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МЕТОДИЧЕСКИЕ УКАЗАНИЯ СТУДЕНТАМ
CLASS 1.
TOPIC: Introduction in a subject. Safety regularions in a chemical
laboratory.
1. Introduction in a subject. (20 min.).
1.1. Control of a student’s initial level in chemistry. (60 min.).
1.2. Ways of expressing concentration of solutions (mass fraction, molarity)
(30 min.).
2. Safety regularions in a chemical laboratory (20 min.).
Safety regularions in a chemical laboratory.
Working in a chemical laboratory requires following certain safety
regulations.
Using small amounts of chemicals reduces occurance of accidents but does
not exclude them. That is why everyone working in a chemical laboratory must
be aware of and keep to safety regulations.
• Be careful. Keep in mind that being careless and inattentive can cause an
accident.
• Do not let metallic K, Na or Li touch water to avoid the inflammation and
explosion of hydrogen that is formed in the traction.
• Alkaline metals, caustic alcalines, concentrated acids exposed to the skin
and mucus cause severe burns. Injuring eyes is particularly hasardous. When
a concentrated acid or alkaline gets onto the skin, wash the burn with the
running water.
• Dilute H2SO4 with water by adding the acid to the water drop by drop
but not vice versa, stirring the solution all the time.
• Be careful when washing the glass with chromium mixture, so as the
mixture does not get onto the skin, clothes or foot wear.
• Wash off the concentrated acids and alkalies with a strong stream of the
running water. Spilled concentrated acids and alcalies mist be cavered with
sand and cleaned away.
• If a thermometre is broken up and mercury spills, it should be collected with
a special trap or a bulb. The minute particles of mercury are collected with a
crush made of white tin. The surface of the floor or the table, contaminated
with mercury is thoroughly soaked with 20% of FeCl3 solution or covered
with sulphur.
• Heat the liquid in the test-tube gradually. Direct the outlet of the test tube
sideway from youself and other students, working next to you as partial
overheating can bring on a gush of the liquid.
• Don’t bend over a test-tube with a boiling liquid. Do not breathe in chemical
substances even if their amounts are very small. To smell the substance you
should direct the vapour or gases towards yourself with the movement of
your hand.
• You must not taste reagents.
• All the experiments with foul-smelling as well as poisonous substances
(aniline, bromine) are done in the exhaust-hood.
• When working with a drain tube, you can take away the burner from under
the test tube with the regent mixture only when the end of the pipe is
removed from the liquid. Other wise the liquid can get sucked into the
overheated test tube and cause the reagent mixture splashing about and
getting on to the skin.
• Work with easily imflammable liquids (ether, benzene, acetone etc.) in the
distance from the open fire. Their heating is done on the water-bath or sandbath.
• Taking liquids (acid and base solutions, salts of heavy metals, volatile
substances) is done with a special pipette.
• The students should wear gouns and hats when in lab classes.
• All experiments are done in clean glass. After finishing work, the glassware
is washed and prepared for the next class.
• Containers with reagents are put in their place and closed with the same
stoppers as they have been closed before. It is forbidden to move containers
with reagents from one table to another.
• Books and bags as well as other students’ belongins are kept in the box of
the laboratory table.
• A student on duty is appointed in every group. He/she is responsible for
- bringing all the necessary things from the lab room;
- keeping the laboratory clean and tidy;
- checking every student’s working place at the end of the class;
- reporting to the lab assistant about the condition of the laboratory.
CLASS 2.
TOPIC: Solutions. Ways of expressing concentration of solutions.
1. SEMINAR. (75 min.).
1.1. Ways of expressing concentration of solutions (mass fraction, molarity,
molality, molar concentration of the equivalent, titer).
1.2. Preparation of a solution of a given concentration.
2. LABORATORY WORK. (35 min.).
2.1. Preparation of a solution from a fixanale.
2.2. Preparation of a standard solution from an initial substance with a given
weight.
2.3. Preparation of a delute solution of a certain concentration from a
concentrated solution of mineral acid.
Literature
1. Lectures.
2. Ebbing D.D. General Chemistry/ D.D.Ebbing , M.S.Wrighton;.- Boston.
Third
edition. Houghton Mifflin Company. – Р. 459 - 466, 471 - 475.
3. Lister T. Chemistry for Advanced Level / T. Lister, J. Renshaw; - Third
edition.
Stanley Thornes (Publishers) Ltd. 2000. – Р. 247-259.
PRACTICAL PROBLEMS
1. What volume (in mL) of 40% H3PO4 solution (ρ=1,25 g/cm3) is necessary to
prepare 400mL of 0,25N of phosphoric acid solution (ρ=1g/cm3)? Calculate the
mole fraction of H3PO4 in the obtained solution.
2. How many grams of Na2CO3·10H2O are necessary to prepare 100mL of 10%
Na2CO3 solution (ρ=1,12 g/cm3)? What is the mole fraction and the molarity of the
obtained Na2CO3 solution?
3. How many grams of K2SO4 are necessary to prepare 100mL of 0,1N K2SO4
solution? Calculate the mass and the mole fraction of K2SO4 in the obtained
solution (density of the solution = 1,0 g/cm3).
4. How many mL of 15% potassium chloride solution (ρ=1,092g/cm3) are
necessary to prepare 200mL of 8% solution (ρ=1,025g/cm3)? Calculate the mole
fraction, molality, normality and titer of the obtained solution.
5. The mass of oxalic acid crystalline hydrate H2C2O4·2H2O (3,1547g) was
dissolved in a volumetric flask of 750mL. Calculate the molality, normality and
titer of this solution.
6. How many grams of crystal CuSO4. 5H2O are necessary to prepare 500 mL of
5% CaCl2 solution with ρ = 1,049 g/cm3. Describe how you would prepare this
solution.
2. LABORATORY WORK
2.1. Preparation of solutions of a given concentration from a fixed weight.
In practice exactly weighed ready made amounts of chemically pure solid
substances or exactly measured volumes of their solutions are used to prepare
titrant of certain normality.
These substances are placed in a special glass ampule and soldered. These
ampules with strictly defined amounts of a substance are called fixed substances.
They usually contain 0,1 mole substance equivalents; just this is the quantity
required for preparing 1 liter of 0,1N solution.
Normality of a given solution can be calculated by the equation:
n( 1 X )
z
1
C( X ) =
z
V ( solution)
Where: n ( 1 z X) is the number of moles of substance equivalent
(it must be shown on the container);
V is a delivery flask volume, liter.
To prepare a required titer solution the glass ampule is broken up over a
funnel with a breaking rod and the substance is transferred into a delivery flask.
The ampule and the funnel are washed with water and some water is added, up to
the mark etched on the neck of the flask. Then it is carefully stirred.
2.2. Preparation of a standard solution from an initial substance with a given
weight.
A standard solution is a solution, the concentration of which is known. The
substances used for the preparation of a standard solution, are called initial or
standard substances. They should meet the following requirements:
• be chemically pure and have a crystal structure;
• be soluble well enough and not hydroscopic;
• the chemical composition of a substance must correspond to its
formula;
• have a relatively big molar mass.
• the solutions of these substances must not change and they must be airresistant when stored;
For preparation of a titrant from a given substance, it is weighed on a platform
balance (the accuracy of ±0,0001g), then it is transferred carefully to a volumetric
flask, and some water is added to bring the solution level to the mark on the neck
of the volumetric flask. Calculate the normality and titer of the obtained solution.
T (X ) =
m( X )
,
V
T ( X ) ⋅ 1000
C( 1 X ) =
z
M(1 X)
z
Where: m(X) is the weight mass, g;
V is the volume of the delivery flask, mL;
M( 1 z X) is the molar mass of the substance equivalent.
Problem 1.
How many grams of oxalic acid should be added to a 100-mL volumetric
flask to prepare 0,1 N oxalic acid solution when the flask is filled up to the mark
with water? Describe, how you would prepare this solution.
Oxalic acid, H2C2O4·2H2O, is used in a neutralization method as a standard
substance to determine normality and titer of alkali, as it meets all the requirements
to initial substances. The prepared initial oxalic acid solution is used as a standard
solution in acid – base volumetric analysis.
Solution
Using the formula
n( 1 X )
m( X )
z
1
, we calculate
C( X ) =
=
z
V ( solution) M ( 1 X ) ⋅ V ( solution)
z
m (H2C2O4·2H2O) = C( 1 2 H2C2O4·2H2O)·M( 1 2 H2C2O4·2H2O)·V;
m (H2C2O4·2H2O) = 0,1mol/L· 63,04g/mol · 0,1L = 0,6304 g.
Preparation of a solution: 0.6304 g C2H2O4. 2H2O is weighed on a
platform balance. The oxalic acid crystals are transferred carefully to a volumetric
flask. Water is added to bring the solution level to the mark on the neck of the 100ml volumetric flask.
Problem 2.
A sample of C2H2O4.2H2O weighing 0.6822g is placed in a 100-mL volumetric
flask. The flask is then filled with water up to the mark on the neck. What is the
normality and titer of the resulting solution?
Solution
The concentration of the obtained solution is calculated by the formula:
m( 1 H 2 C 2 O4 ⋅ 2 H 2 O )
2
1
,
m( H 2 C 2 O4 ⋅ 2 H 2 O ) =
2
M ( 1 H 2 C 2 O4 ⋅ 2 H 2 O) ⋅ V ( solution)
2
C ( 1 H 2 C 2 O4 ⋅ 2 H 2 O ) =
2
0,6822 g
= 0,1082mol / L
g
63,04
⋅ 0,1L
mol
m( 1 H 2 C 2 O4 ⋅ 2 H 2 O) 0,6822 g
2
T ( H 2 C 2 O4 ⋅ 2 H 2 O ) =
=
= 0,006822( g / mL)
V ( solution)
100mL
2.3. Preparation of a delute solution of a certain concentration
from a concentrated solution of mineral acid.
Problem 3
The given concentrated sulfuric acid is an aqueous solution, containing 40%
of H2SO4 by mass. The density is 1.30 g/mL. How many milliliters of concentrated
H2SO4 are required to make 200 mL of 0,1 N sulfuric acid solution?
Solution
We first calculate the mass of H2SO4 that is contained in 250mL of 0,15N
solution:
m( H 2 SO4 )
C ( 1 H 2 SO4 ) =
2
M ( 1 H 2 SO4 ) ⋅ V ( solution)
2
m (H2SO4) = C( 1 2 H2SO4)·M( 1 2 H2SO4)·V = 0,1 mol L ·49 g mol ·0,2 L =0,98 g.
Knowing the mass of H2SO4 we can calculate the mass of the concentrated
solution:
m( solution) =
m( H 2 SO4 ) ⋅ 100% 0,98 ⋅ 100
=
= 2,45( g ) ,
40%
40
then the volume of concentrated solution will be:
V =
m
ρ
=
2,45
= 1,88(mL)
1,30
Preparation of a solution: use a measuring pipet to deliver 3.53 mL of 40%
H2SO4 solution into a 250-mL volumetric flask and then add water up to the mark
on the neck of the flask and stir.
Enter the results of the laboratory work in the record of proceedings.
CLASS 3,4.
TOPIC : Acid –base method of measure analysis. (Neutralization method).
1. SEMINAR. (60 min.).
1.1.
Classification
of
volumetric
analysis
methods.(Neutralization,
oxidation-reduction, precipitation, complex formation methods).
1.2. Basics of the volumetric analysis. The Law of equivalent.
1.3. Theoretical grounds of neutralization method.
1.4. Titration curves and acid-base indicators.
1.5. The ways of calculation of normality and titer of solutions.
1.6. Measuring vessels. Volumetric analysis technique.
2. LABORATORY WORK. (45 min.).
2.1. Determination of normality and titer of alkali solution by a standard
solution of oxalic acid.
2.2. Determination of normality and titer of acid solution by the titrant of
alkali.
Literature.
1. Lectures.
2. Ebbing D.D. General Chemistry/ D.D.Ebbing , M.S.Wrighton;.- Boston.
Third
edition. Houghton Mifflin Company. – Р. 121-127, 666 – 670.
3. Lister T. Chemistry for Advanced Level / T. Lister, J. Renshaw; - Third
edition.
Stanley Thornes (Publishers) Ltd. 2000. – Р. 341 – 348.
PRACTICAL PROBLEMS
1. A 15,0 mL of NaOH solution is titrated with 0.075 N HCl , and 10,0 mL is
required to reach the endpoint.
a. What is the equivalent molar concentration and titer of NaOH solution?
b. If the density of the NaOH solution is 1g/cm3, what is the mass percent of
NaOH in the given solution?
2. Sodium hydroxide solution is usually standardized by titration of a pure sample
of potassium hydrogen phthalate (KHP), an acid with one acidic hydrogen and a
molar mass of 204.22 g. It takes 20.46 mL of a sodium hydroxide solution to titrate
a 0,1082–g sample of KHP. What is the molarity and normality of the sodium
hydroxide?
3. A sample of CH3COOH weighing 2.4 g is placed in a 100-mL volumetric flask.
The flask is then filled with water to the mark on the neck. A 50,0 mL of the
obtained solution is titrated with 14,4 mL of 0,15 N KOH solution. Calculate the
mass fraction of the formic acid in the original solution.
4. A 10-mL sample of vinegar, an aqueous solution of acetic acid (CH3COOH), is
titrated with 0,5062 N NaOH, and 16,58 mL is required to reach the endpoint.
a. What is the molarity of the acetic acid?
b. If the density of the vinegar is 1,006 g/cm3, what is the mass percent of
acetic acid in the vinegar?
EXAMPLE PROBLEMS
Example 1.
A 5,1-mL sample of NaOH solution requires 5,0 mL of 0,1030 N oxalic acid
for complete neutralization. What is the titer and molar concentration of KOH
equivalent. What indicator may be used in this case of titration?
Solution
The molar concentration of substance equivalent may be calculated using
the law of equivalents.
Two substances react completely in quantities,
proportional to their equivalents, the volumes of the two reacting solutions
inversely proportional to their normalities, that is:
n( 1 z X1) = n( 1 z X2);
from where
or
C( 1 z X2) =
C( 1 z X1) · V1 = C( 1 z X2) · V2;
C ( 1 X 1 ) ⋅ V1
z
;
V2
or
C ( 1 H 2 C 2 O4 ) ⋅ V ( H 2 C 2 O4 )
2
(mol / L) ;
V ( KOH )
C ( 1 NaOH ) ⋅ M ( 1 NaOH )
1
1
T ( NaOH ) =
g / mL;
1000
Ñ ( 1 KOH ) =
1
Ñ ( 1 NaOH ) =
1
0,1030 ⋅ 5,0
= 0,1010mol / L ;
5,1
T ( NaOH ) =
0,1010 ⋅ 40
= 0,004040 g / mL;
1000
The titer of substance solution is determined sometimes by using such a
value as the titer of standard solution per the determined substance. It shows
what mass of the substance that should be determined can be titrated by 1 mL of
standard solution. The calculation is carried out by the equation:
1
1
 s tan d .  C ( z s tan d .) ⋅ M ( z det erm.)
T
=
( g / mL)

1000
 det erm. 
T
1
1
H 2 C 2 O4 ⋅ 2 H 2 O C ( 2 H 2 C 2 O4 ⋅ 2 H 2 O) ⋅ M ( 1 NaOH ) 0,1030 ⋅ 40
=
=
= 0,004120( g / mL)
NaOH
1000
1000
It means that 0,004120g of KOH is neutralized by 1 mL of 0,1030 N oxalic
acid solution.
The potassium hydroxide titer solution:
 s tan d .  V ( s tan d .)
T (det erm) = T 
( g / mL) ;
⋅
 det erm.  V (det erm.)
 H C O  V ( H 2 C 2 O4 )
T ( NaOH ) = T  2 2 4  ⋅
( g / mL)
 NaOH  V ( NaOH )
T ( NaOH ) = 0,004120 ⋅
5,0
= 0,004039( g / mL)
5,1
Knowing the titer of a solution, we may calculate its normality :
T (det erm.) ⋅ 1000
C ( 1 (det erm.) =
( mol / L)
z
M ( 1 det erm.)
z
T ( NaOH ) ⋅ 1000 0,004039 ⋅ 1000
C ( 1 NaOH ) =
=
= 0,1010(mol / L)
1
40
M ( 1 NaOH )
1
The mass of the determined substance in the volume to be titrated is found like
 s tan d . 
m(det erm.) = T 
 ⋅ V ( s tan d .)( g )
 det erm. 
this:
m( NaOH ) = T (
H 2 C 2 O4
) ⋅ V ( NaOH ) = 0,004120 ⋅ 5,0 = 0,0206( g )
NaOH
T ( NaOH ) =
m( NaOH ) 0,0206
=
= 0,004039( g / mL)
V
5,1
In this case we use phenolphthalein as an indicator. (See Appendix. Titration
curves. Figure 2.)
Example 2.
A 5,0-mL sample of Na2CO3 solution is titrated with HCl solution, and 5,8
mL is required to reach the endpoint. Calculate the normality and the titer of HCl
solution. The titer of Na2CO3 solution equals 0,005936g/mL. Which of the
indicators could be used for doing that titration?
Solution
A sodium carbonate, Na2CO3, is the substance which may be used as a
standard substance in neutralization method for the determination of mineral acids
titer. In water Na2CO3 is hydrolyzed forming alkali solution:
Na2CO3 + 2 H2O = H2O + CO2 + 2 NaOH
This base, NaOH, is titrated with HCl acid solution. To detect the equivalence
point, we will add an indicator that changes color within the pH range 3-11
(dramatic change of pH on the titration curve of strong acid – strong base. See
Appendix. Figure 1). Calculation of normality and titer of HCl acid solution is the
same as in the previous example.
The molar concentration of sodium carbonate equivalent is calculated by
formula:
T ( Na 2 CO3 ) ⋅ 1000 0,005936 ⋅ 1000
C ( 1 Na 2 CO3 ) =
=
= 0,1120mol / L
2
53
M ( 1 Na 2 CO3 )
2
The molar concentration of HCl substance equivalent is calculated by using
the law of equivalents.
C ( 1 HCl ) =
1
T ( HCl ) =
C ( 1 Na 2 CO3 ) ⋅ V ( Na 2 CO3 ) 0,1120 ⋅ 5,0
2
=
= 0,0965mol / L
V ( HCl )
5,8
C ( 1 HCl ) ⋅ M ( 1 HCl ) 0,0965 ⋅ 36,5
1
1
=
= 0,003522 g / mL
1000
1000
The titer of HCl solution is calculated by using the such a value as titer of
standard solution per detemined.
1
1
 Na CO  C ( 2 Na 2 CO3 ) ⋅ M ( 1 HCl ) 0,1120 ⋅ 36,5
T 2 3  =
=
= 0,004088 g / mL
1000
1000
 HCl 
5,0
 Na CO  V ( Na 2 CO3 )
T ( HCl ) = T  2 3  ⋅
= 0,004088 ⋅
= 0,003524 g / mL
5,8
 HCl  V ( HCl )
T ( HCl ) ⋅ 1000 0,003524 ⋅ 1000
C ( 1 HCl ) =
=
= 0,0965mol / L
1
36,5
M ( 1 HCl )
1
2. LABORATORY WORK
2.1. Determination of normality and titer of alkali solution by a standard solution
of oxalic acid. Phenolphthalein is an indicator.
Fill a buret with alkali solution. Put 5 mL of standard oxalic acid solution
into a beaker by using a measuring pipette. Then add a small amount (2-3 drops) of
phenolphthalein indicator. The base is added drop by drop to the acidic solution in
the beaker during the titration, the indicator changes color, but the color disappears
on mixing. The stoichiometric (equivalence) point is marked by appearance of
light – pink color, that isn’t vanishing for 30 seconds.
Carry out titration to take three concurrent readings. Make a table of the
titration readings. Calculate the average volume, and then determine normality and
titer of alkali solution.
Write the equation of the reaction.
2.2. Determination of normality and titer of the mineral acid solution by
a titrant of base.
Fill a buret with a mineral acid solution. Put 5 mL of base solution into a
beaker by using a measuring pipette. Then add a small amount (2-3 drops) of
methyl orange indicator. In basic solution the color of methyl orange is yellow.
The acid is added drop by drop to the alkali solution in the beaker during the
titration, the indicator changes color. Stoichiometric (equivalence) point is marked
by appearance of yellow color.
Carry out titration to take three concurrent readings. Make a table of the
titration readings. Calculate the average volume, and then determine normality and
titer of alkali solution.
Write the equation of the reaction.
CLASS 5.
TOPIC: Oxidation-reduction measure analysis. (Permanganatometry).
1. SEMINAR. (60 min.).
1.1. The characteristics of oxidation-reduction reactions. Oxidizing and
reducing agents. The oxidation numbers.
1.2. Balancing oxidation-reduction equations. Oxidizing and reducing agent
equivalent.
1.3. Methods of oxidation-reduction titration.
1.4. Oxidation-reduction reactions in the body.
1.5. Volumetric analysis by using potassium permanganate as a standard
solution.
2. LABORATORY WORK. (45 min.).
2.1. Determination of normality and titer of a potassium permanganate
solution by a standard solution of oxalic acid.
2.2. Determination of normality and titer of hydrogen peroxide solution by
the titrant of potassium permanganate.
Literature.
1. Lectures.
2. Ebbing D.D. General Chemistry/ D.D.Ebbing , M.S.Wrighton;.- Boston.
Third
edition. Houghton Mifflin Company. – Р. 514-533.
3. Lister T. Chemistry for Advanced Level / T. Lister, J. Renshaw; - Third
edition.
Stanley Thornes (Publishers) Ltd. 2000. – Р. 149 – 152.
PRACTICAL PROBLEMS
1. Define: oxidation, reduction; oxidizing agent, reducing agent.
2. During an oxidation process in an oxidation-reduction reaction the species
oxidized _________ electrons.
3. During an oxidation process in an oxidation-reduction reaction the species
__________ is the oxidizing agent.
4. During an oxidation process in an oxidation-reduction reaction the species
_________ is the reducing agent.
5. Assign oxidation states to all atoms in each compound.
a) KMnO4,
d) P4O6
b) NiO2
e) Na2C2O4
c) K4[Fe(CN)6]
f) F2O
6. How to calculate a reducing and oxidizing agent equivalent? Is the equivalent of
oxidizing and reducing agents a constant value? Give examples.
7. Balance the following oxidation-reduction reactions using the half-reaction
method. Identify the oxidizing and reducing agents.
a) KMnO4 + Na2C2O4 + H2SO4 → MnSO4 + K2SO4 + Na2SO4 + CO2 + H2O
b) KMnO4 + Na2SO3 + H2O → MnO2 + Na2SO4 + KOH
c) KMnO4 + Na2SO3 + KOH → K2MnO4 + Na2SO4 + H2O
d) HI + HNO3 → I2 + NO + H2O
e) Cr(OH)3 + H2O2 + KOH → K2CrO4 + H2O
f) MnCl2 + H2O2 + KOH → H2MnO3↓ + KCl + H2O
g) K2Cr2O7 + H2SO4 + Na2SO3 → Cr2(SO4)3 + Na2SO4 + K2SO4 + H2O
h) P + HNO3 + H2O → H3PO4 + NO↑
i) HClO3 → HClO4 + ClO2 + H2O
j) K2Cr2O7 + KI + H2SO4 → Cr2(SO4)3 + I2 + K2SO4 + H2O
k) Ag + HNO3 → AgNO3 + NO↑ + H2O
l) CrCl3 + KMnO4 + KOH → K2CrO4 + KCl + MnO2 + H2O
8. Balance the following skeleton equations. The reactions occur in acidic or basic
solution, as indicated in parentheses.
a) MnO4− + S2− → MnO2 + S
(basic)
b) IO3− + HSO3 − → I− + SO42−
(acidic)
c) Cl2 → Cl− + ClO−
(basic)
9. What solutions are used as standard solutions in permanganatometry analysis?
10. What substance is used as an indicator in permanganatometry analysis?
11. A solution of permanganate is standardized by titration with oxalic acid
(H2C2O4). It required 28.97 mL of the permanganate solution to react completely
with 0.1058 g of oxalic acid. The unbalanced equation for the reaction is
MnO4−(aq) + H2C2O4(aq) → Mn2+(aq) + CO2(g)
What is the molarity and normality of the permanganate solution?
12. The iron content of iron ore can be determined by titration with standard
KMnO4 solution. The iron ore is dissolved in HCl, and all the iron is reduced to
Fe2+ ions. This solution is then titrated with KMnO4 solution, producing Fe3+ and
Mn+2 ions in acidic solution, required 38,37 mL of 0,0198 N KMnO4 to titrate a
solution made from 0,6128 g of iron ore. What is the mass percent of iron in the
iron ore?
EXAMPLE PROBLEMS
Example 1.
Balancing oxidation-reduction equations occurring in acidic solution by
the half-reaction method
• Write the equations for the oxidation and reduction half-reactions.
• For each half reaction:
a. balance all the elements except hydrogen and oxygen;
b. balance oxygen using H2O;
c. balance hydrogen using H+;
d. balance the charge using electrons.
• If necessary, multiply one or both balanced half-reactions by integers to
equalize the number of electrons transferred in the two half-reactions.
• Add the half-reactions, and cancel identical species.
• Check to be sure that the elements and charges balance.
Balance the equation
KMnO4 + FeSO4 + H2SO4 → MnSO4 + K2SO4 + Fe2(SO4)3 + H2O
using the half-reaction method.
Solution
The ionic equation is
K++MnO4−+Fe2++SO42−+2H++SO42−→
→ Mn2++SO42−+2K++SO42−+2Fe3++3SO42−+ H2O
MnO4- + Fe2+ + H+→ Mn2+ + Fe3+ + H2O
First, carefully observe products and reactants to detect which ions, molecules,
or atoms are oxidized or reduced. Examination of our unbalanced equation shows
that Fe2+ changes to Fe3+ and that MnO4- changes to Mn2+. We therefore write our
two half-reactions as
and
Fe2+ → Fe3+
(incomplete half-reaction)
MnO4- → Mn2+
(incomplete half-reaction)
Balance each half-reaction.
For the reduction reaction, we have
MnO4− (aq) → Mn2+(aq)
a) The manganese is balanced.
b) We balance oxygen by adding 4H2O to the right side of the equation:
MnO4− (aq) → Mn2+(aq) + 4H2O(l)
c) Next, we balance hydrogen by adding 8H+ to the left side:
8H+(aq) + MnO4− (aq) → Mn2+(aq) + 4H2O(l)
d) All the elements have been balanced, but we need to balance the charge by
using electrons. At this point we have the following charges for reactants and
products in the reduction half-reaction:
8H+(aq) + MnO4− (aq) → Mn2+(aq) + 4H2O(l)
+8
+
-1
+2
+
0
+7
+2
We can equalize the charges by adding five electrons to the left side:
8H+(aq) + MnO4− (aq) + 5e- → Mn2+(aq) + 4H2O(l)
Both the elements and the charges are now balanced, so this represents the
balanced reduction half-reaction.
For the oxidation reaction,
Fe2+(aq) → Fe3+(aq)
The half-reaction for iron is balanced in iron but not in charge. Thus, we add
one electron to the right side.
Fe2+(aq) → Fe3+(aq) + eBecause in Fe2(SO4)3 molecule there are two iron atoms, we may write
oxidation reaction in such a way:
2Fe2+(aq) → 2Fe3+(aq) + 2eThis half-reaction is balanced. Note that electrons are lost, as we expect for
oxidation.
The last step is to equalize the number of electrons lost and gained in
oxidation and reduction reactions. For that multiply the reduction reaction by 2,
and oxidation reaction by 5 before the addition:
8H+(aq) + MnO4− (aq) + 5e- → Mn2+(aq) + 4H2O(l)
x2
2Fe2+(aq) → 2Fe3+(aq) + 2ex5
_____________________________________________________________
16H+(aq) + 2MnO4−(aq) + 10Fe2 +(aq) → 2Mn2+(aq) + 8H2O(l) +10Fe3+(aq)
2KMnO4 + 10FeSO4 + 8H2SO4 → 2MnSO4 + K2SO4 + 5Fe2(SO4)3 + 8H2O
Example 2.
Balancing oxidation-reduction equations occurring in basic solution by
the half-reaction method
• Write the equations for the oxidation and reduction half-reactions.
• For each half reaction:
a. balance all the elements except hydrogen and oxygen;
b. balance oxygen and hydrogen atoms by adding OH- or H2O;
c. balance the charge by using electrons.
• If necessary, multiply one or both balanced half-reactions by integers to
equalize the number of electrons transferred in the two half-reactions.
• Add the half-reactions, and cancel identical species.
• Check to be sure that the elements and charges balance.
Balance the equation
MnCl2 + H2O2 + KOH → H2MnO3 ↓ + H2O + KCl
by using the half-reaction method.
Solution
The ionic equation is:
Mn2+ + 2Cl− + H2O2 + K+ + Cl− → H2MnO3↓ + H2O + K+ + Cl−
Mn2+ + H2O2 → H2MnO3↓ + H2O
Mn2+ → H2MnO3↓
(incomplete half-reaction)
H2O2 → H2O
(incomplete half-reaction)
Balance each half-reaction.
For the oxidation reaction, we have
Mn2+(aq) → H2MnO3 ↓(s)
a) The manganese is balanced.
b) We balance oxygen and hydrogen atoms by adding OH- to the left side and
H2O molecules to the right side of the equation:
Mn2+(aq) + 4OH− (aq) → H2MnO3 ↓(s) + H2O(l)
c) All the elements have been balanced, but we need to balance the charge by
using electrons. At this point we have the following charges for reactants and
products in the reduction half-reaction:
Mn2+(aq) + 4OH− (aq) → H2MnO3 ↓(s) + H2O(l) + 2e+2
+
-4
0
+
0
-2
0
We can equalize the charges by adding two electrons to the right side:
Both the elements and the charges are now balanced, so this represents the
balanced oxidation half-reaction. Note that electrons are lost, as we expect for
oxidation.
