric state of proteins. Isoelectric point and its determining methods. Ionic
state of biopolymers in aqueous solutions.
3.2. Overview
Disperse systems stability is the ability to keep during certain time composition
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and main properties, such as dispersion, concentration, even distribution of
dispersed phase particles in dispersion medium and way of particles interaction
unchangeable. A stable colloidal system is one in which the particles resist
flocculation or aggregation and exhibits a long shelf-life. This will depend upon the
balance of the repulsive and attractive forces that exist between particles as they
approach one another. If all the particles have a mutual repulsion then the dispersion
will remain stable. However, if the particles have little or no repulsive force then
some instability mechanism will eventually take place e.g. flocculation, aggregation
etc.
Kinetic (sedimentation) stability of a disperse system is the conservation of
particles been distributed throughout the whole volume of a system. The main
conditions for kinetic stability are high dispersion and Brownian movement
(motion). This stability increases with temperature increasing and decreasing of
particles size.
Aggregative stability of disperse systems is the ability of a system to counteract
adhesion (aggregation) of particles and keep certain degree of dispersion.
The stability of the lyophobic sols is due to the presence of charged on colloidal
particles. If, somehow, the charge is removed, the particles will come nearer to each
other to form aggregates (or coagulate) and settle down under the force of gravity.
To coagulate lyophilic sols, it is necessary to remove charge as well as the
solvent layer. This is indicated by the fact that after the addition of an electrolyte if
we also add a dehydrating agent, the coagulation sets in at once due to the removal
of water layer. Thus the stability of the colloidal solutions is mainly due to three
reasons:
− Brownian motion;
− presence of similar charge on colloidal particles;
− solvation of colloidal particles.
The process of settling of colloidal particles is called coagulation or
precipitation of the sol. Coagulation is the process of decreasing of system
dispersivity due to the dispersion phase particles enlargement.
The coagulation of the lyophobic sols can be carried out in the following ways:
− by electrophoresis: the colloidal particles move towards oppositely charged
electrodes, get discharged and precipitated;
− by mixing two oppositely charged sols: oppositely charged sols when mixed in
almost equal proportions, neutralise their charges and get partially or
completely precipitated. This type of coagulation is called mutual coagulation;
− by boiling: when a sol is boiled, the adsorbed layer is disturbed due to
increased collisions with the molecules of dispersion medium. This reduces the
charge on the particles and ultimately lead to settling down in the form of a
precipitate;
− by persistent dialysis: on prolonged dialysis, traces of the electrolyte present in
the sol are removed almost completely and the colloids become unstable and
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ultimately coagulate;
− by addition of electrolytes: When excess of an electrolyte is added, the
colloidal particles are precipitated. The reason is that colloids interact with
ions carrying charge opposite to that present on themselves. This causes
neutralisation leading to their coagulation. The ion responsible for
neutralisation of charge on the particles is called the coagulating ion. A
negative ion causes the precipitation of positively charged sol and vice versa.
The most important coagulation factor for soles is electrolytes influence.
Only certain quantity of electrolyte causes coagulation. Minimal concentration
of electrolyte in mmol that may cause 1 l of sole coagulation is called coagulation
threshold (Cthr) or critical concentration of coagulation (Ccr). The beginning of
coagulation first signs are system color changing, appearing of turbidity.
C cr =
Vel ⋅ C el ,
Vsole + Vel
where Vel – volume of the electrolyte (l), that causes coagulation; Cel –
concentration of electrolyte (mol/l), Vsole – volume of colloidal solution (l).
The value that is inverse to coagulation threshold is called coagulation capacity:
Vc = 1/Cthr.
Coagulation capacity is the volume of sol, for whose coagulation 1 mmol of the
electrolyte is needed.
Critical concentration of coagulation value is mainly determined by the valence
rather than the type of the ions with opposite charge sing to the particles. Ions with
the same charge sing than the particles are of secondary importance.
According to the Schulze-Hardy rule the coagulating ability of ion increases
with increasing of its charge, therefore the lowest critical concentration of
coagulation have electrolytes with multi-charged coagulating ion:
Сthr(K+) > Сthr (Ca2+) > Сthr (Al3+)
С thr (Cl–) > С thr (SO42–) > С thr (PO43–).
Coagulation threshold is inversely proportional to ion charge (z) raised to the
power of 6: Сthr = 1/z6.
That’s why the magnitudes of coagulation thresholds for one-, two- and threevalence ions correlate as:
1 1 1
C1 : C 2 : C 3 = 6 : 6 : 6 = 1 : 0.0156 : 0.00137 = 730 : 11.4 : 1 .
1 2 3
For inorganic ions with the same charges their coagulating ability increases with
decreasing their degree of hydratation and increasing of the ion radius.
Lyotropic, or Hofmeister Series:
coagulating ability
+
Li < Na+ < K+ < Rb+ < Cs+
coagulating ability
CI– < Br– < I– < CNS–
degree of hydratation
degree of hydratation
According to the kind of the repulsion force two mechanisms of the colloidal
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stability protection take place:
− electrostatic stabilization of colloids;
− stabilization of colloids with polymers.
Polymeric stabilization of colloids involves polymeric molecules added to the
dispersion medium in order to prevent the aggregation of the colloidal particles. The
protective action of different polymers is expressed in terms of the protective
number.
Protective number is the number of milligrams of dry protective substance
(polymer or hydrophilic colloid) which just prevents the precipitation of 10 ml of
certain sol on the addition of 1 ml of 10% NaCl solution.
A polymer molecule consists of the same repeating units, called monomers, or of
different but resembling units. Figure 1 shows an example of a vinyl polymer, an
industrially important class of polymer. In the repeating unit, X is one of the
monofunctional units such as H, CH3, Cl, and C6H5 (phenyl). The respective
polymers would be called polyethylene, polypropylene, polyvinyl chloride), and
polystyrene. A double bond in a vinyl monomer CH2=CHX opens to form a
covalent bond to the adjacent monomer. Repeating this polymerization step, a
polymer molecule is formed that consists of n repeating units. We call n the degree
of polymerization (DP). Usually, n is very large. It is not uncommon to find
polymers with n in the range of 104–105.
H H
Figure 1. Vinyl polymer
C C
H X
n
Fig. 2 shows three architectures of a polymer molecule: a linear chain (a), a
branched chain (b), and a cross-linked polymer (c). A bead represents a monomer
here. A vinyl polymer is a typical linear polymer. A branched chain has branches,
long and short. A cross-linked polymer forms a network encompassing the entire
system. In fact, there can be just one supermolecule in a container. In the branched
chain, in contrast, the branching does not lead to a supermolecule. A cross-linked
polymer can only be swollen in a solvent. It cannot be dissolved. We will learn
linear chain polymers in detail and about branched polymers to a lesser extent.
Figure 2. Architecture of polymer chain: a linear chain (a), a branched chain (b),
and a cross-linked polymer (c).
Some polymer molecules consist of more than one kind of monomers. An A–B
copolymer has two constituent monomers, A and B. When the monomer sequence
is random, i.e., the probability of a given monomer to be A does not depend on its
neighbor, then the copolymer is called a random copolymer.
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Two main reactions are used for polymers preparation:
− Polymerization: monomers' double-bonds open up to form continuous chain
− Condensation: elimination of smaller molecule when functional groups react.
