1

64. Chemistry Olympiad

Final competition 2018

Theoretical tasks and solutions

TASK 1

The impact of protonation on precipitation equilibria

A laboratory team’s task was to evaluate the solubility product constant for a salt of a weak, diprotic

acid MX. To perform these calculation the team prepared two saturated solution of MX salt, one of

pH=7.0 and the second of pH = 6.0. It was found out that the concentration of M2+

ion in the first

solution was equal to 0.001 mol∙dm−3

while in the second one it was equal to 0.01 mol∙dm−3

. Negative

logarithms of the dissociation constants of the H2X weak acid are equal: pKa1 = 10.0 and pKa2 = 13.0.

Assume that the M2+

cation does not exhibit any acidic properties and does not form any complexes

with neither HX− nor X

2− ions.

Problems:

a. Derive the equation for the total concentration of H+ ions in the solution of H2X as a function of

the total concentration of this acid and its dissociation constants.

In the next step do the justified simplifications and calculate the pH of the H2X solution with the

total concentration of the acid equal to 0.1 mol∙dm−3

.

b. Derive the equation for the solubility product constant of the MX salt as a function of M2+

and

H+ concentrations.

c. Calculate the solubility product constant of the MX salt using the data provided in the

introduction and using both solutions with different pH values.

An interesting example of an almost insoluble salt of a weak, diprotic acid is PbCO3. Negative

logarithm of the solubility product constant for this salt is equal to 12.8, while negative logarithms of

dissociation constants of the carbonic acid are equal to pKa1 = 6.4 and pKa2 = 10.3. The solubility of

this salt depends also on a reaction that can be written down in a simplified form as::

Pb2+

+ H2O ⇄ Pb(OH)

+ + H

+ pKa = 7.9

2

Problems:

d. Derive the equation for the solubility product constant of the PbCO3 taking into account the

different acid-base equilibria for both anions and cations.

e. Calculate the solubility of that salt in water (pH = 7.0) and compare it to the solubility of that

salt calculated without taking into account the acid-base equilibria.

TASK 2

Electrides or not electrides?

Scientists have noticed already at the beginning of the XIX century that alkali metals can be

dissolved in liquid ammonia. The process results in the change of liquid’s colour from colourless

to dark navy. Detailed investigations of the obtained solutions’ structure revealed that at low metal

concentrations there are solvated cations in the solution, e.g. Na(NH3)6

+ in case of sodium

dissolution, and solvated electrons – e(NH3)6

− – localised in relatively large voids of the solvent

structure. Similar phenomena take place upon alkali metals dissolution in some organic solvents

(e.g. primary amines or dimethyl ether). The formed ionic compounds in which electrons are

anions are called electrides.

In other words electrides consist of cations and electrons which are not bound to any atomic cores.

The solvents used for the synthesis of electrides must be able to solvate

free electrons and cannot be prone to reduction (for instance, due to kinetic

hindrances). Organic compounds which form stable complexes with alkali

metal cations are often added to the solvents in order to increase the alkali

metals solubility. 1,4,7,10,13,16-hexaoxocyclooctadecane – a crown ether

denoted 18C6 and whose structure is presented in the scheme is one of

such compounds.

Samples of metallic caesium and 18C6 crown ether mixed in 1:1 molar ration were prepared and

dissolved in dimethyl ether at −15°C. Deeply coloured solution was obtained and a portion of

trimethylamine of similar quantity to that of dimethyl ether was added to it. Subsequently,

dimethyl ether was evaporated from the solution until dark blue, needle-shaped crystals of

compound A began to precipitate at −10°C. Analogous synthesis was carried out but the molar

ration of caesium to 18C6 crown ether was 1:2,3 and black, shiny, plate-like crystals of compound

B were formed. Both compound are stable at room temperature in vacuum.

123 mg of compound A were carefully reacted with water at low temperature. Care was taken so

that the temperature would not rise and the formed organic compounds would not decompose.

3,84 mL of colourless, flammable gas X whose density is almost 15 times smaller than the density

of air were evolved. The volume was measured at 25°C and 1000 hPa. Then, the obtained solut ion

was titrated with hydrochloric acid solution of 0.0100 mol∙L−3

concentration and 31.0 mL of

O

O

O

O

O

O

3

titrant were used. The solution was evaporated to dryness after titration and the residue was

dissolved in 2.0 mL of deuterated water D2O together with 1.0 mmol of tert-butyl alcohol (t-

BuOH). The obtained solution was analysed by 1H NMR at room temperature. No signals coming

from dimethyl ether or trimethylamine were observed in the spectrum and the ratio of peak areas

coming from 18C6 crown ether protons and protons from t-BuOH methyl groups was 0.826. In the

course of analogous analysis of 138 mg of compound B 2.59 mL of gas X were evolved, 20.9 mL

of 0.0100 mol∙L−3

hydrochloric acid solution were used and the intensity ratio of signals coming

from 18C6 crown ether methylene groups and from t-BuOH methyl groups in the 1H NMR

spectrum was 1.11. No peaks coming from other substances were observed in the 1H NMR

spectrum either. There are two signals in the 133

Cs NMR spectrum of solid compound A whereas

there is only one in the spectrum of solid compound B. It was also found that compound A is

diamagnetic while compound B is paramagnetic. The band gap of both compounds was

determined and they were both found to be semiconductors.

Problems:

a. Write down the name and chemical formula of gas X.

b. Write the structural formula of tert-butyl alcohol. How many signals coming from methyl group

protons are there in the 1H NMR spectrum of this alcohol’s solution recorded at room

temperature? How many signals are there in the 1H NMR spectrum of 18C6 crown ether

recorded at room temperature? What would be the intensity ratio of the signals coming from

methyl group protons of tert-butyl alcohol and 18C6 crown ether, if they were mixed in 1:1

molar ratio? Justify your answers.

c. Write down the empirical (sum) formulae of compound A and B. Justify your answer and

confirm it with appropriate calculations.

d. Write down the ions present in compounds A and B. Which of the compounds are electrides?

Justify your answer.

e. Write down the equations in molecular form of chemical reactions of compounds A and B with

water. Write also appropriate half reactions and name the oxidising and reducing agents in each

reaction.

f. Write down the equation in molecular form of the chemical reaction taking place in absence of

oxygen and moisture between compound A and anhydrous gold(III) chloride dissolved in

dimethyl ether. Red colloidal solution is formed as a result of this reaction.

Denote the crown ether molecule 18C6 to simplify your answers.

Use the following values of molar masses in your calculations (g∙mol−1

):

H – 1.008; C – 12.01; O – 16.00; Cs – 132.91.

Gas constant R = 8.3145 J∙mol−1

∙K−1

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TASK 3

Determination of the polymer molecular weight by the measurements of osmotic pressure

The equation describing the dependence of osmotic pressure π on concentration of solution c,

resembles in its form the ideal gas equation (Clapeyron equation). Just as the Clapeyron equation

describing the relationship p(V) is fulfilled only for gases of low density, the equation π(c) describes

well only the very dilute solutions (c → 0). It means that the dependence of osmotic pressure of real

solutions on their concentrations is not linear and shows significant deviations from the well-known

equation.

