In the general case, in accordance with the Bronsted-Lowry protolytic theory, according to equation (4.2), we have for the dissociation of a weak monobasic acid:
True thermodynamic constant TO this balance will be
where all activities are equilibrium. Let's imagine this ratio in the form:
Let us denote, as in the previous case, the product of two constants TO and a(H 2 O) through (H 2 O) = const at T= const. Then
or approximately:
where all concentrations are equilibrium. Here the value TO A called acid dissociation (ionization) constant or simply acidity constant.
For many weak acids the numerical values TO A are very small, so instead of the size TO A apply strength indicator (or simply indicator):
RK A =- lg TO A .
The more TO A(i.e., the less p TO A ), the stronger the acid.
Let the initial concentration of a monobasic acid HB be equal to the degree of its dissociation (ionization) in solution. Then the equilibrium concentrations of the ions [H 3 O + ] and [B - ] will be equal to [H 3 O + ] = [B - ] = αс A , a equilibrium acid concentration [НВ] = With A - α With A = With A(1 - α). Substituting these values of equilibrium concentrations into the expression for the equilibrium constant (4.10), we obtain:
If instead of concentration With A use its inverse V- dilution (dilution), expressed in l/mol, V=1/With A , then the formula for TO A will look like:
This relation and also the expression
describe Ostwald's law of dilution (or dilution law) for a weak binary electrolyte. At a1 (a typical case in many analytical systems)
It is easy to show that, in the general case, for a weak electrolyte of any composition K n A m , decomposing into ions according to the scheme
K n A m = P TO t+ +t A n -
Ostwald's dilution law is described by the relation
Where With- the initial concentration of a weak electrolyte, for example, a weak acid. So, for orthophosphoric acid H 3 PO 4 (P = 3,
T= 1), which totally decays into ions according to the scheme
.
For a binary electrolyte, the relation becomes (4.11). For a1 we have:
Let us find the equilibrium pH value of a solution of monobasic acid NV. Equilibrium concentration of hydrogen ions
Using the notation and we get:
pH = 0.5(p TO A+p With A). (4.12)
Thus, to calculate the equilibrium pH value of a solution of a weak monobasic acid, it is necessary to know the acidity constant of this acid TO A and its initial concentration With A .
Calculate the pH of an acetic acid solution with an initial concentration of 0.01 mol/l.
At room temperature for acetic acid TO A = 1.74 10 -5 and p TO A = 4,76.
According to formula (4.12) we can write:
pH = 0.5(p TO A+p With A) = 0,5(476-0,01) = 0,5(4,76+2) = 3,38.
A similar consideration can be carried out for equilibria in a solution of any weak polybasic acids.
Polybasic acids dissociate into ions stepwise, in several stages, each of which is characterized by its own equilibrium constant stepwise acid dissociation constant. So, for example, in solutions of orthoboric acid H 3 BO 3, equilibria are established (the values of the constants are given for 25 ° C):
H 3 VO 3 + H 2 O = H 3 O + +, TO 1
= 
H 2 O = H 3 O + +, TO 2
= 
H 2 O = H 3 O + +, TO 3 =
The acid dissociation constant of each subsequent stage is less than the dissociation constant of the previous stage - usually by several orders of magnitude.
The product of all stepwise dissociation constants is equal to the total acid dissociation constant K:
TO 1 TO 2 ...TO P =K.
Thus, it is easy to see that for orthoboric acid the value
TO 1
TO 2
TO 3
=K= 
there is a complete acid dissociation constant according to the scheme:
4.3.2 Basicity constant and pH of solutions of weak bases
In accordance with the protolytic theory of acids and bases of Bronsted-Lowry, in the general case, for the ionization of a single acid weak base B in aqueous solutions, one can write:
B + H 2 O = HB + + OH -
If the degree of ionization of the base is a1, then the concentration constant can be taken as the constant of this chemical equilibrium
Proceeding similarly to the previous one, we get:
TO = =K b = const when T= const
as the product of two constants TO=const and [H 2 O] = const.
Let's call the quantity K b , equal, therefore,
K b = , (4.13)
dissociation (ionization) constant of a weak one-acid baseorjust a basicity constant this base, and the magnitude
p K b = K b ,
A strength indicator (or simply an indicator) of the basicity constant.
