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 some cases, when going from aqueous solutions to organic solvents, the degree of dissociation of electrolytes can sharply increase or decrease. In what follows, in the absence of special instructions, we will assume that the solvent is water.
According to the degree of dissociation, electrolytes are conventionally divided into strong(a> 30%), average (3% < a < 30%) и weak(a< 3%).
Strong electrolytes include:
1) some don't organic acids(HCl, HBr, HI, HNO 3, H 2 SO 4, HClO 4 and a number of others);
2) hydroxides of alkaline (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.
Everyone considers weak electrolytes carboxylic acids(except HCOOH) and hydrated forms of aliphatic and aromatic amines. Many inorganic acids (HCN, H 2 S, H 2 CO 3, etc.) and bases (NH 3 ∙ H 2 O) are also weak electrolytes.
Despite some coincidences, in general, one should not equate the solubility of a substance with its degree of dissociation. So, acetic acid and ethanol are infinitely 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
Although the terms "acid" and "base" are widely used to describe chemical processes, there is no unified approach to the classification of substances in terms of classifying them as acids or bases. Currently existing theories ( ionic theory S. Arrhenius, protolithic theory I. Bronsted and T. Lowry and electronic theory G. Lewis) have certain restrictions and, therefore, are applicable only in special cases. Let us dwell on each of these theories in more detail.
Arrhenius's theory.
In the Arrhenius ionic theory, 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 with the formation of OH - ions;
An ampholyte (amphoteric electrolyte) is an electrolyte that dissociates in solutions to form both H + and OH - ions.
For example:
HA ⇄ Н + + А - nH + + MeO n n - ⇄ Ме (ОН) n ⇄ Ме n + + nОН -In accordance with the ionic theory, acids can be both neutral molecules and ions, for example:
HF ⇄ H + + F -
H 2 PO 4 - ⇄ H + + HPO 4 2 -
NH 4 + ⇄ H + + NH 3
Similar examples can be given for reasons:
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, having no hydrogen ions in their composition, even in the absence of water, exhibit acidic properties, i.e. neutralize the bases.
In addition, the behavior of the electrolyte in aqueous solution and in a non-aqueous environment may be the 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 base properties:
HF + CH 3 COOH ⇄ CH 3 COOH 2 + + F -
Research similar types reactions, and especially reactions proceeding in non-aqueous solvents, led to the creation of more general theories acids and bases.
Bronsted and Lowry's theory.
Further development 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 the foundation is significantly expanded, which is confirmed 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, acid and base form a conjugated pair and are linked 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 with respect to each other. This can be written as an equilibrium process:
HA + B ⇄ VN + + A -,
where HA is an acid, B is a base, BH + is an acid conjugated with a base B, A - is a base conjugated with an acid HA.
Examples.
1) in the reaction:
HCl + OH - ⇄ Cl - + H 2 O,
HCl and H 2 O are acids, Cl - and OH - are the corresponding bases conjugated to them;
2) in the 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 the 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 Н 2 О molecule attaches a proton (base), forming a conjugate acid - the hydroxonium ion Н 3 О +, the other gives up a proton (acid), forming a conjugated base ОН -. This process is called autoprotolysis.
It can be seen from the examples given that, in contrast to the ideas of Arrhenius, in the theory of Bronsted 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 protolytic theory considers the concepts of "acid" and "base" not as a property, but as a function that the considered compound performs in a protolytic reaction. One and the same compound can react as an acid under some conditions, and as a base under others. So, in an aqueous solution 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, tin halides.
Lewis theory.
Another approach to the classification of substances in terms of attributing them to acids and bases was the electronic theory of Lewis. Within the framework of 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, acid and base interact with each other to form a donor-acceptor bond. As a result of the attachment of a pair of electrons, an electron with an electron deficit 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 products of the reaction between Lewis acids and bases. So, HCl is a product of the 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.
The reactions between acids and Lewis bases are also illustrated by the following examples:
Lewis bases also include halide ions, ammonia, aliphatic and aromatic amines, oxygen-containing organic compounds type R 2 CO, (where R is an organic radical).
Lewis acids include the halides of boron, aluminum, silicon, tin and other elements.
Obviously, in Lewis's theory, the concept of "acid" includes a wider range of chemical compounds. This is due to 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 in an aqueous solution undergoes complete decomposition into ions. Dissociation equation:
Al 2 (SO 4) 3 + (2x + 3y) H 2 O 2 3+ + 3 2 -,
or (excluding 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 give up protons in both ways:
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 education chemical bonds, which proceeds according to the donor-acceptor mechanism, the Ag + ion, having a free orbital, is an acceptor of electron pairs, and thus exhibits the properties of a Lewis acid.