The osmotic pressure for polymer solutions of known concentrations as measured against a pure

solvent can be used to determine the average molecular weight, Mav, of dissolved polymer. However,

since different results of measured molecular weights are usually obtained for different

concentrations, it is necessary to perform several measurements for a set of concentrations, ci,

calculate the resulting apparent molecular masses, Mi, and extrapolate these values to the

concentration c = 0. The dependence of M(c) is obviously not linear, so, if we do not know the exact

relation of M(c), then a simple extrapolation of the values Mi to c = 0 is not possible.

A reaction of radical polymerization of styrene was carried out to determine the dependence of

polymer molecular weight on duration time of the polymerization. In the flask, under atmosphere of

pure argon, the amount of 10.414 g of degassed and inhibitor-free styrene and small amount of

benzoyl peroxide as the initiator of the radical chain reaction (in a molar ratio of 1:100) were placed.

The reaction was carried out at 120° C and after an appropriate time delay, the reacting mixture was

rapidly quenched in ice to stop the reaction.

The post-reaction mixture was divided into several samples, which were subjected to further testing.

The sample of 5.678 g was evaporated under vacuum at a sufficiently low temperature, yielding 3.456

g of dry polymer. Each of the four successive samples of the post-reaction mixture with identical

polymer concentrations, c, and masses, m, listed below in Table 1 were dissolved in toluene in the

measuring flasks and diluted with toluene to the volume of 100 cm3. The solutions were used for the

measurements of their osmotic pressure against the pure solvent, i.e. toluene. The difference in the

levels of liquids in both arms of the osmometer, h, was measured. The results are summarized in

Table 1.

Table 1

1 2 3 4

m [g] 1.200 0.900 0.600 0.300

h [cm] 3.57 2.26 1.17 0.43

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The experiments were carried out at t = 30.85 °C. The density of toluene at this temperature is

d = 0.8566 g∙cm−3

and it can be assumed that the density of polymer solutions is the same. Due to the

very small diameter of the capillaries in the arms of the osmometer, we omit the effect of diluting the

solution with a solvent passing through the semi-permeable membrane; the h values given in Table 1

are already corrected for the capillarity effect.

Note: benzoyl peroxide is decomposed to benzene during the reaction; in osmometric measurements,

it can be assumed that the molecules of styrene and toluene are of the same size.

In the calculations, take the following values of molar masses (g∙mol−1

): C – 12.010, H – 1.008; the

value of the gravitational acceleration is g = 9.81 m/s2.

Problems:

a. For all four given solutions calculate the concentration, c in g∙dm−3

, of polymer and the osmotic

pressure π, determined in individual measurements. Put the results into Table 2 in the solutions

sheet.

b. Using the grates in sheets, make a plot of the calculated dependence π(c) inserting also the point

at c = 0. Note: on the scale with resolution 1 you can insert points with a precision of ± 0.1.

c. From the equation describing the dependence of π(c), drive the formula for the dependence of the

molecular weight, M, of the polymer on π, c and T.

d. Calculate the molecular weight Mi of the polymer at each of the four experiments and insert the

results in Table 2 on the first page of the solution sheet.

e. Find the function f(M), which is a linear function of the concentration c and allows for an

extrapolation of f(M) to c = 0. Calculate the values of function f(Mi) for each individual

concentration of polystyrene and place them in Table 2.

Note: analysis of the results have shown that the dependence of the osmotic pressure of polystyrene

solutions on their concentration has the form: π(c) = A∙c + B∙c2.

f. Similarly to the b., plot the calculated values of f(Mi) as a function of the concentration c. Draw the

straight line as close as possible to all points and extrapolate this linear dependence to the

concentration c = 0.

g. From the two most distant points on the graph of f(M) plotted as a function of the concentration

c (plot from f.), calculate the parameters "a" and "b" of the straight line approximating

experimental points and used for extrapolation to the concentration c = 0. Calculate the value of

function f(M) for c = 0 and the resulting mean value of the average molecular mass, Mav, of

polystyrene.

Hint: instead of calculating the parameters of the straight line through the two chosen points, you

can apply for all four points the linear regression function available in the calculator.

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h. Calculate the degree of polystyrene polymerization as the ratio of the calculated average molecular

weight of polystyrene Mav to the molecular weight of the monomer Mm. Calculate the extent of the

reaction, λ, achieved in the system at the moment of "freezing" of the polymerization reaction.

Note: the extent of reaction, λ, is defined as the ratio of the number of moles of substrate or product

that has been used or produced in the reaction till the moment when the reaction is stopped to the

appropriate stoichiometric coefficient:

λ = (nz − n0)/z

where: nz - number of moles of the particular reactant z at the moment of stopping the reaction,

n0 - initial number of moles of the reactant (for the product n0 = 0), z - stoichiometric coefficient of

the reactant (for the substrate a negative value should be taken).

TASK 4

Identification of aromatic compounds

Benzene, five-membered and six-membered heterocyclic aromatic compounds belong to most

important structural motifs present in many relevant organic compounds - synthetic as well as

naturally occurring products.

Six aromatic or heteroaromatic compounds (A-F) along with products of further reactions are

described below. Selected information based on the nuclear magnetic resonance (NMR) useful for

identification process are shown in the table:

Compound: A B, B1, B2 C D, D1 D2 E F G H I X

Number of signals in 13

C NMR spectra 4 7 5 7 4 2 2 6 9 13 16

1H NMR spectra (in CDCl3): Signals in the range of 0-5 ppm are present only for compounds B,

D1, D2, F and X.

I) Compounds A and B are hydrocarbons having the same composition. They contain 93.5% of

carbon, and the molecular mass do not exceed 200 g∙mol−1

. A and B do not decolourise bromine

water, however react with bromine in the presence of iron. Moreover, only B can be oxidized when

heated with Na2Cr2O7 or KMnO4 solutions, to form products B1 and B2 with molecular mass of 198

and 216 g∙mol−1

respectively.

a. Determine molecular formula of hydrocarbons A and B.

b. Draw the structures of hydrocarbons A and B.

c. Draw the structures of compounds B1 and B2.

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II) Compounds C and D have a molecular weight of 122 g∙mol−1

and beside carbon and hydrogen

contains also oxygen (26.2%). They can be easily separated by extraction of C with diluted solution

of NaOH. Compound C also reacts with thionyl chloride to form C1. Reaction of D with NaBH4, and

selective reduction with hydrogen on palladium catalyst leads to achiral product D1 with molecular

weight of 126 g∙mol−1

. In 1H NMR spectra of D1 there are three multiplets with relative intensity

1:1:1 in the range of 6-7.5 ppm. Further reduction of D1 with hydrogen on nickel catalyst with acid

additive resulted in formation of product with molecular weight of 130 g∙mol−1

and D2 with 128

g∙mol−1

. The 13

C NMR spectra of D2 shows four peaks, and at proton spectrum only signals with

chemical shifts below 5 ppm are present. In addition, compounds D reacts with cyclopenta-1,3-diene.

d. Determine molecular formula of compounds C and D.

e. Draw the structures of compounds C and C1.

f. Draw the structures of compounds D, D1 and D2.

