The value of a is expressed in fractions of a unit or in % and depends on the nature of the electrolyte, solvent, temperature, concentration and composition of the solution.
The solvent plays a special role: in a number of cases, when passing from aqueous solutions to organic solvents, the degree of dissociation of electrolytes can sharply increase or decrease. In the future, in the absence of special instructions, we will assume that the solvent is water.
According to the degree of dissociation, electrolytes are conditionally divided into strong(a > 30%), medium (3% < a < 30%) и weak(a< 3%).
Strong electrolytes include:
1) some inorganic acids (HCl, HBr, HI, HNO 3 , H 2 SO 4 , HClO 4 and a number of others);
2) hydroxides of alkali (Li, Na, K, Rb, Cs) and alkaline earth (Ca, Sr, Ba) metals;
3) almost all soluble salts.
Medium-strength electrolytes include Mg (OH) 2, H 3 PO 4, HCOOH, H 2 SO 3, HF and some others.
All carboxylic acids (except HCOOH) and hydrated forms of aliphatic and aromatic amines are considered weak electrolytes. Weak electrolytes are also many inorganic acids (HCN, H 2 S, H 2 CO 3, etc.) and bases (NH 3 ∙ H 2 O).
Despite some similarities, in general, one should not identify the solubility of a substance with its degree of dissociation. So, acetic acid and ethyl alcohol are unlimitedly soluble in water, but at the same time, the first substance is a weak electrolyte, and the second is a non-electrolyte.
Acids and bases
Despite the fact that the concepts of "acid" and "base" are widely used to describe chemical processes, there is no single approach to the classification of substances in terms of classifying them as acids or bases. Current theories ( ionic theory S. Arrhenius, protolytic theory I. Bronsted and T. Lowry and electronic theory G. Lewis) have certain limitations and are therefore applicable only in particular cases. Let's take a closer look at each of these theories.
Arrhenius theory.
In the ionic theory of Arrhenius, the concepts of "acid" and "base" are closely related to the process of electrolytic dissociation:
An acid is an electrolyte that dissociates in solutions to form H + ions;
The base is an electrolyte that dissociates in solutions to form OH - ions;
Ampholyte (amphoteric electrolyte) is an electrolyte that dissociates in solutions with the formation of both H + ions and OH - ions.
For example:
ON ⇄ H + + A - nH + + MeO n n - ⇄ Me (OH) n ⇄ Me n + + nOH -In accordance with the ionic theory, both neutral molecules and ions can be acids, for example:
HF ⇄ H + + F -
H 2 PO 4 - ⇄ H + + HPO 4 2 -
NH 4 + ⇄ H + + NH 3
Similar examples can be given for the grounds:
KOH K + + OH -
- ⇄ Al(OH) 3 + OH -
+ ⇄ Fe 2+ + OH -
Ampholytes include hydroxides of zinc, aluminum, chromium and some others, as well as amino acids, proteins, nucleic acids.
In general, the acid-base interaction in solution is reduced to a neutralization reaction:
H + + OH - H 2 O
However, a number of experimental data show the limitations of the ionic theory. So, ammonia, organic amines, metal oxides such as Na 2 O, CaO, anions of weak acids, etc. in the absence of water, they exhibit the properties of typical bases, although they do not contain hydroxide ions.
On the other hand, many oxides (SO 2, SO 3, P 2 O 5, etc.), halides, acid halides, without hydrogen ions in their composition, even in the absence of water, exhibit acidic properties, i.e. bases are neutralized.
In addition, the behavior of an electrolyte in an aqueous solution and in a non-aqueous medium can be opposite.
So, CH 3 COOH in water is a weak acid:
CH 3 COOH ⇄ CH 3 COO - + H +,
and in liquid hydrogen fluoride it exhibits the properties of a base:
HF + CH 3 COOH ⇄ CH 3 COOH 2 + + F -
Studies of these types of reactions, and in particular reactions in non-aqueous solvents, have led to more general theories of acids and bases.
Theory of Bronsted and Lowry.
A further development of the theory of acids and bases was the protolytic (proton) theory proposed by I. Bronsted and T. Lowry. According to this theory:
An acid is any substance whose molecules (or ions) are capable of donating a proton, i.e. be a proton donor;
A base is any substance whose molecules (or ions) are capable of attaching a proton, i.e. be a proton acceptor;
Thus, the concept of base is significantly expanded, as evidenced by the following reactions:
OH - + H + H 2 O
NH 3 + H + NH 4 +
H 2 N-NH 3 + + H + H 3 N + -NH 3 +
According to the theory of I. Bronsted and T. Lowry, an acid and a base form a conjugate pair and are connected by equilibrium:
ACID ⇄ PROTON + BASE
Since the proton transfer reaction (protolytic reaction) is reversible, and a proton is also transferred in the reverse process, the reaction products are acid and base in relation to each other. This can be written as an equilibrium process:
ON + B ⇄ VN + + A -,
where HA is an acid, B is a base, BH + is an acid conjugated with base B, A - is a base conjugated with acid HA.
