Download Experiment 9

Olga V. Kovalchukova, Nasrin
Namichemazi & Rusul Alabada
Experiments in
CHEMISTRY
For the 1-st year students of the «Dentistry»
speciality of the Medical Faculty of the Peoples’
Friendship University of Russia
Ковальчукова О.В., Насрин
Намичемази, Русул Алабада
Лабораторные работы по
химии
Для студентов 1 курса медицинского
факультета специальности «Стоматология»
(на английском языке)
Москва
Издательство Российского университета дружбы народов,
2015
1
Утверждено
РИС Ученого совета
Российского университета дружбы народов
Ковальчукова О.В., Насрин Намичемази, Русул Алабада.
Лабораторные работы по химии. Для студентов I курса
медицинского факультета специальности «Стоматология». М.:
Изд-во РУДН, 2015. – 56 с.
Настоящее учебное пособие представляет собой описание
лабораторных
работ,
предназначенное
для
студентов
медицинского факультета специальности «Стоматология».
Составлено в соответствии с Федеральным государственным
образовательным стандартом и программой курса «Химия».
Предназначено для работы студентов I курса медицинского
факультета специальности «Стоматология» при подготовке к
лабораторным занятиям и к экзамену по курсу «Химия».
Подготовлено на кафедре общей химии.
The manual represents the lab. manual of the Chemistry
discipline delivering for the students of the «Dentistry» speciality of
the Medical Faculty of Peoples’ Friendship University of Russia. It is
prepared in accordance with the Federal state educational standard and
the Program in Chemistry.
Intended for the 1-st year students of the «Dentistry» speciality
of the Medical Faculty of the Peoples’ Friendship University of Russia.
Prepared at the department of General Chemistry.
© Ковальчукова О.В., Насрин Намичемази, Русул Алабада
© Издательство Российского университета дружбы народов, 2015
2
SUMMARY
Page
4
In the Laboratory…………………………………………………
6
Seminar 1 Chemical equivalent. Law of equivalents…………..
Seminar 2 General classes of inorganic compounds…………… 7
8
Seminar 3 Rate of a chemical reaction. Chemical equilibrium..
Seminar 4 Structure of the atom……………………………….. 10
Experiment 1 Preparation of a working solution of HCl…….. 11
Experiment 2 Standardization of the solution of HCl………… 12
13
Experiment 3 Ionic and heterogeneous equilibria in solutions
19
Experiment 4 Ionic product of water. pH. Hydrolysis of salts
Experiment 5 Buffer solutions…………………………………. 24
Experiment 6 Colloid solutions…………………………………. 27
Experiment 7 Oxidation-reduction reactions……………….… 29
33
Experiment 8 Complex (coordination) compounds…………..
37
Seminar 5 Isomerism and nomenclature of hydrocarbons …
Experiment 9 Chemical properties of hydrocarbons………… 39
Experiment 10 Alcohols and phenols………………………….. 43
Experiment 11 Aldehydes and ketones…………………………. 47
Experiment 12 Carboxylic acids. Amino acids…………………. 49
Syllabus "Chemistry"……………………………………………… 51
The cited literature……………………………………………….. 56
3
In the Laboratory
The two most important principles that must be fulfilled
whenever you work in a chemical laboratory are:
1) BE ACCURATE and
2) STRICTLY OBEY SAFETY RULES.
For this reason always bear these principles in mind:
1. Keep your working place in a good order and don’t change it
without the permission from the instructor.
2. Before starting the experiment, read the manual and listen to
the instructor, be quite sure that all the theoretical and experimental
aspects are clear to you.
3. Check that all the equipment and chemicals required are on
your bench. If no, approach the instructor or the technician. Never
take any item from other benches.
4. Make sure that all the equipment you are going to work with
is clean and if necessary, clean it.
5. All the experimental observations and quantitative data
should be recorded immediately.
6. Strictly follow the procedure.
7. After finishing the experiment, clean all the equipment you
used and your working place.
SAFETY RULES
1. Handling the Bunsen burner
a) Don’t work with a burner if the rubber tubing connecting it
to the gas tap is damaged or loose;
b) don’t keep the gas tap open while waiting to the lighter;
c) close the gate valve at the end of the laboratory session;
d) if you smell the gas while the tap is closed, inform the
instructor immediately.
4
2. Handling chemical reagents
a) Be careful with corrosive substances (like acids, bases,
organic solvents), and particularly careful with their concentrated
solutions;
b) if a corrosive substance is dropped on your skin, wash it
immediately with water and inform the instructor without delay;
c) if you dilute concentrated sulfuric acid always pour acid to
water; newer do the opposite;
d) use spatulas for taking solid reagents - not hands;
e) never close a test tube with a finger - use a cork;
f) don’t taste anything in the laboratory;
g) if you need to smell a substance don’t put your nose into the
container; keep the container at some distance and by waving motions
of your palm above the mouth of the container send the vapors of the
substance towards your nose;
h) reactions involving dangerous or unpleasant odors are to be
performed in a hood.
3. Handling glassware
a) Hold the glassware firmly but not squeezing it;
b) do not use broken or cracked glassware;
c) if the glassware is accidently broken, the remainings must be
swept to the waste bin immediately.
4. Heating procedure
a) The mouth of a test tube mustn’t point at yourself or
anybody else;
b) don’t stand too close to the apparatus in which material is
being heated, but never leave it;
c) don’t hold a test tube in your hands while heating, use test
tube holders.
5
Seminar 1
CHEMICAL EQUIVALENT. LAW OF EQUIVALENTS
According to the law of equivalents, all the chemicals react
with each other in the amounts which are proportional to their chemical
equivalents.
Chemical equivalent is the amount of a substance (in moles)
which can react with 1 mole of hydrogen atoms or replace its same
amount from a chemical compound.
Equivalent mass, МE is the mass of one equivalent of a
substance expressed in grams per mole.
Calculation of equivalent masses if compounds of different
classes can be performed using the following formulae:
(i) elements in free state and in chemical compounds:
МE = М / V (V is valency of the element)
(ii) acids and bases: МE = М / n (n is basidity of the acid
or acidity of the base, means number of Н+ or ОН- ions)
(iii) oxides and salts:
МE = М / p q ( p is number od
metallic atoms and q is their valency)
Number of equivalents
sample).
nE = m / ME ( m is mass of the
For gaseous substances
nE = V0 / VE0 (V0 is molar volume of the gas under normal
conditions, V0 = 22.4 l  mole-1)
VE0 is equivalent volume of the gas, means the volume which
is occupied by one equivalent of a gas under normal conditions.
Possible ways of mathematical expression of law of
equivalents:
nE1= nE2 ; M1 / ME 1 = М2 / ME 2 ; V01 / VE01 = V02 / VE02
6
1.
2.
3.
4.
5.
6.
QUESTIONS AND PROBLEMS
A metal hydride contains 2.02 g of hydrogen and 13.88 g of metal.
Calculate equivalent mass of the metal.
While 53.96 g of a metal is oxidized 101.96 g of an oxide is
formed. Calculate equivalent mass of the metal.
4.80 g of Ca and 7.85 g of Zn replace the same amount of hydrogen
from an acid. Calculate equivalent mass of zinc if the equivalent
mass of calcium equals 20.0 gmole-1.
Calculate equivalent masses of a metal and sulfur if 4.86 g of the
metal form 5.22 g of oxide and 5.58 g of sulfide.
While 0.595 g of an unknown substance reacts with 0.275 g of
hydrogen chloride, 0.440 g of a salt is formed. Calculate
equivalent masses of the substance and the salt.
The volume of 2800 ml of hydrogen measured under normal
conditions can reduce 11.75 g of a metal oxide. Calculate
equivalent masses of the metal and its oxide.
Seminar 2
GENERAL CLASSES OF INORGANIC COMPOUNDS
1. Find the mass 21023 molecules of carbon monoxide (IV).
2. How many molecules of hydrogen contained in 56 liters of it ,
under normal conditions?
3. What is the volume occupied under normal conditions of 18.25
g of hydrogen chloride ?
4. What is the amount of hydrogen ( STP ) was isolated by
dissolving 1.5 kg of zinc in hydrochloric acid ?
5. What mass natural limestone containing 90 % by weight of
calcium carbonate required to obtain 7 t quicklime ( calcium
oxide ) ?
6. What is the volume of gas ( STP) is highlighted by the action
of 10.42 g of limestone (containing 4 % did not react with acid
impurities) 36.5 g of 24% hydrochloric acid solution ?
7
7. Calculate the weight of 55 % nitric acid solution prepared from
one ton of ammonia in the NO output if the contactor is 98 %
and the yield 94 % acid- .
NaNaOHNaHCO3Na2CO3Na2SO4NaClNa
Fe  FeCl2  Fe(OH)2  Fe(OH)3  Fe2O3
Al  AlCl3  Na3[Al(OH)6]  Al(NO3)3  Al
S  SO2  Na2SO3  NaHSO3  Na2SO3  Na2SO4
N2  NH3  NO  NO2  HNO3  NH4NO3
СO2  C  CO  CO2  CaCO3  Ca(HCO3)2  CO2
Seminar 3
RATE OF A CHEMICAL REACTION.
CHEMICAL EQUILIBRIUM
The rate of a chemical reaction can be considered as a change
of the amount of reactants or products per a unit time:
v=
C
t
It is affected by the nature of reactants, their concentrations,
temperature, and the presence of a catalyst.
Concentration dependence:
For a simple reaction
mA+nB  pC
m
v = k [A]  [B]n (law of mass action)
where k - rate constant; m and n - reaction orders.
For a simple reaction m and n are stoichiometric coefficients;
for a multistep reaction reaction orders are determined experimentally
from the step with the lowest rate.
Temperature dependence:
8
v(T2) = v(T1)  
T2  T1
10
(Vant Hoff’s equation)
where  - temperature coefficient.
For majority of homogenous reactions the evaluation of
temperature by 10 degrees provides the 2-4 times evaluation of the rate
of the reaction. This significant growth of the rate is due to the
evaluation of the number of active molecules in a geometrical
progression.
The extra energy of molecules which provokes their change
and the appearance of new substances is called the activation energy.
Catalyst dependence:
The rate of a chemical reaction grows in the presence of a
catalyst. The latter causes
unstable intermediates (activated
complexes) to appear whose decomposition leads to the formation of
products. As a result the activation energy of a chemical reaction
decreases, and the rate of the reaction increases.
Reversibility of chemical reactions:
For reversible reactions when the rates of both forward and
reverse processes become the same, a state of a chemical equilibrium
sets in.
The state of a chemical equilibrium can be described in terms
of equilibrium constant K: for a reaction
mA+nB  pC
K=
p
[C ]
[ A ]m [ B ]n
When conditions (temperature, pressure, concentration of one
of the reactants) of the reversible reaction change, the rates of the
forward and reverse processes change differently and the chemical
equilibrium shifts according to Le Chatelier’s principle:
If any change of conditions is imposed on a system in
equilibrium, the equilibrium will shift in such a way as to counteract
the imposed change.
QUESTIONS AND PROBLEMS
9
1.
2.
3.
4.
5.
Find the value of the rate constant for the reaction A + B  AB
if at concentration of substances A and B equal to 0.05 M and
5
0.01 M respectively, the rate of the chemical reaction is 5 x 10
M/min.
How many times will the rate of the reaction 2A + B  A2B
change if the concentration of A is doubled, and that of B is
halved ?
What is the temperature coefficient of the reaction rate if the rate
grows 15.6 times when the temperature is increased by 30
Kelvins?