III) Compounds E and F exhibit basic properties, and beside carbon and hydrogen contains 41.2%mas.

and 25.9% mas. of nitrogen respectively. They molecular masses do not exceed 130 g∙mol−1

, and 13

C

NMR spectra shows only two signals for each compound. Moreover, E is also a weak acid (slightly

stronger then water), and reacts with 2 moles of C1 in the presence of NaOH solution with ring

opening and formation of product G and simple carboxylic acid salt. G contains 10.5% of nitrogen

and has a molecular weight of 266 g∙mol−1

. Compound F heated with excess of B1 forms product H

with molecular weight of 468 g∙mol−1

, which has no basic properties.

g. Determine molecular formula of compounds E and F.

h. Draw the structures of compounds E and F.

i. Draw the structures of compounds G and H.

IV) Compounds A, C and E were utilized in the synthesis of antifungal drug X, illustrated at the

scheme:

SOCl2

AlCl3

NaBH4

SOCl2

C C1A

J

E

X1)

2)

(excess)I

j. Draw the structures of compounds I, J and X.

Use atomic masses of elements (g∙mol−1

): H - 1; C – 12; N – 14; O - 16

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TASK 5

Bacteria with off-on switch

Bioorthogonal reactions are organic reactions that can performed on biomolecules present in living

organisms without perturbing the basic function of those organisms. Bioorthogonal chemistry is focused

on developing reactions that can be carried out in water, at physiological pH (6.5-8), at temperature of 30-

37 °C, and in a selective manner, preferably without catalyst, and without toxic by-products. A typical

bioorthogonal reaction is carried out between two functional groups, which normally do not occur in

Nature. To carry out the reaction in a living bacterial cell, one of the reacting group must be beforehand

introduced into bacterial biomolecule (e.g. polysaccharide or protein). In this problem, you will get

familiar with one of the known biorthogonal reactions and learn how it can be used to control light

emitting properties (fluorescence) of bacteria.

To carry out bioorthogonal metabolic labelling of green fluorescent protein (GFP-1) in bacterial cells, an

unnatural amino acid X was synthesized as depicted (Fig. 1).

Fig.1

The starting point for the synthesis of amino acid X was natural proteinogenic aromatic amino acid A.

Amino acid A subjected to a mixture of concentrated nitric and sulfuric acid gave para-substituted

derivative B. Derivative B was treated with di-tetr-butyl dicarbonate to yield compound C. Compound C

was reacted with hydrogen on a palladium catalyst to give compound D (M=280,32 g∙mol-1

). Compound

D reacts with compound K leading to derivative E. A by-product of this reaction is volatile compound

with pungent odour. Derivative E is deprotected under acidic conditions to yield amino-acid X

(M=274,28 g∙mol-1

).

Fig.2

The first step of the synthesis of compound K (mol. Formula C4H6N4S) is refluxing thiohydrocarbazide in

methanol, to yield monomethylated derivative J (Fig. 2). Condensation of derivative J with

trimethylacetate in ethanol, in the presence of triethylamine, followed by addition of sodium nitrite in the

presence of acid as a mild oxidant, leads to compound K as the main product. Compound K is a cyclic,

aromatic compound, which does not undergo dimerization in the presence of mild oxidants. In the proton

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NMR spectrum of compound K there are two signals (singlets od equal intensities, for both δH is below 4

ppm) and in the carbon spectrum there are 4 signals.

Fig.3

Aminoacid X was added to growth medium, in which genetically engineered E.coli strain was cultured.

The bacteria incorporated single amino acid X into GFP-1 protein that they expressed. Such modified

GFP-1 protein did not emit light (was non-fluorescent; Fig. 3). Next, compound Y was added to the

bacterial culture. Just after 1 minute fluorescence of the bacteria was observed, indicating that a rapid

chemical reaction occurred with participation of the modified GFP-1 protein. The first step of this reaction

is [4+2] cycloaddition between amino acid X present in the protein and compound Y, leading to the

formation of GFP-2 adduct that contains a modified amino acid Z’. GFP-2 protein spontaneously

converts into GFP-3 that contains amino acid Z, and a non-toxic gas is released during this process.

MALDI-MS analysis revealed that conversion of GFP-1 into GFP-3 results in a change of protein

molecular mass by 124 g∙mol-1

. The structure of compound Y, not showing its stereochemistry, is

depicted in Fig. 3. The double bond in compound Y has E configuration, and one of the substituents in the

cyclopropane ring has configuration trans relative to other substituents. Finally, catalytic hydrogenation of

compound Y leads to mezo compound.

Problems:

a. Draw structural formula for compounds B, C, and D. You do not need to show the

stereochemistry.

b. Draw structural formula for compounds E, K, and X. You do not need to show the

stereochemistry.

c. Draw the structural formula of Y clearly showing its stereochemistry.

d. Draw a scheme for the reaction between protein GFP-1 and compound Y. In your scheme, use

skeletal formula of amino acids Z and Z’ (you do not need to show stereochemistry). Depict

the protein schematically in such way that amino acid X is fully shown, including the peptide

bonds it engages in. N-terminal part of the protein label as R1 and C-terminal as R2. If you

cannot draw the protein, draw the same reaction for the free amino acid X Draw the structure

of appropriate isomer of chloronicotinic acid, oxidation product of compound X.

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e. Rapid reaction between amino acid X and compound Y is ensured by the presence of double

bond of E-configuration in compound Y. This is related to (choose one answer):

A. Steric hindrance

B. Electronic effects

C. Angular strain

D. Lack of interactions with the hydroxyl group

f. Reaction between compounds X and Y is often called inverse electron demand Diels-Alder reaction

(iEDDA). This is because:

A. The participating diene is electron-rich, whereas the dienophile is electron-poor.

B. The participating diene is electron-poor, whereas the dienophile is electron-rich.

C. The LUMO orbital of the diene has much higher energy than HOMO of the dienophile.

D. The LUMO orbital of the dienophile has much higher energy than HOMO of the diene.

g. The Diels-Alder reaction is usually reversible. What makes the reaction analyzed in this problem

irreversible, especially, at low reagent concentrations present in the living cells?

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Solutions of theoretical Tasks

SOLUTION OF TASK 1

a.

H2X ⇄ HX− + H

+ pKa1 = 10.0

HX-⇄ X

2− + H

+ pKa2 = 13.0

Ka1 and Ka2 dissociation constants are defined as:

Ka1 = [HX−][H

+] / [H2X]

Ka2 = [X2−

][H+] / [HX

−]

and the total concentration of H+ ions [H

+]c is equal to the concentration of both stages of dissociation:

[H+]c = [H

+]1 + [H

+]2

[H+]1 can be calculated on the basis of the first step of dissociation solving the quadratic equation:

[H+]1

2 + Ka1·[H

+]1 – Ka1·c = 0

[𝐇+]𝟏 =−𝑲𝐚𝟏 + √𝑲𝐚𝟏

𝟐 + 𝟒𝑲𝐚𝟏 · 𝒄

𝟐

[H+]1 = 3.1·10

−6 mol∙dm

−3

Calculating [H+]2 is more complex, as it depends on the first step of dissociation. We can, however,

note that:

[H+]2 = [X

2−]

and the concentration of of [HX−] ion can be written as:

[HX−] = [H

+]1 – [H

+]2

Substituting these equations into the Ka2 equation gives us::

𝐾a2 =([H+]1 + [H+]2)[H+]2

[H+]1 − [H+]2

And finally:

[𝐇+]𝟐 = 𝟏/𝟐(−𝑲𝐚𝟐 − [𝐇+]𝟏 + √(𝑲𝐚𝟐 + [𝐇+]𝟏)𝟐 + 𝟒𝑲𝐚𝟐[𝐇+]𝟏

where:

[𝐇+]𝟏 =−𝑲𝐚𝟏 + √𝑲𝐚𝟏

𝟐 + 𝟒𝑲𝐚𝟏 · 𝒄

𝟐

12

If the difference between [H+]1 and [H

+]2 concentration values is large (meaning that Ka2 is a lot

smaller than [H+]1 or Ka2 is a lot smaller than Ka1), which we can note as [H

+]1 + [H

+]2 ≈ [H

+]1 then

Ka2 ≈ [H+]2,

so:

[H+]c = [H

+]1 + [H

+]2 ≈ [H

+]1 + Ka2

and finally:

[H+]c = 3.1·10

−6 mol∙dm

−3+ 10

−13 mol∙dm

−3 ≈ 3.1·10

−6 mol∙dm

−3

pH = 5.5

b.