Examples.
1) in reaction:
HCl + OH - ⇄ Cl - + H 2 O,
HCl and H 2 O are acids, Cl - and OH - are the corresponding conjugate bases;
2) in reaction:
HSO 4 - + H 2 O ⇄ SO 4 2 - + H 3 O +,
HSO 4 - and H 3 O + - acids, SO 4 2 - and H 2 O - bases;
3) in reaction:
NH 4 + + NH 2 - ⇄ 2NH 3,
NH 4 + is an acid, NH 2 - is a base, and NH 3 acts as both an acid (one molecule) and a base (another molecule), i.e. shows signs of amphotericity - the ability to exhibit the properties of an acid and a base.
Water also has this ability:
2H 2 O ⇄ H 3 O + + OH -
Here, one H 2 O molecule adds a proton (base), forming a conjugate acid - a hydroxonium ion H 3 O +, the other gives a proton (acid), forming a conjugate base OH -. This process is called autoprotolysis.
It can be seen from the above examples that, in contrast to the ideas of Arrhenius, in the theory of Brönsted and Lowry, the reactions of acids with bases do not lead to mutual neutralization, but are accompanied by the formation of new acids and bases.
It should also be noted that the protolytic theory considers the concepts of "acid" and "base" not as a property, but as a function that the compound in question performs in the protolytic reaction. The same compound can react as an acid under certain conditions and as a base under others. So, in an aqueous solution of CH 3 COOH exhibits the properties of an acid, and in 100% H 2 SO 4 - a base.
However, despite its merits, the protolytic theory, like the Arrhenius theory, is not applicable to substances that do not contain hydrogen atoms, but, at the same time, exhibit the function of an acid: boron, aluminum, silicon, and tin halides.
Lewis theory.
A different approach to the classification of substances in terms of classifying them as acids and bases was the electronic theory of Lewis. Within the electronic theory:
an acid is a particle (molecule or ion) capable of attaching an electron pair (electron acceptor);
A base is a particle (molecule or ion) capable of donating an electron pair (electron donor).
According to Lewis, an acid and a base interact with each other to form a donor-acceptor bond. As a result of the addition of a pair of electrons, an electron-deficient atom has a complete electronic configuration - an octet of electrons. For example:
The reaction between neutral molecules can be represented in a similar way:
The neutralization reaction in terms of the Lewis theory is considered as the addition of an electron pair of a hydroxide ion to a hydrogen ion, which provides a free orbital to accommodate this pair:
Thus, the proton itself, which easily attaches an electron pair, from the point of view of the Lewis theory, performs the function of an acid. In this regard, Bronsted acids can be considered as reaction products between Lewis acids and bases. So, HCl is the product of neutralization of the acid H + with the base Cl -, and the H 3 O + ion is formed as a result of the neutralization of the acid H + with the base H 2 O.
Reactions between Lewis acids and bases are also illustrated by the following examples:
Lewis bases also include halide ions, ammonia, aliphatic and aromatic amines, oxygen-containing organic compounds of the R 2 CO type (where R is an organic radical).
Lewis acids include halides of boron, aluminum, silicon, tin and other elements.
Obviously, in the theory of Lewis, the concept of "acid" includes a wider range of chemical compounds. This is explained by the fact that, according to Lewis, the assignment of a substance to the class of acids is due solely to the structure of its molecule, which determines the electron-acceptor properties, and is not necessarily associated with the presence of hydrogen atoms. Lewis acids that do not contain hydrogen atoms are called aprotic.
Problem Solving Standards
1. Write the equation for the electrolytic dissociation of Al 2 (SO 4) 3 in water.
Aluminum sulfate is a strong electrolyte and undergoes complete decomposition into ions in an aqueous solution. Dissociation equation:
Al 2 (SO 4) 3 + (2x + 3y)H 2 O 2 3+ + 3 2 -,
or (without taking into account the process of ion hydration):
Al 2 (SO 4) 3 2Al 3+ + 3SO 4 2 -.
2. What is the HCO 3 ion - from the standpoint of the Bronsted-Lowry theory?
Depending on the conditions, the HCO 3 ion can donate protons:
HCO 3 - + OH - CO 3 2 - + H 2 O (1),
and add protons:
HCO 3 - + H 3 O + H 2 CO 3 + H 2 O (2).
Thus, in the first case, the HCO 3 ion - is an acid, in the second - a base, that is, it is an ampholyte.
