Section “Structural and mechanical properties of dispersed systems. Rheological properties of dispersed systems What types of deformation are characteristic of dispersed systems
In free dispersed systems aх particles of the dispersed phase ns are connected with each other and can freely move independently of each other. In coherently dispersed systems, the particles are interconnected and form a structural network with a fixed relative position of the constituent parts of the body (atoms and molecules) in space. They cannot move freely and only oscillate around the equilibrium position. Structural networks are formed under the action of van der Waals forces and short-range chemical forces in concentrated suspensions, emulsions, and flippers. The structure of dilute aggregatively stable disperse systems is very similar in a number of properties to the structure of true solutions, the difference lies only in the size of particles. An increase in the concentration of the dispersed phase leads to the interaction of its particles, similar to the association of molecules and ions in true solutions. This interaction is accompanied by a change in the properties of dispersed systems, which occurs gradually until coagulation occurs in the system.
In colloidal chemistry, the concept of structure is usually associated with coagulation. In the process of coagulation, a spatial structural network is formed from particles of the dispersed phase, which sharply increases the strength of the system. Structure formation in free-dispersed systems is directly related to the loss of their aggregate stability. As the strength increases, the free-dispersed system transforms into a coherent-dispersed system.
In coherently dispersed systems, the appearance and nature of the resulting structures are determined by the mechanical properties of the system: viscosity, elasticity, strength and plasticity. These properties are called structural-mechanical or rheological, as they are investigated by the methods of rheology - the science of deformation and flow of material systems.
The dispersed system acquires a complex structural and mechanical(rheological) properties characterizing its ability to resist deformation and division into parts, as a result of a radical change in the course of structuring.
The main rheological property is mechanical strength. This property is common to all solids and materials, it determines their role in nature and technology. The regularities of structure formation in dispersed systems, the mechanical properties of structured systems and various materials obtained on their basis, with special attention to the role of physicochemical phenomena at the interface, is being studied by rheology, a separate section of colloidal chemistry. Otherwise, this section is called
is studied by physicochemical mechanics. Rheology studies the mechanical properties of systems by the manifestation of deformation under the influence of external stresses.
It is known that deformation is a relative displacement of points of a system, at which its continuity is not violated. Deformation is divided into elastic (reversible, when the body is fully restored after removing the load) and residual (irreversible). Elastic deformation, in turn, is subdivided into volumetric (tension-compression), shear and torsional deformation. Residual deformation, in which there is no destruction of the body, is called plastic.
When considering the structural and mechanical properties, two types of deformations are important - tension-compression and shear. Quantitatively, both types of deformation are characterized by relative values:
Elongation (here / Q is the initial
sample length; / - its length after the applied load; D / - specimen elongation);
Relative shift or (here at- displacement of the upper
layer, NS is the height over which the displacement occurs).
These concepts are included in the laws of rheology, formulated in the form of two axioms. According to the first axiom of rheology, under all-round uniform (isotropic) compression, all material systems behave in the same way - like ideal elastic bodies, only elastic deformation acts. Isotropic compression does not allow one to reveal qualitative differences in the structure of bodies.
The second axiom of rheology is the statement that any material system has all rheological properties. These properties are manifested during shear deformation. The nature and magnitude of deformation depend on the properties of the material, the shape of the body and the method of applying external forces.
The quantity 8 in rheology is called elastic deformation, the value at- shear deformation or simply deformation (Fig. 8.1).
causing deformation of the body is determined by the ratio of the external force F to a unit of body surface s on which it acts: P = F! S. The voltage unit is Pascal (Pa): [R] = Pa. One Pascal is equal to the pressure caused by a force of 1 N, evenly distributed over a surface normal to it with an area of 1 m 2: 1 Pa = 1 N / m ".Liquids and gases are deformed when minimal loads are applied, they flow under the influence of a pressure difference. Flow is a type of deformation in which the amount of deformation increases continuously under constant pressure (load). Unlike gases, liquids do not compress during flow and their density remains practically constant.
Rice. 8.1. Tensile (s) and shear deformation diagram (b)
In rheology, mechanical properties are presented in the form of models, which are based on the laws relating stress to deformation (Fig. 8.2). Three sections of the curve correspond to three elementary models of idealized materials that correspond to the main rheological characteristics (elasticity, viscosity, plasticity): Hooke's perfectly elastic body, Newton's perfectly viscous body, and Saint-Venant-Coulomb's perfectly plastic body.
Rice. 8.2.
Rice. 8.3.
In section I, the dependence is directly proportional and is described by Hooke's law for bodies with elastic deformation. Hooke's ideal elastic body is represented in the form of a spiral spring (Fig. 8.3).
According to Hooke's law deformation at in an elastic body is proportional to the shear stress R:
where E- elastic modulus (Young's modulus), which is a characteristic of a material and its structure. For molecular crystals, the modulus of elasticity is ~ 10 9 Pa, for covalent crystals and metals ~ 10 "Pa and more. After removing the load, Hooke's ideal elastic body instantly takes its original shape (R= 0; / = 0). The energy spent on deformation of an ideally elastic body returns after the stress ceases, therefore Hooke's body belongs to conservative structures.
In section II in Fig. 8.2 at a voltage value exceeding a certain value P ^, called the elastic limit, that is P> P k
body destruction or permanent deformation may occur (P = 0; u f 0). This is the area of a perfectly viscous Newtonian body, which is represented in the form of a piston with holes placed in a cylinder with a liquid (Fig. 8.4).
During the flow of an ideally viscous fluid, one layer of fluid is shifted relative to another during their plane-parallel movement. This flow describes Newton's law, according to which, in the laminar motion of a fluid, the shear stress R proportional to the gradient of the fluid velocity or strain rate:
where and- fluid flow rate; l; - coordinate; -
velocity gradient associated with velocity 
; dr- time,
for which there is a change in deformation dy accompanying the plane-parallel motion of two layers of liquid: 
; /
Strain rate 
Rice. 8.5.
Rice. 8.4.
The relationship between the flow rate and the deformation rate follows from the above ratios:
In equations (8.1) the proportionality coefficient /; called the coefficient of internal friction, but more often - Newtonian viscosity, dynamic viscosity, or simply viscosity. This is the most important property that characterizes the structure of any dispersed system. Viscosity is a rheological constant and determines the ability of a fluid to resist flow.
The reciprocal of I / 77 viscosity is called fluidity.
The dynamic viscosity in the SI system has the dimension of H-s / m 2. This dimension is called pascal - second (Pass): = Pa s. The passalsskund is equal to the viscosity of the medium, the shear stress in which in a laminar flow and with a difference in the velocities of layers located at a distance of 1 m but the normal to the direction of velocity equal to 1 m / s is equal to 1 Pa:
... For example, the viscosity of water at 20.5 ° C is 1.005 mPas.
In the literature, there is another unit for measuring viscosity - poise (P): 1 P = 0.1 Pa-s. The viscosity of gases is about 50 times less, for high-viscosity liquids, the viscosity values can reach a thousand or millions of times greater importance, for solids a viscosity of more than 10 15 - 10 20 Pa s.