MX ⇄ M2+

+ X2−

KS0 = [M2+

][X2−

]

In the absence of ions of insoluble salt we can note that:

[M2+

] = S and [X2−

] = S, to yield KS0 = S2

S = (KS0)1/2

X2−

ion is, however, a weak base described by the following equilibria:

H2X ⇄ HX−+ H

+ pKa1 = 10.0

HX−⇄ X

2− + H

+ pKa2 = 13.0

described by the following dissociation constants Ka1 and Ka2:

Ka1 = [HX−][H

+] / [H2X] czyli [H2X] = [HX

−][H

+] / Ka1

Ka2 = [X2−

][H+] / [HX

−]

Solving the second equation we get: [HX−] = [X

2−][H

+] / Ka2

Solving the first equation we get: [H2X] = [HX−][H

+] / Ka1 = [X

2−][H

+]

2 / Ka1Ka2.

If the solubility products constant of the MX salt is equal to KS0 = [M2+

][X2−

], then the actual

molar solubility S of this salt can be describes as::

[M2+

] = S

[H2X] + [HX−] + [X

2−] = S

Using the second equation we get:

𝑆 = [X2−] + ([X2−][H+]

𝐾a2

) + ([X2−][H+]2

𝐾a1𝐾a2

) = [X2−] (1 +[H+]

𝐾a2

+[H+]2

𝐾a1𝐾a2

)

[X2−] =𝑆

1 +[H+]𝐾a2

+[H+]2

𝐾a1𝐾a2

And finally the solubility products constant of the MX salt is equal to:

13

𝑲𝑺𝟎 = 𝑺 ·𝑺

𝟏 +[𝐇+]𝑲𝐚𝟐

+[𝐇+]𝟐

𝑲𝐚𝟏𝑲𝐚𝟐

c. Using the equation derived above at pH = 7.0 we get:

𝐾𝑆0 = 𝑆 · (𝑆

1 +[H+]𝐾a2

+[H+]2

𝐾a1𝐾a2

) = 10−3 · (10−3

1 +10−7

10−13 +(10−7)2

10−1010−13

) = 𝟏𝟎−𝟏𝟓

d. Using the same equation at pH = 6.0 we get:

𝐾𝑆0 = 𝑆 · (𝑆

1 +[H+]𝐾a2

+[H+]2

𝐾a1𝐾a2

) = 10−2 · (10−2

1 +10−6

10−13 +(10−6)2

10−1010−13

) = 𝟏𝟎−𝟏𝟓

e. For anions all equation were already derived above. In the case of M2+

cation we find the

following equilibrium:

Pb2+

+ H2O ⇄ Pb(OH)

+ + H

+ pKa = 7.9

Ka = [Pb(OH)+][H

+] / [Pb

2+] czyli [Pb(OH)

+] = Ka·[Pb

2+] / [H

+]

Since lead ions exist in two forms, the molar solubility is equal to the sum of their concentrations:

[Pb2+

] + [Pb(OH)+] = S

Using this equation we get:

𝑆 = [Pb2+] +[Pb2+]𝐾a

[H+] = [Pb2+] (1 +

𝐾a

[H+])

[Pb2+] =𝑆

(1 +𝐾a

[H+])

For anions we have:

[H2X] + [HX−] + [X

2−] = S

To finally yield:

𝑆 = [X2−] + ([X2−][H+]

𝐾a2

) + ([X2−][H+]2

𝐾a1𝐾a2

) = [X2−] (1 +[H+]

𝐾a2

+[H+]2

𝐾a1𝐾a2

)

[X2−] =𝑆

1 +[H+]𝐾a2

+[H+]2

𝐾a1𝐾a2

14

Using the equation derived for the concentrations of M2+

and X2-

ions we can write the final equation

for the solubility product constants as:

𝐾𝑆0 =𝑆

1 +𝐾a

[H+]

·𝑆

1 +[H+]𝐾a2

+[H+]2

𝐾a1𝐾a2

f. Using the equation derived above we get:

𝑆2 =𝐾𝑆0

(1 +𝐾a

[H+]) (1 +

[H+]𝐾a2

+[H+]2

𝐾a1𝐾a2)

𝑆2 =10−12,8

(1 +10−7.9

10−7 ) (1 +10−7

10−10.3 +10−14

10−6.410−10.3)= 4.4 · 10−8

S = 2.1·10-5

mol∙dm−3

Without using the acid-base equilibria for ions the solubility of that salt in water would be equal to:

S = (KS0)

1/2 = 4·10

−7 mol∙dm

−3

SOLUTION OF TASK 2

a. X – hydrogen, H2

b. The structural formula of tert-butyl alcohol:

All methyl groups rotate freely around the C−C bond axis at room temperature in solution and

hydroxyl group also rotates around the C−O bond. Consequently, all methyl group protons are

equivalent and there is only one signal (singlet) coming from them in the 1H NMR spectrum.

Similarly methylene groups of 18C6 crown ether also rotate freely in solution at room

temperature and all protons in the 18C6 molecule are equivalent and give rise to one signal (also

singlet).

When t-BuOH and 18C6 are mixed in a 1:1: molar ratio, the molar ratio of methyl group protons

of t-BuOH and 18C6 ether protons equals (1 ∙ 9)a : (1 ∙ 24) = 0,375. This ratio is equal to the

intensity ratio of respective signals in the 1H NMR spectrum.

a – the number of moles of a particular compound multiplied by the number of equivalent nuclei

of a given type

c. It follows from the information concerning the 1H NMR spectra for samples resulting from

compounds A and B dissolution in water that they consist of caesium and 18C6 crown ether only.

OH

15

One may calculate the amount of crown ether in the sample of compounds A and B basing on the

intensity ration of the signals coming from tert-butyl alcohol protons and crown ether protons.

There is 1.0 mmol of tert-butyl alcohol i.e. 9.0 mmol of methyl group protons in the analysed

solution. Given that the signal intensity ratio is 0.826, there are 0.826 ∙ 9.0 = 7.43 mmol of crown

ether protons in 123 mg of compound A. One 18C6 crown ether molecule contains 24 protons.

Therefore, 123 mg of compound A contains 7.43 / 24 = 0.310 mmol of 18C6 ether. The 18C6

ether molar mass of the formula C12H24O6 equals 264.31 g∙mol−1

which indicated that 123 mg of

compound A contains 0.310 mmol ∙ 264.31 g∙mol−1

= 81.9 mg of 18C6 ether and 41.1 mg of

caesium which amount to 0.309 mmol.