3. Determine what, from the standpoint of the Lewis theory, is the Ag + ion in the reaction:
Ag + + 2NH 3 +
In the process of formation of chemical bonds, which proceeds according to the donor-acceptor mechanism, the Ag + ion, having a free orbital, is an electron pair acceptor, and thus exhibits the properties of a Lewis acid.
4. Determine the ionic strength of the solution in one liter of which there are 0.1 mol of KCl and 0.1 mol of Na 2 SO 4.
The dissociation of the presented electrolytes proceeds in accordance with the equations:
Na 2 SO 4 2Na + + SO 4 2 -
Hence: C (K +) \u003d C (Cl -) \u003d C (KCl) \u003d 0.1 mol / l;
C (Na +) \u003d 2 × C (Na 2 SO 4) \u003d 0.2 mol / l;
C (SO 4 2 -) \u003d C (Na 2 SO 4) \u003d 0.1 mol / l.
The ionic strength of the solution is calculated by the formula:
5. Determine the concentration of CuSO 4 in a solution of this electrolyte with I= 0.6 mol/l.
The dissociation of CuSO 4 proceeds according to the equation:
CuSO 4 Cu 2+ + SO 4 2 -
Let's take C (CuSO 4) for x mol / l, then, in accordance with the reaction equation, C (Cu 2+) \u003d C (SO 4 2 -) \u003d x mol/l. In this case, the expression for calculating the ionic strength will look like:
6. Determine the activity coefficient of the K + ion in an aqueous solution of KCl with C (KCl) = 0.001 mol / l.
which in this case will take the form:
.
The ionic strength of the solution is found by the formula:
7. Determine the activity coefficient of the Fe 2+ ion in an aqueous solution, the ionic strength of which is equal to 1.
According to the Debye-Hückel law:
Consequently:
8. Determine the dissociation constant of the acid HA, if in a solution of this acid with a concentration of 0.1 mol/l a = 24%.
By the magnitude of the degree of dissociation, it can be determined that this acid is an electrolyte of medium strength. Therefore, to calculate the acid dissociation constant, we use the Ostwald dilution law in its full form:
9. Determine the electrolyte concentration, if a = 10%, K d \u003d 10 - 4.
From Ostwald's Dilution Law:
10. The degree of dissociation of monobasic acid HA does not exceed 1%. (HA) = 6.4×10 - 7 . Determine the degree of dissociation of HA in its solution with a concentration of 0.01 mol/l.
By the magnitude of the degree of dissociation, it can be determined that this acid is a weak electrolyte. This allows us to use the approximate formula of the Ostwald dilution law:
11. The degree of dissociation of the electrolyte in its solution with a concentration of 0.001 mol / l is 0.009. Determine the dissociation constant of this electrolyte.
It can be seen from the condition of the problem that this electrolyte is weak (a = 0.9%). That's why:
12. (HNO 2) = 3.35. Compare the strength of HNO 2 with the strength of the monobasic acid HA, the degree of dissociation of which in solution with C(HA) = 0.15 mol/l is 15%.
Calculate (HA) using the full form of the Ostwald equation:
Since (HA)< (HNO 2), то кислота HA является более сильной кислотой по сравнению с HNO 2 .
13. There are two KCl solutions containing other ions. It is known that the ionic strength of the first solution ( I 1) is equal to 1, and the second ( I 2) is 10 - 2 . Compare Activity Factors f(K +) in these solutions and conclude how the properties of these solutions differ from the properties of infinitely dilute solutions of KCl.
The activity coefficients of K + ions are calculated using the Debye-Hückel law:
Activity factor f is a measure of the deviation in the behavior of an electrolyte solution of a given concentration from its behavior at an infinite dilution of the solution.
Because f 1 = 0.316 deviates more from 1 than f 2 \u003d 0.891, then in a solution with a higher ionic strength, a greater deviation in the behavior of the KCl solution from its behavior at infinite dilution is observed.
Questions for self-control
1. What is electrolytic dissociation?
2. What substances are called electrolytes and non-electrolytes? Give examples.
3. What is the degree of dissociation?
4. What factors determine the degree of dissociation?
5. What electrolytes are considered strong? What are medium strength? What are the weak? Give examples.
6. What is the dissociation constant? What does the dissociation constant depend on and what does it not depend on?
7. How are the constant and the degree of dissociation in binary solutions of medium and weak electrolytes related?
8. Why do solutions of strong electrolytes exhibit deviations from ideality in their behavior?
9. What is the essence of the term "apparent degree of dissociation"?
10. What is the activity of an ion? What is an activity coefficient?
11. How does the value of the activity coefficient change with dilution (concentration) of a strong electrolyte solution? What is the limiting value of the activity coefficient at infinite dilution of the solution?