Find the equilibrium constant of the reaction N2O4 
2NO2 if
the initial concentration of the N2O4 was 0.08 M, and by the
moment when equilibrium was established 50% of N2O4 was
dissociated.
In which direction will the following equilibria shift :
a) 2CO + O2


2CO2
H 0  566kJ

H  180 kJ
b) N2 + O2 
2NO
( 1 ) when the temperature is lowered; ( 2 ) when the pressure is
increased; ( 3 ) when the concentration of oxygen is increased ?
0
Seminar 4
The structure of the atom. Periodic Law and the Periodic Table.
Chemical bond
1. The physical meaning and values of quantum numbers.
2. Write the values of the quantum numbers of the orbitals :
a ) 2s; b ) 3d; c ) 2pz.
3. What is the maximum number of electrons can be : a) 2pz orbitals ;
b) on the 5f- sublayer ; c) electronically layer with n = 3 ?
4. Select impossible configurations : 2p8, 3d6, 3s1, 2s2, 2d4, 3d2,
4f10, 2p5, 5p2
5. Formulate the basic rules for filling electronic sublevels
6. Write the electronic formula of atoms of elements
10
№ 15, 18 , 34, 47, and specify the number of unpaired valence email .
7. Write the electronic formula of ions Se2- ; Mg2+ ; Fe2+; Fe3+.
8. Identify the possible valence states of oxygen and sulfur atoms .
9. Which element of the pair has a higher: a) ionization energy ; b) the
energy of the electron affinity ; c) electronegativity (№ 7 and
number 15 ; № 20 and № 38 ; № 6 and number 7)
10. Determine the nature of the relationship and schemes of education
CaS; S2; H2S; РH4+.
Experiment 1
Preparation of a working solution of HCl
Theory: Majority of diluted solutions are not stable systems and tend
to change their concentration within time. That is why chemicals are
usually stored as solids or concentrated solutions dilute when
necessary. Hydrochloric acid is an aqueous solution of a gaseous
hydrogen chloride HCl. With the change of conditions of storage it can
easily evolve from the solution to decrease its concentration. Thus, the
concentration of HCl solutions is not constant and needs to be adjusted.
There is a correlation between concentrations of solutions and their
densities.
Goal: To prepare 250 ml of 0.1 N solution of HCl by diluting of a
concentrated solution of HCl.
Procedure: 1. Using a densimeter, measure the density of an initial
solution of HCl and determine concentration using the table 1.
Table 1
Densities and mass fractions of aqueous solutions of HCl (20oC)
Density,
, g/ml
1,000
1,005
1,010
1,015
1,020
Mass
fraction,
, %
0,360
1,360
2,364
3,374
4,388
Density,
, g/ml
1,035
1,040
1,045
1,050
1,055
Mass
fraction,
, %
7,464
8,490
9,510
10,52
11,52
Density,
, g/ml
1,070
1,075
1,080
1,085
1,090
Mass
fraction,
, %
14,49
15,48
16,47
17,45
18,43
11
1,025
1,030
5,408
6,433
1,060
1,065
12,52
13,50
1,095
1,100
19,41
20,39
2. In case if the measured density doesn’t correspond to any table
datum find the interval inside which your measured value lies
max = 1,085 max = 17,45
x = 1,083 x = ?
min = 1,080 max = 16,47
Using the formula of interpolation, calculate the mass fraction of the
given solution (x).
(x-min)( max -min)
(1,083-1,080)(17,45-16,47)
x =min+ (max-min)
=16,47+ (1,085-1,080)
=17,06
1. Calculate the volume of the given solution of HCl. You are in need
to prepare 250 ml of 0,1 N solution of HCl:
CN=m(HCl)/ME.v
M(HCl) = ME.CN.v=36,46.0,1,0,25=0,912 g
x = m(HCl).100/m(solution)
m(solution) = m(HCl).100/x=0,912.100/17,06=5,346 g
v(solution) = m(solution)/ x = 4,936 ml
2. Using a cylinders, measure the calculated volume of HCl solution,
transfer it to a measured flask and adjust the volume up to 250 ml
using the distilled water.
Experiment 2
STANDARTISATION OF THE SOLUTION OF HCl
Theory: The solution of HCl prepared in the experiment 9 is just of an
approximate concentration. The determination of an exact
concentration of the solution (standardization) is performed with the
help of acid-base titration (NaOH as a working solution).
Goal: To determine an extract concentration of the prepared solution of
HCl.
Procedure:1. Adjust a burette on a stand.
2. Wash the burette first with the tap-water, then with the distilled
water, and after all with the solution of prepared HCl.
12
3. Fill the burette by the solution of HCl (use the funnel!). Take the air
bubbles away from the lowest part of the burette and adjust the volume
of the solution to the zero point. Measurethe volume, using the
lower meniscus of the liquid.
4.Using a pipette or a burette? Measure an aliquot part of 10.0 ml of
working solution of NaOH with a given concentration (CN(NaOH)) in
Erlenmayer (conic) flask.
3. Add 20 or 30 ml of distilled water and 2 or 3 drops of methilorange into the flask filled with NaOH. The indicator turns yellow.
4. Titrate the solution of NaOH by HCl (drowise by stirring) until the
indicator turns orange (not pink).
5. Write down the obtained volume of HCl.
8. Adjust the volume of HCl in burette up to the zero point.
9. Repeat the procedure 4-7 two more times.
10. Arrange your results as a table.
№
V(NaOH), m l
CN(NaOH)
V(HCl), ml
1
10
(Ask your teacher)
…
2
10
…
3…
10
…
11. Using the law of equivalents, calculate the exact concentration of
the prepared solution and its titre.
CN(NaOH).v(NaOH)
CN =
v(HCl)
CN(HCl).ME(HCl)
T(HCl) =
1000
12. Check your results.
Experiment 3
IONIC AND HETEROGENEOUS EQUILIBRIA IN SOLUTIONS
Some substances while being dissolved interact with molecules
of a solvent. As a result they dissociate and form ions. The process of
dissociation can be written as:
13
AnBm


nAm+ + mB n


Na2CO3

2Na+ + CO 23 

Ca(OH)2 
Ca2+ + 20H 
The process of dissociation can be quantitatively characterized
by a degree of dissociation (a) and a dissociation constant (K dis).
=
n d is
C
 dis
n total C total
n - number of molecules (moles)
C – molar concentration
If  > 0.3, the electrolyte is called strong. Strong electrolytes
are salts, strong acids (HCl, H2SO4, HNO3, HClO4 and some others),
hydroxides of alkaline and alkaline-earth metals.
If < 0.03, the electrolyte is called weak. The dissociation of a
weak electrolyte is a reversible and a stepwise process.
Kdis is determined as an equilibrium constant of a reaction of
dissociation of an electrolyte.
[ H  ][ H 2 PO4 ]
H3PO4


H+ + H2PO 4
( K1 
H2PO4-


H+ + HPO 2
4
( K2 

HPO 2

4
+
H +
PO 3
4
( K3 
[ H 3PO4 ]
7.5103 )
[ H  ][ HPO42 ]
[ H 2 PO4 ]
6.3108 )
[ H  ][ PO43 ]
1.31012 )
[ HPO42 ]
K1,K2,K3 - stepwise dissociation constants.
For the overall process of the dissociation of phosphoric acid
 3
3
  PO3 ( K  [ H ] [ PO4 ]  K  K K )
H 3PO4 
3
H