Thus, the Cs:18C6 molar ratio is 0.309 mmol : 0.310 mmol ≈ 1:1 and the empirical (sum) formula

of compound A is Cs(18C6) or Cs(C12H24O6).

In case of compound B analysis there is also 9.0 mmol tert-butyl alcohol methyl group protons in

the solution, but there are 9.0 ∙ 1.11 = 9.99 mmol of 18C6 ether methylene group protons which

amounts to 9.99 / 24 = 0.416 mmol of 18C6 ether. Such an amount of ether weighs 0.416 mmol ∙

264.31 g∙mol−1

= 110 mg. There are 28 mg of caesium in 138 mg of compound B.

Therefore, the Cs:18C6 molar ratio is 0.211 mmol : 0.416 mmol ≈ 1:2 and the empirical (sum)

formula of compound B is Cs(18C6)2 or Cs(C12H24O6)2.

Any other correct reasoning allowing for the unequivocal determination of compounds A and B

empirical formulae will receive maximum number of points.

d. In the course of compound B analysis 2.59 mL of hydrogen were evolved. The number of moles of

hydrogen can be calculated using Clapeyron equation:

𝑝𝑉 = 𝑛𝑅𝑇 ⇒ 𝑛 =𝑝𝑉

𝑅𝑇

𝑛 =100000 Pa ∙ 2.59 ∙ 10−6 m3

8.3145 J ∙ mol−1 ∙ K−1 ∙ 298.15 K= 0.104 mmol

Hydrogen is formed as a result of hydronium cations reduction which proceeds according to the

following half reaction:

2H2O + 2e− → H2 + 2OH

−

0.208 mmol of electrons had to react for 0.104 mmol of gaseous hydrogen to be formed. This

indicates that there is 0.208 mmol/0.211 mmol = 1,0 mol of electrons per 1 mol of caesium in

compound B. Moreover, the presence of only one peak in the 133

Cs NMR spectrum indicates that

there is only one kind of caesium nuclei in this compound and the information about the

paramagnetic properties of compound B shows that it contains unpaired electrons. One may,

therefore, conclude that compound B contains caesium cations complexed by the crown ether:

16

Cs(18C6)2

+ lub Cs(C12H24O6)

2

+) and electrons e

− instead of anions. Compound B is an electride

with formula [Cs(18C6)2

+, e

−].

Analogously, in case of compound A analysis 3.84 mL of hydrogen were formed. This amounts to

𝑛 =100000 Pa ∙ 3.84 ∙ 10−6 m3

8.3145 J ∙ mol−1 ∙ K−1 ∙ 298.15 K= 0.155 mmol

Here 0.310 mmol of electrons had to react for such an amount of hydrogen to be evolved. This

means that there is 0.310 mmol / 0.309 mmol = 1.0 mol of electrons per 1 mol of caesium in

compound A. Compound A is paramagnetic which indicates that all electrons are paired therein.

There are two signals in the 133

Cs NMR spectrum. One must come from nuclei of caesium cations

complexed by crown ether: Cs(18C6)2

+ (or Cs(C12H24O6)

2

+). The second one comes from caesium

nuclei in a different chemical environment. Given that there are electron pairs in the compound,

one may conclude that there is one electron pair per one such caesium atomic core. Thus, the

second signal comes from caeside anions Cs−. In the compound A structure there are no free

electron pairs instead of anions, because there is a second signal in the 133

Cs NMR spectrum in

addition to the signal coming from cations. The signal stemming from caeside anions is displaced

relative to the signal coming from Cs(18C6)2

+ cations because the electron pair is localised on the

6s orbital which implies non-zero probability of finding these electrons on the caesium nucleus of

the caeside anion. Summing up, compound A is not an electride but a salt with formula

[Cs(18C6)2

+, Cs

−].

This is an example of quite a big group of alkali metal alkalides obtained using a series of

complexing agents forming stable chelates with these metal cations.

e. The reaction of compound A with water:

[Cs(18C6)2

+,Cs−](s) + 2H2O(l) → 2CsOH(aq) + 2 18C6(aq) + H2(g)

Half reactions: Agent

oxidation: Cs− → Cs

+ + 2e

− reducing: Cs

−

reduction: 2H2O + 2e− → H2 + 2OH

− oxidising: H2O

The reaction of compound B with water:

2[Cs(18C6)2

+,e−](s) + 2H2O(l) → 2CsOH(aq) + 4 18C6(aq) + H2(g)

(or [Cs(18C6)2

+,e−](s) + H2O(l) → CsOH(aq) + 2 18C6(aq) + ½ H2(g))

Half reactions: Agent

oxidation: isolated electrons from electride are the

reducing agent

reducing: e−

reduction: 2H2O + 2e− → H2 + 2OH

− oxidising: H2O

17

The following equations are also correct:

[Cs(C12H24O6)2

+,Cs−](s) + 2H2O(l) → 2CsOH(aq) + 2C12H24O6(aq) + H2(g)

2[Cs(C12H24O6)2

+,e−](s) + 2H2O(l) → 2CsOH(aq) + 4C12H24O6(aq) + H2(g)

or:

2Cs(C12H24O6)(s) + 2H2O(l) → 2CsOH(aq) + 2C12H24O6(aq) + H2(g)

2Cs(C12H24O6)2(s) + 2H2O(l) → 2CsOH(aq) + 4C12H24O6(aq) + H2(g)

f. Gold(III) cations are reduced by caeside anions to metallic gold according to the following

equation:

2AuCl3(ether) + 3[Cs(18C6)2

+,Cs−](ether) → 2Au(ether) + 3[Cs(18C6)

2

+,Cl

−](ether) + 3CsCl(ether)

or

2AuCl3(ether) + 6Cs(C12H24O6)(ether) → 2Au(ether) + 3Cs(C12H24O6)2Cl(ether) + 3CsCl(ether)

or another equivalent equation.

Using this method one may obtain metal nanoparticles with diameters ranging from 1 to 19 nm.

SOLUTION OF TASK 3

Table 2:

1 2 3 4

c (g∙dm−3

) 7.304 5.478 3.652 1.826

π (Pa) 300 190 98 36

Mi (g∙mol−1

) 61500 72860 93830 127700

M−1

∙105 / (mol∙g

−1) 1.6260 1.3724 1.0658 0.7834

a. From the mass of the post-reaction mixture contained in V = 100 cm3 of the solution and knowing the

ratio of the polymer mass to the mass of the reaction mixture we calculate the concentration c:

c = m∙3.456∙/(V∙5.678). (1)

An osmotic pressure, π, in the present experiments is a hydrostatic pressure of the solution column

raised in a capillary of the osmometr to the height h. We use the formula:

π = d∙g∙h (2)

where: d - density of the solution at a given temperature (here, the density of toluene), g - acceleration

of gravity, and h - difference in levels of the liquid column in the arms of the osmometer. The

calculated values of c and π (after conversion of units) are inserted in Table 2 above.

18

It should be noted that the unreacted molecules of styrene present in the solution (as well as the

negligible amount of benzene produced in decomposition of benzoyl peroxide) will behave like

toluene (solvent) due to their very close molecular sizes. In these measurements they behave as the

substances which are not osmotically active and do not contribute to the measured osmotic

pressure. The osmotic pressure measured is solely due to the presence of large polymer molecules.