According to the Ostwald dilution law in the case under consideration (similar to relation (4.11))
K b =,
where is the degree of ionization of a single acid weak base, and is its initial concentration. Since for a weak base a1, then
Let us find the equilibrium pH value of an aqueous solution of the monoacid base under consideration at room temperature. In accordance with formula (4.7) we have:
pH = p TO w - pOH = 14 - pOH.
Let's determine the value pOH = [OH - ]. Obviously
[OH - ] = = 
Using the indicators pOH = [OH - ], p TO b =K b And
p = , we get: pOH = 0.5(p TO b+ p). Substituting this expression into the above formula for pH, we arrive at the relationship
pH = 14 - pOH = 14 – 0.5 (p TO b+ p).
So, the equilibrium pH value in a solution of a weak single acid base can be calculated using the formula (4.15):
pH = 14 - 0.5(p TO b+ p). (4.15)
Let us calculate the pH in a 0.01 mol/l aqueous solution of ammonia, for which at room temperature TO b= and p TO b = 4,76.
In an aqueous solution of ammonia, an equilibrium is established:
which is mostly shifted to the left, so that the degree of ionization of ammonia is . Therefore, to calculate the pH value, you can use relation (4.15):
pH = 14 - 0.5(p TO b+ p) =
A similar consideration can be carried out for any weak polyacid grounds. True, this results in more cumbersome expressions.
Weak polyacid bases, like weak polybasic acids, dissociate stepwise, and each step of dissociation also has its own stepwise dissociation constant of the base - stepwise basicity constant.
For example, lead hydroxide Pb(OH) 2 in aqueous solutions decomposes into ions in two stages:
The same equilibria can be written in another way, adhering (within the framework of the protolytic theory) to the definition of a base as a substance that attaches a proton, in this case, accepting it from a water molecule:
The stepwise basicity constants can be represented in the form:
With this recording of the indicated equilibria, it is assumed that a proton from a water molecule passes to a hydroxyl group with the formation of a water molecule (), as a result of which the number of water molecules near the lead (II) atom increases by one, and the number of hydroxyl groups associated with the lead (II) atom ), also decreases by one at each dissociation step.
Work TO 1 TO 2 =K=[Pb 2+ ][OH - ] 2 /[Pb(OH) 2 ] =
2.865, where TO- total dissociation constant according to the scheme
or according to a different scheme written down
which ultimately leads to the same result.
Another example is the organic base ethylenediamine, which undergoes ionization in an aqueous solution in two stages. First stage:
Second stage:
Work 
-
total dissociation constant. It corresponds to equilibrium
The numerical values of the equilibrium constants are given above for room temperature.
As in the case of polybasic acids, for a weak polyacid base the dissociation constant of each subsequent step is usually several orders of magnitude less than the dissociation constant of the previous stage.
In table Table 4.2 shows the numerical values of the acidity and basicity constants of some weak acids and bases.
Table 4.2. True thermodynamic ionization constants in aqueous solutions of some acids and bases.
TO A- acidity constant, TO b- basicity constant,
TO 1 - dissociation constant for the first step,
TO 2
- dissociation constant for the second step, etc.
