EntranceLaw of distribution in the extraction method. Basic laws and quantitative characteristics of extraction
If you take two immiscible liquids and add a third component, it will dissolve to varying degrees in both solvents.
“When equilibrium is established, the ratio of concentrations of the resulting solutions is constant at a given temperature”- Nernst distribution law (1.13).
Where 
- concentration of the third component in phases I and II; K - distribution coefficient.
If a solute dissociates or associates in one of the solvents, then the Nernst equation has the form:
= K, (1.14)
To find Kip We take the logarithm of equation (1.14) and obtain the equation of the straight line:
By constructing a straight line in coordinates 
, let's find " P» as the tangent of the angle of inclination of a straight line (by any two points lying on a straight line) tga
=
.
InK can be found from the equation by substituting into it the values of any point located on the line.
Nernst's distribution law underlies the extraction process. Extraction is the removal of a component from one phase to another. Extraction happens solid phase- extraction of substances from the solid phase into the liquid phase (for example, brewing tea, coffee, preparing tinctures, herbal extracts, etc.) and liquid phase- extraction of a solute from a liquid solution extractant. A solution of the extracted substance in the extractant is called extract, and the original solution after extracting the substance from it is called raffinate.
To calculate the efficiency of liquid-phase extraction, use equation (1.15):
,
(1.15)
Where X- the proportion of unextracted substance in the raffinate;
V- volume of the initial solution;
- extractant volume;
TO- distribution coefficient
;
P- number of extractions.
As can be seen from equation (1.15), the more extractions, the less substance remains in the raffinate, that is, unextracted, the more substance is extracted by the extractant. The efficiency of extraction is largely determined by the value of the distribution coefficient: the greater the coefficient in favor of the extractant, the more effective the extraction.
Often, the reference book gives the distribution coefficient as the ratio of concentrations in the raffinate to the extract, that is, the inverse value of what should be in equation (1.14). In this case, you should take the inverse reference value and use it in equation (1.15), or use another equation, where
1.10 Examples of problem solving
Problem 1
50 g aqueous solution FeCl 2 with a molality of 1.97 mol/kg with a density of 1.332 g/cm 3 mixed with 150 ml of 5 wt. % solution FeCl 2 with a density of 1.038 g/cm. Express the composition of the resulting solution in all possible ways.
Given:
Solution 1: Solution 2:
=50r 
= 150 ml
b
=1.97mol/kg 
= 1.332 g/cm 3 
= 1.038 g/cm 3
b 3 = ? W 3 = ? X 3 = ?
Solution:
For each solution we find the mass FeCl 2 And N 2 0 .
Solution 1
:
b
= 1.97 mol/kg, that is, 1 kg of water accounts for 1.97 mol FeCl 2
,
= l.97M FeCl 2
= 1.97∙127.5 =251 g; 
= 1000 g; m p . pa = 1251 g
1251 g (solution) - 251g FeCl 2 ;
50 g solution - 
X FeCl 2
G.
So, in solution 1: 
= 10 g;
= 50 – 10 = 40 g.
Solution 2
: m r. ra =∙V = 1.038 ∙ 50 =156 g; 
= 156-8 = 148 g
Solution 3
: m р- pa = 50 + 156 = 206 g; 
=10 + 8=18g
m H 2 O =40+ 148 = 188 g
b
Problem 2
At 100 °C, tetrachlorethylene has a saturated vapor pressure of 400 mmHg, and bromobenzene - 196 mmHg. Find the composition of a solution boiling at 100 °C under a pressure of 360 mmHg, and the composition of the saturated steam over the solution. Consider the solution ideal.
=400 mmHg
= 196 mmHg
P total = 360 mmHg.
Solution:
Boiling occurs when external pressure and saturated vapor pressure are equal. Therefore, 360 mmHg. - this is both external pressure and the total pressure of saturated vapor above the solution.
;
Problem 3
At 25 °C, the distribution constant of iodine between water and amyl alcohol is 0.00435. How many grams of iodine will remain in three liters of an aqueous solution (C = 1.3 g/l) after double extraction with amyl alcohol, if a total of 400 ml of alcohol is consumed?
