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58 Chemistry Olympiad
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F i n a l c o m p e t i t i o n (31 Mar 2012)
Theoretical tasks and solutions
TASK 1
Complexes with cyclam
1,4,8,11-tetraazacyclotetradecane (cyclam) is a macrocyclic ligand with four
coordination sites at nitrogen atoms.
This ligand can form complexes with heavy metal cations (with 1:1 stoichiometry).
These complexes are applicable as catalysts, they can be also applied in medicine, as
they affect important processes of biological significance.
NH
HN
NH
HN
cyclam
These complexes are stable in pH range limited by disadvantageous effect of
nitrogen atoms protonation (only neutral molecule exhibits complexing properties) and possibility
of heavy metal hydroxides deposition.
This task concerns general conditions of stability of complexes with cyclam, in solutions of
constant pH = 7. Basing on obtained data it would be possible to predict stability of complexes of
specific metal cations. By solving this task, please assume predomination of some protonated forms
of the ligand.
Problems:
a. Calculate concentration of the neutral ligand in solution with total concentration of cyclam
(protonated and non-protonated forms) equal to 0.01 M (pH = 7).
b. Determine the ratio of stability constant, , and so-called „conditional” stability constant, ’,
where: ’ = [ML]/([M]cL), [ML] is concentration of the complex, [M] is concentration of free
metal ions (in symbols M, ML the charge is omitted), cL is total concentration of cyclam
(protonated and non-protonated forms) not bound with metal ions.
c. Derive equation and calculate the minimal value of stability constant, , for which in solution of pH =
7 at least 99.9% of metal ions is complexed, for cL = 0.01 M.
d. Heavy metal ions of charge 2+ can form hydroxide deposits. What is the minimal value of
stability constant, , to avoid hydroxide deposition in solution containing cyclam and metal ions
(at pH = 7). Express the stability constant, , as a function of solubility product of the hydroxide,
Ks0, and other selected equilibrium constants, given in the task, and concentrations ML and cL.
e. Are the conditions from sections (c) and (d) fulfilled for cations Cu2+ i Ni2+ for [ML] = 0.01 M
and cL = 0.01 M. For the case of the complex with Ni(II) ions calculate the concentration of noncomplexed nickel ions. How does this concentration change after 2-fold dilution of the solution ?
Dissociation constants of acids: Ka1 (for H4L4+) = 4.103; Ka2 (for H3L3+) = 3.102;
Ka3 (for H2L2+) = 3.1011; Ka4 (dla HL+) = 3.1012 (L is the form presented in the figure).
Stability constants of complexes with cyclam, ; for Cu2+: 2.1027, for Ni2+: 2.1022.
Solubility products, Ks0; for Cu(OH)2: 3.1019, for Ni(OH)2: 2.1015.
1
TASK 2
Phosphorus oxide – acidic or alkaline?
Compound A is formed in the reaction of white phosphorus with oxygen carried out at lowered
pressure and at a temperature below 50 °C. Reactive compound A can be isolated via vacuum
distillation from the products mixture in the form of white crystalline substance. The compound
melts at ca. 24 °C and can be easily dissolved in a series of organic solvents e.g. in benzene. Only
one signal was found in the 31P NMR spectrum of its benzene solution. It was also observed that the
solution obtained by dissolution of 0.185 g of compound A in 10.00 g of benzene freezes at a
temperature that is by 0.43° lower than melting point of pure benzene. The cryoscopic constant Et
for benzene equals 5.12 K · kg/mol.
Compound A reacts very easily with water to form a solution that contains acid B which forms a
precipitate of an anhydrous barium salt upon addition of the excess of Ba(OH)2 solution.
A sample of compound A with a mass m1 = 1.26 g was reacted with a substantial excess of nickel
carbonyl Ni(CO)4. The reaction was proceeding at room temperature and a colourless gas was being
evolved. It was found that, after compound A had been consumed, the mass of the reaction mixture
decreased by 0.642 g. 4.5 g of pure compound C was obtained after removal of the unreacted
carbonyl. The 31P NMR spectrum of this compound comprised only one signal as well.
Compound A in the solid state (3.08 g) was also reacted with gaseous diborane (B2H6) at room
temperature. After 24 hours products of the reaction were dissolved in n-pentane and left for
crystallisation. 2.67 g of pure crystalline compound D were collected.
It was found that this compound reacts violently with water and in the course of reaction colourless
gas is liberated and the resulting solution contains acids B and E. In a carefully controlled reaction
0.147 g of compound D were used and ca. 80 mL (equivalent for 0 °C and atmospheric pressure) of
gas were given off. Compound D reacts easily with nickel carbonyl too and using an excess of
carbonyl leads to the previously described compound C.
Problems:
a. Determine compound A formula and confirm it with appropriate calculations. Write the
equation of white phosphorus reaction with oxygen.
b. Draw Lewis electron structure of compound A molecule taking into account all of the lone
valence electron pairs. Justify your answer.
c. Write the chemical equation in molecular form (reagent form) of compound A reaction with
water.
d. Write the chemical equation in molecular form of acid B reaction with barium hydroxide. Draw
the molecular structure of anion present in the structure of barium salt. Justify your answer.
e. Determine the formula of compound C and write the equation of its formation reaction. Justify
your answer and confirm it with calculations
f. Draw and describe the structure of nickel coordination sphere in compound C. Justify your answer.
g. Determine the formulae of compounds D and E. Write the equation of compound D reaction
with water. Justify your answer and confirm it with calculations
h. Basing on the appropriate acids and bases definition determine the chemical character of
reagents in the formation reaction of compound D and calculate the reaction yield.
i. Draw and describe the boron coordination sphere in compound D. Justify your answer.
Use the following values of molar masses in your calculations (g/mol):
B – 10.81
C – 12.01
H – 1.008
Ni – 58.69
O – 16.00
P – 30.97
and the molar volume of gases at 0 °C and under atmospheric pressure Vm = 22.41·10−3 m3/mol
2
TASK 3
Determination of stoichiometry and stability of complexes with an NMR method.
Application of approximations in limiting conditions
One of the problems in the 2nd stage of the 58 Chemistry Olympiad was
related to the crown ethers ability to complex metal ions. Now we
present the results of 1 H NMR study on formation reaction of a
12-crown-4 ether (labelled as 12C4) with Na+ cation introduced into the
system in a form of sodium thiocyanate, NaSCN.
Three independent experiments have been carried out: the first and the
second at temperature 23°C, and the third one at temperature –50°C. In
Fig. 1. Molecular
the two first experiments the spectra were recorded at varying NaSCN
formula
of a 12C4
concentration and a single averaged signal coming from both the
crown ether.