The fluids, the flow of which is described by the above equations, are called Newtonian. The magnitude of the deformation of a Newtonian fluid depends on the time of action of the stress. Since or,
then, that is, at constant voltage R deformation is proportional to the duration of this stress.
Newtonian fluids are capable of flowing (deforming) under the influence of very small external loads as long as they act. The energy spent on deformation of an ideally viscous body (Newtonian fluid) turns into heat, therefore the body of Nyotoia belongs to dissipative structures.
The ideally plastic body of Saint-Venant-Coulomb is represented by a rigid body located on a plane, during the movement of which heating is constant and does not depend on the normal, that is, perpendicular to the surface of the force (Fig. 8.5). This model is based on the law of external friction, according to which there is no deformation if the shear stress is less than a certain value Pj, called the yield point, that is, at P deformation is absent: at= 0 and y = 0. In fig. 8.2
this is the III section, showing the possibility of transition of elastic to plastic deformation.
If the stress reaches the yield point, then the developed deformation of an ideally plastic body ns has a limit, the flow occurs at any speed, that is, at P = P.j. deformation is positive:
y> 0 and y> 0.
Yield point P ^ reflects the strength of the body structure. On condition P = P t the structure of an ideal plastic body collapses, after
whereby resistance to voltage is completely absent. The energy spent on deformation of the ideally plastic body of Saint-Venant - Coulomb turns into heat, therefore it belongs to dissipative structures.
STRUCTURAL AND MECHANICAL PROPERTIES OF DISPERSED SYSTEMS
| Parameter name | Meaning |
| Topic of the article: | STRUCTURAL AND MECHANICAL PROPERTIES OF DISPERSED SYSTEMS |
| Rubric (thematic category) | Chemistry |
The formation of structures in colloidal and microheterogeneous systems is a consequence of the coagulation of these systems, and as the concentration of the dispersed phase increases, a wide “spectrum” of states passes - from truly liquid (sol) through structured liquids, gels, to solid-like systems.
Colloidal and microheterogeneous systems with liquid and solid dispersion media, like all condensed systems, have certain mechanical properties - viscosity, often plasticity, elasticity and strength. These properties are associated with the structure of systems, in this regard, they are called structural and mechanical properties, or rheological.
Colloidal and microheterogeneous dispersed systems are divided into free-dispersed and cohesive-dispersed. If the dispersion medium is a liquid, then there are also transition systems, the individual particles of which are connected to each other in loose aggregates, but do not form a continuous structure - structured liquids.
The type of system is greatly influenced by the concentration of the dispersed phase. Colloidal systems in which particles are at sufficiently large distances from each other and practically do not interact are called free-dispersed. With the introduction of a stabilizer into such a system, which prevents the approach of particles and the manifestation of molecular forces between them, it is possible to significantly increase the critical concentration at which bonds arise between the elements of the structural network. It should be noted that, in terms of their properties, such colloidal systems They are very similar to ordinary liquids, their viscosity differs little from the viscosity of the dispersion medium and slightly increases with an increase in the content of the dispersed phase.
In coherently dispersed systems, the concentration of the dispersed phase can reach high values. Particles in such a system are connected to each other by intermolecular forces and, as a result, are not capable of mutual displacement; they form a spatial network or structure. Coherently dispersed systems have, to a certain extent, the properties of solids - the ability to maintain shape, some strength, resilience, elasticity. But due to the low strength of the bond between the individual elements of the structure, they are easily destroyed and these systems acquire the ability to flow.
Structured liquids are systems with a low concentration of the dispersed phase, but with a pronounced tendency of particles to stick together. Οʜᴎ have intermediate properties, these systems are able to flow, but they do not obey the laws of flow of ordinary unstructured fluids.
STRUCTURAL AND MECHANICAL PROPERTIES OF DISPERSED SYSTEMS - concept and types. Classification and features of the category "STRUCTURAL AND MECHANICAL PROPERTIES OF DISPERSED SYSTEMS" 2017, 2018.
The appearance of structures and their nature are usually determined by measuring the mechanical properties of systems: viscosity, elasticity, plasticity, strength. Since these properties are related to structure, they are called structural – mechanical.
Structurally – the mechanical properties of the systems are investigated by rheological methods.
Rheology – the science of deformations and flow of material systems. She studies the mechanical properties of systems by the manifestation of deformation under the influence of external stresses.
The term deformation means the relative displacement of the points of the system, which does not violate its continuity.
External stress – is nothing more than the pressure of R.
In continuum mechanics, it is proved that in the case of incompressible materials, which are the majority of dispersed systems, all types of deformation (tension, compression, torsion, etc.) can be reduced to the basic – shear deformation under the action of shear stress P (N / m 2 =. Pa). The strain rate is the shear rate. Deformation is usually expressed in terms of dimensionless quantities. Strain rate – where t – time.
Studying structurally – mechanical properties of dispersed systems, it is possible to determine whether a structure is formed in the system and what is its nature.
FREE DISPERSION (STRUCTURE-FREE) SYSTEMS
Aggregatively stable sols (structureless systems) obey the laws of Newton, Poiseuille and Einstein.
Newton's law establishes a relationship between strain rate and shear stress:
Figure 11.2. Dependence of strain rate on shear stress
Figure 11.3. Pressure dependence of fluid flow
Rice. 11.4. Dependence of the viscosity of a dispersed system on concentration
where P – shear stress supporting the fluid flow, Pa; – deformation (flow) of the liquid; – deformation rate; – proportionality coefficient, called viscosity coefficient or dynamic viscosity, Pa – with; – the reciprocal of viscosity is called fluidity.
Equation (11.1) is the equation of the straight line shown in Fig. 11.2.
Viscosity – constant value, independent of P.
Poiseuille's law expresses the dependence of the volume of liquid flowing through a pipe or capillary on pressure:
where Q – liquid consumption per unit of time; P is the pressure in the pipe; TO – constant determined by the geometric parameters of the pipe or capillary, (r and l – radius and length of the pipe). From the graph corresponding to Poiseuille's law (Fig. 11.3), it can be seen that the dynamic viscosity does not depend on pressure, and the fluid flow rate is directly proportional to the pressure.
Einstein's law establishes the dependence of the viscosity of a structureless liquid dispersed system on the concentration of the dispersed phase:
where – dynamic viscosity of the dispersion medium; – volumetric concentration of the dispersed phase; – coefficient determined by the shape of the dispersed phase particles. The graph corresponding to Einstein's law is shown in Fig. 11.4.
Thus, the relative increase in viscosity is directly proportional to the relative content of the dispersed phase. The more, the more pronounced the inhibitory effect of particles, the higher the viscosity. Einstein's calculations showed that for spherical particles = 2.5, for particles of a different shape> 2.5. Fluids obeying the considered laws are called Newtonian fluids.
LIQUID STRUCTURED SYSTEMS
In the presence of a structure, the interaction between the particles of the dispersed phase cannot be neglected. The applied shear stress not only causes the fluid to flow, but can also destroy the structure existing in it. This must inevitably lead to a violation of proportionality between the applied stress P and the strain rate, the viscosity of the system becomes a value depending on P. Therefore, for such fluids the laws of Newton, Poiseuille and Einstein are not satisfied. These fluids are called non-Newtonian fluids.
