DANYLO HALYTSKY LVIV NATIONAL MEDICAL UNIVERSITY
DEPARTMENT of GENERAL, BIOINORGANIC, PHYSICAL and
COLLOIDAL CHEMISTRY
MEDICAL CHEMISTRY
STUDY GUIDE
for the 1st year students of medical faculty
(Module 1. Acid-Base Equilibrium and Coordination
Compounds Formation Processes in Biological
Liquids)
L’VIV – 2012
Методичні вказівки з медичної хімії для студентів медичного
факультету (Модуль 1. Кислотно-основні рівноваги та
комплексоутворення в біологічних рідинах)
Методичні вказівки уклали: доценти Роман О.М., Кленіна О.В.,
Огурцов В.В., асистент Маршалок О.І.
За загальною редакцією: доцента Роман О.М.
Методичні вказівки обговорені і схвалені до друку цикловою
методичною комісією з фізико-хімічних дисциплін (протокол № 3 від 4
вересня 2011 р.).
Рецензенти:
проф. Й.Д. Комариця – професор кафедри фармацевтичної,
органічної та біоорганічної хімії ЛНМУ імені Данила Галицького.
проф. О.Я. Скляров – завідувач кафедри біологічної хімії ЛНМУ імені
Данила Галицького.
2
General information on the educational process
organization of “Medical Chemistry” studying within the
credit-module system
The educational process of medicinal chemistry studying is organized according
to the requirements of credit-module system within the Bologna process.
Medical Chemistry as an educational discipline is structured into 2 modules
9 practical classes in each:
Module 1. Acid-Base Equilibrium and the Processes of Coordination
Compounds Formation in Biological Liquids
Thematic modules:
1. The chemistry of bioelements. Coordination compounds formation in
biological liquids
2. Acid-base equilibrium in biological liquids
Module 2. The equilibrium in Biological Systems Occurring on the Interfaces
Thematic modules:
3. Thermodynamical and kinetical regularities of processes and
electrokinetical phenomena in biological liquids
4. Physical-chemistry of surface phenomena. Lyophilic and lyiphobic
disperse systems
Laboratory records are to be kept in a bound notebook. Include in the notebook
the aim of the experiments, a complete description of the work performed, all
reference materials consulted, and ideas that you have related to the work, and the
conclusions.
Forms of the discipline assessment
The maximum number of points assigned to students in each module (credit) 200, including the practice and laboratory educational activity - 120 points, and
the final module control - 80 points.
The assessment of trained knowledge is made on a three-point scale.
For checking the student’s educational achievements are stipulated the following
types and forms of the trained knowledge control:
1) the current control;
2) the practical skills gained and the laboratory experiments carrying out
assessment;
3) the final module control assessment.
The maximal assessment of current progress in a semester makes 60 % from a
final assessment of knowledge on discipline, and the maximal assessment of
examination makes 40 % from a final assessment of knowledge on discipline.
1. The current control is a regular check of educational trained achievements,
spent by the teacher on current employment according to syllabus of the discipline.
It is performed at each practice class according to specific objectives.
3
Theoretical students’ self-preparation control is performed in writing by answering
18 multiple choice questions in the form one-of-five, the correct answer to each is
estimated at 1 point, and two numerical problems, the correct solving being
estimated at 2 points. The minimum number of points that a student must gain for
the crediting the theoretical part is 9 points.
2. The practical skills gained and the laboratory experiments carrying out
assessment is performed after the laboratory work fulfilling by assessing the quality
and fullness of its performance, the ability to interpret the obtained results. For the
practical part of the lesson the student can get:
4 points if laboratory work is completely fulfilled and the student correctly
explains the experiments interpret the results and make conclusions;
2 points if the laboratory work is done with some errors, the student can not fully
explain and summarize the obtained results;
0 points if the laboratory work is not performed or the student can not explain
and summarize the obtained results.
The final score for the class is determined by the sum of the points for the
current theoretical control and the laboratory experiments carrying out points as
follows:
Total points
Grade evaluation
> 21
17 – 21
11 – 16
< 11 points for the current
theoretical control
or
0 points for the laboratory
experiments carrying out
5
4
3
2
Converting into
rating points
13
11
8
0
The maximal number of points a student can get for the module is calculated by
multiplying the number of points that correspond to the grade "5" to the number of
topics in the module with the addition of points for individual independent work (3
points) and is equal to 120 points (13×9+3=120).
The minimal number of points a student can get for the module is calculated by
multiplying the number of points that correspond to the grade "3" to the number of
topics in the module (8×9=72).