Solutions of a high-molecular-weight polymer, even at low concentrations, can
flow only slowly. Addition of a small amount of the polymer to the fluid can make it
viscous, thereby preventing unwanted turbulence in the flow.
When a polymer is added to a given solvent, attraction as well as dispersion
forces begin acting between its segments, according to their polarity, chemical
characteristics, and solubility parameter. If the polymer-solvent interactions are
higher than the polymer-polymer attraction forces, the chain segment start to absorb
solvent molecules, increasing the volume of the polymer matrix, and loosening out
from their coiled shape. We say the segments are now "solvated" instead of
"aggregated", as they were in the solid state.
The whole "solvation-unfolding-swelling" process takes a long time, and given it
is influenced only by the polymer-solvent interactions, stirring plays no role in this
case. However, it is desirable to start with fine powdered material, in order to
expose more of their area for polymer-solvent interactions.
When crystalline, hydrogen bonded or highly crosslinked substances are
involved, where polymer-polymer interactions are strong enough, the process does
stop at this first stage, giving a swollen gel as a result.
If on the contrary, the polymer-solvent interactions are still strongly enough, the
"solvation-unfolding-swelling" process will continue until all segments are solvated.
Thus, the whole loosen coil will diffuse out of the swollen polymer, dispersing into
a solution. At this stage, the disintegration of the swollen mass can be favored by
stirring, which increases the rate of dissolution.
The degree of swelling is thus used as a common measure of the degree of crosslinking. The degree of swelling is measured as follows:
1. Measure the dry weight and/or volume of the polymer before swelling.
2. Swell the polymer in a solvent till equilibrium swelling would be achieved.
3. Measure the weight of the swollen polymer and/or volume of the polymer
after equilibrium swelling.
4. The degree of swelling is measured using the following two swell ratios:
Mass swell ratio i = m − m0 ⋅ 100%,
m0
Volume swell ratio
V − V0
i=
V0
⋅ 100%,
where m0 and V0 are the initial mass nd volume of a polymer sample,
m and V are its mass and volume after the swelling.
The swell ratio is influenced with various factors: the nature of the polymer and
the solvent, temperature, addition of electrolytes and pH. The latter variable is of
the greatest importance. Electrolytes affecting is based on the anions hydratation
ability. Some anions have the low ability for hydratation and can enhance the
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polymers swelling while other anions being well-hydrated would reduce the
swelling.
Lyotropic series of anions:
> Cl– >
CH3COO– > SO42– > C2O42–
CNS– > I– > KBr– > NO3–
no influence
on swelling swelling reducing
swelling enhancing
Proteins are polyelectrolytes whose macromolecules contain acid and basic
functional groups in the amino acids side chains. Amino acids in aqueous solutions
are in the form of bipolar ions:
H2N–R–COOH ⇄ +H3N–R–COO–.
In acidic medium, when the ionization of carboxylic groups decreases, the
protein molecule shows basic properties and gains positive charge:
+
H3N–R–COO– + H+ ⇄ +H3N–R–COOH.
In basic medium vice versa the ionization of amino groups decreases, the protein
molecule shows acidic properties and gains negative charge:
+
H3N–R–COO– + OH– ⇄ H2N–R–COO– + H2O.
So proteins are amphoteric molecules while changes in pH value of the medium
influence whether protein will be positively or negatively charged. Proteins net
charge is the sum of all positive and all negative charges of amino acid side chains
and of amino and carboxyl groups. Isoelectric point (pI) is characteristic pH value
at which proteins net charge is zero. Proteins are positively charged at pH values of
solution below their pI value and negatively charged when pH value of solution is
higher then their pI value.
When there is no electrical field, ampholytes are randomly distributed. By
applying electrical field, ampholites start to move to one of the electrodes. For
example, acidic ampholitic molecules with the lowest pI values will be most
negatively charged and thus will migrate to anode until their net charge reaches
zero. They will concentrate in that region. Since each ampholyte is a good buffer pH
value of the medium is the sum of pI values of ampholites. After concentration of
ampholytes is finished, steady state is reached and there is constant gradient of pH
throughout the gel. If proteins’ pI values are inside the range of pH covered by
ampholites which are used to construct pH gradient, when applied on gel, they will
migrate because of their charge until they reach zone with pH value identical to their
pI (because at that pH value they have no net charge). This means that proteins with
positive net charge will migrate to cathode and will become less and less charged
until they reach their pI. Proteins with net negative charges will under the influence
of an electrical field migrate across pH gradient to anode; again, until their pI value
is reached. Because of this, in the focusing step, concentration of proteins happens
according to their pI values.
The particular behavior shown by polymer molecules, explains the high viscosity
of polymer solutions. Solvent and low molecular weight solutes have comparable
molecular size, and the solute does not swell when dissolving. Since molecular
mobility is not restricted, and therefore intermolecular friction does not increase
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drastically, the viscosity of the solvent and the solution are similar. But the
molecular size of polymer solutes is much bigger than that of the solvent. In the
dissolution process such molecules swell appreciably, restricting their mobility, and
consequently the intermolecular friction increases. The solution in these cases,
becomes highly viscous.
3.3. References
46. V.O. Kalibabchuk, V.I. Halynska, L.I. Hryshchenko et al. Medical
Chemistry. – AUS MEDICINE Publishing. – 2010. – 224 p.
47. Raymond Chang. Chemistry (6th Edition). – WCB/McGraw-Hill. – 1998. – 995
p.
48. Rodney J. Sime. Physical Chemistry. Methods. Techniques. Experiments. –
Saunders College Publishing. – 1990. – 806 p.
49. John McMurry, Robert C. Fay. Chemistry (3rd Edition). – Prentice Hall. – 2001.
– 1067 p.
50. David E. Goldberg. Fundamentals of Chemistry (2nd Edition). – WCB/McGrawHill. – 1998. – 561 p.
51. Theodore L. Brown, H.Eugene LeMay, Bruce E. Bursten. Chemistry. The
Central Science. – Prentice Hall. – 2000. – 1017 p.
52. John Olmsted III, Gregory M. Williams. Chemistry. The Molecular
Science. – Mosby. – 1994. – 977 p.
53. Steven S. Zumdahl. Chemistry (4th Edition). – Houghton Mifflin Company. –
1997. – 1031 p.
54. Gary L. Miessler, Donald A. Tarr. Inorganic Chemistry. – Prentice Hall. – 1991.
– 625 p.
3.4. Self Assessment Exercises
а) Review Questions
1. Define the reasons of colloidal stability. Kinetic (sedimentation) and aggregative
stability of disperse systems.
2. Give the definition of the coagulation process. List the factors which may cause
the coagulation of sols.
3. Define the coagulation threshold (critical concentration of coagulation).
4. State Schulze-Hardy rule. Explain it on some examples.
5. Explain the mutual coagulation of sols, give a few examples.
6. Explain the protective action of polymers or hydrophilic colloids towards
hydrophobic sols. Define the term of the protective number.
7. Describe the protective action significance and applications in biology,
medicine, pharmacy.
8. Explain the terms polymer and monomer.
9. What are natural and synthetic polymers? Give two examples of each type.
10. How can you differentiate between addition and condensation polymerisation?
11. Discrete polymer’s dissolution steps.
123
12. Swelling and dissolution of polymers. Swelling mechanism and stages. Types of
swelling.
13. Swell ratio and its affection with defferent variables.
14. Explain the nature of electrolytes affecting the swelling ratio.
15. Common and distinctive features of polymers solutions, true solutions and
lyiphobic sols.