To describe the relationship between the strain rate and the applied shear stress P, the empirical Ostwald equation is usually used – Weil:
Or, (11.4)
where k and n – constants characterizing a given liquid-like system.
For n – 1 and k = equation (11.4) will turn into Newton's equation. Thus, the deviation of the value of n from unity characterizes the degree of deviation of the properties of non-Newtonian fluids from Newtonian ones. For n< 1 ньютоновская вязкость уменьшается с увеличением напряжения и скорости сдвига. Такие жидкости называются псевдопластическими.
For n> 1, the Newtonian viscosity of the fluid increases with increasing stress and shear rate. Such fluids are called dilatant.
In fig. 11.5 shows the flow curve of a pseudoplastic fluid. There are three characteristic areas on the curve. In section I (OA), the system behaves like a Newtonian fluid with high viscosity. This behavior of the system is explained by the fact that at low flow rates, the structure destroyed by the applied load has time to recover. This flow is called creep.
Figure 11.5. Flow curve of pseudoplastic structured
fluid system
Creep – it is a slow flow with constant viscosity without progressive destruction of the structure.
For semi-structured systems, region I is usually small and almost impossible to detect. For highly structured systems, the range of P values at which creep is observed can be quite significant. The stress P k corresponds to the beginning of the destruction of the structure.
In section II (AB), the dependence on P loses its linear character, while the viscosity decreases. This decrease is due to the destruction of the structure. At point B, the structure is almost completely destroyed. The stress corresponding to this point is called the ultimate shear stress P m. At stresses P> P m, when the structure of the system is destroyed, the system flows like a Newtonian fluid with a viscosity.
The stress P t is called the yield point – this is the minimum shear stress at which the creep of the system transforms into flow. The stronger the structure, the higher the yield strength. Liquid flow rate per unit time Q flowing through the pipe at P< Р m можно рассчитать по уравнению Бингама: . Величина PS характеризует прочность сплошной пространственной сетки.
Figure 11.6. Flow curve of a solid structured system
At P> PS, the flow curve of a solid-like system is similar to the flow curve of a liquid-like system considered above.
For solid-like elastoplastic bodies, it is many orders of magnitude larger than for liquid-like bodies, and when the yield point P T is reached, an avalanche-like destruction of the structure occurs, followed by plastic flow.
In elastic-brittle bodies, no flow is observed, since the stress at which brittle rupture occurs is reached earlier than the yield point.
The structural and mechanical properties include viscosity, elasticity, plasticity, strength. Structural and mechanical properties are investigated by science rheology.
In dilute sols at large distances between colloidal particles, the interaction between them can be neglected. Such systems are called free-dispersed (structureless).
As the concentration of the dispersed phase increases, contacts arise between the particles, and structures are formed due to van der Waals forces. Structured colloidal systems include gels with low strength, in which the particles of the dispersed phase are separated by interlayers of the medium. At certain conditions gels can turn into sols and, conversely, during long-term storage, hydrophilic sols turn into a special "jelly-like" colloidal state. As such, they are called gels. The structure of the gel is such that micelles are not destroyed, but simply bind to each other, forming a kind of cells, inside which the medium, for example, water, is preserved. The gel can be dried by turning it into a solid colloid. An example of a hydrophilic sol is a gelatin sol. When hard gelatin swells in water, a gel forms. When the gel ("jelly") is heated, a sol is formed. All processes are reversible:
Sol D Gel D Solid colloid
Conclusion
Colloidal solutions are of great importance in our life and are widely used in technology, medicine and agriculture... Many constituent parts of living organisms are in a colloidal state: blood, lymph, intracellular fluid. Soil, natural waters, medicinal substances, foodstuffs, perfumery products, pest and weed control agents are in a colloidal state. The exceptional importance of these systems for the life of mankind determines the need to study them.
test questions
1. Describe solutions as homogeneous systems.
2. Give definitions of the ways of expressing the composition of solutions (mass fraction, molar and molar concentration of equivalents).
3. Characterize the properties of dilute molecular solutions (lowering the vapor pressure of the solvent, increasing the boiling point, lowering the freezing point, osmotic pressure).
4. Characterize electrolyte solutions, their differences from molecular solutions.
5. Give the main provisions of S. Arrhenius' theory of electrolytic dissociation and D.I. Mendeleev.
6. Give the definition of the degree of dissociation of electrolytes. How are electrolytes divided according to the degree of dissociation? Give examples of stepwise dissociation of polybasic acids and polyacid bases.
7. Describe the dissociation of weak electrolytes as a reversible process. Give the derivation of the equation of the Ostwald dilution law.
8. Explain why the dissociation constant of strong electrolytes changes with the concentration.
9. What determines the direction of reactions in electrolyte solutions? Give the formulation of Berthollet's rule.
10. Characterize water as a weak electrolyte with a dissociation constant. What is pH? How does he characterize the acidity of the solution?
11. Give the definition of the product of solubility (SR), as an indicator characterizing the equilibrium between the sediment of a poorly soluble substance and a saturated solution.
12. What is a dispersed system, dispersed phase and dispersed medium? Give the classification of dispersed systems according to the state of aggregation of phases. Give examples of dispersed systems.
13. Give a definition and give examples of lyophobic and lyophilic disperse systems.
14. Using the example of considering the liquid-gas interface, explain what free surface energy and surface tension are.
15. Describe surface phenomena - adsorption and desorption. Give definitions to the concepts - adsorbent, adsorbate, adsorptive.
16. Describe the structure of a colloidal particle by the example of AgCl obtained with an excess of AgNO 3.
17. Explain the mechanism of occurrence and structure of the electric double layer on the surface of colloidal particles.
18. Describe the methods of obtaining colloidal solutions.
19. Explain what coagulation and sedimentation of colloidal solutions is, how to cause them and how to prevent them.
20. Describe the optical and kinetic properties of colloidal solutions.
Examples of assignments
Example 1a. Determine the mass of crystalline hydrate CuSO 4 × 5H 2 O required to prepare 50 g of 10% copper sulfate solution. How much water is needed to prepare this solution?
Find the mass of crystalline hydrate containing 5 g of anhydrous salt:
1 mol of CuSO 4 × 5H 2 O contains 1 mol of CuSO 4,
v(CuSO 4 × 5H 2 O) = v(CuSO 4) = 0.031 mol.
M(CuSO 4 × 5H 2 O) = 160 + 5 × 18 = 250 g / mol,
m(CuSO 4 × 5H 2 O) = 250 × 0.031 = 7.8 g.
Find the volume of water required to prepare the solution:
m(H 2 O) = m p - pa - m v - va,
m(H 2 O) = 50 - 7.8 = 42.2 g
, r (Н 2 О) = 1 g / ml,
Answer: m(CuSO 4 5H 2 O) = 7.8 g;
V(H 2 O) = 42.2 ml.
Example 1b. Give the calculation for the preparation of 200 ml 0.3 N. a solution of nickel sulfate from NiSO 4 × 7H 2 O available in the laboratory and water.