3. The final module control is carried out on completion of the module practical
classes. The students fulfilling all types of works included in the curriculum with the
points number on less than 72 are allowed.
The final control is carried out in the standardized form and includes the
theoretical and practical skills assessment. It should be performed in writing as 66
multiple choice questions (1 point for each correct answer) and 6 numerical
problems (2 points for each in the case of being solved correctly).
The maximum points for the final module control are 80. The final module
4
control is supposed to be credited if the student scored at least 50 points.
The discipline assessment
The discipline assessment is possible in the case of all modules credited only.
The total assessment of discipline is shaped as an average of points number of the 2
modules each evaluated by summation of points for current control and
experimental skills and final module control.
The points on medical chemistry conversion into the ECTS scale evaluation and
4-grade evaluation
The points on discipline may be conversed into the ECTS scale evaluation as
follows:
ECTS
scale
А
В
С
D
E
FХ
F
Statistical index
The top 10 % students
Next 25 % students
Next 30 % students
Next 25 % students
The last 10 % students
Repeated making up
The repeated course is required
The points on discipline may be conversed into 4-grade evaluation as follows:
The number of points
on discipline
The 4-grade evaluation
170 and over
140 – 169
122 – 139
less than 122
«5»
«4»
«3»
«2»
5
THEMATIC SCHEDULE
of practice and laboratory studies on medicinal chemistry
Module 1. Acid-Base Equilibrium and Coordination Compounds Formation
Processes in Biological Liquids
№
1.
2.
3.
4.
5.
6.
7.
8.
9.
•
Number
of hours
Solutions. Ways of expressing concentrations of solutions.
2.5
Preparation the solution with known concentration
Colligative properties of solutions. Experimental determination of
2.5
the osmotic concentration of solutions with the method of
cryometry
The equilibrium and processes involving coordination (complex)
2.5
compounds. Preparation and properties of complex and inner
complex compounds. Complexonometry
Bioelements and their classification. Chemical properties and
2.5
biological role of macroelements
2.5
Chemical properties and biological role of microelements
Acid-base
equilibrium.
Calculation
and
experimental
2.5
determination of the рН of solutions
2.5
Protolytical processes in living organisms. The hydrolysis of salts
2.5
Buffer solutions, their classification and the mechanism of the
buffer action. Preparation of buffer solutions. Determination of the
buffer capacity and the pH values of buffer solutions. The
biological role of buffer systems
The basic principles of the volumetric analysis. Acid-base titration.
2.5
Determination of the acidity of stomach liquid
The final control of the acquirement the Module 1 “Acid-Base
2.5
Equilibrium and the Processes of Coordination Compounds
Formation in Biological Liquids”
The topics
6
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
7
leaving the laboratory.
Eating and drinking any type of food are prohibited in the laboratory at all times.
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.
8
Topic 1
Solutions. Ways of expressing concentrations of solutions.
Preparation the solution with known concentration
1. Actuality of the topic
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.Goals and objectives:
– 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. The content of the topic
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. Block of information
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
9
solutions of metals are referred to as alloys. Types of solutions:
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:
10
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. Literature
1. V.A. Kalibabchuk, V.I. Halynska, V.I. Hryshchenko. Medical chemistry. –
Kyiv:AUS Medicine Publishing, 2010. – p. 55–61.
2. Darrel d. Ebbing. General Chemystry. – Boston:Houghton Mifflin company,
1984. – p. 197–215.
3. Raymond Chang. Chemistry.– New York.:McGraw Hill, 2010. – p.103–131.
3.4. Self-control questions
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?
б) Self-control numerical problems.
11
Numerical Problem 1. Calculate the molarity and normality of 49 % H3PO4
solution (d = 1.88 g/mL).
Given:
For solution of such type of numerical problems the
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 г 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 г
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
hydrochloric acid solution?
Given:
Calculate how many grams of HCl are contained in
CP1(HCl) = 40%
100 mL of 0.15 M HCl:
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:
12
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 is the sum of 1st and 2nd solutions as
CP1 = 35%
well as the mass of solute in 3rd solution is the sum of masses of
d1 = 1.15 g/mL 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:
С P ⋅ m 2 (sol.) 25% ⋅120 g
m 2 ( A)
2
C P2 =
⋅100% → m 2 ( A) =
=
= 30 g
m 2 (sol.)
100%
100%
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
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:
13
m = CN⋅E⋅V; E = M/2;
M (Na2CO3⋅10H2O) = 106 + 180 = 286 g/mol
Е = 286/2 =143 g/mol
m = 0,1⋅ 143 ⋅ 0,25 = 3,575 g
Answer: 3,575 g of crystalline soda should be dissolve in measuring flask 250 sm3
by volume.
