If you add a third component to two immiscible liquids, which is soluble in both liquids, then 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 equilibrium, chem. 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 the Nernst distribution law: the distribution of each of the dissolved substances between the 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 conc. Distributed thing in 2 equilibrium liquid phases.
An important consequence of this law is extraction, i.e. extraction of ve-va from the solution with a suitable solvent that does not mix with another component of the solution. Multiple substances can be separated by extraction. To increase the completeness of the extraction of a thing from the aqueous layer with an organic solvent, the extraction is carried out sequentially in small portions of the extractant (υ), while the greater the number of successive stages of extraction (n), the greater the completeness of the 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 - the mass of the component in the initial mixture and after the first extraction there was m 1;
After n extractions, the first solution will contain substances 
When extracting with one total volume equal to nV 2, the mass of the substance that will remain in the extracted solution according to Eq. 
From Ur-th it follows that extraction is n times more efficient than carrying out 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.
The state of the two-component system. determined by 3 parameters (temperature, pressure and concentration of one of the components)
Systems without the formation of chemical compounds
I - liquid melt (C = 2 - 1 + 1 = 2);
II - liquid melt and crystals of set 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 by 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 the crystals of component A and B. This melt is called eutectic, and the mixture of precipitating crystals at T e eutectic ... Eutectic crystallizes at constant temperature - system conditionally is invariant because with a change in pressure, both the melting point and the composition of the eutectic change. During the 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).Melting diagrams for systems with congruently and incongruently melting compounds.
If components A and B can form solid. chem. melting compound AB 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. phase. On both sides of the point C, there are eutectics E and E 1. If the composition of the system is between the pure component A and the chemical compound AB, then at the eutectic temperature T E, the melt of composition at E coexists with crystals A and AB. If the composition of the system lies between chem. conn. AB and component B, then at T E1 the melt of composition at E1 coexists with AB and B. crystals. the considered state diagram is a combination of 2 state 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, since the system can be formed from only one chemical. conn. AB. At T C, crystals AB (Ф = 2) and the number of steps are precipitated from the melt. free C = 1-2 + 1 = 0, i.e. the system is variable and crystallizes at a constant temperature.
Chem. Compounds, upon reaching a certain temperature, begin to melt, decomposing into crystalline and liquid phases, the compositions of which do not coincide. If components A and B form a solid chemical compound AB, melting with decomposition, those. incongruent, then AB is stable only below T C. At the slightest t-ry, this solid compound decomposes and 2 phases are formed: crystals B and a melt of composition at(point C) When cooling the melt of composition M at the point a 0 crystals of component B will begin to precipitate. In the temperature range from a 0 before b 0 sist. 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). In equilibrium there are 3 phases: melt, crystals AB and B. C = 2-3 + 1 = 0 - the constancy of T C, the composition of the solution y (t. C) and the composition of the chemical. compounds at 2 (point D).
So that the composition of the melt does not change, simultaneously with the crystallization of AB, the previously precipitated crystals of B should be dissolved, keeping the compound of component B constant in the melt. T. S name peritectics (transitional), T C - peritectic temperature. At this point, the melt and 2 solid are in equilibrium. phase, but differs from the eutectic, (where 2 TV phases fall out simultaneously) one TV. phase drops out and the other dissolves. A horizontal section (bb ') is observed on the cooling curve.
Cooling process at point b 0 ends with the dissolution of all previously precipitated crystals of B. A 2-phase system remains, consisting of a melt and crystals AB. In this case, each temperature corresponds to a certain composition of the melt (CE). Further cooling is described by the A-AB state diagram 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, natural minerals, to establish the dark boundaries of the existence of many compounds (salts, polymers), to determine the heats of phase transformations, heat transfer, heat capacity ...
The method is based on the automatic recording of thermograms by a differential thermocouple - curves ∆Т - Т, where ∆Т is the temperature difference between the test substance and the standard, heated or cooled under the same conditions; T is the sample temperature or heating / cooling time. The standard is a substance that does not have phase transformations in the investigated temperature range.