complexed molecules and the free ether was monitored. The signal chemical shift, δobs, was
dependent on the ratio of sodium ions [Na+]0 and crown ether [E]0 concentrations. The appearance
of the averaged signal proves that the complexation reaction is a so called fast exchange reaction.
Let us remind that for such reactions the observed chemical shift, δobs, is an average of characteristic
chemical shifts of the ligand, δE, and the complexes, δEnM, multiplied by corresponding
concentration ratios of ligands unbound and bound in complexes to the total concentration of
ligands in the system. By writing the general complex formula as EnM (where E means ether, and
M means metal), the relationship can be presented as follows:
[ E M]
[E]
[EM ]
[ E M]
 obs 
 E 
  EM  2  2   E2M  3  3   E3M          (1)
[E] 0
[E]0
[E]0
[E]0
The total concentration of ligands in the system: [E]0 = [E] + [EM] + 2·[E2M] + 3·[E3M]+ . . . .
Experiment 1 The characteristic chemical shift, δE = 2.761 ppm, has been determined from the
NMR spectrum of the pure crown ether. Then, small amounts of solid sodium thiocyanate were
added to a 12C4 solution (in deuterated methanol, CD3OD) with concentration [E]0 =
0.219 mol/dm3, and the chemical shift, δobs, was recorded as a function of total concentrations
[NaSCN]/[12C4]. The measurement results are shown in Fig. 2, and additionally, the data for points
within the linearity range, i.e., at [NaSCN]/[12C4] ≤ 0,3, are given in Table 1.
The experimental points form a
curve showing two characteristic
bends, which prove that two
complexes, k1 and k2, with different
stoichiometry and different overall
complex formation constants for
metal ion and for ether, βn, are
present in the system Based on the
plot given in Fig. 2 one can
determine the stoichiometry of the
two complexes being formed and
remaining in equilibrium.
complex 2
Fig. 2
3
complex 1
Table 1. Results of experiment 1: [E]0 = 0.219 mol/dm3
[NaSCN]/[12C4]
δobs / ppm
0
0.045
0.096
0.150
0.240
0.300
2.761
2.846
2.937
3.025
3.190
3.282
Experiment 2 In all measurements the initial concentration of the crown ether was kept constant at
[E]0 = 0.050 mol/dm3, while the thiocyanate concentration varied but was much higher than that of
the emerging complex. The sodium ion concentrations, c, and the corresponding resultant chemical
shifts, δobs, are given in Table 2.
Table 2. Results of experiment 2: [E]0 = 0.050 mol/dm3
Total Na+ ion concentration
c / (mol/dm3)
Observed chemical shift
δobs / ppm
4.437
4.558
4.660
4.757
4.821
4.856
0.204
0.271
0.359
0.495
0.646
0.768
Experiment 3 The spectrum of the pure 12C4 ether was recorded again at temperature –50°C, then
a small amount of NaSCN was added, and the spectrum was recorded again. The spectrum showed
distinctly resolved ether and complex peaks (the fast exchange reaction does not occur here). It
turned out that the chemical shift of the pure ether virtually did not change with decreasing
temperature. One can therefore assume that the chemical shift of the k2 complex that is stable in the
presence of a large excess of ether, is the same at both 23°C and at –50°C, and is δk2 = 4.575 ppm.
Using the results of the three experiments described above, determine the stoichiometry of complexes
that are formed in the system, calculate their stability constants and characteristic chemical shift of the
k1 complex, by following the directions given below.
In the solution one should assume the following symbols for the major quantities occurring in the problem:
[E]0
+
– total concentration of the crown ether, constant in each experiment,
[Na ] = c
– total concentration of sodium ions, varying during titration,
δE
– chemical shift of the pure 12C4 crown ether,
δk1, δk2
– chemical shifts of two complexes,
ck1, ck2
– concentrations of EnM complexes, respectively, in equilibrium.
To simplify the notation of equations, the following symbols can be also introduced:
 obs   obs   E ,  k1   k1   E and  k2   k2   E .
4
Problems:
a. Based on the plot in Fig. 2 determine the stoichiometric coefficients of the resulting complexes. Justify
the answer with a short comment.
b. Give chemical equations for reactions taking place in the system and the formulae for stability
constants of the complexes, by writing them down in a form containing the initial ether concentration,
[E]0, and the concentration of sodium ions, c.
c. Using appropriate equations determine complex concentrations ck1 and ck2 by introducing the
simplifications proposed in Note 1.
d. Give an equation describing the dependence of the averaged chemical shift, δobs, on corresponding
chemical shifts of the ether, δE (data), and the two complexes: δk1 (this should be treated as an unknown)
and δk2 (data). Present the equation as a dependence of corresponding differences of chemical shifts, that
is  obs ,  k1 and  k2 . Write the two equations in two forms corresponding to the limiting cases, by
introducing corresponding complex concentrations determined in Direction c.
e. Calculate β2 from the simplified equation from Direction d. for the limiting case [E]0 >> c by
determining the slope of the straight line (δobs–δE) = f(c) from two selected points from Table 1 and using
known value of δk2 (see Note 2).
f. Solve the equation from Direction d. in the high Na+ concentration limit (c), by converting the equation
so as to get a linear dependence of 1/(δobs – δE) on 1/c. Calculate β1 and δk1 constants (see Note 3).
Notes:
1. The equations for equilibrium constants can be simplified if the experiment conditions can
be considered as the limiting ones.
In experiment 1 [E]0 >> c, so the equilibrium is strongly shifted towards the k2 complex and
we can assume that [E]0 >> ck2 >> ck1 ≈ 0.
In experiment 2 [E]0 << c, so the equilibrium is strongly shifted towards the k1 complex and
we assume that c >> ck1 >> ck2 ≈ 0.
Corresponding approximations are best introduced by determining ck1 and ck2 as functions of
complex stability constants β1 and β2.
2. Fig. 3 shows linear dependence of (δobs – δE) on c in the limit [E]0 >> c. The plot should facilitate the
selection of points of which the coordinates allow to calculate the slope of the straight line.
Experiment 1
Fig. 3
5
3. To solve the problem in the high Na+ concentration limit ([E]0 << c), it is necessary to
“linearise” the dependence of the corresponding difference of chemical shifts (δobs–δE) on
varying [Na+] = c concentration. The dependence of 1/(δobs–δE) on 1/c should be linear
(1/(δobs–δE) = a·(1/c) + b) in the limit of large values of the ratio c/[E]0, i.e., in the conditions of
experiment 2. Fig. 4 shows the relationship plotted using the data from Table 2.
Experiment 2
Fig. 4.
TASK 4
Biologically active indole derivatives
Tryptamines are derivatives of indole (a heterocyclic compound consisting of two condensed rings,
a benzene ring and a pyrrole ring), abundantly present in living organisms. To tryptamines belong,
inter alia, neurotransmiters as important as melatonin or serotonin. Due to that also other
tryptamine derivatives are not neutral to human organism, and may have medicinal use or exert
hallucinogenic effects.
The phosphorylated tryptamine derivative F, initially isolated from plant sources, has been obtained
