Extraction of a metal into the organic phase is possible only if the solubility of compounds of this metal in an organic solvent is higher than in water. In real systems, metal exists in the form of various compounds. It should be taken into account that during extraction, forms may be formed that were not present in the original solution. Therefore, first of all, it is necessary to establish in what form the metal is extracted and what its solubility is. The solubility of any compound depends on many factors: the nature of the substance, temperature and pressure. Typically, chemically similar substances dissolve better in each other than in substances of a different structure. At the same time, similarity should not be understood too narrowly, since the presence in molecules of groups that are identical or similar in behavior is often sufficient. Solvent molecules enter into energetic interactions with dissolved molecules, primarily of the electrostatic type, since the molecules of most solvents have electric dipole moments.
One of the conditions for extraction is charge neutralization. Charged compounds cannot pass into an organic solvent. The metal ions present in the solution must be converted into an uncharged complex or into an ionic associate with a suitable ion of the opposite charge. The magnitude of the ion charge plays a significant role in the extraction of ionic associates. In this case, singly charged ions are best extracted into the organic phase, while doubly and especially triply charged ions are extracted worse. In addition, the extracted compound must be hydrophobic and not contain hydrophilic groups, such as hydroxyl or carboxyl groups.
Extraction as a chemical reaction (basic laws and quantitative characteristics).
Basic quantitative characteristics of processes extraction
extraction as a separation method has been used for a long time in analytical chemistry and chemical technology; the theoretical foundations of this method remained unstudied for a long time. In particular, the basic quantitative characteristics of extraction processes remained unstudied for a long time, which was a definite obstacle to the widespread introduction of extraction into practice. To calculate the amount of a substance that is extracted with organic solvents, it is necessary to know the constant and distribution coefficient, the degree of extraction, etc.
M. Berthelot and J. Jungfleisch were the first researchers who, in 1872, based on experimental data, showed that the ratio of the equilibrium concentrations of a substance distributed between two liquid phases is constant. This relationship was derived thermodynamically by W. Nernst, who formulated the distribution law in 1891.
According to the law of distribution, a substance dissolved in two immiscible or limitedly miscible liquids is distributed between them in a constant ratio. This ratio for ideal systems depends only on temperature and the nature of the substance and does not depend on concentration. The distribution law is valid only if the substance being distributed in both phases is in the same form.
Constant distribution substances. A constant quantity expressing a relationship concentrations distributed substances, located in both phases (after the onset of equilibrium) in the same form is called constant distributions:
P 0 = [A] 0 / [A] B (1)
where R o - constant distribution: [A] o -concentration substances in the organic phase solvent, mole/l; [A] B - concentration substances in the aqueous phase, mole/l.
Magnitude constants distribution depends on the nature of the distribution substances, composition and properties of the extractant used, temperature, at which it is produced extraction. This constant does not depend on equilibrium concentrations extractable substances and volumes of aqueous and non-aqueous phases.
Distribution coefficient. When calculating constants distribution substances according to formula (1), you need to be sure that the distributed substance in both phases it is in the same form (in the same molecular state). However, in many extraction systems the above condition is not met. In one of the liquid phases may occur dissociation, association, solvation, hydrolysis distributed substances, formation of complexes, etc. For calculations of extraction equilibria in such systems the form of existence is not taken into account substances in each phase, and only the ratio of total (analytical) concentrations distributed substances in both phases.
On basis determination of total concentrations can not be calculated constant, and the distribution coefficient given substances in the applied system solvents (water- organic solvent). The distribution coefficient is the ratio of the total analytical concentrations substances in the organic phase solvent to the total analytical concentrations this substances in the aqueous phase (without taking into account the form in which it is substance in each phase):
D = C 0 / C B (2)
where D is the distribution coefficient; C about - total analytical concentration substances in the organic phase solvent, mole/l; C B - total analytical concentration substances in the aqueous phase, mole/l.