The type of thermogram of the investigated island depends on the properties of the island itself (composition, structure, thermal conductivity, heat capacity, dispersion, etc.) and on the conditions for taking the thermogram (heating rate, the sample and in the standard, properties of the standard, sensitivity in the differential thermocouple circuit).
If the thermophysical properties of the standard and the investigated island coincide and the latter does not undergo any transformations when heated, then ∆Т = 0, and the thermogram of them. straight line view (1) - zero line;
If the investigated substance differs from the standard in its thermophysical properties, then the thermogram (2) deviates from the zero line and is recorded || the abscissa axis or at an angle to it - baseline.
If during heating in the sample occurs at K.-L. temperature phase transformation or chem. change with the release or absorption of heat, then there is ∆Т, 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 methodology, the deviation is exothermic, ↓ is the endothermic effect.
Exothermic effects(with the release of Q) can be caused by the transition from a nonequilibrium state to an equilibrium state, for example. transition from an amorphous state to a crystalline one.
Endothermic effects(with absorption of 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.
According to the thermogram, it is possible to give a qualitative characteristic to the investigated substance, determine the temperatures of phase transformations or chemical processes, and measure the thermal effect of the process.
| T |
| ∆Т |
| a |
| b |
22.Concepts:
Specific electrical conductivity liquids χ is the electrical conductivity of 1 cm 3 of the solution filling the space between the flat electrodes with the same, och. large area (in cm 2), located at a distance of 1 cm.
Depends on the nature of the electricity and growth, on the concentration of the solution, on T.
When con-tion is weak. elect, χ ↓
Strong when con-tions. elect, first then ↓.
At T, χ.
Equivalent electrical conductivity λ – this is the electrical conductivity of such a volume (cm 3) of a solution in which 1 g-eq of a solute is contained, and the electrodes are 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 ↓ of the con-tions of the solution of the elect, λ;
- λ = max will be at infinity. dilution
Kohlrausch's empirical formula:
λ = λ ∞ -А√с
Ion movement speed (v i, u i - speeds of movement А - and К +, respectively) is determined by the force acting on the ion, the cat. is equal to the product of the ion charge and the gradient of the field potential, and the factor R, characterizing the resistance of the medium, depending on T, the nature of the ion and the solution
z i is the charge of the ion; E / l - field strength, field gradient
Depends on: nature of 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 carried by the ion is equal to the product of the absolute speed of the ions by the Faraday number.
Transfer number ions of the i-th type - the ratio of the number of electr-va q i (depends on z i, conc., speed of movement in an electric field), transferred. this type of ions, to the total number of electr-va q, is transferred. all types of ions in solution t i = q i / q.
Ionic strength of solution (ionic strength) is called the half-sum of the products of the con-tions of each ion per square of the number of its charges z (valence), taken for all ions of the solution.
I = ½ ∑m i z i 2
where m i is molarity (measure of concentration)
The empirical law of ionic strength:
Wednesday ionic coefficient activity γ +/- yavl. universal function of ionic strength I p-ra, i.e. in 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 valence of its ions.
23.Factors affecting the electrical conductivity of solutions, the speed of movement and the mobility of ions.
1.the nature of the ion
3. concentration. (beats.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 e / conductivity and the limit of mobility increase, for metals, vice versa)
5.viscosity of the medium
24. Derivation of the equation connecting the equivalent electrical conductivity with the ion mobility.
I. 
- electrical conductivity,
where ρ is the electrical resistivity
l is the distance between the electrodes
χ - electrical conductivity, [ohm -1 cm -1]
II. 
- equivalent electrical conductivity, [cm 2 / (g-eq ohm)]
where c is the equivalent concentration, [g-eq / l]
III. I = I + + I - - the amount of electricity carried by ions through the solution in 1 sec
- the number of cations passed through the cross section in 1 sec
- current strength, because each g-eq of ions carries according to Faraday's law
F = 96486 K el-va.
- the rate 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
We substitute in this exp-th value of K from (I) and (II), and equating the right-hand sides of equations (III) and (IV), we obtain
Solving the equation for λ, we get
for strong electrolytes, the dissociation of which is considered complete, the ratio is 10 3 s i / s = 1
for weak- 10 3 s i / s = α
V. Considering the mobility of cations and anions,
we obtain a given expression for different degrees of dissociation of electrolytes.
