Definition
Let's put some sparingly soluble salt, for example, AgCl, into a beaker and add distilled water to the precipitate. In this case, Ag + and Cl - ions, being attracted by the surrounding dipoles of water, gradually break away from the crystals and go into solution. Colliding in solution, Ag + and Cl - ions form AgCl molecules and are deposited on the surface of crystals. Thus, two mutually opposite processes occur in the system, which leads to dynamic equilibrium, when as many Ag + and Cl - ions pass into the solution per unit time as they precipitate. The accumulation of Ag + and Cl - ions in the solution stops, it turns out saturated solution. Therefore, we will consider a system in which there is a precipitate of a sparingly soluble salt in contact with a saturated solution of this salt. In this case, two mutually opposite processes take place:
1) The transition of ions from the precipitate to the solution. The rate of this process can be considered constant at a constant temperature: V 1 = K 1 ;
2) Precipitation of ions from solution. The rate of this process V 2 depends on the concentration of Ag + and Cl - ions. According to the law of mass action:
Since the system is in equilibrium, then
V1 = V2
k2 = k1
- = k 2 / k 1 = const (for T = const)
Thus, the product of ion concentrations in a saturated solution of a sparingly soluble electrolyte at a constant temperature is constant value . This value is called the solubility product (PR).
In the given example, PRAgCl = . . In cases where the electrolyte contains two or more identical ions, the concentration of these ions must be raised to the appropriate power when calculating the solubility product.
For example, PRAg 2 S = 2
- ; PRPbI 2 = 2
In the general case, the expression for the solubility product for the electrolyte A m B n
PRA m B n = [A] m [B] n .
The values of the solubility product for different substances are different.
For example, PRCaCO 3 \u003d 4.8
- 10 -9; PRAgCl \u003d 1.56 10 -10.
The PR is easy to calculate, knowing the solubility of the compound at a given t°.
Example 1
The solubility of CaCO 3 is 0.0069 or 6.9
- 10 -3 g/l. Find PRCaCO 3 .
Solution
We express the solubility in moles:
SCaCO 3 = (6,9
· 10 -3 ) / 100,09 = 6.9. 10 -5 mol/l
MCaCO3
Since each CaCO 3 molecule gives one Ca 2+ and CO 3 2- ion when dissolved, then
\u003d [CO 3 2-] \u003d 6.9
- 10 -5 mol/l,
therefore, PRCaCO 3 = = 6.9 10 -5 6.9 10 -5 = 4.8 10 -9
Knowing the value of PR, one can, in turn, calculate the solubility of a substance in mol/l or g/l.
Example 2
Solubility product PRPbSO 4 = 2.2
- 10 -8 g/l.
What is the solubility of PbSO 4 ?
Solution
We denote the solubility of PbSO 4 through X mol / l. Having gone into solution, X moles of PbSO 4 will give X Pb 2+ ions and X SO 4 2- ions, i.e.:
X
PRPbSO 4 = =
- < ПРAgCl - ненасыщенный раствор
- = PRAGCl - saturated solution
- > PRAGCl - supersaturated solution
A precipitate is formed when the product of the ion concentrations of a sparingly soluble electrolyte exceeds the value of its solubility product at a given temperature. When the ion product becomes equal to the value of PR, precipitation stops. Knowing the volume and concentration of the mixed solutions, it is possible to calculate whether the resulting salt will precipitate.
Example 3
Does a precipitate form when mixing equal volumes of 0.2 M solutions of Pb(NO 3) 2 and NaCl.
PRPbCl 2
- 10 -4 .
Solution
When mixed, the volume of the solution doubles and the concentration of each of the substances will decrease by half, i.e. becomes 0.1 M or 1.0
- 10 -1 mol/l. The concentrations of Pb 2+ and Cl - will be the same. Therefore, 2 = 1 10 -1 (1 10 -1) 2 = 1 10 -3 . The value obtained exceeds PRPbCl 2 (2.4 10 -4). Therefore, part of the salt PbCl 2 precipitates. From the foregoing, it can be concluded that various factors influence the formation of precipitation.
Influence of the concentration of solutions
A sparingly soluble electrolyte with a sufficiently large value of PR cannot be precipitated from dilute solutions. For example, PbCl 2 will not precipitate when mixing equal volumes of 0.1 M solutions of Pb(NO 3) 2 and NaCl. When mixing equal volumes, the concentrations of each of the substances will become 0.1 / 2 = 0.05 M or 5
- 10 -2 mol/l. Ionic product 2 \u003d 5 10 -2 (5 10 -2) 2 \u003d 12.5 10 -5. The value obtained is less than PRPbCl 2 , hence no precipitation will occur.
Influence of the amount of precipitant
For the most complete precipitation, an excess of precipitant is used.
For example, we precipitate the BaCO 3 salt: BaCl 2 + Na 2 CO 3 ® BaCO 3 ¯ + 2NaCl. After adding an equivalent amount of Na 2 CO 3 , Ba 2+ ions remain in the solution, the concentration of which is determined by the value of PR.
An increase in the concentration of CO 3 2- ions, caused by the addition of an excess of the precipitant (Na 2 CO 3), will entail a corresponding decrease in the concentration of Ba 2+ ions in the solution, i.e. will increase the completeness of the deposition of this ion.
Influence of the ion of the same name
The solubility of sparingly soluble electrolytes decreases in the presence of other strong electrolytes having similar ions. If a solution of Na 2 SO 4 is gradually added to an unsaturated solution of BaSO 4, then the ion product, which was initially less than PRBaSO 4 (1.1
- 10 -10), will gradually reach the PR and exceed it. Precipitation will begin.