4. Determine the ionic strength of a solution in one liter of which there are 0.1 mol KCl and 0.1 mol Na 2 SO 4.
Dissociation of the presented electrolytes proceeds in accordance with the equations:
Na 2 SO 4 2Na + + SO 4 2 -
Hence: C (K +) = C (Cl -) = C (KCl) = 0.1 mol / l;
C (Na +) = 2 × C (Na 2 SO 4) = 0.2 mol / l;
C (SO 4 2 -) = C (Na 2 SO 4) = 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 a given electrolyte with I= 0.6 mol / l.
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+) = C (SO 4 2 -) = x mol / l. V in this case the expression for calculating the ionic strength will be:
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:
.
We find the ionic strength of the solution by the formula:
7. Determine the activity coefficient of the Fe 2+ ion in an aqueous solution, the ionic strength of which is 1.
According to the Debye-Hückel law:
hence:
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 concentration of electrolyte, if a = 10%, K d = 10 - 4.
From Ostwald's breeding law:
10. The degree of dissociation of the 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 you to use the approximate formula for 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 problem statement 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 full form Ostwald equations:
Since (HA)< (HNO 2), то кислота HA является более strong acid compared to HNO 2.
13. There are two KCl solutions containing other ions as well. 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 rates f(K +) in these solutions and conclude how the properties of these solutions differ from the properties of infinitely dilute KCl solutions.
We calculate the activity coefficients of ions K + using the Debye-Hückel law:
Activity coefficient f is a measure of the deviation in the behavior of an electrolyte solution of a given concentration from its behavior with an infinite dilution of the solution.
Because f 1 = 0.316 deviates more from 1 than f 2 = 0.891, then in a solution with a higher ionic strength, a greater deviation is observed in the behavior of the KCl solution from its behavior at infinite dilution.
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. Which electrolytes are considered strong? What are the average strength? What are the weak ones? Give examples.
6. What is the dissociation constant? What does the dissociation constant depend on and what does it not depend on?
7. What is the relationship between the constant and the degree of dissociation in binary solutions of medium and weak electrolytes?
8. Why do solutions of strong electrolytes show deviations from ideality in their behavior?
9. What is the essence of the term "apparent degree of dissociation"?
10. What is ion activity? What is activity rate?
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 rate calculated? Formulate the Debye-Hückel law.
14. What is the essence of the ionic theory of acids and bases (Arrhenius theory)?
15. What is fundamental difference protolytic theory of acids and bases (the theory of Bronsted and Lowry) from the theory of Arrhenius?
16. How does the electronic theory (Lewis's theory) interpret the concept of "acid" and "base"? Give examples.
Task options for independent decision
Option number 1
1. Write the equation of electrolytic dissociation of Fe 2 (SO 4) 3.
HA + H 2 O ⇄ H 3 O + + A -.
Option number 2
1. Write the equation of electrolytic dissociation of CuCl 2.
2. Determine what the S 2 ion is from the standpoint of the Lewis theory - 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 of 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 of electrolytic dissociation of Ca (OH) 2.
2. Determine what, from the standpoint of the Bronsted theory, is the H2O 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 of electrolytic dissociation 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 equal to 10%.
Option number 6
1. Write the equation of electrolytic dissociation 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 of electrolytic dissociation of Fe (NO 3) 2.
2. Determine what, from the standpoint of the Lewis theory, is the Cl - ion in the reaction:
Cl - + AlCl 3 ⇄ -.
Option number 8
1. Write the equation of electrolytic dissociation of K 2 MnO 4.
2. Determine what the HSO 3 ion is from the point of view of the Bronsted theory - in the reaction:
HSO 3 - + OH - ⇄ SO 3 2 - + H 2 O.
Option number 9
1. Write the equation of 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 of electrolytic dissociation of Ca (NO 3) 2.
2. Determine what, from the standpoint of the Bronsted theory, is the H2O molecule in the reaction:
B + H 2 O ⇄ OH - + BH +.
3. Calculate the concentration of the electrolyte in the solution, if a = 5%, a = 10 - 5.
Option number 11
1. Write the equation of 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 of electrolytic dissociation of Na 2 CO 3.
2. Determine what, from the standpoint of the Bronsted theory, is the H2O 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 rates f(Na +) in these solutions.
Option number 13
1. Write the equation of 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 of electrolytic dissociation 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 of electrolytic dissociation 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 of 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 of electrolytic dissociation (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 equal to 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 of 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:
INFORMATION BLOCK
PH scale
Table 3. The relationship between the concentrations of ions H + and OH -.