25. The main provisions of the theory of strong electrolytes
Debye-Hückel.
These ideas are formulated in the form of the idea of the presence of an ionic atmosphere around each ion of oppositely charged ions. Its density is max. near the center, with a distance from it ↓. At some distance, which can be considered the boundary of the ionic atm., The number of ions of each sign is the same. The theory connects thermodynamic sv-va solutions of electrolytes with the parameters of the ionic atmosphere. - its size and density.
1. The electrolyte in solution is completely dissociated, and the concentration of ions is calculated from the analytical concentration of the electrolyte.
2. The distribution of ions in the ionic atmosphere obeys the classical. static, 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 is assigned the role of a medium with a certain dielectric constant.
4. The dielectric constant of the solution is taken to be equal to the dielectric constant of the pure solvent.
5. Of all the properties of ions, only 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. Such a statistical distribution of ions is established under the influence of 2 factors: 1) electrostatic forces of attraction and repulsion; 2) thermal motion of ions. As a result, a certain intermediate statistical distribution of ions is established around the central ion - ionic atmosphere.
Electrophoretic inhibition. When an electric field is applied to the solution, the central ion and its ionic atmosphere, which have opposite charges, 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 decelerating force, which leads to a decrease in its velocity.
Relaxation braking . The ionic atmosphere has central symmetry. When an ion moves in an electric field, 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 speed over time - relaxation time. As a result, the ionic atmosphere loses its central symmetry, and behind the moving ion there will always be a certain excess of charge of the opposite sign. The resulting forces of electric attraction will slow down the movement of the ion.
The forces of relaxation and electrophoretic. braking
¨ are determined by the ionic strength of the solution, the nature of the solvent and the temperature.
¨ with the concentration of the solution with the constancy of other conditions
27. What is the effect of Wien? What is electrical conductivity dispersion?
Wine effect
In constant electric fields of sufficiently high intensity (10 4 -10 5 V / cm)
Ions move so fast that the ionic atmosphere does not have time to form, as a result of which there are practically no deceleration effects, and λ tends to λ ∞.
In weak electrolytes, the Wien effect is also caused by a shift in dissociative equilibrium in a strong electric field towards the formation of ions.
Dispersion of electrical conductivity (frequency effect) . With an increase in the frequency of the alternating current above a certain value, an increase in electrical conductivity is observed, since at a sufficiently high frequency, the mutual displacements of the ion and the ionic atmosphere are so small that the ionic atmosphere is practically symmetric.
The frequency of alternating current at which an increase in electrical conductivity can be expected is the reciprocal of the relaxation time
Consequently, the effect of relaxation inhibition should disappear.
(The Wien effect occurs when the ionic atmosphere is completely destroyed, and, consequently, both inhibition effects. The frequency effect is explained only by the disappearance of the ionic atmosphere. Experience shows that the latter effect is about 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 the solution component by the equation: 
Index NS indicates that activity refers to a solution in which the concentration of a substance is expressed in molar fractions. Activity is also sometimes referred to as effective or effective concentration.
Activity coefficient- a measure of the deviation of the properties of the solution from sv-in the ideal solution of the same concentration.
, where γ is the molar coefficient. activity,
It is possible to compare the activities of the components with concentrations expressed in other units:
¨ molar-volume concentration (s)
, where f is the molar activity coefficient
¨ molality (m)
where γ 'is the practical activity coefficient
In the general case, the properties of various ions in a solution are not the same, and it is possible to introduce and consider thermodynamic functions for ions of various types:
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:
Taken 1000 g of solvent, n 1 = 1000 / M 1, n 2 = m - molality
Dissocic electrolyte. entirely: 
T, P = const, Þ according to ur-th Gibbs-Duhem (change of chemical potentials of computers of solution when changing the composition of solution):
- subtract the 2nd from the 1st: 
Let us introduce the average ionic activity 
, where 
;
 |
Standard. the state for a 2 is chosen so that const = 1, we get:
Experimentally get a 2 and determine a ± by ur-th;
Introduce wednesday and he. coeff. activityg ± and shit. and he. molalitym ±:
;
- can be expressed practical asset rate
· Determination of the activity of a volatile substance by the partial pressure of its vapor.