4
1 2 3
[ H 3PO4 ]
The dissociation constant of an electrolyte doesn't depend on
concentration but increases with the evaluation of temperature.
14
For an electrolyte AB dissociating into ions A+ and B  the
dissociation constant and the degree of dissociation are related by the
equation (Ostwald's dilution law):
K
 2C
(1 )
For very weak electrolytes and very diluted solutions the
equation may be simplified:
K = 2C
If an ion common to one of the ions formed in the dissociation
process of a weak electrolyte is introduced into the solution,
equilibrium of dissociation is violated and shifts towards the
direction of formation of undissociated molecules so that the degree of
dissociation of the electrolyte diminishes. For instance, the addition of
ammonium ions (for example ammonium chloride) to a solution of
ammonium hydroxide leads to an increase in concentration of NH4+
ions and,
in accordance with Le Chatelier's principle,
the


+
equilibrium of the dissociation NH4OH  NH4 + OH
shifts to
the left.
Ionic concentrations in solutions of strong electrolytes are quite
great so that the forces of interionic interaction manifest themselves
appreciably even at low concentration of an electrolyte. As a result,
the ions are not completely free. This is why the state of ions in a
solution is described, in addition to their concentration, by their
activity, i.e. the conditional (effective) concentration of ions in
accordance with which they act in chemical reactions.
a = f  .C
where a is activity; f is activity coefficient
The activity coefficients depend on the composition and
concentration of the solution, and on the charge and nature of the ion,
and on other conditions. It can be considered approximately that in
diluted solutions the activity coefficient of an ion depends only on
the charge of the ion and ionic strength of the solution:
15
log f = - 0.5  Z2  I1/2
I = 0.5 (C1  z12 + C2  z22 + C3  z32 + ... + Ci  zi2)
where
z - charge of the ion;
I - ionic strength of the solution; C
- concentrations of each ion in the solution
Equilibria in solutions of amphotheric electrolytes
Such hydroxides as Al(OH)3, Cr(OH)3, Zn(OH)2, Pb(OH)2,
Sn(OH)2 and some other can form both H+ and OH  ions while
dissociating in saturated solutions:
2H2O + Zn2+ + 2OH 


Zn(OH)2  2H2O


H2[Zn(OH)4]


Al(OH)3  3H2O


H3[Al(OH)6]


2H+ +
Zn(OH)4]2 
3H2O + Al3+ + 30H 
3


3H+ +
Al(OH)6]
The addition of strong acids and bases shifts the equilibria of
the dissociation process so that only one reaction becomes possible.
Equilibria in solutions with precipitates
In a saturated solution of a sparingly soluble strong electrolyte,
an equilibrium sets between the precipitate (solid phase) of the
electrolyte and its ions in the solution:

Ag3PO4 
3 Ag+ + PO 3
4
This equilibrium state may be characterized by a constant
called a solubility product, Ksp:
Ksp (Ag3PO4) = [Ag+]3 [PO 3
]
4
When the concentration of one of the ions of an electrolyte
in its saturated solution is increased (for instance, by introducing
16
another electrolyte containing common ions), the product of the
concentrations of the electrolyte ions becomes greater than K sp. In this
case the equilibrium between the solid phase and the solution shifts
towards the direction of formation of a precipitate.
Consequently, the condition for formation of a precipitate is
the greater value of the product of the concentrations of the ions
belonging to a sparingly soluble electrolyte in comparison with its
solubility product.
Conversely, if the concentration of one of the ions in a
saturated solution is diminished (for example by combining it with
another ion), the product of the ion concentrations will be lower than
the value of Ksp, the solution will become unsaturated and the
equilibrium between the solid phase and the solution shifts towards the
dissolution of the precipitate.
The value of Ksp can be used to calculate the solubility of
sparingly soluble electrolytes in water and in solutions containing other
electrolytes (solubility can be determined as a molar concentration of
a dissolved substance).
EXPERIMENTAL PART
1. Dissociation of weak electrolytes
a) Fill two test tubes with 3-4 drops of diluted acetic acid and
add 1 drop of an indicator (methyl orange). Add some solid sodium
acetate into one of the test tubes. Compare colors of the solutions in
the test tubes. Explain the results of the experiment.
b) Repeat the experiment with the solution of ammonium
hydroxide. Use phenolphtalein as indicator and add ammonium
chloride.
2. Formation of weak electrolytes
17
a) Take 3-4 drops of a CH3COONa solution in a test tube
and add some diluted HCl. Check the smell before and after the
experiment. Explain the results of the experiment.
b) Repeat the experiment with NH4Cl and NaOH
respectively.
3. Amphoteric hydroxides
a) Take 5-6 drops of a solution of any chromium (III) salt
and add dropwise a diluted solution of sodium hydroxide until the
precipitate of chromium hydroxide is formed. Divide the precipitate of
Cr(OH)3 into two parts and dissolve them in NaOH and HCl
respectively.
b) Repeat the same experiment using any zinc salt.
4. Formation and dissolution of a precipitate
Take 3-4 drops of a solution of any calcium salt into a test tube and
add a solution of ammonium oxalate until the formation of a
precipitate. Dissolve the precipitate in hydrochloric acid. Compare the
values of the dissociation constants of oxalic acid and the solubility
product constant for calcium oxalate and make your conclusion.
1.
2.
18
QUESTIONS AND PROBLEMS
The dissociating constant of butyric acid C3H7COOH is
5
1.5  10 . Calculate the degree of its dissociation in a 0.005 M
solution.
What is the hydrogen ion concentration [H+] in an aqueous
solution of formic acid if  = 0.03?
3.
4.
5.
6.
7.
Calculate the concentration of acetate ions in 0.1 M solution of
acetic acid in presence of 0.01 M HCl.
Calculate the ionic strength and the activities of the ions in a
solution containing 0.01 molel-1 of Ca(NO3)2 and 0.01 mol/l of
CaCl2.
Calculate the solubility product constant of PbBr2 if the solubility
2
of the salt is 1.32  10 molel-1.
Will a precipitate of silver sulfate be formed if a 1M solution of
H2SO4 is added to an equal volume of a 0.02 M solution of
AgNO3?
Calculate the equilibrium constant of the reaction and explain
whether the precipitate of calcium oxalate can be dissolved in
acetic acid.
Experiment 4
IONIC PRODUCT OF WATER. PH.
HYDROLYSIS OF SALTS
Water is a very weak electrolyte and dissociates in a very
small extent, forming hydrogen ions and hydroxide ions:


H2O 
H+ + OH
The equilibrium state of this reaction can be characterized by a
constant called ionic product of water:

 14
KW = [H+]  [OH ] = 10
(at 22oC)

+
Instead of concentrations of H and OH ions, it is more
convenient to use their common logarithms taken with the reverse
sign; these quantities are denoted as pH and pOH:

pH = - log[H+]; pOH = - log [OH ]
Solutions in which concentrations of hydrogen and hydroxide

ions equal each other are called neutral solutions, [H+] = [OH ] =
7

=10 , pH = 7. In acidic solutions, [H+] > [OH ], pH < 7; in

alkaline solutions [H+] < [OH ], pH > 7.
19
When we dissolve a salt comprising an anion of a weak acid or
a cation of a weak base in water, a hydrolysis process occurs, i.e. an
exchange reaction between salt and water, the result of which is the
formation of a weak acid or a weak base.
The process of hydrolysis can be quantitatively described using
the notions of hydrolysis constant (Kh) and degree of hydrolysis (h).
If a salt is formed by a weak acid and a strong base, the anion
of a salt undergoes hydrolysis, hydroxide ions are formed in the
solution:

NaCN + H2O 
NaOH + HCN
CN- + H2O
Kh =


OH- + HCN; pH > 7
[ OH  ][ HCN ] K w

Kacid
[ CN  ]
In the hydrolysis of a salt formed by a strong acid and a weak
base, the cation of the salt becomes hydrolyzed, the solution becomes
acidic:

NH4Cl + H2O 
NH4OH + HCl
NH4+ + H2O


H+ + NH4OH; pH < 7
[ H  ][ NH 4OH ] K w

Kbase
[ NH 4 ]
Kh =
When a salt formed by a weak acid and a weak base reacts
with water, both its cation and its anion become hydrolyzed. In this
case, the reaction of the solution depends on the relative strengths of
the acid and the base forming the salt:
NH4CH3COO + H2O
+
NH4 + CH3COO
20



+ H2O
NH4OH + CH3COOH


NH4OH + CH3COOH
Kh =
[ NH 4OH ][ CH 3COOH ]
Kw

Kacid Kbase
[ NH 4 ][ CH 3COO ]
The hydrolysis of salts formed by weak polybasic acids or
weak polyvalent metal ions proceeds stepwise. Under common
conditions, only first-step hydrolysis process should be taken into
consideration:

K2CO3 + H2O 
KHCO3 + KOH
CO32



+ H2O
AlCl3 + H2O
Al3+ + H2O




HCO 3 + OH

AlOHCl2 + HCl
AlOH2+ + H+
Degree of hydrolysis “h” is defined as a fraction of an
electrolyte that has become hydrolyzed. It is related to the hydrolysis
constant by an equation similar to the Ostwald dilution law for the
dissociation of a weak electrolyte:
h 2C
Kh = 1h
Most often, the hydrolyzed part of a salt is very small, h << 1.
In this case
Kh = h2C
The last equation shows that the degree of hydrolysis of a
given salt increases when its concentration diminishes.
The equilibrium of a reaction of hydrolysis can also be shifted
by change in temperature. As it is an endothermic process, the rise of
temperature leads to the increase of hydrolysis.