б) Problems for solving
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?
CN = 0,1 mol-eq/dm3
Сalculate:
m–?
4. Laboratory Activities and Experiments Section
The content and methods of practice and laboratory studies
4.1. List of practical tasks for laboratory work:
– 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. Instructions for the laboratory work
4.2.1. Preparation of the titrated solution of hydrochloric acid.
Determine the density of initial solution by hydrometer.From the table to find
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
14
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
Colligative properties of solutions. Experimental determination
of the osmotic concentration of solutions with the
method of cryometry
1. Actuality of the topic
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. Goals and objectives:
– 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. The content of the topic
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
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.
15
3.2.Blok of informations
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, ΔP = 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:
ΔTf = Kf Cm,
where Cm is the molality of the solute, [mol/kg], and Kf (or K) is the freezing point
depression constant.
4. 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
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:
16
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. Literature
1. V.A. Kalibabchuk, V.I. Halynska, V.I. Hryshchenko. Medical chemistry. –
Kyiv:AUS Medicine Publishing, 2010.
2. Darrel d. Ebbing. General Chemystry. – Boston:Houghton Mifflin company,
1984.
3. Raymond Chang. Chemistry.– New York.:McGraw Hill, 2010.
Self-control questions
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?
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?
б) Self-control numerical problems
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.
17
Given:
Cp(C7H6O3) = 20%
Tb(C2H5OH) = 78.5 oC
Kb(C2H5OH) = 1.22 oC·kg/mol
Calculate:
Tb(S) - ?
Calculate the molality of the 20% salicic acid
solution:
m(С7 H 6O3 )
Cm =
M(C 7 H 6O3 ) ⋅ m(C 2 H 5OH)
The molar mass of salicic acid:
M(C7H6O3) = 7·12 + 6·1 + 3·16 = 138 g/mol
The mass of salicic acid in solution:
m(Ñ7 H6O3 ) =
ÑP (Ñ7 H6O3 ) ⋅ m(sol.) 20 % ⋅100 g
=
= 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
Cm =
20 g
= 1.81 mol/kg
138 g / mol ⋅ 0.08 kg
Calculate the elevation of the boiling point:
ΔTb = 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)
∆Tf = Kf(H2O) · Cm(A) and С m (A) =
then
Tfr(s) = –0,7 0С
M(A)
⋅ m(H 2O)
0
Kf (H2O) = 1,86 С
K (H O) ⋅ m(A) ⋅1000 .
f 2
Calculate:
M(A) =
∆T ⋅ m(H O)
M–?
2
f
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
18
Cp = 0,9 %
t = –0,55 0С
і–?
α–?
depression ∆Тexpr:
∆Тexpr= 0 – (–0,55) = 0,55 0С
Calculate the theoretical freezing point depression :
K ⋅ m ⋅ 1000 186
, ⋅ 0,9 ⋅ 1000
=
= 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 %.
∆Тtheor= К⋅С =
б) Problems for solution
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.
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.
19
4. Laboratory Activities and Experiments Section
The content and methods of practice and laboratory studies
4.1. List of practical tasks for laboratory work:
– 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.Instructions for laboratory work
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
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:
20
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
K ⋅ m ⋅ 1000
∆Тtheor =
.
L ⋅M
Calculate і = ∆Тexp/∆Тtheor .
i −1
Calculate α =
, where n – number of molecules of electrolyte that dissociate
n −1
to forms ions.
6. Conclusions and Interpretations. Lesson Summary
Topic 3
The equilibrium and processes involving coordination
(complex) compounds. Preparation and properties of
complex and inner complex compounds.
Complexonometry
1. Actuality of the topic
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. In medical practice, using drugs with chelated structure - tetacinum,
unitiol, ferracen.
2. Goals and objectives:
– 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.
3. Self Study Section
3.1. The content of the topic
Complex formation reactions. Werner coordination theory and modern
21
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 an antidotes to remove toxic metal
ions from the organism.
3.2.Blok of information
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
[Ag+ ] ⋅ [NH3 ]2
Kd =
[[Ag(NH3 ) 2 ]+ ]
The greater value of dissociation constant the more stable complex ion. For
22
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 + ]
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. Literature
1. V.A. Kalibabchuk, V.I. Halynska, V.I. Hryshchenko. Medical chemistry. –
Kyiv:AUS Medicine Publishing, 2010.
2. Darrel d. Ebbing. General Chemystry. – Boston:Houghton Mifflin company,
1984.