From equations: p 1 = p * 1 a 1, p 2 = p * 2 a 2, ( where p 1 - the partial pressure of the solvent vapor over the solution, p * 1 - vapor pressure over the liquid solvent; p 2, p * 2 - respectively for the solute) we obtain
(index 1 indicates the number of the selected standard state)
For determining γ solid solute, select the second standard state
(at- auxiliary calculated value, K 2 - Henry's constant)
We get:
(where y = y 0 at x 2 → 0 find graphically).
· Determination of the activity of the solute by the vapor pressure of the solvent.
From the Gibbs-Duhem equation: 
Integration gives: 
,
where a '1 and a' 2 - activity of the solvent and solute in the composition of the solution 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 the solute by coeff. distribution.
(where K is the distribution coefficient, γ 1 2, γ 2 2 - the activity coefficients of the solute in the first and second solvents, x 1 1, x 2 2- the concentration of the solute in the first and second solvents).
a and γ can also be calculated by lowering the freezing point, by increasing the boiling point, by osmotic pressure, and by other properties of solutions.
29. Conductometric measurement of dissociation constant, degree of dissociation, equivalence point during titration.
| a |
| b |
| v |
| R x |
| R M |
| G |
2 - alternator (constant causes electrolysis solution)
R x - measuring cell;
R M - resistance box (known)
v - position of the movable contact (selected so that null-instrument 1 does not show currents 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 the 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 characteristics of the cell - depends on the area of the electrodes, the distance between them, the shape of the vessel, the volume of the solution that conducts current, find experimentally from standard solutions, most often KCl.
The experimental data are used to calculate the values specific
: 
and equivalent electrical conductivity
: 
For calculation the degree of dissociation of a weak electrolyte the equation is used:
, where λ ∞ = l ¥ k + l A ¥- determined by ion mobility
Dissociation constant of binary electrolyte
At conductometric titration the curve of the dependence of the electrical conductivity of the titrated solution on the amount of added titer is plotted. agent. The equivalence point is determined by the kink of the curve of this dependence. A sharp change in electrical conductivity occurs when, in the course of titration, poorly dissociating or poorly soluble compounds are formed (or disappear).
Acid-base titration:
The change in electrical conductivity to the point of equivalence will be determined by the action of 2 mutually opposite tendencies:
After the point of equivalence, a sharp electrical conductivity begins (branch BC), because in the solution, the concentration of Na + and OH - ions will increase, the mobility of the cat. is 199 S × cm 2 / (mol × eq)
Example 1. 0.1 mol of alcohol is distributed between 300 ml of water and 500 ml of CCl 4. Find the concentration of alcohol (mol / L) in equilibrium solutions. The partition coefficient of ethyl alcohol between carbon tetrachloride and water is 0.0244.
Solution: According to the distribution law:
where С 1 is the concentration of the solute in the first solvent (CCl 4);
C 2 - the concentration of the solute in the second solvent (H 2 O).
K - distribution coefficient
The number of moles of alcohol that has passed into carbon tetrachloride is denoted by X, then:
The remaining amount of alcohol, equal to (0.1 - X) mol, will pass into the water, therefore: 
Substituting C 1 and C 2 into the equation, we get 
Whence X = 0.0039 mol
mol / L 
mol / L
Example 2. 0.3 g of crystalline iodine is dissolved in 1 liter of water. Calculate the effective number of stages of extraction of this solution with carbon disulfide if carbon disulfide is taken in portions of 100 ml. The final concentration of iodine in water is 1 × 10-6 g / l? The distribution coefficient of iodine between water and carbon disulfide is 0.0017.