For the reduction reaction
H2O2(l) → H2O(l)
H2O2(l) + 2H+(aq) + 2e- → 2H2O(l)
The last step is to equalize the number of electrons lost and gained in
oxidation and reduction reactions. In our equations they are equal. Add the halfreactions, and cancel identical species:
Mn2+(aq) + 4OH− (aq) → H2MnO3 ↓(s) + H2O(l) + 2eH2O2(l) + 2H+(aq) + 2e- → 2H2O(l)
____________________________________________________________________________________
Mn2+(aq) + H2O2(l) + 2H+(aq) + 4OH− (aq) → H2MnO3 ↓(s) + 3H2O(l)
2H2O + 2OH−
MnCl2 + H2O2 + 2KOH → H2MnO3 ↓ + H2O + 2KCl
Oxidation-reduction reactions are commonly used as a basis for volumetric
analytical procedures. For example, a reducing substance can be titrated with a
solution of a strong oxidizing agent, or vice versa. One of the most frequently used
oxidizing agents is aqueous solution of potassium permanganate.
The reaction that occurs in acidic solution is the one most commonly used:
MnO4− (aq) + 8H+(aq) + 5e- → Mn2+(aq) + 4H2O(l)
Permanganate has the advantage of being its own indicator – the MnO4- ion
is intensely purple, and the Mn2+ ion is almost colorless. As long as some reducing
agent remains in the solution being titrated, the solution remains colorless
(assuming all other species present are colorless), since the purple MnO4- ion being
added is converted to the essentially colorless Mn2+ ion. However, when the whole
of the reducing agent has been consumed, the next drop of permanganate titrant
will turn the solution being titrated light purple. Thus the endpoint (where the color
change indicates the titration should stop) occurs approximately one drop beyond
the stoichiometric point (the actual point at which the whole of the reducing agent
has been consumed completely).
2. LABORATORY WORK
2.1 Determination of a normality and titer of potassium permanganate solution
by a standard sodium oxalate solution.
Fill a buret with a potassium permanganate solution. Using a measuring
pipette transfer 5 mL of standard sodium oxalate, Na2C2O4, solution, and 2 mL of
H2SO4 (1:4) solution into a beaker. The mixture is heated to 70-80°C, till strong
vaporization and then the KMnO4 solution is added drop by drop to the hot
mixture in the beaker. It should be stirred up to the endpoint of the reaction. The
stoichiometric (equivalence) point is marked by appearance of light – purple color
of KMnO4 excess.
Carry out titration to take three concurrent readings. Make a table of the
titration readings. Calculate the average volume, and then determine normality and
titer of potassium permanganate solution. The KMnO4 solution volume is
estimated by the upper meniscus in the buret.
Write the equation of the reaction. Calculate the KMnO4 normality and titer in
two ways.
2.2. Determination of normality and titer of hydrogen peroxide solution by the
titrant of potassium permanganate.
Fill a buret with a potassium permanganate solution. Using a measuring
pipet transfer 5 mL of H2O2 solution (the solution has already contained sulfuric
acid) into a beaker. The KMnO4 solution is added drop by drop to the solution in
the beaker constantly stirring up to the endpoint of the reaction. The stoichiometric
(equivalence) point is marked by appearance of light – purple color of KMnO4
excess.
Carry out titration to take three concurrent readings. Make a table of the
titration readings. Calculate the average volume, and then determine normality and
titer of hydrogen peroxide solution. The KMnO4 solution volume is estimated by
the upper meniscus in the buret.
Write the equation of the reaction. Calculate the normality and titer of H2O2
solution by two ways.
Calculate the equivalent of oxidizing and reducing agents for the given
reaction. Calculate the mass of H2O2 in 100 mL of solution.
CLASS 6.
TOPIC: Oxidation-reduction measure analysis. (Iodometry).
1. SEMINAR. 955 min.).
1.1. Method of iodometry.
1.2. Reactions lying in the base of the iodometry.
1.3. Oxidizing and reducing agents in iodometry.
1.4. Determination of the endpoint in iodometry. Indicators of the method.
1.5. Direction of oxidation - reduction reactions. Redox potential.
2. LABORATORY WORK. (40 min.)
2.1. Determination of normality and titer of sodium thiosulfate solution
by a standard potassium dichromate solution.
2.2. Determination of normality and iodine solution titer by a titrant of
sodium thiosulfate.
3. CONTROL WORK. (10 min.).
Literature.
1. Lectures.
2. Ebbing D.D. General Chemistry/ D.D.Ebbing , M.S.Wrighton;.- Boston.
Third
edition. Houghton Mifflin Company. – Р. 514-533.
3. Lister T. Chemistry for Advanced Level / T. Lister, J. Renshaw; - Third
edition.
Stanley Thornes (Publishers) Ltd. 2000. – Р. 149 – 152.
PRACTICAL PROBLEMS
1. What major reactions are used in iodometry?
2. Continue the red-ox reactions, calculate coefficients by a half-reaction method and calculate the equivalent of
reducing and oxidizing agents.
a) NaBrO3 + KI + H2SO4 →
b) KIO3 + Na2SO3 + H2SO4→
c) NaNO2 + KI + H2SO4 → NO + …….
d) I2 + Cl2 + H2O → HIO3 + …….
e) Ca(OCl)2 + KI + H2SO4 → CaCl2 …..
3. A sample of H2O 50-mL and excess of KI and HCl was titrated by 18.2 mL of
0.01 N Na2S2O3 solution. Calculate the mass of active chlorine in 1 liter of water.
4. What is the mass of iodine (I2) which is nessesary for obtaining of 2 liters of

I

2
 = 0,001575 g / mL .
iodine solution with T 
Na
S
O
 2 2 3
5. What is the volume of 5% I2 solution (ρ= 0,95 g/mL) may be obtained from 10
g of crystalline iodine? Calculate molarity and nomality of 5% solution.
6. Calculate the equivalence factor for reducing agents in the given red-ox
reactions:
a) NaNO2 + KI + H2SO4 → I2 + NO + Na2SO4 + K2SO4 + H2O
b) KClO + H2O2 → O2 + KCl + H2O
7. Calculate the equivalence factor for oxidizing agents in the given red-ox
reactions:
a) NaBrO3 + KI + H2SO4 → NaBr + I2 + K2SO4 + H2O
b) KIO3 + Na2SO3 + H2SO4 → KI + I2 + Na2SO4 + K2SO4 + H2O
8. Triiodide ions are generated in solution by the following (unbalanced) reaction in acidic
solution:
IO3(aq) + I−(aq) → I3−(aq)
Triiodide ion is determined by titration with a sodium thiosulfate (Na2S2O3)
solution. The products are iodide ion and tetrathionate ion (S4O62-).
a. Balance the equation for the reaction of IO3- with I- ions.
b. Write and balance the equation for the reaction of S2O32- with I3- in acidic
solution.
2. LABORATORY WORK
2.1. Determination of normality and titer of sodium thiosulfate solution
by a standard potassium dichromate solution.
Fill a buret with a sodium thiosulfate solution. Put 5 mL: of standard
potassium dichromate, K2Cr2O7, solution by using a measuring pipette, into the
beaker, add 3 mL of 3% potassium iodide solution and 2 mL of sulfuric acid
solution (1:4). Shake the mixture, cover the beaker with glass and allow the
solution to rest for 5 minutes. Then the educed iodine is titrated by sodium
thiosulfate solution till a faint-yellow (straw) color appears. Then add 5 drops of
1% starch solution into the mixture (the solution will get blue) and continue
titration, shaking the solution intensively till a light–blue color, which is to
disappear from one drop of the sodium thiosulfate solution. At the equivalent
point the solution has a greenish shade due to the presence of Cr3+ ions.
Carry out titration to take three concurrent readings. Make a table of the
titration readings. Calculate the average volume, and then determine the normality
and titer of the sodium thiosulfate solution.
Write the equation of the reaction. Calculate the Na2S2O3.5H2O normality and
titer in two ways.
2.2 Determination of normality and titer of iodine solution
by a titrant of sodium thiosulfate.
Fill a buret with a sodium thiosulfate solution. Put 5 mL of iodine solution by
using a measuring pipette into the beaker. Then the iodine solution is titrated
by sodium thiosulfate solution till a faint-yellow (straw) color appears. Then add 5
drops of 1% starch solution into the mixture (the solution will get blue) and
continue titration, shaking the solution intensively till a light–blue color, which is
to disappear from one drop of the sodium thiosulfate solution.
CLASS 7.
TOPIC: Elements of chemical thermodynamics.
1. SEMINAR. (60 min.).
1.1. The main notions of thermodynamics: a thermodynamic system; types
of thermodynamic system (isolated, closed, opened); intensive and extensive
parameters of a system; internal energy; work and heat are the forms of energy
transmission;
thermodynamic
processes
(isothermal,
isobaric,
isochoric);
thermodynamically reversible and irreversible processes ; a standard condition.
1.2. The first law of thermodynamics. Enthalpy. Exo-, and endothermic
reactions. A standard enthalpy of substance formation; standard enthalpy of
substance combustion.
1.3. A chemical reaction standard enthalpy. Hess’s Law, consequences of
Hess’s law. Enthalpy of phase transitions and dissolving of substances.
1.4. The second Law of thermodynamics. Entropy. The standard entropy of
substance. Spontaneous and non spontaneous processes. Gibb’s energy. Prognosis
of the direction of spontaneously proceeding processes in isolated and closed
systems; entropy and enthalpy factors. Thermodynamics conditions of equilibrium.
1.5. The standard Gibbs energy of substance formation. The standard Gibb’s
energy of chemical reactions.
1.6. Examples of exothermic and endothermic processes in the body. The
principle of energetic conjugation.
1.7. Chemical thermodynamics as a theoretical foundation of chemical and
biochemical processes.
2. LABORATORY WORK. (50 min.).
2.1. Determination of neutralization reaction enthalpy.
2.2. Determination of copper sulfate hydration enthalpy.
Literature.
1. Lectures.
2. Ebbing D.D. General Chemistry/ D.D.Ebbing , M.S.Wrighton;.- Boston.
Third
edition. Houghton Mifflin Company. – Р. 710-733.
3. Lister T. Chemistry for Advanced Level / T. Lister, J. Renshaw; - Third
edition.
Stanley Thornes (Publishers) Ltd. 2000. – Р. 393 – 402.
PRACTICAL PROBLEMS
1. Given:
3
O2(g) → SO3 (g)
2
S(s) +
2SO2(g) + O2(g) → 2SO3(g)
∆H = -395,2 kJ
∆H = -198,2 kJ
Calculate ∆H for the reaction S(s) + O2((g) → SO2(g)
2. Hydrogen sulfide gas is a poisonous gas with the odor of rotten eggs. It
occurs in natural gas and is produced during the decay of organic matter, which
contains sulfur. The gas burns in oxygen as follows:
2H2S(g) + 3O2((g)→ 2H2O(l) + 2SO2(g)
Calculate the standard enthalpy change for this reaction by using standard
enthalpies of formation. (See Appendix, Table 3.)
3. Predict the sign of ∆S for each of the following changes:
a) AgCl(s) → Ag+(aq) + Cl-(aq)
b) 2H2(g) + O2(g) → 2H2O(l)
c) 3O2(g) → 2O3(g)
4. For the reaction
2Al(s) + 3Br2(l) → 2AlBr3(s)
∆S0 is equal to -144 J/K. Use this value and the data from Appendix to calculate
the value of S0 for solid aluminum bromide.
5. From the data in Appendix calculate ∆H0, ∆S0, and ∆G0 for each of the
following reactions at 25oC.
a. CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
b. 6CO2(g) + 6H2O(l) → C6H12O6(s) + 6O2(g)
Glucose
6. For the reaction at 298 K,
2NO2(g) ↔ N2O4(g)
The values of ∆H0 and ∆S0 are -58.03 kJ/mol and -176.6 J K-1mol-1, respectively.
What is the value of ∆G0 at 298 K? Assume that ∆H0 and ∆S0 do not depend on
temperature. At what temperature is ∆G0 = O? Is ∆G0 negative above, or below,
this temperature?
LABORATORY WORK.
We can determine the heat associated with a chemical reaction experimentally
by using a device called a calorimeter. Calorimetry, the science of measuring
heat, is based on observing the temperature change when a body absorbs or
discharges energy as heat. Substances respond to heating differently. The heat
required to raise the temperature of a substance is called its heat capacity, C =
q
.
∆t
The heat capacity (C) of a sample of substance is the quantity of heat to raise
the temperature of the sample of substance by one degree Celsius (or one
Kelvin). Changing the temperature of the sample from an initial temperature to a
final temperature tf requires heat equal to q=C·∆t , where ∆t is the change of
temperature and equals tf-ti . Heat capacity is directly proportional to the amount of
substance. Often heat capacities are listed for molar amounts of substances. The
molar heat capacity of a substance is the heat capacity for one mole of substance.
It has the units J K-1mol-1 or J oC-1 mol-1. Heat capacities are also compared for
one-gram amounts of substances. The specific heat capacity (or simply specific
heat) is the quantity of heat required to raise the temperature of one gram of a
substance by one degree Celsius (or one Kelvin) at constant pressure, which
has the units JK-1g-1 or JoC-1g-1.
Although the calorimeters used for highly accurate work are precision
instruments, a very simple calorimeter can be used to examine the fundamentals of
calorimetry. All we need are two nested Styrofoam cups with a Styrofoam cover
through which a stirrer and a thermometer can be inserted, as shown in Figure 1.
This device is called a coffee cup calorimeter. The outer cup is used to provide
extra insulation. The inner cup holds the solution in which the reaction occurs.
The measurement of heat by using a simple calorimeter such as that shown in
Figure 1. is an example of constant-pressure calorimetry. Constant-pressure
calorimetry is used in determining changes in enthalpy occurring in solution.
Under these conditions the enthalpy equals the heat.
Figure 1.
A simple coffee-cup calorimeter.
After reactants are added, the cup
is covered to reduce heat loss by
evaporation and convection. The
heat of reaction is determined by
noting the temperature rise or fall.
2.1. Determination of neutralization reaction enthalpy.
The method of carrying out the experiment.
A group of students defines the heating affect of the neutralization reaction of
a given variant: an acid – a base.
NaOH + HCl = NaCl + H2O;
2NaOH + H2SO4 = Na2SO4 + 2H2O;
NaOH + CH3COOH = CH3COONa + H2O;
NH4OH + HCl = NH4Cl + H2O;
∆H1;
∆H2;
∆H3;
∆H4;
Put 100 mL of 1N acid solution in a calorimeter. In 5 minutes take the
temperature of the solutions up to 0,1°C accuracy (ti). Then add 100 mL of 1N
base solution. Stir the solution with a stirring rod without stopping. When these
reactants (both originally at the same temperature) are mixed, the temperature of
the mixed solution is observed to increase (tf). Thus the chemical reaction must be
releasing energy as heat. This increases the random motions of the solutions
components, which in turn increases the temperature. The quantity of energy
released can be determined from the temperature increase, the mass of the solution,
and the specific heat capacity of the solution.
Working up of the experimental data.
For an approximate result we will assume that the calorimeter does not absorb
or leak any heat and that the solution can be treated as if it were pure water with a
density of 1,0 g/mL.
We also need to know the heat required to raise the temperature of a given
amount of water by 1oC. The specific heat capacity (C) of water is 4,18 J 0C-1g-1.
1. From these assumptions and definitions we can calculate the heat (change in
enthalpy) for the neutralization reaction using the formula:
q = C · m · (tf – ti),
where m – the mass of the reaction mixture,
the density of the reaction mixture is equals to 1g/mL.
2. Calculate the number of moles of the water, formed in the result of the
neutralization reaction, then recalculate the heating effect of neutralization per 1
mole of water.
3. Record into the Table 1. results of the experiments carried out by other
students and compare the data of heat of neutralization by various variants of
reactions acid – base: ∆H1; ∆H2; ∆H3; ∆H4.
Table 1.
reaction, J
∆H of neutralization reaction, kJ
Heat,Q, of neutralization
Heat effect, q of experiment, J
Mass of mixture
Volume of mixture
Change of temperature, ∆T
Final temperature, Tf
Initial temperature, Ti
Volume of base solution, mL
solution, mL
Volume of acid
Results of the experiments
2.2. Determination of copper sulfate (II) hydration enthalpy.
The method of carrying out the experiment.
a. Put 100mL of distilled water in a calorimeter. In 5 minutes take the
temperature with 0,1°C accuracy (ti). Transfer a 3,2-g sample of a waterless copper
(II) sulfate, CuSO4, into the calorimeter, stir it quickly with a stirring rod and mark
the maximum temperature.
The pure water has a density of 1,0 g/mL. The specific heat capacity (C) of
water is 4,18 J 0C-1g-1.
Working up of the experimental data.
1. Calculation of the heat for dissolving of waterless CuSO4 according to the
equation:
q = c · m1 · (t2 – t1),
Where m1 – is the mass of CuSO4 solution, g.
2. Calculate the number of CuSO4 moles which were dissolved in water.
3. Recalculate the heat of waterless copper (II) sulfate dissolving per 1 mol of
the substance (∆H1).
∆Η 1 = −
q
n1
b. Put 100mL of distilled water in a calorimeter. In 5 minutes take the
temperature with 0,1°C accuracy (ti). Transfer a 5.0-g sample of a copper (II)
sulfate pentadydrate, CuSO4 5H2O , into the calorimeter, stir it quickly with a
stirring rod and mark the maximum temperature.
4. Calculation of the heat for dissolving of CuSO4.5H2O according to the
equation:
q = c · m2 · (tf’ – ti),
Where m2 – is the mass of CuSO4 .5H2O solution, g.
5. Calculate the number of CuSO4.5H2O moles which were dissolved in
water.
6. Recalculate the heat of copper (II) sulfate pentahydrate dissolving per 1
mol of the substance (∆H2).
∆Η 2 = −
q
n2
7. The enthalpy of CuSO4 hydration is calculated by the formula:
∆Hhydration = ∆H1-∆H2
Table 2.
Results of the experiments
∆Hhydration = ∆H1 - ∆H2
CuSO4.5H2O
∆H2 – heat of dissolving
temperature at dissolving
tf’ – final (maximum)
Solution mass, g (m2)
CuSO4
Sample CuSO4·5H2O, g
∆H1 – heat of dissolving
dissolving CuSO4
tf – maximal temperature of
ti – initial temperature
Repeat the experiment
with a 5-g sample of copper
sulfate pentahydrate,
Solution mass, g (m1)
СLASS 8.
TOPIC: Elements of chemical and biochemical reaction kinetics.
1. SEMINAR. 960 min.).
1.1. The subject and basic notions of chemical kinetics.
1.2. Rate of reaction, the average rate and instantaneous rate. Classification
of reactions: homogeneous, heterogeneous, simple and complex.
1.3. Molecularity of a reaction. The order of a reaction. Dependence of a
reaction rate on concentration. Kinetic equations of zero order reactions, the first
order and the second order reactions. A rate constant. A half - life of a reaction.
The experimental determination of a reaction rate, rate constant and order of a
reaction.
1.4. Dependence of reaction rate on temperature. Temperature coefficient of
a reaction rate. Arrhenius equation. The activation energy. Energy profile of a
reaction. Activated complex theory. Catalysis. Homogeneous and heterogeneous
catalyst. Energy profile of a catalyzed reaction. Fermentative catalysis.
2. LABORATORY WORK. (50 min.).
2.1. Determination of rate and order of sodium thiosulfate decomposition
reaction.
2.2. Determination of order of iodide–ions oxidation reaction with Fe3+ ions
(the Clock Reaction).
Literature.
1. Lectures.
2. Ebbing D.D. General Chemistry/ D.D.Ebbing , M.S.Wrighton;.- Boston.
Third
edition. Houghton Mifflin Company. – Р. 547-585.
3. Lister T. Chemistry for Advanced Level / T. Lister, J. Renshaw; - Third
edition.
Stanley Thornes (Publishers) Ltd. 2000. – Р. 158 – 167, 369 – 375.
PRACTICAL PROBLEMS
1. Write the rate law for the following reactions:
a) N2(g) + O2(g) → 2 NO(g)
b) H2(g) + O2(g) → 2H2O(l)
c) CO2(g) + C(s) →2CO(g)
d) Zn(s) + Cl2(g) → ZnCl2(s)
2. The reaction of nitrogen monooxide and chlorine is the third order. The
reaction proceeds according to the equation: 2 NO(g) + Cl2(g) → 2 NOCl. How
long would it take a) for the concentration of NO to increase in double;
b) for the concentration of Cl2 to decrease in double:
c) for the concentration of both reactants to increase in double?
3. For the reaction of nitric oxide, NO, with chlorine,
2NO (g) + Cl2(g) → 2 NOCl
the observed rate law is Rate = k[NO]2 [Cl2]. What is the reaction order with
respect to nitric oxide, with respect to Cl2? What is the overall order?
4. The temperature of the reaction increases by 400. How does the constant rate
increase? (γ=3,2).
5. Calculate the temperature coefficient (γ), if you know that the constant rate
at 1200C equals 5,88.10-4/s, and at 1700C it equals 6,7.10-2/s.?
6. Methyl isocyanide, CH3NC, isomerizes when heated to give acetonitrile,
CH3CN: CH3NC(g) → CH3CN(g). The reaction is the first order. At 2300C, the
rate constant for the isomerization is 6,3.10-4/s. What is the half-life? How long
would it take for the concentration of CH3NC to decrease to 25% of its initial?
7. Sulfuryl chloride, SO2Cl2, decomposes when heated.
SO2Cl2(g) → SO2(g) + Cl2(g)
In an experiment, the initial concentration of sulfuryl chloride was 0,0248
mol/l. If the rate constant is 2.2.10-5/s, what is the concentration of SO2Cl2 after 4,5
hr? The reaction is the first order.
8. The rate constant for the formation of hydrogen iodide from the elements
H2(g) + I2(g) → 2HI(g)
is 2,7.10-4 l/mol.s at 600 K and 3,5.10-3l/mol.s at 650 K.
a. Find the activation energy Ea.
b. Then calculate the rate constant at 700 K.
LABORATORY WORK.
2.1. Determination of rate and order
of sodium thiosulfate decomposition reaction.
Sodium thiosulfate decomposes in sulfuric acid solution by the equation:
Na2S2O3 + H2 SO4 = Na2SO4 + SO2 + S↓ + H2O
The reaction goes on in two stages:
a. Na2S2O3 + H2SO4 = Na2SO4 + H2S2O3;
b. H2S2O3 = SO2 + S + H2O
The first step proceeds very fast, the second one proceeds slowly, that’s why the
rate of the overall process is defined by the second stage.
Put the mixture of volume (A) of 0, 1M Na2S2O3 solution and volume (B) of
distilled water into 5 test-tubes in ratio, given in the Table 3:
Table 3.
Results of the experiments
№
1
2
3
4
5
Time of the
Volume (number of drops)
Concentra1
solution dimness
tion of
Na2S2O3
H2O
H2SO4
R = (sec −1 )
stativity τ (sec).
Na2S2O3
τ
(A)
(B)
(С)
5
10
15
20
25
20
15
10
5
0
25
25
25
25
25
0,01
0,02
0,03
0,04
0,05
Put 25 drops of 1M H2SO4 solution into other 5 test-tubes. Then mix each
sodium thiosulfate solution with H2SO4 solution; switch on the stop – watch (you
can use a watch with a second hand) and fix the time (τ) when white sulfur appear.
Carry out the same experiment with the other pairs of solutions. Enter the reading
in the table. Calculate conditional rate of the reaction and make up a graph of
conditional rate dependence on sodium thiosulfate concentration.
Evaluate the order of the studied reaction. It is known that for the zero, the
first and the second order reactions, the rate dependence on the concentration is
expressed as follows:
R
R
R
The first order
Zero orger
C
The second order
C
2.2. Determining the order of the reaction from the rate law.
Oxidation of iodide – ions are by Fe3+ ions (the Clock Reaction)
C
Iodide –ions oxidation proceeds by the equation:
2 KI + 2 FeCl3 = 2FeCl2 + I2 + 2KCl
The iodide formation is fixed by the appearance of blue-colored complex in
the presence of starch. Carry out experiments to study the reaction rate dependence
on FeCl3 concentration.
Place the components of reactional mixture in 4 test-tubes according to the
table.
Table 4.
Results of the experiments
№
Of
the
testtube
The reagents volume
(number of drops)
FeCl3 HNO3 H2O
0,02 М 0,1 М
starch
(а)
(b)
(c)
Concentration of
FeCl3
solution,
mol/L
1
16
6
0
2
10
2
8
6
8
2
5
3
4
4
2
6
6
12
14
2
2
2,5
1,25
Time of
starchiodine
complex
appearance
τ(sec)
R=
1
τ
(sec )
−1
Then add 8 drops of 0,02M KI into the first test-tube and fix the time when the
solution gets blue. Carry out the same experiment with the other reactants. Enter
the results in the table; calculate conditional rate of the reaction and make up a
graph of rate dependence on the concentration of FeCl3 solution. Make a
conclusion about a particular order of a reaction by Fe3+ ion.
The reaction order with respect to a given reactant species equals the
exponent of the concentration of that species in the rate law, as determined
experimentally.
The initial method is a simple way to obtain reaction orders. It consists of
doing a series of experiments in which the initial, or starting, concentration of
reactants is varied. Then the initial rates are compared, from which the reaction
orders can be deduced. In experiments we calculate the rate of I2 formation with
respect to each reactant species (concentration of one of the reactant is changed,
and the second reactant concentration remains constant).
The rate law for a given reaction is:
− n2
Rate 1 = k[I ]
⋅[Fe3+]1n1 = kobserved.⋅[Fe3+]1n1.
− n2
Rate 2 = k[I ]
⋅[Fe3+]2n1 = kobserved.⋅[Fe3+]2n1.
.
That is
3+ n 1
Rate = k [Fe ]
Applying logarithms, we get:
ln Rate = lnk + n1ln[Fe3+]
It is possible to determine the order of a reaction by graphcal plotting of the
data for a particular experiment. The value of the reaction order (n1) is the tangent
of angle α in the Figure 2. A particular order by I- ions (n2) can be obtained in the
same way.
lnR
tgα =
a
α
a
= n1
b
b
lnC
Figure 2. A plot of lnR versus lnC.
СLASS 9.
TOPIC: Thermodynаmic and kinetic conditions of chemical equilibrium.
1. SEMINAR. (25 min.).
1.1. Reversible and irreversible chemical reactions.
1.2. Chemical equilibrium. Kinetics conditions of chemical equilibrium.
Equilibrium constant (Kc). Equilibrium constant for homogeneous and heterogeneous reactions. Dependence of equilibrium constant on the nature of reactants
and temperature.
1.3. Thermodynamic conditions of chemical equilibrium. The relationship
between the free energy and the equilibrium constant. Predicting the direction of a
reaction.
1.4. Changing the reaction equilibrium; Le Chatelier’s principle.
1.5. Homogeneous and heterogeneous equilibrium in the body.
2. LABORATORY WORK. (20 min.).
2.1. Effect of changing the concentration of reactants on a chemical
equilibrium
2.2. Effect of changing the temperature on chemical equilibrium.
3. CONTROL WORK/ (70 min.).
Literature.
1. Lectures.
2. Ebbing D.D. General Chemistry/ D.D.Ebbing , M.S.Wrighton;.- Boston.
Third
edition. Houghton Mifflin Company. – Р. 597-625.
3. Lister T. Chemistry for Advanced Level / T. Lister, J. Renshaw; - Third
edition.
Stanley Thornes (Publishers) Ltd. 2000. – Р. 124 – 135, 319 – 335.
PRACTICAL PROBLEMS
1. Write the equilibrium-constant expression Kc for
a) catalytic methanation: CO(g) + 3H2(g) ↔ CH4(g) + H2O(g)
b) the reverse of the reaction in (a) – that is:
CH4(g) + H2O(g) ↔ CO(g) + 3H2(g)
c) the synthesis of ammonia N2(g) + 3H2(g) ↔ 2 NH3(g)
d) 3Fe(s) + 4 H2O(g) ↔ Fe3O4(s) + 4 H2(g)
e) CaCO3(s) ↔ CaO(s) + CO2(g)
f) H2O(l) ↔ H2O(g)
2. At a particular temperature a 3.00-L flask contains 3.50 mol of HI, 4.10mol of
H2, and 0.30 mol of I2 at equilibrium. Calculate Kc at this temperature for the
reaction H2(g) + I2(g) ↔ 2 HI(g)
3.At a particular temperature 8.0 mol of NO2 are placed into a 1.0-L container and
the NO2 dissociates by the following reaction:
2NO2(g) ↔ 2 NO(g) + O2(g)
At equilibrium the concentration of NO(g) is 2.0 mol/L. Calculate the value of Kc
for this reaction.
4. For the reaction 2NH3(g) ↔ N2(g) + 3 H2(g).
What is the relationship between the equilibrium concentrations of NH3, N2 and H2
for the following situation: the initial concentration of NH3 is 6 mol/L (60%
reacts).
5. Consider the reaction
Fe+3 (aq) + SCN-(aq) ↔ FeSCN2+(aq)
How will the equilibrium position shift if
a) water is added, doubling the volume?
b) AgNO3(aq) is added? (AgSCN is insoluble)
c) NaOH (aq) is added? [Fe(OH)3 is insoluble.]
d) Fe(NO3)3(aq) is added?
6. The equilibrium constant Kc, for the reaction
N2(g) + 3 H2(g) ↔ 2 NH3(g)
at 450oC is 0.159. Calculate the equilibrium composition when 1.00 mol N2 is
mixed with 3.00 mol H2 in a 2.0–L vessel.
7. Cells use hydrolysis of adenosine triphosphate (ATP) as a source of energy.
Symbolically, this reaction can be represented as
ATP(aq) + H2O(l) → ADP(aq) + H2PO4-(aq)
Where ADP represents adenosine diphosphate. For this reaction ∆G0= -30,5
kJ/mol. Calculate Kc at 250C.
2. LABORATORY WORK.
2.1. Effect of changing the concentration of reactants on a chemical equilibrium.
The concentration is changed by adding or removing a substance.