Let us find the mass of the NiSO 4 × 7H 2 O salt. Since 1 mol of crystalline hydrate contains 1 mol of anhydrous salt, when preparing solutions of a given normal or molar concentration, the calculation is based on the molar mass of the crystalline hydrate.
m in-va = C n × M× f × V solution,
M= 59 + 32 + 64 + 18 × 7 = 281 g / mol, f = 1/2,
m v-islands = 0.3 × 281 × 0.5 × 0.2 = 8.43 g.
Solutions of molar and normal concentration are prepared in volumetric flasks of a given volume, therefore the volume of added water is not calculated.
Answer: m(NiSO 4 × 7H 2 O) = 8.43 g.
Example 1c. What volume of a 15% sodium carbonate solution (r = 1.10 g / ml) is required to prepare 120 ml of a 0.45 M solution?
Let's find in what mass of a 15% solution contains 5.72 g of sodium carbonate:
,
Answer: V 15% solution = 34.67 ml.
Example 1d. What mass of salt is needed to prepare 200 ml of a solution with a molar concentration equivalent of 0.3.
Answer: m(AlCl 3) = 26.7 g.
Example 2... Write the expressions for the dissociation constants for carbonic and sulfurous acids. Use Table 7 to determine which is the weaker electrolyte. Calculate the degree of dissociation of sulfurous acid in a 0.001 M solution.
H 2 CO 3 D H + + HCO 3 - 
HCO 3 - D H + + CO 3 2– 
K total = K 1 × K 2 = 4.5 × 10 –7 × 4.7 × 10 –11 = 2.1 × 10 –17
H 2 SO 3 D H + + HSO 3 - 
HSO 3 - D H + + SO 3 2– 
K total = K 1 × K 2 = 1 × 10 –9
From a comparison of the values of the dissociation constants of carbonic and sulfurous acids, it can be seen that carbonic acid is weaker ( K(H 2 CO 3)< K(H 2 SO 3)).
Example 3... Make up the expressions for the PR of magnesium and iron (II) hydroxides, using table 6, compare their solubility.
In saturated aqueous solutions of sparingly soluble substances, heterogeneous equilibria are established between the precipitate and electrolyte ions in the solution:
Mg (OH) 2 (q) D Mg 2+ + 2OH - PR (Mg (OH) 2) = 2 = 2.3 × 10 –13
Fe (OH) 2 (c) D Fe 2+ + 2OH - PR (Fe (OH) 2) = 2 = 7.9 × 10 –16
PR (Mg (OH) 2)> PR (Fe (OH) 2).
The solubility of magnesium hydroxide is greater than the solubility of iron (II) hydroxide.
Example 4... The solubility product of AgIO 3 is 3.2 × 10 –8 at 25 ° C. Calculate the molar concentration of AgIO 3 in its saturated solution at a given temperature.
Answer: C M = 1.79 × 10 –4 mol / L.
Example 5... Calculate the concentration of OH - ions, as well as the pH and pOH of the solution, indicate the reaction of the medium, if = 8.32 × 10 –4 mol / l.
pOH = –lg;
pOH = –lg1.2 × 10 –11 = - (log1.2 + log10 –11) = - (0.08 - 11) = 10.92
Check: pH + pOH = 14; 3.08 + 10.92 = 14
Answer: = 1.2 × 10 –11 mol / l; pOH = 10.92; pH = 3.08; acidic environment.
Example 6... Which of these salts undergo hydrolysis: lithium nitrate, chromium (III) sulfate, sodium orthophosphate, chromium (III) sulfide? Give the ionic and molecular equations of hydrolysis, indicate the reaction of the medium and the conditions for the displacement of equilibrium.
Those soluble salts that are formed by a weak base (hydrolysis by cation) or weak acid (hydrolysis by anion) undergo hydrolysis, because only the interaction of their ions leads to the formation of a weak electrolyte (binding of ions) and a shift in the ionic equilibrium of dissociation of water.
Algorithm for drawing up the equations of reversible hydrolysis:
1. Make a short ionic equation for the interaction of one ion of a weak electrolyte with one molecule of water. Specify the pH of the salt solution (> or< 7), а также влияние подкисления или подщелачивания среды на смещение равновесия гидролиза (в соответствии с принципом Ле-Шателье).
2. Make a complete ionic equation by assigning to the left and right sides of the equation the formulas of strong electrolyte ions that do not participate in hydrolysis.
3. Make up the molecular equation of salt hydrolysis according to the first stage, if necessary add the coefficients.
4. If hydrolysis proceeds via a triply charged ion, make up the hydrolysis equations for the second stage, repeating steps 1-3 and taking the product of the first stage as the initial salt. Note that the number of stages of hydrolysis with respect to a multiply charged ion that actually occur under normal conditions is, as a rule, one less than the charge of the ion, because each subsequent stage would lead to the formation of a stronger electrolyte.
Solution.
LiNO 3 - the salt is formed by a strong base (LiOH) and a strong acid (HNO 3), does not undergo hydrolysis, because ions Li + and NO 3 - do not form weak electrolytes with water, pH = 7.
Cr 2 (SO 4) 3 - the salt is formed by a weak base (Cr (OH) 3), and a strong acid (H 2 SO 4), cationic hydrolysis.
Stage I Cr 3+ + HOH D CrOH 2+ + H + pH< 7
Cr 3+ + SO 4 2- + HOH D 2+ + H + + SO 4 2–
Cr 2 (SO 4) 3 + 2HOH D 2SO 4 + H 2 SO 4
II stage 2+ + HOH D + + H +
2+ + SO 4 2– + HOH D + + H + + SO 4 2–
2SO 4 + 2HOH D 2 SO 4 + H 2 SO 4
There is practically no hydrolysis in the third stage, because this would lead to the formation of a stronger electrolyte Cr (OH) 3 than Cr (OH) 2 +.
Hydrolysis can be enhanced by binding protons to water, i.e. adding alkali, and heating and diluting the solution. The addition of acid will shift the hydrolysis equilibrium to the left.
Na 3 PO 4 - the salt is formed by a strong base (NaOH) and a medium-strength acid (H 3 PO 4), the hydroanions of which are weak electrolytes (anion hydrolysis).
Stage I PO 4 3– + HOH D HPO 4 2– + OH - pH> 7
PO 4 3– + Na + + HOH D HPO 4 2– + Na + + OH -
Na 3 PO 4 + HOH D Na 2 HPO 4 + NaOH
II stage HPO 4 2– + HOH D H 2 PO 4 - + OH -
Na + + HPO 4 2– + HOH D H 2 PO 4 - + Na + + OH -
NaHPO 4 + HOH D NaH 2 PO 4 + NaOH
Enhanced hydrolysis is possible by adding acid, heating, dilution. Alkalinization of the medium will shift the equilibrium to the left.
Under normal conditions, hydrolysis practically does not proceed in the third stage, because H 3 PO 4 is a stronger electrolyte than H 2 PO 4 -.
Cr 2 S 3 - the salt is formed by a weak polyacid insoluble base Cr (OH) 3 and a weak polybasic volatile acid - H 2 S. Such salts undergo complete irreversible hydrolysis. In this case, the final products of hydrolysis are immediately formed - a precipitate of a weak base and a volatile acid.