Solution:
The formula for calculating the extraction process: 
where g 0 is the initial amount of the substance undergoing extraction;
V 0 - the volume of the solution in which the extractable substance is located;
V e - the volume of solvent (extractant) consumed for one extraction;
n is the total number of extraction stages; K is the distribution coefficient.
lg g n = lg g 0 + n lg KV 0, whence
91 - 100. Mixed 1 mole of alcohol with a volume V of 1 ml of carbon tetrachloride and a volume of V of 2 ml of water. Determine the concentration of ethyl alcohol solutions in carbon tetrachloride and water. The partition coefficient of ethyl alcohol between carbon tetrachloride and water is 0.0244.
101 - 110. It is necessary to extract a certain amount of% (wt.) Acid from 1 ml of acid solution in ether. How much water is needed for this as an extractant with the number of extraction stages equal to n, if the distribution coefficient of the acid between water and ether is K?
111 - 120. Calculate how much HgBr 2 can be extracted from a volume V of 1 ml with a concentration of C m mol / L of an aqueous solution using a volume V of 2 ml of benzene by n-fold extraction. The partition coefficient of HgBr 2 between water and benzene is 0.893.
Topic 2.
Solutions are physical and chemical systems.
Colligative properties of solutions.
Solubility
Example 1. Determination of the mass of a gas in a solution by its solubility.
How much hydrogen chloride HCI will dissolve in 100 liters of water at 40 ° C and a pressure of 98625 Pa, if the solubility of HCI at this temperature and pressure 1.0133 × 10 5 Pa is 386 m 3 per 1 m 3 of water?
Solution.
Solubility (or coefficient of solubility) is expressed by the mass of the substance (g) that can be dissolved in 100 g of solvent at a given temperature.
Determine the volume of HCI contained in 100 liters of water at 40 ° C and a pressure of 1.0133 × 10 5 Pa:
1000 l H 2 O - 386 m 3
100 l H 2 O - NS m 3 NS= 
m 3
We calculate the HCI mass according to the Mendeleev-Cliperon equation;
M(HCI) = 36.46 g / mol. Then:
m == 
53.4 kg.
Example 2... Determination of the composition of the gas mixture by the solubility of gases.
A gas mixture containing 21% O 2 and 79% N 2 was passed through water at 0 ° C and a pressure of 1.0133 × 10 5 Pa. Calculate the volume fractions φ of the gas mixture dissolved in water if the solubility of oxygen and nitrogen in water at this temperature and pressure, respectively, is 0.048 and 0.0236 m 3 per 1 m 3 of water.
Solution.
According to Henry's law, solubility ( R) gas in water is proportional to its partial pressure in the mixture. Let us determine the partial pressure of gases in the mixture:
p О 2 = 1.0133 × 10 5 × 0.21 = 0.2128 × 10 5 Pa;
p= 1.0133 × 10 5 × 0.79 = 0.8005 × 10 5 Pa.
Taking into account the partial pressures, we determine the solubility of gases:
R= 
= 0.0104 m 3;
R= 
= 0.0189 m 3.
The total volume of nitrogen and oxygen; 0.0104 + 0.0189 = 0.0293 m 3. Then the volume fraction of gases and mixture will be (%):
φ = 0.0104 × 100 / 0.0293 = 35.49; φ = 100.00 - 35.49 = 64.51.
Tasks for independent solution
121 - 130. A gas mixture consisting of O 2, N 2 and Cl 2 is dissolved in water at 20 ° C and a total pressure of 2.5 × 10 5 Pa. The volume fractions of these gases in the mixture are, respectively, ω (О 2), ω (N 2) and ω (Cl 2)%. Solubility of gases in 1 m 3 of water (m 3): PO 2 = 0.031; PH 2 = 0.016; PCl 2 = 2.299. Determine the volume fractions of gases in a gas mixture dissolved in water.
| options | Task number | |||||||||
| ω (О 2)% | ||||||||||
| ω (N 2)% | ||||||||||
| ω (Cl 2)% |
131 - 140. In a certain volume V 1 l of water, the volume V 2 l of substance A is dissolved at a temperature of t ºС and a pressure of P Pa. Determine the mass fraction of substance A in the resulting solution.