K = 0.00435 = 
=1.3g/l
= 400/2 = 200 ml = 0.2 l
=
?
Solution:
The extractant is amyl alcohol. Therefore, this distribution constant is the inverse of that required in the equation for calculating extraction efficiency:
We calculate the distribution constant:
- the proportion of iodine remaining in the raffinate.
Mass of dissolved iodine in the original solution:
m = m 0 x = 3,9 . 0.0037 = 0.014 g - mass of iodine remaining in the raffinate
Problem 4
A closed container with a volume of 5.0 L contains gas N 2 S at a temperature of 20 0 C and a pressure of 740 mm. rt. Art. What volume of water must be added so that the partial pressure N 2 S dropped to 500 mm. rt. Art.? Absorption coefficient N 2 S(α) is equal to 2.58.
= 5.0 l = 5.0∙10 -3 m 3
T 1 = T 2 = 293 K
P = 740 mm Hg = 0.984∙10 5 Pa
- ?
Solution:
The decrease in pressure occurs due to the dissolution of part N 2 S in water. Let's calculate the initial quantity N 2 S and residual quantity N 2 S after dissolving it in water.
We neglect the change in gas volume due to the addition of water, that is, V 2 = V 1.
Let's calculate the amount of H 2 S dissolved in water:
Let's calculate V 0 - the volume of H 2 S, reduced to normal conditions (P = 1.0133∙10 5 Pa, T = 273.15 K), which is dissolved in water:
Problem 5
Methane melts at a temperature of 90.5 K, its heat of fusion is 70.7 J/mol. Determine the solubility of methane in liquid nitrogen at a temperature of 50 K.
Solution:
We usually know methane as a gas, but in this case we are talking about solid methane at a temperature below its melting point. Therefore, it is necessary to find the solubility of a solid in a liquid, which can be done using the Schroeder equation:
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 chemical structure 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 ones 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, and many substances during processing food products(passivation of vegetables), 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.
If a third liquid capable of dissolving is added to a system of two mutually insoluble liquids, then the added liquid will be distributed between both liquid layers.
The addition of a third component can increase or decrease the mutual solubility of the components. This property of binary systems can be used for analytical purposes. Big practical significance have three-component mixtures in which the solute is distributed between two immiscible solvents.
In a state of equilibrium, the chemical potential of the dissolved distributed substance (m) in both immiscible layers should be the same, i.e.
It is known that m = m 0 + RTlna,
where m 0 is the standard chemical potential of the solute,
a is the activity of the component in solution.
Then the equilibrium condition can be written
m 3 0 ¢ + RTlna 3 ¢ = m 3 0 ¢¢ + RTa 3 ¢¢,
where one stroke characterizes the first solvent, two strokes characterize the second.
From here 
Considering that m 3 0 ¢ And m 3 0 ¢¢ when T – const are constant, then the last expression can be written as follows:
where K 0 is the thermodynamic distribution constant.
This equation is an expression of the Nernst-Shilov distribution law: the third component added to a system consisting of two mutually insoluble or limitedly soluble liquids is distributed between both liquid layers in a certain ratio, constant at a given temperature.
Because a = c . g
The distribution law equation can be written in another form:
,
where K is the distribution coefficient.
The distribution coefficient is the ratio of the total concentration of a substance in one liquid phase to its concentration in the second liquid phase under equilibrium conditions:
With a change in the concentration of the distributed substance in two equilibrium liquid phases, the distribution coefficient changes.
When c 3 ¢ ® 0 and c 3 ¢¢ ® 0 g 3 ¢ ® 1 and g 3 ¢¢ ® 1, then 
.
This law applies if the substance being distributed in both solvents has the same molecular weight.
Let us recall the cryoscopic and ebullioscopic laws, which require the introduction of an isotonic coefficient, which takes into account the dissociation of molecules or their association.
In such cases, another expression should be used:
if the particle size of the substance being distributed in one solvent (¢) is two times less than in the other (¢¢), then:
,
those. when the substance being distributed in the first liquid layer remains in molecular form, and in the second it dissociates into ions:
AB A + + B – .
And if , then the general expression takes the form: .
Extraction . The distribution law underlies the extraction method of extracting a substance. Extraction is the process of extracting a solute from a solution using a second solvent that is practically immiscible with the first.