3. Raymond Chang. Chemistry.– New York.:McGraw Hill, 2010.
3.4. Self-control questions
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.
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.
б) Self-control numerical problems.
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:
[Cd 2 + ] ⋅ [CN − ]4
= 7,8 ⋅10 −18
[Cd (CN ) 4 2 − ]
2. Under the secondary dissociation complex ion gives the central metal ion and the
ligands ions or molecules according to the following equation:
Кd=
23
[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.
б) Problems for solutions
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.
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.
24
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
Topic 4
Bioelements and their classification. Chemical properties and
biological role of macroelements
1. Actuality of the topic
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.
2. Goals and objectives:
– know the chemical properties of the macro, based on their position in the
periodic system;
25
– know the biological role of macro and use of their compounds in medical
practice;
– to carry out reactions that characterize the acid-base and redox properties of
macroelements;
– to carry out qualitative reactions of identification of Na+, K+, Mg2+, Ca2+, CO32–
PO43–, NO3–, NO2– ions.
3. Self Study Section
3.1. The content of the topic
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-and p-elements. Typical chemical
properties of s-and p-elements and their compounds (reactions without changing of
oxidation charge). 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– ions.
3.2.Blok of informations
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
26
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).
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
27
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
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.
3.3. Literature
1. V.A. Kalibabchuk, V.I. Halynska, V.I. Hryshchenko. Medical chemistry. –
Kyiv:AUS Medicine Publishing, 2010.
2. Darrel d. Ebbing. General Chemystry. – Boston:Houghton Mifflin company,
1984.
3. Raymond Chang. Chemistry.– New York.:McGraw Hill, 2010.
3.4. Self-control questions
a) Review Questions.
28
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?
б) Self-control numerical problems.
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.
According to the problem, with 100 g of bone obtain 27
Given:
g of ash.
Cp(P) = 2,12 %
Cp(Ca)=7,56 %
So, mass percentage of the elements will be:
Cp(P) = 2,12 : 27 = 0,0785 or 7,85 %
Cp(Mg)=1,51 %
Cp(ash) = 27 %
Cp(Ca) = 7,56 : 27 = 0,28 or 28 %
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 %.
c) Self-control numerical problems.
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. List of practical tasks for laboratory work:
– reaction of macroelements oxides with water;
– Hydrolysis of salts ;
– reactions of identification of Na+ ions ;
– reactions of identification of K+ ions ;
– reaction of identification of Са+2 ions;
– color of the flame of Na+, K+, Ca2+, Li2+ ions;
– reactions of identification of РО43– ions ;
–
– reactions of identification of NO2 ions;
2–
– reactions of identification of S ions .
4.2.Instructions for laboratory work
4.2.1. Reactions of oxides of macro-elements with water.
29
Contain 0.1 g of powder oxides of СаО, МgО, Р2О5 into the test-tubes and add
3-5 ml of distilled water. What is the pH in the prepared solutions? Write the
equations of the reactions and name the products. Can the oxides dissolve in acids
and alkalis solutions?
4.2.2. Reaction of identification of Na+ ion.
To 5–6 drops of a sodium salt solution in a test-tube add the identical volume
of potassium dihydrogenantimonate KH2SbO4 solution. Cool the test-tube in the
steam of water and rub up the test-tube with a glass stick at the same time. What is
observed? Write the equation of the reaction.
4.2.3. Reaction of identification of K+ ion.
To 5–6 drops of a potassium salt solution in a test-tube add the identical
volume of sodium hydrogentartrate NaHC4H4O6. Cool the test-tube in the steam of
water and rub up the test-tube with a glass stick at the same time. What is observed?
Write the equation of the reaction.
4.2.4. Reaction of identification of 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.
4.2.5. Reaction of identification of Mg2+ ion.
To 2–3 drops of a magnesium salt solution add 3–4 drops of 2 N solution of
HCl and 2–3 drops of sodium hydrogenphosphate Na2HPO4 solution. Then add the
diluted solution of NH4OH and mix the prepared solution with a glass stick. The
precipitate of MgNH4PO4 forms. What is the color of the precipitate? Write the
equation of the reaction.
4.2.6. 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.
4.2.7. Reaction of identification of 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.
4.2.8. Reaction of identification of nitrite- ion.
To 2–3 ml of sodium nitrite NaNO2 solution in a test-tube add the identical
volume of 1М solution of acetic acid and 5–6 drops of КІ solution. What is
observed? Write the equation of the reaction.
4.2.9. Reaction of identification of sulfide- ion.
To 2–3 ml of a sodium sulfide Na2S solution in two test-tube add the
30
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.