Add 2 drops of saturated FeCl3 solution and 2 drops of saturated KSCN
solution into the beaker containing 10 mL of distilled water. The color of the
reaction mixture changes from colorless to red. Stir the initial solution.
FeCl3 + 3 KSCN ↔ Fe(SCN)3 + 3 KCl
Fe3+(aq) + 3 SCN-(aq) ↔ Fe(SCN)3
This solution is divided into 4 parts (in 4 test-tubes).
The first test-tube is standard.
Then add 2 drops of concentrated FeCl3 solution into the second test-tube;
Add 2 drops of concentrated KSCN solution into the third test-tube;
Add some KCl crystals into the forth test-tube.
Explain the change of color in the three test-tubes, using the Le Chatelier’s
principle.
2.2. Effect of changing the temperature on chemical equilibrium.
3-4 drops of 0,1N iodine solution, I2, are added to 5 mL of starch solution. As
a result, starch-iodine complex forms, which has a blue color. Stir the reactional
mixture. Then the obtained solution is poured into 2 test-tubes.
The first test-tube is standard.
Put the second test-tube into the water-bath and heat it for 2-3 minutes, then cool it.
Explain the change of color in the second test-tube.
Is the reaction of starch-iodine complex formation exothermic or endothermic?
СLASS 10.
TOPIC: Properties of Molecular and Ionic Solutions.
1. SEMINAR. (70 min.).
1.1. The role of water and solutions in the body. Physical and chemical
properties of water. The phase diagram of water. Dependence of substance
solubility in water on the ratio of hydrophilic and hydrophobic properties. Effect of
external conditions on solubility. Thermodynamics of solubility.
1.2. The notion of ideal solution. Solubility of gases in liquid. Henry’s
Law, its medical and biological importance.
1.3. Diffusion in solutions. Osmosis. Osmotic pressure. Vant-Hoff’s Law.
Measuring of osmotic pressure. Osmometry. The role of osmosis and osmotic
pressure in biological systems. Osmotic homeostasis. Isotonic, hypotonic and
hypertonic solutions.
1.4. Raoult’s law. Consequences of Raoult’s law. Ebulliometry and cryometry. Deviation of properties of diluted solutions of electrolytes according to
Raoult’s and Vant-Hoff’s Laws. Isotonic coefficient.
2. LABORATORY WORK. (40 min.).
Determination of substances molar mass by Rast’s method (on the example
of organic compounds).
Literature.
1. Lectures.
2. Ebbing D.D. General Chemistry/ D.D.Ebbing , M.S.Wrighton;.- Boston.
Third
edition. Houghton Mifflin Company. – Р. 459-486.
3. Lister T. Chemistry for Advanced Level / T. Lister, J. Renshaw; - Third
edition.
Stanley Thornes (Publishers) Ltd. 2000. – Р. 320 – 322.
PRACTICAL PROBLEMS
1. What is the boiling point of a solution of 0.152 g of glycerol, C3H8O3 , in 20.0 g
of water? What is the freezing point of a solution?
2. An aqueous solution of a molecular compound freezes at – 0.0860C. What is
the molality of the solution?
3. The osmotic pressure of blood at 370C is 7.7 atm. What should be the molarity
of glucose solution which is isotonic with blood?
4. A 0.0182 g sample of unknown substance was dissolved in 2.135 g of benzene.
Molality of this solution is 0.0698m. What is the molecular weight of the unknown
substance?
5. How many atoms does the molecule of sulfur consist of, if 4.455 g of it was
dissolved in 50 g of benzene? Boiling-point elevation is 0.8910C. Boiling-pointelevation constant of benzene is 2.60C/m.
6. Safrole is contained in oil of sassafras and was once used to flavor root beer. A
2.39 mg sample of safrole was dissolved in 103.0 mg of diphenyl ether. The
solution had a melting point of 25.700C. Calculate the molecular weight of safrole.
The freezing point of pure diphenyl ether is 26.840C, and the freezing - point –
depression constant is 8.000C/m.
7. What is the freezing point of 0.0095 m aqueous sodium phosphate solution?
8. Maltose, C12H22O11 , is a sugar produced by malting (sprouting) grain. A
solution of maltose at 250C has an osmotic pressure of 5.61 atm. What is the
molarity of maltose?
9. Which aqueous solution has the lower freezing point: 0.10 m CaCl2 or 0.10 m
glucose?
10. What is the freezing point of 0.0085 m aqueous calcium chloride, CaCl2
solution ? (α=0.8).
11. In a mountainous location, the boiling point of pure water is found to be 950C .
How many grams of sodium chloride must be added to 1 kg of water to bring the
boiling point to 1000C? (α=0.5).
2. LABORATORY WORK.
Determination of nonelectrolyte molar mass by Rast’s method
(on the example of an organic compound).
Colligative properties are used to obtain molecular weights. Although the
mass spectrometer is now often used for routine determinations of the molecular
weight of pure substances, colligative properties are still employed to obtain
information about the spesies in solution. Freezing-point depression is often used
because it is simple to determine a freezing point or a metting point.
Freezing-point lowering helps to obtain the molecular weight but only for
relatively large quantities of a substance. From melting-point depression we can
obtain the molecular weight for a small quantities of a substance.
Camphor is a white solid that melts at 179.50C. It has been used to determine
the molecular weights of organic compounds, because of its unusually large
freezing-point-depression constant (400 C/m), which allows to use ordinary
thermometers. The organic substance with unknown molar mass ( M ) by mass (m)
is dissolved in melted camphor by mass (q) . Then the melting point of the solution
is determined.
For example a 1.07-mg sample of a organic compound was dissolved in 78.1
mg of camphor. The solution melted at 176.00C. The freezing-point lowering is
∆T f = (Tcamphor − Tmixture ) = (179.5 − 176.0) 0 C = 3.5 0 C
Using the formula ∆T f = K f ⋅ C m we calculate the molality of the solution C m
Cm =
∆T f
Kf
=
3 .5 0 C
= 0.088m
40 0 C / m
From this we compute the moles of the compound that are dissolved in 1.00 g of
camphor C m =
mole( subst.)
; Hence, for the given solution:
kg ( solvent )
0.088
mole( subst.)
mol ( subst.)
=
kg (camphor ) 78.1 ⋅ 10 − 6 kg (camphor )
Or, rearranging this equation, we get
mol ( subst.) = 0.088
mol ( subst.)
⋅ 78.1 ⋅ 10 −6 kg (camphor ) = 6.9 ⋅ 10 −6 mol
kg (camphor )
The molar mass of the substance equals the mass of the substance divided by the
number of moles
M =
1.07 g
= 1.6 ⋅ 10 2 g / mol
−6
6.9 ⋅ 10 mol
The total formula for molar weight calculation is M =
K f ⋅ m ⋅ 1000
∆T ⋅ q
.
СLASS 11.
TOPIC: Equilibrium in Ionic Solutions.
1. SEMINAR. (70 min.).
1.1. Arrhenius concept of acids and bases.
1.2. Brǿnsted-Lowry concept of acids and bases.
1.3.
Lewis concept of acids and bases (electron-donor compounds and
electron acceptor compounds).
1.4. Self-ionization of water. The pH of a solution. Solutions of a strong
acids or bases.
1.5. Degree of electrolytic dissociation. Factors effecting the degree of
electrolytic dissociation.
1.6.
Solutions of weak acid or base. Acid-ionization equilibria. Acid-
ionization constant, base-ionization constant.
1.7. Oswald’s dilution law.
1.8. The main principles of the theory of strong electrolytes solution. Activity
and ionic activity coefficient. Ionic strength of a solution. Seeming degree of
dissociation.
1.9. Calculation of pH for strong and weak acids and bases. Delta of pH
values for biological liquids.
1.10. Acid-base properties of salt solutions; hydrolysis of salts.
1.10.1. Hydrolysis mechanism on cation.
1.10.2. Hydrolysis mechanism on anion.
1.10.3. Degree of hydrolysis. Hydrolysis constant.
1.10.4. Hydrolysis equilibrium shift.
1.10.5. Medical and biological role of hydrolysis.
1.11. Heterogeneous reactions in solutions of electrolytes. The solubility
product constant (Ksp). Conditions of precipitate formation and dissolution.
2. LABORATORY WORK. (45 min.).
2.1. Effect of the same ion on dissociation degree of weak electrolytes.
2.2. Determination of pH of a solution with the help of universal indicator and
ion meter.
2.3. Shift of ionic equilibrium.
2.4. Hydrolysis of salts.
Literature.
1. Lectures.
2. Ebbing D.D. General Chemistry/ D.D.Ebbing , M.S.Wrighton;.- Boston.
Third
edition. Houghton Mifflin Company. – Р. 503-513, 635-660, 682-688.
3. Lister T. Chemistry for Advanced Level / T. Lister, J. Renshaw; - Third
edition.
Stanley Thornes (Publishers) Ltd. 2000. – Р. 138 – 145. 339 – 341.
PRACTICAL PROBLEMS
1. What are the concentrations of H+, OH- and the pH in each of the following?
0.025 M HCl
0.0015M HNO3
0.0035 M Ca(OH)2
0.01 M NaOH
2. A sample of vinegar has a hydrogen-ion concentration of 7.5·10-3. What is the
pH of the vinegar?
3. A 0.5 L aqueous solution contained 0.29 g of sodium hydroxide .What was the
pH of the solution at 250C?
4. A detergent solution has a pH of 11.63 at 250C. What is the hydroxide-ion
concentration?
5. The solution of washing soda (sodium carbonate, Na2CO3) has molarity of 0.03
M. What is the pH of the solution at 250C?
6. Find pH for the following solutions:
0.01 M NH4Cl
0.0001 M CH3COONa
0.025 M C5H5NHCl
0.0035 M C6H5ONa
7. What is the pH of a 0.35 M solution of methylammonium chloride,
[CH3NH3]Cl?
8. The following solutions are mixed: 0.5L of 0.0001M NaOH and 0.5L of 0.002
MgSO4. Is a precipitate expected? Explain your answer.
9. What is the molar solubility of BaSO4?
2. LABORATORY WORK
2.1. Effect of some ion on dissociation degree of weak electrolytes.
2.1.1. Take two flasks, put 5-6 drops into each of a dissolved acetic acid and
into each flask, drop indicator (methyl orange solution). Leave one flask for
comparison, insert with a micro palette-knife some crystals of sodium acetate into
the other flask and shake it thoroughly. Compare the coloring of the solutions in
both the flasks and explain its changes.
2.1.2. Take two flasks and put 5-6 drops of ammonia and add of 1 drop of
phenolphthalein solution into each flask. Insert with a micro palette-knife some
crystals of ammonia chloride salt into one of the flasks and shake it thoroughly.
Explain the changes of the solution color at inserting ammonia chloride salt.
2.2. Determination of pH of a solution
with the universal indicator and pH meter.
Determine pH of acid, alkali and salt solutions and also running water. Put a
drop of the studied solution on a phenolphthalein paper stripe, saturated with the
universal indicator. Compare the obtained color with the one of the standard.
Write down the readings into the table 5.
Table 5.
Results of the experiments
SOLUTION
pH value
Hydrochloric acid, HCl
Acetic acid, CH3COOH
Sodium hydroxide, NaOH
Ammonium hydroxide, NH3 .H2O
Water, H2O
2.3. Hydrolysis of salts.
2.3.1. Put 1 drop of the studied 1N salt solution (sodium sulfate, sodium
carbonate, ammonium chloride, sodium acetate, borax solution). Mark pH of the
solution. Give the equations of the salts hydrolysis in ionic and molecular forms.
Explain the conditions of the medium reaction in each case.
2.3.2. Pour out sodium carbonate and sodium hydrogen carbonate solutions
into two test-tubes, add two drops of phenolphthalein into each. Explain why the
color of phenolphthalein is different in these solutions.
2.3.3. Temperature effect on hydrolysis.
a. Put 1N solutions of sodium acetate, ammonium chloride, zinc chloride,
aluminum sulfate into 5 flasks, by 3 drops into each flask. Add 3 drops of the
universal indicator and heat when they are about to boil. What effect does
temperature have on hydrolysis?
b. Pour out 3ml aluminum sulfate solution and 5 ml sodium acetate solution
into a flask and heat the mixture up to boiling. Explain the precipitate appearance.
c. Pour out iron chloride solution into 2 test-tubes. The first test-tube is the
standard, solution of the other test-tube is boiled. Explain the observed changing.
2.3.4. Shift of hydrolysis equilibrium.
a. Pour out 3 ml of antimony chloride solution into a test-tube, add water by
drops. A precipitate will fall. The precipitate is dissolved by addition of
hydrochloric acid and falls again if we dilute solution by water. Explain this
phenomenon.
SbCl3 + H2O SbOCl + 2HCl
b.
Carry out an analogous experiment with tin chloride
solution.
СLASS 12.
TOPIC: Buffer solutions and their properties. Buffer systems of an body.
1. SEMINAR. (60 min.).
1.1. Acidic-basic buffer solutions (definition).
1.2. The value of buffer solutions in chemistry, biology, medicine.
1.3. Chemical composition of buffer systems. Acidic, basic, amphoteric buffer
systems.
1.4. Henderson –Hasselbalch equation.
1.5. Preparation of buffer solutions. Buffer curves.
1.6. Buffer system action on addition of acid or base.
1.7. Blood buffer systems. Buffer action – is the main mechanism of protolytic
homeostasis of an organism.
1.8. Buffer capacity: dependence on various factors, methods of determination.
2. LABORATORY WORK. (50 min.).
2.1. Preparing buffer solutions.
2.2. Studying the action of the buffer solutions.
2.3. Determination of pH buffer solutions.
Literature.
1. Lectures.
2. Ebbing D.D. General Chemistry/ D.D.Ebbing , M.S.Wrighton;.- Boston.
Third
edition. Houghton Mifflin Company. – Р. 660-666.
3. Lister T. Chemistry for Advanced Level / T. Lister, J. Renshaw; - Third
edition.
Stanley Thornes (Publishers) Ltd. 2000. – Р. 348 – 350.
PRACTICAL PROBLEMS
1. Draw a scheme of a phosphate buffer, NaH2PO4/Na2HPO4 , action.
2. Calculate the pH for buffers:
a. 0.01 M CH3COOH / 0.01 M CH3COONa
0.01 M CH3COOH / 0.1 M CH3COONa
0.001 M CH3COOH / 0.01 M CH3COONa
b. 0.01 M NH3 ·H2O / 0.01 M NH4Cl
0.01 M NH3 ·H2O / 0. 1 M NH4Cl
0.001 M NH3 ·H2O / 0.01 M NH4Cl
3. The normal pH of blood is 7.41. What is a CO2/HCO3− concentration ratio in
blood?
4. A solution is made by adding 10 mL of 0.1 M CH3COOK to 5 mL of a 0.05 M
HCl. What is the pH of the final solution? Is the solution a buffer?
5. How many moles of hydrofluoric acid, HF, must be added to 500.0 mL of 0.3
M sodium fluoride to give a buffer of pH 3.5? Ignore the volume change due to the
addition of hydrofluoric acid.
2. LABORATORY WORK
2.1. Determination of the buffer solution action.
Place the following solutions into 3 flasks:
1. 5.0 mL of 0.1M CH3COOH solution
2. 5.0 mL of 0.1M CH3COONa solution
3. 2.5 mL of 0.1 M CH3COOH + 2.5 ml of 0.1M CH3COONa (buffer mixture).
a. Measure pH of the solutions with the universal indicator or methyl red.
b. The contents of every flask dilute with distilled water by two times.
Shake the solutions throughly and fix a pH change. Will the color of the indicator
in the third flask change? Comment on the obtained data.
c. Add some drops of a strong acid and alkali solutions in to each flask. Will
the color of the indicator in the third flask change? Why?
2.2. Preparing buffer solutions and studying their action.
a. Using buffer curves prepare 25 mL of a buffer solution with the a given
pH0.
b. Add 1.0 mL of 0.01M HCl solution to a 5.0 mL of the prepared solution and
measure pH1. Why is pH0=pH1? Explain the action of the prepared buffer system.
Write down the equations of chemical reactions.
c. Add 1.0 mL of 0.01M NaOH solution to 5.0 mL of a buffer solution with
pH0. Shake the solution and determine the medium of the mixture. Comment on
the readings of the experiment and write down the equations of the chemical
reactions.
Usage of buffer systems
Defining a pH value of the medium has a great importance for many reactions
in vitro and biochemical processes. A buffer solution resists changes in pH through
its ability to combine with both H+ and OH- ions. Blood is a buffer solution.
Each liquid medium of the body is characterized by a strictly definite pH
value. (For example, saliva pH is 6.8, artery blood pH is 7.44, and vein blood pH is
7.35). Changing of pH value in the acidic way (acidosis) or the basic way
(alkalosis) can cause pathological changes in the body. It is connected with the fact
that, firstly, the stability of the protein, that composes protoplasm cells, greatly
depends on the active reaction of the medium. For any kind of protein there are pH
values, at which the given protein is the least stable and the least active. Secondly,
pHs shift from the normal. Strongly effected conjugated biochemical reactions, are
going on under enzymes (biocatalysts). Each enzyme has an optimum pH, under
which it is the most active. Changing of a pH value depresses the enzyme activity
and its action can be even stopped. Thus, the optimum action of pepsin is marked
at 1.5−2.0 pH, saccharose at 4.2 pH, catalase – 7.0 pH, lipase at 7.0−7.5 pH. Due
to the fact, that each enzyme acts at a definite pH value it is possible to carry out
chemical transformations of various substances in cells and tissues of the body. In
the vital processes, especially under physical work, about 20-30 liters of 1M acid
(the total acidity of hydrochloric, lactic, pyruvic, carbonic and other acids) are
formed. At a big height (more than 3 km), substances of a basic character are
formed in the body. Hence, a healthy organism copes with the above mentioned
phenomena; pH of liquid media of the body becomes normal again. The stable pH
value in the body is supported by the buffer systems of the body.
Buffer mixtures are widely used in a qualitative analysis, for example, in
carrying out red-ox reactions, in complexometry method. Complete reactions of
complex compound formation often depend on the acidity of the medium.
Ammonia and acetate buffer systems are often applied in complexometry.
СLASS 13.
TOPIC: Chemistry of dispersed systems. Phenomena on the surface. Colloidal
solutions.
1. SEMINAR. (60 min.).
1.1.
Dispersed
systems.
Dispersing
medium
and
dispersed
phase,
Classification of dispersed systems.
1.2. Surface energy. Surface tension. Surface phenomena on the boundary
wean: liquid – gas, liquid-liquid, solid-gas, solid-liquid.
1.3. Adsorption. Adsorbent and adsorbate. The nature of adsorptial forces.
Specific adsorption.
1.4. Surface-inactive and surface-active (surfactant) agents.
1.5. Medical and biological role of adsorption.
1.6. Chromatography. The major notions of chromatography. Classification of
chromatography. Chromatography in biology and medicine.
1.7. Colloids and colloidal dispersions. Hydrophobic and hydrophilic colloids.
The structure of a colloidal micelle. Properties of sols. Stability and coagulation of
dispersed systems. The role of colloidal solutions in biology and medicine.
2. LABORATORY WORK. (45 min.).
2.1. Determination of Pb2+ and Hg2+ ions in a mixture by the method of
column chromatography (demonstration).
2.2. Preparation of colloidal solutions.
2.3. Purification of colloidal solutions by the dialysis technique.
2.4. Determination of the change of the particles of colored sols.
Literature.
1. Lectures.
2. Ebbing D.D. General Chemistry/ D.D.Ebbing , M.S.Wrighton;.- Boston.
Third
edition. Houghton Mifflin Company. – Р. 487 - 491.
2. LABORATORY WORK
2.1.
Determination of Pb2+ and Hg2+ ions in a mixture
by column chromatography method of (Demonstration).
Chromatography and chromatograpfy/mass spectrometry are versatile and
powerful analytical tools used in the clinical laboratory for the separation and
quantification of a variety of clinically relevant analytes.
Chromatography is a physical method of separation in which the
components (solutes) of a sumple mixture are separated by their differential
distribution between two phases-stationary and mobile. In the laboratory these
solutes are the analytes to be measured. During the chromatographic process of
separation the mobile phase carries the sample through a bed, layer, or column
containing the stationary phase.
Those solutes with lower affinities for the
stationary phase reside more in the mobile phase than those with greater affinities
for the stationary phase. Thus the lower affinity solutes travel faster and separate
from solutes having grater affinities for stationary phase. Strongly bound solutes
subsequently are displaced from the stationary phase by changing the physical or
chemical nature of the mobile phase.
The two basic forms of chromatography are planar and column. In planar
chromatography the stationary phase is coated on a sheet of paper or bound to
glass or plastic plate. For paper chromatography the stationary phase is a layer of
water or a polar solvent coated onto the paper fibres. In thin-layer chromatography
(TLC), a thin layer of particles of a material such as silica gel is spread uniformly
on a glass plate or a plastic sheet.
In column chromatography support particles, on which the stationary phase
is coated or chemically bonded, are “packed” into a tube or the stationary phase is
coated onto the inner surface of the tube. The technique is either gas
chromatography (GC) or liquid chromatography (LC).
Chromatographic separations are classified by the chemical or physical
mechanisms used to separate the solutes. These include adsorption, affinity, ionexchange, partition, and steric-exclusion mechanisms.
The thechnique depends on the observation that certain powdered-solid
substances adsorb various other substances and hold them. But some substances
are held more tightly than others.
A solution, containing a mixture of Pb2+ and Hg2+ ions (10 mL) passes
through a glass column filled with powdered Al2O3 (adsorbent). Liquid is poured
through the column and the components of the mixture are carried along down the
column. But the substances are not held with equal strength. The substance most
strongly held moves very slowly, while the substance least strongly held moves
rapidly down the column. The components of the mixture are adsorbed by solid.
The result is that the various components become separated into different zones
(layers).
Because the adsorbed Pb2+ and Hg2+ ions are colorless, we cannot determine
the place of the layers. Therefore 10 mL of KI solution is poured through the
column to display color:
Pb2+ + 2I- → PbI2 ↓ (orange)
Hg2+ + 2I- → HgI2↓ (yellow)
For isolation of pure substances the excess of KI solution (eluent) is poured
through the column and elution of orange and yellow layers is observed. At first a
substance with less adsorptivity is removed, then with a bigger one.
HgI2 + 2KI = K2[HgI4]
PbI2 + 2KI = K2[PbI4]
Then separated solutions can be investigated by instrumental methods (for
example FEC) for the quantative contents of ions [HgI4]2- and [PbI4]22.2. Preparation of colloidal solutions (sols).
2.2.1. Formation of iron hexacyanoferrate(II), Fe4[Fe(CN6)]3, sol.
a. Put 2-3 mL of 0.1 N FeCl3 solution into the test-tube, add 2-3 drops of 10%
K4[Fe(CN6)] solution constantly stirring it. A dark-blue solution of Prussian blue
sol is formed.
b. Put 2-3 mL of 0,1% K4[Fe(CN6)] solution into a test-tube, add 2-3 drops of
10% FeCl3 solution constantly stirring it. A dark-blue solution of Prussian blue sol
is formed.
2.2.2. Formation of iron trihydrixide, Fe(OH)3, sol.
a. Method of peptization.
Put 5-6 drops of saturated FeCl3 solution into a test-tube. Add (NH4)2CO3
solution drop by drop till a red-brown Fe(OH)3 sol is formed.
b. Hydrolysis of Fe(OH)3
Add 3-4 drops of saturated FeCl3 solution to 25 mL of boiling water. As a
result of hydrolysis a red-brown Fe(OH)3 sol is obtained.
2.3. Purification of colloidal solutions by the method of dialysis.
The simplest dialyzer is a glass tube in which one end is closed by a
cellophane film (semipermeable membrane). Put 5-10 mL of Fe(OH)3 sol into the
tube. Place the dialyzer into a glass with distilled water. Part of water molecules
and Cl- ions pass through a membrane into distilled water, but large molecules of
sol are retained in the tube. After 20 minute of the experiment check the presence
of Cl- ions in the distilled water. Pick up a sample (1-2 mL) of the distilled water
and add some drops of AgNO3 solution into it. A white AgCl precipitate is formed.
The presence of Cl- -ion in the sample shows that the process of dialysis is going
on normally.
2.4.
Determination of the change of the colored sols particles.
You can determine the change of the colored sols particles by the method of
capillary analysis, applying by the dependence of sol’s adsorption on the surface
change sign of the adsorbent. It is better to use filter paper as a changed surface.
On watering cellulose walls of filter paper capillaries are negatively charged, and
adjoing to them water is positively charged.
Put a drop of the studied sol (in experiments 2.3) on a sheet of filter paper
with the help of a capillary. The spot colored in the center and colorless at sides
shows that the change of the particles is positive. A uniformly colored spot shows
that the change of the particles is negative.
Knowing the charge of sols, write down the formulas of micelles.
СLASS 14.
TOPIC: “SOLUTIONS”.
1. CONTROL WORK. (120 min.).
СLASS 15.
TOPIC: S-Elements and Their Compounds.
1. SEMINAR. (60 min.).
1.1. General observations of the s-elements. Electronic structure of s-elements
atoms. Properties of Group IA elements.
1.2. Isotopes of hydrogen. Major hydrogen containing compounds. Water.
Hydrogen peroxide. Its red-ox and acid-base properties.
1.3. The most important compounds, containing K and Na. Biological action
of K+ and Na+ ions.
1.4. Lithium and beryllium. Their structures and properties.
1.5. Properties of Group IIA elements: calcium, magnesium, barium. Major
compounds, containing Ca, Mg and Ba. Their biological role. Analytical reactions
for ions: Ca2+, Mg2+ and Ba2+.
1.6. Application of alkali metal compounds and alkali earth compounds in
medicine.
2. LABORATORY WORK (45 min.).
2.1. Chemical test-reactions for the following ions:
2.1.1. Potassium (K+ ) a. with sodium hydrogen tartrate, NaHC4H4O6.
b. with complex compound Na2Pb[Cu(NO2)6].
2.1.2. Sodium (Na+)
with potassium dihydrogen antimonate, NaH2SbO4
2.1.3. Magnesium (Mg2+) with sodium dihydrogen phosphate, NaH2PO4, in
presence of NH4OH and NH4Cl mixture.
2.1.4. Calcium (Ca2+)
with ammonium oxalate, (NH4)2C2O4.
2.1.5. Barium (Ba2+)
with potassium chromate, K2CrO4.
2.2. Analysis of the solution, containing one of s-elements of group IA and
IIA of the Periodic System of Elements.
Literature
1. Lectures.
2. Ebbing D.D. General Chemistry/ D.D.Ebbing , M.S.Wrighton;.- Boston.
Third
edition. Houghton Mifflin Company. – Р. 835−856.
3. Lister T. Chemistry for Advanced Level / T. Lister, J. Renshaw; - Third
edition.
Stanley Thornes (Publishers) Ltd. 2000. – Р. 190 – 195.
PRACTICAL PROBLEMS
1. Identify more metallic elements in each of the following pairs:
a. Na or Mg; b. Na or Be; c. Ga or Cs. Explain your answers.
2. Show the relashionship between the chemical properties of alkali metals and
earth metals, and their compounds and the radius and ionization energy of the
atom.
3. Why does the precipitation of potassium hydrogen tartrate and sodium
dihydrogen antimonate proceed in neutral medium?
4. Calculate the solubility product constant, Ksp,, for barium carbonate, BaCO3.
Solibility of barium carbonate in pure water at 200C equals 7.14.10-5 mol/L.
2. LABORATORY WORK
2.1. Test - reactions for K+.
a. Action of potassium hydrogen tartrate, NaHC4H4O6.
Add 0.5 mL of NaHC4H4O6 solution to 0.5 mL of KCl solution. Cool the
mixture stirring it with a glass stick. The uncolored crystals of KHC4H4O6 are
formed. Divide the obtained crystals into two portions and place them into 2 testtubes and check them for solubility in HCl and NaOH solutions.
This reaction is carried out in neutral medium
KCl + NaHC4 H 4O6 = KHC4 H 4O6 ↓ + NaCl
−
K + + HC 4 H 4O6 = KHC4 H 4O6 ↓
because in acidic solution the solid KHC4H4O6 dissolves with the formation of
weak tartaric acid
KHC4 H 4O6 + HCl = H 2 C4 H 4O6 ↓ + KCl
KHC4 H 4O6 + H + ↔ H 2C 4 H 4O6 + K +
in basic solution the solid KHC4H4O6 dissolves with the formation of neutral salt.
KHC4 H 4O6 + KOH = K 2C 4 H 4O6 ↓ + H 2O
KHC4 H 4O6 + OH − ↔ C4 H 4O6
2−
+ 2 K + + H 2O
b. Action of complex compound Na2Pb[Cu(NO2)6] (Microcrystalloscopic
reaction).
Put 1 drop of colorless KCl solution and 1 drop of green Na2Pb[Cu(NO2)6]
solution on a pure thick glass. Mix them. The black cubic crystals of
K2Pb[Cu(NO2)6] are formed. We observe this crystals under microscope.
2.2. Test - reactions for Na+.
a. Action of potassium dihydrogen antimonat, KH2SbO4.
Add 0.5 mL of KH2SbO4 solution to 0.5 mL of NaCl solution. The mixture is
cooled and mixed with a glass stick. Uncolored crystals of KH2SbO4 are formed.
Divid the obtained crystals into two portions and place them into 2 test-tubes and
check them for solubility in HCl and NaOH solutions.
This reaction is carried out in neutral medium
NaCl + KH 2 SbO4 = NaH 2 SbO4 ↓ + KCl
−
Na + + H 2 SbO4 ↔ NaH 2 SbO4 ↓
because in acidic solution solid NaH2SbO4 transforms into white amorphous
precipitate of antimonic acid, HSbO3
NaH 2 SbO4 + HCl = HSbO3 ↓ + NaCl + H 2O
NaH 2 SbO4 + H + ↔ HSbO3 ↓ + Na + + H 2O
in basic solution solid NaH2SbO4 dissolves with the formation of neutral salt
NaH 2 SbO4 + 2 NaOH = Na3 SbO4 + H 2O
NaH 2 SbO4 + 2OH − ↔ Na + + SbO4
3−
+ 2 H 2O
2.3. Test - reactions for Mg 2+.
a. Action of sodium hydrogen phosphate, Na2HPO4, in ammonia buffer
solution.