This method is widely used in the pharmaceutical industry, non-ferrous metallurgy and rare metals, in nuclear technology and radiochemistry, as well as in nanotechnology.
To increase the completeness of extraction of a substance from the aqueous layer with an organic solvent, extraction is carried out sequentially in small portions of the extractant (n), while larger number successive stages of extraction (n), the greater the completeness of extraction with the same amount of extractant taken. Extraction efficiency can be calculated using the following reasoning and conclusions:
Let's assume that in V 0(l) of aqueous solution is g 0 kg of substance to be extracted. If after the first extraction in aqueous solution remains g 1 kg of unextracted substance, then when equilibrium is established, (g 0 – g 1) kg of substance will pass into the extract. Distribution coefficient.
If a third component, soluble in both liquids, is added to two immiscible liquids, it will be distributed between them in a certain quantitative ratio. This ratio is a constant value and is called the thermodynamic distribution coefficient K:
At chemical equilibrium. The potential of the third liquid in 2 phases will be the same
The chemical potential depends on the activity of the 3rd component in the solution.
μ 3 = μ º 3 + RT ln a 3
μ º(I) 3 + RT ln a (I) 3 = μ º(II) 3 + RT ln a (II) 3
,
Where 
- equilibrium concentration of distributed substances in both the first and second liquid phases.
The equation is called Nernst's distribution law: the distribution of each dissolved substance between two phases is determined by the distribution coefficient, the value of which does not depend on the presence of other substances.
Coef. distribution changes with changing concentration. Distribution of a substance in 2 equilibrium liquid phases.
An important consequence of this law is extraction, i.e. extracting a substance from a solution with a suitable solvent that does not mix with another component of the solution. Extraction can be used to separate multiple substances. To increase the completeness of extraction of a substance from an aqueous layer with an organic solvent, extraction is carried out sequentially in small portions of the extractant (υ), and the greater the number of successive stages of extraction (n), the greater the completeness of extraction with the same amount of extractant taken.
where V 1 is the volume of the extracted solution; V 2 is the volume of the extractant; m 0 is the mass of the component in the initial mixture and m 1 remains after the first extraction;
After n extractions, substances will remain in the first solution 
During extraction with one total volume equal to nV 2, the mass of the substance that will remain in the extracted solution according to Eq. 
From Eq it follows that extraction is n times more efficient than performing one extraction with the same total volume of extractant.
19.Fusibility diagram of 2-component systems.
The fusibility diagram expresses the dependence of the melting temperatures of mixtures on their composition. A special case of fusibility diagrams are solubility diagrams, which represent the dependence of the solubility of solids in a liquid on temperature.
Status of two-component system. determined by 3 parameters (temperature, pressure and concentration of one of the components)
Systems without formation of chemical compounds
I - liquid melt (C=2-1+1=2);
II - liquid melt and crystals of comp. A (C=
III - liquid melt and crystals
Component B (C=2-2+1=1);
IV - crystals A and B (C=2-2+1=1);
Line aEb is called line liquidus– compositions of liquid melts, when cooled to a given temperature, crystallization of pure component A or B begins from the melt.
CED line solidus, below it the liquid cannot exist.
Point E is called eutectic point– it corresponds to a melt that is simultaneously in equilibrium with crystals of components A and B. This melt is called eutectic, and the mixture of precipitating crystals at T e eutectic . The eutectic crystallizes at a constant temperature - system conditionally invariant, because When the pressure changes, the melting temperature and the composition of the eutectic also change. During crystallization of the eutectic, the composition of the liquid melt differs from the composition of each of the solid phases in equilibrium with it.
20(1).Fusibility diagrams for systems with congruently and incongruently melting compounds.
If components A and B can form solid. chem. compound AB, melting without decomposition, those. congruently, then on the phase diagram the liquidus curve forms maxima at point C, when the composition of the crystalline phase coincides with the composition of the liquid. phases. On both sides of point C there are eutectics E and E 1. If the composition of the system is between pure component A and the chemical compound AB, then at the eutectic temperature T E, the melt of composition E coexists with crystals A and AB. If the composition of the system lies between chemical conn. AB and component B, then at T E1 the melt of composition E1 coexists with crystals AB and B. T.o. The considered phase diagram is a combination of 2 phase diagrams with the eutectic A - AB and AB - B.