12. What is the ionic strength of a solution?
13. How is the activity coefficient calculated? Formulate the Debye-Hückel law.
14. What is the essence of the ionic theory of acids and bases (Arrhenius theory)?
15. What is the fundamental difference between the protolytic theory of acids and bases (the theory of Bronsted and Lowry) and the theory of Arrhenius?
16. How does the electronic theory (Lewis theory) interpret the concepts of "acid" and "base"? Give examples.
Variants of tasks for independent solution
Option number 1
1. Write the equation for the electrolytic dissociation of Fe 2 (SO 4) 3 .
ON + H 2 O ⇄ H 3 O + + A -.
Option number 2
1. Write the equation for the electrolytic dissociation of CuCl 2 .
2. Determine what, from the standpoint of the Lewis theory, is the S 2 ion - in the reaction:
2Ag + + S 2 - ⇄ Ag 2 S.
3. Calculate the molar concentration of the electrolyte in the solution if a = 0.75%, a = 10 - 5.
Option number 3
1. Write the equation for the electrolytic dissociation of Na 2 SO 4 .
2. Determine what, from the standpoint of the Lewis theory, is the CN ion - in the reaction:
Fe 3 + + 6CN - ⇄ 3 -.
3. The ionic strength of the CaCl 2 solution is 0.3 mol/l. Calculate C (CaCl 2).
Option number 4
1. Write the equation for the electrolytic dissociation of Ca(OH) 2 .
2. Determine what, from the standpoint of the Bronsted theory, is the H 2 O molecule in the reaction:
H 3 O + ⇄ H + + H 2 O.
3. The ionic strength of the K 2 SO 4 solution is 1.2 mol/l. Calculate C(K 2 SO 4).
Option number 5
1. Write the equation for the electrolytic dissociation of K 2 SO 3 .
NH 4 + + H 2 O ⇄ NH 3 + H 3 O +.
3. (CH 3 COOH) = 4.74. Compare the strength of CH 3 COOH with the strength of the monobasic acid HA, the degree of dissociation of which in solution with C (HA) = 3.6 × 10 - 5 mol / l is 10%.
Option number 6
1. Write the equation for the electrolytic dissociation of K 2 S.
2. Determine what, from the standpoint of the Lewis theory, is the AlBr 3 molecule in the reaction:
Br - + AlBr 3 ⇄ - .
Option number 7
1. Write the equation for the electrolytic dissociation of Fe(NO 3) 2 .
2. Determine what, from the standpoint of the Lewis theory, is the ion Cl - in the reaction:
Cl - + AlCl 3 ⇄ - .
Option number 8
1. Write the equation for the electrolytic dissociation of K 2 MnO 4 .
2. Determine what, from the standpoint of the Bronsted theory, is the HSO 3 ion - in the reaction:
HSO 3 - + OH - ⇄ SO 3 2 - + H 2 O.
Option number 9
1. Write the equation for the electrolytic dissociation of Al 2 (SO 4) 3 .
2. Determine what, from the standpoint of the Lewis theory, is the Co 3+ ion in the reaction:
Co 3+ + 6NO 2 - ⇄ 3 -.
3. 1 liter of solution contains 0.348 g of K 2 SO 4 and 0.17 g of NaNO 3. Determine the ionic strength of this solution.
Option number 10
1. Write the equation for the electrolytic dissociation of Ca(NO 3) 2 .
2. Determine what, from the standpoint of the Bronsted theory, is the H 2 O molecule in the reaction:
B + H 2 O ⇄ OH - + BH +.
3. Calculate the electrolyte concentration in the solution if a = 5%, a = 10 - 5.
Option number 11
1. Write the equation for the electrolytic dissociation of KMnO 4 .
2. Determine what, from the standpoint of the Lewis theory, is the Cu 2+ ion in the reaction:
Cu 2+ + 4NH 3 ⇄ 2 +.
3. Calculate the activity coefficient of the Cu 2+ ion in a CuSO 4 solution with C (CuSO 4) = 0.016 mol / l.
Option number 12
1. Write the equation for the electrolytic dissociation of Na 2 CO 3 .
2. Determine what, from the standpoint of the Bronsted theory, is the H 2 O molecule in the reaction:
K + + xH 2 O ⇄ + .
3. There are two NaCl solutions containing other electrolytes. The values of the ionic strength of these solutions are respectively equal: I 1 = 0.1 mol/l, I 2 = 0.01 mol/l. Compare Activity Factors f(Na +) in these solutions.
Option number 13
1. Write the equation for the electrolytic dissociation of Al(NO 3) 3 .
2. Determine what, from the standpoint of the Lewis theory, is the RNH 2 molecule in the reaction:
RNH 2 + H 3 O + ⇄ RNH 3 + + H 2 O.