Degree extraction. Degree extraction(percent extraction) is the ratio of the amount of extracted substances to the total (initial) amount of this substances in water solution:
R = A 100 / N (3)
where R is the degree extraction substances, %; A - quantity substances, which was extracted with organic solvent; N - total (initial) quantity substances in water solution.
Classification of extraction processes.
Extraction systems are very diverse. The classification of extractable compounds according to the type of compound passing into the organic phase is considered. Classification according to this criterion was proposed by Zolotov Yu.A.:
Extraction of non-polar and low-polar substances,
Extraction of complex metal acids,
Extraction of intracomplex compounds (ICCs).
VKS are formed during the interaction of metal catons with organic
reagents, one of the active groups of which must contain a mobile
hydrogen atom replaced by metal during complex formation, second
(third, etc.) can also be acidic or, more often, basic.
The extraction of VKS is influenced by such factors and parameters as the acidity of the aqueous phase, the concentration of the reagent, the distribution constants and dissociation constants of the reagent, the stability constants and distribution constants of the complex, competing reactions in the aqueous phase, the presence of electrolyte salts, element concentration, temperature, and solvent.
A convenient classification is one that takes into account the type of compound in which the extracted element passes into the organic phase. Moreover, all compounds can be divided into two large categories - non-ionized compounds and ionic associates. These compounds differ not only in their chemical composition, but also in the mechanism of their formation and transition to the layer of organic diluent. Some of them pre-exist or are predominantly formed in an aqueous solution, so pure diluents are usually used to extract them. Others, on the contrary, are formed during the extraction process itself due to the interaction of the reagent located in the organic phase with metal cations contained in the aqueous solution.
The considered classification of extraction products allows us to identify certain groups of extractants and solvents that are similar in electronic structure and the nature of electronic interaction in extraction systems. According to this classification, extractants can be divided into three groups:
1. Hydrocarbons: saturated aliphatic (hexane, octane), unsaturated aliphatic (pentene, hexene), aromatic (benzene, toluene).
2. Compounds whose molecules contain one functional group of atoms: alcohols, ethers, esters, ketones, nitro compounds, halogen derivatives of hydrocarbons (chloroform, carbon tetrachloride, chlorobenzene), sulfur-containing compounds (carbon disulfide, thiophene).
3. Compounds that contain more than one functional group of atoms: diethyldithiocarbamates, 8-hydroxyquinoline, etc.
Possibilities of practical use of extraction.
The separation of mixtures of elements is carried out primarily using selective extractants. For example, it is not difficult to separate mercury and bismuth in the form of dithizonates from zirconium and aluminum, since neither zirconium nor aluminum react with dithizone at all. A more typical case is when the shared elements are, in principle, all extracted, but not equally. In this case, another separation technique is used, which is based on varying concentration conditions: pH, concentrations of system components, including the extractant. Separation is also achieved by changing the oxidation state of the elements. For example, when separating gallium and iron with amines, the effect is achieved by reducing iron to a non-extractable divalent state. Gallium then passes into the organic phase. To improve separation during extraction, masking agents are introduced into the aqueous phase.
Extraction is widely used in chemical, oil refining, food, metallurgical, pharmaceutical and other industrial fields, as well as in analytical chemistry and chemical synthesis.
Conclusion.
The applications of extraction are expanding rapidly. Currently, we can name analytical chemistry, radiochemistry, nuclear technology, technology of non-ferrous and rare metals. In addition, it is necessary to note the great importance of extraction for preparative and analytical purposes in scientific research, for example, when studying complex formation processes and the state of substances in solutions. The development of extraction methods has reached such a stage that it is now possible to extract any element or separate any pair of elements by using certain extraction systems or choosing appropriate extraction conditions. To predict the extraction ability of various compounds, the achievements of thermodynamics, coordination chemistry, solution theory, and organic chemistry are used. Therefore, the study of extraction systems contributes to the development of chemistry in general.