Fig.1. Dependence of osmotic pressure on concentration of the polystyrene solutions.

b. The graph π(c) should look as shown above in Figure 1 (competitors do not have to plot a

fitted curve describing the relationship π(c) = A∙c + B∙c2).

c. The data given in the problem are: concentration of polystyrene, c [g∙dm−3

], osmotic pressure π [Pa], and

temperature t [° C]. Temperature of the experiment in Kelvin scale is T = 30.85 ° C + 273.15 K = 304.00 K.

The osmotic pressure π as a function of molar concentration, cm, is described by the formula:

π = cm∙R∙T (3)

The molar concentration, cm, we calculate as: cm = c/M, where M is the molecular weight (in

[g∙mol−1

]) of the polystyrene. Inserting above expression into formula (3) we get:

π = c∙R∙T/M (4)

Transformation of equation (4) to the form „M as a function of c” leads to:

M = c∙R∙T/π (5)

d. Using equation (5), we calculate the "apparent" molecular weight, Mi, of the polystyrene and

introduce them into the Table 2.

e. Taking advantage of the given note: π(c) = A∙c + B∙c2 we can write an equality of this relationship

and the relationship described by equation (4):

π = c∙R∙T/M = A∙c + B∙c2 (6)

After dividing of both sides of the equation (6) by c (with the assumption that c ≠ 0) it can be

noticed that the inverse of the molecular weight must be described by a straight line:

19

M−1

= a + b∙c , (7)

where: a = A/R∙T and b = B/R∙T .

The function we were looking for is f(M) = M−1

(c). We calculate the inverse values of Mi and

place them in Table 2.

f. We make a graph f(M) = M−1

(c). The calculated points show a linear dependence on the

concentration of polystyrene. We extrapolate a straight line through experimental points to the

value of M−1

(c = 0), as shown in Figure 2 below.

Fig. 2. Dependence of the inverse "apparent" molecular weight 1/M(c) on the concentration of

polystyrene

g. The equation of the straight line calculated with a least squares method has the form:

M−1

= a + b∙c (8)

and the parameters of this equation are: a = 0.5033∙10−5

mol∙g−1

and b = 0.1553∙10−5

mol∙g−1

∙K−1

.

The value of M−1

at the concentration c = 0 is equal to the constant „a”, which gives the average

value of the actual molecular weight of polystyrene equal to:

Mav = 198700 g∙mol−1

.

The straight line calculated from the two extreme points In Fig. 2 has the parameters a = 0.5025∙10−5

mol∙g−1

i b = 0.1538∙10−5

mol∙g−1

∙K−1

. The reciprocal of the "a" parameter gives an average

molecular weight of 199000 g∙mol−1

. The small difference between these two calculations

indicates good linearity M−1

(c).

h. The reaction of polymerization can be written as:

C8H8 + 2 H* → H−(C8H8)−H

The molecular weight of styrene is Mm = 104.144 g∙mol−1

. The degree of styrene polymerization is

therefore:

20

= Mav/Mm = 198700/104.144 ≈ 1908

The extent of the reaction, λ, we calculate from the definition taking into account the number of

moles of styrene, which has reacted until the reaction was frozen:

λ = (nz – n0)/ (10)

where: nz - number of moles of styrene at the moment of stopping the reaction, n0 - initial number

of moles of styrene, - stoichiometric coefficient of the reactant (here: styrene for which we take a

negative value as for the substrate). The number of moles used for the reaction is

n0 = 10.414g/104.144 g∙mol−1

= 0.100 mole.

The remaining amount of unreacted styrene is calculated as:

nz = 10.414g∙(1−3.456/5.678)/104.144 g∙mol−1

= 0.039 mol

The extent of the reaction is therefore:

λ = – (0.039 −0.100)/1908 = 3.19∙10−5

mole.

SOLUTION OF TASK 4

a. Molecular formula of A and B – C12H10

b. and c.

CO2HCO

2H

O OO

A B B1 B2

(biphenyl) (acenaphthene)

d. Molecular formula of C and D – C7H6O2

e. and f.

CO2H

OCHO

COCl

OOH

O

O

C

C1

D

D1

(benzoic acid) ((E)-3-(2-furyl)acrolein)

(or isomer Z)

D2

21

g. Molecular formula of E – C3H4N2 Molecular formula of F – C6H8N2

h.

N

NH

NH2

NH2

E F

(imidazole) (1,4-diaminobenzene)

i.

NH

NH

OO

NN

O

O

O

O

HG

(or isomer-E)

j.

OCl

N

N

I J X (Bifonazole)

SOLUTION OF TASK 5

a. i b.

22

c.

d.

e. C.

f. B.

g. The irreversibility of the reaction ensures spontaneous conversion of the Z 'adduct present in

GFP-2 into the Z adduct present in GFP-3 with the disconnection of the volatile N2 molecule,

which shifts the equilibrium towards the final product (Z).

Seitchik et al. J. Am. Chem. Soc., 2012, 134 (6), pp 2898–2901

23

64 Chemistry Olympiad

Final competition 2018

Practical tasks and solutions

TASK 1

Malonyl dichloride purity determination

Acyl chlorides are synthesised in the substitution reaction of hydroxyl group by chloride ions. The

reaction involves heating an acid with thionyl chloride and proceeds according to the following

equation:

RCOOH + SOCl2 RCOCl + SO2 + HCl

The final product may be contaminated with the unreacted organic acid or with hydrochloric acid. The

potentiometric titration curve of an acetone-aqueous solution containing the products of pure malonyl

dichloride (propanedioic acid dichloride) hydrolysis with a standard sodium hydroxide solution is

shown in the figure.

You receive a 200-mL volumetric flask denoted with letter P with a solution of technical malonyl

dichloride in water. Malonyl dichloride is highly hygroscopic which makes it difficult to prepare a

precisely weighed amount of it.

You have a standard solution of hydrochloric acid at your disposal with the concentration given on the

bottle and sodium hydroxide solution with a concentration of ca. 0.1 mol∙L−1

. There is also burette

24

with a funnel, 25-mL volumetric pipette, two Erlenmeyer flasks, two beakers and graduated cylinder

on your table.

All the participants have methyl orange and phenolphthalein at their disposal.

The data for a few points from a conductometric titration of a sample with an identical composition as

yours have been given in a table on your answer sheet. The titrant concentration was exactly the same

as that of yours.

REMARK: Assume that the sample of technical malonyl dichloride contains only one substance

treated as a contamination.

Problems:

a. Using the potentiometric titration curve and your knowledge in organic chemistry, write down the

names of the substances present in the titrated solution. Write down the equation of malonyl

dichloride hydrolysis.

b. Determine the sodium hydroxide solution concentration with a precision of four significant digits

using the given procedure. Justify your choice of indicator.

c. Using the available glassware and substances determine the total number of acids’ millimoles

in the flask P expressing it as the amount of hydronium ions.

d. Sketch the conductometric titration curve using the data given in the table on your answer

sheet. Determine the equations of lines corresponding to the titration of particular components

and find the lines’ intersection point. Calculate the number of millimoles of the substances in

flask P.

e. Basing on the results of titration answer what was the contamination in the malonyl dichloride

sample. Determine the millimoles number of the contaminant in flask P.

f. Calculate the weight percentage of malonyl dichloride in the hydrolysed sample.