6. Conclusions and Interpretations. Lesson Summary
Topic 5
Chemical properties and biological role of microelements
1. Actuality of the topic
The biological role of chemical elements in animals and humans very important.
Trace elements are part of a large number of enzymes, certain vitamins and
hormones. They take part in the processes of hematopoiesis, reproduction,
metabolism. 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. Goals and objectives:
– know the major types of reactions that characterize the chemical properties of
minerals;
– Know the biological role of microelements and medicines, which contain
molecules of these elements;
– Know toxic effects of some xenobiotics;
– to carry out reactions of identification.
3. Self Study Section
3.1. The content of the topic
Electronic structure and electronegativity of d-elements. Typical chemical
properties of d-elements and their compounds (oxidation and reduction reactions,
complex formation reactions). The biological role. Uses in medicine. Toxic effects
of d-elements and their compounds.
Reactions of identification of MnO4–, Fe3+, Cu2+, Ag+ ions.
3.2. Blok of information
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.
31
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.
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
32
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
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
33
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. Literature
1. V.A. Kalibabchuk, V.I. Halynska, V.I. Hryshchenko. Medical chemistry. –
Kyiv:AUS Medicine Publishing, 2010.
2. Darrel d. Ebbing. General Chemystry. – Boston:Houghton Mifflin company,
1984.
3. Raymond Chang. Chemistry.– New York.:McGraw Hill, 2010.
3.4. Self-control questions
а) Review Questions
1. 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?
2. 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.
3. What are the chemical properties indicate biological role of microelements
manganese, copper, chromium and zinc?
5. What are the structure units of hemoglobin? Describe the mechanism of binding
oxygen with hemoglobin and explain the toxic action of carbon monoxide CO.
6. Specify effects of lowering of iron in human blood. How it can be prevented?
7. Describe the biological role of microelements copper, zinc, manganese and
molybdenum and specify their containing in the body and daily requirement.
8. What is the structure of vitamin B12 molecule and what is its biological role?
9. What are the most important medicines from a family of d-elements specify their
34
chemical composition, the conditions under which they are assigned to treat.
b) Self-control numerical problems.
Numerical problem 1. 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
%
100⋅Cp(Н2О)blood and this mass contains water of plasma
Cp (Н2О)cells= 64 % 60⋅0,91 and water of cells 40⋅0,64.
Cp (Н2О) blood – ?
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 2. 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
m (Zn) = 1,5 g
Calculate the mass of pure zinc in commercial. According
to the equation:
V = 0,448 l
from 65 g of Zn we obtain 22.4 l of H2
Cp (Zn) – ?
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%
1.5 g
Answer: Cp (Zn) = 86,7 %
c) Problems for solving
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.
Answer: 128 tones.
35
4. Laboratory Activities and Experiments Section
4.1. List of practical tasks for laboratory work:
– preparation and properties of copper, zinc and chromium hydroxides;
– the hydrolysis of the iron and copper salts;
– reduction properties of the ions of d-elements in the lowest oxidation states;
– oxidation properties of the ions of d-elements in the highest oxidation states;
– preparation and properties of coordination compounds;
– reactions of identification of Fe(II) and Fe(III) ions;
– reactions of identification of Ni(ІІ) and Co(ІІ) ions;
– reaction of identification of chromate-ion;
– reaction of identification of Mn(II) ions;
– reaction of identification of Zn2+ ions.
4.2.Instructions for laboratory work
І. Acid-Base Properties
4.2.1. Preparation and properties of copper, zinc and chromium hydroxides.
To 2-3 ml of solutions of copper, zinc and chromium 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.
а) 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.
б) To the solution of chromium (III) salt add an alkali solution till the precipitate
will dissolve. Then add bromine water Br2. Does the color of the solution change?
Write the equations of the reactions of transforming of sodium chromite into sodium
chromate.
в) To 2-3 ml of manganese (II) salt add 1 ml of 0.1 N solution of KOH. Add
some bromine water Br2 to the 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
а) Oxidizing properties of chromium (VI) compounds.
To 4-5 drops of potassium dichromate K2Cr2O7 add the same volume of sulfuric
acid and the solution od sodium nitrite NaNO2. What is observed? Write the
equations of the reactions.
36
б) 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. Reactions of formation complexes
4.2.4. Preparation and properties of complex compounds.
a) To 1-2 ml of copper (II) sulfate CuSO4 solution add concentrated ammonium
solution. What is the color of solution. Write the equations of the reactions.