Literature
Zolotov Yu.A. Extraction in inorganic analysis. M.: Moscow State University Publishing House, 1988. 82 p.
Zolotov Yu.A. Extraction of intracomplex compounds. M.: Nauka, 1968. 313 p.
www.pereplet.ru/obrazovanie/stsoros/790.html
Task
Write the equations for the anodic and cathodic processes, as well as the overall equations for the electrolysis of sodium chloride solution and melt. What substances will be formed if the process is carried out with stirring? Write the reaction equation, draw up an electron-ion balance. How long does it take to carry out electrolysis with a current of 2000 A to obtain 5 kg of sodium, the current efficiency is 87%? How many liters of chlorine will be released? Calculate the isotonic coefficient of a 15% sodium chloride solution if it freezes at -10.4 o C, and its degree of dissociation.
aCl Na + + C -K(-): Na + +е=Na 0 - oxidizing agent
A(+): 2Cl - – 2е= Cl 2 - reducing agent
2Na + + 2Cl - = 2Na + Cl 2
aCl Na + + Cl -K(-): 2H 2 O +2е= H 2 +2 OH - - oxidizing agent
A(+): 2Cl - - 2е = Cl 2 - reducing agent
2H 2 O + 2NaCl = H 2 + Cl 2 + 2NaOH
With stirring: Cl 2 + 2NaOH = NaClO + NaCl + H 2 O
When mixing two liquids they can be:
Unlimitedly soluble, i.e. dissolve in each other in any proportion;
Practically insoluble;
Limited soluble.
Mutual solubility depends on the chemical structure of liquids, which in turn are divided into polar and non-polar.
It was already noted by alchemists that “like dissolves in like,” i.e. Polar liquids dissolve polar liquids well, and nonpolar liquids dissolve nonpolar liquids well.
For this reason, water, a polar solvent, dissolves polar liquids (acetic acid, ethanol) well and does not dissolve non-polar liquids at all (benzene, hexane, kerosene, gasoline, vegetable oil, etc.).
If liquids differ slightly in polarity, then they dissolve in each other to a limited extent, forming two-layer systems, for example, water - aniline.
If a third substance capable of dissolving in each of them is introduced into a system consisting of two practically insoluble liquids, then the solute will be distributed between both liquids in proportion to its solubility in each of them.
It follows from this distribution law, Whereby the ratio of concentrations of a substance distributed between two immiscible liquids at a constant temperature remains constant, regardless of the total amount of solute.
WITH 1 /WITH 2 = k,
Where WITH 1 and WITH 2 – concentration of the dissolved substance in the 1st and 2nd solvents;
k– distribution coefficient.
The distribution law is widely used in processes extraction – extracting a substance from a solution with another solvent that is immiscible with the first. The distribution law allows you to calculate the amount of substance extracted and remaining in solution after single or multiple extraction of a given volume with a solvent at a given temperature:
Where m 1 – mass of the substance remaining in solvent 1 after its single extraction with solvent 2;
m o – initial amount of substance in solvent 1.
V 1 and V 2 – volume of solvents 1 and 2;
When extracted multiple times, Equation 1 becomes:
Where n– number of extractions.
When extracting, it is never possible to completely extract the substance completely. But the completeness of extraction will be greater if the solution is treated repeatedly with small portions of the solvent, each time separating the resulting extract, than if the solution is treated once with a large portion of the solvent.
Extraction is used in many areas of technology and laboratory research. Extraction is based on the extraction of sugar from beets, oils from seeds, many substances in food processing (passivation of vegetables), and the production of pharmaceuticals. Thus, penicillin and a number of other antibiotics cannot be concentrated by evaporation, since they are destroyed when heated. To obtain concentrated solutions of antibiotics, extraction is carried out with butyl or ethyl acetate.