Temperature effect
PR is a constant value at a constant temperature. As the temperature increases, the PR increases; therefore, precipitation is best carried out from cooled solutions.
Dissolution of precipitation
The solubility product rule is important for transferring sparingly soluble precipitates into solution. Let's assume that it is necessary to dissolve the precipitate of BaCO 3 . The solution in contact with this precipitate is saturated with respect to BaCO 3 .
This means that = PRBaCO 3 .
If an acid is added to the solution, then the H + ions will bind the CO 3 2- ions present in the solution into molecules of fragile carbonic acid:
2H + + CO 3 2- ® H 2 CO 3 ® H 2 O + CO 2 -
As a result, the concentration of the CO 3 2- ion will sharply decrease, the ion product will become less than the value of PRBaCO 3 . The solution will be unsaturated with respect to BaCO 3 and part of the BaCO 3 precipitate will go into solution. With the addition of a sufficient amount of acid, the entire precipitate can be brought into solution. Consequently, the dissolution of the precipitate begins when, for some reason, the ion product of a sparingly soluble electrolyte becomes less than the PR value. In order to dissolve the precipitate, an electrolyte is introduced into the solution, the ions of which can form a slightly dissociated compound with one of the ions of a sparingly soluble electrolyte. This explains the dissolution of sparingly soluble hydroxides in acids.
Fe(OH) 3 + 3HCl ® FeCl 3 + 3H 2 O
OH - ions bind into slightly dissociated H 2 O molecules.
Table. Solubility product (SP) and solubility at 25°C of some sparingly soluble substances
| Formula | Solubility | PR mol/l |
| AgBr | 7,94 . 10 -7 | 6,3 . 10 -13 |
| AgCl | 1,25 . 10 -5 | 1,56 . 10 -10 |
| AgI | 1,23 . 10 -8 | 1,5 . 10 -16 |
| Ag2CrO4 | 1,0 . 10 -4 | 4,05 . 10 -12 |
| BaSO4 | 7,94 . 10 -7 | 6,3 . 10 -13 |
| CaCO3 | 6,9 . 10 -5 | 4,8 . 10 -9 |
| PbCl2 | 1,02 . 10 -2 | 1,7 . 10 -5 |
| PbSO4 | 1,5 . 10 -4 | 2,2 . 10 -8 |
HARDLY SOLUBLE ELECTROLYTES
The law of mass action is applicable to a heterogeneous equilibrium system consisting of crystals of a sparingly soluble electrolyte (salt, base, acid) and its ions in a saturated solution. Consider the equilibrium observed in a saturated solution of some hardly soluble substance, for example, CaSO 4 . In this system, the precipitate is in equilibrium with a saturated solution of this substance:
СaSO 4 Ca 2+ + SO 4 2–
precipitate solution
Under the established heterogeneous ionic equilibrium, as many ions pass into the solution per unit time as they again return to the precipitate (due to the negligible solubility, we assume that the degree of ionization of the electrolyte in the solution is 1). The equilibrium constant for the precipitate dissolution process is as follows:
whence, K∙ TV = ∙
The concentration of a solid is a constant value:
TV = const.
Taking this into account, K∙ TV as a product of two constant values can also be considered a constant value, some constant characteristic of a given substance. This constant is called the solubility product. It is denoted by PR:
For a saturated solution of calcium sulfate, the solubility product at 25°C was found to be 3.72∙10 -5 g-ion 2 /l 2 .
The solubility product characterizes the solubility of electrolytes. For binary electrolytes, the solubility is numerically equal to
s = √PR.
In the general case, if a poorly soluble electrolyte has a complex composition A n B m and more than two ions are formed during its dissociation:
A n B m ↔ nA a + + mB b – ,
then in the expression of the equilibrium constants, the concentrations of ions are written in powers equal to the stoichiometric coefficients n And m.
PR \u003d n [B b - ] m
Hence, for a saturated aqueous solution of a sparingly soluble electrolyte, the product of the equilibrium molar concentrations of its ions in powers equal to stoichiometric coefficients at a given temperature is a constant value, called the solubility product.
Comparing the values of the solubility products of sparingly soluble salts, one can see which of them dissolves better (Table 15).
PR values are used in general chemistry, analytical chemistry, hydrochemistry, ocean chemistry, ecology, etc., since they allow quantifying:
conditions for the formation and dissolution of precipitation;
From the value of PR follows the condition for the formation and dissolution of the precipitate:
If n [B b - ] m = PR, then the precipitate is in equilibrium with the solution (saturated solution);
If n [B b - ] m > PR, then the precipitate precipitates (supersaturated solution);
If n [B b - ] m< ПР, то осадок растворяется (ненасыщенный раствор).
Table 15
The solubility product of some
sparingly soluble electrolytes at 25 ° C
|
Electrolyte |
Electrolyte |
Electrolyte | |||
Thus, if during a particular chemical reaction, the product of the concentrations of the ions participating in it becomes greater than the solubility product, then a precipitate of a sparingly soluble substance precipitates. And vice versa, if the product of the concentrations of the ions of a given electrolyte in its saturated solution as a result of a particular reaction becomes less than the solubility product for the ions of this electrolyte, then the precipitate goes into solution.
From the foregoing, it can be concluded that various factors influence the formation of precipitation.