Problem solving standards
1. The concentration of hydrogen ions in the solution is 10 - 3 mol / l. Calculate the values of pH, pOH and [OH -] in this solution. Determine the environment of the solution.
Note. For calculations, the following ratios are used: lg10 a = a; 10 lg a = a.
The solution medium with pH = 3 is acidic, since the pH< 7.
2. Calculate the pH of the solution of hydrochloric acid 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 (to-you) = C (to-you), we can write:
3. To 10 ml of acetic acid solution 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:
Hence:
2) Adding 90 ml of water to 10 ml of acid solution corresponds to 10-fold dilution of the solution. That's why.
SOLUTIONS
THEORY OF ELECTROLYTIC DISSOCIATION
ELECTROLYTIC DISSOCIATION
ELECTROLYTES AND NONELECTROLYTES
Electrolytic dissociation theory
(S. Arrhenius, 1887)
1. When dissolved in water (or melted), electrolytes decompose into positively and negatively charged ions (undergo electrolytic dissociation).
2. Under the action of an electric current, cations (+) move to the cathode (-), and anions (-) to the anode (+).
3. Electrolytic dissociation is a reversible process (the reverse reaction is called molarization).
4. The degree of electrolytic dissociation ( a ) depends on the nature of the electrolyte and solvent, temperature and concentration. It shows the ratio of the number of molecules decayed into ions ( n ) To the total molecules introduced into the solution ( N).
a = n / N 0< a <1
Mechanism of electrolytic dissociation of ionic substances
When dissolving compounds with ionic bonds ( e.g. NaCl ) the hydration process begins with the orientation of the water dipoles around all the projections and faces of the salt crystals.
Orienting around the ions of the crystal lattice, water molecules form either hydrogen or donor-acceptor bonds with them. During this process, a large amount of energy is released, which is called hydration energy.
The energy of hydration, the value of which is comparable to the energy of the crystal lattice, is used to destroy the crystal lattice. In this case, the hydrated ions pass layer by layer into the solvent and, mixing with its molecules, form a solution.
The mechanism of electrolytic dissociation of polar substances
Substances, the molecules of which are formed according to the type of polar covalent bond (polar molecules), dissociate in a similar way. Around each polar molecule of matter ( e.g. HCl ), the water dipoles are oriented in a certain way. As a result of interaction with water dipoles, the polar molecule becomes even more polarized and turns into ionic, then free hydrated ions are easily formed.
Electrolytes and non-electrolytes
The electrolytic dissociation of substances, which occurs with the formation of free ions, explains electrical conductivity solutions.
It is customary to write the process of electrolytic dissociation in the form of a diagram, without revealing its mechanism and omitting the solvent ( H 2 O ), although he is the main contributor.
CaCl 2 «Ca 2+ + 2Cl -
KAl (SO 4) 2 "K + + Al 3+ + 2SO 4 2-
HNO 3 "H + + NO 3 -
Ba (OH) 2 «Ba 2+ + 2OH -
It follows from the electroneutrality of the molecules that the total charge of cations and anions should be zero.
For example, for
Al 2 (SO 4) 3 ––2 (+3) + 3 (-2) = +6 - 6 = 0
KCr (SO 4) 2 ––1 (+1) + 3 (+3) + 2 (-2) = +1 + 3 - 4 = 0
Strong electrolytes
These are substances that, when dissolved in water, almost completely disintegrate into ions. As a rule, strong electrolytes include substances with ionic or strongly polar bonds: all readily 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 strong electrolyte solution, the solute is found mainly in the form of ions (cations and anions); undissociated molecules are practically absent.
Weak electrolytes
Substances partially dissociating 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 slightly water-soluble salts, bases and ammonium hydroxide(Ca 3 (PO 4) 2; Cu (OH) 2; Al (OH) 3; NH 4 OH);
4) water.
They are bad (or almost never) electricity.
CH 3 COOH "CH 3 COO - + H +
Cu (OH) 2 "[CuOH] + + OH - (first stage)
[CuOH] + "Cu 2+ + OH - (second stage)
H 2 CO 3 «H + + HCO - (first stage)
HCO 3 - "H + + CO 3 2- (second stage)
Non-electrolytes
Substances, aqueous solutions and melts of which do not conduct electric current. They contain covalent non-polar or low-polarity bonds that do not decay into ions.
Gases, solids (non-metals), organic compounds (sucrose, gasoline, alcohol) do not conduct electric current.