If we introduce a reagent combining with the H+ or OH ions
formed in hydrolysis into a solution of a hydrolyzing salt, in
accordance with Le Chatelier’s principle equilibrium will shift
towards the direction of an intensification of hydrolysis; as a result,
hydrolysis may go up to the end, leading to formation of its products.
21

The H+ (or OH ) ions can be combined together to form water
molecules by introducing another salt whose hydrolysis leads to the


accumulation of OH (or H+) ions in the solution; the H+ and OH
ions will neutralize one another, this causing the mutual intensification
of the hydrolysis of both salts and the formation of the hydrolysis
products.
EXPERIMENTAL PART
1. Determination of pH of solutions of some salts
1. Take some small amounts of solid salts Na2CO3,
Al2(SO4)3, Na3PO4, Na2B4O7, NH4Cl, (NH4)2CO3 or others into
different test tubes.
2. Add some distilled water into each test tube and stir until
the salts are dissolved.
3. Using a universal indicator, determine pH values of each
solution including pH of distilled water itself.
4. Write down the values of pH and the reactions of hydrolysis
of the given salts.
2. Shift of the equilibrium of hydrolysis
1. Take 4 drops of antimony chloiride solution and add some
distilled water. Pay attention to the precipitation of antimony
oxochloride SbOCl. (The formation of SbOCl is due to the
decomposition of Sb(OH)2Cl).
2. Add some drops of HCl until the precipitate is dissolved.
3. Write down two steps of hydrolysis of SbCl3 and explain
the shift of the equilibrium of the process.
22
3. Irreversible hydrolysis
1. Take 4-5 drops of aluminum sulfate into a test tube and add
the same amount of sodium carbonate solution.
2. Make sure that the obtained precipitate is not aluminum
carbonate but aluminum hydroxide (use amphoteric properties of
Al(OH)3).
4. Temperature dependence of hydrolysis
Take 4-5 drops of a 1M solution of sodium acetate into a test
tube and add 2 drops of an indicator phenolphtalein. Heat the
solution. Explain the change in color of the indicator.
1.
2.
3.
4.
5.
6.
QUESTIONS AND PROBLEMS
Calculate
pH of 0.1 M solution of NaOH (assume the
dissociation to be complete).
Calculate pH of a 0.01 M solution of acetic acid if the degree of
dissociation of the electrolyte equals 0.042.
Calculate pH of an ammonium buffer solution prepared by mixing
of equal volumes of 0.1 M solution of NH4OH and 0.01 M
solution of NH4Cl.
Which of the salts listed below undergo hydrolysis? Write the net
ionic equations and indicate whether aqueous solutions of salts are
neutral, acidic or basic. NaCN, KNO3, K2S, ZnCl2, NH4NO2,
MgSO4.
Calculate hydrolysis constant and degree of hydrolysis in 0.1 M
solutions of: a) NH4Cl; b) Na2CO3 (only the first step of
hydrolysis should be taken into consideration).
When aqueous solutions of Cr(NO3)3 and Na2S are mixed
together, a precipitate is formed and a gas is evolved. Write the
molecular and net ionic equations of the reaction.
23
Experiment 5
BUFFER SOLUTIONS
Buffers are used to maintain a constant pH value in the sample
solution upon addition thereto of small amounts of strong acids, strong
bases, or dilution.
As the buffer solutions, a mixture solution of weak acids or
weak bases and their salts or mixtures of salts of polybasic acids with
different degrees of substitution usually used.
Buffer systems can bind both H + and ОН ions at the addition
of strong acids and bases to produce weak electrolytes that slightly
alterin the pH of the solution.
Example: Acetate buffer solution comprises a mixture of
CH3COOH and CH3COONa. The dissociation of a weak electrolyte acetic acid – is reflected by the reaction equation:
CH3COOH
CH3COO- + H+
and is described by the equilibrium constant:
K a=
[CH 3 COO  ]  [ H  ]
= 1.8 10 5
[CH3 COOH ]
By adding sodium acetate, the concentration of CH3COO- ions
increases and is determined by the concentration of the salt:
[CH3COO-]=Csalt.
The dissociation of a weak electrolyte is reduced by the
introduction of a common ion, so
[CH3COOH]= Cacid.
Thus,
C
K a=
24
 [H  ]
salt
C
acid
;
[H+] = Ka 
C
acid ;
C
salt
C
pH = –log[H+] = pKa – log ( acid ),
C
salt
where pKa = - log Ka.
Thus, pH of a buffer solution is not dependent on the
concentrations of the components but is determined by their mole ratio.
By adding small amounts of a strong acid or a strong base
(alkali), the components of the buffer solution translating them into
weak electrolytes:
Examples.
1. At addition of NaOH, it can react with the qacetic acid:
CH3COOH + NaOH = CH3COONa + H2O
(the salt concentration increases accordantly to the concentration of the
added amount of the alkali, and the concentration of the weak acid
decreases to the same value):
pH = –log[H+] = pKa - log (
С
acid
NaOH ).
C
C
salt
NaOH
C
2. The addition of a strong acid provokes its interaction with
the sodium acetate:
CH3COONa + HCl = CH3COOH + NaCl,
C
pH = –log[H+] = pKa - log ( acid
С
NaOH ).
C
C
salt
NaOH
Since the change of the ratio of the concentrations is much less
than their sum or the difference, the total the pH value changes
insignificantly.
25
EXPERIMENTAL PART
Determination of pH of buffer solutions
Prepare acetate buffer solutions containing different
concentrations of the components according to the table presented
below (ask the technician about the concentrations of the reagents):
Buffer solution
1
10
10
2
18
2
3
6
14
4
2
18
Volume of the acid, ml
Volume of the salt, ml
Cacid/Csalt=Vacid /Vsalt
pH (measured)
pH (calculated)
2. Measure pHs of the prepared buffer solutions with the help
of the рН-meter, and put the results into the table.
3. Using theoretical formulae, calculate pH values of the
solutions.
4. Compare the theoretical and experimental data.
QUESTIONS AND PROBLEMS
1.
Write the reactions of the processes in the ammonium
buffer solution after addition of small amounts of HCl (a)
and NaOH (b).
2. Calculate рН of a formic buffer solution 1 liter of which
contains 0,1 mole of formic acid and 0.01 mole of sodium
formiate.
26
3. Calculate рН of an ammonia buffer solution, 500 ml of
which contains 5.25 g of ammonium hydroxide and 4.01 g
of ammonium chloride.
4. Calculate рН of an acetate buffer solution composed of
0.25 mole/l of acetic acid and 0.12 mole/l of sodium
acetate after addition of 0.02 mole/l of potassium
hydroxide.
Experiment 6
COLLOID SOLUTIONS
Colloid solutions are ultra-microheterogenous systems
consisting of two phases: a dispersion medium and a dispersed phase.
The dispersed phase comprises colloidal solutions of molecular
aggregates of a size of 10-7 – 10-5 cm.
Colloidal particles are small in size and can be suspended
indefinitely. This determines the kinetic stability of colloidal systems .
On the other hand, a large colloidal particle surface defines excessive
surface tension and the tendency to "sticking" of colloidal particles,
which reduces the energetic stability of colloidal systems.
Stabilization of colloidal systems is due to the adsorption of the
dispersion medium or the electrolyte ions present in the solution. On
the border of the colloidal particles, the electric double layer is formed.
Thus. colloidal particles acquire the same charge , which prevents
"sticking".
Ions that form the charge of colloidal particles are called
potential forming ions. The- counterions of the opposite charge are
located in the adsorption and diffusion layers of the micelle.
27
The ion exchange occurs between the adsorption and diffusion
layers. The colloidal particles are charged and the micelle electrically
neutral.
Example. The structure of the micelle of the colloidal solution
of silver iodide .
The formation of silver iodide can be represented as :
KI (excess) + AgNO3 = AgI + KNO3
The micelle structure : {m[AgI] nI- (n-x)K+}n- x K+
where "m" is the number of molecules of silver iodide in the core; "N"
is the number of anions in the first adsorption layer of the colloidal
particle ; n + x = m; typically m >> n.
EXPERIMENTAL PART
Preparation of colloid solutions by different methods
1. Ionic exchange
Mix in beakers at stirring: (a) 10 ml of a 0.05 M solution of KI
and 8 ml ml of a 0.05 M solution of AgNO3; (b) 10 ml of a solution of
AgNO3 and 8 ml of a solution of KI. Compare color of solutions in the
passing and reflected light.
Pass the colloid solutions into U-tubes and connect electrodes