Physicochemical properties
If you take two immiscible liquids and add a third component, then it will dissolve to varying degrees in both solvents. When equilibrium is established, the concentration ratio of the resulting solutions will be constant at a given temperature
= K, this is an expression of the Nernst distribution law, where
Concentrations of the third component in phases I and II;
K is the distribution coefficient.
If the solute dissociates or associates in one of the solvents, then the Nernst equation has the form:
To find Kip We take the logarithm of the equation and obtain the equation of the straight line
Constructing a straight line in coordinates 
, we find “ NS»As the tangent of the angle of inclination of a straight line (at any two points lying on a straight line) tga =
InK can be found from the equation by substituting the values of any point on the line into it.
This law is at the heart of the extraction process. Extraction is the extraction of a component from one phase to another. Extraction happens solid phase- the 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... The solution of the extracted substance in the extractant is called extract, and the initial solution after extracting the substance from it is called raffinate.
To calculate the efficiency of liquid-phase extraction, use the equation
(3.31)
Where NS- the proportion of unrecovered substance in the raffinate;
V- the volume of the initial solution;
Extractant volume;
TO- distribution coefficient
NS- the number of extractions.
As can be seen from the equation, the more extractions, the less substance remains in the raffinate, that is, unrecovered, the more substance is extracted by the extractant. The efficiency of extraction is largely determined by the value of the distribution coefficient: the larger the coefficient in favor of the extractant, the more efficient the extraction.
Lecture 10. Distribution of matterbetween the two phases. Extraction.
.
2. Extraction, its types. Extraction equation.
Principles of obtaining tinctures, decoctions.
Lecturer: Cand. ped. Sci., Associate Professor Grigorieva Marina
Victorovna
1. Nernst distribution law
If any substance is soluble in twoimmiscible liquids, then when it
dissolving in a mixture of two such liquids
it is distributed between them according to
Nernst distribution law:
The concentration ratio of the third
component between two
immiscible liquids at
constant temperature is
constant K = C1 / C2, where C1 is the concentration of the component in solution 1
liquid, C2 is the concentration of the component in
solution of 2 liquid, K - coefficient
distribution.
1. Nernst distribution law
For example, if you shake iodine with waterand carbon tetrachloride, part of it
dissolve in water, and part in
tetrachloromethane CCl4. Finally
installed in the system
dynamic balance.
1. Nernst distribution law
Regardless of,how many
iodine is used in
experiment,
final
attitude
concentrations
turns out
permanent.
1. Nernst distribution law
The distribution law is satisfied only forcertain conditions, namely:
1) at constant temperature;
2) with sufficient dilution of both
solutions;
3) provided that the solute is not
reacts, does not associate and does not
dissociates in both solvents.
The distribution law underlies the important
and very common in
laboratory and industrial practice
a process called extraction.
2. Extraction
Extraction is extractionfrom a solution of one or more
solutes with the help of another
solvent (extractant), not
mixing with the first.
To carry out the extraction, it is necessary
so that the extractable substance is better
dissolved in a second solvent than in
first. The purpose of the extraction is
increased concentration of any
desired substance or release
solvent from the impurities present in it,
or replacement of the solvent.
2. Extraction
Extraction can be:single, when the extractant
added in one go,
fractional - addition of extractant
is carried out in portions of several
receptions.
Extraction is widely used in
pharmacy to extract from
vegetable raw materials of essential oils,
alkaloids and other physiologically
active substances.
2. Extraction
Extraction laboratories oftenuse separating funnels,
for example, when extracting with ether. For
this aqueous solution with the component
combine with ether in a fission
funnel. The solution is shaken and after
settling layers are separated. Ether
evaporated to give pure product.
2. Extraction
Dividing linesfunnels
2. Extraction
Under chemical andpharmaceutical production
widely used apparatuses
extractors, the action of which
based on different principles
mixing liquids and their
fragmentation. Are used
poppet, vibrating,
centrifugal and other types
extractors.
2. Extraction
Derivation of the extraction equation:Let in a solution, the volume of which is V0 ml,
is m0 of the extractable
substances. To this solution is added
V ml of another solvent, which with
it does not mix. Let's pretend that
after the first extraction in the first
solvent left t1 g
extractable substance, then
concentration in it will be C1 = m1 / V0, and
concentration in the second solvent
C2 = (m0 - m1) / V.