The extraction process can almost always be described as a normal, albeit two-phase, chemical reaction. Extraction reactions are almost always reversible, so the law of mass action can be applied to extraction processes. It becomes possible to talk about the equilibrium constant of the extraction reaction, which in this case is called the extraction constant. The extraction process of substances capable of dissociating can be represented as follows:
The subscripts denote the organic (o) and aqueous (a) phases. The organic reagent (HA) is soluble in an organic solvent. The concentration equilibrium constant of this reaction (extraction constant - Kex)
The expression for the extraction constant can be written only if the composition of the extracted compound of the original components is known. Along with the law of mass action, the distribution law is applicable to extraction systems, according to which, at constant temperature and pressure, the ratio of the equilibrium concentrations of a substance in two immiscible phases is a constant value that does not depend on the total concentration of the substance. This quantity is called the distribution constant:
where [A]о and [A]в are the equilibrium concentrations of the substance in the organic and aqueous phases. However, the extractable substance is usually present in different forms. Compounds can participate in various chemical transformations: complexation, hydrolysis, polymerization, dissociation, etc. In this regard, the ratio of the total concentrations of the substance in the organic and aqueous phases is not constant. For each individual form of a substance, the distribution law must be satisfied, that is, [A]o / [A]b = . It follows from this that quantitative characteristics are needed that could be measured in direct experiment. This is primarily the distribution coefficient D, which is the ratio of the total concentration of a substance in the organic phase to its total concentration in water:
The value of D can be calculated by determining any suitable method of concentration and. When extracting metals, the methods of atomic emission and atomic absorption spectrometry, voltammetry, spectrophotometry, and the method of radioactive indicators are used for this purpose. In contrast to the distribution constant, in the case of the distribution coefficient there is no requirement for constancy and equality of the forms of existence of the substance in general phases and no requirement for the system to be in equilibrium.
The distribution coefficient can also be written as follows
The distribution coefficient describes the ability of a substance to be extracted, but does not determine the actual completeness of extraction, which depends on the ratio of the volumes of the organic and aqueous phases. At the same distribution coefficient, the more completely the substance is extracted, the larger the volume of the organic phase (at a constant volume of aqueous). The proportion of extracted substance is expressed by the degree of extraction:
where R is the degree of extraction of a substance from the aqueous phase into the organic phase; Cw and Co - the amount of substance in the organic and aqueous phases. The recovery rate is most often expressed as a percentage.
where Vo and Vв are the volumes of organic and aqueous phases. In the case of equality of phase volumes (Vо = Vв) we obtain
For good separation, it is not enough that the separation coefficient be high. It is also necessary that the product of their distribution constants be close to unity. In practice, the concentration factor (S AB) is also used:
Important factors affecting extraction include phase contact time. The practical importance of the issue is associated, first of all, with the fact that in many extraction systems equilibrium is not achieved immediately. The extraction rate depends on the rate of chemical reactions occurring in the system, in particular on the rate of mass transfer of the substance between the two phases. In this case, to speed up extraction it is necessary to use various factors. If mass transfer is the slowest, the phase mixing speed should be increased. The rate of chemical reactions can be influenced by increasing the concentrations of interacting components, suppressing interfering reactions, such as hydrolysis and polymerization.
Thus, in practice, when developing extraction methods, it is necessary to study the reaction rate. To do this, you need to set the time to reach equilibrium. This is usually done using kinetic curves, for example, the dependence of the distribution coefficient on the time of phase contact (Fig. 1). Studying the dependence of the degree of extraction on the time of phase contact may not provide correct information about the establishment of equilibrium if the distribution coefficients are sufficiently high, as can be seen from Fig. 2. Although the partition coefficients continue to increase, that is, there is no equilibrium, the recovery rate is 100% and the system may appear to be in equilibrium.
Rice. 1.
Rice. 2.
One of the interesting questions in the theory of extraction is the question of where the extracted compound is formed - in the aqueous phase, in the organic phase, or at their interface. This may be different in different systems, but studying the extraction rate, for example with different solvents, allows us to resolve this issue.