16. Explain the mechanism of gels formation in polymers solutions. What factors do
influence gels formation?
17. Polyelectrolytes. Amphoteric properties of proteins. The charge sign of a ptotein
macromolecule as a function of pH value of the medium.
18. Isoelectric point of proteins and its determining methods.
19. Properties of polymers solutions (viscosity, osmotic pressure, thixotropy).
20. Explain the salting out effect of biopolymers. Define coacervation as the process
of phase separation and its role in biological systems.
b) Types of Numerical Problems and Their Solving Strategies
Numerical problem 1. Calculate the critical concentration of coagulation for AgI
sol, if 0.4 ml of 0.05 M BaCl2 solution should be added for
2 ml of sole coagulation.
Steps to Solution:
1. Calculate the number of BaCl2 moles in 0.4 ml of 0.05 М solution:
n = 0.05 mol/l · 0.4·10–3 l = 0.02·10–3 mol = 0.02 mmol.
2. Calculate the volume of solution after coagulation:
V = 2 + 0.4 = 2.4 ml= 2.4·10–3 l.
3. Calculate the critical concentration of coagulation – the minimal number of
BaCl2 mmol which is necessary for coagulation of 1 l of sol:
Ccr =
n 0.02 mmol
=
= 8.3 mmol/l
V
2.4 ⋅ 10 − 3 l
Answer: the critical concentration of coagulation is 8.3 mmol/l.
Numerical problem 2. How many millilitres of 1 M NaCl solution should be added
to 500 ml of Fe(OH)3 sol to cause its coagulation if the
critical concentration of coagulation is 104·10–3 mol/l?
Steps to Solution:
From the equation for critical concentration of coagulation:
n
V ⋅C
Ccr = el = el el
V
Vsol + Vel
→
Vel
С
104 ⋅ 10 −3 mol/l
= cr =
= 104 ⋅ 10 −3 .
Vsol + Vel С el
1 mol/l
Calculate the volume of NaCl solution solving the following equation:
Vel = 104 ⋅ 10 −3 ⋅ (Vsol + Vel ) , Vel − 104 ⋅ 10 −3Vel = 104 ⋅ 10−3 ⋅ Vsol ,
52
0.896Vel = 52 , Vel =
= 58 ml .
0.896
124
Answer: the volume of 1 M NaCl solution is 58 mmol/l.
Numerical problem 3. How many times the coagulation threshold of As2S3 sol will
decrease if 0.5 М NaCl solution firstly been used for 10 ml
of sol coagulation (1.2 ml were used for the coagulation)
would be substituted with 0.036 М MgCl2 solution (0.4 ml
were needed) or 0.001 М AlCl3 solution (0.1 ml were
needed) for the same volume of sol (10 ml) coagulation.
Steps to Solution:
1. Calculate the number of mmoles of each electrolyte needed for 10 ml of Ag2S3
sol coagulation:
n1(NaCl) = С(NaCl)⋅V(NaCl) = 0.5 ⋅ 1.2 = 0.6 mmols NaCl;
n2(MgCl2) = С(MgCl2)⋅V(MgCl2) = 0.036 ⋅ 0.4 = 0.0144 mmols MgCl2;
n3(AlCl3) = С(AlCl3)⋅V(AlCl3) = 0.01 ⋅ 0.1 = 0.001 mmols AlCl3.
2. Calculate the total volumes of obtained solutions containing sol and coagulating
electrolyte (V):
V1 = Vsol + V(NaCl) = 10 + 1.2 = 11.2 ml = 11.2⋅10–3 l;
V2 = Vsol + V(MgCl2) = 10 + 0.4 = 10.4 ml = 10.45⋅10–3 l;
V3 = Vsol + V(AlCl3) = 10 + 0.1 = 10.1 ml = 10.1⋅10–3 l.
3. Calculate the coagulation threshold values for each electrolyte:
n
0.6
Сcr1 = 1 =
= 53.6 mmol/l NaCl;
V1 11.2 ⋅ 10 − 3
Сcr2 =
Сcr3 =
n2
V2
n3
= 0.0144 = 1.4 mmol/l MgCl2;
10.4 ⋅ 10 −3
=
V3
0.001
10.1 ⋅ 10 − 3
= 0.1 mmol/l AlCl3.
4. How many times the coagulation threshold values of As2S3 sol will be lower for
MgCl2 and AlCl3 than for NaCl:
Ccr1
Ccr 2
Ccr1
Ccr 2
=
53.6 = 38.3 times for MgCl ;
2
1.4
=
53.6
= 536 times for AlCl3.
0 .1
Numerical problem 4. A sol of AgCl was prepared by adding of 10 ml of 0.03 %
NaCl solution to 25 ml of 0.002 M AgNO3 solution. Which
electrolyte through the listed ones will have the lowest
value of the critical concentration of coagulation for this
sol: KBr, Ba(NO3)2, K2CrO4, MgSO4, AlCl3, K3PO4?
125
Steps to Solution:
1. Calculate the molar concentration of 0.03 % NaCl solution, accepting the
density being equal 1 g/ml and М(NaCl) = 58.5 g/mol using the following
equation:
ω ⋅ d ⋅ 10 0.03 ⋅ 1 ⋅ 10 = 0.00513 mmol/ml.
СM =
M ( NaCl)
=
58.5
2. Calculate the volume of mixed solutions:
V = V1 + V2 = 10 + 25 = 35 ml.
3. Calculate the number of mmoles of each reactant:
n1(NaCl) = С1⋅V1 = 0.00513 ⋅ 10 = 0.0513 mmol;
n2 (AgNO3) = С2⋅V2 = 0.001 ⋅ 25 = 0.025 mmol.
4. According to the equation of reaction:
NaCl + AgNO3 → AgCl + NaNO3
0.0513
0.025
NaCl is in excess, so it is the stabilizer of the sol AgCl. The formula of the micelle
is:
{[m(AgCl) ⋅ nCl–]n– ⋅ (n – x) Na+}x– ⋅ xNa+.
+
2+
5. Cations K , Ba , Mg2+, Al3+ will cause the coagulation of this sol. According to
the Schulze-Hardy rule the coagulating ability of ion increases with increasing of
its charge. Therefore, the lowest value of critical concentration of coagulation
has AlCl3. KBr, K2CrO4 and K3PO4 solutions of the same normality have the
least coagulating ability.
Numerical problem 5. Calculate the protective number and the coagulation
threshold for Al2S3 sol with negatively charged granules
if 3 ml of 0.5 % dextrin aqueous solution were added to
5.5 ml of the sol and the first signs of coagulation were
observed with 1.5 ml of 2.5 M NaCl solution adding.
Steps to Solution:
1. Calculate the mass of dextrin in 3 ml of 0.5 % diluted solution (accepting its
density being 1 g/ml):
3 ⋅ 0 .5
= 0.015 g = 15.0 mg of dextrin.
100
2. Calculate the mass of dextrin needed for its protective action toward 10 ml of sol
instead of 5.5 ml:
for 5.5 ml of sol protection 15 mg of dextrin are needed
for 10 ml
–
х mg
15 ⋅ 10
х=
= 27.3 mg.
5 .5
3. Find out the mass of NaCl in 1.5 ml (0.0015 l) of its 2.5 M solution, the molar
mass of NaCl is 58.5 g/mol:
126
m(NaCl) = CΜ⋅V(NaCl)⋅M(NaCl) = 2.5 ⋅ 0.0015 ⋅ 58.5 = 0.219 g NaCl.