The process of cooling the melt specified by point M. With this composition, the number of independent computers = 1, because a system can be formed from only one chemical. conn. AB. At T C, AB crystals (Ф = 2) and the number of degrees fall out of the melt. free C=1-2+1=0, i.e. the system is variable and crystallizes at a constant temperature.
Chem. When a certain temperature is reached, the compounds begin to melt, breaking up into crystalline and liquid phases, the compositions of which do not match. If components A and B form a solid chemical compound AB, melting with decomposition, those. incongruently, then AB is stable only below T C. At at the slightest t-ry this solid compound disintegrates and 2 phases are formed: crystals B and melt composition at(point C). When cooling the melt of composition M at point a 0 the release of crystalline component B will begin. In the temperature range from a 0 before b 0 syst. yavl. 2-phase and single-variant: C=2-2+1=1. At the point b 0 at T C the crystallization of compound AB begins and continues, the composition of which corresponds to at 2(Vol. D). There are 3 phases in equilibrium: melt, crystals AB and B. C = 2-3 + 1 = 0 - constancy of T C, solution composition y (t. C) and chemical composition. connections at 2 (t.D).
To ensure that the composition of the melt does not change, simultaneously with the crystallization of AB, the previously precipitated crystals of B must dissolve, maintaining a constant connection of component B in the melt. T. S called peritectics (transitional), T S - peritectic temperature. At this point the melt and 2 solids are in equilibrium. phase, but differs from the eutectic, (where 2 solid phases simultaneously fall out) one solid. phase precipitates and the other dissolves. The cooling curve exhibits a horizontal section (bb’).
Point cooling process b 0 ends with the dissolution of all previously precipitated crystals B. A 2-phase system remains, consisting of a melt and crystals AB. In this case, each temperature corresponds to a certain melt composition (CE). Further cooling is described by the phase diagram A-AB with eutectic.
21. Differential Thermal Analysis (DTA)
DTA is the determination of the relationship between the melting point of a substance and its properties. The method allows you to study the phase composition of Me systems and natural minerals, establish the boundaries of the existence of many compounds (salts, polymers), determine the heat of phase transformations, thermal conductivity, heat capacity .
The method is based on the automatic recording of thermograms by a differential thermocouple - curves ∆T - T, where ∆T is the temperature difference between the test substance and the standard, heated or cooled under the same conditions; T – sample temperature or heating/cooling time. The standard is a substance that does not have phase transformations in the temperature range under study.
The type of thermogram of the substance under study depends on the properties of the substance itself (composition, structure, thermal conductivity, heat capacity, dispersion, etc.) and on the conditions for taking the thermogram (heating rate, sample size, packing density of the substance in the crucible, position of the junction in sample and in the standard, properties of the standard, sensitivity in the differential thermocouple circuit).
If the thermophysical properties of the standard and the substance under study coincide and the latter does not experience any transformations when heated, then ∆T = 0, and the thermogram named after. straight line view (1) – zero line;
If the substance under study differs from the standard in its thermophysical properties, then the thermogram (2) will deviate from the zero line and || abscissa axis or at an angle to it – baseline.
If during the heating process in the sample occurs at r.l. temperature phase transformation or chemical. change with the release or absorption of heat, then ∆T arises, proportional to the amount of absorbed/released heat. The resulting temperature difference is recorded by the deviation of the DTA curve up or down from the baseline (3) – thermal effect. According to the method, deviation is an exothermic effect, ↓ is an endothermic effect.
Exothermic effects(with Q highlighted) may be due to the transition from a nonequilibrium state to an equilibrium one, for example. transition from an amorphous to a crystalline state.
Endothermic effects(with absorption Q) are associated with phase transformations (melting, evaporation, sublimation, polymorphic transformations) or chemical processes (oxidation, decomposition, dehydration, dissociation, etc.). When most substances are heated, several transformations are observed, which are recorded on the DTA curve at appropriate temperatures by thermal effects characteristic of a given substance.