ІV. Tests on the ions
4.2.5. 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.
в) reaction with ammonia tiocianate or potassium tiocianate.
To 4-5 drops of iron (III) chloride FeCl3 add 2-3 drops of ammonia tiocianate
NH4SCN solution. What is the color of the solution? Write the equations of the
reactions, name the product compound.
4.2.6. Tests on Ni(ІІ) and Co(ІІ) ions.
а) To 1–2 сm3 salt of nicol(ІІ) add 1 сm3 sodium hydroxide solution till
precipitate will forms. To prepared precipitate add 5 drops of bromine water and
worm it. How wiil change the color of the precipitate. Write the equations of the
reactions.
4.2.7. 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.
4.2.8. Tests on Mn(II) ion.
To1 drop of manganis(II) sulfate add PbO2 solid, 1–2 ml 30 % HNO3 solution
and heat ittillthe boiling. Dilute the solution with the water.What is the color of
solution? Write the equations of the reactions.
4.2.9. Tests on Zn (II)-ion.
To 4-6 drops of zinc solution add the same volume of H2S. What is the color of
precipitate? Divide it to two test-tubes and add some hydrochloric acid to the first
one, and some acetic acid to the second one. In what test-tube the precipitate
dissolves? Write the equations of the reactions.
37
6. Conclusions and Interpretations. Lesson Summary
Topic 6
Acid-base equilibrium. Calculation and experimental
determination of the рН of solutions
1. Actuality of the topic
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
2. Goals and objectives:
– know the theory of weak and strong electrolytes;
– know the pH for fluids of the human body in norm and pathology;
– know about the role of hydrolysis in biochemical processes;
– be able to experimentally determine the pH value;
– be able to solve situational problems on this topic.
3Self Study Section
3.1. The content of the topic
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.
3.2.Blok of information
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:
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
38
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
3.3. Literature
1. V.A. Kalibabchuk, V.I. Halynska, V.I. Hryshchenko. Medical chemistry. –
Kyiv:AUS Medicine Publishing, 2010.
2. Darrel d. Ebbing. General chemystry. – Boston:Houghton Mifflin company,
1984.
3. Raymond Chang. Chemistry.– New York.:McGraw Hill, 2010.3.4.
3.4. Self-control questions
а) Review Questions.
1. Give the definition of the dissociation process. What compounds are called
electrolytes?
39
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?
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?
b) Self-control numerical problems.
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
CM
0,3
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
CM =
=
= 0,4109 mol/dm3.
Calculate:
(
)
36
,
5
M
HCl
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.
c) Problems for solutions
1. Calculate α and [H+] of 0,1 М hypochloric acid solution HClO.
Answer: 7⋅10–4; 7⋅10–5 mol/l.
40
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. Laboratory Activities and Experiments Section
The content and methods of practice and laboratory studies.
4.1. List of practical tasks for laboratory work:
– reactions of formation of feebly ionised compounds;
– shifting the equilibrium in a solution of ammonia;
4.2. Instructions for laboratory work
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
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.
6. Conclusions and Interpretations. Lesson Summary
Topic 7
Protolytical processes in living organisms. The hydrolysis of
salts
1. Actuality of the topic
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.
41
2. Goals and objectives:
– know about the role of hydrolysis in biochemical processes;
– to solve typical problems on this topic.
3. Self Study Section
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. Blok of information
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
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
Kh =
, h=
, [OH–] = K ⋅ [salt ]
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=
, [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=
Kh ,
[OH–] =
K H 2O ⋅ K acid
K base
3.3. Literature
1. V.A. Kalibabchuk, V.I. Halynska, V.I. Hryshchenko. Medical chemistry. –
Kyiv:AUS Medicine Publishing, 2010.
2. Darrel d. Ebbing. General chemystry. – Boston:Houghton Mifflin company,
1984.
3. Raymond Chang. Chemistry.– New York.:McGraw Hill, 2010.
3.4. Self-control questions
а) Review Questions.
42
1. What factors affect the hydrolysis equilibrium shifting? Give the examples.
2. Write formulas for calculation of the constants, degree of hydrolysis and pH of
solutions of different types of salts.
3. The role of hydrolysis in biochemical processes.
b) Self-control numerical problems.
Numerical problem 1. 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
СМ(NH4OH)=0.1 М
and a strong acid, therefore:
–5
KH O
Кd(NH4OH)= 1.77⋅10
− 14
2 = 10
Kh =
= 0.565 ⋅10− 9
Кh – ?
Kbase 1.77 ⋅10− 5
α–?
Equations of NH4Cl hydrolysis:
pH – ?