4. Calculate the protective number that is the mass of dextrin (in mg) needed for
just preventing the precipitation of 10 ml of sol on the addition of 1 ml of 10%
NaCl solution (or ≈ 0.1 g of NaCl):
for 0.219 g of NaCl
23.3 mg of dextrin re needed,
for 0.1 g
–
х mg
27.3 ⋅ 0.1
x=
= 12.5 (mg of dextrin).
0.219
5. Calculate the coagulation threshold for given sol:
C ⋅V1
Сcr =
,
V
where:
С – the initial concentration of the coagulating electrolyte, mol/l;
V1 – the volume of electrolyte needed for the coagulation, ml;
V – the total volume of solution, ml .
2.5 ⋅ 1.5
C ⋅ V1
Сcr =
=
= 0.375 mol/l.
Vsol + Vdextrin + V1
5.5 + 3.0 + 1.5
Numerical problem 6. A natural rubber sample with the volume of 1.094⋅10–4 m3
(V0) was placed into carbon disulfide. After 48 hours
exposure at 293 K the sample volume becomes 9.204⋅10–4
m3 (V). Calculate the volume swell ratio of the polymer (%).
Steps to Solution:
The volume swell ratio of a polymer may be calculated by the equation:
α=
α=
9.204 ⋅ 10 −4 − 1.094 ⋅ 10 −4
1.094 ⋅ 10
−4
V − V0
∆V
⋅ 100 % =
⋅ 100 %
V0
V0
⋅ 100 % =
8.11⋅ 10 −4
1.094 ⋅ 10 −4
⋅ 100 % = 7.4132 ⋅ 100 % = 741.32 %
Numerical problem 7. Calculate the molecular weight of ethylcellulose (in kg/mol)
if its solution in aniline with the concentration of 4.9 kg/m3
has the osmotic pressure magnitude as 250.9 Pa. The second
virial coefficient b2 is 0.52 Pа⋅m6/kg2.
Steps to Solution:
Osmotic pressure is a colligative property, which means that it is proportional to
the concentration of solute. The van’t Hoff equation is often presented in
introductory chemistry for calculating osmotic pressure from the moles of solute
(nsolute) that occupy a given volume (V) and the absolute temperature (T) of the
solution:
π=
nsolute ⋅ R ⋅ T (1)
V
Since all non-volatile, non-electrolytic solutions approach ideal behavior in the
dilute limit, equation (1) is actually a limiting law, and should be written in the
127
form:
lim
nsolute →0
n
⋅ R ⋅T
π = solute
V
(2)
It is more convenient to express concentration in terms of the 'grams' of solute
per liter (or kg/m3), whereby we can make the following substitution in equation (1):
nsolute C ,
=
V
M
where M is the molecular weight of the solute. In this fashion, and with minor
rearrangement, equation (2) can be written as:
C ⋅ R ⋅ T (3)
lim π =
nsolute →0
M
Since equation (3) is only exact in the dilute limit, we can recognize this
relationship as the first term in a more general power series expansion in C:
C ⋅ R ⋅T
π=
+ b2 C 2 + b3C 3 + ... (4)
M
where b2 and b3 are called the second and third virial coefficients, respectively.
These coefficients are empirically determined constants for a given solute-solvent
system, and also depend on temperature. According to statistical mechanical
solution theory, b2 represents the interaction of a single solute particle with the
solvent, and higher order virial coefficients are associated with correspondingly
larger number solute particle cluster interactions with the solvent.
For the given problem solving we’ll use the equation (4) just with the second
virial coefficient:
π=
C ⋅ R ⋅T
+ b2 C 2
M
where: π is the osmotic pressure of solution, Pa;
С– the polymer solution concentration, kg/m3;
R = 8.314 J/mol⋅К; T– temperature, К;
M – the molecular weight of the polymer.
CRT
M
4.9 ⋅ 8.314 ⋅ 313
π − b2 C 2 =
M =
CRT
π − b2 C 2
=
250.9 − 0.52 ⋅ 4.9 2
= 53.5.
Answer: the molecular weight of ethylcellulose is 53.5.
Numerical problem 8. Blood serum albumin (рНpI = 4.64) was placed into the
buffer solution with the hydrogen ions concentration [H+] =
10–6 mol/l. Towards which electrode the albumen perticles
will migrate under electrophoresis?
Steps to Solution:
1. Calculate the pH value of the buffer solution in which [H+] = 10–6 mol/l:
pH = –log [H+] = –log10–6 = 6.
128
2. As pH of the buffer is higher than pHpI, then the albumen particles would gains
negative charge:
+
NH3– R–COO– + OH– → NH3OH–R–COO–.
That is why the albumen particles will migrate in the direction of positively
charged electrode – anode during the electrophoresis.
Answer: albumen particles will migrate towards anode.
Numerical problem 9. The efflux time for the egg albumin solution with the
density of 1200 kg/m3 at 293 K from Ostwald viscometer is
27 sec. while the efflux time for the distilled water is 15
sec. Calculate the kinematic viscosity of the albumin
solution if the density of water is 1000 kg/m3, water
viscosity being equal 1.005⋅10–3 Pa⋅sec.
Steps to Solution:
The viscosity ηx of a solution (density dx) may be obtain the from the
measurement of the efflux time τx, if the efflux time τ0 of a liquid with a known
viscosity η0 and density d0 is measured for the same viscometer:
d ⋅τ
η =η x x ,
x
0
d0 ⋅ τ 0
where: η0 and ηx – kinematic viscosities of water and solution, respectively, Pa⋅sec;
τ0 and τx – efflux times for water and solution, sec.;
d0 and dx – the densities of water and solution, kg/m3.
η x = 1.005 ⋅ 10 −3 ⋅
27 ⋅ 1200
= 2.17 ⋅ 10 −3 Pa⋅sec.
15 ⋅ 1000
Answer: the kinematic viscosity of the albumin solution is 2.2⋅10–3 Pa⋅sec.
c) Problems to Solve
1. How many times the coagulation threshold of AgI sol will decrease if 1 М КNO3
solution firstly been used for 10 ml of sol coagulation (1.5 ml were used for the
coagulation) would be substituted with 0.1 М Сa(NO3)2 solution (0.5 ml were
needed) or 0.01 М Al(NO3)3 solution (0.2 ml were needed) for the same volume
of sol (10 ml) coagulation.
Answer: the coagulation threshold would decrease 27.4 times and 665.2 times,
respectively
2. The coagulation of Al2S3 sol with negatively charged granules was carried out
by the following electrolytes adding: KNO3, MgCl2 і AlCl3. Their critical
concentrations of coagulation are 50.0, 0.72, and 0.093, respectively. Calculate
the coagulation capacities ration for the cations with the different valences.
Answer: 1 : 69.4 : 537.6
129
3. Calculate the protective number for Fe(OH)3 sol if 5 mg of powder starch were
added to 2 ml of the sol and the first signs of coagulation were observed with 0.2
ml of 10 % NaCl solution adding.
Answer: 125 mg
4. Gelatine with рНpI of 4.7 is placed into the solution with the hydrogen ions
concentration [H+] 1000 times greater than in water. What charge gelatine
particles will gain in this solution?