2. Extraction
Using the distribution law, you canwrite:
m1
V0
C1
m1V
TO
С 2 m0 m1 V0 (m0 m1)
V
KV0
m1 m0
V KV0
2. Extraction
If the first solvent with the remainingre-treat it with the same substance
volume V of the second solvent, then,
repeating the previous calculation, we get:
KV0
m2 m1
V KV0
2. Extraction
Replacing t1 in the latterequation, we find:
If you repeat
extraction with the same
volume of the second
solvent n times, then
amount remaining
in the first solvent
substances will
KV0
m2 m0
V KV0
2
KV0
mn m0
V KV0
n
2. Extraction
Calculations using the receivedequations show that extraction
will be more complete if we divide
the entire volume of the solvent per portion,
than to extract all at once
volume of solvent.
2. Extraction
Extraction methods are especially widespreadused in the analysis of plant
medicinal raw materials, as well as for obtaining
infusions, decoctions, tinctures, extracts
medicinal substances. In this case, under
extract means a dosage form,
obtained by the extraction method in accordance
with certain requirements.
According to the State Pharmacopoeia of the XI edition
infusions and decoctions are liquid medicinal
forms representing aqueous extracts
from medicinal plant materials, as well as
aqueous solutions of dry or liquid extracts
(concentrates).
2. Extraction
Extracts - concentrated extractsfrom medicinal plant materials.
Tinctures - colored alcohol or
hydroalcoholic extracts from
medicinal plant materials,
obtained by extraction without heating.
When receiving infusions and decoctions, carry out
extraction of medicinal substances from
crushed medicinal raw materials with water,
when receiving extracts - water, ethanol and
other extractants.
2. Extraction
For the preparation of infusions and decoctions, extractioncarried out as follows. To shredded
medicinal plant raw materials are added
required volume of water at room temperature,
the mixture is kept in a boiling water bath at
stirring (infusion -15 min, decoctions - 30 min),
cooled at room temperature (infusions - 45
min, decoctions - 10 min), filtered and get
filtrate, which is diluted if necessary
water.
The content of pharmacologically active substances in
obtained infusions and decoctions determine
various analytical methods, re
recommended in the relevant pharmacopoeial
articles.
When mixing two liquids, they can be:
Infinitely soluble, i.e. dissolve in each other in any ratio;
Practically insoluble;
Limitedly soluble.
Mutual solubility depends on the chemical structure of liquids, which in turn are divided into polar and non-polar.
Even alchemists noticed that "like dissolves in like", ie. polar liquids dissolve polar liquids well, and non-polar liquids dissolve non-polar ones.
For this reason, water, a polar solvent, dissolves well polar liquids (acetic acid, ethanol) and does not dissolve non-polar liquids (benzene, hexane, kerosene, gasoline, vegetable oil, etc.) at all.
If liquids differ in polarity insignificantly, then they dissolve to a limited extent in each other, 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 dissolved substance will be distributed between both liquids in proportion to its solubility in each of them.
Hence follows distribution law, Whereby the concentration ratio 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 - the concentration of the solute 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 not miscible with the first... The distribution law makes it possible to calculate the amount of the substance extracted and remaining in the solution after a single or multiple extraction with a solvent of a given volume at a given temperature:
where m 1 - the mass of the substance remaining in solvent 1 after a single extraction of it with solvent 2;
m o - the initial amount of the substance in the solvent 1.
V 1 and V 2 - the volume of solvents 1 and 2;
When retrieved multiple times, Equation 1 becomes:
where n- number of extractions.
Extraction never succeeds in completely removing the substance completely. But the completeness of the extraction will be greater if the solution is treated repeatedly with small portions of the solvent, separating each time the extract obtained, than with a single processing of the solution 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 (passing vegetables), obtaining pharmaceuticals. Thus, penicillin and a number of other antibiotics cannot be concentrated by evaporation, since they are destroyed by heating. To obtain concentrated solutions of antibiotics, extraction is carried out with butyl or ethyl acetate.
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