Influence of concentration of solutions. A sparingly soluble electrolyte with a sufficiently large value of PR cannot be precipitated from dilute solutions. For example, PbCl 2 will not precipitate when mixing equal volumes of 0.1 M solutions of Pb(NO 3) 2 and NaCl. When mixing equal volumes, the concentrations of each of the substances will become 0.1 / 2 = 0.05 M or 5 10 -2 mol / l. Ionic product 2 = 5 10 -2 (5 10 -2) 2 = 12.5 10 -5. The value obtained is less than PR(PbCl 2), therefore, precipitation will not occur.
Influence of the amount of precipitant. For the most complete precipitation, an excess of precipitant is used. For example, we precipitate the BaCO 3 salt:
BaCl 2 + Na 2 CO 3 → BaCO 3 ↓ + 2NaCl.
After an equivalent amount of Na 2 CO 3 is added, Ba 2+ ions remain in the solution, the concentration of which is determined by the PR value. An increase in the concentration of CO 3 2- ions, caused by the addition of an excess of the precipitant (Na 2 CO 3), will entail a corresponding decrease in the concentration of Ba 2+ ions in the solution, i.e. will increase the completeness of the deposition of this ion.
The influence of the ion of the same name. The solubility of sparingly soluble electrolytes decreases in the presence of other strong electrolytes having similar ions. If a solution of Na 2 SO 4 is gradually added to an unsaturated solution of BaSO 4, then the ion product, which was initially less than PR (BaSO 4) (1.1 10 -10), will gradually reach PR and exceed it. Precipitation will begin.
This is used, for example, in the deposition of valuable metals. For example, PR AgCl in water =1.6×10 -10. The concentration of silver in such a solution over AgCl will be
Is it a lot or a little? That's 1.4 milligrams of silver poured out with every liter of washing liquids at the film factory. If we wash not with water, but with a 0.1 N NaCl solution, then
PR/ = 1.6×10 -9 mol/l,
those. the concentration of silver carried away in solution will decrease by a factor of 10,000.
The effect of temperature. PR is a constant value at a constant temperature. As the temperature increases, the PR increases; therefore, precipitation is best carried out from cooled solutions.
Dissolution of precipitates. The solubility product rule is important for transferring sparingly soluble precipitates into solution. Let's assume that it is necessary to dissolve the precipitate of BaCO 3 . The solution in contact with this precipitate is saturated with respect to BaCO 3. This means that
= PR (BaCO 3).
If an acid is added to the solution, then the H + ions will bind the CO 3 2- ions present in the solution into molecules of fragile carbonic acid:
2H + + CO 3 2- → H 2 CO 3
As a result, the concentration of the CO 3 2- ion will sharply decrease, the ion product will become less than the value of PR (BaCO 3). The solution will be unsaturated with respect to BaCO 3 and part of the BaCO 3 precipitate will go into solution. With the addition of a sufficient amount of acid, the entire precipitate can be brought into solution. Consequently, the dissolution of the precipitate begins when, for some reason, the ion product of a sparingly soluble electrolyte becomes less than the PR value. In order to dissolve the precipitate, an electrolyte is introduced into the solution, the ions of which can form a slightly dissociated compound with one of the ions of a sparingly soluble electrolyte. This explains the dissolution of sparingly soluble hydroxides in acids.
Fe(OH) 3 + 3HCl → FeCl 3 + 3H 2 O
OH‾ ions bind into slightly dissociated H 2 O molecules.
Knowing the PR, it is possible to explain why some substances dissolve, while others do not. Conversely, it is easy to explain why some substances precipitate and others do not.
For example, FeS dissolves in hydrochloric acid, but CuS does not:
FeS + 2HCl → FeCl 2 + H 2 S
The values of PR (FeS) = 3.7 10 -19, PR (CuS) = 8.5 10 -45. From this it follows that in the case of copper sulfide in a saturated solution there are very few S 2– ions, and H molecules 2 S will not be formed, and, therefore, the equilibrium in a saturated solution of copper sulfide will not be disturbed. The precipitate will not dissolve. In the case of iron (II) sulfide, sulfide ions are sufficient to form H 2 S molecules, and the equilibrium shifts to the right. The iron(II) sulfide precipitate dissolves.
Another example: FeS does not precipitate out of solution with H 2 S and precipitates with a solution of (NH 4) 2 S:
FeCl 2 + H 2 S ≠
FeCl 2 + (NH 4) 2 S \u003d FeS ↓ + 2NH 4 Cl
Hydrosulfuric acid is weak (K 2 = 1 10 -14). Hence, there are not enough S 2- ions to fulfill the condition
> PR,
and there is no residue.
Ammonium sulfide is a strong electrolyte, and sulfide ions are sufficient to fulfill the above condition. And this leads to precipitation.
The solubility product can be used to selectively separate ions by precipitation from solutions.
For example, consider the precipitation Ba 2+ and Sr 2+ ions from a solution containing 0.010 mol/l BaCl 2 and 0.020 mol/l SrCl 2 using concentrated Na 2 SO 4 solution.
The PR of barium and strontium sulfates is determined by the relations
PR = = 1.5 10 -9;
PR \u003d 7.6 10 -7.
Therefore, precipitation of barium sulfate in the presence of 0.010 mol/l Ba 2+ ions will not occur until the concentration of SO 4 2- ions reaches the value
1.5 10 -9 / 0.010 \u003d 1.5 10 -7 mol / l.
The precipitation of strontium sulfate will begin at a sulfate ion concentration equal to
7.6 10 -7 / 0.020 = 3.8 10 -5 mol / l.