Add 10 drops of NH4Cl solution and 6 drops of NH4OH solution to 6 drops of
MgCl2 solution. Add 5 drops of Na2HPO4 solution to the obtained mixture. After
mixing a white precipitate is formed. Check the solubility of magnesiumammonium phosphate salt in mineral (HCl) and organic (CH3COOH) acids.
4Cl
MgCl 2 + NH 4 OH + Na 2 HPO4 NH

→ MgNH 4 PO4 ↓ +2 NaCl + H 2 O
Mg 2+ + HPO4
2−
+
NH 4
+ NH 4OH 
→ MgNH 4 PO4 ↓ + H 2O
The precipitate dissolves in mineral acid
MgNH 4 PO4 + 3HCl = MgCl 2 + NH 4 Cl + H 3 PO4
+
MgNH 4 PO4 + 3H + ↔ Mg 2+ + NH 4 + H 3 PO4
and organic acid
2 MgNH 4 PO4 + 4CH 3 COOH = Mg (CH 3 COO) 2 + 2CH 3COONH 4 + Mg ( H 2 PO4 ) 2
b. Action of ammonium carbonate,(NH4)2CO3.
Put 0.5 mL of MgCl2 solution into a test-tube, add 5 drops of distilled water
and 0,5 mL of (NH4)2CO3 solution. Heat the reaction mixture. A white precipitate
is formed.
Check the solubility of the forming magnesium carbonate salt in
mineral and organic acids.
MgCl 2 + ( NH 4 ) 2 CO3 = MgCO3 ↓ +2 NH 4 Cl
Mg 2+ + CO3
2−
↔ MgCO3 ↓
The precipitate dissolves in mineral acid
MgCO3 + 2 HCl = MgCl 2 + CO2 + H 2 O
MgCO3 + 2 H + ↔ Mg 2+ + CO2 + H 2O
and organic acid
MgCO3 + 2CH 3COOH → Mg (CH 3 COO) 2 + CO2 + H 2 O
2.4. Test - reactions for Ca 2+
a. Action of ammonium oxalate, (NH4)2C2O4.
Put 0.5 mL of CaCl2 solution into a test-tube, add 0.5 mL of (NH4)2C2O4
solution. The formed white precipitate is divided into three portions and placed
into three test-tubes. Check the solubility of calcium oxalate salt in mineral (HCl)
and organic (CH3COOH) acids.
The precipitate in one test-tube is control.
CaCl 2 + ( NH 4 ) 2 C 2 O4 = CaC 2 O4 ↓ +2 NH 4 Cl
Ca 2+ + C2O4
2−
↔ CaC2O4 ↓
The precipitate dissolves in mineral acid with the formation of weak oxalic acid
CaC 2 O4 + 2 HCl = CaCl 2 + H 2 C 2 O4
CaC2O4 + 2 H + ↔ Ca 2+ + H 2 C 2O4
and cannot be dissolved in organic acid because Ka of oxalic acid is higher then Ka
of acetic acid.
b. Action of potassium sulfate, K2SO4.
Put 0.5 mL of CaCl2 solution into a test-tube, add 6 drops of K2SO4
solution. The formed white precipitate is divided into three portions and placed
into three test-tubes. Check the solubility of calcium sulfate salt in mineral (HCl)
and organic (CH3COOH) acids.
The precipitate in one test-tube is control.
CaCl 2 + K 2 SO4 = CaSO4 ↓ +2 KCl
Ca 2+ + SO4
2−
↔ CaSO4 ↓
The precipitate does not dissolve in mineral and organic acids.
2.5. Test - reactions for Ba 2+
a. Action of potassium chromate, K2CrO4.
Put 0.5 mL of BaCl2 solution into a test-tube, add 6 drops of K2CrO4
solution. The formed yellow precipitate is divided into three portions and placed
into three test-tubes. Check the solubility of barium chromate salt in mineral (HCl)
and organic (CH3COOH) acids.
The precipitate in one test-tube is control.
BaCl 2 + K 2 CrO4 = BaCrO4 ↓ +2 KCl
Ba 2+ + CrO4
2−
↔ BaCrO4 ↓
The precipitate dissolves in mineral acids with the formation of dissolving barium
dichromate
2 BaCrO4 + 2 HCl = BaCr2 O7 + BaCl 2 + H 2 O
2 BaCrO4 + 2 H + ↔ 2 Ba 2+ + Cr2O7
The precipitate does not dissolve in organic acids.
b. Action of potassium sulfate, K2SO4.
2−
+ H 2O
Put 0.5 mL of BaCl2 solution into a test-tube, add 6 drops of K2SO4
solution. The formed white precipitate is divided into three portions and placed
into three test-tubes. Check the solubility of calcium sulfate salt in mineral (HCl)
and organic (CH3COOH) acids.
The precipitate in one test-tube is control.
BaCl 2 + K 2 SO4 = BaSO4 ↓ +2 KCl
Ba 2+ + SO4
2−
↔ BaSO4 ↓
The precipitate is insoluble in acids and bases.
СLASS 16.
TOPIC: Coordination Compounds.
1. SEMINAR. (60 min.).
1.1. The structure of coordination compounds. The inner and outer spheres of
coordination compounds. The central atom; coordination number of metals.
Ligand. Its dentation.
1.2. Classification of coordination compounds.
1.3. Nomenclature of coordination compounds.
1.4. The nature of the chemical bond in coordination compounds.
1.5. The ability of atoms and ions of various elements to form coordination
compounds. The notion of hard and soft central and donor atoms.
1.6. Dissociation of a complex in solutions. The instability constant and
stability constant of a coordination compound.
1.7. Chemical properties of complexes. Exchange reactions and oxidation –
reduction reactions with keeping and destroying complex ions.
1.8. Inner complex compounds (chelates).Complex compounds, their
application in medicine. Complexometry.
1.9. The biological importance of coordination complexes.
2. LABORATORY WORK (45 min.).
2.1. Complexometry. Determination of hardness of water.
2.2. Determination of Fe3+ ion concentration.
Literature.
1. Lectures.
2. Ebbing D.D. General Chemistry/ D.D.Ebbing , M.S.Wrighton;.- Boston.
Third
edition. Houghton Mifflin Company. – Р. 934-968.
3. Lister T. Chemistry for Advanced Level / T. Lister, J. Renshaw; - Third
edition.
Stanley Thornes (Publishers) Ltd. 2000. – Р. 469 – 276.
PRACTICAL PROBLEMS
1. Give the oxidation state, the charge of complex ion, the coordination number of
the metal atom of the following complexes.
a. [Ni(NH3)6](ClO3)2
f. [Pt(NH3)4][PtCl4]
b. [Cu(NH3)4]SO4
g. K[Ag(CN)2]
c. K2[Ni(CN)4]
h. [Ag(NH3)2]Cl
d. [Co(NH3)4(H2O)2]Cl3
i. H2[SnCl6]
e. [Cr(H2O)4Cl2]Cl
k. K4[Fe(CN)6]
2. Give the name for each of the following coordination compounds.
a. [Cu(NH3)4](OH)2
d. [Fe(CO)5]
b. K[Al(OH)4(H2O)2]
e. Ca[Zn(CN)4]
c. [Cr(NH3)4SO4]Cl
f. [Zn(NH3)4]SO4
3. Give formulas for the following.
a. pentaammineiodochromium(III) iodide
b. trisethylenediaminenikel(II) bromide
c. potassium tetracyanonikelate(II)
d. tetraamminedichloroplatinum(IV) tetrachloroplatinate(II)
e. potassium hexacyanomanganate(III)
f. sodium tetracyanozincate(II)
g. tetraamminedichlorocobalt(III) nitrate
h. hexaamminechromium(III) tetrachlorocuprate(II)
4. Give formulas for following complex ions.
a. tetrachloroferrate(III) ion
b. pentaammineaquaruthenium(III) ion
c. iodopentakispyridineplatinum(IV) ion
d. amminetrichloroplatinate(II) ion
5. Describe bonding in [Co(H2O)6]2+, and [Co(CN)6]4- complexes using the valence
bond theory.
6. Give the schema of the following complex compound dissociation:
a. K[BiI4]
c. [Cr(H2O)6]Cl
b. Cu2[Fe(CN)6]
d. [Pt(NH3)4Cl2]Cl
7. Give the instability constant expression for the following complex ions:
a. [Co(NO2)6]3-
c. [Fe(CN)6]4-
b. [Ag(NH3)2]+
d. [Cr(H2O)6] 3+
8. Write the reaction of [Zn(NH3)4]SO4 with KCN in molecular and ionic forms.
Why is the complex compound [Zn(CN)4]SO4 formed? Explain. (See Appendix.
Table 6.).
9. Calculate the concentration of Ag+ ions in 0.1M [Ag(NH3)2]Cl.
10. Calculate the concentration of Ni2+ ions in 1L of 0.1M K2[Ni(CN)4]. How does
the concentration of Ni2+ change when 1L of 0.5M NH4CN is added to the initial
solution of the complex compound.
10. Write the chemical equations in molecular and ionic forms (See Appendix.
Table 5, and Table 6.):
a. K[Ag(CN)2] + K2S →
b. [Cd(NH3)4]Cl2 + NaOH →
11. Write the chemical equations of K4[Fe(CN)6] with
a. KMnO4, b. Cl2.
2. LABORATORY WORK
2.1. Determination of general hard water by complexometry.
Fill a buret with 0,05N of ethylenediaminetetraacetate (EDTA) solution.
Using a graduated cylinder, put 25 mL of water into a beaker, add 2 mL of
ammonia buffer, and 3-4 drops of eriochrome black T indicator. The EDTA is
added drop by drop to the reaction mixture in the beaker till the color of the
indicator changes from wine-red to blue.
Carry out titration to take three concurrent readings. Make a table of the
titration readings. Calculate the average volume, and then determine hardness of
the water ( in mmol/L) using the equation:
V ( Na 2 H 2Y ) ⋅ C ( 1 Na 2 H 2Y )
2
Hardness =
⋅ 1000 ,
V ( H 2 O)
where
V ( Na 2 H 2Y )
is the volume of EDTA , mL;
C ( 1 Na 2 H 2Y ) is the normality of EDTA solution, mol/L;
2
V ( H 2 O)
is the volume of water, mL.
2.2 . Determination of Fe3+ ion concentration by complexometry.
Fill a buret with 0,05N of ethylenediaminetetraacetate (EDTA) solution.
Using a measuring pipette transfer 5 mL of analyzed FeCl3 into a beaker, add 1 mL
of acetate buffer solution, and 5−7 drops of alcohol solution of salicylic acid as
indicator. The EDTA is added to the reaction mixture drop by drop till the color of
indicator changes from purple to yellow (essential for complex of iron with
EDTA).
Carry out titration to take three concurrent readings. Make a table of the
titration readings.
Calculate the average volume, and then calculate the normality of Fe3+ ions in
FeCl3 solution(in mol/L) using the equation:
C ( 1 Fe 3+ ) =
2
C ( 1 Na 2 H 2Y ) ⋅ V ( Na 2 H 2Y )
2
;
V ( FeCl 3 )
Determine the mass of Fe3+ ions in 100 ml of analyzed solution by formula
m( Fe 3+ ) = V ⋅ C ( 1 Fe 3+ ) ⋅ M ( 1 Fe 3+ ) ,
2
2
Where
3+
M ( 1 Fe 3+ ) is the molar weight of Fe equivalent.
2
It is equal to half of the molar mass of Fe because Fe3+ ion substitutes two ions of
hydrogen in H2Y2-:
Fe 3+ + H 2Y 2− ↔ FeY − + 2 H +
СLASS 17.
TOPIC: d-Elements and their compounds.
1. SEMINAR. (60 min.).
1.1. General observation of d-elements.
1.2. Properties of elements of the Groups IB and IIB from the point of view of
their position in the Periodic Table. Structure of atoms of the elements. Their acid
− base and red −ox properties.
1.2.1. The major compounds, containing Cu, Ag, Au, Zn, Hg. Their biological
action. Toxicity of these metals.
1.2.2. Test-reaction for Cu2+, Ag+, Zn2+, Hg22+, Hg2+ ions.
1.2.3. Application of copper, silver, gold, zinc and mercury containing
compounds in medicine.
1.3. Properties of elements of the Groups VIB and VIIB from the point of view
of their positions in the Periodic Table. The structure of atoms of elements. Their
acid − base and red −ox properties.
1.3.1. The major compounds, containing Cr(III), Cr(VI), Mo(VI), Mn(II).
Their biological role.
1.3.2. Test-reaction for Cr2+, Cr3+, MoO42-, Mn2+ ions.
1.3.3. Application of chromium, molybdenum, and manganese containing
compounds in medicine.
1.4. Properties of elements of the Groups VIIIB from the point of view of their
position in the Periodic Table. The structure of atoms of elements. Their acid −
base and red −ox properties.
1.4.1. The major compounds, containing Fe(II), Fe(III), Co(II), Co(III), Ni(II).
Their biological action.
1.4.2. Test-reaction for Fe2+, Fe3+, Cr2+, Cr3+, Ni2+ ions.
1.4.3. Application of iron, cobalt, nickel containing compounds in medicine.
2. LABORATORY WORK (45 min.).
2.1. Chemical test-reactions for the following ions:
2.1.1. Copper (Cu2+) a. with ammonium hydroxide, NH4OH (equivalent
quantity and eccess);
2.1.2. Silver (Ag+)
b. with potassium hexacyanoferrate(II), K4[Fe(CN)6].
a. with hydrochloric acid, HCl, followed by solving of
the precipitate, AgCl, in ammonium hydroxide, NH4OH, and decomposition of the
obtained salt by nitric acid, HNO3;
b. with potassium chromate, K2CrO4.
2.1.3. Mercury (Hg22+) and (Hg2+) Demonstration by teacher.
a. with hydrochloric acid, HCl, followed by action with
ammonium hydroxide, NH4OH;
b. with potassium iodide, KI (equivalent quantity and
eccess).
2.1.4. Zinc (Zn2+)
quantity and eccess);
a. with sodium or potassium hydroxides (equivalent
b. with potassium hexacyanoferrate (II), K4[Fe(CN)6];
2.1.5. Chromium(Cr3+) a. with sodium or potassium hydroxides (equivalent
quantity and eccess);
b. with hydrogen peroxide, H2O2, in strong base solu-
tion of NaOH.
2.1.6. Molibdenum (MoO42−)
a. with sodium hydrogen phosphate, Na2HPO4, in solution of nitric acid, HNO3.
2.1.7. Manganese (Mn2+ )
a. with sodium or potassium hydroxides;
b. with hydrogen peroxide, H2O2 , in strong base solution of NaOH.
2.1.8. Iron (Fe2+ )
a. with sodium or potassium hydroxides;
b. with potassium hexacyanoferrate(III), K3[Fe(CN)6].
2.1.9. Iron (Fe3+ )
a. with sodium or potassium hydroxides;
b. with potassium hexacyanoferrate(III), K4[Fe(CN)6];
c. with potassium rhodanide, KNCS.
2.2. Analysis of the preparation, containing one of the cations Cu2+, Ag+,Cr3+,
Zn2+, Fe2+, Fe3+, Mn2+.
Literature.
1. Lectures.
2. Ebbing D.D. General Chemistry/ D.D.Ebbing , M.S.Wrighton;.- Boston.
Third
edition. Houghton Mifflin Company. – Р. 923-942.
3. Lister T. Chemistry for Advanced Level / T. Lister, J. Renshaw; - Third
edition.
Stanley Thornes (Publishers) Ltd. 2000. – Р. 464 – 492.
2. LABORATORY WORK
2.1. Test - reactions for Cu2+ ion.
a. Action of ammonium hydroxide, NH4OH.
Put 0.5 mL of CuSO4 solution into a test-tube, add 0,5 mL of NH4OH solution.
The formed blue-green precipitate of Cu2(OH)2SO4 is divided into two portions
and placed into two test-tubes. Check the solubility of the given solid compound
in excess of ammonium hydroxide.
The precipitate in one test-tube is control.
2CuSO4 + 2 NH 4 OH = Cu 2 (OH ) 2 SO4 ↓ +( NH 4 ) 2 SO4
2−
2Cu 2+ + SO4 + 2 NH 4OH ↔ Cu2 (OH ) 2 SO4 ↓ +2 NH 4
+
The precipitate dissolves in excess of ammonium hydroxide solution with the
formation of dark blow solution of complex compound.
Cu 2 (OH ) 2 SO4 + 10 NH 4 OH = 2[Cu ( NH 3 ) 4 ](OH ) 2 + ( NH 4 ) 2 SO4 + 8 H 2 O
+
Cu2 (OH ) 2 SO4 + 6 NH 4OH + 2 NH 4 ↔ 2[Cu( NH 3 ) 4 ]2+ + SO4
2−
+ 8 H 2O
b. Action of potassium hexacyanoferrate(II), K4[Fe(CN)6].
Put 0.5 mL of CuSO4 solution into a test-tube, add 0,5 mL of K4[Fe(CN)6]
solution. The formed red-brown precipitate of Cu2[Fe(CN)6] is divided into three
portions and placed into three test-tubes. Check the solubility of the given
compound in excess of ammonium hydroxide, NH4OH and sodium hydroxide,
NaOH.
The precipitate in one test-tube is control.
2CuSO4 + K 4 [ Fe(CN ) 6 ] = Cu 2 [ Fe(CN ) 6 ] ↓ +2 K 2 SO4
2Cu 2+ + [ Fe(CN ) 6 ]4− ↔ Cu2 [ Fe(CN ) 6 ] ↓
The precipitate dissolves in excess of ammonium hydroxide solution with the
formation of dissolving ammonium hexacyanoferrate(II).
Cu 2 [ Fe(CN ) 6 ] + 12 NH 4 OH = 2[Cu ( NH 3 ) 4 ](OH ) 2 + ( NH 4 ) 2 [ Fe(CN ) 6 ] + 8 H 2 O
Cu2 [ Fe(CN ) 6 ] + 8 NH 4OH ↔ 2[Cu( NH 3 ) 4 ]2+ + [ Fe(CN ) 6 ]4− + 8 H 2O
The given solid complex compound reacts with NaOH solution transforming in
less soluble Cu(OH)2 .
Cu 2 [ Fe(CN ) 6 ] + 4 NaOH = 2Cu (OH ) 2 ↓ + Na 4 [ Fe(CN ) 6 ]
Cu2 [ Fe(CN ) 6 ] + 4OH − ↔ 2Cu(OH ) 2 ↓ +[ Fe(CN ) 6 ]4−
2.2. Test - reactions for Ag+ ion.
a. Action of hydrogen chloride acid.
Put 0.5 mL of AgNO3 solution into a test-tube, add 2-3 drops of HCl solution.
The formed white precipitate of AgCl is divided into two portions and placed into
two test-tubes. Check the solubility of the given solid compound in excess of
ammonium hydroxide.
The precipitate in one test-tube is control.
AgNO3 + HCl = AgCl ↓ + HNO3
Ag + + Cl − ↔ AgCl ↓
The precipitate dissolves in excess of concentrated ammonium hydroxide solution
with the formation of dissolving complex compound.
AgCl + 2 NH 4 OH = [ Ag ( NH 3 ) 2 ]Cl + 2 H 2 O
AgCl + 2 NH 4OH ↔ [ Ag ( NH 3 ) 2 ]+ + Cl − + 2 H 2O
Complex compound is destroyed by nitric acid with the formation of solid AgCl.
[ Ag ( NH 3 ) 2 ]Cl + 2 HNO3 = AgCl ↓ +2 NH 4 NO3
[ Ag ( NH 3 ) 2 ]+ + Cl − + 2 H + ↔ AgCl ↓ +2 NH 4
+
b. Action of potassium chromate, K2CrO4.
Put 0.5 mL of AgNO3 solution into a test-tube, add 3-4 drops of K2CrO4
solution. The formed white precipitate of Ag2CrO4 is divided into three portions
and placed into three test-tubes. Check the solubility of thegiven solid compound
in excess of ammonium hydroxide solution, and nitric acid solution.
The precipitate in one test-tube is control.
2 AgNO3 + K 2 CrO4 = Ag 2 CrO4 ↓ +2 KNO3
2 Ag + + CrO4
2−
↔ Ag 2CrO4 ↓
The precipitate dissolves in excess of concentrated ammonium hydroxide solution
with the formation of dissolving complex compound.
Ag 2 CrO4 + 6 NH 4 OH = 2[ Ag ( NH 3 ) 2 ](OH ) + ( NH 4 ) 2 CrO4 + 4 H 2 O
Ag 2CrO4 + 4 NH 4OH ↔ 2[ Ag ( NH 3 ) 2 ]+ + CrO4
2−
+ 4 H 2O
The precipitate dissolves in nitric acid solution with the formation of chromic
acid
2 Ag 2 CrO4 + HNO3 = 4 AgNO3 + H 2 Cr2 O7 + H 2 O
2 Ag 2CrO4 + H + ↔ 4 Ag + + Cr2O7
2−
+ H 2O
2.3. Test - reactions for Hg22+ and Hg2+ ions.
2.3.1. Test - reactions for Hg22+
a. Action of hydrogen chloride acid, HCl, or its salts.
Put 0.5 mL of Hg2(NO3)2 solution into a test-tube, add 5-6 drops of HCl
solution. The formed white precipitate of Hg2Cl2 (horn mercury, calomel) is
divided into two portions and placed into two test-tubes. Check the action of horn
mercury with excess of ammonium hydroxide solution till the Hg black formation.
The precipitate in one test-tube is control.
Hg 2 ( NO3 ) 2 + 2 HCl = Hg 2 Cl 2 ↓ +2 HNO3
Hg 2
2+
+ 2Cl − ↔ Hg 2Cl2 ↓
The precipitate reacts with the excess of ammonium hydroxide solution with the
formation of Hg.
2 Hg 2 Cl 2 + 2 NH 4 OH = HgNH 2 Cl ↓ + Hg ↓ + NH 4 Cl + H 2 O
+
2 Hg 2Cl2 + 2 NH 4OH ↔ HgNH 2Cl ↓ + Hg ↓ + NH 4 + Cl − + H 2O
b. Action of potassium iodide, KI.
Put 5 drops of Hg2(NO3)2 solution into a test-tube, add 5-6 drops of KI
solution. The formed green precipitate of Hg2I2 is divided into two portions and
placed into two test-tubes. Check the solubility of precipitate with excess of
potassium iodide solution.
Hg 2 ( NO3 ) 2 + 2 KI = Hg 2 I 2 ↓ +2 KNO3
Hg 2
2+
+ 2 I − ↔ Hg 2 I 2 ↓
The precipitate does not react with the excess of potassium iodide.
2.3.2. Test - reactions for Hg2+
a. Action of hydrogen chloride acid, HCl, or its salts.
Put 0.5 mL of Hg(NO3)2 solution into a test-tube, add 5-6 drops of HCl
solution. The formed colorless substance, HgCl2, is comparatively poorly soluble
in cold water. Divide the given solution on two test-tube. Check the action of
HgCl2 with excess of ammonium hydroxide solution till the white precipitate of
HgNH2Cl forms.
The precipitate in one test-tube is control.
Hg ( NO3 ) 2 + 2 HCl = HgCl 2 ↓ +2 HNO3
Hg 2+ + 2Cl − ↔ HgCl2 ↓
The precipitate reacts with the excess of ammonium hydroxide solution with the
formation of white solid HgNH2Cl.
HgCl 2 + 2 NH 4 OH = HgNH 2 Cl ↓ + NH 4 Cl + 2 H 2 O
+
HgCl2 + 2 NH 4OH ↔ HgNH 2 Cl ↓ NH 4 + H 2O
b. Action of potassium iodide, KI.
Put 5 drops of Hg(NO3)2 solution into a test-tube, add 5-6 drops of KI
solution. The formed orange precipitate of HgI2 is divided into two portions and
placed into two test-tubes. Check the solubility of the precipitate with excess of
potassium iodide solution.
Hg ( NO3 ) 2 + 2 KI = HgI 2 ↓ +2 KNO3
Hg 2+ + 2 I − ↔ HgI 2 ↓
The precipitate reacts with the excess of potassium iodide with the formation of
soluble complex compound.
HgI 2 + 2 KI = K 2 [ HgI 4 ]
HgI 2 + 2 I − ↔ [ HgI 4 ]2−
2.4. Test - reactions for Zn2+ ion.
a. Action of alkaline metals hydroxides, NaOH or KOH
Put 0.5 mL of ZnCl2 solution into a test-tube, add NaOH or KOH solution
drop by drop till the formation of white precipitate of Zn(OH)2. The precipitate is
divided into three portions and placed into three test-tubes. Check the solubility of
the solid compound in excess of sodium hydroxide solution, and hydrochloric acid
solution.
The precipitate in one test-tube is control.
ZnCl 2 + 2 NaOH = Zn(OH ) 2 ↓ +2 NaCl
Zn 2+ + 2OH − ↔ Zn(OH ) 2 ↓
The precipitate dissolves in excess of sodium hydroxide solution with the
formation of dissolving complex compound.
Zn(OH ) 2 + 2 NaOH = Na 2 [ Zn(OH ) 4 ]
Zn(OH ) 2 + 2OH − ↔ [ Zn(OH ) 4 ]2−
The precipitate dissolves in nitric acid solution.
Zn(OH ) 2 + 2 HCl = ZnCl 2 + 2 H 2 O
Zn(OH ) 2 + 2 H + ↔ Zn 2+ + 2 H 2O
b. Action of ammonium hydroxide, NH4OH.
Put 0.5 mL of ZnCl2 solution into a test-tube, add NH4OH solution drop by
drop till the formation of white precipitate of Zn(OH)2. The precipitate is divided
into two portions and placed into two test-tubes. Check the solubility of the solid
compound in excess of ammonium hydroxide solution.
The precipitate in one test-tube is control.
ZnCl 2 + 2 NH 4 OH = Zn(OH ) 2 ↓ +2 NH 4 Cl
Zn 2+ + 2OH − ↔ Zn(OH ) 2 ↓
The precipitate dissolves in excess of concentrated ammonium hydroxide solution
with the formation of dissolving complex compound.
Zn(OH ) 2 + 6 NH 4 OH = [ Zn( NH 3 ) 4 ](OH ) 2 + 2 NH 4 Cl + 4 H 2 O
Zn(OH ) 2 + 4 NH 4OH ↔ [ Zn( NH 3 ) 4 ]2+ + 4 H 2O
c. Action of potassium hexacyanoferrate(II), K4[Fe(CN)6].
Put 0.5 mL of ZnCl2 solution into a test-tube, add 0,5 mL of K4[Fe(CN)6]
solution. The formed white precipitate of K2Zn3 [Fe(CN)6] is divided into two
portions and placed into two test-tubes. Check the solubility of the complex
compound in excess potassium hydroxide, KOH.
The precipitate in one test-tube is control.
3ZnCl 2 + 2 K 4 [ Fe(CN ) 6 ] = K 2 Zn3 [ Fe(CN ) 6 ] 2 ↓ +6 KCl
3Zn 2+ + 2 K + + 2[ Fe(CN ) 6 ]4− ↔ K 2 Zn3 [ Fe(CN ) 6 ]2 ↓
Precipitate dissolves in excess of potassium hydroxide solution.
K 2 Zn3 [ Fe(CN ) 6 ]2 + 12 KOH = 2 K 4 [ Fe(CN ) 6 ] + 3K 2 ZnO2 + 6 H 2 O
−
K 2 Zn3 [ Fe(CN ) 6 ]2 + 12OH − ↔ 2 K + + 2[ Fe(CN ) 6 ]4− + 3ZnO2 + 6 H 2O
2.5. Test - reactions for Cr3+ ion.
a. Action of alkaline metals hydroxides, NaOH or KOH
Put 0.5 mL of CrCl3 solution into a test-tube, add NaOH or KOH solution
drop by drop till the formation of light green precipitate of Cr(OH)3. The
precipitate is divided into three portions and placed into three test-tubes. Check the
solubility of the solid compound in excess of sodium hydroxide solution, and
hydrochloric acid solution.
The precipitate in one test-tube is control.
CrCl 3 + 3 NaOH = Cr (OH ) 3 ↓ +3 NaCl
Cr 3+ + 3OH − ↔ Cr (OH ) 3 ↓
The precipitate dissolves in excess of sodium hydroxide solution with the
formation of dissolving complex compound.
Cr (OH ) 3 + 3 NaOH = Na 3 [Cr (OH ) 6 ]
Cr (OH ) 3 + 3OH − ↔ [ Zn(OH ) 6 ]3−
The precipitate dissolves in hydrochloric acid solution.
Cr (OH ) 3 + 3HCl = CrCl 3 + 3H 2 O
Cr (OH ) 3 + 3H + ↔ Cr 3+ + 3H 2O
b. Oxidation of Cr3+ by hydrogen peroxide,H2O2, in basic solution.
Put 0.5 mL of CrCl3 solution into a test-tube, add excess of NaOH or KOH
solution till the formation of green solution of complex compound. The solution is
divided into two portions and placed into two test-tubes. Add 0.5 mL of hydrogen
peroxide solution into one test-tube and heat the reaction mixture till yellow
colored solution of K2CrO4 forms.
The solution in one test-tube is control.
2CrCl 3 + 10 KOH + 3H 2 O2 = 2 K 2 CrO4 + 6 KCl + 8 H 2 O
2Cr 3+ + 10OH − + 3H 2O2 ↔ 2CrO4
2−
+ 8 H 2O
2.6. Test - reactions for MoO42- ion.
a. Action of sodium hydrogen phosphate, Na2HPO4 in nitric acid solution.