of a direct current. Notice the direction of migration of colloid particles
(electrophoresis) and determine a particle charge sign.
Write the reactions of formation of colloid solutions and the
schemes of the micelles.
2. Peptization method
28
Take 1.5 ml of a 20% solution of K4[Fe(CN)6] and add 0.5 ml
of a saturated solution of FeCl3. Divide the dark-blue precipitate of
KFe[Fe(CN)6] and put it onto two folded filter papers in funnels. Wash
one part of the precipitate with the distilled water, and the another part
with the 0.1 М solution of H2C2O4. Observe the peptization of the
precipitate in the second case (C2O42- ions adsorb on the surface of the
nucleus of the colloid particle, and the hydrogen cations act as counterions).
Compare color of solutions in the passing and reflected light.
Write the reaction of formation of a colloid solution and the
scheme of the micelle.
QUESTIONS AND PROBLEMS
1. After uncompleted titration of a solution of KBr by AgNO3, a
colloid solution is formed. Write the reaction of formation of a
colloid solution and the scheme of the micelle.
2. After uncompleted titration of a solution of AgNO3 by KBr, a
colloid solution is formed. Write the reaction of formation of a
colloid solution and the scheme of the micelle.
3. Write the reaction of full hydrolysis of FeCl3 and the scheme of the
micelle of the colloid particle.
4. Write the reaction of interaction of excess of K4[Fe(CN)6] with
FeCl3 and the scheme of the micelle of the colloid particle.
29
5. The solution of CuSO4 was added to the S-sole with positively
charged micelle particles. Which ions will provoke coagulation?
6. The solutions of KCl, K2SO4, K3PO4 and MgSO4 of the same molar
concentrations were separately added to Au-sole with positively
charged micelle particles. Which of the added electrolytes will
possess the least coagulation threshold value?
Experiment 7
OXIDATION-REDUCTION REACTIONS
In oxidation-reduction reactions (redox-reactions), the
oxidation number of one or more elements in the reacting substances
changes. The loosing of electrons by an atom attended by an increase
in its oxidation number is called oxidation; the gaining of electrons by
an atom attended by a decrease in its oxidation number is called
reduction.
A substance containing an element that undergoes oxidation is
called a reducing agent. These are almost all metals and some nonmetals (C, H2 and others, negatively charged ions of non-metals (S2  ,
I  , N3  and others), cations in intermediate oxidation numbers (Sn2+,
Fe2+ and others), ions containing elements in intermediate oxidation
numbers (SO32  , NO2  , SnO22  and others). In laboratories, such
reducing agents as H2, SO2, KI, H3PO3, H2S, HNO2 are usually used.
A substance containing an element that undergoes reduction is
called an oxidizing agent. These are atoms and molecules of some nonmetals of high activity (O2, O3, Cl2 and others) positively charged
metallic ions (Fe3+, Cu2+, Hg2+ and others), particles containing ions in
their highest oxidation numbers (MnO4  , NO3  , SO42  , Cr2O72  ,
ClO3  and others). The strongest oxidizing agent is electrical current
(oxidation on anode). In laboratories, such oxidizing agents as KMnO4,
K2Cr2O7, HNO3, H2SO4 (conc.), H2O2, PbO2 are used.
30
Oxidizing and reducing properties of substances are described
with the help of electrode potentials of systems.
The standard electrode potential (is defined as the
potential of a given electrode at concentrations (activities) of all the
substances participating in the electrode process equal unity.
The dependence of an electrode potential on concentrations of
substances participating in electrode processes and on temperature is
expressed by the Nernst equation:
RT
[ Ox]
 = o + 2.3 nF log [Re d ]
where R - the molar gas constant; T - absolute temperature; F - the
Faraday’s constant; n - number of electrons participating in the
electrode process; [Ox] - concentration of the oxidized form of a
substance; [Red] - concentration of the reduced form of a substance.
In case if T = 297 K (25oC),
 = o +
0.059
n
[ Ox ]
log [Re d ]
The more is the absolute value of redox potential, the stronger
are oxidizing properties of the oxidized form.
The possibility of a redox-reaction can be determined from the
electromotive force of the reaction (E):
E = (ox) - (red)
In case if E > 0, the direct redox-reaction is possible. In case if
E < 0, the direct redox-reaction is impossible and the reaction proceeds
in the backward direction.
EXPERIMENTAL PART
1. Transfer of an ion to a higher oxidation state
Take 6-8 drops of chromium (III) nitrate and add excess of
NaOH. When the precipitate of chromium hydroxide dissolves add 3-4
drops of 3% H2O2 solution. Heat the mixture until the color turns
yellow. Write down and balance a corresponding redox-reaction.
31
2. Redox properties of hydrogen peroxide
a). Take 3 drops of KI solution, add 2 drops of diluted sulfuric
acid and 3% H2O2 solution. Add some starch solution to indicate the
evaluation of iodine. Write down and balance a corresponding redoxreaction.
b). Take 6 drops of KMnO4 solution, add 2 drops of diluted
sulfuric acid and dropwise 3% H2O2. Observe the evaluation of oxygen.
Write down and balance a corresponding redox-reaction.
3. Oxidizing properties of potassium permanganate
Fill 3 test tubes with 5 drops of KMnO4 each. Add 2 drops of
diluted sulfuric acid into the first test tube, 2 drops of distilled water
into the second, and 2 drops of sodium hydroxide into the third one.
Add Na2SO3 solution into each test tube until the change of their color.
Write down and balance corresponding redox-reactions.
4. Oxidation of cations of d-elements
Take 2 drops of a manganese (II) salt in a test tube, then add 56 drops of nitric acid and some crystals of NaBiO3. Observe the
appearance of a pink color of HMnO4.. Write down and balance a
corresponding redox-reaction.
5. Reducing properties of p-elements
Take 3-4 drops of a solution of SnCl2 in a test tube, add
NaOH until the formed precipitate dissolves, and 2-3 of a bithmuth (III)
salt solution. Observe the black precipitate of elementary bithmuth.
Write down and balance a corresponding redox-reaction.
32
QUESTIONS AND PROBLEMS
1. Complete the equations of the following reactions and
balance them:
(a) K2S + KMnO4 + H2SO4 = S + ....
(b) KI + K2Cr2O7 + H2 SO4 = I2 + ...
( c) K MnO4 + H2O2 = ...
2. Indicate the direction in which the following reactions can
proceed spontaneously:
(a) H2O2 + HClO = HCl + O2 + H2O
(b) H3PO4 + 2HI = H3PO3 + I2 + H2O
3. Can a salt of iron (III) be reduced to a salt of iron (II) in an
aqueous solution by (a) potassium bromide, (b) potassium iodide?
4. Using the table of standard electrode potentials, calculate the
equilibrium constants for the following reactions:
(a) Zn + CuSO4 = Cu + ZnSO4
(b) Sn + Pb(CH3COO)2 = Sn(CH3COO)2 + Pb
Experiment 8
COMPLEX (COORDINATE) COMPOUNDS
By complex (coordinate) compounds are meant definite
chemical compounds formed by a combination of individual
components without formation of new pairs of electrons.
Examples of complex compounds are Na3[Co(NO2)6],
[Cu(NH3)4]SO4, K4[Fe(CN)6].
In a molecule of a complex compound, one of the atoms,
generally positively charged, occupies the central site (central ion or
complexing agent). Oppositely charged ions or neutral molecules
called ligands are coordinated around the central ion. The complexing
agent and the ligands form inner sphere of a complex compound. It is
33
characterized by coordinate bonds which are formed while overlapping
of empty p- and d-orbitals of a central ion and orbitals containing lone
electron pairs of ligands. The ions in the outer sphere are mainly
bonded to the complex ions by forces of electrostatic interaction (ionic
bonds).
The total number of coordinate bonds formed by the
complexing agent with ligands is known as coordination number of
the central ion. In accordance with the number of coordinate bonds
formed by a ligand with the central ion, the ligand may be a
monodentate, bidentate, or polydentate.
Majority of complex compounds are electrolytes. In solutions
they dissociate and form both simple and complex ions (outer and inner
spheres). This type of dissociation is an irreversible process (complex
compounds are strong electrolytes).