Sodium hydroxide concentration determination

Fill the burette with sodium hydroxide solution. Transfer 25.00 mL of hydrochloric acid to

Erlenmeyer flasks. Dilute it with ca. 45 mL of water and add a few drops of an appropriate indicator.

Titrate until there is a noticeable and permanent change of indicator’s colour. Note the titration results

and calculate the millimoles number of the titrant used for the titration. Repeat the titration until you

get reproducible results. After the titration wash the burette with tap water and then with deionised

water. Calculate the sodium hydroxide solution concentration (with a precision of four significant

digits).

ATTENTION! Take a close look at the answer sheet. Plan your answers and write them in such

a way that they fit in the given space. Explain the abbreviations that you use.

Use the solution economically as you will not be given additional amounts.

25

TASK 2

Solvent composition influence on solution’s colour. Spectrophotometric determination of water

Thiocyanate anions form coloured complexes with ions such as Fe(III), Co(II), Mo(V) and others.

Identification of cobalt(II) is an interesting case. Blue cobalt(II) complex is formed in aqueous

solutions for high concentrations of added thiocyanate anions. The method’s sensitivity depends on

the thiocyanate ions concentration. Introduction of acetone to the solution greatly increases the

reaction sensitivity. For low thiocyanate ion concentration aqueous solution is practically colourless

and the addition of acetone gives rise to intense blue colour. This testifies to the significant influence

of the solvent on the colour intensity (solvatochromism). Similar phenomena can be observed for

cobalt(II) chloride.

In the attached figures (attachment to problem 2) there are absorption spectra of:

- anhydrous cobalt(II) chloride solutions in ethanol with variable water content

- anhydrous cobalt(II) chloride solutions in ethanol with variable methanol content at room

temperature and after heating (og)

- thiocyanate cobalt(II) complex solutions in a mixture of acetone and ethanol (1:1) with

variable water content

- CI 42051 dye solutions in ethanol with variable water content

The influence of hydrochloric acid addition to the solutions on the spectra is also shown.

The following solutions are placed in ampoules denoted 1, 2 and 3: cobalt(II) chloride in ethanol,

thiocyanate cobalt(II) complex in ethanol-acetone mixture and CI 42051 dye in ethanol.

There is a vegetable sample in ampoule denoted S which contains water and ethanol. Sample mass

(without ethanol) is given on the ampoule. There is a sample of ethanol-acetone mixture containing

water in the graduated test tube denoted R.

You have the following solvents at your disposal: denatured anhydrous ethanol, methanol and

ethanol-acetone mixture. There are also a beaker, stirring rod, small funnel with filter paper, 25-mL

volumetric flask, 5 mL graduated pipette, test tube with a stopper, six test tubes and 6 polyethylene

Pasteur pipettes.

Attention! All the operations in this problem have to be carried out using dry glassware. Wet

glassware should be washed with denatured anhydrous ethanol.

Problems

a. Basing on the information from absorption spectra given in the attachment and at least two

experiments, identify the solutions in ampoules 1, 2 and 3.

b. Prepare a calibration curve for the relative absorbance decrease of a thiocyanate cobalt(II)

complex solution in water-ethanol-acetone mixture as a function of water content in the

26

solution (use the given procedure). Determine the range in which the function is linear and find

the line equation. Repeat this for cobalt(II) chloride solution in ethanol.

c. Determine the method sensitivity and its quantification limit, assuming that the error associated

with preparing the water solution in organic solvent is negligible and the measurement error is

0.02 when absorbance equals 1.

d. Determine the mass of water in sample R and percentage of moisture in sample S, using the

given procedure.

e. Is selective determination of methanol in ethanol feasible? Justify your answer using the

attached figures.

Procedure

Add denatured ethanol to the sample in ampoule S, mix it with a mixing rod and leave until the

precipitate falls to the ampoule bottom. Transfer the solution from above the residue to the 25-mL

volumetric flask. Add ca. 7 mL of ethanol to the ampoule with precipitate, mix it and, after the

precipitate has fallen, transfer the solution to the flask. Repeat the residue washing. Fill the volumetric

flask up to the graduation mark with denatured ethanol. Filter the solution from the volumetric flask

to the dry graduated test tube. Throw away the first filtrate portion. Collect 5 mL of filtrate, add 2 mL

of cobalt(II) chloride ethanol solution and add ethanol up to 10 mL. Measure the solution absorbance

at 625 nm (due to technical reasons)*. The measured solution absorbance Asolution should be compared

to the absorbance of solution that does not contain water (the Aanhydr will be given by the person

operating the spectrophotometer). Calculate the relative absorbance decrease:

∆𝐴rel =𝐴anhydr − 𝐴sol

𝐴anhydr

Determine the percentage of water in the studied solution using the calibration curve.

Add 2 mL of cobalt(II) thiocyanate complex in aceton-ethanol solution to the test tube R and fill with

acetone-ethanol mixture up to 15 mL. Measure the absorbance at 625 nm. Calculate the relative

decrease of absorbance according to the formula given above. Determine the percentage of water in

the studied solution using the calibration curve.

* One obtains the highest measurement sensitivity when measuring absorbance at max. When a peak

is broad one does not lose much sensitivity when measuring close to the maximum. In case of the

experiments described here max is 620 and 655 nm, respectively. Due to technical reasons tuning the

light wavelength on the available spectrophotometers would take too much time and the

measurements are carried out at an intermediate wavelength of 625 nm.

Remember about safety rules when doing experiments!

The total number of points for the laboratory tasks is 60.

Practical examination duration: 300 minutes.

27

Solutions of laboratory Problems

SOLUTION OF LABORATORY TASK 1

Ad a. Malonyl dichloride hydrolyses in water yielding hydrochloric acid and maleic acid which is a

dicarboxylic acid. It follows from the potentiometric titration curve that the dissociation constants of

this acid Ka1 and Ka2 differ significantly. There are two abrupt changes in the titration curve, the first

one, poorly visible, at ca. 39 mL and the second 52 mL (the total titrant volume consumed to titrate

both acids). The difference between the first and second abrupt change is 13 mL which is 25% of the

total amount of titrant used. Summing up, there are two acids in the mixture – hydrochloric acid

(strong one) and maleic acid (weak one).