Dissociation degree. Dissociation constant
The concentration of ions in solutions depends on how completely a given electrolyte dissociates into ions. In solutions of strong electrolytes, the dissociation of which can be considered complete, the concentration of ions can be easily determined from the concentration (c) and the composition of the electrolyte molecule (stoichiometric indices), for example :
The concentration of ions in solutions of weak electrolytes is qualitatively characterized by the degree and constant of dissociation.
Dissociation degree (a) is the ratio of the number of molecules decayed into ions ( n ) to the total number of dissolved molecules ( N):
a = n / N
and is expressed in fractions of one or in% ( a = 0.3 - the conditional border of division into strong and weak electrolytes).
Example
Determine the molar concentration of cations and anions in 0.01 M solutions KBr, NH 4 OH, Ba (OH) 2, H 2 SO 4 and CH 3 COOH.
Dissociation of weak electrolytes a = 0.3.
Solution
KBr, Ba (OH) 2 and H 2 SO 4 - strong electrolytes dissociating completely(a = 1).
KBr “K + + Br -
0.01 M
Ba (OH) 2 «Ba 2+ + 2OH -
0.01 M
0.02 M
H 2 SO 4 «2H + + SO 4
0.02 M
[SO 4 2-] = 0.01 M
NH 4 OH and CH 3 COOH - weak electrolytes(a = 0.3)
NH 4 OH + 4 + OH -
0.3 0.01 = 0.003 M
CH 3 COOH "CH 3 COO - + H +
[H +] = [CH 3 COO -] = 0.3 0.01 = 0.003 M
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 increases ( H 2 O ) per molecule of solute. According to Le Chatelier's 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 by knowing the degree of dissociationaand the initial concentration of the substancec in solution.
Example
Determine the concentration of undissociated molecules and ions in a 0.1 M solution NH 4 OH if the degree of dissociation is 0.01.
Solution
Molecular concentration NH 4 OH , which by the moment of equilibrium decay into ions, will be equal toac... Ion concentration NH 4 - and OH - - will be equal to the concentration of dissociated molecules and equalac(according to the equation of electrolytic dissociation)
|
NH 4 OH |
NH 4 + |
OH - |
||
|
c - a c |
A c = 0.01 0.1 = 0.001 mol / L
[NH 4 OH] = c - a c = 0.1 - 0.001 = 0.099 mol / l
Dissociation constant ( K D ) is the ratio of the product of equilibrium ion concentrations in the power of the corresponding stoichiometric coefficients to the concentration of undissociated molecules.
It is the equilibrium constant of the electrolytic dissociation process; characterizes the ability of a substance to decay into ions: the higher K D , the greater the concentration of ions in the solution.
Dissociation of weak polybasic acids or polyacid bases proceeds in steps, respectively, for each step there is its own dissociation constant:
First stage:
H 3 PO 4 «H + + H 2 PO 4 -
K D 1 = () / = 7.1 10 -3
Second stage:
H 2 PO 4 - "H + + HPO 4 2-
K D 2 = () / = 6.2 10 -8
Third step:
HPO 4 2- "H + + PO 4 3-
K D 3 = () / = 5.0 10 -13
K D 1> K D 2> K D 3
Example
Get an equation relating the degree of electrolytic dissociation of a weak electrolyte ( a ) with the dissociation constant (Ostwald dilution law) for a weak monobasic acid ON .
HA «H + + A +
K D = () /
If the total concentration of a weak electrolyte is indicatedc, then the equilibrium concentrations H + and A - are equal ac, and the concentration of undissociated molecules HA - (c - a c) = c (1 - a)
K D = (a c a c) / c (1 - a) = a 2 c / (1 - a)
In the case of very weak electrolytes ( a £ 0.01)
K D = c a 2 or a = \ é (K D / c)
Example
Calculate the degree of dissociation of acetic acid and the concentration of ions H + in 0.1 M solution, if K D (CH 3 COOH) = 1.85 10 -5
Solution
We use the Ostwald dilution law
\ é (K D / c) = \ é ((1.85 10 -5) / 0.1)) = 0.0136 or a = 1.36%
[H +] = a c = 0.0136 0.1 mol / l
Solubility product
Definition
Put some insoluble salt in a beaker, e.g. AgCl and add distilled water to the sediment. In this case, the ions Ag + and Cl - , experiencing attraction from the side of the surrounding water dipoles, gradually detach from the crystals and pass into solution. Colliding in solution, ions Ag + and Cl - form molecules AgCl and are deposited on the surface of the crystals. Thus, two mutually opposite processes occur in the system, which leads to dynamic equilibrium, when the same number of ions pass into the solution per unit time Ag + and Cl - how many are precipitated. Accumulation of ions Ag + and Cl - stops in solution, 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 occur:
1) Transfer of ions from sediment to 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 speed of this process V 2 depends on ion concentration Ag + and Cl -. According to the law of action of the masses:
V 2 = k 2
Because this system is in a state of equilibrium, then
V 1 = V 2
k 2 = k 1
K 2 / k 1 = const (at T = const)
Thus, the product of ion concentrations in a saturated solution of a sparingly soluble electrolyte at a constant temperature is constant size... This quantity is calledsolubility product(NS ).