One of the solutions can be to study the kinetics of extraction of intracomplex compounds with various solvents. Solvents must be selected in such a way that the distribution constants of the reagent () in them differ. The extraction conditions must be such that the rate of chemical interaction is kinetically determining. In this case, the pH and concentration of the reagent must be the same. If the compound is formed in the aqueous phase, the higher the equilibrium concentration of the organic reagent in the aqueous phase, the higher the extraction rate. The lower, the greater the concentration. In other words, the less soluble the extractant is in an organic solvent, the greater the extraction rate. If the compound is formed at the boundary, then the better the reagent is soluble in the organic solvent, the greater the extraction rate will be. The formation of a compound in the organic phase is hardly possible, since for this to happen the extracted element must pass into the organic solvent in some other way.
The article provides a comparative description of the extraction properties of some polar organic solvents (ethylene glycol, DMF, DMSO, acetonitrile) and ionic liquids (N-methylbutylimidazolium acetate and N-butylpyridinium tetrafluoroborate). The introduction outlines the advantages of ionic liquids as solvents - practically zero vapor pressure and extremely low solubility in aliphatic hydrocarbons. And since most ionic liquids contain an organic cation capable of coordinating electron-donating particles, they can be used as extractants of polar and electron-donating organic substances. In this work, a multicomponent standard solution of PAHs was studied. They were separated by HPLC. The value of the distribution constant of substances P was calculated using the following formula
where and is the peak area in the chromatogram of the equilibrium octane solution and the initial chromatogram;
and - volumes of polar and octane phases.
It was found that the distribution constant decreased with increasing number of aromatic rings.
Despite all the advantages, it is irrational to use ionic liquids for the extraction of PAHs, since they are inferior in properties to DMF and DMSO. DMF and DMSO are widely used in extraction. They have electron-deficient centers in their molecules, which contributes to the effective solvation of PAH molecules. As a result, p-complexation and solvation of nuclear protons occur. Due to this property, DMF and DMSO are used to purify various extracts.
The separation efficiency is also affected by the addition of salts. The work examines the extraction of aromatic hydrocarbons. To study the extractability of PAHs at 293 K, extraction systems such as n-heptane - a solution of N,N"-butylmethylimidazolium chloride in methyl alcohol with organic salt concentrations of 1, 2, 3, 4, 5 mol/l were studied. With increasing concentration of N,N chloride "-butylmethylimidazolium increases the structure of the polar phase (the increment of the methylene group increases by approximately 0.1 units). PAHs behaved as follows. With increasing salt concentration, there is a significant drop in the distribution constants of aromatic hydrocarbons. In this case, the distribution constants of multi-ring arenes drop most sharply, and the series of extractability of hydrocarbons with increasing number of aromatic rings (the more rings, the smaller the distribution constants) corresponds to the series for the most active and selective solvents of aromatic hydrocarbons - DMF and DMSO.
The observed effects are due, on the one hand, to specific solvation effects between the salt and the arenes, which can be caused by p-complexation, hydrogen bonds, etc. and the effect of pushing out hydrophobic hydrocarbons by the polar phase, on the other hand. The presence of two oppositely directed effects explains the presence of weakly expressed minima in the dependence of the distribution constants of PAHs on the concentration of N,N"-butylmethylimidazolium chloride in methyl alcohol.
Based on the results of this work, we can conclude that with an increase in the concentration of N,N"-butylmethylimidazolium chloride in methyl alcohol, there is a significant increase in the ratio of the distribution constant of n-heptane to the distribution constants of PAHs, which leads to an increase in the separation capacity of this extraction system.
Third year students of the Faculty of Technology
To perform laboratory work
In physical chemistry on the topic:
“The law of distribution. Liquid extraction"
GOAL OF THE WORK: study of the liquid extraction process of one of the main processes of food technology.
OBJECTIVE OF THE WORK: acquisition of practical skills in conducting laboratory research, processing experimental data using analytical and graphical methods.