3. Compare the activity coefficients of cations in a solution containing FeSO 4 and KNO 3, provided that the electrolyte concentrations are 0.3 and 0.1 mol/l, respectively.
Option number 14
1. Write the equation for the electrolytic dissociation of K 3 PO 4 .
2. Determine what, from the standpoint of the Bronsted theory, is the H 3 O + ion in the reaction:
HSO 3 - + H 3 O + ⇄ H 2 SO 3 + H 2 O.
Option number 15
1. Write the equation for the electrolytic dissociation of K 2 SO 4 .
2. Determine what, from the standpoint of the Lewis theory, is Pb (OH) 2 in the reaction:
Pb (OH) 2 + 2OH - ⇄ 2 -.
Option number 16
1. Write the equation for the electrolytic dissociation of Ni(NO 3) 2 .
2. Determine what, from the standpoint of the Bronsted theory, is the hydronium ion (H 3 O +) in the reaction:
2H 3 O + + S 2 - ⇄ H 2 S + 2H 2 O.
3. The ionic strength of a solution containing only Na 3 PO 4 is 1.2 mol / l. Determine the concentration of Na 3 PO 4.
Option number 17
1. Write the equation for the electrolytic dissociation of (NH 4) 2 SO 4 .
2. Determine what, from the standpoint of the Bronsted theory, is the NH 4 + ion in the reaction:
NH 4 + + OH - ⇄ NH 3 + H 2 O.
3. The ionic strength of a solution containing both KI and Na 2 SO 4 is 0.4 mol / l. C(KI) = 0.1 mol/L. Determine the concentration of Na 2 SO 4.
Option number 18
1. Write the equation for the electrolytic dissociation of Cr 2 (SO 4) 3 .
2. Determine what, from the standpoint of the Bronsted theory, is a protein molecule in the reaction:
BLOCK OF INFORMATION
pH scale
Table 3 The relationship between the concentrations of H + and OH - ions.
Problem Solving Standards
1. The concentration of hydrogen ions in the solution is 10 - 3 mol/l. Calculate the pH, pOH and [OH - ] values in this solution. Determine the medium of the solution.
Note. The following ratios are used for calculations: lg10 a = a; 10 lg a = a.
The medium of a solution with pH = 3 is acidic, since pH< 7.
2. Calculate the pH of a hydrochloric acid solution with a molar concentration of 0.002 mol/l.
Since in a dilute solution of HC1 » 1, and in a solution of a monobasic acid C (k-you) \u003d C (k-you), we can write:
3. To 10 ml of a solution of acetic acid with C(CH 3 COOH) = 0.01 mol / l was added 90 ml of water. Find the difference between the pH values of the solution before and after dilution, if (CH 3 COOH) = 1.85 × 10 - 5.
1) In the initial solution of a weak monobasic acid CH 3 COOH:
Consequently:
2) Adding 90 ml of water to 10 ml of acid solution corresponds to a 10-fold dilution of the solution. That's why.
Measurement of the degree of dissociation of various electrolytes showed that individual electrolytes at the same normal concentration of solutions dissociate into ions very differently.
The difference in the values of the degree of dissociation of acids is especially great. For example, nitric and hydrochloric acids in 0.1 N. solutions almost completely decompose into ions; carbonic, hydrocyanic and other acids dissociate under the same conditions only to a small extent.
Of the water-soluble bases (alkalis), ammonium oxide hydrate is weakly dissociating, the remaining alkalis dissociate well. All salts, with a few exceptions, also dissociate well into ions.
The difference in the values of the degree of dissociation of individual acids is due to the nature of the valence bond between the atoms that form their molecules. The more polar the bond between hydrogen and the rest of the molecule, the easier it is to split off, the more the acid will dissociate.
Electrolytes that dissociate well into ions are called strong electrolytes, in contrast to weak electrolytes, which form only a small number of ions in aqueous solutions. Solutions of strong electrolytes retain high electrical conductivity even at very high concentrations. Conversely, the electrical conductivity of solutions of weak electrolytes rapidly decreases with increasing concentration. strong electrolytes include acids such as hydrochloric, nitric, sulfuric and some others, then alkalis (except NH 4 OH) and almost all salts.