Put 0.5 mL of ammonium molybdate, (NH4)6Mo7O24 . 4H2O, in nitric acid
solution into a test-tube, add 0,5 mL of Na2HPO4 solution. The reaction mixture is
heated till the formation of yellow precipitate of the composition
(NH4)3PO4 . 12MoO3 . 6H2O.
12( NH 4 ) 2 MoO4 + Na2 HPO4 + 23HNO3 = ( NH 4 ) 3 PO4 ⋅ 12 MoO3 + 2 NaNO3 + 21NH 4 NO3 + 12 H 2O
2.7. Test - reactions for Mn2+ ion.
a. Action of alkaline metals hydroxides, NaOH or KOH
Put 0.5 mL of MnCl2 solution into a test-tube, add NaOH or KOH solution
drop by drop till the formation of light brown precipitate of Mn(OH)2. The
precipitate is divided into three portions and placed into three test-tubes. Check the
solubility of the solid compound in excess of sodium hydroxide solution, and
hydrochloric acid solution.
The precipitate in one test-tube is control.
MnCl 2 + 2 NaOH = Mn(OH ) 2 ↓ +2 NaCl
Mn 2+ + 2OH − ↔ Mn (OH ) 2 ↓
The precipitate does not dissolve in excess of sodium hydroxide solution, and
dissolves in hydrochloric acid solution.
Mn(OH ) 2 + 2 HCl = MnCl 2 + 2 H 2 O
Mn (OH ) 2 + 2 H + ↔ Mn 2+ + 2 H 2O
b. Oxidation of Mn2+ by hydrogen peroxide,H2O2, in basic solution.
Put 0.5 mL of MnCl2 solution into a test-tube, add excess of NaOH or KOH
solution till the formation of precipitate. The precipitate is divided into two
portions and placed into two test-tubes. add 0.5 ml of hydrogen peroxide solution
into one test-tube and the dark brown solid compound, H2MnO3 is formed.
The solution in one test-tube is control.
MnCl 2 + 2 KOH + H 2 O2 = H 2 MnO3 ↓ +2 KCl + H 2 O
Mn 2+ + 2OH − + H 2O2 ↔ H 2 MnO3 ↓ + H 2O
2.8. Test - reactions for Fe2+ ion.
a. Action of alkaline metals hydroxides, NaOH or KOH
Put 0.5 mL of FeSO4 solution into a test-tube, add NaOH or KOH solution
drop by drop till the formation of light green precipitate of Fe(OH)2. The
precipitate is divided into three portions and placed into three test-tubes. Check the
solubility of the solid compound in excess of sodium hydroxide solution, and
hydrochloric acid solution.
The precipitate in one test-tube is control.
FeSO4 + 2 NaOH = Fe(OH ) 2 ↓ + Na 2 SO4
Fe 2+ + 2OH − ↔ Fe(OH ) 2 ↓
The precipitate does not dissolve in excess of sodium hydroxide solution, and
dissolves in hydrochloric acid solution.
Fe(OH ) 2 + 2 HCl = FeCl 2 + 2 H 2 O
Fe(OH ) 2 + 2 H + ↔ Fe 2+ + 2 H 2O
b. Action of potassium hexacyanoferrate(III), K3[Fe(CN)6].
Put 0.5 mL of FeSO4 solution into a test-tube, add 0,5 mL of K3[Fe(CN)6]
solution. The forming dark blow precipitate of Fe3 [Fe(CN)6] 2 is divided into two
portions and placed into two test-tubes. Check the solubility of the complex
compound in excess of potassium hydroxide, KOH.
The precipitate in one test-tube is control.
3FeCl 2 + 2 K 3 [ Fe(CN ) 6 ] = Fe3 [ Fe(CN ) 6 ] 2 ↓ +6 KCl
3Fe 2+ + 2[ Fe(CN ) 6 ]3− ↔ Fe3 [ Fe(CN ) 6 ]2 ↓
The precipitate dissolves in excess of potassium hydroxide solution with the
Fe(OH)2 formation.
Fe3 [ Fe(CN ) 6 ]2 + 6 KOH = 3Fe(OH ) 2 ↓ +2 K 3 [ Fe(CN ) 6 ]
Fe3 [ Fe(CN ) 6 ]2 + 6OH − ↔ 3Fe(OH ) 2 ↓ +2[ Fe(CN ) 6 ]3−
2.9. Test - reactions for Fe3+ ion.
a. Action of alkaline metals hydroxides, NaOH or KOH
Put 0.5 mL of FeCl3 solution into a test-tube, add NaOH or KOH solution
drop by drop till the formation of red-brown precipitate of Fe(OH)3. The
precipitate is divided into three portions and placed into three test-tubes. Check the
solubility of the solid compound in excess of sodium hydroxide solution, and
hydrochloric acid solution.
The precipitate in one test-tube is control.
FeCl 3 + 3 NaOH = Fe(OH ) 3 ↓ +3 NaCl
Fe 3+ + 3OH − ↔ Fe(OH ) 3 ↓
The precipitate does not dissolve in excess of sodium hydroxide solution, and
dissolves in hydrochloric acid solution.
Fe(OH ) 3 + 3HCl = FeCl 3 + 3H 2 O
Fe(OH ) 3 + 3H + ↔ Fe 3+ + 3H 2O
b. Action of potassium hexacyanoferrate(II), K4[Fe(CN)6].
Put 0.5 mL of FeCl3 solution into a test-tube, add 0,5 mL of K4[Fe(CN)6]
solution. The formed dark blow precipitate of Fe4 [Fe(CN)6]3 is divided into two
portions and placed into two test-tubes. Check the solubility of complex compound
in excess of potassium hydroxide, KOH.
The precipitate in one test-tube is control.
4 FeCl 2 + 3K 4 [ Fe(CN ) 6 ] = Fe4 [ Fe(CN ) 6 ]3 ↓ +12 KCl
4 Fe 2+ + 3[ Fe(CN ) 6 ]4− ↔ Fe4 [ Fe(CN ) 6 ]3 ↓
The precipitate dissolves in excess of potassium hydroxide solution with the
Fe(OH)3 formation.
Fe4 [ Fe(CN ) 6 ]3 + 12 KOH = 4 Fe(OH ) 2 ↓ +3K 4 [ Fe(CN ) 6 ]
Fe4 [ Fe(CN ) 6 ]3 + 12OH − ↔ 4 Fe(OH ) 2 ↓ +3[ Fe(CN ) 6 ]4−
c. Action of potassium or ammonium rhodanide, KNCS or (NH4)NCS.
Put 0.5 ml of FeCl3 solution into a test-tube, add 1-2 drops of KNCS solution.
The garnet-red iron(III) thiocyanide, Fe(SCN)3, is formed.
FeCl 3 + 3KNCS = Fe( SCN ) 3 + 3KCl
Fe 3+ + 3NCS − ↔ Fe( SCN ) 3
СLASS 18.
TOPIC: p-Elements and their compounds.
1. SEMINAR. (60 min.).
1.1. General observation of p-elements.
1.2. Properties of elements of the Groups IIIA and IVA from the point of view
of their positions in the Periodic Table. Structure of their atoms. Their acid −
base properties.
1.2.1. The major compounds, containing boron and aluminum (boric acid and
borax). Their properties and biological action.
1.2.2. The major compounds of group IVA elements: oxides,
hydroxides,
acids, salts; their properties.
1.2.3. Carbon monoxide, carbon dioxide, their biological activity. Carbonic
acid and its salts.
1.2.4. Silicon dioxide and silicates. Application of a silica glass in medicine.
1.2.5. The major compounds, containing C, Si, Sn, Pb. Their biological action.
Toxicity of lead.
1.2.6. Test-reactions for CO32−, HCO3−, C2O42−, CH3COO− ions
1.3. Properties of elements of the Groups VA from the point of view of their
positions in the Periodic Table. Structure of element’s atom. Their acid − base and
red −ox properties.
1.3.1. The major compounds of nitrogen. Ammonia. Ammonium salts.
Nitrogen oxides, nitrous acid, nitric acid. Nitrates, their toxicity.
1.3.2. The major compounds of phosphorus. Phosphorus oxides and their
properties. Ortho-, metha- and pyrophosphoric acids. Phosphates.
1.3.3. Test-reactions for NO3−, PO43−, HPO42−, AsO43− ions.
1.3.4. The major compounds, containing N, P, As, Sb, Bi. Their biological
action.
1.4. Properties of elements of the Groups VIA from the point of view of their
positions in the Periodic Table. Structure of elements atom. Their acid − base and
red −ox properties.
1.4.1. Oxygen. Molecular oxygen and ozone.
1.4.2. The major compounds of sulfur. Sulfur oxides, sulfurous acid and
sulfuric acid.
1.4.3. Test-reactions for S2− , SO42−, S2O32−, SCN−.
1.4.4. The biological role of oxygen and sulfur. Application of oxygen and
ozone in medicine.
1.5. Properties of elements of the Groups VIIA from the point of view of their
positions in the Periodic Table. Structure of elements atom. Their acid − base and
red −ox properties.
1.5.1. The halogens. Hydrogen halides. Halogen oxoacid.
1.5.2. Test-reactions for Cl −, Br−, I−.
1.5.3. The biological role of F, Cl, Br, I containing compounds.
2. LABORATORY WORK. (45 min.)
2.1. Chemical test-reactions for the following ions:
2.1.1. Carbonate ion (CO32−, HCO3−)
a. with barium salts, BaCl2;
b. with mineral acids.
2.1.2. Acetate ion (CH3COO−)
a. with sulfuric acid, H2SO4;
b. with salts of iron (III), FeCl3 .
2.1.3. Oxalate ion (C2O42−)
a. with barium salts, BaCl2;
b. with acidic solution of potassium permanganate,
KMnO4.
2.1.4. Sulfate ion (SO42−)
a. with barium salts, BaCl2;
b. with antimony acetate, Pb(CH3COO)2.
2.1.5. Tiosulfate ion (S2O32−)
a. with barium salts, BaCl2;
b. with iodine solution, I2.
2.1.6. Rhodanide ion (CNS−)
a. with silver nitrate, AgNO3;
b. with iron(III) chloride, FeCl3 .
2.1.7. Phosphate ion (PO43−, HPO42−)
a. with magnesium chloride in ammonium buffer
solution (MgCl2 + NH4OH + NH4Cl);
b. with ammonium molybdate, (NH4)2MoO4 .
2.1.8. Nitrate ion (NO3−)
a. with copper in concentrated sulfuric acid;
b. with excess of iron(II) sulfate, FeSO4.
2.1.9. Chloride ion (Cl−)
a. with silver nitrate, AgNO3.
2.1.10. Bromide ion (Br−)
with chlorine water, Cl2.
2.1.11. Iodide ion (I−)
a. with chlorine water, Cl2 ;
b. with antimony acetate, Pb(CH3COO)2.
2.2. Analysis of a sample, containing one of the anion with p-element.
3. CONTROL WORK.
Literature.
1. Lectures.
2. Ebbing D.D. General Chemistry/ D.D.Ebbing , M.S.Wrighton;.- Boston.
Third
edition. Houghton Mifflin Company. – Р. 874-911.
3. Lister T. Chemistry for Advanced Level / T. Lister, J. Renshaw; - Third
edition.
Stanley Thornes (Publishers) Ltd. 2000. – Р. 429 – 462.
2. LABORATORY WORK
2.1. Chemical test-reactions for the following ions:
2.1.1. Test - reactions for CO32− and HCO3− ions.
a. Action of barium chloride,BaCl2.
Put 0.5 mL of Na2CO3 solution into a test-tube, add 4-5 drops of BaCl2
solution. The formed white precipitate of BaSO4 is divided into three portions and
placed into three test-tubes. Check the solubility of the given solid compound in
mineral and organic acids.
The precipitate in one test-tube is control.
Na 2 CO3 + BaCl 2 = BaCO3 ↓ +2 NaCl
2 Na + + CO3
2−
+ Ba 2+ + 2Cl − ↔ BaCO3 ↓ +2 Na + + 2Cl −
CO3
2−
+ Ba 2+ ↔ BaCO3 ↓
The precipitate dissolves in mineral and organic acids to evolve bubbles of carbon
dioxide.
BaCO3 + 2 HCl = CO2 ↑ + H 2 O + BaCl 2
BaCO3 + 2CH 3 COOH = CO2 ↑ + H 2 O + Ba (CH 3COO ) 2
BaCO3 + 2 H + ↔ CO2 ↑ + H 2O + Ba 2+
b. Action of mineral and organic acids.
Put 0.5 mL of Na2CO3 solution into a test-tube, add 4-5 drops of HCl or
CH3COOH solution. The formation of carbon dioxide bubbles is observed.
Na 2 CO3 + 2 HCl = CO2 ↑ + H 2 O + 2 NaCl
2 Na + + CO3
2−
+ 2 H + + 2Cl − ↔ CO2 ↑ + H 2O + 2 Na + + 2Cl −
CO3
2−
+ 2 H + ↔ CO2 ↑ + H 2O
2.1.2. Test - reactions for CH3COO− ion.
a. Action of sulfuric acid,H2SO4 .
Put 0.5 mL of CH3COONa solution into the test-tube, add 0,5 mL of dilute
H2SO4 solution. The mixture is heated and we feel a specific smell of acetic acid
vapor.
t
2CH 3 COONa + H 2 SO4 
→
2CH 3COOH ↑ + Na 2 SO4
2CH 3 COO − + 2 Na + + 2 H + + SO4
2−
t

→
2CH 3 COOH ↑ +2 Na + + SO4
2−
CH 3COO − + H + ↔ CH 3COOH ↑
b. Action of iron (III) chloride, FeCl3 .
Put 0.5 mL of CH3COONa solution into a test-tube, add 5-6 drops of FeCl3
solution. The formed red-brown solution Fe(CH3COO)3 is hydrolyzed under the
heat with the formation of red-brown precipitate of Fe(OH)2(CH3COO).
3CH 3COONa + FeCl 3 = Fe(CH 3 COO ) 3 + 3 NaCl
t
Fe(CH 3COO ) 3 + 2 H 2 O 
→
Fe(OH ) 2 (CH 3COO ) ↓ +2CH 3COOH
2.1.3. Test - reactions for C2O42− ions.
a. Action of barium chloride, BaCl2.
Put 0.5 mL of (NH4)2C2O4 solution into a test-tube, add 0,5 mL of BaCl2
solution. The formed white precipitate is divided into three portions and placed
into three test-tubes. Check the solubility of barium oxalate salt in mineral (HCl)
and organic (CH3COOH) acids.
The precipitate in one test-tube is control.
( NH 4 ) 2 C 2 O4 + BaCl 2 = BaC 2 O4 ↓ +2 NH 4 Cl
C 2O 4
2−
+ Ba 2+ ↔ BaC 2O4 ↓
The precipitate dissolves in mineral acid with the formation of weak oxalic acid
BaC 2 O4 + 2 HCl = BaCl 2 + H 2 C 2 O4
BaC 2O4 + 2 H + ↔ Ba 2+ + H 2C 2O4
and cannot be dissolved in organic acid because Ka of oxalic acid is higher then Ka
of acetic acid.
b. Action of acidic solution of potassium permanganate, KMnO4.
Put 5-6 drops of (NH4)2C2O4 solution into a test-tube, add the same volume of
0.1M H2SO4 solution. The mixture is heated slightly. Add 3-5 drops of KMnO4
solution to the reaction mixture. The violet color of KMnO4 solution disappears.
5( NH 4 ) 2 C 2 O4 + 8H 2 SO4 + 2 KMnO4 = 2MnSO4 + 10CO2 + 8H 2 O + 5( NH 4 ) 2 SO4 + K 2 SO4
2.1.4. Test - reactions for SO42− ion.
a. Action of barium chloride, BaCl2.
Put 0.5 mL of Na2SO4 solution into a test-tube, add 3-4 drops of BaCl2
solution. The forming white precipitate is divided into three portions and placed
into three test-tubes. Check the solubility of barium sulfate salt in mineral (HCl)
and organic (CH3COOH) acids.
The precipitate in one test-tube is control.
Na 2 SO4 + BaCl 2 = BaSO4 ↓ +2 NaCl
2 Na + + SO4
2−
+ Ba 2+ + 2Cl − ↔ BaSO4 ↓ +2 Na + + 2Cl −
SO4
2−
+ Ba 2+ ↔ BaSO4 ↓
b. Action of antimony acetate, Pb(CH3COO)2.
Put 0.5 mL of Na2SO4 solution into a test-tube, add 0.5 mL of Pb(CH3 COO)2
solution. The forming white precipitate is divided into three portions and placed
into three test-tubes. Check the solubility of antimony sulfate salt in mineral (HCl)
acid and the excess of alkaline metals hydroxides, KOH or NaOH.
The precipitate in one test-tube is control.
Na 2 SO4 + Pb(CH 3COO ) 2 = PbSO4 ↓ +2CH 3COONa
2 Na + + SO4
2−
+ Pb 2+ + 2CH 3COO − ↔ PbSO4 ↓ +2 Na + + 2CH 3COO −
SO4
2−
+ Pb 2+ ↔ PbSO4 ↓
The presipitate dissolves in excess of alkaline metals hydroxides with the
formation of complex salt.
PbSO4 + 4 NaOH = Na 2 [ Pb(OH ) 4 ] + Na 2 SO4
PbSO4 + 4OH − = [ Pb(OH ) 4 ] 2− + SO4
2−
and does not dissolve in mineral acids.
2.1.5. Test - reactions for S2O32− ion.
a. Action of barium chloride, BaCl2.
Put 0.5 mL of Na2S2O3 solution into a test-tube, add the same volume of
0.1M BaSO4 solution. The formed white precipitate is divided into two portions
and placed into two test-tubes. Check the solubility of barium thiosulfate salt in
mineral (HCl) acid under the heat.
Na 2 S 2 O3 + BaCl 2 = BaS 2 O3 ↓ +2 NaCl
2 Na + + S 2O3
2−
+ Ba 2+ + 2Cl − ↔ BaS 2O3 ↓ +2 Na + + 2Cl −
S 2 O3
2−
+ Ba 2+ ↔ BaS 2O3 ↓
The precipitate dissolves in mineral acid
BaS 2 O3 + 2 HCl = H 2 S 2 O3 + BaCl 2
BaS 2O3 ↓ +2 H + ↔ H 2 S 2O3 + Ba 2+
But the forming thiosulfuric acid under the heat disintegrates according the
reaction
H 2 S 2 O3 → S ↓ + SO2 ↑ + H 2 O
b. Action of iodine, I2.
Put 5-6 drops of Na2S2O3 solution into a test-tube, add iodine solution drop by
drop. The red-brown color of iodine solution disappears.
Na 2 S 2 O3 + I 2 = 2 NaI + Na 2 S 4 O6
S 2 O3
2−
+ I 2 ↔ 2 I − + S 4O6
2−
2.1.6. Test - reactions for CNS− ion.
a. Action of barium chloride, silver nitrate, AgNO3.
Put 0.5 mL of KCNS solution into a test-tube, add the same volume of 0.1M
AgNO3 solution. The formed white precipitate is divided into two portions and
placed into two test-tubes. Check the solubility of silver rhodanide salt in mineral
(HNO3) acid and in the excess of ammonium hydroxid.
KNCS + AgNO3 = AgCNS ↓ + KNO3
CNS − + Ag + ↔ AgCNS ↓
The precipitate dissolves in the excess of ammonium hydroxide with the formation
of complex compound
AgCNS + 2 NH 4 OH = [ Ag ( NH 3 ) 2 ]CNS + 2 H 2 O
The precipitate does not dissolve in mineral acid.
b. Action of iron(III) chloride, FeCl3.
Put 0.5 mL of KNCS solution into a test-tube, add 1-2 drops of FeCl3 solution.
The garnet-red iron(III) thiocyanide, Fe(SCN)3, is formed.
FeCl 3 + 3KNCS → Fe( SCN ) 3 + 3KCl
Fe 3+ + 3NCS − ↔ Fe( SCN ) 3
2.1.7. Test - reactions for PO43− (HPO42−) ions.
a. Action of magnesium chloride in ammonium buffer solution (MgCl2 +
NH4OH + NH4Cl). (See class 14, Lab. work. 2.3).
b. Action of ammonium molybdate, (NH4)2MoO4. (See class 15, Lab. work.
2.6).
2.1.8. Test - reactions for NO3− ion (demonstration).
a. Action of copper in concentrated sulfuric acid.
Put 5-6 drops of sodium or potassium nitrate, NaNO3 or KNO3, solution into a
test-tube, add 6-7 drops of concentrated H2SO4 solution, and a little bit of Cu
metal. The mixture is heated on the flame. The formation of reddish nitrogen
dioxide, NO2, gas is observed.
3Cu + 8 NaNO3 + 4 H 2 SO4 = 3Cu ( NO3 ) 2 + 4 Na 2 SO4 + 2 NO ↑ +4 H 2 O
In normal conditions the forming NO is oxidized by the oxygen with the NO2
2 NO + O2 = 2 NO2
formation:
b. Action of iron(II) sulfate, FeSO4.
Put 5-6 drops of FeSO4 saturated solution into a test-tube, add 3-4 drops of
investigated NaNO3 solution and the reaction mixture gets mixed. Then add into
the mixture 5-6 drops of concentrated H2SO4 solution placing it on the test-tube
wall carefully. A bilayer is formed ( FeSO4 is above, H2SO4 is below). On the
surface of the two liquids contact a brown ring of a complex compound
[FeNO]SO4 is observed.
6 FeSO4 + 2 NaNO3 + 4 H 2 SO4 = 3Fe2 ( SO4 ) 3 + Na 2 SO4 + 2 NO ↑ +4 H 2 O
The NO reacts with the excess of Fe2+ ions with the formation of brown unstable
complex ions
NO + FeSO4 = [ FeNO]SO4
NO + Fe 2+ ↔ [ FeNO ]2+
2.1.9. Test - reaction for Cl− ion.
Action of silver nitrate, AgNO3. (See class 16, Lab. work. 2.2).
2.1.10. Test - reaction for Br− ion.
Action of chlorine water, Cl2 .
Put 0,5ml of potassium chloride KBr, solution into a test-tube, add 6-7 drops of
chlorine water, and 5-6 drops of chloroform. The reaction mixture is shaken
intensively. After that the organic layer is colored by bromine in yellow color.
2 KBr + Cl 2 → 2 KCl + Br2
2.1.11. Test - reaction for I− ion.
a. Action of chlorine water, Cl2 .
Put 0,5ml of potassium chloride KI, solution into a test-tube, add 6-7 drops of
chlorine water, and 5-6 drops of chloroform. The reaction mixture is shaken
intensively. After that the organic layer is colored by iodine in pink color.
2 KI + Cl 2 = 2 KCl + I 2
b. Action of antimony acetate, Pb(CH3COO)2.
Put 6-7 drops of Pb(CH3COO)2 solution into a test-tube, add KI solution drop
by drop till the formation of yellow PbI2 precipitate. The formed precipitate is
divided into three portions and placed into three test-tubes. Check the solubility of
precipitate with the excess of potassium iodide solution.
The precipitate in one test-tube is control.
Pb(CH 3 COO ) 2 + KI = PbI 2 ↓ +2CH 3 COOK
2CH 3COO − + Pb 2+ + 2 K + + 2 I − ↔ PbI 2 ↓ +2 K + + 2CH 3COO −
Pb 2+ + 2 I − ↔ PbI 2 ↓
The precipitate dissolves in hot water, and the excess of potassium iodide with the formation of
a complex salt.
PbI 2 + 2 KI = K 2 [ PbI 4 ]
PbI 2 + 2 I − ↔ [ PbI 4 ]2−
2.2. Analysis of a sample, containing one of the anions with p-element.
СLASS 19.
TOPIC: Electrochemistry. Electrical processes in the body.
1. SEMINAR. (60 min.).
1.1. The subject of electrochemistry. Liquids and tissues of the body as
electrical conductors of the IId kind.
1.2. Electro-conductivity of ionic solutions. Electrical and molar conductivity
of ionic solutions. Limited molar conductivity. Anions and cations movement. The
Law of independent migration of ions. (Kol’raush’s Law).
1.3. Conductomety titration. Determination of ionization constant and degree
of dissociation of weak electrolytes by conductometry method. Application of
conductometry in medicine and biology.
1.4. The double electrical layer. Electrodes and redox potentials. A standard
electrode potential (standard reduction potential).
1.5. Electrochemical cells (concentration cell, bimetallic cell, isometallic
cell).
1.6. The relationship between the EMF (electromotive force) of a cell and
solution concentrations. The Nernst equation.
1.7. The maximum electrical work. The relationship between EMF and
Gibbs energy, EMF and equilibrium constants for redox reactions. Determining the
direction of a spontanetity from an electrode potentials.
1.8. Types of electrodes. Glass electrode (hydrogen electrode). Silver-silver
chloride, and calomel electrodes; ion - selective membrane electrodes.
1.9. Potentiometry. Determination of pH by using potentiometry. Potentiometry titration.
2. LABORATORY WORK. (45 min.).
2.1. Determination of pH value for biological liquids by potentiometry.
2.2. Potentiometric titration.
Literature.
1. Lectures.
2. Ebbing D.D. General Chemistry/ D.D.Ebbing , M.S.Wrighton;.- Boston.
Third
edition. Houghton Mifflin Company. – Р. 747-774.
3. Lister T. Chemistry for Advanced Level / T. Lister, J. Renshaw; - Third
edition.
Stanley Thornes (Publishers) Ltd. 2000. – Р. 354 – 362.
PRACTICAL PROBLEMS
1. Describe the difference between a voltaic cell and an electrolytic cell.
2. Give the definition cathode and anode. Describe the migration of cations and
anions in a voltaic cell.
3. How are standard electrode potentials defined?
4. Give the mathematical relationship between the members of each possible
pair of the three quantities: ∆G0, EMF and K.
5. Write the cell reaction for the voltaic cell
Zn(s)│Zn2+(aq)║Fe3+(aq),Fe2+(aq)│Pt
6. The EMF of a particular voltaic cell with the cell reaction
Hg22+(aq) + H2(g) ↔ 2Hg(l) + 2H+(aq)
is 0.650V. Calculate the maximum electrical work of this cell when 0.500 g H2
is
consumed.
7. Consider, the reaction
Zn2+(aq) + 2Fe2+ (aq) ↔ Zn(s) + 2Fe3+(aq)
Does the reaction go spontaneously in the direction indicated, under the
standard
conditions?
8. The standard EMF for the following voltaic cell is 1.10 V.
Zn(s)│Zn2+(aq)║Cu2+(aq)│Cu(s)
Calculatee the equibrium constant Kc for the reaction
Zn(s) + Cu2+(aq) ↔ Zn2+(aq) + Cu(s)
2. LABORATORY WORK.
2.1. Potentiometric determination of pH value for biological liquids.
Potentiometry is the measurement of the electrical potential difference
between
tween two electrodes in an electrochemical cell. A galvanic electrochemical
electrochem
cell (Figure 3) consists of two
electrodes (electron or metallic conductors)
connected by an electrolyte solution (ion conductor). An electrode,
ele
or half-cell,
consists of a single metallic conductor in contact with an electrolyte solution. The
ion conductors may be composed of one or more phases that are either in direct
contact with each ither or separated by membranes
me branes permeable only to specific
cations or anions.
ns. One electrolyte solution is the unknown, or test, solution, which
may be replaced with an appropriate reference
re erence solution for calibration purposes.
A salt solution or bridge may be interposed in the cell to reduce any liquid-liquid
liquid
junctionpotential present.
sent. By convention the left electrode (ML) is the reference
electrode; the right electrode (MR) is the indicator (measuring) electrode.
ele
Figure 3
Schematic diagram of an electro-chemical
electro
cell.
The liquid junction between S1 and S2 may be an open contact
between the two solutions, a porous membrane, or a fiber
junction. The two electrodes (ML and MR) are connected
externally via a high-impedance
impedance potentiometer. The
electromotive force
of the cell then is read directly. ML and MR, Two metallic
conductors; S1, S2, S3, and S4, variable number of ion-conductive phases; S1 may be a saturated ЛСд solution, S2, a
given test solution, S3, an ion-selective membrane (not necessarily a thin membrane), and S4, a given reference
solution.
The EMF is defined as the maximum difference in the potential between two
electrodes obtained when the cell current is zero. The cell potential is measured
with a potentiometer, of which the common pH meter is a special type. The directreading potentiometer is a voltmeter that measures the potential between two
electrodes; however, to obtain accurance potential measurements, the current must
not flow through the cell.
The potentiometric method of analysis is based on using the dependence of
EMF of an electrochemical cell on the ion activity (concentration), can be
determined. This dependence is expressed by the Nernst equation:
Ε = Ε0 −
Where
2 ⋅ 303RT
a Re d
log
nF
aOx
E is the electrode potential of the half-cell;
E0 is the standard electrode potential when a Re d aOx = 1 ;
n is the number of electrons involved in the reduction
reaction;
F is Faraday constant (= 96,487C .mol-1) C=Coulomb;
R is Gas constant (=8.314J . mol-1 ·K-1);
T is Absolute temperature (unit: K [kelvin]);
a Re d
a
aOx
is product of mass action for reduction reaction;
is activity.
Determination of pH is widely used in biological studies and also in clinical
and pharmacological practice for acid-basic characteristics of various biological
media and medical preparations. The range of pH of physiological liquids variants
is broad.
Potentiometric method of pH determination has a number of advantages in
comparison with a usual acidic-basic titration:
• it is more accurate;
• is able to determine the temperature of the solution;
• it gets the ability to determine the pH of colored and turbid solutions.