Cr 2 S 3 + 6H 2 O = 2Cr (OH) 3 ¯ + 3H 2 S pH "7.
Example 7a. Determine the vapor pressure of a solution at 40 ° C containing 3.6 g of glucose in 250 g of water. The vapor pressure of water at the same temperature is 55.32 mm Hg. Art.
Answer: R 1 = 55.24 mm Hg. Art.
Example 7b. Determine the freezing point of a solution containing 0.8132 g of naphthalene in 25.46 g of benzene. The freezing point of benzene is 5.5 ° C, and its cryoscopic constant is 5.12.
Example 7c. Determine the mass fraction of urea in an aqueous solution if its boiling point is 100,174 ° C. Ebullioscopic water constant 0.512.
Self-help assignments
Exercise 1. Solve the design problem.
1. How much g of Na 2 CO 3 is contained in 500 ml of 0.25 N solution?
2. What mass of KCl is required to prepare 250 ml of 0.15 M solution?
3. Find the mass of NaNO 3 required to prepare 150 ml of a 2 M solution.
4. What volume of 0.1 n CuSO 4 contains 8 g of anhydrous salt?
5. What volume of 0.3 M NaCl solution contains 2 g of salt?
6. Determine the molar concentration of a solution containing 11.2 g of KOH in 200 ml of solution.
7. How much g Na 2 CO 3 is contained in 500 ml of 0.25 N solution?
8. Determine the molar concentration of a Na 2 SO 4 solution containing 42 g of salt in 300 ml of solution.
9. Calculate the molar concentration of Ba (OH) 2 in the solution if 2 l contains 2 g of alkali.
10. How many grams of NaBr is required to prepare 700 ml of 0.15 N salt solution?
11. What mass of KCl is required to prepare 250 ml of 1.15 M salt solution?
12. Find the mass of NaOH required to prepare 2 liters of 0.3 N solution.
13. What volume of 2 M NaCl solution contains 5 g of salt?
14. What is the concentration of a solution containing 9 g of CaCl 2 in 500 ml of solution?
15. What volume of 0.1 N CuCl 2 solution contains 5 g of salt?
Assignment 2. Write the dissociation equations for the compounds, the formulas of which are given below. Where necessary, provide the equations of stepwise dissociation. Write the expressions for the dissociation constants of the proposed acids. Which of the two is the weaker? Calculate the degree of dissociation of one of the acids in a 0.01 M solution.
Assignment 3... Make up the expressions for the PR of the indicated substances, compare their solubility. Calculate the concentration of cations and anions in a saturated solution of one of these substances.
| Option | ||||||||
| Substances | AgBr AgCl | Mg (OH) 2 MgS | CaSO 4 BaSO 4 | AgI AgCl | FeS CuS | SrSO 4 BaSO 4 | CdS CuS | AgBr AgI |
| Option | ||||||||
| Substances | CaCO 3 CaSO 4 | HgS CuS | FeS Fe (OH) 2 | CuI AgI | CuCl CuI | Cu (OH) 2 CuS | Zn (OH) 2 ZnS |
Assignment 4. Calculate the concentration of OH - ions, as well as the pH of the solution, indicate the reaction of the medium if the concentration of H + ions is:
| Option | , mol / l | Option | , mol / l | Option | , mol / l |
| 3.82 × 10 -12 | 9.12 × 10 -5 | 7.55 × 10 -7 | |||
| 2.85 × 10 –2 | 6.38 × 10 -10 | 4.52 × 10 –4 | |||
| 2.88 × 10 -6 | 8.85 × 10-11 | 3.33 × 10 –3 | |||
| 5.28 × 10 -13 | 8.32 × 10 -6 | 1.86 × 10 -11 | |||
| 7.56 × 10 -1 | 6.25 × 10 -9 | 8.84 × 10 -12 |
Assignment 5... Give the molecular and ionic equations of those reactions that are possible in solution, explain their direction.
| Option | Reaction schemes | Option | Reaction schemes |
| Cu (OH) 2 + Na 2 SO 4 ® Na 2 SO 4 + BaCl 2 ® Fe (OH) 3 + H 2 SO 4 ® Al (OH) 3 + NaOH ® Pb (NO 3) 2 + H 2 S ® | NaOH + CaCO 3 ® CaSO 4 + SrCl 2 ® 2 SO 4 + HCl ® Cu (NO 3) 2 + H 2 S ® CuCl 2 + AgNO 3 ® | ||
| H 2 CO 3 + Ca (NO 3) 2 ® NaCl + LiNO 3 ® Cr (OH) 3 + KOH ® Na 2 CO 3 + HCl ® Ca (HCO 3) 2 + Ca (OH) 2 ® | KCl + H 2 SO 4 ® 2 SO 4 + HCl ® Pb (NO 3) 2 + Cu (OH) 2 ® Na 2 CO 3 + Ca (NO 3) 2 ® NH 4 NO 3 + NaOH ® | ||
| CaCO 3 + LiCl ® K 2 S + HCl ® Mg (OH) 2 + Na 2 SO 4 ® 2 SO 4 + H 2 SO 4 ® Pb (OH) 2 + HNO 3 ® | Pb (OH) 2 + KOH ® H 2 S + FeCl 2 ® ZnSO 4 + Cu (OH) 2 ® NaH 2 PO 4 + NaOH ® CaCl 2 + Na 2 CO 3 ® | ||
| Cu (NO 3) 2 + BaSO 4 ® Ba (OH) 2 + H 2 SO 4 ® CuSO 4 + KCl ® Be (OH) 2 + KOH ® Ba (HCO 3) 2 + NaOH ® | HNO 3 + K 2 SO 4 ® Al (OH) 3 + KOH ® KOH + Na 2 SO 4 ® CaCO 3 + H 2 O + CO 2 ® BaS + CuSO 4 ® | ||
| Be (OH) 2 + KOH ® FeOHCl 2 + HCl ® Na 2 SO 4 + K 2 CO 3 ® NaNO 3 + Zn (OH) 2 ® CaCl 2 + H 2 SO 4 ® | Na 2 SiO 3 + HCl ® Cr (OH) 3 + HNO 3 ® CuCl 2 + KNO 3 ® CrOHSO 4 + H 2 SO 4 ® AgNO 3 + CuCl 2 ® | ||
| Ba (NO 3) 2 + Na 2 SO 4 ® CuOHCl + NaOH ® KNO 3 + Na 2 CO 3 ® Al (OH) 3 + KCl ® Na 2 CO 3 + H 2 SO 4 ® | CaS + HCl ® CoCl 2 + KOH ® Pb (NO 3) 2 + KCl ® Be (OH) 2 + Ba (OH) 2 ® NaNO 3 + H 2 SO 4 ® | ||
| BaCl 2 + H 2 SO 4 ® Na 3 PO 4 + H 2 S ® NaOH + AlCl 3 ® PbOHNO 3 + HNO 3 ® Na 2 S + KCl ® | K 2 SO 4 + H 2 SO 3 ® BaS + CuSO 4 ® Al (OH) 3 + NaCl ® SrSO 4 + CaCl 2 ® Cr (OH) 3 + NaOH ® | ||
| Be (OH) 2 + KOH ® AgNO 3 + CaCl 2 ® HNO 3 + CaS ® KBr + Zn (OH) 2 ® NaHSO 4 + NaOH ® |
Assignment 6... Write molecular reaction equations corresponding to the following short ionic.