Polyoonic acids and polyacid bases dissociate in steps. So, for example, sulfuric acid molecules first of all dissociate according to the equation
H 2 SO 4 ⇄ H + HSO 4 '
or more precisely:
H 2 SO 4 + H 2 O ⇄ H 3 O + HSO 4 '
Elimination of the second hydrogen ion according to the equation
HSO 4 ‘⇄ H + SO 4 »
or
HSO 4 '+ H 2 O ⇄ H 3 O + SO 4 "
it is already much more difficult, since it has to overcome the attraction from the doubly charged ion SO 4 ”, which, of course, attracts the hydrogen ion to itself more strongly than the singly charged ion HSO 4 '. Therefore, the second stage of dissociation or, as they say, secondary dissociation occurs in a much smallerdegree than the primary one, and ordinary sulfuric acid solutions contain only a small number of SO 4 ions "
Phosphoric acid H 3 RO 4 dissociates in three steps:
H 3 PO 4 ⇄ H + H 2 PO 4 '
H 2 PO 4 ⇄ H + HPO4 »
HPO 4 » ⇄ H + PO 4 »’
H 3 RO 4 molecules strongly dissociate into H and H 2 RO 4 ions. Ions H 2 PO 4 ' behave like a weaker acid, and dissociate into H and HPO 4 "to a lesser extent. HPO 4 ions, on the other hand, dissociate as a very weak acid, and almost do not give H ions
and PO four "'
Bases containing more than one hydroxyl group in the molecule also dissociate in steps. For example:
Va(OH) 2 ⇄ BaOH + OH'
VaOH ⇄ Va + OH'
As for salts, normal salts always dissociate into metal ions and acid residues. For example:
CaCl 2 ⇄ Ca + 2Cl 'Na 2 SO 4 ⇄ 2Na + SO 4 "
Acid salts, like polybasic acids, dissociate in steps. For example:
NaHCO 3 ⇄ Na + HCO 3 '
HCO 3 ‘⇄ H + CO 3 »
However, the second stage is very small, so that the acid salt solution contains only a small number of hydrogen ions.
Basic salts dissociate into ions of basic and acid residues. For example:
Fe(OH)Cl 2 ⇄ FeOH + 2Cl"
The secondary dissociation of ions of the main residues into metal and hydroxyl ions almost does not occur.
In table. 11 shows the numerical values of the degree of dissociation of some acids, bases and salts in 0 , 1 n. solutions.
Decreases with increasing concentration. Therefore, in very concentrated solutions, even strong acids are relatively weakly dissociated. For
Table 11
Acids, bases and salts in 0.1 N.solutions at 18°
| Electrolyte | Formula | Degree of dissociation in % |
| acids | ||
| Salt | HCl | 92 |
| Hydrobromic | HBr | 92 |
| Hydroiodide | HJ | . 92 |
| Nitrogen | HNO3 | 92 |
| sulfuric | H 2 SO 4 | 58 |
| sulphurous | H 2SO3 | 34 |
| Phosphoric | H 3 PO 4 | 27 |
| Hydrofluoric | HF | 8,5 |
| Acetic | CH3COOH | 1,3 |
| Coal | H2 CO3 | 0,17 |
| Hydrogen sulfide | H 2 S | 0,07 |
| hydrocyanic | HCN | 0,01 |
| Bornaya | H 3 BO 3 | 0,01 |
| Foundations | ||
| barium hydroxide | Ba (OH) 2 | 92 |
| caustic potash | con | 89 |
| Sodium hydroxide | NaON | 84 |
| ammonium hydroxide | NH4OH | 1,3 |
| salt | ||
| Chloride | Kcl | 86 |
| Ammonium chloride | NH4Cl | 85 |
| Chloride | NaCl | 84 |
| Nitrate | KNO 3 | 83 |
| AgNO3 | 81 | |
| acetic acid | NaCH 3 COO | 79 |
| Chloride | ZnCl 2 | 73 |
| sulfate | Na 2 SO 4 | 69 |
| sulfate | ZnSO4 | 40 |
| Sulfate |
Electrolytes are classified into two groups depending on the degree of dissociation - strong and weak electrolytes. Strong electrolytes have a degree of dissociation greater than one or more than 30%, weak ones - less than one or less than 3%.
Dissociation process
Electrolytic dissociation - the process of disintegration of molecules into ions - positively charged cations and negatively charged anions. Charged particles carry electric current. Electrolytic dissociation is possible only in solutions and melts.
The driving force of dissociation is the disintegration of covalent polar bonds under the action of water molecules. Polar molecules are pulled away by water molecules. In solids, ionic bonds are broken during the heating process. High temperatures cause vibrations of ions in the nodes of the crystal lattice.
Rice. 1. The process of dissociation.
Substances that readily decompose into ions in solutions or melts and therefore conduct electricity are called electrolytes. Non-electrolytes do not conduct electricity, tk. do not decompose into cations and anions.
Depending on the degree of dissociation, strong and weak electrolytes are distinguished. Strong ones dissolve in water, i.e. completely, without the possibility of recovery, decompose into ions. Weak electrolytes decompose into cations and anions partially. The degree of their dissociation is less than that of strong electrolytes.
The degree of dissociation shows the proportion of decomposed molecules in the total concentration of substances. It is expressed by the formula α = n/N.