| Dissociation constants of weak acids |
||
| Acid | TO A | R TO A=-lg TO A |
| Nitrogenous Aminoacetic Benzoinaya Boric (orthoboric) Tetraboric Types of protolytic reactions. MU "Solutions" pp. 52-55 Autoprotolysis of water. Ionic product of water.MU "Solutions"» page 56 A small part of water molecules is always in the ionic state, although it is a very weak electrolyte. Ionization and further dissociation of water, as already mentioned, is described by the equation of the protolytic reaction of acid-base disproportionation or autoprotolysis. Water is a very weak electrolyte, hence the resulting conjugate acid and conjugate base are strong. Therefore, the equilibrium of this protolytic reaction is shifted to the left. The constant of this equilibrium K equals = The quantitative value of the product of water ion concentration × is ionic product of water. It is equal to: × = K equal. × 2 = 1 × 10 - 14 Therefore: KH 2O = × = 10 – 14 or simplified KH 2O = × = 10 – 14 K H 2 O is the ionic product of water, the constant of autoprotolysis of water, or simply the constant of water. KH2O depends on temperature. It increases with increasing temperature. In chemically pure water = = = 1×10 – 7. This is a neutral environment. The solution may contain > – the medium is acidic or< – среда щелочная = ; = pH value To quantitatively express the acidity of solutions, use hydrogen ion concentration indicator pH. Hydrogen index is a value equal to the negative decimal logarithm of the concentration of free hydrogen ions in a solution. pH = – log ⇒ = 10 – pH In a neutral environment pH = 7 At acidic pH< 7 In alkaline pH > 7 To characterize the basicity of the medium, the hydroxyl index pOH is used. рОН = – log [ОH - ] ⇒ [ОH - ] = 10 – рОН pH + pOH = 14 Þ pH = 14 – pOH and pOH = 14 – pH Formulas for calculating pH for solutions of acids and bases. pH = – lg
Calculate the pH of a HCl solution with C(HCl) = 0.1 mol/l under the condition of its complete dissociation. C(HCl) = 0.1 mol/l; pH \u003d - lg 0.1 \u003d 1 2. Strong bases: [ОH - ] = С(1/z base) Calculate the pH of the NaOH solution under the same conditions. C(NaOH) = 0.1 mol/l; = = 10 – 13; pH = – log 10 – 13 = 13 3. Weak acids Calculate the pH of a solution of acetic acid with a molar concentration of 0.5 mol/L. K CH 3COOH = 1.8×10 – 5. 3×10 - 3 pH = – log 3×10 – 3 = 2.5 4. Weak foundations Calculate the pH of an ammonia solution with a molar concentration of 0.2 mol/l. K NН 3 = 1.76×10 – 5 1.88×10 - 3 0.53 × 10 - 11; pH = – log 0.53×10 – 11 = 11.3 5. C(H +) = [H + ] = 10 – pH At pH = 7, [H + ] = 10 – 7 There are various methods for determining pH: using indicators and ionomer devices. The value of pH for chemical reactions and biochemical processes in the body. Many reactions require a strictly defined pH value to proceed in a certain direction. Normally, in a healthy body, the reaction of the environment of most biological fluids is close to neutral. Blood – 7.4 Saliva – 6.6 Intestinal juice – 6.4 Bile – 6.9 Urine – 5.6 Gastric juice: a) at rest – 7.3 b) in a state of digestion – 1.5-2 Deviation of pH from the norm has diagnostic (definition of the disease) and prognostic (course of the disease) significance. Acidosis is a shift in pH to the acidic side, the pH decreases, the concentration of hydrogen ions increases. Alkalosis is a shift in pH to the alkaline region, the pH increases, and the concentration of hydrogen ions decreases. A temporary deviation of blood pH from the norm by tenths leads to serious disturbances in the body. Long-term deviations in blood pH can be fatal. Deviations in blood pH can be 6.8 - 8; changes outside this range in any direction are incompatible with life. Combined and isolated protolytic equilibria. Protolytic processes are reversible reactions. Protolytic equilibria are shifted towards the formation of weaker acids and bases. They can be considered as competition between bases of different strengths for the possession of a proton. They talk about isolated and combined equilibria. If several simultaneously existing equilibria are independent of each other, they are called isolated. A shift in equilibrium in one of them does not entail a change in the equilibrium position in the other. If a change in equilibrium in one of them leads to a change in equilibrium in the other, then we speak of combined (conjugate, competing) equilibria. The predominant process in systems with combined equilibrium is the one characterized by a larger value of the equilibrium constant. The second process will be predominant, because its equilibrium constant is greater than the equilibrium constant of the first process. The equilibrium in the second process is shifted to the right to a greater extent, because methylamine is a stronger base than ammonia, NH 4 + is a stronger acid than CH 3 NH 3 +. Conclusion: The stronger base suppresses the ionization of the weaker base. Therefore, when a small amount of hydrochloric acid is added to a mixture of ammonia and methylamine, it will be mainly the methylamine that undergoes protonation. And also: the strongest acid suppresses the ionization of weak acids. Thus, hydrochloric acid found in gastric juice suppresses the ionization of acetic acid (coming from food) or acetylsalicylic acid (medicinal substance). ______________________________________________________________