CH2(COCl)2 + 2H2O → CH2(COOH)2 + 2HCl

Ad b. I will use standard hydrochloric acid solution and phenolphthalein as an indicator to determine

the exact NaOH solution concentration. The hydrochloric acid concentration is 0.1057 mol/L. I will

carry out the titration until the first noticeable change of colour occurs. The colour will be compared

to a solution of HCl with the added indicator. The change of indicator colour should occur as close to

the equivalence point as possible which due to the possibility of NaOH solution contamination with

carbonates will be above pH=7. Methyl orange is a two-coloured indicator changing its colour in the

pH range 3.1 – 4.4 from red to yellow-orange. Addition of too much titrant do not cause a change in

colour and one is unable to say if an appropriate amount of titrant was added. The use of

phenolphthalein makes such observation possible (after addition of an excessive titrant amount the

solutions colour will be more intense). I repeat the titrations until I get similar results. I used on

average 28.20 mL of NaOH solution. I will calculate the NaOH concentration from the following

formula:

𝑐NaOH =𝑐HCl ∙ 𝑉HCl

𝑉0NaOH=

0.1057 ∙ 25.00

28.20= 0.09371 mol ∙ L−1

Ad c. Using a volumetric pipette I transferred 25,00 mL of solution from flask P (1/8 of the total

sample) to an Erlenmeyer flask. I diluted it with water from a graduated cylinder and have approx. 70

mL of solution in the flask. I added a few drops of phenolphthalein and titrated the resulting solution

with NaOH solution until slight pink colour appeared. The volume of titrant used is denoted VNaOH. I

repeated the titration a few times until I got reproducible results. I calculated the average VNaOH which

equals to 32.70 mL. I calculated the number of hydronium ions millimoles in flask P according to the

following formula:

nH+ = 8∙cNaOH∙VNaOH = 8∙0.09371∙32.7 = 24.51 mmol

28

Ad d. I sketch the conductometric titration curve:

Example data

Titrant volume, mL Conductivity, mS

4.1 11.94

12.2 7.49

20.4 4.84

28.6 4.8

I calculate the x (titrant volume) coordinate of the two lines intersection:

-0.5494∙x + 14.192= -0.0049∙x +4.94 x = 16.99 mL

The first part of the curve (squares) corresponds to hydrochloric acid titration, whereas the second one

(diamonds) to maleic acid. Therefore, there are 16.99·8·cNaOH hydronium ions coming from

hydrochloric acid in flask P:

8∙cNaOH∙xNaOH = 8∙0.09371∙16.99 = 12.74 mmol of hydrochloric acid

Taking into account the total number of hydronium ions millimoles determined from the titration with

phenolphthalein one may calculate that in flask P there are also

8∙cNaOH∙(VNaOH – xNaOH)/2 = 4∙0.09371∙(32.70-16.99) = 5.89 mmol of maleic acid

formed as a result of malonyl dichloride hydrolysis.

Ad e. Basing on the results from point d. one may conclude that the sample of technical malonyl

dichloride contains hydrochloric acid as a contamination because 2·xNaOH > VNaOH. If 2·xNaOH = VNaOH

the sample would be pure malonyl dichloride and for 2·xNaOH < VNaOH the sample would be polluted

with maleic acid. The number of hydrochloric acid additive millimoles was equal to:

29

nHCldom = 8∙cNaOH∙(2∙xNaOH - VNaOH) = 8∙0.09371∙(2∙16.99-32,70) = 0.96 mmol

If the maleic acid had been the additive, the number of such an additive millimoles would have been:

nmal dom = 4∙cNaOH∙( VNaOH – 2·xNaOH)

Ad f. In case of 2·xNaOH > VNaOH the number of maleic acid millimoles is equal to the number of

malonyl dichloride millimoles in the hydrolysed sample. The mass of technical malonyl dichloride

sample equals::

nClmal∙ MClmal+ nHCladd∙MHCl = 5.89∙140.92 + 0.96∙36.46 = 865.0 mg

The weight percentage of malonyl dichloride in the technical sample amounts to 95,95%.

If 2·xNaOH < VNaOH the number of malonyl dichloride moles is equal to the half of the hydrochloric

acid moles number in flask P.

SOLUTION OF LABORATORY TASK 2

An exemplary set of substances

Ampoule number 1 2 3

Coloured mixture CI 42051 dye Ethanol CoCl2 Ethanol-acetone Co(SCN)n

Ad a. The coloured mixtures presented in the attached figures exhibit varied sensitivity of colour

change with respect to the solvent composition. The system containing the CI 42051 dye is the least

sensitive to water content, the mixture containing thiocyanate cobalt(II) complex is more sensitive and

the ethanol solution of cobalt(II) chloride is the most sensitive. The last solution is also sensitive to

the methanol content but this solution changes colour to pink at room temperature and to blue after

heating (this is an example of thermochromism).

I take ca. 1 mL of every solution to dry test tubes and add a few drops of water to every test tube. The

first solution does not change colour despite the addition of ca. 1mL of water. The second one is

discoloured after addition of 1 drop of water, while the third one after addition of 5 drops of water.

The second solution is the only one to be discoloured after addition of ca. 1mL of methanol (it

becomes faintly pint) and after heating in hot water the blue colour comes back. One of the solutions

changes colour upon hydrochloric acid addition which confirms the identification of CI 42051 dye.

The observed phenomena allow one to conclude that there is CI 42041 dye solution in ampoule 1,

cobalt(II) chloride solution in ampoule 2 and thiocyanate cobalt(II) complex solution in ampoule 3.

Ad b. The calibration curve graphs for the measurements of absorbance at 625 nm for the thiocyanate

cobalt(II) complex in ethanol-acetone solution and cobalt(II) chloride in ethanol are shown in the

figure below.

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For the cobalt(II) chloride in ethanol system the dependence of relative absorbance decrease as a

function of water content is not linear in the whole studied range, but it is linear for water content in

the range 0-1%. The slope of the line is 0.5464. For the thiocyanate cobalt(II) complex in ethanol-

acetone mixture system the calibration curve is linear in the range 0-15% with a slope of 0.061.

Ad c. The slope of the calibration curve determines the spectrophotometric method sensitivity. The increase

in water content by 1% causes a relative decrease of absorbance of 0.546 unit for the cobalt(II) chloride in

ethanol system and 0.061 unit for the thiocyanate cobalt(II) complex in ethanol-acetone system. The

quantification limit in the linear range of the curve is 0.05% water content for ethanol with cobalt(II)

chloride and a much higher value of 0.71% water content for the thiocyanate cobalt(II) complex in ethanol-

acetone mixture system.

Ad d. A sample of 2.23 g was mixed with a few portions of ethanol and the resulting solution was

collected in a 25-mL volumetric flask. The flask was filled with ethanol up to graduation mark and filtered

to a 10-ml test tube. 5 mL of filtrate were collected, ethanol solution of cobalt(II) chloride was added and

ethanol was added up to 10 mL. After mixing the solution absorbance was measured at 625 nm.

Absorbance of the anhydrous solution was 0.785 and that of the studied solution 0.572. The relative

decrease of absorbance was 0.271. I calculate the water concentration in the solution basing on the

calibration curve equation and obtain 0.48%. Therefore, there are 48 mg of water in 10 mL of solution.

There are 25/5 as much water in the whole sample which is 240 mg. The water content in the whole

sample is 0.240·100%/2.23 = 10.8%.

I will determine the water content in sample R by measuring the absorbance of a solution resulting from

the addition of 2 mL of cobalt(II) thiocyanate solution and complementing the solution with acetone-

ethanol mixture up to 15 mL. The absorbance of anhydrous solution was 1.035, while that of the studied

solution 0.572. The relative absorbance decrease was 0.355. The water content calculated from the

calibration curve is 6.2% and the water mass in sample R is 0.062·15 = 0.93 g.

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Ad e. The decrease in cobalt(II) chloride ethanol solution absorbance caused by the presence of

methanol could serve as a basis for the methanol content determination in ethanol. However, the result

would be strongly affected by water. Heating of the solution could be a way to check whether that

absorbance decrease was caused by water. In case of methanol the blue colour returns which does not

take place in the presence of water.