Therefore, barium ions will begin to precipitate first. When the concentration of sulfate ions reaches 3.8·10 -5 mol/l, the precipitation of strontium sulfate will begin. By that time, the solution will remain
1.5 10 -9 / 3.8 10 -5 \u003d 3.9 10 -5 mol / l.
This will amount to only 0.39% of the initial amount of barium ions. The remaining 99.6% of the barium ions will precipitate before strontium sulfate begins to precipitate.
Page 1
ACTIVITY #
Subject: Heterogeneous equilibria. Solubility constant.
Motivation for studying the topic. In biological liquid media, not homogeneous, but heterogeneous equilibria predominate. The study of these equilibria and the direction of their shifts is of great practical importance for understanding the patterns of formation and dissolution of sparingly soluble substances in the body.
Scientists studying biological evolution believe that the different solubility of natural compounds in water has had a great influence on their content in living organisms. There is a close relationship between the solubility of compounds in water and the toxic effect of ions of a number of elements. For example, the introduction of Al 3+ ions into the body due to the formation of poorly soluble AlPO 4 leads to rickets.
Precipitation reactions form the basis of the precipitation method, which is widely used in the quantitative analysis of pharmaceuticals. The precipitation method is used in the clinical analysis of chlorides in urine, gastric juice, blood; in sanitary and hygienic analysis - in the analysis of drinking water.
Target: To study heterogeneous equilibria in saturated solutions of sparingly soluble electrolytes.
Study objectives:
1. Acquire the skills of calculating the solubility constant in saturated solutions of sparingly soluble electrolytes.
2. Acquire the skills of calculating the solubility of the electrolyte by the value of the solubility constant of the electrolyte.
3. Acquire skills to predict the formation and dissolution of precipitation.
Lesson duration- 165 minutes (135 study time and 30 minutes break).
Location of the lesson- educational workshop (Department of General Chemistry)
Tasks for independent work of the student during extracurricular time (self-study).
A. Control questions
1. Heterogeneous equilibria in saturated solutions of sparingly soluble electrolytes.
2. Solubility constant.
3. Conditions for the formation and dissolution of precipitates. Influence of ions of the same name on the solubility of a sparingly soluble electrolyte.
1. General chemistry. Biophysical chemistry. Chemistry of biogenic elements: A textbook for universities / Yu.A. Ershov, V.A. Popkov, A.S. Berland and others - 2nd ed. - M.: VSh, 2000.
2. Workshop on general chemistry. Biophysical chemistry. Chemistry of biogenic elements / ed. V.A. Popkova, A.V. Babkova - M.: VSh, 2006.
3. Puzakov S.A., Popkov V.A., Filippova A.A. Collection of tasks and exercises in general chemistry. - M.: VSh, 2007.
4. Workshop on General and Bioorganic Chemistry / ed. Popkova V.A. - 3rd ed.-M .: publishing center "Academy", 2008.-240s
B. Teaching material.
In saturated solutions of sparingly soluble electrolytes, the product of the concentration of ions of these electrolytes, raised to the power of their stoichiometric coefficients, is a constant value at a certain temperature and is called the solubility constant (Ks).
Based on the value of Ks, it is possible to predict the formation and dissolution of electrolyte precipitates:
An electrolyte precipitates when the product of its ion concentrations in solution is greater than Ks.
The electrolyte precipitate dissolves when the product of the electrolyte ion concentrations in the solution is less than the Ks value of this electrolyte.
Note. If the solubility of an electrolyte is expressed in mol/L, this is the molar concentration of the electrolyte.
Task #1 The solubility of BaF 2 in water at 18°C is 7.5·10 -3 mol/l. Calculate the solubility constant of BaF 2 at this temperature.
Given: Solution.
C (Ba F 2) \u003d 7.5 10 -3 mol / l 1. BaF 2 ⇄ Ba 2+ + 2F -
Ks-? Ks (BaF 2) \u003d [Ва 2+] 2
During the dissociation of 1 mol of BaF 2, 1 mol of Ba 2+ ions and 2 mol of F - ions are formed;
The BaF 2 solubility constant is:
Answer: Ks (BaF 2) = 1.69 10 -6
Task number 2. The solubility constant of manganese sulfide MnS at 25°C is 2.5 x 10 -10 . Calculate the solubility of manganese sulfide in water (in mol/l) at this temperature.
Given: Solution.
Ks (MnS) \u003d 2.5 10 -10 1. MnS ⇄ Mn 2+ + S 2-
C(MnS) - ? Ks (MnS) = ·
upon dissociation of 1 mol of MnS, 1 mol of Mn 2+ and S 2- ions are formed, therefore, the concentrations of Mn 2+ and S 2- ions are equal to the concentration (solubility in mol/l) of the MnS electrolyte. Denoting the concentrations of Mn 2+ and S 2- ions through X, we get:
X = 
Answer: the solubility of manganese sulfide in water at 25 ° C is 1.58 10 -5 mol / l.
Task number 3. Ks (CdS) = 7.1 10 -28. Will a precipitate of cadmium sulfide CdS precipitate if the same volume of 0.01 N solution of sodium sulfide Na 2 S is added to 1 liter of a 0.1 N solution of cadmium nitrate Cd (NO 3) 2? The degree of dissociation of the initial electrolytes is taken equal to unity.