INTRODUCTION
One of the special cases of three-component systems is the following: two liquids are mutually insoluble, and the third component is able to dissolve in both. When equilibrium is reached in the system, the third component is distributed between two immiscible liquid phases in a certain ratio, qualitatively characterized by a distribution constant.
The most famous example of the application of the distribution law is extraction, i.e., the extraction of a substance from a solution with a suitable solvent that does not mix with the first and at the same time dissolves the extracted substance in a larger quantity than the first. For many organic substances such a solvent is ether, and for inorganic substances it is water. To complete the extraction, the extracted substance is transferred to the molecular state in which it is found in both phases. For example, when extracting a weak organic acid, it is advantageous to reduce the degree of its dissociation by adding a mineral acid. In this case, undissociated organic acid molecules are extracted more completely. The solubility of organic substances in water is significantly reduced in the presence of salts - the effect of salting out.
Extraction is one of the most common processes in the chemical, pharmaceutical, food and other industries. Extraction is widely used to extract essential oils from plant materials and purify them. In winemaking, the extraction process is used to obtain raw materials for the production of vermouth and the processing of grape pomace.
I.THEORETICAL PART
1.1. Law of distribution.
If a substance is soluble in two immiscible liquids, then when it is dissolved in a mixture of two such liquids, it is distributed between them in the ratio determined by the Nernst-Shilov law:
The solute is distributed between two immiscible liquids in a constant ratio of concentrations, independent of the amount of solute added.
K=C1/C2(1), where C1 and C2 are the concentrations of the substance in the 1st and 2nd solvents.
The law is satisfied at a constant temperature, sufficient dilution of solutions, and in the absence of interaction of the solute with the solvent.
In real conditions, when substances are dissolved, processes of association and dissociation of molecules of the dissolved substance occur.
Case 1. The substance is associated in one of the solvents due to the formation of hydrogen bonds with the appearance of dimers:
Or polymers
The distribution law in this case will look like this:
K=C1N/C2 or 2)
Here N=M1/M2– the ratio of the average molecular weights of a substance in one and another solvent.
If we transform equation (2) and take logarithm, we get
Lg K = n lg C1 – lg C2 or
Lg C2 = n lg C1 – lg K (3)
- this is a direct relationship LgC2 =F(LgC1)
The equation allows you to graphically determine N And TO from experimental data.
Case 2. In one of the solvents (most often water) the substance is dissociated, and in the other (organic) it is associated.
The distribution law takes the form:
A-degree of dissociation.
Example 1. When distributing phenol between water and benzene, the following data were obtained:
C1(H2O), kmol/dm3 0.0316 0.123 0.327 0.750
C2(C6H6), kmol/dm3 0.0077 0.159 0.253 0.390
Calculate value TO And N graphically.
The different solubility of particles of one phase in another is due to the forces of intermolecular interaction. If the particles in the phases are connected to each other by identical intermolecular forces, then they dissolve when mixed without limit (for example, benzene - alcohol); in the case of different intermolecular nature of the bonds between particles (for example, covalent in one and ionic in the other), the solubility of the phases is either limited (benzene - water) or practically impossible (water - mercury).
For three-component liquid systems in which two components are insoluble or sparingly soluble, the third component is distributed between both insoluble liquid phases at a constant temperature so that the ratio of its concentrations in these two phases remains constant. This position is the essence of the distribution law established in 1890 by Nernst and supplemented by the research of the Soviet scientist N.A. Shilov.
This law, which characterizes the equilibrium state of the system, can be derived as follows. Let us assume that at the same temperature there are two immiscible liquids (for example, water and chloroform), we introduce a third component into this system - iodine, which will dissolve in both phases.