Using a thermogram, you can give a qualitative characteristic of the substance under study, determine the temperatures of phase transformations, or chemical processes, measure the thermal effect of the process.
| T |
| ∆T |
| A |
| b |
22.Concepts:
Electrical conductivity liquids χ - this is the electrical conductivity of 1 cm 3 of solution filling the space between flat electrodes with the same, very high. large area (in cm 2), located at a distance of 1 cm.
Depends on the nature of the electron and plant, on the concentration of the solution, on T.
When congested, it is weak. elect-ta, χ ↓
When con-tion is strong. electa, first then ↓.
At T, χ.
Equivalent electrical conductivity λ – this is the electrical conductivity of such a volume (cm 3) of a solution that contains 1 g-equiv of a dissolved substance, and the electrodes are located at a distance of 1 cm from each other. [cm 2 /g-eq*ohm]
where φ – dilution, [cm 3 /g-eq]
с – equivalent concentration, [g-eq/l]K
At ↓ concentration of the electricity solution, λ ;
- λ=max will be at infinity. dilution
Empirical Kohlrausch formula:
λ=λ ∞ -А√с
Ion speed (v i ,u i – movement speeds A - and K + respectively) determined by the force acting on the ion, cat. equal to the product of the ion charge and the field potential gradient, and the factor R, characterizing the resistance of the medium, depending on T, the nature of the ion and the solution
z i – ion charge; E/l – field strength, field gradient
Depends on: the nature of the ions, E\l, concentration, temperature, viscosity of the medium.
Similarly for u i
Absolute. Movement speed . ions used when comparing ion velocities if field strength = 1 V/cm
v=ez i /R (same for u)
Ion mobility – The number of electrons transferred by an ion is equal to the product of the absolute speed of the ions and the Faraday number.
Carry number ions of the i-th type - the ratio of the number of electrons q i (depends on z i, conc., speed of movement in the electric field), transferred. this type of ions is transferred to the total number of electrons q. all types of ions present in the solution t i =q i /q.
Ionic strength of the solution (ionic strength) is called the half-sum of the products of the concentrations of each ion by the square of the number of its charges z (valency), taken for all ions of the solution.
I = ½ ∑m i z i 2
where m i is molarity (measure of concentration)
Empirical law ionic strength:
Avg. ion coefficient activity γ +/- yev. universal function of ionic strength I solution, i.e. in a solution with a given ionic strength, all dissociating substances named after. activity coefficients that do not depend on the nature and concentration of a given substance, but depend on the number and valency of its ions.
23.Factors affecting the electrical conductivity of solutions, movement speeds and mobility of ions.
1. nature of the ion
3. concentrations. (specific e\conductivity on the graph ae=f(s,mol\l) - rainbow; eq.e\conductivity λ=c,g*eq\l - a slide concave to 0, down.
4. temperature (with it the specific conductivity and mobility limit increase, for metals it’s the other way around)
5. Viscosity of the medium
24. Derivation of an equation relating equivalent electrical conductivity to ion mobilities.
I. 
- electrical conductivity,
where ρ – electrical resistivity
l – distance between electrodes
χ – electrical conductivity, [ohm -1 cm -1 ]
II. 
- equivalent electrical conductivity, [cm 2 /(g-eq ohm)]
where c – equivalent concentration, [g-equiv/l]
III. I = I + +I - - the amount of electricity transferred by ions through the solution in 1 second
- the number of cations passing through the cross section in 1 second
- current strength, because Each g-eq of ions carries according to Faraday's law
F=96486 K el.
- speed of cations,
where u is the absolute mobility of cations, [cm 2 /sec*v]
Similar formulas for anions (v, v’, c - , n - , I -)
We get:
IV. 
- Ohm's law
Let us substitute the value of K from (I) and (II) into this expression, and equate the right-hand sides of equations (III) and (IV), we obtain
Having solved the equation for λ, we get
For strong electrolytes, the dissociation of which is considered complete, the ratio is 10 3 c i /c = 1
For weak- 10 3 s i /s = α
V. Taking into account the mobility of cations and anions,
we obtain the given expression for different degrees of dissociation of electrolytes.
25. Basic principles of the theory of strong electrolytes
Debye-Hückel.