Answer: positive charge
5. The efflux time for gelatine solution with the density of 1050 kg/m3 at 298 K
from Ostwald viscometer is 25 sec. while the efflux time for the distilled water is
10 sec. Calculate the kinematic viscosity of the gelatine solution if the density of
water is 977.04 kg/m3, water viscosity being equal 0.8937⋅10–3 Pa⋅sec.
Answer: 2.6 Pa⋅sec.
6. Calculate the efflux time of nitro cellulose solution from Ostwald viscometer.
The density of the solution is 1120 kg/m3, its kinematic viscosity being 1.52⋅10–3
Pa⋅sec. The efflux time for water of the same volume is 12 sec. The kinematic
viscosity of water is 1.11⋅10–3 Pa⋅sec, its density is 998.94 kg/m3.
Answer: 14.7 sec.
4. Laboratory Activities and Experiments Section
4.1. Practical Skills and Suggested Learning Activities
– to improve the skills of micelles structure composition for sols with different
charges of granules;
– to get practical skills in the critical threshold calculations;
– to state and interprete Schulze-Hardy rule;
– to measure experimentally the swelling degree of a polymer sample;
– to learn the electrolytes nature affection the polymer swell ratio;
– to determine gelatine pI and pHpI by swelling;
– to learn the pH value affection the viscosity of a protein solution. To determine
gelatine pI and pHpI as the minimum of viscosity;
– to determine gelatine pI and pHpI as its maximal precipitation.
4.2. Experimental Guidelines
4.2.1. The coagulation threshold determining for iron (III) hydroxide sol
Iron (III) sol Fe(OH)3 with positively charged granules was prepared in advance
under the reaction of FeCl3 hydrolysis. Compose the micelle structure for this sol.
For Fe(OH)3 sol coagulation electrolytes with the different valence anions are
used, for example KCl, K2SO4, K3PO4 or K3[Fe(CN)6] their concentrations in mol/l
are: 2.5; 0.1; 0.01, respectively.
In three test tubes pour with the pipette 5 ml of the Fe(OH)3 sol each. The
respective electrolyte solution should be added from burettes drop be drop into each
test-tube until the first signs of coagulation would be observed: the turbidity
130
appearance, color change (darkening), precipitation. Record the experimental data
into the table below.
Table 1
The experimental data obtained while coagulation threshold determining
Test
tube
No
Electrolyte
added for
coagulation
The electrolyte
Electrolyte
solution used
solution
Coagulating ion
for the
concentration and its valence
coagulation V,
С, mol/l
ml
Coagulation
threshold
Сcr =
C el ⋅ Vel
Vsol +V el
1.
2.
3.
Make the conclusions which of electrolytes and ions used have the highest
coagulating ability. Explain why. Calculate the coagulation capacities ration for the
anions with the different valences.
4.2.2. The protective action of a polymer towards hydrophobic sol coagulation
studying
In four test tubes pour with the pipette 2 ml of Fe(OH)3 sol with positively
charged granules each. Study the coagulation of sol in two test tubes causing it by
two electrolytes adding with the anions of the different valence (two and three,
respectively). The respective electrolyte solution should be added from burettes
drop be drop into each test-tube until the first signs of coagulation would be
observed. To the test tubes No 3 and 4 add firstly 1 ml of 0.5 % polymer solution
(gelatine, starch etc.) and then study their coagulation in the same way as for the test
tubes No 1 and 2. Compare the volumes of expended electrolytes solutions. Record
the experimental data into the table below.
Table 2
The experimental data for the polymer protective action studying
Electrolyte
Concentratio
n of the
electrolyte
solution
С, mol/l
Used for coagulation
V, ml
without a
in the
polymer
presence of
polymer
Coagulation threshold
Ccr, mmol/l
without a
in the
polymer
presence of
polymer
Calculate the coagulation threshold values for each test tube by the equation:
Сcr =
Cel ⋅ Vel
Vsol + Vpolymer + Vel ,
where: Сel – concentration of the electrolyte, mol/l;
Vsol, Vpolymer, Vel – the volumes of a sol, polymer and electrolyte, respectively, l.
Explain the obtained results and make the conclusion about the protective action
of polymer towards sol coagulation.
4.2.3. Mutual coagulation of sols
131
Pour 2 ml of Fe(OH)3 sol with positively charged granules into the test tube
(compose the micelle structure for this sol if FeCl3 was used as a nucleating agent).
Add 2 ml of Berlin blue sol with negatively charged granules. Compose the micelle
structure for the obtained sol with K4[Fe(CN)6] being used as stabilizer. What
process was observed after the two sols were mixed? Record and explain your
observation.
4.2.4. Swell ratio of gels determining with the mass-volumetric method
Fill the vials of the device for swelling experiments (Fig.3) with pure water so
that when you turn the device enclosed vial down the water level reaches about mid
graduated tube. Record the level (V0).
Figure 1. The device for the swell ratio of
gels determining with the mass-volumetric
method
Then introduce a piece of dry gelatin (or other gel) with the mass m tied to
filament into the opened vial, shut the stopper, carefully turn the device into a
horizontal position and mark the swelling starting time. Every 15 minutes within 1
hour again pour the liquid into the enclosed vial and determine the loss of water
volume recording its level on the graded tube (V). Calculate the swelling degree of
gel (і).
Table 3
Experimental data of swell ratio determining
Time from
No of
The initial level The loss of water
the, τ ,
of water, V0, ml volume V, ml
record
minutes
1
2
3
4
5
The solvelt volume
been swollen with
the gel sample
during the time τ
∆V=(V0–V), ml
Swell ratio
і =∆V/m,
ml/g
0
15
30
45
60
Draw the plot of swell ratio as a function of time. Make the conclusion about the
type of swelling more typical for this polymer at a room temperature.
4.2.5. Determining the electrolytes nature affection swell ratio of gelatine
Measure 10-15 ml of the following electrolytes 0.1 M solutions: K2SO4,
CH3COOK, KCl, KI, KCNS. Pour each into one of five marked vials. Weigh five
gelatine slabs, record the masses. Place the slabs into the vials and leave for 1 hour.
Then take out the gelatine slabs, dry gently with filter paper and reweigh.
Record the experimental data into the table and calculate mass swell ratio for
132
each electrolyte solution.
Table 4
Experimental data of electrolytes nature affection swell ratio of gelatine
No of
vial
Gelatine mass, g
Electrolyte
before the
swelling, m0
after the
swelling, m
The solvelt mass
been swollen with
the gel sample, g
∆m = m – m0
Mass swell
ratio
і = ∆m/m0
Make the conclusion about the nature of anions affecting gelatine swelling.
Write down the Lyotropic series of anions and explain which anions can enhance
the polymers swelling while other anions reduce the swelling.
4.2.6. Gelatine pI and pHpI by swelling determining
Pipette 10 ml of one of the acetate buffer solutions with different pH each into
one of five marked vials. Weigh five gelatine slabs, record the masses (mo, g). Place
the slabs into the vials and leave for 1 hour. Then pour out the solutions, dry
gelatine slabs gently with filter paper and reweigh (m, g).
Record the experimental data into the table and calculate mass swell ratio for
each buffer solution.
Table 5
Experimental data of pH affection swell ratio of gelatine
No of
vial
Gelatine mass, g
before the
after the
swelling, m0 swelling, m
The solvelt mass been swollen
with the gel sample, g
∆m = m – m0
Mass swell ratio
і = ∆m/m0
1
2
3
4
5
Draw the plot of swell ratio і (Y-axis) as a function of pH (x-axis). Determine pI
and pHpI for gelatine using the plot. Make the conclusion about the swell ratio
magnitude of a polymer in its isoelectric point.