through the reaction sequence depicted in Scheme 1. The starting material for the synthesis was 4hydroxyindole. It is known that the most reactive position of indole is the C-3 within the pyrrole
ring, which easily undergoes aromatic electrophilic substitution, but may also be involved in
reactions typical for enamines.
First, the phenol group of 4-hydroxyindole was protected by reacting it with benzyl chloride in the
presence of sodium methanolate. This led to compound A, which was then subjected to FriedelCrafts acylation with oxalyl chloride leading to compound B. Subsequently, B without isolation,
was reacted with dimethylamine, which resulted in compound C. In the next step, C was reduced to
D (molecular formula C19H22N2O) by means of LiAlH4 and then catalytically hydrogenated to yield
compound E and toluene as a by-product. In the last step, E was converted into F by applying
following reagents: 1. n-buthyllithium 2. tetrabenzyl pyrophosphate (TBPP) 3. catalytic
hydrogenation (Pd/C, H2).
A homologue of compound E, known as 4-hydroxygramine and having molecular formula of
C11H14N2O, may be obtained in a multi-component reaction (Mannich reaction), by reacting
compound A with formaldehyde and dimethylamine, followed by a catalytic hydrogenation.
Exploiting a similar multi-component reaction, but from different starting materials, one can obtain
compound X, as a mixture of several stereoisomers (Scheme 2 depicts only one of them).
6
O
Cl
X
Cl
H3C
OH
O
Cl
NH
[B]
A
N
H
C
MeOH/MeOLiAlH4
TBPP =
BnO
O
O Bn
P
P
BnO
O Bn
O
O
F
1. n-BuLi
2.TBPP
3. H2, Pd-C
E
Scheme 1
Chiral
O
H2, Pd-C
D
C19H22N2O
1
O
H3C
2
3
NH
4
CH3
N O
O
Scheme 2
Problems:
a. Draw structural formulas for compounds A-F and 4-hydroxygramine.
b. Draw structural formulas of the starting materials for the synthesis of compound X.
c. Compound X could be obtained from the same starting materials, but in a stereochemically pure
form (as depicted in Scheme 2). Choose, which conditions should be applied to achieve this:
I. the reaction temperature should be increased;
II. the reagents should be mixed in a certain order;
III. the reaction should be carried out in the presence of L-proline;
IV. the reaction mixture should be stirred always in the same direction.
d. Determine the absolute configuration of all stereogenic centers in compound X.
e. Explain the reactivity of the C-3 position of indole by means of appropriate mesomeric structures.
TASK 5
The Diels-Alder Reaction
Cycloaddition reactions, including Diels-Alder reaction, in most cases are stereospecific, what
means that geometry of substrates (e.g. dienophile or diene) determine the structure of the formed
product.
+
X
X
Y
Y
General Scheme of the Diels-Alder Reaction
I. The carboxylic acid with molecular formula C4H4O4 as well as its dimethyl esters (A , B and C)
exist as two geometric isomers and one regioisomer, and all of them undergo the Diels-Alder
reaction. The reaction of geometric isomers A and B mixture with diene D, composed entirely of
carbon and hydrogen atoms, provides diastereomeric products E1, E2 and F. Ozonolysis of E and F
mixture with zinc dust form product G (diastereomeric mixture) and subsequent reduction with
LiAlH4 excess leads to the mixture of diastereomers of polihydroxylic alcohol H.
7
D
AiB
E1 + E2
i
F
(diastereomers)
1) O3
2) Zn
1) O3
2) Zn
HO
OH
LiAlH4
G
HO
mixture of three
diastereomers
OH
H
mixture of three
diastereomers
D
I
J
Compounds A and B after ozonolysis with zinc dust form the same product I, which also undergo
cycloaddition reaction with diene D. The reaction leads to the cyclic ether J. The molecular mass of
J is equal to the sum of mass of reactants D and I.
II. Stereochemistry in Diels-Alder reaction: Isomer A in the reaction with diene D forms the
mixture of two achiral diastereomers E1 and E2. The reduction of diastereomeric mixture of E1 and
E2 with hydrogen on palladium catalyst leads to only one product K.
The Diels-Alder reaction of isomer B with diene D provides exclusively racemic mixture of
cycloadduct F. The product F after ozonolysis with zinc dust and subsequent reduction with LiAlH4
excess leads to one diastereomer of polihydroxylic alcohol H (as a racemic mixture).
Problems:
I
a. Draw the structure of geometric isomer A or B and structure of regioisomer C
b. Draw the structure of diene D
c. Determine general structure of Diels-Alder product E/F (no stereochemistry required)
d. Draw the general structure of ozonolysis product G (no stereochemistry required)
e. Draw the structure of compound I and general structure of J (no stereochemistry required)
II
a. Determine unambiguously geometry of isomers A and B
b. Draw stereochemical structures of compounds E (E1 and E2) and their reduction product K
c. Draw the stereochemical structure of compound F and stereochemical structure of
corresponding diastereomer of polihydroxylic alcohol H
d. Draw all possible stereoisomers of the product J. Determine enantiomers and diastereomers.
SOLUTIONS
SOLUTION OF TASK 1
a. The total concentration of cyclam, cL = [L] + [HL+] + [H2L2+] + [H3L3+] + [H4L4+]. Using acidic
dissociation constants, after rearrangement one can write:
cL = [L]{1 + [H+]/Ka4 + [H+]2/(Ka4 Ka3) + [H+]3/(Ka4 Ka3 Ka2) + [H+]4/(Ka4 Ka3 Ka2 Ka1)}.
Taking into account the values of dissociation constants, it can be assumed that at pH = 7 the
form H2L2+ predominates in the solution, i.e. the above equation can be simplified to the form:
cL = [L][H+]2/(Ka3 Ka4), thus [L] = cL Ka3 Ka4 /[H+]2. After introducing numerical results, for
cL = 0.01 M and [H+] = 107 M, the obtained [L] = 9.1011 M.
8
b. Because stability constant  = [ML]/([M][L]), the ratio /’ = cL/[L]. Basing
on previous data, cL/[L] = 0.01 / 9.1011 = 1.1.108.
P
O
c. Because [ML]/[M] = 99.9 / 0.1, ’ = [ML]/([M]cL) = 99.9 / (0.1 0.01) = 1 10 .
.
.
Thus,  = ’ 1.1 10 = 1.1 10 .
.
.
8
.
13
O
O
5
P
O
P
P
O
O
d.  = [ML]/([M][L]) = [ML][H ] /([M]cL Ka3 Ka4). Assuming that a deposit of
hydroxide does not precipitate: Ks0 = [M][OH]2, the above equation can be written in the form: 
= [ML][H+]2[OH]2/(cL Ks0 Ka3 Ka4) = KW2 ([ML]/cL) /( Ks0 Ka3 Ka4).
+ 2
e. For [ML] = cL = 0.01 M, using appropriate constants, the minimal values of  fulfilling the
conditions from (d) are: 4.1012 for the Cu2+ complex and 6.108 for the Ni2+ complex. Comparison
of these results with experimental data shows that Cu2+ and Ni2+ complexes obey the conditions
from (c) and (d).
Because for Ni 2+ - cyclam complex  ’ = [ML]/([M]cL), for [ML] = cL, [M] = 1/’ is obtained.
For the Ni 2+ complex  ’ = 2 . 1022 / 1.1 . 108 = 2 . 10 14 . Thus, [M] = [Ni 2+ ] = 1/ (2 . 10 14 ), i.e.
[Ni2+] = 5.1015 M.
2-fold dilution results in the same value of the [ML]/cL ratio, therefore Ni2+ ions concentration does
not change as well.
SOLUTION OF TASK 2
a. White phosphorus is built of P4 molecules. Depending on the reaction conditions, it may form
P4On (n = 6, 7, 8, 9 or 10) oxides in the reaction with oxygen whose structures stem from P 4
molecules structure. Freezing point depression of benzene allows one to calculate the molality of
oxide A solution (cmA) and consequently its molar mass:
cmA 
mA
Tt 0.43
0.185