| Var. | Brief ionic equation | Var. | Brief ionic equation |
| H 2 PO 4 - + OH - = HPO 4 2– + H 2 O Pb 2+ + SO 4 2– = PbSO 4 ¯ + + 2H + = Cr 3+ + 2H 2 O | 3Ag + + PO 4 3– = Ag 3 PO 4 ¯ HCO 3 - + OH - = CO 3 2– + H 2 O BaCO 3 + 2H + = Ba 2+ + CO 2 + H 2 O | ||
| + + 2H + = Fe 3+ + 2H 2 O H 2 S + 2Ag + = Ag 2 S¯ + 2H + CO 3 2– + 2H + = CO 2 + H 2 O | HAsO 4 2– + OH - = AsO 4 3– + H 2 O 2+ + H + = Al 3+ + H 2 O 2H + + S 2– = H 2 S | ||
| HCO 3 - + H + = H 2 O + CO 2 Ni (OH) 2 + 2H + = Ni 2+ + 2H 2 O CO 3 2– + Ca 2+ = CaCO 3 ¯ | FeOH 2+ + 2OH - = Fe (OH) 3 ¯ MgCO 3 + 2H + = Mg 2+ + CO 2 + H 2 O Cu 2+ + S 2 - = CuS¯ | ||
| SO 3 2– + 2H + = SO 2 + H 2 O + + H + = Cu 2+ + H 2 O 3Ca 2+ + 2PO 4 3 - = Ca 3 (PO 4) 2 ¯ | H 2 S + Pb 2+ = PbS¯ + 2H + BaSO 3 + 2H + = Ba 2+ + SO 2 + H 2 O + + OH - = Fe (OH) 3 ¯ | ||
| H 2 PO 4 - + 2OH - = PO 4 3– + 2H 2 O Pb 2+ + 2I - = PbI 2 ¯ CaCO 3 + 2H + = Ca 2+ + CO 2 + H 2 O | Pb 2+ + 2Br - = PbBr 2 ¯ HS - + OH - = S 2– + H 2 O BaSeO 3 + 2H + = Ba 2+ + SeO 2 + H 2 O | ||
| HSO 3 - + OH - = SO 3 2– + H 2 O Ag + + I - = AgI¯ Co (OH) 2 + 2H + = Co 2+ + 2H 2 O | + + OH - = Fe (OH) 2 ¯ Sr 2+ + SO 4 2+ = SrSO 4 ¯ SO 3 2– + 2H + = H 2 O + SO 2 | ||
| 3Ba 2+ + 2PO 4 3- = Ba 3 (PO 4) 2 ¯ HSe - + H + = H 2 Se Ag + + Br - = AgBr¯ | + + H + = Cu 2+ + H 2 O H 2 S + 2Ag + = Ag 2 S¯ + 2H + H 2 AsO 4 - + 2OH - = 2H 2 O + AsO 4 3– | ||
| Hg 2+ + S 2– = HgS¯, + + 2H + = Al 3+ + 2H 2 O, H + + OH - = H 2 O |
Assignment 7. Calculate the concentration of H + and OH - ions in a solution, the pH of which is:
| Option | ||||||||
| NS | 4,3 | 10,8 | 2,6 | 12,5 | 8,3 | 11,6 | 1,8 | 6,7 |
| Option | ||||||||
| NS | 9,4 | 3,8 | 9,3 | 2,3 | 13,2 | 5,6 | 1,1 |
Z Adania 8. Which of these salts are hydrolyzed? Give the molecular and ionic equations of hydrolysis, indicate the reaction of the medium and the conditions for the displacement of the equilibrium of hydrolysis.
| Option | Salt names |
| Potassium phosphate, copper (II) sulfate, rubidium chloride, aluminum sulfide | |
| Ammonium acetate, chromium (III) nitrate, lithium carbonate, cesium sulfate | |
| Sodium nitrate, barium sulfide, iron (III) chloride, iron (III) sulfide | |
| Barium chloride, sodium sulfide, copper (II) sulfate, chromium (III) carbonate | |
| Potassium sulfite, cobalt (II) nitrate, potassium nitrate, chromium (III) sulfide | |
| Lithium iodide, sodium sulfite, aluminum nitrate, ammonium carbonate | |
| Sodium acetate, iron (II) sulfate, aluminum carbonate, potassium bromide | |
| Lead (II) nitrate, ammonium nitrite, potassium sulfate, sodium carbonate | |
| Zinc chloride, barium nitrate, aluminum sulfide, copper (II) acetate | |
| Calcium nitrite, nickel (II) nitrate, iron (III) carbonate, rubidium chloride | |
| Potassium bromide, sodium sulfide, copper (II) chloride, chromium (III) sulfide | |
| Copper (II) nitrate, calcium sulfide, strontium nitrate, ammonium sulfide | |
| Copper (II) chloride, cesium phosphate, potassium sulfate, iron (II) sulfide | |
| Sodium nitrate, beryllium chloride, potassium phosphate, ammonium sulfite | |
| Aluminum sulfate, barium nitrite, potassium chloride, aluminum sulfide |
Assignment 9... Give the molecular and short ionic-molecular reaction equations that can be used to carry out the following transformations, indicate the conditions for their occurrence:
| Var | Transformations | Var | Transformations |
| BaCO 3 ® Ba (HCO 3) 2 ® BaCO 3 Fe 2 O 3 ® Fe (OH) 3 Cu ® CuSO 4 Cu (NO 3) 2 ® CuOHNO 3 ® Cu (NO 3) 2 Na 2 SO 4 ® NaCl | ZnO ® Zn (OH) 2 Ca 3 (PO 4) 2 ® CaHPO 4 ® Ca 3 (PO 4) 2 FeCl 3 ® FeOHCl 2 ® FeCl 3 KNO 3 ® HNO 3 Fe ® FeCl 2 | ||
| Hg ® Hg (NO 3) 2 ZnSO 4 ® (ZnOH) 2 SO 4 ® ZnSO 4 Na 3 PO 4 ® NaCl K 2 SO 3 ® KHSO 3 ® K 2 SO 3 MgO ® Mg (OH) 2 | KCl ® KNO 3 Cu ® CuCl 2 PbO ® Pb (OH) 2 MgCO 3 ® Mg (HCO 3) 2 ® MgCO 3 AlCl 3 ® Al (OH) 2 Cl ® AlCl 3 | ||
| Fe ® Fe (NO 3) 2 CuSO 4 ® CuCl 2 Al 2 O 3 ® Al (OH) 3 FeCl 2 ® FeOHCl ® FeCl 2 K 3 PO 4 ® KH 2 PO 4 ® K 3 PO 4 | K 2 S ® KHS ® K 2 S CuO ® Cu (OH) 2 Cr 2 (SO 4) 3 ® CrOHSO 4 ® Cr 2 (SO 4) 3 KCl ® HCl Al ® Al 2 (SO 4) 3 | ||
| Cr (NO 3) 3 ® CrOH (NO 3) 2 ® Cr (NO 3) 3 SnO ® Sn (OH) 2 K 2 CO 3 ® KHCO 3 ® K 2 CO 3 K 3 PO 4 ® KNO 3 Cu ® Cu ( NO 3) 2 | Fe ® FeCl 3 Ba 3 (PO 4) 2 ® Ba (H 2 PO 4) 2 ® Ba 3 (PO 4) 2 FeO ® Fe (OH) 2 AlCl 3 ® AlOHCl 2 ® AlCl 3 K 2 SO 4 ® KCl | ||
| MnO ® Mn (OH) 2 Al 2 (SO 4) 3 ® 2 SO 4 ® Al 2 (SO 4) 3 BaCl 2 ® Ba (NO 3) 2 CaCO 3 ® Ca (HCO 3) 2 ® CaCO 3 Mg ® Mg (NO 3) 2 | ZnCl 2 ® ZnOHCl ® ZnCl 2 Na 2 SO 3 ® NaHSO 3 ® Na 2 SO 3 Cr 2 O 3 ® Cr (OH) 3 CuSO 4 ® Cu (NO 3) 2 Fe ® FeSO 4 | ||