In the example given NS AgCl = [Ag +] [Cl -] ... In cases where the electrolyte contains two or more identical ions, the concentration of these ions, when calculating the product of solubility, must be raised to the appropriate power.
For example, PR Ag 2 S = 2; PR PbI 2 = 2
V general case solubility product expression for electrolyte A m B n
OL A m B n = [A] m [B] n.
The values of the solubility product are different for different substances.
For example, PR CaCO 3 = 4.8 10 -9; PR AgCl = 1.56 10 -10.
NS easy to calculate knowing ra c the solubility of the compound for a given t °.
Example 1
The solubility of CaCO 3 is 0.0069 or 6.9 10 -3 g / l. Find PR CaCO 3.
Solution
Let us express the solubility in moles:
S CaCO 3 = ( 6,9 10 -3 ) / 100,09 = 6.9 10 -5 mol / l
M CaCO 3
Since each molecule CaCO 3 gives, upon dissolution, one ion at a time Ca 2+ and CO 3 2-, then
[Ca 2+] = [CO 3 2-] = 6.9 10 -5 mol / l ,
hence, PR CaCO 3 = [Ca 2+] [CO 3 2-] = 6.9 10 -5 6.9 10 -5 = 4.8 10 -9
Knowing the value of PR , you can in turn calculate the solubility of the substance in mol / l or g / l.
Example 2
Solubility product PR PbSO 4 = 2.2 10 -8 g / l.
What is solubility PbSO 4?
Solution
Let's denote solubility PbSO 4 via X mol / l. Going into solution X moles of PbSO 4 will give X Pb 2+ ions and X ionsSO 4 2- , i.e .:
= = X
NSPbSO 4 = = = X X = X 2
X =\ é(NSPbSO 4 ) = \ é(2,2 10 -8 ) = 1,5 10 -4 mol / l.
To go over to the solubility, expressed in g / l, we multiply the found value by molecular weight, after which we get:
1,5 10 -4 303,2 = 4,5 10 -2 g / l.
Precipitation formation
If
[ Ag + ] [ Cl - ] < ПР AgCl- unsaturated solution
[ Ag + ] [ Cl - ] = OLAgCl- saturated solution
[ Ag + ] [ Cl - ]> OLAgCl- oversaturated solution
A precipitate is formed when the product of the concentration of ions of a poorly soluble electrolyte exceeds the value of its product of solubility at a given temperature. When the ionic product becomes equal toNS, the 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 the precipitate precipitate when mixing equal volumes 0.2MsolutionsPb(NO 3
)
2
andNaCl.
NSPbCl 2
= 2,4 10
-4
.
Solution
When mixing, the volume of the solution doubles and the concentration of each of the substances is halved, i.e. becomes 0.1 M or 1.0 10 -1 mol / l. Such are there will be concentrationsPb 2+ andCl - ... Hence,[ Pb 2+ ] [ Cl - ] 2 = 1 10 -1 (1 10 -1 ) 2 = 1 10 -3 ... The resulting value exceedsNSPbCl 2 (2,4 10 -4 ) ... Therefore, some of the saltPbCl 2 precipitates. From all that has been said above, 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 valueNScannot be precipitated from dilute solutions.For example, sedimentPbCl 2 will not drop out when mixing equal volumes 0.1MsolutionsPb(NO 3 ) 2 andNaCl... When mixing equal volumes, the concentration of each of the substances will become0,1 / 2 = 0,05 Mor 5 10 -2 mol / L... Ionic product[ Pb 2+ ] [ Cl 1- ] 2 = 5 10 -2 (5 10 -2 ) 2 = 12,5 10 -5 .The resulting value is lessNSPbCl 2 therefore no precipitation will occur.
Influence of the amount of precipitant
For the most complete precipitation, an excess of the precipitant is used.
For example, precipitating saltBaCO 3 : BaCl 2 + Na 2 CO 3 ® BaCO 3 ¯ + 2 NaCl. After adding an equivalent amountNa 2 CO 3 ions remain in the solutionBa 2+ , the concentration of which is due to the valueNS.