4.2.7. The pH value affection the viscosity of a protein solution. Gelatine pI and
pHpI as the minimum of viscosity determining
Pipette 10 ml of one of the acetate buffer solutions with different pH: 3.7; 3.9;
4.2; 4.7; 5.2 and pour each into one of five marked test tubes. Add 5 ml of 3 %
gelatine solution into each tst tube and mix.
Determine the viscosity of solutions with Ostwald viscometer.
133
Figure 2. Capillary Ostwald viscometer
1 – container for measuring the volume of liquid flowing
through the capillary;
2 – capillary;
3 – container for the liquid collecting.
To measure the efflux time of solvent (τ0) and gelatine solution (τх) from
capillary Ostwald viscometer using stopwatch. Firstly the efflux time for water
should be fixed, then the efflux time for each solution. The time measurements
should be performed three times for each case and the average value for each
solution should be recorded. Calculate the relative viscosity for each solution by the
equation:
τ
η x = η0 x ,
τ0
where:η0– the relative viscosity of water at given temperature (1⋅10–3 Pa⋅sec at 298
K);
ηх – the relative viscosity of the analyte solution.
Record the measurements into the table.
Table 6
Experimental data for the relative viscosity determining
No of the test tube
pH of solution
Efflux time τ, sec.
Relative viscosity
1
3.7
2
3.9
3
4.2
4
4.7
5
5.2
Н2О
Plot the relative viscosity values against pH. What pH corresponds to the
isoelectric point? Explain the effect of pH on the viscosity of protein solutions.
Conclusions and Interpretations. Lesson Summary
134
Appendixes
Appendix A
∆
8
9
02
S
,
,
kJ/mol
,
J/(mol·К)
kJ/mol
G
8
9
02
∆
8
9
02
Substances
H
Standard Thermodynamical Functions or Some Substances at 298 К
0
28.3
0
–1676.0
50.9
–1582.0
0
5.7
0
–135.4
214.4
–64.6
СО (g)
–110.5
197.5
–137.1
CO2 (g)
–393.5
213.7
–394.4
СаСО3 (s)
–1207.0
88.7
–1127.7
СаF2 (s)
–1214.6
68.9
–1161.9
Ca3N2 (s)
–431.8
105.0
–368.6
СаО (s)
–635.5
39.7
–604.2
Са(ОН)2 (s)
–986.6
76.1
–896.8
0
222.9
0
Сl2O (g)
76.6
266.2
94.2
СlO2 (g)
105.0
257.0
122.3
Сl2O7 (l)
251.0
–
–
Cr2O3 (s)
–1440.6
81.2
–1050.0
CuO (s)
–162.0
42.6
–129.9
Al (s)
Al2O3 (s)
С (graphite)
ССl4 (l)
Cl2 (g)
Fe (s)
FeO (s)
0
27.2
0
–264.8
60.8
–244.3
0
130.5
0
–36.3
198.6
–53.3
HCN (g)
135.0
113.1
125.5
HCl (g)
–92.3
186.8
–95.2
HF (g)
–270.7
178.7
–272.8
HI (g)
26.6
206.5
1.8
HN3 (l)
294.0
328.0
238.8
H2O (g)
–241.8
188.7
–228.6
H2O (l)
–285.8
70.1
–237.3
H2S (g)
–21.0
205.7
–33.8
H2 (g)
HBr (g)
135
∆
8
9
02
S
,
,
kJ/mol
,
J/(mol·К)
kJ/mol
G
8
9
02
8
9
02
∆
H
Substances
KCl (s)
–435.9
82.6
–408.0
KClO3 (s)
–391.2
143.0
–289.9
MgCl2 (s)
–641.1
89.9
–591.6
Mg3N2 (s)
–461.1
87.9
–400.9
MgO (s)
–601.8
26.9
–569.6
0
191.5
0
HN3 (l)
294.0
328.0
238.8
NH3 (g)
–46.2
192.6
–16.7
NH4NO2 (s)
–256.0
–
–
NH4NO3 (s)
–365.4
151.0
–183.8
82.0
219.9
104.1
NO (g)
90.3
210.6
86.6
N2O3 (g)
83.3
307.0
140.5
NO2 (g)
33.5
240.2
51.5
N2O4 (g)
9.6
303.8
98.4
N2O5 (g)
–42.7
178.0
114.1
NiO (s)
–239.7
38.0
–211.6
О3 (g)
142.3
237.7
163.4
O2 (g)
0
205.0
0
–1492.0
114.5
–1348.8
PbO (s)
–219.3
66.1
–189.1
PbO2 (s)
–276.6
74.9
–218.3
N2 (g)
N2O (g)
P2O5 (s)
0
31.9
0
SO2 (g)
–296.9
248.1
–300.2
SO3 (g)
–395.8
256.7
–371.2
SiH4 (g)
34.7
204.6
57.2
SiO2 (quartz)
–910.9
41.8
–856.7
SnO (s)
–286.0
56.5
–256.9
SnO2 (s)
–580.8
52.3
–519.3
S (s)
0
30.6
0
TiCl4 (l)
–804.2
252.4
–737.4
TiO2 (s)
–943.9
50.3
–888.6
WO3 (s)
–842.7
75.9
–763.9
ZnO (s)
–350.6
43.6
–320.7
Ti (s)
136
Appendix B
∆
8
9
02
S
,
,
kJ/mol
,
J/(mol·К)
kJ/mol
G
8
9
02
∆
8
9
02
Formule and State
H
Standard Thermodynamical Functions or Some Organic
Substances at 298 К
СН4 (g)
–74.9
186.2
–50.8
С2Н2 (g)
226.8
200.8
209.2
С2Н4 (g)
52.3
219.4
68.1
С2Н6 (g)
–89.7
229.5
–32.9
С6Н6 (l)
49.0
124.5
172.8
СН3ОН (l)
–238.7
126.8
–166.3
С2Н5ОН (l)
–277.6
160.7
–174.8