Thus, according to this theory An acid is any substance whose molecules (including ions) are capable of donating a proton, i.e. be a proton donor; A base is any substance whose molecules (including ions) are capable of attaching a proton, i.e. be a proton acceptor; An ampholyte is any substance that is both a donor and an acceptor of protons. This theory explains the acid-base properties of not only neutral molecules, but also ions. An acid, giving up a proton, turns into a base, which is the conjugate of this acid. The terms "acid" and "base" are relative concepts, since the same particles - molecules or ions - can exhibit both basic and acidic properties, depending on the partner. During protolytic equilibrium, acid-base pairs are formed. According to the proton theory, hydrolysis, ionization and neutralization reactions are not considered as a special phenomenon, but are considered to be the usual transfer of protons from an acid to a base. Particle A formed after the separation of a hydrogen ion | ||
| is called the conjugate base of a given acid, because it is capable of reattaching the H + ion to itself. | ||
| According to the protolytic theory, acids and bases can be of three types: neutral, anionic and cationic. The first ones are neutral molecules capable of donating or attaching the H + ion, for example: HCl, H 2 SO 4, HNO 3 (acids); NH 3, CH 3 –O–CH 3 (bases). Anionic bases and acids are negatively charged ions, for example: HSO 4 –, HPO 4 2–, HS – (acids); OH – , Cl – , NO 3 – (bases). In the role cationic bases and acids positively charged ions appear, for example: NH 4 +, H 3 O + (acids); H 2 N–NH 3 + , H 2 N–(CH 2) 2 –NH 3 + (bases). Many particles (both molecules and ions) have amphoteric properties, i.e. depending on the conditions, they can act both as an acid and as a base, for example: H 2 O, NH 3, HSO 4 -, H 2 N–NH 3 +, etc. These compounds are called amphiprotic or ampholytes. Although the Bronsted-Lowry theory is more perfect than the Arrhenius theory, it also has certain drawbacks and is not comprehensive. So, it is not applicable to many substances that exhibit the function of an acid, but do not contain H + ions in their composition, for example: BCl 3, AlCl 3, BF 3, FeCl 3, etc. | ||
where: K a – acidity constant; K p – equilibrium constant.
The acid there is stronger, the higher the acidity constant. pK a values are often used. The lower the pKa value, the stronger the acid.
pK a = -logK a
For example, pK a of phenol = 10, pK a of ethanol = 16. This means that phenol is six orders of magnitude (million times) a stronger acid than ethyl alcohol.
Basicity can be expressed in terms of pK b.
RKb = 14 - pKa
It is important to remember that the pKa of water is 15.7. All substances that have pKa greater than water are not able to exhibit acidic properties in aqueous solutions. Water, as a stronger acid, inhibits the dissociation of weaker acids. Since most organic compounds have acidic properties that are many times weaker than those of water, a polarographic approach to assessing their acidity has been developed (I.P. Beletskaya et al.). It allows you to evaluate acidity up to pK a = 50, although for very weak acids pK a values can only be estimated very approximately.
Qualitative assessment of acidity both in the series of substances with similar structures and for compounds of different classes is extremely important. The ability of an acid to donate a proton is related to the stability of the resulting anion. The more stable the resulting anion, the less its tendency to capture the proton back and turn into a neutral molecule. Several factors must be taken into account when assessing the relative stability of an anion.
The nature of an atom that donates a proton. The more easily an atom loses a proton, the higher its electronegativity and polarizability. Therefore, in the series of acids, the ability to dissociate decreases as follows:
S-h>O-H>-N-h>C-H
This series corresponds perfectly to the properties of atoms known from the periodic table.
The influence of the environment. If substances that are similar in structure are compared, the assessment is carried out by comparing the electron density on the atom that donated the proton. All structural factors that contribute to a decrease in charge stabilize the anion, and an increase in charge destabilize it. Thus, all acceptors increase acidity, all donors decrease it.
This occurs regardless of what effect of electron transfer (inductive or mesomeric) is responsible for the redistribution of electron density.
Solvation effect. Solvation (interaction with solvent molecules) increases the stability of the anion due to the redistribution of excess electron density between the anion and solvent molecules. In general, the pattern is as follows:
The more polar the solvent, the stronger the solvation;
The smaller the ion, the better it is solvated.
Basicity according to Brønsted is the ability of a substance to provide its pair of electrons for interaction with a proton. As a rule, these are substances containing nitrogen, oxygen and sulfur atoms in the molecule.
The weaker the basic center holds a pair of electrons, the higher the basicity. In a row
R3-N >R 2O>R 2S
basicity decreases. This sequence is easy to remember using the mnemonic rule “NOS”.