Given: Solution:
Ks (CdS) \u003d 7.1 10 -28 1. CdS⇄ Cd 2+ + S 2-
V (Cd (NO 3) 2) \u003d V (Na 2 S) \u003d 1 l CdS precipitate is formed if [ S 2- ]> Ks (CdS)
C f equiv (Cd (NO 3) 2) = 0.1 mol / l 2. Cd (NO 3) 2 ⇄ Cd 2+ + 2NO 3 -
C f equiv (Na 2 S) \u003d 0.01 mol / l C (Cd 2+) \u003d C (Cd (NO 3) 2)
α(Cd(NO 3 ) 2 ) = α(Na 2 S) = 1 C (Cd (NO 3) 2) \u003d ½ C f equiv (Cd (NO 3) 2)
does CdS- precipitate form? C (Cd (NO 3) 2) \u003d ½ 0.1 \u003d 0.05 mol / l
C (Cd 2+) \u003d 0.05 mol / l (before mixing solutions)
3. Na 2 S⇄ 2Na + + S 2-
C(S 2-) = C(Na 2 S); C(Na 2 S) = ½ C f equiv (Na 2 S)
C (Na 2 S) \u003d ½ 0.01 \u003d 0.005 mol / l; C(S 2-) = 0.005 mol/l (before mixing solutions).
4. after mixing 1 l Cd (NO 3) 2 and 1 l Na 2 S, the volume of the solution doubled, and the concentration of ions per unit volume will decrease by half and become equal:
C (Cd 2+) \u003d 0.05: 2 \u003d 0.025 mol / l \u003d 2.5 10 -2 mol / l
C(S 2-) \u003d 0.005: 2 \u003d 0.0025 mol / l \u003d 2.5 10 -3 mol / l
5. [S 2-] = 2.5 10 -2 2.5 10 -3 = 6.25 10 -5
6. [ S 2- ]> Ks (CdS)
6.25 10 -5 > 7.1 10 -28 therefore CdS will precipitate
D. Tasks for independent solution.
Problem 1. In 3 l of a solution of lead (II) sulfate PbSO 4 saturated at room temperature, 0.132 g of salt is contained. Calculate Ks(PbSO 4). (Answer: 2.1 10 -8).
Task 2 . Solubility constant of silver iodide AgI at 25°C 1.5·10 -16 . Calculate the solubility of AgI in water at this temperature (in mol/l). (Answer: the solubility of AgI is 1.22·10 -8 mol/l).
3. Will silver bichromate Ag 2 Cr 2 O 7 precipitate when mixing equal volumes of 0.05 N solutions of AgNO 3 and K 2 Cr 2 O 7 ? Ks (Ag 2 Cr 2 O 7) \u003d 2 10 -7 (25 ° C). The degree of dissociation of AgNO 3 and K 2 Cr 2 O 7 is taken equal to one. (Answer: Ag 2 Cr 2 O 7 will precipitate).
Page 1
6. Solubility constant. Solubility.
In a system consisting of a solution and a precipitate, there are two processes - the dissolution of the precipitate and precipitation. The equality of the rates of these two processes is the equilibrium condition.
A saturated solution is a solution that is in equilibrium with the precipitate.
Let us apply the law of mass action to the equilibrium between the precipitate and the solution.
AgCl(solid) Ag + + Cl -
Since = const, we have:
K = K SAgCl = In general, we have:
A m B n (solid) mA + n + nB -m and K S (A m B n) = m n .
Solubility constant K S (or solubility product PR) - the product of ion concentrations in a saturated solution of a sparingly soluble electrolyte - is a constant value and depends only on temperature.
Examples of K S values for a number of salts:
K S (AgCl) \u003d 1.610 -10
K S (Ag 2 CrO 4) \u003d 1.110 -12
K S (Ag 2 S) \u003d 610 -50.
Solubility sparingly soluble substance s can be expressed in moles per liter. Depending on the value of s, substances can be divided into poorly soluble - s< 10 -4 моль/л, среднерастворимые - 10 -4 моль/л s 10 -2 моль/л и хорошо растворимые s >10 -2 mol/l.
The solubility of compounds is related to their solubility constant. For AgCl we have:
AgCl(solid) Ag + + Cl -
Solubility s - molar concentration of AgCl in solution:
s = = m/(MV) = s = = .
Hence K S AgCl = = s 2 and s= 
.
In general, for A m B n we have: A m B n (solid) mA +n + nB -m
K S (A m B n) = m n = (ms) m (ns) n = m m n n s m+n .
Example. Find the solubility of AgCl (K S = 1.610 -10) and Ag 2 CrO 4 (K S = 1.210 -12).
Solution. AgCl Ag + + Cl - ,
K S \u003d s 2, s \u003d 
\u003d 1.3410 -5 mol / l.
Ag 2 CrO 4 2Ag + + CrO 4 2-
K S = (2s) 2 s = 4s 3 , s = 
\u003d 6.510 -5 mol / l.
Although usually the solubility is lower, the lower K S , in this case for compounds of different types s(AgCl)< s(Ag 2 CrO 4), хотя K S (AgCl) >K S (Ag 2 CrO 4).
Precipitation and dissolution condition
For equilibrium between the precipitate and the solution - a saturated solution - we have in the case of AgCl:
AgCl Ag + + Cl - = K S .
The precipitation condition is written as: > K S
During precipitation, the ion concentrations decrease until equilibrium is established.
The condition for the dissolution of a precipitate or the existence of a saturated solution is written as: < K S .
During the dissolution of the precipitate, the concentration of ions increases until equilibrium is established.
Common ion effect
The addition of a common ion reduces the solubility of sparingly soluble compounds.