| Na 3 PO 4 ® NaH 2 PO 4 ® Na 3 PO 4 NiO ® Ni (OH) 2 Al (NO 3) 3 ® Al (OH) 2 NO 3 ® Al (NO 3) 3 Ag ® AgNO 3 NaNO 3 ® Na 2 SO 4 | CoO ® Co (OH) 2 Al 2 (SO 4) 3 ® 2 SO 4 ® Al 2 (SO 4) 3 KCl ® K 2 SO 4 Al ® AlCl 3 Na 2 CO 3 ® NaHCO 3 ® Na 2 CO 3 | ||
| CrCl 3 ® Cr (OH) 2 Cl ® CrCl 3 BeO ® Be (OH) 2 Na 2 SO 4 ® NaNO 3 Zn ® Zn (NO 3) 2 Mg 3 (PO 4) 2 ® Mg (H 2 PO 4) 2 ® Mg 3 (PO 4) 2 | Na 2 S ® NaHS ® Na 2 S Fe 2 O 3 ® Fe (OH) 3 Pb (NO 3) 2 ® PbOHNO 3 ® Pb (NO 3) 2 KNO 3 ® K 2 SO 4 Al ® Al 2 (SO 4) 3 | ||
| PbO ® Pb (OH) 2, K 3 PO 4 ® KCl, Pb ® Pb (NO 3) 2 CaSO 3 ® Ca (HSO 3) 2 ® CaSO 3, Fe 2 (SO 4) 3 ® FeOHSO 4 ® Fe 2 ( SO 4) 3 |
Task 10. Solve the design problem.
1. A solution of 7.252 g of glucose in 200 g of water freezes at –0.378 ° C. The cryoscopic constant of water is 1.86. Determine the molecular weight of glucose.
2. The vapor pressure of water at 80 ° C is 355.1 mm Hg. Calculate the vapor pressure over a solution containing 6 g of urea CO (NH 2) 2 in 180 g of water.
3. At what temperature will a solution of 0.022 mol of glucose in 100 g of water freeze? The cryoscopic constant of water is 1.86.
4. Determine the relative decrease in vapor pressure over a solution containing 4.14 g of salicylic acid C 7 H 6 O 3 in 100 g of ethyl alcohol.
5. Determine the freezing point of a solution of 0.625 grams of urea CO (NH 2) 2 in 50 grams of water. Cryoscopic water constant 1.86.
6. The vapor pressure of benzene (C 6 H 6) at 20 ° C is equal to 75.18 mm Hg. Calculate the vapor pressure over a solution containing 0.514 g of diphenylamine (C 6 H 5) 2 NH in 50 g of benzene at this temperature.
7. Determine the freezing point of a solution containing 1.205 · 10 23 molecules of non-electrolyte in 1 liter of water. Cryoscopic water constant 1.86.
8. What is the mass fraction of fructose С 6 Н 12 О 6 in an aqueous solution that freezes at –0.524 ° С? Cryoscopic water constant 1.86.
9. The vapor pressure of water at 50 ° C is 92.5 mm Hg. How many moles of glucose must be dissolved in 270 g of water so that the vapor pressure above the solution drops by 0.5 mm Hg?
10. Calculate the vapor pressure of water over 5% aqueous solution glycerol C 3 H 8 O 3 at 25 ° C. The vapor pressure of water at this temperature is 23.76 mm. rt. Art.
11. Determine the freezing point of a solution of 0.8 grams of urea CO (NH 2) 2 in 80 g of water. Cryoscopic water constant 1.86.
12. Calculate the boiling point of an aqueous one percent solution of glycerin C 3 H 8 O 3. Ebullioscopic water constant 0.512.
13. Calculate the molecular weight of benzaldehyde if a solution of 1.612 ° g of it in 100 g of C 4 H 10 O ether boils at 34.806 ° C. The boiling point of ether is 34.5 ° C, and its ebullioscopic constant is equal to 2.
14. The vapor pressure of a solution containing 0.425 g of aniline in 100 g of ether (C 4 H 10 O) at 20 ° C is 440.9 mm Hg. The vapor pressure over pure ether at the same temperature is 442.4 mm Hg. Art. Determine the molecular weight of aniline.
15. The vapor pressure of benzene (C 6 H 6) at 20 ° C is equal to 75.18 mm Hg. Calculate the vapor pressure over a solution containing 0.4 g of aniline C 6 H 7 N in 50 g of benzene at this temperature.
Test tasks for self-control
1 ... The number of moles of a substance per unit volume of a solution -
1) molar concentration;
2) normal concentration;
3) mass fraction.
2 ... The ratio of the mass of the solute to the mass of the solution is
1) mass fraction;
2) normal concentration;
3) molar concentration.
3 ... The solubility of a gas is proportional to its partial pressure above the solution -
1) Henry's Law;
2) Raoult's law;
3) Van't Hoff's law.
4 ... Does it belong to strong electrolytes?
5 ... In solution, the reaction is not possible:
1) Be (OH) 2 + 2KOH = K 2;
2) KOH + HNO 3 = KNO 3 + H 2 O;
3) KOH + NaNO 3 = KNO 3 + NaOH.
6 ... Negative logarithm of the concentration of hydrogen ions in solution:
1) K instability;
1) S. Arrhenius;
2) D.I. Mendeleev;
3) N.K. Kurnakov.
8 ... Electrolytes, during the dissociation of which only OH - - ions are formed as anions, are:
1) acids;
2) grounds;
9 ... Molecules undergo stepwise dissociation in solution ...
8. Equation K q = α 2 with is an expression of the law:
2) Van't Hoffa;
3) Ostwald.
11 ... The dispersed system is:
1) sugar solution;
2) mineral water;
3) milk.