These ideas are formulated in the form of the idea that around each ion there is an ionic atmosphere of oppositely charged ions. Its density is max. near the center, with distance from it ↓. At a certain distance, which can be considered the boundary of the ionic atmosphere, the number of ions of each sign is the same. The theory connects the thermodynamic properties of electrolyte solutions with the parameters of the ionic atmosphere. – its size and density.
1. The electrolyte in the solution is completely dissociated, and the ion concentration is calculated from the analytical concentration of the electrolyte.
2. The distribution of ions in the ionic atmosphere obeys the classical. statics, and the ionic atmosphere itself is considered as a continuous medium.
3. Of all types of interaction, taking into account only the electrostatic interaction of ions. The solvent plays the role of a medium with a certain dielectric constant.
4. The dielectric constant of the solution is assumed to be equal to dielectric constant pure solvent.
5. Of all the properties of ions, only the charge is taken into account.
26.What is ionic atmosphere, relaxation and electrophoretic inhibition?
Ionic atmosphere
If you mentally isolate one central ion in a dilute solution of a strong electrolyte, then ions of the opposite sign will be more often observed near it than ions with the same charge. This statistical distribution of ions is established under the influence of 2 factors: 1) electrostatic forces of attraction and repulsion; 2) thermal movement of ions. As a result, some intermediate statistical distribution of ions is established around the central ion - ionic atmosphere.
Electrophoretic inhibition. When an electric field is applied to a solution, the central ion and its ionic atmosphere, which have charges of opposite signs, move in opposite directions. Since the ions are hydrated, the movement of ions, the movement of the central ion occurs in a medium moving towards it. Therefore, the moving ion is under the influence of an additional braking force, which leads to a decrease in its speed.
Relaxation inhibition . The ionic atmosphere has central symmetry. When an ion moves in an electric field, the symmetry is broken, which is associated with the destruction of the atmosphere in one position of the ion and its formation in another, new one. This process occurs at a finite rate over time - time of relaxation. As a result, the ionic atmosphere loses its central symmetry, and behind the moving ion there will always be some excess charge of the opposite sign. The resulting forces of electrical attraction will inhibit the movement of the ion.
Relaxation and electrophoretic forces. braking
¨ are determined by the ionic strength of the solution, the nature of the solvent and temperature.
¨ with the concentration of the solution at constant other conditions
27. What is the Wien effect? What is electrical conductivity dispersion?
Wine effect
In constant electric fields sufficiently high voltage (10 4 -10 5 V/cm)
The ions move so fast that the ionic atmosphere does not have time to form, as a result of which there are practically no braking effects, and λ tends to λ ∞.
In weak electrolytes, the Wien effect is also caused by a shift in the dissociative equilibrium in a strong electric field towards the formation of ions.
Electrical conductivity dispersion (frequency effect) . When the frequency of alternating current increases above a certain value, an increase in electrical conductivity is observed, because at a sufficiently high frequency, the mutual displacements of the ion and the ionic atmosphere are so small that the ionic atmosphere is practically symmetrical.
The alternating current frequency at which an increase in electrical conductivity can be expected is the reciprocal of the relaxation time
Therefore, the effect of relaxation inhibition should disappear.
(The Wien effect occurs when the ionic atmosphere is completely destroyed, and therefore both inhibition effects. The frequency effect is explained only by the disappearance of the ionic atmosphere. Experience shows that the latter effect is approximately 3 times weaker than the Wien effect, i.e. the electrophoretic effect is 2 times stronger than the relaxation effect)
28. Activities and activity coefficients of electrolytes. Methods for their determination and calculation.
Activity is related to the chemical potential of a solution component by the equation: 
Index X indicates that the activity refers to a solution in which the concentration of the substance is expressed in mole fractions. Potency is also sometimes called effective or active concentration.
Activity factor– a measure of the deviation of the properties of a solution from the properties of an ideal solution of the same concentration.