 0.084 mol/kg, czyli M A 

 220 g/mol.
Et
5.12
mb  c mA 0.010  0.084
This value corresponds to P4O6 molar mass (219.88 g/mol). Thus, compound A is
phosphorus(III) oxide, P4O6. Phosphorus reaction with oxygen proceeds according to the
following equation:
P4 + 3O2  P4O6
b. The P4O6 molecule exhibits a cage structure stemming form the structure of P4 molecule. There is
one lone electron pair on every phosphorus atom and oxygen atoms have two lone electron pairs.
This structure agrees with 31P NMR spectrum indicating all phosphorus nuclei are chemically
equivalent
c. P4O6 is phosphoric(III) acid anhydride and reacts with water according to the following equation:
P4O6 + 6H2O  4H3PO3
d. Phosphoric(III) acid contains P–H bond and is a diprotic acid of an average strength. It forms,
therefore, two series of salts and sparsely soluble barium salt containing HPO32− anion will be
formed with the excess of barium hydroxide:
H3PO3 + Ba(OH)2 
2−
3
3
+ 2H2O
HPO3 anion exhibits tetrahedral structure (sp hybridisation of phosphorus). There are three
oxygen ligands and hydrogen atom in the vertices of a deformed tetrahedron.
9
2
H
O
O
2
H
P
O
or
O
O
P
O
e. Due to the presence of lone electron pairs on phosphorus atom P4O6 oxide can act as Lewis base
in chemical reactions. Therefore, Lewis adduct is formed in the reaction with nickel carbonyl in
which phosphorus forms a chemical bond with nickel replacing one of the CO ligands. Assuming
that the only gaseous product of the reaction is CO, the molar ratio of the evolved CO to the used
P4O6 equals
0.642 g
1.26 g

 0.02292 : 0.00573  4 :1 which indicates that four nickel atoms
28.01 g/mol 219.88 g/mol
were bound by a P4O6 molecule. The reaction proceeds according to the following equation:
4Ni(CO)4 + P4O6  P4O6·4Ni(CO)3 + 4CO
The formula of compound C is P4O6·4Ni(CO)3 (or P4O6[Ni(CO)3]4). The molar mass of
compound C equals 790.76 g/mol. Such a reaction course is confirmed by the approximate molar
mass of compound C, calculated from the reaction product mass:
MC 
mC
4.5

 785 g/mol
nP4O6 0.00573
f. It follows from the fact that all of phosphorus nuclei are chemically
equivalent in the molecule of compound C that every phosphorus atom is
bound to one nickel atom and three oxygen atoms. The coordination
centres in the form of nickel atoms satisfy the 18 electron rule (10 valence
electrons in 3d orbitals + 6 electrons from CO ligands + 2 electrons
coming from phosphorus). Therefore, the structure of nickel coordination
sphere is tetrahedral and ligands bound to nickel atom are localised in
vertices of a slightly deformed tetrahedron.
O
O
O
C
P
O
Ni
C
C
O
O
Nickel coordination sphere
g. Similarly as in the reaction with nickel carbonyl, P4O6 phosphorus oxide serves as a Lewis base
in the reaction with B2H6 forming an adduct with acid (BH3). As a result of the reaction with
water, the adduct is decomposed and its hydrolysis leads to phosphoric(III) acid (acid B) and
H3BO3 boric acid (acid E). The liberated gas is molecular hydrogen and its quantity depends on
the amount of reacted diborane.
nB2H6 
VH2
Vm  6

80
 0.595 103 mole which correspond to B2H6 mass equal to
3
22.4110  6
mB2H6  nB2H6 M B2H6  0.595 103  27.67  0.0165 g. The molar ratio of P4O6 to B2H6 in the Lewis
adduct equals:
(0.147  0.0165) g
: 0.595 103  0.594 : 0.595  1:1
219.88 g/mol
Only two phosphorus atoms are involved in bond formation with BH3 and the formula of
compound D is P4O6 · 2BH3 (or P4O6 · B2H6). The hydrolysis reaction proceeds according to the
following equation:
P4O6 · 2BH3 + 12H2O  4H3PO3 + 2H3BO3 + 6H2
10
h. The reaction between P4O6 and B2H6 leading to the formation of compound D:
(P4O6 + B2H6  P4O6·2BH3) is an acid-base reaction according to the Lewis definition.
Phosphorus oxide, being electron pair donor, is a Lewis base and B2H6 (acceptor) is a Lewis acid.
The reaction product, compound D, is a Lewis adduct. The yield of compound D formation
reaction equals:
w% 
nD
mD / M D
2.67 / 247.55
100% 
100% 
100%  77.0%
nP4O6
mP4O6 / M P4O6
3.08 / 219.88
i. There is a donor-acceptor bond formed between phosphorus and boron atoms in adduct D.
Ligands bound to boron atom (sp3 hybridisation) are localised in vertices of a slightly deformed
tetrahedron:
H
H
O
B
H
H
H
O
B
P
O
O
or
H
P
O
O
SOLUTION OF TASK 3
a.
The experimental curve δobs as a function of the concentration ratio [NaSCN]/[12C4] shows
clear changes of the slope at [NaSCN]/[12C4] = c/[E]0 ≈ 0.5 and at c/[E]0 ≈ 1.0. This means
that complexes with E/M = 2 and E/M = 1, i.e., E2M and EM, are formed in the system. In the
range c/[E]0 >> 1.0 (thiocyanate excess), the equilibrium is shifted towards the EM complex,
and in the range c/[E]0 << 0.5 towards the E2M complex.
b.
The complex formation reactions occurring at the equilibrium are as follows:
(k1): E + M  EM and
(k2): 2E + M  E2M.
The complex formation equilibrium constants, k1 and k2, also referred to as the complex
stability constants, can be written as:
[ET ]
ck1
,
(1)
β1 