12 ... According to the theory of the structure of colloidal solutions, a micelle is a _______ particle:
1) electrically neutral;
2) negatively charged;
3) positively charged.
13 ... The movement of particles of the dispersed phase to one of the electrodes is called:
1) electrophoresis;
2) electroosmosis;
3) coagulation.
14 ... The process of particle enlargement, leading to the loss of sol stability, is called:
1) sedimentation;
2) coagulation;
3) electrophoresis.
15 ... Light scattering by particles of the dispersed phase of a colloidal solution, which is a consequence of the colloidal degree of dispersion of these particles, discovered:
2) Tyndall;
Answers to test tasks
| Question number | |||||||||||||||
| Answer No. |
Laboratory work No. 6
Part I. Strong and weak electrolytes
Objectives of the work: to study the electrical conductivity of electrolyte solutions and the influence of various factors on it.
Reagents: 1 n. solutions of hydrochloric acid, acetic acid, sodium hydroxide and ammonium hydroxide.
Equipment: dc power supply
The structure is understood as the relative position of body parts. The structure of dilute aggregatively stable lyosols is similar to the structure of true solutions. An increase in the concentration of particles leads to their aggregation, and then to coagulation. The emergence of a structure in dispersed systems is always associated with the concept of coagulation. The formation of the structure goes through the following stages:
sol structured liquid gel solid-like systems.
Structuring leads to a change in the nature of the flow or complete solidification of the liquid and a change in all its properties. Disperse systems acquire the ability to resist the load, the nature of their flow changes, etc. The features of the behavior of various systems during flow and deformation are studied by rheology - the science of deformation and flow of bodies. The structural and mechanical properties of dispersed systems are studied by physical and chemical mechanics, which is a section of the course on surface phenomena and dispersed systems. Structural and mechanical properties include: viscosity, plasticity, elasticity and strength.
Free dispersed state of lyosols. If the particles do not interact with each other and are able to move freely in a dispersion medium, then such a state of lyosols is called free-dispersed. Free-dispersed systems flow like any liquid. The resistance to external pressure during flow is characterized by viscosity. But the nature of their flow is different from Newtonian fluids. Newtonian fluids are those that obey Newton's law. Newton found that the force of internal friction (F), equal in magnitude, but opposite in direction to the force applied from the outside, is proportional to the area of the layer (S) to which this force is applied and the change in deformation over time (deformation rate dх / dτ):
F = η S (dх / dτ) = ηSγ
The proportionality coefficient η is called the coefficient of viscosity or viscosity of a liquid. The ratio F / S = P is called shear stress. Free-dispersed systems and Newtonian fluids flow at any shear stress.
η = P / γ. (XIII.1)
For Newtonian fluids, viscosity is a constant value at a given temperature and does not depend on the shear stress (Fig. XIII.1).
Lyosols have a number of features. They do not obey Newton's law. The viscosity of the sol is always higher than the viscosity of the dispersion medium. Due to the presence of particles of the dispersed phase, the flow of sols is characterized by early turbulence (i.e., the Reynolds number Re for them is less than for Newtonian fluids). The viscosity of the sols depends on the measurement method and the velocity gradient, i.e. is not constant. Therefore, colloidal systems are characterized by effective viscosity η *. Newton's law for them will be written in the form
P = η * γ. (XIII.2)
The dependence of the viscosity of free dispersed sols on the concentration of the dispersed phase is described by various forms of the Einstein equation
(η - η o) / η o = Kφ;
η / η about = 1+ Kφ;
η = η о (1+ Кφ),
where η is the viscosity of the sol;
η about - viscosity of the dispersion medium;
η / η about - the relative viscosity of the sol; K is a coefficient that depends on the shape of the particles;
φ is the volume fraction of the dispersed phase (V dis) in the total volume of the system (V) (φ = V dis / V).
Figure XIII.1 - Dependence of strain rate on shear stress for Newtonian fluids (1) and free-dispersed lyosols (2).
For spherical particles with a volume fraction of the dispersed phase ≤6%, the Einstein equation takes the form: η = η о (1+ 2.5φ), with a volume fraction of the dispersed phase ≤30%, this equation is written as
η = η о (1+ 2.5φ +14.7 φ 2). (XIII.3)
The dependence of the viscosity of the sol on the concentration of the dispersed phase is shown in Fig.
Sol structuring ... With an increase in concentration or as a result of coagulation, a spatial structure is formed in the sols. The structure is a spatial framework formed by particles of the dispersed phase, which are interconnected. Such structured dispersed systems are called bound dispersed systems. They are characterized by a new set of properties, showing strength, plasticity, elasticity, and fragility.
Classification of structures according to P.A. Rebinder ... Depending on the nature of the forces acting in the structured system, P.A. Rebinder proposed to distinguish between two main types of structures: coagulation (reversibly collapsing) and condensation-crystallization (irreversibly collapsing).
Figure XIII.2- Dependence of the viscosity of a Newtonian liquid (curve 1), free-dispersed lyosol (curve 2), structured sol (curve 3) on the concentration of the dispersed phase
Coagulation structures arise as a result of the loss of the aggregate stability of the system and the interaction of particles in the far energy minimum of the energy curve. In this case, the particles do not stick together completely, but only weakly interact with each other with certain parts, on which the stability factor is removed. The particles form a spatial network, and gelation occurs in the system. In this case, the solution changes its mechanical properties. The diagram of the resulting structure is shown in Fig. XIII.3.
Disperse systems in which the formation of a coagulation structure has occurred are called gels. Gelation is a reversible process. It is facilitated by an increase in the concentration of the dispersed phase, an increase in the degree of dispersion, the addition of electrolytes, the asymmetry of the dispersed phase particles, a decrease in temperature, and the addition of surfactants.
Figure XIII.3- Scheme of the gel structure
Gels exhibit a number of characteristic properties. Spontaneous recovery of the gel after its mechanical destruction is called thixotropy. There are strength thixotropy, which is associated with the destruction and formation of a spatial network, and viscous thixotropy, which is associated with the destruction and formation of particle aggregates.
Syneresis is characteristic of gels. This is a spontaneous decrease in the size of the gel with the simultaneous release of a dispersion medium from it. The essence of this phenomenon is that during storage, the particles in the gel are rearranged, the bonds between them increase and they come closer to each other. This causes the dispersion medium to be squeezed out.
Gels tend to dry out to form a xerogel and swell when a dispersion medium is added.
Gels are characterized by structural viscosity. In the presence of a coagulation structure, the flow of the gel begins only after its destruction. In this case, the voltage R exceeds the critical shear stress Θ required to destroy the structure. The quantity Θ is called the yield point, and the flow of gels is called plastic flow. To describe the properties of such systems, the Bingham-Shvedov equation is used:
P- Θ = η'γ,
where η 'is the plastic viscosity.
The rheological curve for the gel is shown in Fig.
Figure XIII.4 - Gel rheology curve
Condensation-crystallization structures arise as a result of chemical interaction between particles and the formation of a rigid bulk structure. This process corresponds to coagulation in the near potential minimum of the energy curve. These are typical structures for bound dispersed systems. Their destruction is irreversible. They do not swell and exhibit elastic-brittle properties.