, where γ is the molar coefficient. activity,
It is possible to compare the activities of components with concentrations expressed in other units:
¨ molar-volume concentrations (s)
, where f is the molar activity coefficient
¨ molality (m)
where γ’ is the practical activity coefficient
In the general case, the properties of different ions in a solution are not the same, and thermodynamic functions for ions of different types can be introduced and considered:
1. Consider a 2-component solution consisting of a solvent and a salt, dissociating according to the equation:
2. Chem. potential: -
(2 equations K + and A -)
3. Let's make assumptions:
1000g of solvent is taken, n 1 =1000/M 1, n 2 =m – molality
Electrolyte dissoc. completely: 
T,P = const, Þ according to the Gibbs-Duhem equation (measurement of the chemical potentials of solution components when changing the composition of the solution):
- subtract 2 from 1: 
Let us introduce the average ionic activity 
, Where 
;
 |
Standard. the state for a 2 is chosen so that const=1, we get:
A 2 is obtained experimentally and a ± is determined by ur;
Enter avg. and he. coefficient activityg ± And sren. and he. molalitym±:
;
- can be expressed practical coefficient active
· Determination of the activity of a volatile substance by its partial vapor pressure.
From the equations: р 1 =р* 1 а 1 , р 2 =р* 2 а 2 , ( Where p 1 – partial pressure of the vapor of the solvent above the solution, p* 1 – vapor pressure over liquid solvent; p 2, p* 2 – respectively for the dissolved substance) we obtain
(index 1 indicates the number of the selected standard state)
For determining γ solid solute, choose the second standard state
(at– auxiliary calculated value, K 2 – Henry's constant)
We get:
(Where y=y 0 at x 2 →0 found graphically).
· Determination of the activity of a solute by the vapor pressure of the solvent.
From the Gibbs-Duhem equation: 
Integration gives: 
,
Where a’ 1 and a’ 2 – activity of solvent and solute in solution composition x" 2, they must be known.
You can also integrate the Gibbs-Duhem equation, expressed in terms of γ :
Activity a 1 determined by the vapor pressure over solutions of different compositions. The integral is calculated graphically.
· Determination of the activity of a solute by coefficient. distributions.
(where K is the distribution coefficient, γ 1 2 , γ 2 2 - activity coefficients of the solute in the first and second solvents, x 1 1, x 2 2- concentration of the dissolved substance in the first and second solvents).
A And γ can also be calculated by the decrease in freezing point, the increase in boiling point, osmotic pressure and other properties of solutions.
29. Conductometric measurement of the dissociation constant, degree of dissociation, equivalence point during titration.
| a |
| b |
| V |
| Rx |
| R M |
| G |
2 – alternating current generator (constant causes electrolysis of the solution)
R x – measuring cell;
R M – resistance store (known)
V - the position of the movable contact (selected so that the zero tool 1 does not show the current or shows min, then
R x = R M (R 1 / R 2) = R M ( av/vb)
The actual electrical conductivity of the solution is determined by the concentration of the solution, the nature of the components and temperature. The true electrical conductivity of the solution c is proportional to the experimentally measured value c': c=kc', where k is the constant of the vessel - the characteristic of the cell - depends on the area of the electrodes, the distance between them, the shape of the vessel, the volume of the solution conducting current is found experimentally from standard solutions, most often KCl.
Based on experimental data, the values are calculated specific
: 
And equivalent electrical conductivity
: 
For calculation degree of dissociation of a weak electrolyte the equation used is:
, Where λ ∞ =l ¥ to +l A ¥- determined by ion mobilities
Dissociation constant of a binary electrolyte
At conductometric titration a curve is constructed depending on the electrical conductivity of the titrated solution on the amount of added titer. agent. The equivalence point is determined by the break in the curve of this dependence. A sharp change in electrical conductivity occurs when poorly dissociating or poorly soluble compounds form (or disappear) during the titration process.
Acid-base titration:
The change in electrical conductivity up to the equivalence point will be determined by the action of 2 mutually opposite trends: ¯ due to ¯ H+, which has a mobility of 350 S cm 2 /(mol-equiv), which is much higher than the mobility of the Na + ion
After the equivalence point, sharp electrical conductivity begins (BC branch), because in the solution the concentration of Na + and OH - ions will increase, the mobility of the cat. is 199 cm×cm 2 / (mol×eq)
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 for a long time remained unexplored. In particular, the main 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, metallurgy, pharmaceutical and other industrial fields, as well as 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.
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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