[E]  [T] ([E]0  ck1  2ck2 )  (c  ck1  ck2 )
[E 2 T]
ck2
.
(2)
β2 

2
[E]  [T] ([E] 0  ck1  2ck2 ) 2  (c  ck1  ck2 )
c. From the equations for the equilibrium constants β1 and β2, ck1 and ck2, respectively, are to be
determined by introducing the simplifications suggested in Note 1. In equation (1), we assume
that c >> ck1 >> ck2, and consequently c−ck1−ck2 ≈ c. We can further neglect subtraction of ck2
in the first term of the denominator and get:
c k1
.
(3)
β1 
([E]0  ck1 )  c
In equation (2) we assume that [E]0 >> ck2 >> ck1 and we can assume that [E]0 − ck2 − 2ck1 ≈ [E]0,
and in the second term of the denominator we can neglect subtraction of ck1 and we get:
ck2
.
(4)
β2 
2
[E]0  (c  ck2 )
Then, from equations (3) and (4) we determine:
[E]0  c  β1
(5)
ck1 
1  c  β1
11
and
d.
ck2 
[E]02  c  β 2
.
1  [E]02  β 2
(6)
The resultant chemical shift resulting from chemical shifts of the crown ether, δE, and the
complexes, δk1 and δk2, can be written as follows:
[E]0  ck1  2ck2
c
c
(7)
δobs 
 δE  k1  δk1  2  k1  δk2 ,
[E]0
[E]0
[E]0
remembering that [E]0 = [E] + ck1 + 2 · ck2 .
By performing corresponding multiplications, reducing terms in equation (7) and introducing
substitutions to simplify the notation:  obs   obs   E ,  k1   k1   E and  k2   k2   E ,
we obtain:
 obs 
ck1
c
  k1  2  k2   k2 .
[E]0
[E]0
(8)
Equation (8) should be written as two equations for limiting conditions, that is by substituting
equations (5) and (6) to corresponding terms of equation (8) we obtain:
c  β1
 obs_k1 
  k1
(9)
1  c  β1
and
e.
 obs_k2  2 
[E]0  c  β 2
  k2 .
1  [E]02  β 2
(10)
In the case when [E]0 >> c and the equilibrium is strongly shifted towards the k2 complex
(E2M) (experiment 1) we solve equation (10), written as a function of c:
[E]  β  
 obs_k2  2  0 2 2 k2  c .
(11)
1  [E]0  β 2
One can easily notice that the factor in front of c is equal to the slope Δδobs_k2 = m·c. The factor
m can be determined from the point that is farthest away from the {0,0} point but still
belonging to the line, that is point 5 in Table 1. One only has to recalculate the coordinates
of the point to {c, Δδobs_k2}. For point {0.24, 3.190} we obtain the values c = 0.0526 mol/dm3
and Δδobs = 0.429 ppm as well as m = 8.155 ppm/mol/dm3. From equation (11) we determine
β2 and having substituted the data we get β2 = 20.1 (mol/dm3)-2.
f. Solution in the case c >> [E]0 (experiment 2) boils down to the case solved in Folder B of the
58 Chemistry Olympiad, i.e., the solution of equation (9). To convert equation (9) into a linear
function one must write the equation as an equality of reciprocals, as suggested by Note 4 (by
rising both sides of the equation to the power −1):
1
 obs_k1   E

 1

1  c  β1
1
1

 
 1 
,
c  β1  k1   E  c  β1   k1   E
And convert to the form:
11
1
1
1
.