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 2. 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:
43
[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 for solving
1. Write the molecular and ionic equations of the hydrolysis of the following salts:
KCN, Li3PO4, Cr2(SO4)3, CuCl2, CH3COONH4. Indicate the pH.
2. Calculate the pH of the solution, constant of hydrolysis and percent of hydrolysis
of ammonia bromide NH4Br in its 0.01 M solution.
3. Calculate the pH of 0.36 M CH3COONa solution. Kd(CH3COOH) = 1.75·10–5.
4. Calculate the pH of 0.42 M NH4Cl solution. Kd(NH4OH) = 1.8·10–5.
5. 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.
6. 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 List of practical tasks for laboratory work:
– 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;
– a complete hydrolysis.
4.2..Instructions for laboratory work
4.2.1. 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
44
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.
4.2.2. 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.
4.2.3. 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.
4.2.4. 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
Topic 8
Buffer solutions, their classification and the mechanism of the
buffer action. Preparation of buffer solutions. Determination of
the buffer capacity and the pH values of buffer solutions. The
biological role of buffer systems
1. Actuality of the topic
Constant value of pH of biological liquds is provided by several physiological
45
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. Goals and objectives:
– know the concept of buffer systems, their composition, types and mechanism of
their action;
– know the formula for calculatiion of the pH of buffer systems;
– know the definition of buffer capacity and be able to carry out calculations of
value;
– be able to carry out calculations related with the preparation of buffer solutions;
– be able to prepare buffer solutions and determine the pH value.
3. Self Study Section
3.1. The content of the topic
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. Blok of information
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.
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 ]
46
[base]
[ 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
[mmol/ml–1],
BC =
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. Literature
3. V.A. Kalibabchuk, V.I. Halynska, V.I. Hryshchenko. Medical chemistry. –
Kyiv:AUS Medicine Publishing, 2010.
4. Darrel d. Ebbing. General Chemystry. – Boston:Houghton Mifflin company,
1984.
3. Raymond Chang. Chemistry.– New York.:McGraw Hill, 2010.
for any basic buffer solution pH = 14 – pKd(base) + lg
3.4. Self-control questions
а) Review Questions.
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) Self-control numerical problems.
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
47
buffer solution:
С(Н2СО3) = 0.1 mol/dm3
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
−
7
3
V(NaHCO3) = 60 cm
= 2.93 ⋅10
mol/dm
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
C(CH3COOH)= 0.1 mol/dm3 acidic buffer solution:
C(CH3COONa)=0.1 mol/dm3 [H+] = antilog (–5.24) = 10–5.24 = 5.76·10–6 mol/dm3
V(buffer sol.) = 200 cm3
2. The concentration of hydrogen ions in acidic
рН = 5.24
buffer solution is:
С (acid ) ⋅ V (acid )
К(CH3COOH )=1.75⋅10–5
[H+] = Kd(acid)·
V(salt) – ?
C ( salt ) ⋅ V ( salt )
V(acid) –?
V (acid )
If C(acid)=C(salt), then:
[H+] = Kd·
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.
48
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:
Cbase Vbase
; if Сbase = Сsalt, then
Csalt Vsalt
V
x
[OH–] = К⋅ base = K ⋅
.
Vsalt
Vbuff − x
[OH–] = К⋅
Hence we find the volumes of the base and salt needed to prepare basic buffer.
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 – ?
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 for solutions
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.
49
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. List of practical tasks for laboratory work:
– 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.Instructions for laboratory work
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
Phosphate
Ammonium
CH3COOH
H2CO3
NaH2PO4
NH4OH
CH3COONa
NaHCO3
Na2HPO4
NH4Cl
1,8⋅10–5
4,3⋅10–7
6,3⋅10–8
1,8⋅10–5
3,7-5,6
5,5-7,4
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),
Csalt Vsalt
and 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:
[H+] = K ⋅
x
;
50 − x
x=
50 ⋅ [ H + ]
K д + [H + ]
=
50 ⋅ 5,25 ⋅ 10 −5
(1,8 + 5,25) ⋅ 10 −5
= 37,2
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.
50
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.
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 C base
∆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.
6. Conclusions and Interpretations. Lesson Summary
51
Topic 9
The basic principles of the volumetric analysis. Acid-base
titration. Determination of the acidity of stomach liquid
1. Actuality of the topic
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. Without knowledge of the composition of different
biological liquids of living organisms is important 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.
2. Goals and objectives:
– know the theoretical basis of the method of acid-base titration;
– know the methods of calculations in titration analysis;
– know the methods of standardization of titrated solutions;
– be able to choice an indicator for the method of neutralization titration;
– be able to determine the acidity of gastric juice.