Example. Find the solubility of AgCl in 0.1 M NaCl solution.
Solution. AgCl Ag + + Cl -
K S AgCl \u003d = s 0.1 \u003d 1.610 -10, s \u003d 1.610 -9 mol / l.
The solubility of AgCl in water (see above) is 1.3410 -5 mol/l, the solubility of AgCl in 0.1M NaCl is 1.610 -9 mol/l, i.e. 10 4 times less.
salt effect
Increasing the ionic strength of a solution increases the solubility of poorly soluble compounds.
Since the concentrations of ions formed during the dissociation of poorly soluble compounds, as well as the ionic strength of the resulting solutions, are small, it turns out that it is possible to use the concentrations of ions, rather than their activities, in the K S expressions. In cases where strong electrolytes are present in the solution, which determine the high ionic strength of the solution, it is necessary to substitute the activity of the ions in the expression for KS.
Determine the solubility of AgCl in 0.1 M NaCl, taking into account the ionic strength of the solution
AgCl Ag + + Cl -
For 0.1M NaCl = 0.1 and f Ag+ = f Cl - = 0.78.
K S = a Ag+ a Cl - = f Ag+ f Cl - = 0.78s0.780.1 = 1.610 -10,
s = 1.610 -9 /(0.78) 2 = 2.610 -9 M, i.e. 1.64 times more than without taking into account the ionic strength of the solution. The salt effect is much less than the effect of the ion of the same name.
Example. The solubility of Mg(OH) 2 is 0.012 g/L. Find K S .
Solution. M (Mg (OH) 2) \u003d 58 g / mol, s \u003d 0.12 g / l / 58 g / mol \u003d
2.0710 -4 M.
Mg (OH) 2 Mg 2+ + 2OH -
K S = 2 = s(2s) 2 = 4s 3 = 4(2.0710 -4) 3 = 3.610 -11.
Example. Does PbCl 2 precipitate when mixing equal volumes of solutions of 0.1M Pb (NO 3) 2 and 0.1M NaCl, if K S (PbCl 2) =
Solution. After mixing the solutions, the ion concentrations will decrease by half, i.e. we have: = = = 0.05M, a = 0.1 M. Find the ionic strength of the solution:
= 1/2(0,052 2 + 0,11 2 + 0,051 2 + 0,051 2) = 0,2.
We find the activity coefficients: f Pb2+ = 0.24 and f Cl - = 0.70.
Having for PbCl 2 Pb +2 + 2Cl -
K S PbCl2 = a Pb2+ a Cl - 2 , we calculate the value of a Pb2+ a Cl - 2 for our solution:
a Pb2+ a Cl - 2 = f Pb2+ (f Cl -) 2 = 0.240.050.70 2 0.05 2 = 1.4710 -5, which is less than PR PbCl2 (1.610 -5), so no precipitate is formed.
7. Redox reactions
Redox reactions- these are reactions that go with a change in the oxidation states of elements. The oxidation state is the conditional charge of an atom in a molecule, where all polar bonds are considered ionic.
Oxidation is the process of donating electrons.
Recovery is the process of adding electrons.
Oxidizer is an atom, molecule or ion that accepts electrons and lowers its oxidation state, i.e. is being restored.
Reducing agent- this is an atom, molecule or ion that donates electrons and increases its oxidation state, i.e. oxidized.
Typical reducing and oxidizing agents
Reclaimers: a) metals - the lower the ionization potential, the stronger the reducing properties; b) compounds of elements in lower oxidation states (NH 3 , H 2 S, HBr, HI, etc.), in which all orbitals are filled and can only donate electrons.
Oxidizers: a) non-metals (F 2, Cl 2, Br 2, O 2, etc.) - the greater the electron affinity, the stronger the oxidizing properties; b) metal ions in high oxidation states (Fe 3+, Sn 4+, Mn 4+, etc.); c) compounds of elements in higher oxidation states (KMnO 4, K 2 Cr 2 O 7, NaBiO 3, HNO 3, H 2 SO 4 (conc.), etc.), in which all valence electrons have already been given away and can only be oxidizing agents .
Compounds of elements in intermediate oxidation states (HNO 2 , H 2 SO 3 , H 2 O 2 , etc.) can exhibit oxidizing and reducing properties depending on the redox properties of the second reagent.
H 2 SO 3 + 2H 2 S \u003d 3S + 3H 2 O
oxid. restore
H 2 SO 3 + Br 2 + H 2 O \u003d H 2 SO 4 + 2HBr
restore oxid.
reducing agent (strong) |
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oxidizing agent (weak) |
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reducing agent (weak) |
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oxidizer (strong) |
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reducing agent |
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reducing agent, oxidizing agent |
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reducing agent, oxidizing agent |
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oxidizer |
Oxidizing agents, by accepting electrons, that is, by being reduced, pass into the reduced form:
F 2 + 2e 2F -
oxid. restore
Reducing agents, donating electrons, that is, being oxidized, pass into the oxidized form:
Na 0 - 1e Na +
restore oxid.
Thus, both oxidizing agents and reducing agents exist in oxidized (with a higher oxidation state of the element) and reduced (with a lower oxidation state of the element) forms. At the same time, oxidants are more likely to change from the oxidized to the reduced form, while reducing agents are characterized by the transition from the reduced to the oxidized form. Reverse processes are not characteristic, and we do not consider, for example, that F - is a reducing agent, and Na + is an oxidizing agent.