 
 obs_k1   E β1   k1   E  c  k1   E
(12)
(13)
1
= a·(1/c)+b, so comparing
 obs_k1   E
corresponding parameters of the straight line with equation (13) we get the following
relationships:
The plot in Fig. 4 shows ideal linearity of data
12
a
1
β1  ( k1   E )
and b 
1
.
 k1   E
For two extreme points from Table 2 we calculate corresponding reciprocals, as given in the
Table below.
c / (mol/dm3)
0.204
0.768
(1/c) / (mol/dm3)-1
4.902
1.302
δobs / ppm
4.437
4.856
(1/(δobs – δE) / ppm-1
0.597
0.477
Then, we calculate parameters of the straight line: a = 0.0332 ppm-1·(mol/dm3) and b = 0.434 ppm-1,
and substitute them to formulae: δk1 =1/b+ δE and β1= b/a, obtaining: δk1 = 5.065 ppm and
β1= 13.1 (mol/dm3)-1.
SOLUTION OF TASK 4
a.
A
C
B
E
D
F
4-hydroksygramina
b. Reactants for the synthesis of X:
c. Conditions for
stereoselective
reaction: III
d. Absolute
configuration of X:
3S, 4S
e. Reactivity of position C-3 of indole comes from high electron density in this position
In
contrary to position 2 (structure III) localization of negative charge in position 3 (structure II) is
not connected with dearomatization of the benzene ring, which would be energetically unfavorable.
struktura I
struktura II
13
struktura III
SOLUTION OF TASK 5
Part I
CO2Me
CO2Me
Ia
MeO2C
or
CO2Me
CO2Me
MeO2C
C
(A or B)
Ia
Ic
D
Ib
Ie
Id
O
CO2Me
CO2Me
or
CO2Me
O
O
OMe
CO2Me
I
CO2Me
CO2Me
O
E or F
G
Ie
O
O
or
CO2Me
CO2Me
J
Part II
II a
II b
CO2Me
CO2Me
CO2Me
CO2Me
MeO2C
CO2Me
A
CO2Me
CO2Me
or
(E)-isomer
(Z)-isomer
+
B
H
II b
CO2Me
CO2Me
H
H
CO2Me
CO2Me
+
H
or
CO2Me
CO2Me
K
CO2Me
CO2Me
CO2Me
CO2Me
+
CO2Me
CO2Me
IIc
two diastereomers (achiral)
CO2Me
or
CO2Me
CO2Me
II c
HO
CO2Me
H
or
cycloadducts cis E (E1 i E2)
(from ester A)
MeO2C
OH
MeO2C
H
H
CO2Me
H
HO
CO2Me
CO2Me
MeO2C
CO2Me
MeO2C
OH
or
HO
HO
enantiomers
OH
OH
diastereomer of compound H
obtained from cykloadduct F
cycloadduct trans F
(obtained from ester B)
14
58 Chemistry Olympiad
F i n a l c o m p e t i t i o n (30th March 2012)
Practical tasks and solutions
TASK 1
Aluminium polychloride analysis
Suspensions are removed during the purification of water and sewage using the so-called
coagulants. They form sols having high surface which detain suspension particles and after addition
of flocculants they form precipitates. Aluminium polychloride (PAC) is one of the coagulants and it
is obtained via a reaction between aluminium hydroxide and hydrochloric acid, but only some of its
hydroxyl groups are substituted with chloride ions and, as a result, its composition is variable.
A PAC sample, whose mass (m1) is given on a 100 mL volumetric flask labelled A, was dissolved
in water with a small amount of nitric acid and a clear solution was obtained. 50,00 mL of EDTA
solution, whose concentration is given on a bottle, was added. The resulting solution’s pH was
adjusted to ca. 4.5 using methyl orange as an indicator. Afterwards, the solution was heated and
boiled for 10 minutes, while the pH was corrected using diluted ammonia solution. After cooling,
the solution was transferred to flask A and water was added to the graduation mark.
A second PAC sample, whose mass (m2) is given on a 100 mL volumetric flask labelled B, was
dissolved in 10 mL of 2 mol/L nitric acid solution. The solution was transferred to flask B and
water was added to the graduation mark.
Glassware and reagents at your disposal:
burette
two Erlenmeyer flasks with ST
beaker
25 mL volumetric pipette
graduated cylinder
10 mL volumetric pipette
small funnel
wash bottle with distilled water
ca. 0.050 mol/L KSCN solution
ca. 0.025 mol/L MgCl2 solution
Reagents at disposal of all participants:
EDTA solution (concentration given on a bottle)
ca. 0.05 mol/L AgNO3 solution
eriochrome black T mixed with NaCl and a spatula
ammonium buffer with pH = 10
10 % NH4Fe(SO4)2 solution
chloroform with a pipette
Attention: The ratio of AgNO3 solution volume to KSCN solution volume equals S and is given in
your answer sweet.
Additional information: The EDTA complex of aluminium exhibits lower conditional stability
constant that the MgY2− complex, but the aluminium complex is inert. The eriochrome black T
magnesium complex is much less stable than the MgY2− complex. The AgNO3 solution is acidified
with nitric acid (its concentration is 0.5 mol/L). The AgCl solubility product is higher than that of
AgSCN which is higher than AgBr Kso.
Problems:
a. Propose an analysis plan to determine the percentage of aluminium in the PAC sample using the
procedures given below and the information contained in the problem.
b. Propose an analysis plan to determine the percentage of chloride ions in the PAC sample using
the information contained in the problem, the procedures given below and the available reagents.
c. Write the equations (at least 6) of chemical reactions taking place during sample analysis as well
as of the reactions carried out before and described in the problem.
d. Derive the formulae to calculate the concentrations of titrants necessary for the completion of
tasks a. and b.
e. Give the determined concentrations of solutions form task d.
f. Determine the percentage of aluminium in PAC.
g. Determine the percentage of chloride ions in PAC.
h. Determine the stoichiometric formula of PAC, Al(OH)xCly.
i. Why does the solution of the PAC sample with EDTA turn red upon heating (as in the problem
description)?
j. Why is chloroform introduced into the flask during chloride ions determination? Is it necessary
to do it when determining bromide ions? Justify your answer.
Procedures
Complexometric determination of aluminium
Transfer an accurately measured portion of aluminium-ions-containing solution into an Erlenmeyer
flask. Introduce a precisely known amount of EDTA solution to the flask, excessive with respect to
the expected amount of aluminium. Add one drop of methyl orange and diluted ammonia solution
until the colour changes to yellow-orange. Heat the resulting solution and boil it for 10 minutes,
correcting its pH with diluted ammonia solution. After cooling, add 5 mL of ammonium buffer
solution with pH = 10, a pinch of eriochrome black T and titrate the navy solution quickly with
MgCl2 solution until the colour changes to violet-blue.
Repeat the titration.
Attention! The titration has to be carried out quickly and as soon as a noticeable change of colour is
observed the titrant’s volume has to be read. The titration must not be continued even though the
analyte solution may recover the blue colour.
Argentometric determination of chloride ions
Transfer an accurately measured portion of chloride-ions-containing solution into an Erlenmeyer
flask with ST. Add a precisely known volume of AgNO3 solution, excessive with respect to the
expected amount of chloride ions. If necessary add nitric acid so that its concentration lies in the
range 0.2 – 0.5 mol/L. Add ca. 2 mL of chloroform, ca. 1 mL of NH4Fe(SO4)2 solution, close the
flask with a stopper and shake it. Open the flask, wash the stopper with distilled water and titrate the
obtained solution with KSCN solution until slightly orange colours is obtained (when the precipitate
settles down).
Repeat the titration.
2
TASK 2
Distinguishing surface active agents (surfactants)
Owing to their characteristic structures and specific behaviour in aqueous solutions, surface active
agents (surfactants) are used diversely in analytics. The interactions between cationic surfactants
and dyes chelating metal ions bring about significant colour changes (batochromic shift) and if the
chelates are not very stable (e.g. complexes with metal titration indicators), they may be
decomposed. Metal chelates which are insoluble in water may be dissolved in aqueous solutions of
non-ionic surfactants due to the fact that they form micelles at appropriate concentrations. Cationic
and anionic surfactants may form precipitates that are soluble in the excess of surfactant with
suitable dyes (acidic or alkaline). The mentioned phenomena make it possible to differentiate
between cationic, anionic and non-ionic surfactants.
Test tubes A-J
The solutions of substances given in the table below are placed in test tubes labelled A-J. The
solutions concentrations are also given.
Substance
Concentration
Iron(III) chloride
210-5 mol/L
Mercury(II) nitrate
210-5 mol/L
Dithizone, HDz
210-4 mol/L
Eriochrome cyanine R, ECR
210-4 mol/L
Safranin T, SFT
210-4 mol/L
Rose Bengal, RB
210-4 mol/L
Potassium palmitate, PK
110-2 mol/L
Sodium dodecyl sulphate, SDS
110-2 mol/L
Triton X-100, TX
110-2 mol/L
Cetyltrimethylammonium chloride, CTA
110-2 mol/L
Dithizone is present in a slightly alkaline aqueous solution and is used for extractionspectrophotometric determination of mercury or silver (it forms orange and yellow chelates,
respectively, in strongly acidic solutions). In slightly acidic solutions., eriochrome cyanine R forms
violet complexes with aluminium and iron(III) ions, which turn blue under the influence of cationic
surfactants. Safranin T is an alkaline, whereas rose Bengal acidic non-chelating dye. Triton X-100,
poly(ethelene oxide) substituted octylphenol, is a non-ionic surfactant.
Glassware and reagents at your disposal:
8 empty test tubes
sulphuric acid, 1mol/L
6 polyethylene Pasteur pipettes
indicator paper
wash bottle with distilled water
You may use the solutions for problem 1.
3
Problems:
a. Give the probable arrangement of substances in test tubes A-J taking into account solutions’
colour and their pH and carrying out simple tests for the presence of surfactants.
b. Derive a plan that will allow you to identify substances in the solutions.
c. Identify the substances present in the solutions in test tubes A-J, using the available reagents and
the given procedure.
d. Give justification for your identification, confirming it with two observations.
e. Can magnesium be determined by titration with EDTA and eriochrome black T as an indicator,
if the analysed solution contains cationic surfactants? Justify your answer with appropriate
observations.
Investigating the influence of surfactants on coloured systems:
Transfer ca. 1 mL of the analysed dye solution into a test tube and introduce the same amount of
metal solution. Add surfactant solution drop by drop, shake the test tube and carefully watch what is
taking place in the solution. Carry out a blind test for comparison.
USE YOUR SOLUTIONS ECONOMICALLY. DO NOT USE PORTIONS LARGER THAN 1 ML FOR THE TESTS.
Eriochrome cyanine R, ECR
Rose Bengal, RB
Safranin T, SFT
Dithizone, H2Dz
4
SOLUTION OF PROBLEM 1
a. Analysis plan for the determination of aluminium percentage in PAC sample
Upon heating of the m1 sample with EDTA solution, according to the procedure, aluminium ions
reacted with some of the EDTA forming AlY  . The remaining EDTA has
to be titrated with MgCl2 solution. In order to determine the number of aluminium moles in the
sample, a portion of 25.00 mL from flask A has to be titrated with MgCl2 solution using V1 mL of
the titrant. Taking into account the fact that metal ions react with EDTA in a 1:1 molar ratio and
the relative volumes of the flask and pipette, one may write:
n Al  nEDTA  4  nMgCl2  50  c EDTA  4  V1  cMgCl2
b. Analysis plan for the determination of chloride ions percentage in PAC sample
Upon dissolution of m2 sample in nitric acid chloride ions are transferred to the solution and the
resulting mixture in flask B is acidic. In order to determine the amount of chloride
ions, one has to take 25.00 mL of the solution from flask B, add 25.00 mL of acidified AgNO3
solution with a known concentration (one gets the appropriate concentration of acid according to
the given procedure) and determine chloride ions according to the procedure using V2 mL of
KSCN solution. Taking into account the relative volumes of the flask and pipette, one gets the
following relationship:
n Cl  8  (n AgNO3  n KSCN )  8  (25  c AgNO3  V2  c KSCN )
c. Equations of the reactions taking place during the chemical analysis