2.2 Potentiometric titration of an acid solution by a titrant of base.
The cell potential is sensitive to the concentrations of the reactants and
products involved in the cell reaction, measured potentials can be used to
determine the concentration of an ion. A pH meter is a familiar example of an
instrument that measures concentration from an observed potential. The pH meter
has three main components: a standard electrode of known potential, a special
glass electrode that changes potential depending on the concentration of H+ ion in
the solution into which it is dipped, and a potentiometer that measures the potential
between the two electrodes. The cell is
Pt│H2 (1 atm)│H+(test solution)║H+(1 M)│H2(1 atm)│Pt
The EMF of this cell equals the potential developed by the test solution half-cell,
whose half-reaction is
H2(1 atm) ↔2 H+ (test solution) +2 eThen, according to the Nernst equation for this electrode at 250C
Ε 0 cell = Ε 0 −
where
2,303RT
a Re d m +
log
a (H )
nF
aOx
+
a m ( H + ) is the activity of H3O ions
m is a stoichometric coefficient befor H+ in half-reaction
Supposing that a(Ox) = a(Red) the equation changes
Ε 0 cell = Ε 0 −
2,303RT
log( H + )
nF
Ε 0 cell = −0.0592 log[ H + ] = 0.0592 pH
Solving for pH, we get
pH =
E cell
0.0592
This equation shows the dependence of EMF on the pH of a solution.
The potentiometer reading is automatically converted electronically to a direct
reading of the pH of the solution being tested.
Methods of an experiment
Fill a buret with a 0.1N solution of NaOH. Put 20 mL of HCl solution into a
100-mL chemical glass. Dip two electrodes into the glass, and then add NaOH
solution by fractions constantly stirring the reactional mixture. Notice the number
of pH after the addition of each portion of NaOH solution. At the end point of the
reaction (the equivalence point) a dramatic change of pH is observed. Continue
the addition of NaOH solution by fracrions up to stop pH change.
Enter the obtained readings into the table, and draw a graph illustrating that
the pH is dependent of the volume of NaOH added solution.
The titration curve of a strong acid with a strong base is represented in the
figure. The equivalence point of the reaction lies in the middle of a dramatic
change of pH on a titration curve.
The concentration of acid solution is calculated according to the Equivalence
law (See Class 3.).
СLASS 20.
Topic: Organic Chemistry. The theory of the chemical structure of organic
compounds. Classification, nomenclature. Isomers. (135 min.)
1. SEMINAR. (50 min.)
1.1. The general principles of Butlerov's theory of the chemical structure of organic
compounds.
1.2. Structure of carbon atom. Orbital hybridization (sp3, sp2, sp).
σ-, π- bonds, bond lengths, bond strengths, and bond angles.
1.3. Organic compounds classification. Nomenclature (IUPAC). Functional
groups.
1.4. Isomerism. Straight-chain and branched-chain compounds, isomers of the
position of double and triple bonds, stereoisomers.
1.5. The ways of demolish covalent bonds. Radicals, electrophiles, nucleophiles.
1.6. Alkanes. Reactions of alkanes. Radical substitution of alkanes ( SR). Radical
stability.
2. LABWORK. (50 min.).
Literature
1. Lectures.
2. Fundamentals of bioorganic chemistry. S.E. Zurabyan. M.:GEOTAR-MED
2003. P. 19-26, 40-44, 94-97.
3. General Chemistry. D.D.Ebbing , M.S.Wrighton. Houghton Mifflin Company.
Boston. P. 971-976.
4. New understanding Chemistry for Advanced Level. T. Lister, J.Renshaw.
Stanley Thornes . Publishers Ltd. P. 247-259.
PRACTICAL PROBLEMS (30 min.)
1. Give the IUPAC name for each of the following compounds:
a)
CH3 CH2CH2
b)
CHCH2CH2 CH3
CH3
CH3-C-CH3
CH3CH2 CH2 CH2
CH2
CH3
c)
CH3 CHCHCH3
CH3
d) CH3 CHCHCH2 CH3
CH3
CH3
CH2
CH3
2. Write the structural formulas of alkanes given in the IUPAC names.
a) 2,5-dimethyl-3-ethylhexane;
b) 3-methyl-3-ethylpentane;
c) 2-metyl-4-isopropylheptane;
d) 2,2,4- trimethyloctane.
3. Write isomers and give the IUPAC names for hexane.
4. Show the initiation, propagation, and termination steps for the chlorination of: a)
propane, b) 2-methylbutane.
5. Arrange the given radicals in order of increasing their stability: methyl, ethyl,
sec.-butyl, isopropyl, tert.-butyl, isobutyl.
2. LABORATORY WORK (20 min.)
Saturated hydrocarbons. Alkanes.
2.1 Ethane obtaining. Relation of saturated hydrocarbons to bromine water and
potassium permanganate water solution.
Put dehydrated sodium propionate and sodium hydroxide to a dry test-tube in
ratio 1:2. Close with a gas-leader-tube. Heat the mixture on flame.
t
H3CCH2COONa
+ NaOH ----> H3C-CH3 + Na2CO3
Set fire to the opening of the gas-leader-tube. Write down your observations.
Then pass the forming gas through bromine water solution and potassium
permanganate water solution. What do you observe?
СLASS 21.
Topic: Mutual influence of atoms in the molecules of organic substances.
Alkenes, alkynes reactivity.
1. SEMINAR. (50 min.)
. 11. Inductive electron donation, inductive electron withdrawal (+I). Mesomeric
electron effect (M). π,π- and p,π-conjugation.
1.2. Electrophilic addition reactions (AE), Markovnikov's rule for unsaturated
hydrocarbons.
1.3. Alkynes. Chemical properties. CH - acidity.
1.4. Carbocation stability.
1.5. Radical addition reactions (AR).
1.6. Oxidation of unsaturated hydrocarbons.
1.7. Conjugated double bonds. The structure of butadiene-1,3. Electrophilic
addition reactions for conjugated dienes.
1.8. Polymerization of unsaturated hydrocarbons.
1.9. Terpenes (Limonene, Pinene, Camphor). The structure, properties.
β-Carotene.
2. LABWORK. (45 min.).
Literature
1. Lectures.
2. Fundamentals of bioorganic chemistry. S. E. Zurabyan. M.: GEOTAR -MED
2003. P. 40-44, 98-104.
3. General Chemistry. D.D.Ebbing , M.S.Wrighton. Houghton Mifflin Company.
Boston. P. 977-980.
4. New understanding Chemistry for Advanced Level. T.Lister , J. Renshaw.
Stanley Thornes . Publishers Ltd. P. 260-277.
PRACTICAL PROBLEMS (45 min.)
1. Give the IUPAC name for each of the following compounds:
a) CH3CH2CHCH
CHCH2CH2 CHCH3 ;
CH3
CH2CH3
C
C
CH ;
C
CH3
CH3
CH3
c)
b) CH2
;
d)
CH3CHC
CCH3 ;
C
CH3CH2
C2 H5
CH2 CH2CHCH3
e) HC
CCHCH2CHCH2CH2 .
CH3
C 2H 5
CH3
2. Give the structures and the IUPAC names for all alkenes with the molecular
formula C6H12.
3. Draw the structure for each of the following hydrocarbons:
a) 2,2,4-trimethylheptene-3;
b) 3-methyl-3-isopropylhexyne-1;
c) pentadiene-1,2;
d) 1,3,5-tribromo-2-pentene;
e) trans-3-methyl-2-heptene;
f) di-t-butylacetylene;
g) t-butyl-s-butylacetylene.
4. What is the major product in the reaction of 2-methyl-2-butene with each of the
following reagents?
a) HBr ;
d) Br2/CCl4 ;
b) H2/Pd ;
e) (O) KMnO4 (H2O) ;
c) H2 + trace HCl ;
f) Stronger oxidation.
5. Indicate, which member of each of the following pairs is more stable.
a)
+
CH3 CCH3
or
+
CH3 CHCH2CH3
or
+
CH3 CHCH2Cl
CH3
b)
+
CH3CHCH3
c)
.
CH3CH2CH2CHCH3
or
.
CH3CH2CH2 CH2CH2
d)
CH2
+
CHCH2
or
CH2
+
CHCH2CH2
6. Give the mechanism for the following reaction. (Show all arrows)
CH2CH2CH
CH2 + Br2
AlBr3
?
7. What will the major product of the reaction of 1 mol of 2-butyne and each of the
following reagents be?
a) HBr (1mol) ;
d) excess H2/Pt ;
b) HBr (2 mol) ;
e) H2O/ Hg+2 .
c) Br2 (2 mol) ;
2. LABORATORY WORK (20 min.)
Unsaturated hydrocarbons. Chemical properties.
2.1. Using the name of the given hydrocabon draw the structural formula, write all
isomers and give their IUPAC names.
2.1.1. Oxidation reaction.
Add 5 drops of 0.1N KMnO4 solution to 5 drops of alkene. Mix it. The
solution becomes uncoloured.
2.1.2. Addition reaction.
Add an equal volume of Br2(H2O) to 5 drops of alkene, mix it. The red-brown
colour of the Br2 is immediately disappeared.
2.2. Terpenes hydrocarbon properties.
2.2.1. α - Pinene oxidation.
Add 5 drops 0.1 N KMnO4 solution to 5 drops of pinene. Mix it. The solution
becomes uncoloured.
CH3
(O),KMnO4 (H2O)
HO
HO
CH3
Pinene
2.2.2 Limonene bromination. Add 1 ml of bromine water solution to 5 drops of
limonene, mix it. The red-brown colour of the Br2 is disappeared.
СLASS 22.
Topic: Cyclic conjugated systems. Aromaticity. A general mechanism
for electrophilic aromatic substitution reactions (SE).
1. SEMINAR. (70 min.)
1.1. Benzene and its homologues. Nomenclature of substituted benzenes.
Isomerism.
1.2. Criteria for aromaticity. Huckel's rule. Aromaticity, antiaromaticity.
1.3. Chemical properties of benzene and its homologues: electrophilic substitution
reactions ( halogenation, nitration, sulfonation, alkylation); oxidation.
1.4. The effect of substituents on orientation ( ortho-and para-directing substituent
and meta-directing substituent).
1.5. Representatives of aromatic substances: benzene, toluene, xylenes, styrene.
1.6.
Polycyclic
benzenoid
hydrocarbons
(naphthalene,
anthracene,
phenanthrene, pyrene). Their properties.
2. LABWORK. (25 min.)
Literature
1. Lectures.
2. Fundamentals of bioorganic chemistry. S. E. Zurabyan. M.: GEOTAR-MED.
2003. P. 104-111.
3. General Chemistry. D. D. Ebbing , M. S. Wrighton. Houghton Mifflin
Company. Boston. P. 981-986.
4. New understanding Chemistry for Advanced Level. T. Lister , J. Renshaw.
Stanley Thornes . Publishers Ltd. P. 496-516.
PRACTICAL PROBLEMS (20 min.)
1. Write structural formulas for
a) 1,1,1-triphenylethane; b) o-ethylmethylbenzene.
2. Write structural formulas and isomers for
a) 1,2,3,-trimethylbenzene; b) p-diethylbenzene.
3. Write isomers for a) C10H14; b) C8H10.
4. Benzene, toluene bromination. See the mechanism for electrophilic aromatic
substitution.
5. Toluene, o-xylene, m-xylene sulfonation. See the SE mechanism.
6. Nitration of benzoic acid (mechanism).
7. Alkylation of benzene. Write the scheme of the oxidation:
a) toluene; b) propylbenzene; c) p-methylisopropylbenzene.
8. Ortho-, para-directing substituents, and meta-directing substituents.
9. List the following compounds in order of decreasing reactivity toward
electrophilic substitution:
a) benzene, toluene, nitrobenzene, bromobenzene, phenol;
b) dichloromethylbenzene, difluoromethylbenzene, toluene, chloromethylbenzene.
10. Which of the following compounds are aromatic?
a)
b)
c)
d)
Problem.
Explaine why the halogens have the relative order of reactivity shown in
Appendix Table 3.
Solution.
Fluorine is the least deactivating of the halogen substituents, while iodine is
the most deactivating. Fluorine is the most electronegative, which means that it is
best at withdrawing electrons inductively. Fluorine is also best at donating
electrons by resonance its 2p orbital, compared to the 3p orbital of chlorine, the 4p
orbital of bromine, or the 5p orbital of iodine, can better overlap with the
2porbitals of carbon. So the fluorine substituent is best at withdrawing electrons
inductively and best at donating electrons by resonance. The fact that, of the
halogens, it is the weakest deactivator means that electron donation by resonance is
the most important factor in determining the relative order of reactivity of
halosubstituted benzenes.
2. LABORATORY WORK (25 min.)
Chemical properties of arenes.
2.1. Substitution reaction. Toluene nitration.
Add 10 drops of sulfuric acid to a dry test-tube with 5 drops of concentrated
nitric acid carefully. The mixture is cooled, than add 5 drops of toluene mixing
continuously. Obtaining nitro toluene (in sub layer) is characterized by specific
smell and colour.
2.2. Oxidation of benzene and substituted benzenes.
2.2. Oxidation of benzene and substituted benzenes.
5 drops of benzene, toluene, o-xylene, ethylbenzene, some crystals of
naphthalene put to each 5 test-tube correspondently. Then, add 5 drops of
0.1 N KMnO4 solution and 1 drop of 2 N H2SO4 solution to each test-tube. Mix
and put them to the water-bath. The violet colour of potassium permanganate is
disappeared except benzene. Write all equations.
СLASS 23.
Topic: General principles of the structure, isomerism and reaction ability of
hydrocarbons. Halogenated hydrocarbons.
1. SEMINAR (30 min.)
1. Pass an exam on the topic "General principles of the structure, isomerism and
the reactive ability of hydrocarbons". (80 min.)
2. LABWORK. (25 min.)
Literature
1. Lectures.
2. Fundamentals of bioorganic chemistry. S. E. Zurabyan. M.: GEOTAR -MED.
2003. P. 112-118.
3. General Chemistry. D. D. Ebbing , M. S. Wrighton. Houghton Mifflin
Company. Boston. P. 971-986.
4. New understanding Chemistry for Advanced Level. T.Lister , J.Renshaw.
Stanley Thornes . Publishers Ltd. P. 278-294.
PRACTICAL PROBLEMS
1. Classification of halide derivatives.
2. Structure, isomerism and nomenclature of halogenated hydrocarbons.
3. What product would be formed from the SN2 reaction of 2-bromobutane and
hydroxide ion?
4. Chemical properties of mono- , di- , polyhalide derivatives.
5. General ways of the synthesis of halogenated hydrocarbons.
6. Representatives of halogenated hydrocarbons: chloroatane, fluoroatane,
chloroform, tetrachloroethylene, chlorobenzene, tetrafluoroethylene.
2. LABORATORY WORK (25 min.)
Formation of chloroform. Test-reaction on chloroform.
Determination of chloroform purity.
2.1 Chloroform formation.
Put 0,3 g of chloral hydrate and 1,5 ml of distilled water in to a dry test-tube.
For obtaining solution add 1 ml of 10% NaOH solution and intensivly shake. After
finishing of the reaction the hard liquid with specific sweet smell is formed on the
bottom of the test-tube.
O
Cl3C
+
C
NaOH
HCCl3
+
HCOONa
H
2.2 Test-reaction on chloroform.
2.2.1 To a half part of the previously obtained chloroform emulsion add some
resorcinol crystals and heat on water-bath. The red colour of reaction solution is
appeared.
2.2.2 To the remaining part of chloroform emulsion add 6 drops of 0.1 N KMnO4
solution. The appearance of green colour of reaction solution is observed.
HCCl3 + 3 NaOH
t
HCOOH+ 2 KMnO4 + 2 NaOH
3 NaCl + HCOOH + H2O
CO2 + 2H2O + K2MnO4 + Na2MnO4
2.3 Determination of chloroform purity.
During a long time under the influence of light and oxygen air chloroform is
decomposed with the formation of toxic compounds (hydrochloride, chlorine and
phosgen).
O2,hν
HCCl3
O2,hν
HCl + COCl2 ;
COCl2
CO2 + Cl2
2.3.1 Test-reaction on hydrochloride.
Add 4 ml of distilled water to 2 ml of dirty chloroform mix it. Separate two
layers. Add AgNO3 solution to high water layer. The white precipitate is formed.
2.3.2 Test-reaction on chlorine.
To remaining liquid add 6 drops of potassium iodide water solution, shake
intensively. The iodine chloroform solution is coloured in pinc.
СLASS 24.
Topic: Alcohols, phenols, thiols, amines and their derivatives. Basic-acidic
properties of these compounds. Nucleophilic substitution reactions. (135 min.)
1. SEMINAR. (45 min.)
1.1 Alcohols, phenols, thiols, amines and ethers, nomenclature and isomers.
1.2 The structure of functional groups (-OH, -SH, -NH2).
1.3. Acidic-basic properties of alcohols, phenols, thiols, amines, ethers.
1.3.1 Brǿnsted acids and Brǿnsted bases. Basicity and acidity of organic
compounds.
1.3.2 The factors which determine the acidity of organic compounds:
a) electronegativity of the atom,
b) size of the atom,
c) electronic effects of the substituents.
1.3.3 C-H, S-H, N-H acidity.
1.3.4 Properties of mono- and polyatomic alcohols. The reaction with Cu(OH)2.
1.3.5 Acidic-basic properties of ethers and alkyl sulfides.
1.3.6 Properties of phenol in comparison with ones of alcohols.
1.4 Oxidation reactions.
1.4.1 Oxidation reactions of primary, secondary, tertiary alcohols.
1.4.1 Oxidation reactions of primary, secondary, tertiary alcohols.
1.4.2 Oxidation of phenols by strong and mild oxidizing agents.
1.5 Nucleophylic substitution reactions (SN1, SN2).
1.5.1 Action of HHal on primary and secondary alcohols (SN2).
1.5.2 Action of HHal on tertiary alcohols (SN1).
1.5.3 Dehydration of alcohols.
1.6 Glycerol nitration.
1.7 Phenol as aromatic substance.
1.8 Derivatives: glycoles, glycerol, ethylmercaptane, diethylsulfide, benzyl
alcohol, cresols, resorcinol, catechol, hydroquinone, α- and β- naphthols.
2. TEST. (30 min.)
3. LABWORK. (25 min.)
Literature
1. Lectures.
2. Fundamentals of bioorganic chemistry. S. E. Zurabyan. M.: GEOTAR – MED,
2003. P. 118-127.
3. General Chemistry. D.D.Ebbing , M.S.Wrighton. Houghton Mifflin Company.
Boston. P. 988-993.
4. New understanding Chemistry for Advanced Level. T. Lister , J. Renshaw.
Stanley Thornes . Publishers Ltd. P. 295-312; 517-529.
Problem 1.
List the following compounds in order of decreasing acidity.
CH 2 CH 2OH , CH3CH 2 OH , CH 3 CHOH , CH 2 CH 2OH
F
F
1
2
Cl
3
4
Solution. In each case an O-H bond must be broken. Number (2) is the
weakest acid because it does not have an electron-withdrawing halogen atom.
Number (3) is the strongest acid because the electron-withdrawing halogen
atom is closest to the bond that must be broken. The halogen is equally far
away from the O-H bond in number (1) and (4), but number (1) is a stronger acid
because F is more electronegative than Cl.
The order is 3 > 1 > 4 > 2.
PRACTICAL PROBLEMS
1. Write the structural formulas for:
a) 2-ethylbutan-1-ol ;
b) 2,4-dimethyl-3-ethylhexan-2-ol ;
c) 2-methylpropan-1-ol ;
d) prim.-butyl alcohol .
2. Write isomers for alcohol with common formula C5H12O.
3. List the following compounds in order of increasing acidity:
a) ethanol, ethylenglycol, propanol-2;
b) phenol, o-cresol ;
c) ethylmercaptane, ethanol ;
d) ethanol, phenol.
4. List the following compounds in order of increasing basicity: ethanol,
ethylamine, methylphenylamine.
5. List the following compounds in order of increasing acidity:
ethanol, ethane amine, phenol, p-cresol, p-flourophenol.
6. List the following compounds in order of increasing acidity: methanol, tertiary
butyl alcohol, primary propyl alcohol, secondary butyl alcohol.
7. Give the major product that is formed when each of the following alcohols is
heated in the presence of H2SO4:
a) 2,3-dimethylpentan-3-ol;
b) 2,2-dimethylhexan-3-ol.
8. Which is a stronger base, RO- or RS- ?
9. Arrange the following alkyl chlorides in order of decreasing reactivity in an SN1
reaction: isopropyl bromide, propyl bromide, t-butyl bromide, methyl bromide?
10. Arrange the following alkyl halides in order of decreasing reactivity in an SN1
reaction: 2-bromopentane, 2-chloropentane, 1-chloropentane, 3-bromo-3methylpentane.
11. Give the major product obtained from an E2 reaction of each of the following
alkyl halides with hydroxide ion.
F
a) CH3CHCH2CH3
b) CH3CHCHCH2 CH3
CH3
Cl
12. Give the products of the oxidation: a) glycerol, b) ethylene glycol.
13. Write the mechanism: a) 2-iodo-2-methylbutane with sodium hydroxide water
solution; b) phenol with nitric acid.
14. Give the product formed from the reaction of each of the following alcohols
with an acidic solution of sodium dichromate:
a) 3-pentanol ; b) 1-pentanol ; c) 2-methyl-2-pentanol ; d) 2,4-hexanediol.
3. LABORATORY WORK (25 min.)
Chemical properties of alcohols and phenols.
3.1 Interaction of alcohols with sodium.
Put 5 drops of ethanol and a little piece of sodium into a dry test-tube. After
finishing of the reaction add a drop of phenolphthalein. What do you see? Write
equations of corresponding reactions.
3.2 Glycerol reaction with copper dihydroxide.
3 drops of 5% solution of CuSO4 and 3 drops of 10% solution sodium
hydroxide are mixed. For obtaining precipitate add 5 drops of glycerol. Mix it.
What do you see?
Make the same reaction with ethanol. Explain the results.
3.3 Phenol acidity.
Add 10% solution of sodium hydroxide to 5 drops of phenol emulsion up to
the formation of transparent reaction mixture. For separating of phenol this
solution is neutralized by several drops of 10% solution of sulfuric acid.
3.4 Test reactions on phenols.
Into 3 test-tube pour out 5 drops of phenol, hydroquinone and resorcinol
solutions. Add 2-3 drops of FeCl3 solution to each of them. What do you see?
3.5 Phenol bromination.
Add drops by drops bromine water solution to 2 drops of 2% phenol solution
up to the formation of white precipitate of tribromophenol. Write the equation of
this reaction.
3.6 Oxidation of alcohols.
Put in 6 drops of propanol-1, propanol-2 and benzyl alcohol, into 3 test-tubes,
correspondently. Add on 3 drops of strong oxidizing agent – mixture of
potassium dichromate in concentrated sulfuric acid to each test-tube.
The obtaining reaction mixture is mixed and heated for 1-2 min on waterbath. Colour changes from orange to green.
3.7 Oxidation of phenols.
3.7.1 Oxidation by strong oxidizing agents.
Add hydroquinone carefully (without mixing) to 5-6 drops of potassium
dichromate in concentrated sulfuric acid. Than you'll see quinhydrone green
crystals [C6H4O2 . C6H4(OH)2].
3.7.2 Oxidation of mild oxidizing agent.
3 drops of 5% solution of CuSO4 and 3 drops of 10% solution sodium hydroxide
are mixed. For obtaining of Cu(OH)2 precipitate add 5-6 drops of hydroquinone
and heat the mixture on water-bath up to orange Cu2O precipitate appearance.
2Cu(OH) 2+ C6H4(OH) 2
O
+
O
2 CuOH
Cu2O
+
2 H 2O
H2O
Orange
СLASS 25.
Topic: Carbonyl compounds. Aldehydes and ketones.
1. SEMINAR. (40 min.)
1.1. The structure of the carbonyl group. Nomenclature and isomers of aldehydes
and ketones.
1.2. Chemical properties of aldehydes and ketones.
1.2.1. Relative reactivities of aldehydes and ketones.
1.2.2. Nucleophilic addition to aldehydes and ketones (AN):
a) reactions of carbonyl compounds with carbon and hydrogen nucleophiles:
addition of hydrogen cyanide, hydride ion (reduction of carbonyl compounds),
b) reactions of aldehydes and ketones with oxygen and nitrogen nucleophiles:
addition of water, alcohols (formation of hemiacetals and acetals), hydrolysis of
hemiacetals in acidic solutions; addition of primary amines and derivatives of
primary amines (formation of Schiff base-oxime, hydrazone, semicarbazone,
phenylhydrazone).
1.2.3. Oxidation of aldehydes and ketones (Tollens'test "silver mirror" reaction).
1.2.4. Cannizzaro's reaction.
1.2.5. Polymerization and polycondensation of aldehydes. Aldol condensation,
cyclotrimerization.
1.2.6. Halogenation of the α-carbon of aldehydes and ketones.
1.2.7. Analytical reactions:
a) aldehydes with fuchsinsulfurous acid,
b) ketones with sodium nitroprusside Na2[Fe(NO)(CN)5].
c) Triidomethane reaction for acetaldehyde and acetone.
2. TEST. (35 min.)
3. LABWORK. (25 min.)
Literature
1. Lectures.
2. Fundamentals of bioorganic chemistry. S. E. Zurabyan. M.: GEOTAR –MED,
2003. P.133-148.
3. General Chemistry. D. D. Ebbing , M. S. Wrighton. Houghton Mifflin
Company. Boston. P. 989-990.
4. New understanding Chemistry for Advanced Level. T. Lister , J. Renshaw.
Stanley Thornes . Publishers Ltd. P. 530-540.
PRACTICAL PROBLEMS
1. Give the structure for each of the following:
a) isobutyraldehyde ;
b) 4-hexenal ;
c) diisopentyl ketone ;
d) 3-methylcyclohexanone ;
e) 4-bromo-3-heptanone ;
f) meta-formylbenzaldehyde.
2. List the following compounds in order of decreasing reactivity:
acetaldehyde,
butyraldehyde,
acetone,
formaldehyde,
methylethylketone,
benzaldehyde, benzophenone.
1. Give two names for each of the following compounds.
O
a)
; b) CH3CH2CH2 CCH2 CH2CH 3 ;
CH3 CH2CHCH2C
CH3
H
O
O
c)
O
CH3 CHCH2 CCH2CH2CH 3 ; d)
C 6H5 CH2CH2CH2 C
;
H
CH3
O
O
e)
; f)
CH3CH2CHCH2 CH2 C
C2 H5
CH2
CHCCH2 CH2CH 2CH3 .
H
4. Oxidation of formaldehyde by mild oxidizing agent.
5. Cannizzaro reaction for formaldehyde and benzaldehyde.
6. Cyclotrimerisation of formaldehyde.
7. Write the equation of the obtaining of urotropine
8. Reaction of butyraldehyde and acetone with ethanol.
9. Reaction of propionaldehyde with hydrogen cyanide.
10.Reaction of acetaldehyde with hydroxylamine.
3. LABORATORY WORK (25 min.)
Chemical properties of aldehydes and ketones.
3.1 Action of the strong oxidizing agent.
Put 10 drops of 0.1 N water solution of KMnO4 and 5 drops of 10% H2SO4
solution to each of two test-tubes. Then add 10 drops of formaline to one test-tube
and 10 drops of acetone to the other. Both test-tubes are heated on water-bath.
Write the equations.
3.2 Action of the mild oxidizing agent - Cu(OH)2 .
10 drops of 10% NaOH solution and 5 drops of 5% CuSO4 solution are
mixed. For obtaining precipitate add 10 drops of formaline and reactional mixture
is heated on the water-bath.
Make the same reaction with acetone. Explain the results.
3.3 Fuchsinsulfurous acid is an indicator on aldehydes.
When 2-3 drops of colourless fuchsinsulfurous acid are added to formaline,
the red-violet colour appears.
3.4 Triiodomethane reaction for acetone.
10 drops of acetone and 5 drops of 2% NaOH solution are mixed. Iodine
solution is added to the mixture drops by drops. The yellow precipitate of iodoform
with characteristic sharp smell is formed.
I2 + NaOH
CH3 CCH3 + 3 HOI
O
I3CCCH3 + NaOH
NaI + HOI
I3CCCH 3 + 3 H2O
O
CHI3 + CH3COONa
O
3.5 Nitroprusside is an indicator on ketones.
When 5 drops of nitroprusside are added to the mixture of 5 drops of acetone
and 4 drops of 10% NaOH solution, the red colouring is appeared.
3.6. Addition of sodium hydrosulfite (NaHSO3)
The mixture of 10 drops of acetone and 5 drops of NaHSO3 solution is cooled,
mixing with glass stick constantly. The white solid precipitate is formed.
СLASS 26.
Topic: Carboxylic acids.
1. SEMINAR. (35 min,)
1.1 Classification of carboxylic acids. Nomenclature, isomerism.
1.2 The structure of carboxylic group.
1.3 Chemical properties of carboxylic acids.
1.3.1 Acidic properties. Salts of carboxylic acids.
1.3.2 General mechanism for nucleophilic acyl substitution reactions. Formation of
acyl halides, acid anhydrides, esters, and amides. Acid-catalyzed ester hydrolysis.
1.3.3 Halogenation of the α-carbon of carboxylic acids. Aromatic carboxylic acids
derivatives.
1.4 Representatives of carboxylic acids:
a) formic, acetic, propionic, butyric, valeric, caproic acids;
b) oxalic, malonic, succinic acids;
c) Acrylic acid and its derivatives, crotonic, fumaric, maleic acids;
d) palmitic, stearic, oleic, linoleic, linolenic acids;
e) benzoic, phtalic acids.
1.5 Oxidation of carboxylic acids.
2. TEST. (35 min.)
3. LABWORK. (25 min.)
Literature
1. Lectures.
2. Fundamentals of bioorganic chemistry. S.E.Zurabyan. M.: GEOTAR-MED.
2003. P. 149-160.
3. General Chemistry. D.D.Ebbing , M.S.Wrighton. Houghton Mifflin Company.
Boston. P. 991-994.
4. New understanding Chemistry for Advanced Level. T.Lister , J.Renshaw.
Stanley Thornes . Publishers Ltd. P. 553-574.
Problem 1.
Is the acid-catalyzed hydrolysis of acetyl amide a reversible or an irreversible
reaction? Explain the mechanism for acid-catalyzed hydrolysis of an acetyl amide.