Chemical potentials of substances in two equilibrium phases
oh, they should be the same. This means that if in one phase δ (water) the chemical
The potential of dissolved iodine is equal to:
and in the other phase β (chloroform):
then µ () µ ()
|
|
|
|
|
|
|
|
o() o()
Standard chemical potentials at a given temperature are
We are permanent. But in different phases they naturally have different meanings
). The ratio of active iodine concentrations in different phases is
is a constant value equal to:
o() o()
The ratio of active concentrations of the third component (in our example, iodine) in different immiscible phases is denoted by the letter k and is called the distribution coefficient, i.e.
() kor concentration ratio:
|
|
The distribution coefficient depends on the nature of the substances, composition
lying the system, and temperature.
These simple relationships are valid only in cases where the addition of a third substance to a system of insoluble or limitedly soluble liquids does not change the solubility of the liquids to each other, and also if the solute in each of the equilibrium phases is in the same and in the same state (in the form of the same particles). If the substance distributed between two phases dissociates or associates, then a complex equilibrium is established between simple and associated molecules or ions within each phase and between phases. In general, to describe the distribution of substances
between equilibrium liquid phases they use the equation of N.A. Shi-
Lova and L.K. Lepin
where n is the association coefficient.
For example, if n = 2, this means that in the δ phase the substance exists in the form of single molecules, and in the β phase it forms dimers. When taking the logarithm of this equation, it takes the following form:
logk = n loga (δ) – loga(β).
To find the constants n and k, construct a graph in coordinates: lga(δ) – lga(β). The angular coefficient of the straight line is equal to k, and at the point of intersection of the straight line with the ordinate axis - the value of lgk.
Equations relating the distribution coefficient to concentration -
mi of the third component are derived using simple reasoning.
The third component dissociates in one of the phases. Let the substance CA, consisting of the K+ cation and the Aˉ anion, be distributed between two solvents δ and β (Fig. 40, a). In solvent δ, the substance CA dissociates into K+ and Aˉ ions, and in solvent β remains in the form of molecules. The concentration of ions and undissociated molecules in the δ phase are related by the equation of the law of mass action. If the analytical concentration of a substance CA in phase δ is equal to C, and its degree of dissociation is equal to α, then the concentration of K+ and Aˉ ions is equal to αC, and the undissociated part of CA is: (C – αC). Dissociation constant for bi-
normal electrolyte is equal to
|
|
K dis. C (1 ) . (3.12)
The undissociated part of the CA substance in the δ phase is in equilibrium with the CA substance in the β phase. This equilibrium is characterized by the coefficient
distributions (3.9)
where C (δ) and C (β) are the concentrations of the CA substance in the δ and β phases, respectively.
C (δ) = C (1 – α),
k C (1 ) . (3.13)
According to the expression of the dissociation constant, the concentration
substance of the spacecraft in phase δ will be equal to
C () C (1 )
and the distribution coefficient:
|
|
|
|
|
|
Association of the third component in one of the phases. Let the substance M be distributed between phases δ and β. But in the β phase it is in the associated state M2, i.e., the molecules combine into a particle consisting of two molecules (see Fig. 40, b). Concentration of associated molecules and
single molecules M are interconnected by a simple relation of the law
β
β
Fig.40. Equilibria between different forms of distributed substance:
a – in the case of dissociation in one of the phases; b – in case of association.
acting masses. If α is the degree of dissociation of M2 into M molecules, and the analytical concentration of the substance M2 in this phase is C, then the concentration of M2 is C(1 – α), and the concentration of M is 2αC. And then the dissociation constant
Kdis. C (1 ) ,
the concentration of simple molecules is equal to
()
С 2С
To dis.С(1 ) .
The concentrations of simple molecules in the δ and β phases are related by the distribution coefficient:
Kdis.C(1)
Similar reasoning should be applied when deriving equations for cases of simultaneous association in one phase and dissociation in another or in the interaction of a solute with a solvent (chemical interaction).
An experimental study of distribution coefficients can serve to determine the degree of association or dissociation of a dissolved substance in a particular solvent, the equilibrium constant of a reaction occurring in one of the phases, etc.