3

PAC reaction with nitric acid: Al(OH)x Cl y  xH  Al  yCl  xH 2 O
Aluminium ions reaction with EDTA: Al3  H 2 Y 2  AlY   2H 
Titration of EDTA excess with magnesium chloride solution:
H 2 Y 2  Mg2  MgY 2  2H 
H   OH   H 2 O
Mg2  HIn  MgIn   H 
H   NH 4 OH  NH 4  H 2 O
Reactions taking place during chloride ions determination:
Cl   Ag   AgCl 
SCN   Ag   AgSCN 
m1 = 0.2718 g
S = 0.800
m2 = 0.5895 g
EDTA concentration
0.04990 mol/L
SCN   Fe3  FeSCN 2
Titration
Conc. of MgCl2
V0
Determination of Al
V1 10.00; 9.90
average 9.95
Conc. of AgNO3 and
KSCN
V3 18.70; 18.80
average 18.75
Determination of Cl
V2 16.05; 16.15
average 16.10
5
Titrant volumes, mL
20.00; 19.90
average 19.95
d. Derivation of the formulae for solutions concentrations
Titration of 10.00 mL of EDTA solution, ammonium buffer solution, eriochrome black T, using V0 mL of
n MgCl2 10  c EDTA
MgCl2 solution: c MgCl2 

V0
V0
Titration of 10.00 mL of MgCl2 solution + 25.00 mL of AgNO3 solution + chloroform + iron(III)
solution, using V3 mL of KSCN solution:
n AgNO3  n Cl  n KSCN
25  c AgNO3  2  n MgCl2  n KSCN  2  10  c MgCl2  V3  c KSCN
From the information in the problem:
25 
c
25