Solution.
In the first step of the acid-catalyzed hydrolysis of an acetamide, the acid
protonates the carbonyl oxygen, increasing its susceptibility to nucleophilic attack.
Nucleophilic attack by water on the carbonyl carbon leads to tetrahedral
intermediate I. Reprotonation can occur either on oxygen to remform tetrahedral
intermediate I or on nitrogen to form tetrahedral intermediate II. Protonation on
nitrogen is favored because nitrogen is a stronger base than oxygen. Tetrahedral
intermediate II has two leaving groups, -OH and NH3. Because NH3 is weaker
base, it is eliminated, forming the carboxylic acid as the final product. (Because the
reaction is being carried out in an acidic solution, NH3 will be protonated after it is
eliminated from the tetrahedral intermediate.) .
+_
OH
_
_
CH3-C-NH2 + H 2O
_
_
OH
_
CH3-C-NH2
_
CH3-C-NH2
H+
_
_
_
_
O
-H+
_
OH
H
tetrahedral intermediate I
_
OH
_
_
H+
+ NH
CH3-C-OH
_
3
H+
_
_
OH
_
_
OH
_
CH3-C-NH2
CH3 -C-NH3
_
OH
_
_
OH
_
tetrahedral intermediate II
_
_
O
H+
_
CH3-C-OH
_
PRACTICAL PROBLEMS
1. Write isomers and give the IUPAC names for butyric and valeric acids.
2. Name the following compounds:
a) CH3CH2CH2CH2CCl ; b) CH3CH2COCH3 ; c) C6H4(COOH)2 .
O
O
3. List the following compounds in order of decreasing acidity: acetic acid, 3chloro-propanoic acid, propionic acid, 2-chloro-propanoic acid.
4. Give the mechanism for bromination of benzoic acid. List benzene and benzoic
acids in order of decreasing reactivity toward electrophilic substitution.
5. Give the mechanism for esterification of propanoic acid and ethanol.
6. Write the scheme for polymerization of acrylic acid, 2- methyl- propenoic acid.
7. List the following compounds in order of increasing acidity:
a) oxalic acid, formic acid;
b) 2-bromopropanoic acid, 3-bromopropanoic acid;
c) acetic acid, acrilyc acid.
8. Acrilyc acid as unsaturated one.
9. Acid-catalyzed ester hydrolysis of ethyl acetate.
10. Give the scheme of the succinimide formation .
11. Give the scheme of the acetamide formation from the acetyl chloride and
ammonia.
2. LABORATORY WORK (25 min.)
Chemical properties of saturated and unsaturated
mono- and dicarbonic acids.
2.1 Formation of formic acid.
The mixture of 5 drops of chloroform and 10 drops of 2% NaOH solution are
heated on water bath. For obtaining formic acid reactional solution is oxidized by 4
drops of acidic KMnO4 solution. The KMnO4 violet colour is disappeared.
2.2 Dissociation of carbonic acids.
Put 5 drops of acetic and oxalic acids to each of 2 test-tubes, respectively than
add 3 drops of indicator – methylorange. What do you see? Write the scheme of
dissociation of these acids.
2.3 Formation and hydrolysis of iron (III) acetate.
Add some crystals of sodium acetate, 5 drops of distilled water and 5 drops of
10% FeCl3 solution to the test-tube. The red-brown solution is obtained. Heat part
of this solution on water-bath. The formation of precipitate is observed. Write the
equations.
2.4 Formation of calcium oxalate.
Put 5 drops of 10% CaCl2 solution to 5 drops of water solution of oxalic acid.
The white crystals of calcium oxalate are formed.
2.5 Formation of potassium benzoate.
Put some crystals of benzoic acid to the test-tube with 10 drops of water than
to this mixture add 5 drops of 10% KOH solution. Crystals are dissolved. For
obtaining solution put several drops of HCl solution up to the formation of white
precipitate of benzoic acid.
2.6 Investigation of unsaturated carbonic acids properties.
2.6.1 Oxidation reaction.
Add 5 drops of 0.1N KMnO4 solution to 5 drops of oleic acid. Mix it. The
solution becomes uncoloured.
2.6.2 Addition reaction.
Add equal volume of Br2 (H2O) to 5 drops of oleic acid, mix it. The redbrown colour of the Br2 is immediately lost.
2.7 Formation of amyl acetate (demonstration).
The mixture consisted of 5 drops of amyl alcohol, 5 drops of concentrated
sulfuric acid and 5 drops of acetic acid is heated up to boiling point. Than after
obtaining ester add 2 ml of distilled water carefully. Two conjugate layers are
observed. Amyl acetate has the smell of pear essence.
Write the mechanism for esterification reaction.
СLASS 27.
Topic: Saponifiable lipids. (135 min.)
1. SEMINAR. (35 min.)
1.1 Classification of lipids.
1.2 Simple lipids: fats, oils, waxes. Structure and composition.
1.3 Saturated and unsaturated fatty acids. Configurational isomers of fatty acids.
1.4 Chemical properties of triacylglycerols:
a) hydrolysis (acidic, basic, enzymic). Soaps;
b) addition of hydrogen (hydrogenation), iodine, bromine;
c) oxidation ( rancidity and peroxide oxidation )of lipids.
1.5 Complex lipids: phosphatidic acid, phospholipids (cephalin, lecithin, phosphatidylserine), sphingolipids (sphingomyelin, cerebrosides). Composition and
structure.
2. TEST (35 min.)
3. LABWORK (25 min.)
Literature
1. Lectures.
2. Fundamentals of bioorganic chemistry. S.E.Zurabyan. M.: GEOTAR-MED” P.
238-246.
3. General Chemistry. D. D. Ebbing , M. S. Wrighton. Houghton Mifflin
Company. Boston. P. 994-996.
4. New understanding Chemistry for Advanced Level. T. Lister , J. Renshaw.
Stanley Thornes . Publishers Ltd. P. 553-574.
PRACTICAL PROBLEMS
1. Write the structural and configurational formulas of fatty acids: palmitic, stearic,
oleic, linoleic, linolenic acids.
2. The common formula of simple and mixed triacylglycerols.
3. Which of the substanse has a higher melting point; dipalmitooleoglycerol or
tripalmitoglycerol?
4. Write the hydrogenation reaction for oleo-palmito-lenoleoglycerol.
5. Prove the unsaturation of oleic acid.
6. Write the scheme of hydrolysis for dipalmito-stearo-glycerol, linoleo-oleopalmito-glycerol in acidic аnd basic solutions.
7. How many different triacylglycerols are there in which one of the fatty acid
components is lauric acid and two are myristic acid?
8. How many different triacylglycerols are there in which one of the fatty acid
components is lauric acid, one is mytistic acid, and one is palmitic acid?
9. Give the structural formula for cephalin, lecithin containig parts of stearic and
linolenic acids. What are the products of hydrolysis of these ones.
10. Give the structural formula for sphingosine.
11. What are the products of sphingomyelin hydrolysis?
12. Give the formula for glucocerebroside.
3. LABORATORY WORK (25 min.)
Chemical properties of saponifiable simple lipids.
3.1 Investigation of oils properties.
3.1.1 Oxidation reaction.
Add 5 drops of 0.1N KMnO4 solution to 5 drops of olive oil. After intensive
mix the solution becomes uncoloured.
3.1.2 Addition reaction.
Add equal volume of Br2(H2O) to 5 drops of olive oil. After intensive mixing
the red-brown colour of the Br2 is lost immediately.
3.2 Hydrolysis of fats and oils.
The mixture consisting of 10 drops of olive oil, 15 drops of alcohol solution
of potassium hydroxide is heated on water-bath during 25 min. Then the obtaining
products of hydrolysis are investigated.
3.2.1 Obtaining of glycerol.
3 drops of 5% CuSO4 solution and 3 drops of 10% solution of sodium
hydroxide are mixed. For obtaining precipitate add a half part of previously
obtained products of hydrolysis. Mix. What do you see? Write the equation of the
reaction.
3.2.2 Obtaining of fatty acids.
To the remaining part of hydrolyze oil add 5 drops of 10% water solution of
H2SO4. Oily layer of fatty acids is observed.
3.3 Determination of the iodine number of fats and oils.
0.1 g of a sample of oil is placed in beaker and is diluted by 10 ml of
chloroform. 15 ml of 0.1 N solution of iodine in alcohol is added and mixed for 3-4
min. Excess of iodine is titrated by the 0.1N solution of sodium thiosulfate in the
presence of starch. During the titration the indicator changes the color from dark
blue to colorless. Formula for calculating of iodine number:
K=
(A-B) · 0,0127 · 100
C
A - volume of 0.1 N solution of iodine (15 ml)
B - volume of 0.1 N solution of Na2S2O3 needed for titration of iodine excess
(ml)
C - a sample of oil or fat (g)
0.0127 - mass of iodine which is equivalence to 1 ml of 0.1 N iodine solution.
СLASS 28.
Topic: Heterofunctional compounds. Oxy-, phenolo-, ketoacids. (135 min.)
1. SEMINAR. (60 min.)
1.1 Classification of heterofunctional compounds.
1.2 Oxyacids. Structure, nomenclature, isomers. Stereogenic center. Enantiomers
and diastereomers of glycerolaldehyde, lactic acid, tartaric acid. Racemic mixture.
Separation of enantiomers.
1.3 α-, β-, γ- Oxyacids. Chemical properties. Dehydration reactions.
1.4 Some representatives: lactic, oxybutyric, malic, tartaric, citric, succinic acids.
1.5 Ketoacids. Structure, nomenclature, isomers. Keto-enol tautomerism.
1.6 Chemical properties of ketoacids, the most important acids: acetoacetic,
oxaloacetic, α-ketoglutaric acids. Ascorbic acid (vitamine C).
1.7 Phenoloacids. Structure, nomenclature, isomers. Chemical properties. Salicylic
acid, its derivatives: sodium salicylate, phenylsalicylate, acetylcalicylate (aspirine),
p-aminocalycylic acid.
2. LABWORK (35 min.)
Literature
1.
Lectures.
2.
Fundamentals of bioorganic chemistry. S.E.Zurabyan. M.: GEOTAR-
MED, 2003. P. 161-171, 61-72.
Problem 1.
Heterofunctional compounds in chemotherapy. Derivatives of salicylic acid.
Solution.
The salicylates were among the first to achieve recognition as analgesics.
Salicylates, in general, exert their antipyretic action in febrile patients by
increasing heat elimination of the body through the mobilization of water and
consequent dilution of the blood. This brings about perspiration, causing cutaneous
dilatation. This does not occur with normal temperatures. The antipyretic and
analgesic action are believed to occur in the hypothalamic area of the brain.
The derivatives of salicylic acid are of two types (I and II a, b).
O
O
R
C
O
O
C
OH
O
H
OH
C
O
O
R
R
C
O
I
II a
II b
Type I represents those that are formed by modifying the carboxyl group (e.g.,
salts, esters, or amides).
Type II represents those that are derived by substitution on the hydroxyl group of
salicylic acid.
The derivatives of salicylic acid were introduced in an attempt to prevent the
gastric symptoms and the undesirable taste inherent in the common salts of
salicylic acid. Hydrolysis of type I takes place to a greater extent in the intestine,
and most of the type II compounds are absorbed unchanged into blood-stream.
Compounds of type I. The alkyl and aryl esters of salicylic acid are used
externally, primarily as counterirritants, where most of them are well absorbed
through the skin. This type of compounds are of little value as an analgesic.
Sodium salicylate.
This salt is the one of choice for salicylate medication and usually is administered
with sodium bicarbonate to lessen gastric distress. The use of sodium bicarbonate
is ill-advised, because it decreases the plasma levels of salicylate and increases the
excretion of free salicylate in the urine.
Phenyl salicylate (Salol).
In salol, two toxic substances (phenol and salicylic acid) were combined into an
ester that, when taken internally, will slowly hydrolyzed in the intestine to give the
antiseptic action of its components.
Compound of type II b.
Aspirin (o-acetylsalicylic acid) was introduced into medicine by Dreser in 1899,
who named it aspirin by taking the a from acetyl and adding it to spirin, an old
name for salicylic or spiric acid, derived from its natural source of spirea plants.
Aspirin is used as antipyretic, analgesic, and antirheumatic. There is some
anesthetic action when applied locally, especially in powder form in tonsillitis
or pharyngitis and in ointment form for skin itching and certain skin diseases.
p-Aminosalicylic acid (PAS-acid). Once available as the sodium salt (Tubasal,
P.A.S. Sodium) for use as an analgesic, the free acid is bring offered as an
orphan drug in the treatment of ulcerative colitis.
Salsalate. Salicyilsalicylic acid. (Amigesic; Disalcid) is the ester formed between
two salicylic acid molecules to which it is hydrolyzed following absorption. It is
said to cause less gastric upset than aspirin because it is relatively insoluble in the
stomach and is not absorbed until it reaches the small intestine.
PRACTICAL PROBLEMS (30 min.)
1. Which of the following compounds has a stereogenic center?
CH3
a) CH3CH2CHCH3 ; b) CH3CH2CCH2CH 2CH3 ; c) CH3CH2CHCH2CH3 ;
Br
Cl
Br
d) CH3CH2CHCH3 ; e) CH3CH2OH ;
f) CH2 CH2CHCH3 .
CH3
NH2
2. Which of the compounds in Problem 1 can exist as a pair of enantiomers?
3. The following compound has one stereogenic center. Why does it have four
stereoisomers?
*
CH3CH2 CHCH2
CHCH3
Br
4. Draw all stereoisomers for each of the following compounds:
a) 1-bromo-2-methylbutane ;
b) 1-chloro-3-methylpentane ;
c) 2-methyl-1-propanol ;
d) 2-chloropentane ;
e) 3-chloropentane ;
f) 2,4-dichloropentane.
5. Write the isomers of hydroxybutyric acid.
6. What are enantiomers? Give the enantiomers of lactic acid. Determine the
configuration of these structures.
7. What are diastereomers? Write the Fischer projection of diastereomers of Dtartaric acid.
8. What is the racemic mixture? Is it optically active or not?
9. What compounds are called as optically active compounds or optical isomers?
10. What isomers are called as levorotatory and dextrorotatory?
11. Is the meso-tartaric acid optically active?
12. Give the scheme of dehydration of α-, β-, γ- oxyacids.
13. Write the scheme of destruction of α-oxyacids on heating them in the presence
of mineral acids.
14. What is tautomerism? Give the tautomeric forms of ethyl acetoacetate and
oxaloacetic acid.
15. Write the scheme of decarboxylation of acetoacetic acid and give the name of
the product.
16. Write the scheme of oxydation of lactic acid and β-oxybutyric acid, give the
names of the products.
2. LABORATORY WORK (35 min.)
Chemical properties of some heterofunctional compounds
2.1 Determination of lactic acid.
Add 3-4 drops of FeCl3 solution to 5 drops of phenol. The solution is colored
into violet. Then to the colored solution add 3-4 drops of lactic acid, the solution
changes its color into yellow - green, because the iron (III) lactate is formed. Write
the equations of the reactions.
2.2 Tartrates and hydrotartrates obtaining.
Add 2 drops of 2% KOH solution to 4-5 drops of tarteric acid. After rubbing
in the test-tube with a glass-stick the white crystals of potassium hydrotartrate
appear. To one half of precipitate to add KOH solution up to the whole dissolving.
Write the equations of the reactions.
2.3 The evidence of the presence of hydroxyl groups in tartaric acid and its salts.
3 drops of 5% CuSO4 solution and 3 drops of 10% solution sodium
hydroxyde are mixed. For obtaining precipitate add a part of previously obtained
potassium tartrate. Mix it.
What do you see? Write the equation of the reaction.
2.4 Test-reactions on ethyl acetoacetate.
Add 2 drops of FeCl3 solution to 5 drops of ethyl acetoacetate. The solution is
coloured into violet. This reaction is the test on enol form of ethyl acetoacetate.
Add 2 drops of sodium nitroprusside solution and 2 drops of NaOH solution
to 5 drops of ethyl acetoacetate. The solution is coloured into red. This is the testreaction on keto-form of ethyl acetoacetate.
2.5 The action of FeCl3 solution on phenol, salicylic acid, its derivatives: sodium
salicylate, phenylsalicylate, acetylsalicylate (aspirine).
Add 3-4 drops of FeCl3 solution to 5 drops of phenol. The solution is coloured into violet. Then repeat analogous experiments with salicylic acid,
phenylsalicylate and aspirin solutions. What do you see?
Add 1 ml of water to some aspirin crystals and heat for 10 min on waterbath. After hydrolysis and cooling add 2 drops of FeCl3. Write the equations
of the reactions.
СLASS 29.
Topic: Carbohydrates. Monosaccharides. (135 min.)
1. SEMINAR. (60 min.)
1.1 Classification of carbohydrates.
1.2 The D and L notation. Fischer projections. Epimers.
1.3 Cyclic structure of monosaccharides: hemiacetal formation. Haworth
projection. Anomers. Anomeric carbon.
1.4 Chemical properties of monosaccharides.
1.4.1 Redox reactions of monosaccharides.
1.4.2 Osazone formation.
1.4.3 Formation of esters and ethers.
1.4.4 Formation of glycosides.
1.4.5 Analitical reactions on aldopentoses and hexoses.
1.5 Some representatives: ribose, deoxyribose, xylose, glucose, mannose,
galactose, fructose. Amino sugars. Ascorbic acid (vitamin C).
2. LABWORK (35 min.)
Literature
1. Lectures.
2. Fundamentals of bioorganic chemistry. S.E.Zurabyan M.: GEOTAR-MED,
2003. P. 188 -199.
3. General Chemistry. D. D. Ebbing, M. S.Wrighton. Houghton Mifflin Company.
Boston. P. 1018-1019.
4. New understanding Chemistry for Advanced Level. T.Lister, J.Renshaw.Stanley
Thornes Publishers Ltd. P.540-544.
Problem 1.
Some derivatives of monosaccharides. Ascorbic acid.
Solution.
Ascorbic acid, vitamin C (L-ascorbic acid) serves as a reducing agent in several
important hydroxylation reactions in the body. L-Ascorbic acid (ascorbate) is the
enol form of 2-oxo-L-gulofuranolactone. Ascorbic acid can be synthesized by
nearly all living organisms, plants, and animals from D-glucose via the lactones of
D-glucoronic and L-gulonic acid; but primates,
guinea pigs, bats, and some other species lack L-gulonolactone oxidase, the
anzyme that catalyzes the formation of 2-keto-L-gulonolactone, which
spontaneously tautomerizes to L-ascorbic acid.
OH
OH
HO
O
O
- 2H
HO
O
O
+ 2H
O
HO
OH
Ascorbic acid
O
Dehydroascorbic acid
Because humans are one of the few animal species that cannot synthesize ascorbic
acid, the vitamin has to be available as a dietary component. The best dietary
sources of the vitamin are citrus fruits, berries, melons, tomatoes, green pipers, raw
cabbage, and leafy, green vegetables.
Ascorbic acid performs important metabolic functions, as evidenced by the
severe manifestations of its deficiency in humans. It has been demonstrated that
this vitamin is involved in metabolic hydroxylations in numerous important
metabolic processes ( e.g., the synthesis of steroids and of neurotransmitters and in
collagen and drug metabolism). It is well known that ascorbic acid is an effective
reducing agent and antioxidant.
PRACTICAL PROBLEMS (30 min.)
1. Draw Fischer projections for L-glucose and L-fructose.
2. How many stereoisomers are possible for:
a) a ketoheptose; b) an aldoheptose; c) a ketotriose?
3. Are D-ribose and L-ribose a pair of enantiomers or a pair of diastereomers?
4. Are D-glucose and D-mannose a pair of enantiomers or a pair of diastereomers?
5. Classify the following monosaccharides.
HOH2C
H
O
C
O
C
H
O
C
HO
C
H
HO
C
H
H
C
OH
H
C
OH
HO
C
H
H
C
OH
H
C
OH
H
C
OH
H
C
OH
H
C
OH
H
C
OH
CH2OH
D-ribose
CH2OH
CH2OH
D-sedoheptulose
6. What sugars are the C-3 epimers for: a) D-glucose , b) D-fructose ?
7. Give the IUPAC name for - a) D-glucose , b) D-fructose.
D-mannose
8. What are the products of the reduction of D-fructose?
9. What other monosaccharide is redused only to the same alditol obtaned from the
reduction of D-galactose?
10. Show the mechanism for the base-catalyzed conversation of D-fructose into Dglucose and D-mannose.
11. Give the products and the amount of each product obtained by the oxidation of
1 mol of each of the following monosaccharides with the excess of periodic acid
(HIO3). a) D-ribose, b) L-sorbose.
12. Give the scheme of oxidation of D-glucose and D-galactose by the action of
mild (Br2) and strong (HNO3) oxidizing agents.
13. Give the structure of the cyclic hemiacetal formed by each of the following:
a) 4-hydroxybutanal;
b) 5-hydroxypentanal;
c) 4-hydroxypentanal;
d) 4-hydroxyheptanal.
14. Draw the following sugars using Haworth projections.
a) β-D-galactopyranose;
b) α-D-glucofuranose;
c) α-D-fructofuranose;
d) β-D-ribofuranose.
15. What OH groups are in axial position ina) β-D-galactopyranose;
b) β-D-fructofuranose;
c) α-D-ribofuranose?
16. Write the structures: a) penta-O-acetyl-β-D-glucopyranose;
b) β-D-glucopyranose pentamethyl ether; c) ethyl β-D-glucopyranoside;
d) ethyl α-D-fructofuranoside.
17. Give the examples of reducing and nonreducing sugars.
2. LABORATORY WORK (35 min,)
Chemical properties of monosaccharides.
2.1 The evidence of the presence of hydroxylic groups in monosaccharides.
3 drops of 5% CuSO4 solution and 3 drops of 10% solution sodium hydroxide
are mixed. For obtaining precipitate add 6 drops of glucose solution. Mix it. What
do you see? Write the equation of the reaction.
2.2 Properties of reducing sugars ( oxidation by the mild oxidizing agents)
2.2.1 Fehling's reaction.
10 drops of 10% NaOH solution and 5 drops of 5% CuSO4 solution are mixed.
For obtaining precipitate add 10 drops of glucose solution and reaction mixture
heat on the water-bath up to the formation of the red coloured Cu2O precipitate.
2.3 Test-reaction on hexoses (on fructose).
Add 1 ml of fructose solution to 1 ml of Selivanov's reagent (0.5% resorcinol
solution in 20% HCl solution) and heat the mixture on water-bath up to red-wine
colouring appearance.
2.4 Test-reaction on pentoses (on ribose).
6 drops of acidic ribose solution and 4 drops of aniline are mixed. The
brightly pink colour is appeared.
CH2OH
HO
H
C
O
C
H
C
OH
resorcinol
HCl, toC
- 3H2O
H
CH2
HO
H
C
O
C
O
OH
CH2OH
D-fructose
5-Hydroximethylfurfurol
The product
of condansation reaction
H
O
C
H
C
OH
HCl, toC
H
C
OH
- 3H2O
H
C
H
O
aniline
C
O
OH
The product
of condansation reaction
CH2OH
D - ribose
Furfurol
СLASS 30.
Topic: Carbohydrates. Di-and polysaccharides. (135 min.)
1. SEMINAR. (30 min,)
1.1 Classification of disaccharides. Reducing (maltose, cellobiose, lactose) and
nonreducing (sucrose) sugars.
1.2 α-1,4 and β-1,4 - Glycosidic linkage.
1.3 Chemical properties of reducing disaccharides.
1.4 Chemical properties of sucrose. Hydrolysis of sucrose. Inversion of sugar cane.
1.5 Polysacharides. Classification. Starch, glycogen, cellulose.
2. LABWORK (35 min.)
3. TEST (45 min.)
Literature
1. Lectures.
2. Fundamentals of bioorganic chemistry. S.E.Zurabyan. M.: GEOTAR- MED,
2003. P. 199 – 210.
3. General Chemistry. D. D. Ebbing, M. S. Wrighton. Houghton Mifflin Company
Boston. P. 1020-1022.
4. New understanding Chemistry for Advanced Level. T. Lister, J. Renshaw.
Stanley Thornes Publishers Ltd. P. 545.
PRACTICAL PROBLEMS (30 min.)
1. Draw the following sugars using Haworth projections: maltose, lactose,
cellobiose, sucrose.
2. What is the main structural difference betweena) amylose and cellulose;
b) amylose and amylopectin;
c) amylopectine and glycogen;
d) cellulose and chitin?
3. Pectin is a polysaccharide obtained from fruits that is used as a jelling agent in
making jams and jellies. It can be synthesized by reacting amylose with nitric acid.
Draw a short segment of pectin.
4. O- and N-glycosides.
5. Tautomerism reducing and nonreducing disaccharides.
6. What disaccharides is the mutarotation characterized for?
7. Write the equation of hydrolysis of sucrose.
8. A disaccharide forms a silver mirror with Tollen’s reagent and reacts with a βglycosidase. When the disaccharide is treated with the excess of methyl iodide in
the presence of Ag2O and then is hydrolyzed with water under acidic conditions,
two products are formed: 2,3,4-tri-O-methylmannose and 2,3,4,6-tetra-Omethylgalactose. Draw the structure of the disaccharide.
9. Hyaluronic acid is a component of connective tissue and is the fluid that
lubricates the joints. It is an alternating polymer of N-acetyl-D-glucosamine and
D-glucuronic acid joined by β-1,3'-glycosidic linkages. Draw a short segment of
hyaluronic acid.
2. LABORATORY WORK (35 min.)
Chemical properties of di-and polysaccharides.
2.1 The evidence of polyatomicity of disaccharides.
In each from 3 test-tubes, consisting of Cu(OH)2 precipitate, add 1 ml of 2%
solutions of maltose, lactose and sucrose correspondently. Mix it. Explain the
reason of dissolving of copper dioxide. Write the equations.
2.2 Reducing ability of the disaccharides.
In each from 3 test-tubes, consisting of Cu(OH)2 precipitate, add 1 ml of 2%
solutions of maltose, lactose and sucrose correspondently. Mix it and put them to
the boiling water-bath for 5 min. What do you observe? Explain these phenomena.
Write all equations.
2.3 Hydrolysis of sucrose.
Add 2-3 drops of 10% H2SO4 solution to 1 ml of sucrose solution. Mix it
and put this mixture to heat on water-bath for 10 min. Hydrolysis products are
separated into two test-tubes. Contents of the first test-tube add to the Cu(OH)2
precipitate and prove the presence of the glucose.
Contents of the second test-tube add to Selivanov's reagent proving the presence of
fructose.
2.4 Hydrolysis of starch.
Add 2-3 drops of 10% H2SO4 solution to 1 ml of starch solution. Mix it and
put this mixture to heat on water-bath for 20 min. Hydrolysis products are
separated into two test-tubes. Contents of the first test-tube is neutralized by 8
drops of 10% NaOH solution, than add 2 drops of 2% CuSO4 solution and heat it
again. What do you observe?
Repeat this reaction with unhydrolyzed starch and write all equations.
2.5 Test-reactions on starch, glycogen, dextrin (with iodine).
Into each of 3 test-tubes, consisting of 1% solutions of starch, dextrin,
glycogen add 2 drops of iodine solution correspondently. You observe the
changing of the color of these solutions. When iodine solution is added to the
solution of starch, a starch-iodine complex is formed, which has a blue colour.
2.6 Reducing ability of the polysaccharides: starch and dextrin.
Into each of 3 test-tubes, consisting of Cu(OH)2 precipitate, add 1 ml of 2%
solutions of starch and dextrin correspondently. Mix it and put them to the boiling
water-bath for 10 min. What do you observe? Explain these phenomena.
СLASS 31.
Topic: Nitrogenated organic compounds. (135 min.)
1. SEMINAR. (60 min.)
1.1 Amines. Classification, nomenclature, isomerism.
1.1.1 Acidic-basic properties of aliphatic and aromatic amines.
1.1.2 Chemical properties of primary, secondary and tertiary amines (action of
nitrous acid).
1.1.3 Alkylation and acylation of amines.
1.1.4 Aniline and its derivatives: sulfanilic acid sulfanilamide, p-aminobenzoic
acid, p-acetamidobenzenesulfonamide.
1.2 Amino alchogols: colamine, choline, acetylcholine. Amino phenoles:
norepinephrine (noradrenaline), epinephrine (adrenaline).
1.3 Amides. Acid-catalyzed amides hydrolysis. Urea. Synthesis and properties.
1.4 Amino acids.
1.4.1 Structure, classification, nomenclature, isomerism.
1.4.2 Acid-base properties of amino acids. Types of salts. A zwitterion. Isoelectric
point (pI) of amino acids.
1.4.3 α-, β-, γ- Amino acids. Dehydration reactions.
1.4.4 Deamination, decarboxylation, transamination reactions for amino acids in
vivo.
2. LABWORK (35 min.)
Literature
1. Lectures.
2. Fundamentals of bioorganic chemistry. S.E.Zurabyan. M.: GEOTAR-MED,
2003. P. 128-132; 211-218;
3. General Chemistry. D. D. Ebbing,
M. S. Wrighton. Houghton Mifflin
Company. Boston. P. 1011-1013.
4. New understanding Chemistry for Advanced Level. T. Lister, J. Renshaw.
Stanley Thornes Publishers Ltd. P. 576-590; 600.
Problem 1. The scheme of catecholamine biosynthesis.
Solution.
Catecholamines are group of similar compounds that consist of monoamines
attached to a benzene ring bearing two hydroxyl groups (catechol). The naturally
occurring catecholamines are epinephrine, norepinephrine and dopamine.
Epinephrine is produced primarily by the adrenal medulla, whereas norepinephrine
is a neurotransmitter produced in the central nervous system.
Dopamine is present in highest concentration in localized regions of the brain and
also functions in the peripheral organs.
Both dopamine and norepinephrine are primary amines, whereas epinephrine is
secondary amine.