[Cu(NH3)4]SO4  [Cu(NH3)4]2+ + SO42
The inner sphere of a complex compound dissociates reversibly
and stepwise (complex ions are weak electrolytes):

[Cu(NH3)4]2+ 
[Cu(NH3)3]2+ + NH3
[Cu(NH3)3]2+


[Cu(NH3)2]2+ + NH3
[Cu(NH3)2]2+


[Cu(NH3)]2+ + NH3

[Cu(NH3)]2+ 
Cu2+ + NH3
Each of the above processes can be characterized by an
equilibrium constant (stepwise instability constants of a complex ion).
The equilibrium constant of the overall process is called an overall
instability constant of a complex ion. The less the value of the overall
instability constant is, the more stable is the complex ion.
The shifting of dissociation equilibrium in systems containing
complex ions follows the same rules as in solutions of simple (noncomplex) electrolytes, namely: equilibrium shifts towards the direction
of the most complete binding of the complexing agent or ligand so that
the concentrations of these particles remaining unbound in the solution
take on the minimum possible values in these conditions. The
34
equilibrium also shifts towards the formation of a more stable complex
ion.
According to the character of dissociation of complex
compounds, cationic, anionic and neutral complexes are distinguished.
In cationic complexes ligands are usually neutral molecules so
that the inner sphere is charged positively:
[Cr(H2O)6]Cl3,
[Co(NH3)6]Cl3.
In anionic complexes ligands are usually negatively charged:
K2[HgI4], Na[Sb(OH)6].
Neutral complexes have both anions and neutral molecules as
ligands: [Pt(NH3)2Cl2]. They have no outer sphere and don’t dissociate
in aqueous solutions.
EXPERIMENTAL PART
1. Formation of complex compounds
Put 4-5 drops of silver nitrate into a test tube. Add ammonium
hydroxide until the precipitate is formed. Add excess of NH4OH and
observe the formation of a complex compound.
Repeat the procedure with CuSO4, NiCl2, Zn(NO3)2. Write
down the reactions. Find the values of overall instability constants and
compare stabilities of the above complex compounds.
Notice: (a) instead of Cu(OH)2, (CuOH)2SO4 precipitate is
formed;
Zn2+
(b) coordination number of silver ion is 2; for Cu2+ and
it is 4; for Ni2+ - 6.
2. Destruction of complex compounds
Take solutions of chloride diammine silver (I) and sulphate
tetraammine copper (II), obtained from the previous experiment, add
35
dropwise diluted (1:1) nitric acid until the complex compounds are
destroyed.
3. Formation and properties of Cd2+ and Hg2+ complexes
Take two test tubes filled with solutions of Cd2+ and Hg2+ salts
respectively. Add a saturated solution of Na2SO3. Observe the
formation of sulfide precipitates and their further dissolution with the
formation of complex compounds Na2[M(SO3)2], where M= Cd2+ or
Hg2+.
Add some drops of NaOH to the solutions of complex
compounds. Do metal hydroxides precipitate? Why?
4. Dissociation of complex compounds
a). Fill 3 test tubes with 3-5 drops of a double salt
NH4Fe(SO4)2. Add 3-4 drops of NaOH into the first test tube and heat
the solution. Smell the vapors. Write down your observations.
Add 2-3 drops of BaCl2 to the second test tube and 2-3 drops of
KSCN to the third one. Make your conclusion about the dissociation
of double salts in aqueous solutions. Write down the reactions of
dissociation of a double salt and reactions of ionic exchange.
b). Fill 2 test tubes with 3-5 drops of K3[Fe(CN)6]. Add 2-3
drops of Na3[Co(NO2)6] and 2-3 drops of KSCN to the second one.
Write down your observations and the reactions of dissociation of a
complex compound and ionic exchange. (Notice: in presence of K +
ions, the yellow precipitate of K2Na[Co(NO2)6] is formed.)
1.
36
QUESTIONS AND PROBLEMS
For the following complex compounds indicate: (a) inner and
outer spheres; (b) central ion, its charge and coordinating number;
(c) ligands ; (d) write the reaction of dissociation of complex
2.
3.
4.
compounds and complex ions; (e) express overall instability
constants; (f) name the following complex compounds.
[Cd(NH3)4]Cl2 ; K2[Cd(CN)4]
Which of the above mentioned complexes is more stable?
Calculate the concentration of the Ag+ ions in a 0.1 M solution
containing an excess of 1 mole  l-1 of NH3.
In which case will a reaction occur between solutions of the
electrolytes indicated below (exchange of ligands)? Write the
equations of these reactions in molecular and net ionic forms and
calculate their equilibrium constants:
(a) K2[HgI4] + KCN =
(b) K[Ag(CN)2] + NH3 =
Seminar 5
ISOMERISM AND NOMENCLATURE OF
HYDROCARBONS
The theory of the structure of organic compounds was
formulated by AM Butlerov in 1861 and includes the following
provisions:
a)
All the atoms in the molecule are linked together in a
strict sequence according to their valences. The chemical structure
determines the way of connecting the atoms in a molecule.
b)
Properties of organic compounds depend not only on
the qualitative and quantitative composition of substances , but also on
the chemical structure of the molecule.
c)
Atoms in the molecule have a mutual influence on each
other, i.e., properties of atomic groups in the molecule may vary
depending on the nature of the other atoms in the molecule. A group of
atoms that determines the chemical properties of organic molecules is
called a functional group.
d)
Each organic compound having only one chemical
formula. Knowing the chemical formula, we can predict the properties
37
of the compound, and by studying its properties, in practice, to
establish the chemical formula.
Isomers are compounds having the same composition but
different chemical structure. Isomers have different chemical
properties.
The special features of organic compounds can also be
considered as the existence of a homologous series, in which each
successive term can be generated from the previous addition of one
specific for a given number of groups of atoms. Homologues have
similar chemical properties, while the physical properties change
somewhat with increasing molar mass compounds.
1.
2.
3.
4.
5.
6.
38
QUESTIONS AND PROBLEMS
Write the structural formulas of all the connections and
determine which ones are isomers: а) 2,2,3- trimethylbutane;
b) 2,3-dimethylbutane; c) 2,3,4-trimethyl-3-etilpentan; d) 3methylhexane.
Write the structural formulas of all the connections and
determine which of them are homologous: а) 2,3,4-trimethyl-2;
b) 3,4- dimethyl-2;
c) Isopropyl
cyclopentane; d) 4-methylpentene-2.
Write the structural formulas of all the connections and define
the types of hybridization of carbon atoms and the spatial
structure of molecules:
а) 2,6-dimethyl-3,5-diethyl-4izopropilgeptan;
b)
2,3,4-trimethyl-1;c)
2,3-1,3dimetilpentadien; г) butyn-1.
Write the structural formulas and name all isomeric
hydrocarbon compounds: С5Н11Cl, С4Н6, С6Н10, С9Н12.
Can the geometric isomerism in the following compounds: а)
2-methylpropene; b) 1,1-dichloroethylene; c) 2-pentene; d)
2,3-Dimethyl-2?
Is it possible to have an optical isomer of the following
compounds: а) chlorobromomethane; b) 1,2-dichloroethane;
c) 1,2-dichloropropane; d) 1,2-dichloropropene?
Write the structural formula 2,2,4,5-tetrametilgeksana. Specify
the number of primary, secondary, tertiary and quaternary
carbon atoms.
Experiment 9
CHEMICAL PROPERTIES OF HYDROCARBONS
Hydrocarbons are organic compounds that contain only carbon
and hydrogen atoms. Their characteristic feature is the lack of
functional groups. Properties of hydrocarbons determined by the
structure of the hydrocarbon radical.
Alkanes (CnH2n+2, acyclic, all carbon atoms are in sp3hybridization condition).
Chemical properties. Low polar covalent bonds of alkanes and
very durable, so saturated hydrocarbons in chemical reactions are
inactive. The main type of chemical reactions of alkanes is radical
substitution reaction, SR.
Alkenes (CnH2n are acyclic compounds with the carbon atoms
in the states of sp3- and sp2 hybridization with one double bond).
Typical reactions are addition reactions (electrophilic mechanism, A E),
and polymerization.
Aromatic
hydrocarbons
or
arenes
(CnH2n-6
are
cycliccompounds with the carbon atoms in sp2-hybridization state, a
molecule contains conjugated system of double bonds).
Conjugated π- electron system of benzene and its homologues
is an energetically favourable state, so its destruction occurs with great
difficulty. Addition reaction to arenes are not typical, more common
are electrophilic substitution reactions (SE). They occur in presence of
catalysts: Fe, trivalent metal salts - Lewis acids.
EXPERIMENTAL PART
39
1. Methane production and study of its properties.
а) production of methane. A mixture of anhydrous sodium
acetate and soda lime in a ratio of 6:10 stir thoroughly, and place a tube
in a horizontally fixed tripod foot. Close the vial stopper with a gas
outlet tube. Methane is produced by heating the reaction mixture in
vitro by the reaction
b) The chemical properties of methane. To study the chemical
properties of methane pour 0.5 ml of a 2% solution of potassium
permanganate and bromine water into two test tubes. Bubble the
previously formed methane for 1-2 minutes into a solution of
potassium permanganate, and then - into bromine water. Note whether
the color of the solutions in test tubes changes. Make a conclusion
about the interaction of methane with aqueous solutions of an oxidizer
and a halogen.
c) Combustion of methane. Ignite methane venting at the end
of the tube, then stop heating tubes. Write down the reaction equation
and note the color of the flame of the burning methane.
2. Preparation of ethylene and study of its properties.
а) Preparation of ethylene. A dry test tube, place 1 ml of
ethanol, 3 ml of concentrated sulfuric acid and a few boiling chips
porcelain required to boil when heated evenly. Close the vial stopper