3. Self Study Section
3.1. The content of the topic
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.
3.2. Blok of information
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;
52
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.
Table 1. Properties of Some Indicators
Color
Indicators
Effective pH range
Acid Form
Base Form
Methyl violet
0.0–3.0
Yellow
Violet
Methyl orange
3.1–4.1
Red
Yellow
Methyl red
4.2–6.2
Red
Yellow
Bromothymol blue
6.0–7.8
Yellow
Blue
Phenolphthalein
8.0–10.0
Colorless
Red
Lithmus
5.0–8.0
Red
Blue
Methods of neutralization are widely used in clinical laboratories to determine
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.
3.3. Literature
5. V.A. Kalibabchuk, V.I. Halynska, V.I. Hryshchenko. Medical chemistry. –
Kyiv:AUS Medicine Publishing, 2010.
6. Darrel d. Ebbing. General chemystry. – Boston:Houghton Mifflin company,
1984.
3. Raymond Chang. Chemistry.– New York.:McGraw Hill, 2010.
3.4. Self-control questions
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?
53
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?
b) Self-control numerical problems.
Numerical problem 1. How many grams of borax Na2B4O7·10H2O are necessary
to prepare 400 ml of 0.05 N solution?
Calculations
1. The mass of borax Na2B4O7·10H2O could be
Given:
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
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 2 CO 3 ) 106
M E (Na 2 CO 3 ) =
=
= 53 (g/mol-eq),
2
2
3. Calculate the mass percentage of Na2CO3 in the anhydrous soda:
54
C p ( Na 2 CO 3 ) =
m(Na 2 CO 3 ) ⋅ 100% 0.1649 ⋅ 100
=
= 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,
C (alkali) ⋅ V(alkali) ⋅ 100
V(0.1 N alkali) = N
V(g.j.) ⋅ 0.1
Calculate acidity of gastric juice:
C (alkali) ⋅ V(alkali) ⋅ 100
0.0485 ⋅13.6 ⋅100
V(0.1N alkali) = N
=
= 66 cm3
V(g.j.) ⋅ 0.1
10 ⋅ 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.
1. From the formula of normality of the solution the mass of
Given:
H2SO4 in 1.5 N solution could be calculated:
Сp(H2SO4) = 55 %
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
Calculate:
m(H2SO4) = CN(H2SO4)∙ME(H2SO4)∙V2(sol.)
V1(sol.) – ?
where molar mass of H2SO4 equivalent is:
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:
55
m (sol) 66.82 g
V ( sol ) = 1
=
= 60.75 ml
1
d (sol) 1.1 g/ml
1
c) Problems for solving
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.
4. Laboratory Activities and Experiments Section
4.1. The content and methods of practice and laboratory studies.
– to choice of the indicator in titration of strong acid with alkali;
– to choice of the indicator in titration of weak acid with alkali;
– to determine the total acidity of gastric juice.
56
4.2. List of practical tasks for laboratory work:
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 to 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 a
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 to 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
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.
57
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:
V(0,1 N base) = V(base) ⋅ СN (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
58
Appendixes
Appendix A
Standard Thermodynamical Functions or Some Substances at 298 К
G
∆
S
,
,
kJ/mol
,
J/(mol·К)
kJ/mol
8
9
02
8
9
02
H
∆
8
9
02
Substances
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)
59
G
∆
S
,
,
kJ/mol
,
J/(mol·К)
kJ/mol
8
9
02
8
9
02
H
∆
8
9
02
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)
60
Appendix B
Standard Thermodynamical Functions or Some Organic
Substances at 298 К
G
∆
S
,
,
kJ/mol
,
J/(mol·К)
kJ/mol
8
9
02
8
9
02
H
∆
8
9
02
Formule and State
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
61
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
–
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
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
–
e
–
NO3 +2H
3+
H2O2
–
MnO2 + 4OH
–
e
+
O2 + 2H
+
+ 8H
2–
2–
Fe
0.77
3e
0
S4O6
0.78
2+
–
MnO4 + 2H2O
2–
0.84
NO2 + H2O
–
+
MnO4
–
2e
0
O2 + 2H
SO4
+ 7H2O
0
–
NO3 +2H
SO4
1.36
3+
–
1.39
–
+
NO3 + 4H
І2
–
–
NO2 + 2H
Fe
10e
1.78
2+
–
ClO4 + 8H
е ,V
+
+ 2H
0.68
–
–
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
62
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
63
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
64
–5