L-Tyrosine is the precursor of the catecholamines. The first step in
biosynthesis of catecholamines is hydroxylation of tyrosine by tyrosine
hydroxylase. The product, dihydroxyphenylalanine (DOPA), is converted in a
variety of tissues to dopamine through the action of aromatic L-amino acid
decarboxylase. Than dopamine through the action of dopamine-β –hydroxylase,
an anzyme responsible for final conversion of dopamine to norepinephrine.
Another
anzyme,
phenylathanolamine-N-methyl
transferase
converts
norepinephrine into epinephrine.
HO
NH2
NH2
tyrosine hydroxylase
COOH
COOH
HO
HO
Tyrosine
HO
HO
NH2
HO
NH2
dopamine-hydroxylase
HO
HO
Norepinephrine
Dopamine
PNMT
OH
HO
NH
CH3
HO
Epinephrine
PRACTICAL PROBLEMS (30 min.)
1. Give the structure for each of the following compounds:
a) 2 -methyl-N-propyl-1-propanamine;
b) N-ethylethanamine;
c) 3-ethyl-1-(methylamino)pentane;
d) 4-amino-2-methylhexane;
e) 5-methyl-1-hexanamine;
f) methyldipropylamine;
g) N,N-dimethyl-3-pentanamine;
h) cyclohexylethylmethylamine.
2. Give two IUPAC names and the common name (for those that have common
names) for each of the following compounds. Indicate whether the names are
primary, secondary, or tertiary.
a)
CH3CHCH2CH2CH2CH2CH2NH2
b)
(CH3CH2)2NCH3
CH3
CH3
c)
CH3CH2CH2NHCH2CH2CHCH3
CH3
d)
H 3C
NH2
3. Write isomers for amines with common formula C4H11N.
4. List the following compounds in order of increasing basicity: C2H5NH2,
(C2H5)2NH, (C2H5)3N, C6H5NH2.
5. Explain why propylamine is classified as a Brǿnsted-Lowry base.
6. Explain why propylamine is stronger base than ammonia.
7. Name the organic products of the following reactions:
a) ethylmethylamine with dilute acid;
b) propylamine plus ethanol chloride;
c) methylamine plus bromoethane.
8. Write the equations proving basic properties of diethylamine, colamine, urea,
pentandiamine-1, 5.
9. Reactions of primary, secondary, tertiary amines with nitrous acid.
10. What is the systematic name of valine?
11. The amino acid isoleucine has two chiral carbons. What are they?
12. Write all isomers for valine and give their names.
13. Write the equations proving the amphoteric properties of lysine, aspartic acid.
14. Write the equations of the formation of all types of salts for alanine.
15. Give the scheme of dehydration of a) glycine;
b) γ- amino butyric acid; c) β-amino isobutyric acid.
16. Give the schemes of decarboxylation of histidine, aspartic acid, glycine, lysine.
17. Give the schemes of deamination of valine, aspartic acid.
18. Give the scheme of transamination between pyruvic acid and glutamic acid.
19. What is the isoelectric point?
20. Draw the form in which glutamate exists at- a) pH = 0, b) pH = 3,
c) pH = 6, d) pH = 11.
21. Draw the form in which the following amino acids predominantely exist at
physiological pH (7.2).
a) aspartate;
b) histidine;
c) glutamine;
d) lysine;
e) arginine;
f) tyrosine.
2. LABORATORY WORK (35 min.)
Chemical properties of amines, amides and amino acids.
2.1 Comparison of acidic-basic properties of aliphatic and aromatic amines.
2.1.1 To 10 drops of water solution of ethylamine add 1 drop of indicator phenolphthalein. What do you see? Write the equation.
2.1.2 To 10 drops of aniline water solution
add 1 drop of indicator-
phenolphthalein. What do you see? Write the equation.
2.1.3 To 10 drops of ethylamine water solution add 5 drops of 10% FeCl3 solution.
The red-brown precipitate is observed. Write the equation.
2.1.4 To 10 drops of aniline emulsion add 5 drops of 10% H2SO4 solution. The
white precipitate is formed.
2. 2 Aniline bromination.
To 15 drops of bromine water solution add drops by drops aniline solution to
the formation of tribromoaniline white precipitate. Write the equation of this
reaction.
2.3 Interaction of amino acids with formaldehyde.
To 10 drops of 1% glycine water solution add 1 drop of methyl-red. You
observe the changing of indicator color up to yellow (neutral environment). To
forming mixture to add equal volume of neutral formaline. You observe the
changing of indicator’s color up to red ( acidic environment).
2.4 Reaction of amino acids with copper oxide (II).
To some amount of copper oxide (II) powder add 10 drops of 0.2 N glycine
solution and heat the reaction mixture on flame. After settling of CuO black
powder during 3 min the formation of dark-blue solution of copper complex salt is
observed.
2.5 Urea properties.
2.5.1 Urea nitrate formation.
To 10 drops of urea saturated water solution add 2-3 drops of concentrated
nitric acid. The white crystals formation is observed.
2.5.2. Biuret formation.
Into dry test-tube put 0.2 g of urea and heat it on flame up to melting. You
observe the urea decomposition with the formation of ammonia and biuret. To
obtaining melt add 15 drops of distilled water, 6 drops of 10% NaOH solution
and 2-3 drops of 5% CuSO4 solution. The formation of red-violet copper
complex salt solution is observed.
O
O
+
C
H2N
NH2
O
NH3
C
H2N
NH2
+
O
C
H2N
C
N
NH2
H
Biuret
H2N
O
H2
N
N
Cu
O
N O
H2
O
N
NH2
Copper complex salt
2.5.3 Urea-formaldehyde plastic.
Into a dry test-tube put 1 g of urea than add 10 drops of 40% formaldehyde and
heat this mixture for 5 min on water-bath. After cooling the formation of
white solid product of polycondensation reaction is observed.
СLASS 32.
Topic: Peptides and Proteins. (135 min.)
1. SEMINAR. (20 min.)
1.1 Peptides. Formation of di-, tri- and polypeptides.
1.1.1 Structure of peptide bonds and disulfide bonds.
1.1.2 Some interesting peptides: enkephalins - leucine enkephalin, methionin
enkephalin; peptide hormones-bradykinin, vasopressin, oxytocin; antibiotic gramicidin S; glutathione.
1.2 Protein structure. Determination of the primary structure of protein.
1.2.1 Secondary, tertiary and quaternary structure of proteines.
1.2.3 Protein denaturation.
2. LABWORK (20 min.)
3. TEST (45 min.)
Literature
1. Lectures.
2. Fundamentals of bioorganic chemistry. S.E.Zurabyan. M.: GEOTAR-MED,
2003. P. 218-224.
3. General Chemistry. D. D. Ebbing, M. S. Wrighton. Houghton Mifflin Company.
Boston. P. 1010-1016.
4. New understanding Chemistry for Advanced Level. T. Lister, J. Renshaw.
Stanley Thornes Publishers Ltd. P. 603-608.
PRACTICAL PROBLEMS (20 min.)
1. Draw a peptide bond in a cis-conformation. The structure of peptide bond.
2. Write formulas of tripeptides forming by lysine, alanine serine.
3. Glutathione is a tripeptide of glutamate, cysteine, and glycine. Write the
structural formula of this one.
4. Write the structural formula of enkephalins: leucine enkephalin (Tyr-Gly-GlyPhe-Leu) and methionin enkephalin (Tyr-Gly-Gly-Phe-Met).
5. Write formulas of tripeptides: Val-Phe-Leu, Lys-Gly-Ile, Asp-Ser-Ala. In what
region of pH values the isoelectric point of these ones takes place?
6. Draw the hydrogen bonding in an α-helix, in a parallel β-pleated sheet and
antiparallel β-pleated sheet.
2. LABORATORY WORK (20 min.)
Test-reactions on proteins
2.1 Biuret test on peptide bond.
To 1 ml of agg’s protein solution add 1 ml of 10% sodium hydroxide solution
and 2 drops of CuSO4 solution. The red-violet colour is appeared.
2.2 Action on protein by phenol and formaline.
To 5 drops of agg’s protein add 3 drops of 3% phenol solution. Destroying of
protein (formation of solid) is observed. Make the same reaction with
formaline.
2.3 Protein xanthanation.
To 10 drops of agg protein solution add 2 drops of concentrated nitric acid.
Heat reactional mixture on flame shaking up constantly. The solution and
precipitate are colored in yellow.After cooling add drops by drops 10% NaOH
solution up to appearance of orange colour.
2.4 Reaction on sulfur consisting α-amino acids.
To 10 drops of agg’s protein solution add double volume of 10% NaOH
solution. Shake up the reactional mixture and heat it on flame for 1-2 min up to
the boiling. To obtained solution add 5 drops of 10% (CH3COO)2Pb solution
and boil it on the flame. The dark-brown colour is appeared.
СLASS 33.
Topic: Biologically important heterocycles with one heteroatom. (135 min.)
1. SEMINAR. (45 min.)
1.1 Heterocycles. Classification.
1.2 Five- and six - membered ring heterocycles (pyrrole, furan, thiophene,
pyridine, quinoline, isoquinoline and their derivatives ).
1.2.1 Aromaticity of heterocycles.
1.2.2 Electronic structures of nitrogen in pyrrole and pyridine.
1.3 Chemical properties of five-and six-membered heterocycles.
1.3.1 Acid-base properties of heterocycles.
1.3.2 Electrophilic substitution reactions.
1.3.3 Hydrogenation reactions for heterocycles.
1.3.4 Pyridine derivatives: nicotinic acid (niacin), nicotinamide (niacinamide);
NAD+ and NADP+.
1.3.5 Some representatives of heterocyclic compounds: indole, benzofuran,
benzothiophene, tryptophan, proline, porphyrin ring system, heme.
2. LABWORK (35 min.)
Literature
1. Lectures.
2. Fundamentals bioorganic chemistry. S.E.Zurabyan. M.: GEOTAR-MED,
2003. P. 172-180.
Problem 1. Tryptophan, serotonin and 5-hydroxyindoleacetic acid.
Tryptophan – essential amino acid; metabolites found in carcinoid disease;
indole ring system, precursor of serotonin and melatonin.
Serotonin is a physiologically important amine and functions as a signal
transmitter in the CNS and peripheral organs, such as the gastrointestinal tract. It
functions also as a powerful smooth-muscle stimulant and vasoconstrictor. It is
transported in the blood by platelets. Serotonin (5-hydroxytryptamine [5-HT]) is
found in various animals, from coelenterates to vertebrates, in bacteria, and many
plants. In humans, it is synthesized in the intestinal chromaffin cells in central or
peripheral neurons and is found in high concentrations in many body tissues,
including the intestinal mucosa, pineal gland, and CNS. The formation and
breakdown of serotonin is depicted on following scheme. As indicated, tryptophan
first is hydroxylated to form 5-hydroxytryptophan (5-HTP). Approximately 1% to
3% of dietary tryptophan normally is metabolized by this pathway. The 5-HTP
then is decarboxylated to 5-HT. The enzymatic decarboxylation is very active in
carcinoid tumors. Pharmacologically, 5-HT is the most active indole amine;
however, its biological activity apparently is lost when it is bound to tissues or
platelets. It rapidly undergoes oxidative deamination in tumor or in the blood after
release frome a tumor.
The oxidative deamination of serotonin by the enzyme MAO (monoamine
oxidase) leads to the formation of 5-hydroxyindoleacetic acid
(5-HIAA), which is quantitatively the most significant metabolite of the 5hydroxyindole pathway.
HO
COOH
tryptophan hydroxylase
COOH
NH2
N
H
NH2
N
H
5-Hydroxytryptophan (5-HTP)
Tryptophan
aromatic
aminoacid
HO
decarboxylase
HO
H
COOH monoamine oxidase
NH2
N
H
N
H
5-Hydroxyindoleacetic acid
(5-HIAA)
Serotonin
5-Hydroxytryptamine; 5-HT
PRACTICAL PROBLEMS (15 min.)
1. Synthesis of 5-membered heterocycles on Yuriev's reaction.
2. Write the schemes of bromination of pyrrol, thiophen, furan. Explain the
mechanism SE.
3. What type of hybridization of nitrogen atom is in pyrrole?
4. Write the equation proving the ampholitic character of pyrrole and indole; basic
properties of pyrrolidine and pyrroline, pyridine and quinoline.
5. Write the sheme of nitration and sulfonation reactions for pyridine. Show the
mechanism of these ones.
6. Compare the activity of α-aminopyridine and pyridine for SE reactions. Give the
mechanism of these reactions.
7. Write the shemes of obtaning: a) nicotinic acid from the β-picoline,
b) N,N- diethylnicotinamide from pyridine.
8. What do pyrrol's and pyridin's nitrogen atoms mean?
9. Write the schemes of tryptofane transformations in body.
10. Give the formulas of tripeptides: a) Gys-Pro-Try; b) HO-Pro-Ala-Gys;
c) Lys-Try-Pro. In what region of pH is the isoelectric point situated in?
2. LABORATORY WORK (35 min.)
Chemical properties of heterocycles with one heteroatom.
2.1 Reactions of furacillin and furadonin with sodium hydroxide.
To 5 drops of NaOH solution add 2 drops of 5% furacillin.The orange-red
colour is appeared. Make the same reaction with furadonin. The red-brown colour
is appeared.
O
H2N
O 2N
O
N
CH
C
NH
O
O 2N
Furacillin
O
N
CH
N
NH
CH2
Furadonin
O
2. 2 Pyridine properties.
2. 2.1 Basic properties (interaction with water).
To water pyridine solution add indicator - phenolphtaleine. What do you
observe? Write the equation.
2. 2. 2 Reaction with iron (III) chloride.
To 2-3 drops of pyridine water solution add 2-3 drops of 10% FeCl3 solution.
The red-brown precipitate, Fe(OH)3, is formed. Write the equation.
2. 2. 3 Oxidation reaction.
To 3-4 drops of pyridine solution add 3-4 drops of 0.1N KMnO4 solution and 12 drops of 10%H2SO4 solution. Mix it. The violet colour of potassium
permanganate is not changed.
2. 3 N,N-diethyl nicotinamide (coramine) properties.
2.3.1 Reaction with copper sulphate and ammonia rhodanide.
To 1 ml of coramine solution add 10 ml of distilled water and 1ml of 5% CuSO4
solution. The dark blue color is appeared. To the obtained solution add 1 ml of 5%
NH4SCN solution. The formation of light green precipitate is observed.
2.3.2 Reaction with sodium hydroxide
0.5 ml of 1% coramine solution and 5ml of 10% NaOH solution are mixed and
is heated on flame up to boiling point. Forming vapor pass through the lacmus
paper moisten with distilled water. Red colour of lacmus paper is changed into
blue.
O
O
C 2H 5
C
N
+ NaOH,toC
C
ONa
+
(C2H5)2NH
C 2H 5
N
N
Coramine
(C2H5)2NH + H2O
[(C2H5)2NH]+OH-
N,N-Diethylnicotinamide (Coramine) appears to act by facilitating excitatory
processes rather then by depressing inhibitory ones. The overall effect resembles
that of an amphetamine more that of a drug such as picrotoxin. It is possible to
stimulate respiration with the drug without inducing generalized central nervous
system (CNS) stimulation. However, selectivity is still very low. The drug is
obsolete in managing poisoning from sedative-hypnotic drugs. It may have a very
limited place in treating acute respiratory insufficiency in chronic obstructive
pulmonary disease (COPD). It also may have value in correcting respiratory
depression caused by oxygen therapy in COPD.
СLASS 34.
Topic: Biologically important heterocycles with two heteroatoms. (135 min.)
1. SEMINAR. (70 min.)
1.1 Five-membered ring heterocycles with two heteroatoms (imidazole, pyrazole,
thiazole). Structure and properties.
1.2 Imidozole ring consisting compounds: histidine,histamine.
1.3 Pyrazole ring consisting compounds: amidopyrine, antipyrine, butadion,
analgin.
1.4 Six-membered ring heterocycles with two heteroatoms (pyrimidine, pyrazine,
pyridazine). Structure and properties.
1.5 Substituted pyrimidines: uracil, 5-bromouracil, 5-fluorouracil, thymine,
cytosine. Barbituric acid.
1.6 Purin and substituted purins: hypoxanthine, xanthine, uric acid, adenine,
guanine.
1.7 Alkaloids: nicotine, caffeine, theophylline, theobromine.
2. LABWORK (25 min.)
Literature
1. Lectures.
2. Fundamentals of bioorganic chemistry. S.E.Zurabyan. M.: GEOTAR-MED,
2003. P. 180-187.
Problem 1. To compare the neutral amines (pyrrolidine), pyrrole and
imidazole basicity.
Solution. Neutral pyrrole is less basic than saturated neutral amines because
the nitrogen in pyrrole is sp2 hybridized which is more electronegative than the
sp3 nitrogen of a saturated amine. In addition, the
nitrogen in pyrrole donates electrons into the ring increasing the acidity of N-H
bond (pKa = 17). Because of the presence of the second ring nitrogen, neutral
imidazole is a stronger acid (pKa = 14.4) than neutral pyrrole.
Problem 2. Barbituric acid and its derivatives.
Solution. The barbiturates are 5,5-disubstituted barbituric acids. The following
scheme shows how the 5,5-dialkyl compounds are synthesized.
H
COOC2H5
C
H
COOC2H5
R-X
NaOC 2H5
Diethylmalonate
R
COOC2 H5
R'-X
NaOC2 H5
C
COOC2 H5
H
Monoalkyl diethylmalonate
O
R'
O
R
COOC2H 5
C
R'
COOC2H 5
Dialkyl diethylmalonate
R
H 2NCNH 2
NaOC 2H5
C
C
NH
C
O
C
N
ONa
Sodium 5,5-dialkylbarbiturate
O
O
R'
R'
R
O
C
C
C
NH
C
C
N
+
H
R
C
NH
C
ONa
Sodium 5,5-dialkylbarbiturate
O
C
N
H
O
5,5-dialkylbarbituric acid
A consideration of the structure of 5,5-disubstituted barbituric acids
reveals their acidic character.
The first sedative-hypnotic barbiturate, 5,5-diethylbarbituric acid, was
introduced in 1903.
The batbiturates have a low therapeutic index and a relatively high abuse
potential.
The short-to intermediate-acting barbiturates are used as sedative-hypnotics
(amobarbital, butabarbital, pentobarbital, secobarbital) and are abuse most
commonly.
The
longer-acting
barbiturates
(mephobarbital, phenobarbital), used
primarily for their anticonvulsant properties, are abused rarely.
Street names of barbiturates include barbs, sleepers, downers, yellow
jackets, and rainbow.
Barbiturates suppress CNS neuronal activity and thus have sedative
and hypnotic properties. The major manifestations of barbiturate intoxication are
CNS, cardiovascular, and respiratory depression. Severe intoxication results in
coma, hypothermia, hypotension and cardiorespiratory arrest.
Barbital-(5,5-diethylbarbituric acid), although discontinued as sedativehypnotic, is interesting because of the biologic consequence of its low lipid/water
partition coefficient. It is slowly eliminated, mostly in the intact form, by the
kidney.
Metharbital - (5,5-diethyl-1-methylbarbituric acid) (Gemonil). This methyl
derivative of barbital finds a limited employment in a variety of epilepsics.
Phenobarbital – (5-ethyl-5-phenylbarbituric acid) (Luminal). The
compound is a long-acting sedative and hypnotic. It is also a valuble
anticonvulsant especially in generalized tonic-clonic and partial seizures.
PRACTICAL PROBLEMS (20 min.)
1. Prove the aromatic character of pyrazole.
2. Explain the amphoteric properties of imidazole.
3. Compare the acidic and basic properties of imidazole with pyrrole ones.
Imidazole boils at 257oC, while N-methylimidazole boils at 199oC. Explain the
difference in boiling points.
5. Prove the basicity of pyridine, pyrimidine, purine.
6. Write the structure of 5-pyrazolone and antipyrine.
7. List the following compounds in order of decreasing reactivity to
electrophilic substitution: pyrrol, pyrimidine, pyridine, benzene.
8. Write the equation of antipyrine bromination (SE).
9. What nitrogen atom will be protonated by strong acid firstly: in antipyrine or
amidopyrine?
10. Give the scheme of the of barbituric acid formation. Write lactim-lactam
tautomeric forms of these ones.
11. Give the scheme of lactim-lactam transformation for uracil, thymine, cytosine,
adenine, guanine, uric acid.
2. LABORATORY WORK (25 min.)
Some chemical properties of some heterocycles with two heteroatoms.
Derivatives of 5-pyrazolone
Several pyrazoline substitution products are used in medicine as analgesic agents.
Many of them are derivatives of 5- pyrazolone.
H5C6
H5C6
.
CH3
2,3-Dimethyl-1-phenyl-3- pyrazolin-5-one
Antipyrine; Phenazone.
H5C6
CH3
N
N
CH3
(H3C)2N
2,3-Dimethyl-4-amino-1-phenyl-3-pyrazolin-5-one
Aminopyrine; Amidopyrine
H5C6
N
O
H3C
N
O
N
O
.
N
CH3
N
CH3
CH3
O
C6H5
N
O
N
CH2SO3Na
Dipyrone; Methampyrone
Analgine
2.1 Analytical reactions for antipyrine.
H9C4
4-Butyl-1,2-diphenyl-3,5- pyrazolidinedione
Phenylbutazone; Azolid
Butazolidin; Phenylzone-A.
Locally, antipyrine exerts a paralytic action on the sensory and the motor
nerves, resulting in some anasthesia and vasoconstraction, and it also exerts a
feeble antiseptic effect.
2.1.1 Action of FeCl3 on antipyrine.
When 2-3 drops of FeCl3 solution are added to antipyrine solution, the red
colouring is appeared.
2.1.2 Action of nitrous acid.
To 1 ml of 1% antipyrine solution add 1 ml of 10% NaNO2 solution and 0.5 ml
of 10% H2SO4 solution. The brightly green product is appeared.
H5C6
H5C6
CH3
N
N
HO - N = O
N
CH3
N
O
O
CH3
CH3
O
N
Antipyrine
This substance is the important intermediate product in the synthesis of
aminopyrine and dipyrone.
H 5 C6
N
O
N
N
N
4 H+
CH3
O
H 5C 6
CH3
- H2O
O
N
N
2 CH3Cl
CH3
H2N
H 5C 6
CH3
- 2 HCl
O
CH3
N
CH3
(H 3C) 2N
Aminopyrine
2.2 Analytical reactions for aminopyrine.
Aminopyrine has been employed as an antipyretic and analgesic, as is
antipyrine, but is somewhat slower in action. However, it seems to be much more
powerful, and its effects last longer.
2.2.1 Action of FeCl3 on aminopyrine.
To 1 ml of 1% amidopyrine water solution add 0.5 ml of 1% K3[Fe(CN)6] and
1 ml of 2% FeCl3 solution. The dark blue colour of the intermediate product is
appeared and after some time it is disappeared.
2. 2. 2 Action of nitrous acid.
To 1 ml of 1% amidopyrine solution add 0.5 ml of 10% H2SO4 solution and 0.5
ml of 10% NaNO2 solution by drops. The unstable violet colour is appeared.
2.3 Analytical reactions for dipyrone.
Dipyrone is used as an analgesic, an antipyretic, and an antirheumatic. It is
available as injectable for veterinary use.
2.3.1 Action of FeCl3 on dipyrone.
To 1 ml of 1% dipyrone water solution add 1 ml of 2% FeCl3 solution. The dark
blue colour is appeared.
2.4 Analytical reactions for phenylbutazone.
2.4.1 Action of FeCl3 basic solution on phenylbutazone.
H5C6
N
O
C6H5
H5C6
N
N
O
O
H9C4
N
OH
H9C4
tautomeric forms of phenylbutazone
C6H5
H5C6
N
H5C 6
C 6H 5
N
O
N
NaOH
OH
- H2O
ONa
H9C 4
H5C6
3
O
N
O
H9C 4
N
C 6H5
H 5C 6
C 6H 5
N
N
+ FeCl3
ONa
- 3 NaCl
H9C 4
O
C6H 5
N
O
-
H 9C 4
Fe3+
3
Brown sediment
2.4.2 Action of nitrous acid.
To 1 ml of 1% phenylbutazone solution add 0.5 ml of 10% H2SO4 solution
and 0.5 ml of 10% NaNO2 solution drops by drops. The orange colour is appeared
and than it is changed to red colour.
2.5 Uric acid properties.
In humans, uric acid is the major product of the catabolism of the purine
nucleosides, adenosine and guanosine. Purines from catabolism of dietary nucleic
acid are converted into uric acid directly. However, the bulk of purines ultimately
excreted as uric acid in the urine originated from
degradation of endogenous nucleic acids. The daily synthesis rate of uric acid is
approximately 400 mg; dietary sources contribute another 300 mg. Men who
consume purine-free diets have an estimated total body pool of exchangeable urate
of 1200 mg; in women, this value is estimated as 600 mg. In contrast, individuals
with gouty arthritis and tissue deposition of urate may have urate pools as large as
18,000 to 30,000 mg.
Gout occurs when monosodium urate precipitates from supersaturated body
fluids; the deposits of urate are responsible for the clinical signs and symptoms.
Gouty arthritis may be associated with urate crystals in joint fluid,
as well as with deposits of crystals (topthi) in tissues surrounding the joint. The
deposits may also occur in other soft tissues, and wherever they occur, they elicit
an intense inflammatory response mediated by polymorphonuclear leukocytes and
macrophages.
O
N
H
N
N
HN
N
O
N
H
Purine
O
NH
NH
Uric acid
2,6,8- trioxypurine
Unsubstituted purine and pyrimidine are not found in nature. Substituted
purines (adenine and guanine) and pyrimidines (cytosine, uracil, thymine) were
some of the first compounds to be studied by organic chemists. Adenine can be
deaminated to give hypoxanthine, and guanine can be deaminated to give
xanthine. Hypoxanthine and xanthine are both enzymatically oxidized
uric acid.
into
O
O
N
HN
NH
N
N
HN
O
Hypoxanthine
NH
NH
Xanthine
The methylxanthines have interesting stimulatory properties.
O
O
H3C
O
N
N
CH3
N
H3C
N
O
CH3
Caffeine
O
N
N
N
NH
CH3
Theophelline
N
HN
O
N
N
CH3
CH3
Theobromine
Caffeine enjoys wide use as CNS stimulant. Theophylline has some use as a
CNS stimulant; its CNS-stimulant properties are encountered more often as side
effects, sometimes severe and potentially lifethreatening, of its use in bronchial
asthma therapy. Theobromine has very little CNS activity.
СLASS 35.
Topic: Nucleic bases. Nucleic acids. (135 min.)
1. SEMINAR. (25 min.)
1.1 Nucleic bases. Substituted purines (adenine and guanine) and substituted
pyrimidines (cytosine, thymine and uracil). Lactime-lactame tautomeric forms.
1.2 Nucleosides as N-glycosides. Ribonucleosides and deoxyribonucleosides. The
names of the nucleosides. Hydrolysis of nucleosides.
1.3 Nucleotides. Ribonucleotides and deoxyribonucleotides. The names of the
nucleotides.
1.4 Nucleic acids. The primary structure of nucleic acid.
1.5 Composition of DNA and RNA. Basic and acidic hydrolysis of DNA and
RNA.
1.6 Secondary structure of DNA. Base pairing in DNA: adenine and thymine;
cytosine and guanine. Chargaff's rules.
1.7 ATP: The carrier of chemical energy. The "High-Energy" character of
phosphoanhydride bonds.
1.8 The structure of NAD+ and NADP+. The system NAD+-NADH, NADPNADPH.
2. TEST. (85 min.)
Literature
1. Lectures.
2. Fundamentals of bioorganic chemistry. S.E.Zurabyan. M.: GEOTAR- MED,
2003. P. 224-237.
3. General Chemistry. D. D. Ebbing, M. S. Wrighton. Houghton Mifflin
Company. Boston. P. 1022-1027.
PRACTICAL PROBLEMS
1. Write the structure of thymine, cytosine and their base pairing.
2. Give the structures of nucleosides: uridine, deoxyguanosine. What is the
tautomeric form of nucleic bases in these nucleosides?
3. Give the structure for each of the following:
a) dCDP; b) guanosine- 5'-triphosphate; c) UMP; d) UDP;
e) dTTP.
4. Give the structure of deoxythymidilic acid, uridilic acid, deoxyguanilic acid;
give their names as phosphates of nucleosides.
5. Write the scheme of total acidic hydrolysis of 2'-deoxyadenosine-5'phosphate.
6. Write the scheme of total hydrolysis of guanilic acid. What are the conditions of
hydrolysis of glycosidic linkage and phosphoanhydride bonds?
7. Write the scheme of hydrolysis of ATP to AMP. What are "high-energy bonds"?
СLASS 36.
Topic: Terpenes. Steroids. (135 min.)
1. SEMINAR. (60 min.)
1.1 Terpenes. Terpenes as hydrocarbons and terpinoids. Isopren rule. Mono- and
bicyclic terpenes. Limonene, pynene, squalene, geraniol, citronellol, citronellal.
1.2 Carotenoids. β-Caroteneretinol (vitamin A), retinal. The chemistry of vision.
1.3 Steroids. Steroid nomenclature, stereochemistry, and numbering.
5α-Cholestane, 5α-androstane, 5α-pregnane, 5α-estrane. Cholesterol as the
precursor of vitamin D.
1.4 Adrenal cortical steroids. Cortisone, aldosterone.
1.5 Sex hormones. Androgens (testosterone, 5-α-dihydrotestosterone); estrogens
(estradiol, estrone).
1.6 Bile acids. Cholic acid, taurocholic acid, glycocholic acid, chenodeoxycholic
acid.
2. PRACRICAL WARK (60 min.).
Literature
1. Lectures.
2. Fundamentals of bioorganic chemistry. S.E.Zurabyan. M.: GEOTAR-MED,
2003. P. 249-258.
3. Wilson and Gisvold's textbook of organic medicinal and pharmaceutical
chemistry.-10th ed./edited by Jaime N.Delgado and William A. Remers.
P. 727 -799.
СLASS 37. Pass to exam.