Rice. 2. Degree of dissociation.
Weak electrolytes
List of weak electrolytes:
- dilute and weak inorganic acids - H 2 S, H 2 SO 3, H 2 CO 3, H 2 SiO 3, H 3 BO 3;
- some organic acids (most organic acids are non-electrolytes) - CH 3 COOH, C 2 H 5 COOH;
- insoluble bases - Al (OH) 3, Cu (OH) 2, Fe (OH) 2, Zn (OH) 2;
- ammonium hydroxide - NH 4 OH.
Rice. 3. Table of solubility.
The dissociation reaction is written using the ionic equation:
- HNO 2 ↔ H + + NO 2 - ;
- H 2 S ↔ H + + HS -;
- NH 4 OH ↔ NH 4 + + OH -.
Polybasic acids dissociate in steps:
- H 2 CO 3 ↔ H + + HCO 3 -;
- HCO 3 - ↔ H + + CO 3 2-.
Insoluble bases also break down in stages:
- Fe(OH) 3 ↔ Fe(OH) 2 + + OH – ;
- Fe(OH) 2 + ↔ FeOH 2+ + OH - ;
- FeOH 2+ ↔ Fe 3+ + OH -.
Water is classified as a weak electrolyte. Water practically does not conduct electricity, because. weakly decomposes into hydrogen cations and hydroxide ion anions. The resulting ions are reassembled into water molecules:
H 2 O ↔ H + + OH -.
If water easily conducts electricity, then it contains impurities. Distilled water is non-conductive.
The dissociation of weak electrolytes is reversible. The formed ions are reassembled into molecules.
What have we learned?
Weak electrolytes include substances that partially decompose into ions - positive cations and negative anions. Therefore, such substances do not conduct electricity well. These include weak and dilute acids, insoluble bases, sparingly soluble salts. The weakest electrolyte is water. The dissociation of weak electrolytes is a reversible reaction.
Electrolytes are substances, alloys of substances or solutions that have the ability to electrolytically conduct galvanic current. To determine which electrolytes a substance belongs to, you can use the theory of electrolytic dissociation.
Instruction
- The essence of this theory is that when melted (dissolved in water), almost all electrolytes are decomposed into ions, which are both positively and negatively charged (which is called electrolytic dissociation). Under the influence of an electric current, negative (anions "-") move towards the anode (+), and positively charged (cations, "+") move towards the cathode (-). Electrolytic dissociation is a reversible process (the reverse process is called "molarization").
- The degree (a) of electrolytic dissociation depends on the nature of the electrolyte itself, the solvent, and on their concentration. This is the ratio of the number of molecules (n) that have decayed into ions to the total number of molecules introduced into the solution (N). You get: a = n / N
- Thus, strong electrolytes are substances that completely decompose into ions when dissolved in water. Strong electrolytes, as a rule, include substances with highly polar or ionic bonds: these are salts that are highly soluble, strong acids (HCl, HI, HBr, HClO4, HNO3, H2SO4), as well as strong bases (KOH, NaOH, RbOH, Ba (OH)2, CsOH, Sr(OH)2, LiOH, Ca(OH)2). In a strong electrolyte, the substance dissolved in it is mostly in the form of ions (anions and cations); there are practically no molecules that are undissociated.
- Weak electrolytes are substances that only partially dissociate into ions. Weak electrolytes, along with ions in solution, contain undissociated molecules. Weak electrolytes do not give a strong concentration of ions in solution. The weak ones include:
- organic acids (almost all) (C2H5COOH, CH3COOH, etc.);
- some of the inorganic acids (H2S, H2CO3, etc.);
- almost all salts, slightly soluble in water, ammonium hydroxide, as well as all bases (Ca3 (PO4) 2; Cu (OH) 2; Al (OH) 3; NH4OH);
- water. They practically do not conduct electric current, or conduct, but poorly.
Weak electrolytes Substances that partially dissociate into ions. Solutions of weak electrolytes, along with ions, contain undissociated molecules. Weak electrolytes cannot give a high concentration of ions in solution. Weak electrolytes include:
1) almost all organic acids (CH 3 COOH, C 2 H 5 COOH, etc.);
2) some inorganic acids (H 2 CO 3 , H 2 S, etc.);
3) almost all water-soluble salts, bases and ammonium hydroxide Ca 3 (PO 4) 2 ; Cu(OH) 2 ; Al(OH) 3 ; NH4OH;
They are poor conductors (or almost non-conductors) of electricity.
Ion concentrations in solutions of weak electrolytes are qualitatively characterized by the degree and dissociation constant.
The degree of dissociation is expressed in fractions of a unit or as a percentage (a \u003d 0.3 is the conditional division boundary into strong and weak electrolytes).