The equilibrium between the oxidized and reduced forms is characterized by the redox potential, which depends on the concentrations of the oxidized and reduced forms, the reaction of the environment, temperature, etc. It can be calculated from Nernst equation:
E = E o + 
where is the molar concentration of the oxidized form;
[Restore] is the molar concentration of the reduced form;
n is the number of electrons involved in the half-reaction;
E 0 - standard value of the redox potential; E \u003d E 0 if [Restore] \u003d [Ok] \u003d 1 mol / l;
The values of the standard electrode potentials E 0 are given in the tables and characterize the oxidizing and reducing properties of the compounds: The more positive the value of E 0, the stronger the oxidizing properties, and the more negative the value of E 0, the stronger the reducing properties.
For example:
F 2 + 2e 2F - E 0 \u003d 2.87 in - a strong oxidizing agent
Na + + 1e Na 0 E 0 \u003d -2.71 in - a strong reducing agent
(the process is always recorded for reduction reactions).
Since the redox reaction is a combination of two half-reactions, oxidation and reduction, it is characterized by the value of the difference in standard electrode potentials of the oxidizer (E 0 ok) and reducing agent (E 0 restore) - electromotive force (emf) E 0 :
E 0 \u003d E 0 ok - E 0 restore,
emf reaction Е 0 is related to the change in Gibbs free energy G: G = -nFЕ 0 , and on the other hand, G is related to the equilibrium constant K of the reaction by the equation G = -2.3RTlnK.
From the last two equations follows the relationship between emf. and the equilibrium constant of the reaction:
Е = (2.3RT/nF)lnK.
emf reactions at concentrations other than standard (i.e. not equal to 1 mol / l) Е is equal to:
Е = Е 0 - (2.3RT/nF)lgK or Е = Е 0 - (0.059/n)lgK.
In the case of equilibrium G = 0 and therefore E =0. From where Е = (0.059 / n) lgK and K = 10 n Е / 0.059.
For a spontaneous reaction to proceed, the requirement must be met: G<0 или К>>1, which corresponds to the condition Е 0 >0. Therefore, to determine the possibility of a given redox reaction, it is necessary to calculate the value of Е 0 . If Е 0 0, the reaction is going on. If Е 0 0, there is no reaction.
Example 1 Determine the possibility of a reaction
2FeCl 3 + 2KI 2FeCl 2 + 2KCl + I 2
Solution: We find that the oxidizing agent is the Fe +3 ion, which is reduced to Fe +2, and the reducing agent is I -, which is oxidized to I 2. We find from the table the values of standard electrode potentials: E 0 (Fe +3 / Fe +2) = 0.77 V and E 0 (I 2 / 2I -) = 0.54 V. We calculate Е 0:
Е 0 \u003d E 0 ok - E 0 restore \u003d 0.77 - 0.54 \u003d 0.23 in 0.
Example 2. Determine the possibility of a reaction
2 KMnO 4 + 16 HCl 2 KCl + 2 MnCl 2 + 5 Cl 2 + 8 H 2 O.
Solution. We find that the oxidizing agent is the permanganate ion MnO 4 -, passing into Mn +2, and the reducing agent is the chloride ion, passing into gaseous chlorine Cl 2. We determine their potentials from the table: E 0 (MnO 4 - / Mn + 2) = 1.51 V and E 0 (Cl 2 / 2Cl -) = 1.36 V. Calculate
Е 0 \u003d E 0 ok - E 0 restore \u003d 1.51 - 1.36 \u003d 0.15 in 0.
The reaction is possible, since Е 0 0.
OVR classification
1. Intermolecular oxidation-reduction reactions - an oxidizing agent and a reducing agent are part of different substances:
2Fe + 3Cl 2 = 2FeCl 3
restore oxid.
2. Disproportionation reactions - an element in an intermediate oxidation state is an oxidizing agent and a reducing agent:
2KOH + Cl 2 \u003d KCl + KClO + H 2 O
3HNO 2 \u003d HNO 3 + 2NO + H 2 O
In these reactions, chlorine and N +3 are oxidizing and reducing agents.
3. Intramolecular oxidation-reduction reactions - an oxidizing agent and a reducing agent are part of one substance:
2KClO 3 
2KCl + 3O 2
NH 4 NO 3 N 2 O + 2H 2 O
In these reactions, O -2 , Cl +5 and N -3 , N +5 are, respectively, reducing agents and oxidizing agents.
It should be noted that the direction of the OVR and the nature of the reaction products depend on the redox properties of the reactants and the nature of the medium (acidic, neutral, or alkaline). For example, potassium permanganate KMnO 4, which exhibits only oxidizing properties, forms various reduction products upon transition from an acidic to a neutral and alkaline environment and its redox potential decreases:
pH< 7: MnO 4 - + 5e Mn +2 (бесцветный) Е 0 = +1,51 в
pH \u003d 7: MnO 4 - + 3e MnO 2 (brown) E 0 \u003d + 0.60 in
pH > 7: MnO 4 - + 1e MnO 4 -2 (green) E 0 = +0.56 in
Chromium (VI) compounds are strong oxidizing agents in an acidic environment (E 0 \u003d +1.33 c), are reduced to Cr +3, and chromium (III) compounds in an alkaline medium show a reducing ability (E 0 \u003d -0.13 c) , oxidizing to chromium (VI) compounds.
pH 7: Cr 2 O 7 2- + 6e 2Cr +3 (blue)
pH > 7: CrO 4 2- + 3e 3- (green)
Hydrogen peroxide H 2 O 2 containing oxygen in an intermediate oxidation state of -1 exhibits oxidizing or reducing properties, and depending on the pH of the solution, its electrode potential and the products to which it is reduced or oxidized change.