С3Н8О3 (l)
–669.1
204.6
–479.4
СН3СООН (l)
–484.4
159.9
–389.6
СО(NH2)2 (s)
СО(NH2)2 (l)
–333.0
–317.7
104.7
175.7
–196.9
–202.7
–1273.0
–1263.1
–2220.9
–2215.8
212.1
264.0
360.2
403.8
–910.5
–914.5
–1544.3
–1551.4
С6Н12О6 (s)
С6Н12О6 (l)
С12Н22О11 (s)
С12Н22О11 (l)
Appendix C
Instability constants of complexes ions at 25 °С
Complex ion
–
[Ag(NO2)2]
–
[Ag(CN)2]
+
[Ag(NH3)2]
3–
[Ag(S2O3)2]
2–
[Cd(CN)4]
2+
[Cd(NH3)4]
2–
[Co(CNS)4]
2+
[Co(NH3)6]
–
[Cu(CN)2]
3–
[Cu(CN)4]
2+
[Cu(NH3)4]
4–
[Fe(CN)6]
β
Complex ion
–3
1,8·10
–21
1,0·10
–8
5,9·10
–13
1,00·10
–18
7,66·10
–8
7,5·10
–3
5,50·10
–5
4,07·10
–24
1,00·10
–31
5,13·10
–13
9,3·10
–37
1,00·10
137
3–
[Fe(CN)6]
2–
[HgCl4]
2–
[Hg(CN)4]
2–
[Hg(SCN)4]
2–
[HgI4]
2–
[Ni(CN)4]
2+
[Ni(NH3)6]
2–
[PbI4]
2–
[Zn(CN)4]
2–
[Zn(CNS)4]
2+
[Zn(NH3)4]
2–
[Zn(OH)4]
β
–42
1,00·10
–16
6,03·10
–42
3,02·10
–22
1,29·10
–30
1,38·10
–22
1,00·10
–9
9,77·10
–5
9·10
–16
1,00·10
–2
5,00·10
–9
2,00·10
–16
7,08·10
Appendix D
The main Half-Reactions and Standard Red-Ox Potentials Values
Half-Reaction
Oxidized Form
nе
+
H2O2 + 2H
+
PbO2 + 4H
–
+
MnO4 + 8H
–
+
ClO +2H
–
+
–
+
ClO3 + 6H
ClO4 + 8H
Cl2
0
Cr2O7
2–
+
+ 14H
–
+
2NO3 + 12 H
0
+
O2 + 4H
Br2
0
2H2O
2e
–
Pb
5e
–
Mn
2e
–
Cl + H2O
6e
–
8e
–
Cl + 4H2O
2e
–
2Cl
6e
–
2Cr
10e
1.78
2+
+ 2H2O
1.69
2+
+ 4H2O
1.51
–
1.49
–
1.45
Cl +3H2O
–
1.36
3+
–
1.39
–
+ 7H2O
0
1.24
1.23
4e
–
2H2O
2e
–
2Br
–
–
1.35
N2 + 6H2O
1.07
e
NO + H2O
1.00
+
3e
–
NO + 2H2O
0.96
–
NO2 + H2O
–
+
2e
–
+
e
–
NO2 + H2O
e
–
Fe
NO3 +2H
3+
H2O2
–
MnO2 + 4OH
–
e
+
O2 + 2H
+
+ 8H
2–
2–
0.77
3e
0
S4O6
0.78
2+
–
MnO4 + 2H2O
2–
0.84
2e
+
MnO4
–
–
0
O2 + 2H
SO4
–
–
NO3 +2H
SO4
2e
е ,V
+
NO3 + 4H
І2
0
Reduced Form
–
NO2 + 2H
Fe
–
+
+ 2H
0.68
–
–
MnO4
2–
0.57
0.54
2e
–
2І
4e
–
4OH
–
0.40
6e
–
0
S + 4H2O
0.36
2e
–
2S2O3
2e
–
SO3 + H2O
–
0.54
2–
138
2–
0.22
0.20
Appendix E
Solubility Product Conctant Ksp Values for Feebly Soluble Electrolytes
at 25 °С
Electrolyte
–13
AgBr
6·10
AgCl
1.8·10
Ag2CrO4
4·10
Ag2S
6·10
–10
–12
1.1·10
AgI
Electrolyte
Ksp
–16
–50
Ag2SO4
2·10
–5
BaCO3
5·10
–9
BaCrO4
1.6·10
–10
BaSO4
1.1·10
–10
Ksp
Fe(OH)3
3.7·10
–40
FePO4
1.3·10
–22
–18
FeS
5·10
HgS
1.6·10
MgCO3
2.1·10
–52
–5
Mg(OH)2
1.3·10
–11
MnS
2.5·10
–10
PbBr2
9.1·10
PbCl2
2·10
–6
–5
–39
PbCrO4
1.8·10
–14
–14
Ba3(PO4)2
6·10
CaCO3
5·10
–9
PbCO3
7.5·10
CaC2O4
2·10
–9
PbI2
8.0·10
CaF2
4·10
–11
PbS
2.5·10
PbSO4
1.6·10
SrCO3
1.1·10
CaSO4
6.3·10
Ca3(PO4)2
1·10
–5
–29
–9
–27
–8
–10
Appendix F
Dissociation constants of some weak electrolytes
К1
Formula
К2
К3
Asids
HNO2
4,0⋅10
–4
HAlO2
4,0⋅10
–13
H3BO3
5,8⋅10
–10
HOBr
2,1⋅10
–9
H2CO3
4,45⋅10
H2SiO3
1,8⋅10
–13
–7
4,5⋅10
–11
2,2⋅10
–10
1,6⋅10
–12
H3AsO4
5,6⋅10
–3
1,7⋅10
–7
H3AsO3
5,7⋅10
–10
3,0⋅10
–14
HAsO2
5,8⋅10
–10
H2O2
2,6⋅10
–12
1,2⋅10
–2
5,0⋅10
–8
–3
H2SeO4
1⋅10
H2SeO3
3,5⋅10
–3
139
2,9⋅10
–12
Formula
H2Se
H2SO3
H2S
К1
К2
–4
1,7⋅10
–2
1,6·10
8,9·10
–8
1,0⋅10
–11
6,3⋅10
–8
1,3⋅10
–13
HOCl
5,0⋅10
–8
H3PO4
7,5⋅10
–3
6,3⋅10
–8
H3PO3
1,0⋅10
–2
3,0⋅10
–7
H3PO2
9,0⋅10
–2
HF
6,6⋅10
–4
HCN
7,2⋅10
–10
C6H5COOH
6,3⋅10
–5
HCOOH
1,8⋅10
–4
C2H5COOH
1,34⋅10
–5
CH3COOH
1,75⋅10
–5
C3H7COOH
1,54⋅10
–5
5,4⋅10
–5
CH2ClCOOH
1,4⋅10
–3
H2C2O4
5,4⋅10
–2
К3
1,3⋅10
–12
1,4⋅10
–9
1,4⋅10
–12
1,0⋅10
–10
Bases
Al(OH)3
Fe(OH)2
Fe(OH)3
Cd(OH)2
Mg(OH)2
Cu(OH)2
NH4OH
Pb(OH)2
Be(OH)2
Cr(OH)3
Zn(OH)2
–4
1,3⋅10
–11
1,8⋅10
–3
5,0⋅10
–3
2,5⋅10
–7
3,4⋅10
–5
1,8⋅10
–4
9,6⋅10
–8
3,0⋅10
–11
5,0⋅10
4⋅10
140
–5
Appendix G
∆
8
9
02
,
,
kJ/mol
,
J/(mol·К)
kJ/mol
G
S
Al2O3 (s)
8
9
02
Al (s)
∆
8
9
02
Substances
H
Standard Thermodynamical Functions or Some Substances at 298 К
0
28.3
0
–1676.0
50.9
–1582.0
0
5.7
0
ССl4 (l)
–135.4
214.4
–64.6
СО (g)
–110.5
197.5
–137.1
С (graphite)
–393.5
213.7
–394.4
СаСО3 (s)
–1207.0
88.7
–1127.7
СаF2 (s)
–1214.6
68.9
–1161.9
Ca3N2 (s)
–431.8
105.0
–368.6
СаО (s)
–635.5
39.7
–604.2
Са(ОН)2 (s)