Increased ion concentrationCO 3 2- caused by the addition of an excess of precipitant(Na 2 CO 3 ) , will entail a corresponding decrease in the concentration of ionsBa 2+ in solution, i.e. will increase the completeness of the deposition of this ion.
Influence of the ion of the same name
The solubility of poorly soluble electrolytes decreases in the presence of other strong electrolytes with ions of the same name. If to an unsaturated solutionBaSO 4 add a little solutionNa 2 SO 4 , then the ionic product, which was initially less NSBaSO 4 (1,1 10 -10 ) will gradually reachNSand will exceed it. Precipitation will begin.
Influence of temperature
NSis constant at constant temperature. With increasing temperature NS increases; therefore, precipitation is best carried out from cooled solutions.
Dissolution of precipitation
The solubility product rule is important for converting sparingly soluble precipitates into solution. Suppose you want to dissolve the precipitateBaWITHO 3
... The solution in contact with this precipitate is saturated relativelyBaWITHO 3
.
It means that[
Ba 2+
] [
CO 3
2-
] = OLBaCO 3
.
If you add acid to the solution, then the ionsH + will bind the ions present in the solutionCO 3 2- into fragile carbonic acid molecules:
2H + + CO 3 2- ® H 2 CO 3 ® H 2 O + CO 2
As a result, the concentration of the ion will sharply decreaseCO 3 2- , the ionic product will become less thanNSBaCO 3 ... The solution will be unsaturated relativeBaWITHO 3 and part of the sedimentBaWITHO 3 will go into solution. By adding 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 ionic product of the poorly soluble electrolyte becomes less than the valueNS... In order to dissolve the precipitate, such an electrolyte is introduced into the solution, the ions of which can form a poorly dissociated compound with one of the ions of the sparingly soluble electrolyte. This explains the dissolution of sparingly soluble hydroxides in acids
Fe (OH) 3 + 3HCl® FeCl 3 + 3H 2 O
JonahOH - bind to poorly dissociated moleculesH 2 O.
Table.The product of solubility (PR) and solubility at 25AgCl
1,25 10 -5
1,56 10 -10
AgI
1,23 10 -8
1,5 10 -16
Ag 2 CrO 4
1,0 10 -4
4,05 10 -12
BaSO 4
7,94 10 -7
6,3 10 -13
CaCO 3
6,9 10 -5
4,8 10 -9
PbCl 2
1,02 10 -2
1,7 10 -5
PbSO 4
1,5 10 -4
2,2 10 -8
Electrolytes are classified into two groups depending on the degree of dissociation - strong and weak electrolytes. Strong electrolytes have a degree of dissociation of more than one or more than 30%, weak electrolytes - less than one or less than 3%.
Dissociation process
Electrolytic dissociation is the process of decomposition of molecules into ions - positively charged cations and negatively charged anions. Charged particles carry an electric current. Electrolytic dissociation is possible only in solutions and melts.
The driving force behind dissociation is the disintegration of covalent polar bonds under the action of water molecules. Polar molecules are pulled back by water molecules. In solids, ionic bonds are destroyed during heating. 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 an electric current are called electrolytes. Non-electrolytes do not conduct electricity. 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 reduction, decay into ions. Weak electrolytes are partially decomposed into cations and anions. The degree of their dissociation is less than that of strong electrolytes.
The degree of dissociation shows the proportion of disintegrated molecules in the total concentration of substances. It is expressed by the formula α = n / N.
Rice. 2. The 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. Solubility table.
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 decompose 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 electric current, because weakly decomposes into hydrogen cations and gyroxide ion anions. The resulting ions are collected back into water molecules:
H 2 O ↔ H + + OH -.
If water conducts electricity easily, it means there are impurities in it. Distilled water is non-conductive.
Dissociation of weak electrolytes is reversible. The resulting 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 electric current well. These include weak and dilute acids, insoluble bases, poorly soluble salts. The weakest electrolyte is water. Dissociation of weak electrolytes is a reversible reaction.
Hydrolysis of salts
By hydrolysis are called reactions of interaction of a substance with water, leading to the formation of weak electrolytes (acids, bases, acidic or basic salts). The result of hydrolysis can be regarded as a violation of the equilibrium of water dissociation. Compounds of various classes are susceptible to hydrolysis, but the most important case is the hydrolysis of salts. Salts are usually - strong electrolytes, which undergo complete dissociation into ions and can interact with water ions.