H 2 O 2 - oxidizing agent:
pH> 7: H 2 O 2 + 2H + + 2e 2H 2 O
pH 7: H 2 O 2 + 2e 2OH -
H 2 O 2 - reducing agent:
pH> 7: H 2 O 2 - 2e O 2 + 2H +
pH 7: H 2 O 2 + 2OH - - 2e O 2 + 2H 2 O
Thus, for the correct spelling of OVR, it is necessary to take into account the conditions for the occurrence of this reaction.
Ion-electronic method for compiling OVR
The ion-electron method (or the half-reaction method) is used to compile OVR occurring in solutions. It is based on the compilation of individual half-reactions for the processes of reduction and oxidation in the form of ion-molecular equations. In this case, it is necessary to take into account the rules for writing ionic reaction equations: strong electrolytes are written in the form of ions, and weak electrolytes and poorly soluble substances are written in the form of molecules.
The procedure for compiling the RIA
1. The left side of the OVR molecular equation is written, the oxidizing agent and reducing agent are determined.
2. Half-reactions are written separately for the processes of reduction and oxidation in the form of ion-molecular equations, in which the left and right parts (excess or lack of oxygen in the left part) are equalized, taking into account the pH of the solution, using H 2 O molecules, H + or OH - ions :
pH< 7: избыток [O] 2H + + [O] H 2 O
pH = 7: excess [O] H 2 O + [O] 2OH -
lack of [O] H 2 O - [O] 2H +
pH > 7: excess [O] H 2 O + [O] 2OH -
lack of [O] 2OH - - [O] H 2 O.
3. The sums of charges in the left and right parts of the half-reactions are equalized by adding or subtracting electrons. After that, the factors for the half-reactions are selected.
4. The total ion-molecular OVR equation is written taking into account the multipliers.
5. The right side of the OVR molecular equation is added and the coefficients from the ionic-molecular equation are transferred into it.
Example 1 Interaction of KMnO 4 with Na 2 SO 3 in an acidic environment.
1. We write down the left side of the equation, we determine the oxidizing agent and reducing agent:
KMnO 4 + Na 2 SO 3 + H 2 SO 4 =
oxid. restore Wednesday
2. We compose half-reactions for the processes of reduction and oxidation, taking into account the acidic environment. In an acidic environment, the permanganate ion MnO 4 - is reduced to the Mn 2+ ion, and the sulfite ion SO 3 2- is oxidized to the sulfate ion SO 4 2-:
MnO 4 - Mn + 2 - we bind the excess of oxygen with H + ions,
SO 3 2- SO 4 2- - we take the missing oxygen from the water and H + ions are released.
We get the following half-reactions:
MnO 4 - + 8H + Mn +2 + 4H 2 O
SO 3 2- + H 2 O SO 4 2- + 2H +
3. We consider the sums of charges in the left and right parts of both half-reactions and equalize the charges with the help of electrons, select the factors:
5 SO 3 2- + H 2 O - 2e SO 4 2- + 2H +
4. We write the total ion-molecular OVR equation taking into account the factors:
2 MnO 4 - + 8H + + 5e Mn +2 + 4H 2 O
2MnO 4 - + 16H + + 5SO 3 2- + 5H 2 O 2Mn + 2 + 8H 2 O + 5SO 4 2- + 10H +
We reduce hydrogen ions and water molecules and get:
5. We add the right side of the molecular equation and transfer the coefficients of the ion-molecular equation into it. The final equation will look like this:
2KMnO 4 + 5Na 2 SO 3 + 3H 2 SO 4 = 2MnSO 4 + 5Na 2 SO 4 + K 2 SO 4 + 3H 2 O
2 MnO 4 - + 8H + + 5e Mn +2 + 4H 2 O
5 SO 3 2- + H 2 O - 2e SO 4 2- + 2H +
2MnO 4 - + 6H + + 5SO 3 2- 2Mn +2 + 3H 2 O + 5SO 4 2-
Example 2 Oxidation of chromium (III) nitrate with hydrogen peroxide in an alkaline medium is a qualitative reaction to the Cr 3+ ion. In an alkaline environment, the Cr 3+ ion is oxidized to the CrO 4 2- chromate ion, which has a yellow color.
2Cr(NO 3) 3 + 3H 2 O 2 + 10KOH 2K 2 CrO 4 + 6KNO 3 + 8H 2 O
2Cr 3+ + 8OH - - 3e CrO 4 2- + 4H 2 O
3H 2 O 2 + 2e 2OH -
2Cr 3+ + 10OH - + 3H 2 O 2 2CrO 4 2- + 8H 2 O
8. Coordination compounds
Coordinating(complex) connections are compounds with a donor-acceptor bond. Coordination compounds consist of ions of the inner and outer spheres. In the formula of a complex compound, the ions of the inner sphere are enclosed in square brackets. Ions of the inner sphere - complex ions - consist of a complexing agent (central ion) and ligands. The number of ligands in the inner sphere of the complex is called the coordination number. The denticity of a ligand is the number of bonds by which a given ligand is connected to a complexing agent.
Example: K 3
1. Ions of the outer sphere - 3K +
2. Ion of the inner sphere - 3-
3. Complexing agent - Fe 3+
4. Ligand - CN - , its denticity - 1
5. Coordination number - 6
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