–986.6
76.1
–896.8
CO2 (g)
0
222.9
0
Сl2O (g)
76.6
266.2
94.2
СlO2 (g)
105.0
257.0
122.3
Cl2 (g)
251.0
–
–
Cr2O3 (s)
–1440.6
81.2
–1050.0
CuO (s)
–162.0
42.6
–129.9
0
27.2
0
–264.8
60.8
–244.3
0
130.5
0
HBr (g)
–36.3
198.6
–53.3
HCN (g)
135.0
113.1
125.5
HCl (g)
–92.3
186.8
–95.2
HF (g)
–270.7
178.7
–272.8
26.6
206.5
1.8
HN3 (l)
294.0
328.0
238.8
H2O (g)
–241.8
188.7
–228.6
H2O (l)
–285.8
70.1
–237.3
H2S (g)
–21.0
205.7
–33.8
KCl (s)
–435.9
82.6
–408.0
KClO3 (s)
–391.2
143.0
–289.9
MgCl2 (s)
–641.1
89.9
–591.6
Mg3N2 (s)
–461.1
141
87.9
–400.9
Сl2O7 (l)
Fe (s)
FeO (s)
H2 (g)
HI (g)
∆
8
9
02
S
,
,
kJ/mol
,
J/(mol·К)
kJ/mol
G
8
9
02
8
9
02
∆
H
Substances
–601.8
26.9
–569.6
0
191.5
0
HN3 (l)
294.0
328.0
238.8
NH3 (g)
–46.2
192.6
–16.7
MgO (s)
N2 (g)
NH4NO2 (s)
–256.0
–
–
NH4NO3 (s)
–365.4
151.0
–183.8
N2O (g)
82.0
219.9
104.1
NO (g)
90.3
210.6
86.6
N2O3 (g)
83.3
307.0
140.5
NO2 (g)
33.5
240.2
51.5
N2O4 (g)
9.6
303.8
98.4
N2O5 (g)
–42.7
178.0
114.1
NiO (s)
–239.7
38.0
–211.6
О3 (g)
142.3
237.7
163.4
O2 (g)
0
205.0
0
P2O5 (s)
–1492.0
114.5
–1348.8
PbO (s)
–219.3
66.1
–189.1
PbO2 (s)
–276.6
74.9
–218.3
0
31.9
0
SO2 (g)
–296.9
248.1
–300.2
SO3 (g)
–395.8
256.7
–371.2
SiH4 (g)
34.7
204.6
57.2
SiO2 (quartz)
–910.9
41.8
–856.7
SnO (s)
–286.0
56.5
–256.9
SnO2 (s)
–580.8
52.3
–519.3
0
30.6
0
TiCl4 (l)
–804.2
252.4
–737.4
TiO2 (s)
–943.9
50.3
–888.6
WO3 (s)
–842.7
75.9
–763.9
ZnO (s)
–350.6
43.6
–320.7
S (s)
Ti (s)
142
∆
,
kJ/mol
,
J/(mol·К)
kJ/mol
8
9
02
S
,
G
8
9
02
8
9
02
∆
Formule and State
H
Appendix H
Standard Thermodynamical Functions or Some Organic Substances at 298 К
СН4 (g)
–74.9
186.2
–50.8
С2Н2 (g)
226.8
200.8
209.2
С2Н4 (g)
52.3
219.4
68.1
С2Н6 (g)
–89.7
229.5
–32.9
49.0
124.5
172.8
СН3ОН (l)
–238.7
126.8
–166.3
С2Н5ОН (l)
–277.6
160.7
–174.8
С3Н8О3 (l)
–669.1
204.6
–479.4
СН3СООН (l)
–484.4
159.9
–389.6
СО(NH2)2 (s)
СО(NH2)2 (l)
–333.0
–317.7
104.7
175.7
–196.9
–202.7
С6Н12О6 (s)
С6Н12О6 (l)
–1273.0
–1263.1
212.1
264.0
–910.5
–914.5
С12Н22О11 (s)
С12Н22О11 (l)
–2220.9
–2215.8
360.2
403.8
–1544.3
–1551.4
С12Н22О11 (l)
–2232.4
394.1
–1564.9
С6Н6 (l)
Appendix I
Surface Tension of Water at Different Temperatures
t, °C
15
16
17
18
19
20
σ
×10 ,
–3
Н 2О
t, °C
2
J/m
73.49
73.34
73.19
73.05
72.90
72.75
21
22
23
24
25
26
143
σ
×10 ,
–3
Н2О
2
J/m
72.59
72.44
72.28
72.13
71.97
71.80
Appendix J
The main Half-Reactions and Standard Red-Ox Potentials Values
Half-Reaction
Oxidized Form
nе
+
H2O2 + 2H
+
PbO2 + 4H
–
+
MnO4 + 8H
–
+
ClO +2H
–
0
Reduced Form
2e
–
2H2O
2e
–
Pb
5e
–
Mn
2e
–
–
1.49
–
Cl +3H2O
1.45
Cl + 4H2O
+ 2H2O
1.69
2+
+ 4H2O
1.51
Cl + H2O
+
ClO3 + 6H
6e
–
–
+
8e
–
2e
–
2Cl
6e
–
2Cr
Cl2
0
Cr2O7
2–
+
+ 14H
–
+
2NO3 + 12 H
0
+
O2 + 4H
Br2
0
0
1.35
N2 + 6H2O
1.24
4e
–
2H2O
1.23
2e
–
2Br
–
–
1.07
e
NO + H2O
1.00
+
3e
–
NO + 2H2O
0.96
–
NO2 + H2O
0.84
NO2 + H2O
0.78
–
+
2e
–
+
e
–
e
–
NO3 +2H
3+
2e
–
–
3e
–
–
e
0
+
O2 + 2H
MnO4 + 2H2O
MnO4
0
+
O2 + 2H
2–
S4O6
SO4
+ 7H2O
–
NO3 +2H
SO4
1.36
3+
–
1.39
–
+
NO3 + 4H
І2
–
–
NO2 + 2H
Fe
10e
1.78
2+
–
ClO4 + 8H
е ,V
+
+ 8H
2–
2–
+
+ 2H
–
Fe
2+
0.77
0.68
H2O2
–
MnO2 + 4OH
–
MnO4
2–
0.57
0.54
2e
–
2І
4e
–
4OH
–
0.40
6e
–
0
S + 4H2O
0.36
2e
–
2S2O3
2e
–
2–
–
0.54
2–
SO3 + H2O
144
0.22
0.20
Appendix K
Solubility Product Conctant Ksp Values for Feebly Soluble
Electrolytes at 25 °С
Electrolyte
Ksp
AgBr
6·10
AgCl
1.8·10
Ag2CrO4
AgI
4·10
Electrolyte
–13
–10
–12
Ksp
Fe(OH)3
3.7·10
–40
FePO4
1.3·10
–22
–18
FeS
5·10
HgS
1.6·10
–50
MgCO3
2.1·10
1.1·10
–16
–52
–5
Ag2S
6·10
Ag2SO4
2·10
–5
Mg(OH)2
1.3·10
–11
BaCO3
5·10
–9
MnS
2.5·10
–10
BaCrO4
1.6·10
–10
PbBr2
9.1·10
1.1·10
–10
BaSO4
PbCl2
2·10
–6
–5
–39
PbCrO4
1.8·10
–14
5·10
–9
PbCO3
7.5·10
–14
CaC2O4
2·10
–9
PbI2
8.0·10
CaF2
4·10
–11
PbS
2.5·10
PbSO4
1.6·10
SrCO3
1.1·10
Ba3(PO4)2
CaCO3
6·10
CaSO4
6.3·10
Ca3(PO4)2
1·10
–5
–29
145
–9
–27
–8
–10