The most important cases of salt hydrolysis:
1. Salt is formed by a strong base and a strong acid. For example: NaCl is a salt formed by a strong base NaOH and a strong acid HCl;
NaCl + HOH ↔ NaOH + HCl - molecular equation;
Na + + Cl - + HOH ↔ Na + + OH - + H + + Cl - - complete ionic equation;
HOH ↔ OH - + H + is an abbreviated ionic equation.
As can be seen from the abbreviated ionic equation, the salt formed by a strong base and a strong acid does not interact with water, that is, it does not undergo hydrolysis, and the medium remains neutral.
2. Salt is formed by a strong base and a weak acid. For example: NaNO 2 is a salt formed by a strong base NaOH and a weak acid HNO 2, which practically does not dissociate into ions.
NaNO 2 + HOH NaOH + HNO 2;
Na + + NO 2 - + HOH ↔ Na + + OH - + HNO 2;
NO 2 - + HOH ↔ OH - + HNO 2.
In this case, the salt undergoes hydrolysis, and hydrolysis proceeds along the anion, and the cation practically does not participate in the hydrolysis process. Since an alkali is formed as a result of hydrolysis, there is an excess of OH - anions in the solution. A solution of such a salt acquires an alkaline medium, i.e. pH> 7.
Stage I Na 2 CO 3 + HOH ↔ NaOH + NaHCO 3;
CO 3 2 - + HOH ↔ OH - + HCO 3 -;
II stage NaHCO 3 + HOH ↔ NaOH + H 2 CO 3;
HCO 3 - + HOH ↔ OH - + H 2 CO 3.
At standard conditions and with moderate dilution of the solution, the hydrolysis of the salts proceeds only through the first stage. The second is suppressed by the products that are formed at the first stage. The accumulation of OH ions - entails a shift in equilibrium to the left.
3. Salt is formed by a weak base and a strong acid. For example: NH 4 NO 3 is a salt formed by a weak base NH 4 OH and a strong acid HNO 3.
NH 4 NO 3 + HOH ↔ NH 4 OH + HNO 3;
NH 4 + + HOH ↔ H + + NH 4 OH.
In this case, the salt undergoes hydrolysis, and hydrolysis proceeds along the cation, and the anion practically does not participate in the hydrolysis process. A solution of such a salt becomes acidic, i.e. NS< 7.
As in the previous case, salts of multiply charged ions are hydrolyzed in stages, although the second stage is also suppressed.
I stage Mg (NO 3) 2 + HOH ↔ MgOHNO 3 + HNO 3;
Mg 2+ + HOH MgOH + + H +;
II stage MgOHNO 3 + HOH ↔ Mg (OH) 2 + HNO 3;
MgOH + + HOH ↔ Mg (OH) 2 + H +.
4. The salt is formed by a weak base and a weak acid. For example: NH 4 CN is the salt formed by the weak base NH 4 OH and the weak acid HCN.
NH 4 CN + HOH ↔ NH 4 OH + HCN;
NH 4 + + CN - + HOH ↔ NH 4 OH + HCN.
In this case, both cations and anions are involved in hydrolysis. They bind both hydrogen cations and hydroxo anions of water, forming weak electrolytes (weak acids and weak bases). The reaction of a solution of such salts can be either weakly acidic (if the base formed as a result of hydrolysis is weaker than the acid), or slightly alkaline (if the base turns out to be stronger than the acid), or it can be neutral (if the base and acid are of the same strength) ...
In the hydrolysis of a salt of multiply charged ions, stage I does not suppress the subsequent ones, and the hydrolysis of such salts proceeds completely even at room temperature.
Stage I (NH 4) 2 S + HOH ↔ NH 4 OH + NH 4 HS;
2NH 4 + + S 2 - + HOH ↔ NH 4 OH + NH 4 + + HS -;
II stage NH 4 HS + HOH ↔ NH 4 OH + H 2 S;
NH 4 + + HS - + HOH ↔ NH 4 OH + H 2 S.
Instructions
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 electric current, negative ("-") to the anode (+), and positively charged (cations, "+"), move to the cathode (-). Electrolytic dissociation is reversible process(the reverse process is called "molarization").
The degree (a) of electrolytic dissociation depends on the electrolyte itself, the solvent, and on their concentration. This is the ratio of the number of molecules (n) that 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, are substances with strongly polar or bonds: these are salts that are highly soluble (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 (); there are practically no molecules that are not dissociated.
Weak electrolytes are substances that only partially dissociate into ions. Weak electrolytes, together with ions in solution, contain undissociated molecules. Weak electrolytes do not give a strong concentration of ions in solution.