with a gas outlet tube and that no liquid poured out, attach to a tripod at
an angle to the surface of the table. Ethylene obtained by heating the
reaction mixture by the reaction:
40
b) Chemical properties of ethylene. To study the chemical
properties of the ethylene as in experiment 1b prepare two test tubes
with solutions of potassium permanganate, bromine, and 0.5 mL of
water. The tube is to be gently heated and the evolved gas to be passed
through a solution of potassium permanganate, then at the same time
through the bromine water. Write the reactions, note whether the color
of the solution changed. Conclude about the interaction of ethylene
with an oxidizer and water solutions of halogens.
3 Interaction of benzene and toluene with potassium
permanganate.
А) Fill the tube with 1 ml of benzene, and add 1 ml of an
aqueous solution of potassium permanganate. was Vigorously shake
the mixture for 20-30 seconds. Note whether the discoloration of the
solution takes place Make a conclusion, explaining the reason.
B) Toluene oxidation with potassium permanganate. Mix in a
test tube 0.5 ml of toluene, 2 ml of an aqueous potassium permanganate
solution and 2 ml of 10% sulfuric acid solution. The mixture should be
shaken before bleaching potassium permanganate. Record observation
and write the reaction of oxidation of toluene to benzoic acid. Explain
why, unlike previous experience, the oxidation occurs.
QUESTIONS AND PROBLEMS
1. Using a reaction can be distinguished from propene propyne?
41
2. How to get from methane, benzene? Write the equations for all
reactions and specify the conditions of their occurrence.
3. What hydrocarbons can be obtained by the action of sodium
metal to a mixture of 2-chloroethane and chloropropane?
4. Write the reaction of polymerization of 2-methylbutadiene-1.3.
4.
5. What is obtained bromo derivative is preferably bromination of
ethylbenzene in the presence of metallic iron?
6. 6. Using a reaction can be distinguished from benzene
cyclohexene?
7. How can I get nitrobenzene from calcium carbide? Write the
equations for all reactions and specify the conditions of their
occurrence.
7.
42
O2
KOH, C2H5OH
CH3 CH2 CH2 CH2Br
HBr
A
0
8.
KOH, H2O
C
D
HBr, H2O2
KMnO4, H2O
G
B
Br2, 500 C
E
F
Experiment 10
ALKOHOLS AND PHENOLS
Monohydric alcohols contain one hydroxyl group bonded to a
saturated hydrocarbon radical. The general formula: R-OH, CnH2n+1OH.
For alcohols, regioisomers of the functional group and isomers of the
carbon skeleton are known.
Characteristics of the chemical properties of alcohols are
defined by the presence of a functional group OH. Electronegative
oxygen atom pulls on shared electron pairs, thus acquiring a negative
charge, and the neighboring carbon and hydrogen atoms produce
positive charges. Displacement of the electron pairs leads to the
formation of polar bonds which are less stable and more reactive.
Polyhydric alcohols contain several hydroxyl groups attached
to different carbon atoms. They possess more acidic properties
compared with the monohydric analogues.
Phenols are aromatic organic compounds in which the
hydroxyl groups of the molecules are linked to carbon atoms of the
aromatic ring. The benzene ring increases the acidic properties of
hydroxy group. Thus, phenols possess properties of weak acids.
43
EXPERIMENTAL PART
1. Alcohols solubility in water.
Take the three tubes and fill them with 5 ml of distilled water.
First , add 1 ml of ethanol , the second - 1ml butyl alcohol , in the third
- 1 ml of amyl alcohol . Shake each tube. Record your observations and
draw conclusions about the solubility of alcohols , depending on where
in the homologous series . Divide each aqueous-alcoholic solution into
two test tubes and add one of them to 2-3 drops of phenolphthalein ,
and in another - 2-3 drops of methyl orange . Do you see the color
change indicators? Draw conclusions about the acid-base properties of
alcohols.
2. Oxidation of alcohols.
а) Burning of alcohols (performed in a fume hood!). In 3
porcelain cups, pour 1 ml of ethyl, butyl, and amyl alcohols. In turn
ignite the contents. Note the nature of the flame in all three cups.
Record the reactions and observations.
b) oxidation of ethanol. Mix 2 drops of ethanol and 2 drops of
a solution of potassium bichromate and 1 drop of 10% sulfuric acid
solution in a test tube. Gently heat the reaction mixture over the flame
until the color change. Observe the formation of a green colored
chromium(III) salts. Smell the cooled test tube gently: the resulting
acetaldehyde has a characteristic odor of green apples.
44
3. Preparation of chelates of polyhydric alcohols (glycolates).
Prepare a fresh solution of copper hydroxide (II). For this, mix
5 drops of 2 % solution of copper sulfate (II) and 5 drops of 10 %
sodium hydroxide solution. Mold half of the resulting copper
hydroxide (II) into a second test tube. Add 1 drop of ethylene glycol to
the first test tube and shake, and- 1 drop of glycerin to the second one.
Whether the dissolution of the precipitate takes place? What color is
the resulting solution? Record the observations. Can this reaction be
considered as a qualitative reaction for polyhydrous alcohols?
4. Bromination of the phenol.
Fill the test tube with 1 ml of 3% aqueous phenol solution and
add it to 1 ml of bromine water. Shake the contents of the tube. Note
the appearance of white flakes of tribromophenol. Write down the
reaction equation, specify its mechanism. Explain why the bromination
of phenol, unlike benzene bromination occurs without the catalyst.
QUESTIONS AND PROBLEMS
1. 1. Write down all the possible isomers of compounds С4Н7ОН.
2. 2. Write the reaction of obtaining glycol pent-2 ..
3. 3. As can be discerned methylpropanol and 2-methylpropanol1?
4.Make the conversion:
45
а)
b)
Experiment 11
ALDEHYDES AND KETONES
Aldehydes and ketones are carbonyl containing compounds:
O
C
.
Carbon atom of the carbonyl group is in the sp2-hybridization
(flat fragment). Electrons of the double bond are strongly biased
toward the more electronegative oxygen atom (C=O bond is polar).
Redistribution of charges in the carbonyl group has an effect on the
polarity of the C-H bonds adjacent to the carbonyl group carbon atom
(-position).
EXPERIMENTAL PART
1. Oxidation of the formaldehyde and acetone with diamminosilver
hydroxide
46
Tollens reagent preparation. Pour 0.5 ml of a 1% silver nitrate
solution and 0.5 ml of 10% sodium hydroxide solution into a clean test
tube. Add dropwise a 10% solution of ammonia to dissolve the formed
precipitate. Write down the reactions.
Split the resulting solution into two test tubes. Into one of
them, add 1 ml of formalin (40% formaldehyde solution in water), and
into another - 1 mL of acetone. Place both test tubes for 5 minutes in a
water bath heated to 50C. In any case, silver metal is deposited on the
walls of the test tubes in the form of shiny plaque (reaction "silver
mirror"). Compare reducing activity of formaldehyde and acetone.
2. Oxidation of the formaldehyde and acetone copper hydroxide
(II)
Prepare a fresh solution of copper hydroxide (II). For this, mix
1 ml of 2% solution of copper sulfate (II), and 1 ml of 10% sodium
hydroxide solution in a test tube. Mold half of the resulting solution to
precipitate copper hydroxide (II) into a second tube. Then add to the 10
drops of a solution of formalin to the 1 st test tube, and 10 drops of
acetone to the 2nd one. Gently heat the test tubes in the flame of the
burner until boiling. Note and explain the observations
Cu(OH)2
CuOH
Cu2O
Cu
blue
yellow
red
«copper
mirror»
3. Preparation of hydrogen sulfite derivatives of aldehydes
and ketones
а) Reaction of benzaldehyde with sodium hydrogen sulphite.
Pour 1 ml of a saturated sodium hydrogen sulfite solution and 0.5 ml of
benzaldehyde into a test tube. Shake the mixture and cool under cold
running water. Put the crystals on a glass plate and examine in a
microscope. Write down the corresponding reaction.
b) Reaction o0f acetone with sodium hydrogen sulphite.
47
Mix 1 drop of a saturated solution of sodium hydrosulfite and 1
drop of acetone on a glass plate. Examine the formed crystals in a
microscope. Write down the corresponding reaction. Compare the
shape of the crystals in both experiments.
QUESTIONS AND PROBLEMS
1. Give examples of the synthesis of aldehydes from alkene,
alkane, alkyne. Specify the reaction conditions.
2. In a test tube, there is a mixture of heptane, heptanol and
heptanal. Describe how you can carry out the separation of
these substances.
3. Write the equation reactions of 2-methyl butanal with ammonia
solution of silver oxide.
4. Write the equation of chemical reactions between:
а)
benzaldehyde with copper hydroxide (II); b) benzaldehyde
diamine silver hydroxyde; c) methyl ethyl ketone with
hydrazine; d) propanone with sodium hydrosulphite; e)
crotonic condensation of propionaldehyde.
Experiment 12
CARBOXYLIC ACIDS. AMINOACIDS
Carboxylic acids compose a class of organic compounds
containing a carboxyl functional group
O .
C
OH
In its structure, the carboxyl group can be associated with only
one carbon atom (end group), so the isomerism of carboxylic acids is
48
associated with the isomerism of the carbon skeleton or position of
multiple bonds.
The reaction centers in the molecules of carboxylic acids are:
A. OH of the carboxyl group.
B. C-OH of the carboxyl group.
C. R-COOH.
D. C-H bond of the hydrogen atom in the -position
EXPERIMENTAL PART
1. Dissociation of carboxylic acids.
а) Determination of pH of the solution of carboxylic acid. In
three test tubes respectively prepare solutions of acetic acid, benzoic
acid and oxalic acid in the following ways. A. Take 0.5 ml of 2N acetic
acid solution and add the same amount of distilled water. B. Dissolve a