The degree of dissociation depends on the concentration of the weak electrolyte solution. When diluted with water, the degree of dissociation always increases, because the number of solvent molecules (H 2 O) increases per solute molecule. According to the Le Chatelier principle, the equilibrium of electrolytic dissociation in this case should shift in the direction of product formation, i.e. hydrated ions.
The degree of electrolytic dissociation depends on the temperature of the solution. Usually, with increasing temperature, the degree of dissociation increases, because bonds in molecules are activated, they become more mobile and easier to ionize. The concentration of ions in a weak electrolyte solution can be calculated knowing the degree of dissociation a and the initial concentration of the substance c in solution.
HAn = H + + An - .
The equilibrium constant K p of this reaction is the dissociation constant K d:
K d = . / . (10.11)
If we express the equilibrium concentrations in terms of the concentration of a weak electrolyte C and its degree of dissociation α, then we get:
K d \u003d C. α. C. α/C. (1-α) = C. α 2 /1-α. (10.12)
This relationship is called Ostwald's dilution law. For very weak electrolytes at α<<1 это уравнение упрощается:
K d \u003d C. α 2. (10.13)
This allows us to conclude that, at infinite dilution, the degree of dissociation α tends to unity.
Protolytic equilibrium in water:
,
,
At a constant temperature in dilute solutions, the concentration of water in water is constant and equal to 55.5, ( 
)
, (10.15)
where K in is the ionic product of water.
Then =10 -7 . In practice, due to the convenience of measuring and recording, a value is used - the pH value, (criterion) of the strength of an acid or base. Similarly 
.
From equation (11.15): 
.
At pH = 7 - the reaction of the solution is neutral, at pH<7 – кислая, а при pH>7 - alkaline.
Under normal conditions (0°C):
, then
Figure 10.4 - pH of various substances and systems
10.7 Solutions of strong electrolytes
Strong electrolytes are substances that, when dissolved in water, almost completely decompose into ions. As a rule, strong electrolytes include substances with ionic or highly polar bonds: all highly soluble salts, strong acids (HCl, HBr, HI, HClO 4, H 2 SO 4, HNO 3) and strong bases (LiOH, NaOH, KOH, RbOH, CsOH, Ba (OH) 2, Sr (OH) 2, Ca (OH) 2).
In a solution of a strong electrolyte, the solute is found mainly in the form of ions (cations and anions); undissociated molecules are practically absent.
The fundamental difference between strong and weak electrolytes is that the dissociation equilibrium of strong electrolytes is completely shifted to the right:
H 2 SO 4 \u003d H + + HSO 4 -,
and therefore the constant of equilibrium (dissociation) turns out to be an indeterminate quantity. The decrease in electrical conductivity with increasing concentration of a strong electrolyte is due to the electrostatic interaction of ions.
The Dutch scientist Petrus Josephus Wilhelmus Debye and the German scientist Erich Hückel postulated:
1) the electrolyte completely dissociates, but in relatively dilute solutions (C M = 0.01 mol. l -1);
2) each ion is surrounded by a shell of ions of the opposite sign. In turn, each of these ions is solvated. This environment is called the ionic atmosphere. In the electrolytic interaction of ions of opposite signs, it is necessary to take into account the influence of the ionic atmosphere. When a cation moves in an electrostatic field, the ionic atmosphere is deformed; it thickens before him and thins behind him. This asymmetry of the ionic atmosphere has the more inhibitory effect on the movement of the cation, the higher the concentration of electrolytes and the greater the charge of the ions. In these systems, the concept of concentration becomes ambiguous and should be replaced by activity. For a binary singly charged electrolyte KatAn = Kat + + An - the activities of the cation (a +) and anion (a -), respectively, are
a + = γ + . C + , a - = γ - . C - , (10.16)
where C + and C - are the analytical concentrations of the cation and anion, respectively;
γ + and γ - - their activity coefficients.
|
It is impossible to determine the activity of each ion separately, therefore, for singly charged electrolytes, the geometric mean values of the activities i
and activity coefficients:
The Debye-Hückel activity coefficient depends at least on temperature, solvent permittivity (ε) and ionic strength (I); the latter serves as a measure of the intensity of the electric field created by ions in solution.
For a given electrolyte, the ionic strength is expressed by the Debye-Hückel equation:
The ionic strength, in turn, is equal to
where C is the analytical concentration;
z is the charge of the cation or anion.
For a singly charged electrolyte, the ionic strength is the same as the concentration. Thus, NaCl and Na 2 SO 4 at the same concentrations will have different ionic strengths. Comparison of the properties of solutions of strong electrolytes can be carried out only when the ionic strengths are the same; even small impurities dramatically change the properties of the electrolyte.
Figure 10.5 - Dependency