The weak include:
- organic acids (almost all) (C2H5COOH, CH3COOH, etc.);
- some of the 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.
note
Though pure water conducts electric current very poorly, it still has measurable electrical conductivity, due to the fact that water dissociates slightly into hydroxide ions and hydrogen ions.
Most electrolytes are corrosive substances, so when working with them, be extremely careful and follow safety rules.
Strong base - inorganic chemical compound formed by the hydroxyl group -OH and alkaline (elements of group I periodic system: Li, K, Na, RB, Cs) or alkaline earth metal (elements of group II Ba, Ca). They are written in the form of the formulas LiOH, KOH, NaOH, RbOH, CsOH, Ca (OH) ₂, Ba (OH) ₂.
You will need
- evaporating cup
- burner
- indicators
- metal bar
- Н₃РО₄
Instructions
Strong foundations show common to all. The presence in the solution is determined by the color change of the indicator. Add phenolphthalein to the sample with the test solution or omit the litmus test. Methyl orange gives a yellow color, phenolphthalein gives a purple color, and litmus paper gives a blue color... The stronger the base, the more intensely the indicator is colored.
If you need to find out which alkalis are presented to you, then conduct a qualitative analysis of the solutions. The most common strong bases are lithium, potassium, sodium, barium, and calcium. Bases react with acids (neutralization reactions) to form salt and water. In this case, Ca (OH) ₂, Ba (OH) ₂ and LiOH can be distinguished. When with acid, insoluble are formed. The rest of the hydroxides will not give precipitation, because all K and Na salts are soluble.
3 Ca (OH) ₂ + 2 H₃PO₄ - → Ca₃ (PO₄) ₂ ↓ + 6 H₂O
3 Ва (ОН) ₂ +2 Н₃РО₄ - → Ва₃ (PO₄) ₂ ↓ + 6 H₂О
3 LiOH + Н₃РО₄ - → Li₃PO₄ ↓ + 3 H₂О
Strain and dry. Add dried sediment to the burner flame. By changing the color of the flame, it is possible to qualitatively determine the ions of lithium, calcium and barium. Accordingly, you will determine where is what hydroxide. Lithium salts give the burner a carmine red color. Barium salts - into green, and calcium salts - into raspberry.
The remaining alkalis form soluble orthophosphates.
3 NaOH + H₃PO₄-- → Na₃PO₄ + 3 H₂O
3 KOH + Н₃РО₄-- → K₃РО₄ + 3 H₂О
It is necessary to evaporate the water to a dry residue. Introduce the evaporated salts on a metal rod into the burner flame one by one. There, sodium salt - the flame will turn bright yellow, and potassium - pink-purple. Thus, having a minimum set of equipment and reagents, you have determined all the strong bases given to you.
An electrolyte is a substance that is a dielectric in a solid state, that is, it does not conduct an electric current, however, in a dissolved or molten state it becomes a conductor. Why is there such a sharp change in properties? The fact is that electrolyte molecules in solutions or melts dissociate into positively charged and negatively charged ions, due to which these substances in this state of aggregation are able to conduct electric current. Most salts, acids and bases have electrolytic properties.
Instructions
What substances are strong? Such substances, in solutions or melts of which almost 100% of the molecules are exposed, and regardless of the concentration of the solution. The list includes the absolute majority of soluble alkalis, salts and some acids, such as hydrochloric, bromic, iodic, nitric, etc.
And how do weak electrolytes? Firstly, they dissociate to a very small extent (no more than 3% of the total number of molecules), and secondly, the higher the concentration of the solution, the worse and slower they are. These electrolytes include, for example, (ammonium hydroxide), most organic and inorganic acids (including hydrofluoric acid - HF) and, of course, the familiar water. Since only a negligible fraction of its molecules decomposes into hydrogen ions and hydroxyl ions.
Remember that the degree of dissociation and, accordingly, the strength of the electrolyte depends on factors: the nature of the electrolyte itself, the solvent, and the temperature. Therefore, this division itself is to a certain extent arbitrary. After all, the same substance can at different conditions be both a strong electrolyte and a weak one. To assess the strength of the electrolyte, a special value was introduced - the dissociation constant, determined on the basis of the law of mass action. But it only applies to weak electrolytes; strong electrolytes they do not obey the law of the masses.
Sources:
- strong electrolytes list
Salt- this is chemical substances consisting of a cation, that is, a positively charged ion, a metal and a negatively charged anion - an acid residue. There are many types of salts: normal, acidic, basic, double, mixed, hydrated, complex. It depends on the compositions of the cation and anion. How can you determine base salt?