 c KSCN
c AgNO3  KSCN
S
S
2  10  c MgCl2 2  10  S  c MgCl2
c KSCN 

25
25 - S  V3
- V3
S
25  c AgNO3 
c KSCN
 2  10  c MgCl2  V3  c KSCN 
S
e. Concentrations of MgCl2, KSCN and AgNO3 solutions
c MgCl2  0.02501 mol/L
cKSCN = 0.04002 mol/L
c AgNO3  0.05003 mol/L
f. Determination of aluminium percentage in PAC
PAC mass in flask A equals m1 = 0.2718 g
mAl = (500.04990 - 49.950.02501)26.98 = 40.47 mg
Al percentage = 14.89%
g. Determination of chloride ions percentage in PAC
PAC mass in flask B equals m2 = 0.5895 g
mCl = 8(250.05001 – 16.100.04003)35.45 = 172.0 mg
Cl percentage = 29.18%
h. Stoichiometric formula of PAC
PAC formula – Al(OH)xCly
x+y=3
molar ratio f Al: 14.89/26.98 = 0.5517
molar ratio of Cl: 29.18/35.45 = 0.8232
y = 0.8232/0.5517 =1.492
x = 3 – 1.49 = 1.508
Al (OH)1.51Cl1.49
i. Explanation of the colour change during heating of PAC with EDTA
In the reaction of disodium salt of ethylenediaminetetraacetic acid with aluminium ions oxonium
cations are formed, acidifying the solution: H 2 Y 2  Al3  AlY   2H 
H+ ions have to be neutralised by adding ammonia solution, so that the reaction equilibrium is
shifted towards AlY  complex formation.
j. Justification of chloroform usage
Chloroform is added to separate AgCl precipitate from the solution. Owing to the fact that AgCl is
more
soluble
than
AgSCN,
the
unwanted
reaction
could
take
place:
AgCl  SCN   AgSCN  Cl  . When determining bromide ions, it is not necessary to
introduce chloroform as AgBr is less soluble than AgSCN.
6
SOLUTION OF PROBLEM 2
An exemplary arrangement of solutions:
Substance
Substance
A
Potassium palmitate, PK
F
Iron(III) chloride
B
Sodium dodecyl sulphate, SDS
G
Mercury(II) nitrate
C
Triton X-100, TX
H
Eriochrome cyanine R, ECR
D
Cetyltrimethylammonium chloride, CTA
I
Safranin T, SFT
E
Dithizone, HDz
J
Rose Bengal, RB
a. Probable arrangement of substance in test tubes A-J.
Colourless solutions, having nearly neutral pH may contain metal ions or surfactant solutions.
They are in test tubes A, B, C, D, F and G.
Soap solution i.e. potassium palmitate solution may be slightly opalescent and alkaline– test tube
A. Dithizone, eriochrome cyanine R, Rose Bengal and safranin T are orange-red (test tubes E, H, I
and J) and the HDz solution is alkaline – test tube E. Surfactant solutions froth upon shaking
which is visible for test tubes B, C and D.
b. Identification plan
Reaction with sulphuric acid: with frothing solutions–PK identification (palmitic acid insoluble in
water – the only one from surfactants); with coloured solution dithizonate anion reacts with acid to form dark precipitate of H2Dz, Rose Bengal changes its
colour to light yellow.
Mercury(II) ions may be identified with dithizone (one of the colourless, not frothing solutions) –
formation of the red-orange precipitate soluble in any surfactant.
The other colourless, not frothing solution, containing Fe(III) ions (may be identified with
potassium thiocyanate solution from problem 1) allows one to identify ECR – formation of violet
solution. This allows one to identify CTA – colour change to violet. CTA gives precipitate with
Rose Bengal as a result of ion-pair adduct formation. The remaining coloured solution (SFT) may
be used for SDS identification basing on precipitate formation with a small amount of surfactant.
Remaining TX does not form precipitates with any of the dyes.
7
c. and d. identification of substances in test tubes A-J and justification.
Identification
A
B
C
D
E
PK
Justification
Opalescent and slightly alkaline solution,
Froths when shook with distilled water
+ K → precipitate, the only one of surfactants
+ MgCl2 or running water → white precipitate
SDS
Colourless and neutral solution,
Upon shaking with distilled or running water froths abundantly
+ K → no changes,
SFT (tt I) + SDS → precipitate, + SDS → precipitate dissolution
RB (tt J) + SDS → no changes; Fe-ECR + SDS → no changes
TX
Colourless and neutral solution,
Upon shaking with distilled or running water froths abundantly
+ K → no changes,
Dissolves dithizone, mercury and silver dithizonates precipitates
SFT (tt I) or RB (tt J) + TX → no changes
CTA
Colourless and neutral solution,
Upon shaking with distilled or running water froths abundantly
+ K → no changes,
Fe-ECR (violet) + CTA → blue,, the only one of surfactants
RB (tt J) + CTA→ precipitate, + CTA → precipitate dissolution
HDz
Orange and slightly alkaline solution,
+ K → brown precipitate, soluble in TX, (SDS, CTA)
Hg(II) + HDz +K → orange↓, soluble in TX (SDS, CTA)
Ag(I) + HDz +K → orange↓, soluble in TX (SDS, CTA)
Colourless and slightly acidic solution, does not froth
+ KSCN (problem 1) → orange solution
F
FeCl3
+ ECR (tt H) → violet solution + CTA → blue solution
+ HDz +K → brown precipitate, soluble in TX, (SDS, CTA)
Colourless and slightly acidic solution, does not froth
+ KSCN (problem 1) → no changes
G
Hg(NO3)2
+ ECR (tt H) → no changes
+ HDz +K → orange precipitate, soluble in TX, (SDS, CTA)
8
Identification
H
I
J
Justification
ECR
Orange and neutral solution
+ K → no changes
+ Fe(III) → vilet solution, + CTA → blue solution
+ Al(III) (diluted flask B) → violet solution, + CTA → blue solution
SFT
Red and neutral solution
+ K → no changes
+ SDS (1-2 drops)→ precipitate, + SDS → precipitate dissolution
+ CTA → no changes
+ TX → no changes
RB
Red and neutral solution
+ K → turns colourless
+ SDS → no changes
+ CTA (1-2 drops)→ precipitate, + CTA→ precipitate dissolution
+ TX → no changes
e. Complexometric determination of Mg(II) in the presence of CTA
The presence of cationic surfactant renders magnesium determination via EDTA titration with
eriochrome black T as an indicator impossible. In the ammonium buffer solution the violet
magnesium complex with eriochrome black T decomposes upon addition of CTA and the solution
colour changes to blue, just as at the end of magnesium titration with EDTA solution.
Used abbreviations: + K – addition of sulphuric acid,
+ CTA – addition of cetyltrimethyl ammonium solution,
+ SDS – addition of sodium dodecyl sulphate solution ,
+ Fe(III) – addition of iron(III) chloride solution,
+ Hg(II) – addition of mercury(II) nitrate solution,
+ TX – addition of Triton X-100 solution,
tt – test tube
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Comments to the solution of task 2
b. Addition of sulphuric acid allows one to identify potassium palmitate (palmitic acid is sparsely
soluble in water), this is the only surfactant solution that becomes cloudy. Dithizonate anion, one of
the orange solutions, forms dark precipitate of H2Dz upon reaction with acid. The second of the
orange dyes, Rose Bengal, turns practically colourless upon reaction with acid. The remaining dyes
do not change their colour upon acidification. One may identify mercury(II) ions (one of the not
frothing colourless solutions) using dithizone which form complexes with them in acidic solutions,
as opposed to iron(III) ions. Dithizone and mercury dithizonate precipitates are dissolved upon
addition of surfactants. The remaining clear, not frothing solution contains iron(III) ions which can
be identified in a reaction with potassium thiocyanate solution from problem 1 (reddish colour).
Using iron solution and ECR one may identify CTA (violet-red solution of Fe(III) with ECR turns
violet-blue upon addition of CTA). One may find the acidic dye using CTA, thanks to the forming
precipitate – Rose Bengal. The other red-orange, alkaline dye, safranin T, can be identified as it
forms precipitate with SDS and remains in the solution upon addition of non-ionic surfactant –
Triton X-100.
c., d. The presence of potassium palmitate in test tube A is confirmed by the precipitation of
magnesium palmitate (with MgCl2 solution from problem 1) or the precipitation of magnesium and
calcium palmitate precipitates from tap water. Other surfactants do not give such precipitates. The
presence of dithizone in test tube E can be confirmed by the reaction with diluted AgNO 3 solution –
formation of yellow chelate precipitate, soluble in surfactants (or chloroform). Silver nitrate
solution allows one to identify CTA and FeCl3 by the precipitation of faint AgCl precipitate. Triton
X-100 is the only of the clear surfactant solutions that does not form precipitate with any of the
dyes. SDS does not form precipitate with Rose Bengal, neither does it affect iron(III) or aluminium
complexes (from flask B after substantial dilution with water). Mercury(II) dithizonate is not
decomposed upon addition of sulphuric acid and chloride anions (MgCl2) as opposed to silver
dithizonate.
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