few crystals of benzoic acid in 1 ml of water. To accelerate the
dissolution of benzoic acid , heat the test tube in a water bath. C.
Dissolve a few crystals of oxalic acid in 1 ml of water. Plunge a clean
glass rod into each solution, and touch the universal indicator paper
slide. Compare the color of the indicator strip with the scale , write
down the equation of dissociation of acids, indicate the color of the
universal indicator and pH of the solutions. Leave the solutions for the
experiment 1b
b) Strengths of carboxylic acids. Add 1 ml of a saturated
sodium carbonate solution into each solution of the experiment 1a.
Observe the changes? Write down the reactions, calculate the
equilibrium constants of the exchange reactions Kdis.СН3СООН =
1,75.10-5; Кdis.С6Н5СООН = 1,26.10-5; Кdis.Н2С2О4 = 5,4.10-2;
Кdis.НС2О4- = 5,4.10-5; Кdis.Н2СО3 = 4,5.10-7; Кdis.НСОО- = 4,7.10-11).
49
Conclude the strengths of carboxylic acids comparing with the
inorganic acids.
2. Oxidation of oxalic acid with potassium permanganate.
Prepare an aqueous solution of oxalic acid as described in the
experiment 1a. Add 2-3 dropsof 2N sulfuric acid solution and 1-2 drops
of a 2% solution of potassium permanganate. Heat the mixture on a
water bath. Which gas is released by the reaction? Record your
observations write down the reaction equation and indicate the
oxidating and reducing agents.
3. Oxidation of oleic acid with potassium permanganate. To 1
ml of oleic acid (СН3(СН2)7-СН=СН-(СН2)7СООН), add 1 ml of a 2%
solution of potassium permanganate. Vigorously shake the contents of
the tube. Oleic acid is oxidized to the dihydroxy acid and potassium
permanganate
is reduced to manganese dioxide. Record your
observations and reaction equation. Specify a class of organic
compounds, which is characterized by this reaction.
4. Preparation of the copper (II) complex compound with αaminopropionic acid.
Pour 10 ml of a 2% solution of alanine (α-aminopropionic acid)
into a conical flask and add 0.5 g of CuCO3. Boil the mixture. Write
down the scheme for the formation of the complex salt and note its
color.
H 2N
CH 3
NH 2
Cu 2+
O C O
O C O
H 3C
50
QUESTIONS AND PROBLEMS
1. 1. Write the formula of carboxylic acid composition C5H10O2 .
2. Write the equation of chemical reactions:
а) formic acid with ammoniacal silver oxide;
b) linoleic acid with bromine water;
c) boiling solution of oxalic acid;
d) benzoic acid with bromine in the presence of iron;
e) producing succinic acid chloride.
f) Preparing poly methacrylic acid;
g) phthalic acid with heating;
h) oxalic acid, sodium carbonate;
i) oxalic acid, sodium carbonate
j) Stearic acid with ammonia;
k) Obtaining dipeptide -aminobutyric acid
Syllabus "Chemistry"
The purpose of studying chemistry course is to build the system
of knowledge about the structure of matter, the basic laws of chemical
reactions, patterns in the chemical behavior of the main classes of
inorganic and organic compounds in relation to their structure to use
this knowledge as a basis for the study of the processes occurring in
living organisms, and base materials used in dental practice.
Course content
51
№
1.
2.
Name of section
discipline
structure of Matter
Thermodynamics and
kinetics of chemical
reactions
Table of contents
Wave-particle duality of the material
world. The wave function. Electronic
configuration of atoms and ions.
Periodic law of DI Mendeleev. Chemical
bonding. Valence bond method.
Hybridization of orbitals. The spatial
configuration of molecules. Chemical
bond in complex compounds. Ionic
bond.
Basics of thermochemistry. Enthalpy.
Hess's law. Entropy. Gibbs free energy.
Terms of spontaneous reaction.
The speed of a chemical reaction. Order
of reaction. Chemical equilibrium. The
rate constant and the equilibrium
constant. Displacement of the chemical
equilibrium.
The concepts
catalysis.
3.
52
Chemical reactions in
solutions
of
adsorption
and
General concepts of disperse systems.
Ways of expressing concentration of
solutions: mass fraction, the title, the
mole, the normal concentration.
The theory of electrolytic dissociation.
Dependence of the acid-base properties
of the electrolyte on the nature of their
dissociation. Amphoteric electrolytes
(ampholytes). Ionic reaction. Terms
reactions of ion exchange.
4.
Chemical equilibrium Weak electrolytes. Dilution law.
in solutions
Common ion effect. Strong electrolytes.
Activity and activity coefficient. Ionic
strength. The ionic product of water.
Hydrogen index. Buffer solutions.
Hydrolysis salts. Hydrolysis constant.
The dependence of the hydrolysis
temperature and the concentration of
solutions.
Solubility constant. Solubility. Terms of
dissolution and precipitation.
Electrolytic
dissociation
constant
instability and complex compounds.
Colloidal solutions.
5.
Classes of inorganic
compounds
The main classes of inorganic
compounds. Double oxides. Ceramic
materials.
6.
General properties of
metals
General
physical
and
chemical
properties of metals. Alloys. Alloys. The
microstructure and phase diagrams of
different types of alloys.
7.
electrochemical
processes
Electrochemical
processes.
The
emergence of the electric double layer at
the metal-electrolyte interface. Electrode
potential
methods
of
measuring.
Electrochemical voltage series of metals.
The
principle
of
operation
of
electrochemical cells. Electrochemical
corrosion.
53
8.
Organic chemistry as a branch of science
that studies the structure and functioning
of biologically active molecules from the
standpoint of organic chemistry. Basics of
the theory of chemical structure. Isomers.
Classes of organic compounds.
Mechanisms of organic reactions.
9. Hydrocarbons.
Saturated and unsaturated hydrocarbons:
the main types of chemical reactions of
alkanes and alkenes. Conjugated dienes.
1,2- and 1,4-addition to the conjugated
diene. Polymerization of conjugated
dienes.
Aromatics. Electrophilic substitution
reactions on the aromatic ring.
10. Alcohols and phenols Alcohols (Alcohol and alkanols). Atomic
alcohols. Hydrogen bond. Reactivity of
alcohols.
Phenol. The acidic properties of phenol.
11. Organic
carbonyl Aldehydes and ketones. Electronic
compound
structure of the carbonyl group. Oxo
compounds by reaction of the carbonyl
group and -position. Dialdehydes and
diketones. Acetylacetone. Keto-enol
tautomerism.
54
Introduction to
organic chemistry
12. Carboxylic acids and Carboxylic acid. The structure of the
their derivatives
carboxyl group. Derivatives of carboxylic
acids to salts , halides , anhydrides ,
amides , nitriles , esters . Methods of
production and properties . Natural higher
fatty acids (IVH ) : palmitic, stearic ,
oleic, linoleic, linolenic , arachidonic .
Lipids and phospholipids. Enzymatic
hydrolysis of fats . Oxidation of acid in
the body. Fragments of the nucleic acid in
phosphoric acid and adenosine phosphate
. Phosphatides . Lecithin and cephalin .
Hydroxy acids . Structure and
nomenclature of hydroxy acids . Lactic
acid formation during lactic acid
fermentation and in the muscles. The
conversion of lactic acid to pyruvic acid
by metabolism . Malic, tartaric and citric
acid. Examples optical isomers of lactic
and tartaric acids .
13. amines
Amines. The basic properties of amines.
Diamines. Ethylenediamine, putrescine,
cadaverine, hexamethylenediamine biological
significance
and
their
applications.
14. Amino Acids
The amino acids that make up proteins:
classification, structure, nomenclature,
stereoisomerism, acid-base properties
(formation of a bipolar ion). The
chemical properties of amino acids.
Biologically important reactions
-amino acids: deamination (oxidative and
non-oxidative),
hydroxylation,
decarboxylation
-amino
acids
(education
kolamina,
histamine,
tryptamine).
55
15. Peptides and proteins
16. carbohydrates
17. heterocyclic
compounds
Peptides and proteins. The primary
structure of proteins. Partial and complete
hydrolysis. The concept of complex
proteins. Glycoproteins, lipoproteins,
nucleoproteins, fosfoproteidy.
Carbohydrates. Carbohydrates in nature.
The
value
of
carbohydrates.
Photosynthesis. Monosaccharides. Cyclochain tautomerism. D- and L- series.
Reactions of monosaccharides functional
groups. Glucose, mannose, galactose,
fructose, ribose and deoxyribose; finding
in nature and biological significance.
Ascorbic acid. Reducing and nonreducing
disaccharides
sucrose,
maltose,
cellobiose, lactose.
Biologically important heterocyclic
systems.
Nucleic acid. Nucleobases.
The rating system of assessing students on the course:
Work in the semester
The maximum number of points scored in the semester - 100
Number of
Number of lump sum
Type of work
entries
points
scores
laboratory
14
2
28
experiments
Examinations
10
2
20
2
10
20
colloquia
1
12
12
Total for the
80
semester
final certification
20
56
TOTAL
100 scores
points
estimates
ECTS grades
95 - 100
86-94
69-85
61-68
51-60
31-50
0-30
5+
5
4
3+
3
2+
2
А
В
С
D
Е
FX
F
The cited literature
1. Rupert Wentworth and Barbara H. Munk. Experiments in
General Chemistry, Lab Manual. Amazon Press, 2012.
2. Irene G. Cesa. Flinn Scientific Laboratory Experiments for
General, Organic and Biological Chemistry. Flinn Scientific
Inc., 2014.
3. Ebbing, Darrell; Gammon, Steven D. Student Solutions
Manual for Ebbing/Gammon's General Chemistry, 10th
edition. Cengage Learning, 2012.
4. Glinka N.L. Problems and questions in General Chemistry.
Integral Press, 2005 and other editions.
57
Oльга Владимировна Ковальчукова,
Насрин Намичемази,
Русул Алабада
Лабораторные работы по химии. Для студентов I
курса медицинского факультета специальности
«Стоматология»
(на английском языке)
Зав. редакцией Т.О. Сергеева
Техн. редактор И.М.Любавская
Тематический план 2015 г., №
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