There is a relationship among Brønsted bases: anions are stronger bases than the corresponding neutral molecules. For example, the hydroxide anion (–OH) is a stronger base than water (H2O). When a base interacts with a proton, onium cations can be formed:
· R 3 O + - oxonium cation;
· NR 4 + - ammonium cation;
· R 3 S + - sulfonium cation.
Qualitative assessment of the basicity of substances with similar structures is carried out using the same logic as the assessment of acidity, but with the opposite sign.
Therefore, all acceptor substituents reduce their basicity, and all donor substituents increase their basicity.
Lewis acids and bases
Lewis bases are electron pair donors, just like Brønsted bases.
Lewis's definition for acids differs markedly from the usual one (according to Brønsted). A Lewis acid is any molecule or ion that has a vacant orbital that can be filled with an electron pair as a result of interaction. If, according to Brønsted, an acid is a proton donor, then according to Lewis, the proton itself (H +) is an acid, since its orbital is empty. There are a lot of Lewis acids: Na +, Mg 2+, SnCl 4, SbCl 5, AlCl 3, BF 3, FeBr 3, etc. Lewis theory allows many reactions to be described as acid-base interactions. For example:
Often, in reactions with Lewis acids, organic compounds that donor a pair of p-electrons participate as bases:
In organic chemistry the following is accepted:
· if the term “acid” is used, it means Brønsted acid;
· if the term “acid” is used in the Lewis sense, they say “Lewis acid”.
Lecture No. 5
hydrocarbons
Alkanes
· Homologous series, nomenclature, isomerism, alkyl radicals. Electronic structure of alkane molecules, sp 3 -hybridization, s-bond. Lengths of C-C and C-H bonds, bond angles, bond energies. Spatial isomerism of organic substances. Methods for depicting the spatial structure of molecules with sp 3 hybridized carbon atoms. Spectral characteristics of alkanes. Physical properties of alkanes and patterns of their changes in the homologous series.
Alkanes (saturated acyclic compounds, paraffins)
Alkanes are hydrocarbons with an open chain of atoms, corresponding to the formula C n H 2 n + 2, where the carbon atoms are interconnected only by σ-bonds.
The term “saturated” means that each carbon in the molecule of such a substance is bonded to the maximum possible number of atoms (four atoms).
The structure of methane is described in detail in lecture No. 2.
Isomerism, nomenclature
The first three members of the homologous series (methane, ethane and propane) exist as one structural isomer. Starting with butane, the number of isomers is growing rapidly: pentane has three isomers, and decane (C 10 H 22) already has 75.
Chapter 20. Quantitative description of chemical equilibrium
20.1. Law of mass action
You became acquainted with the law of mass action by studying the equilibrium of reversible chemical reactions (Chapter 9, § 5). Recall that at constant temperature for a reversible reaction
a A+ b B d D+ f Fthe law of mass action is expressed by the equation
You know that when applying the law of mass action, it is important to know in what state of aggregation the substances involved in the reaction are located. But not only this: the number and ratio of phases in a given chemical system is important. Based on the number of phases, reactions are divided into homophasic, And heterophasic. Among heterophasic ones there are solid phase reactions.
| Homophasic reaction– a chemical reaction in which all participants are in the same phase. |
This phase can be a mixture of gases (gas phase) or a liquid solution (liquid phase). In this case, all particles participating in the reaction (A, B, D and F) are able to perform chaotic movement independently of each other, and the reversible reaction occurs throughout the entire volume of the reaction system. Obviously, such particles can be either molecules of gaseous substances, or molecules or ions that form a liquid. Examples of reversible homophase reactions are reactions of ammonia synthesis, combustion of chlorine in hydrogen, the reaction between ammonia and hydrogen sulfide in an aqueous solution, etc.
If at least one substance participating in the reaction is in a different phase than the other substances, then the reversible reaction occurs only at the interface and is called a heterophase reaction.
| Heterophasic reaction– a chemical reaction whose participants are in different phases. |
Reversible heterophasic reactions include reactions involving gaseous and solid substances (for example, the decomposition of calcium carbonate), liquid and solid substances (for example, precipitation from a solution of barium sulfate or the reaction of zinc with hydrochloric acid), as well as gaseous and liquid substances.
A special case of heterophasic reactions are solid-phase reactions, that is, reactions in which all participants are solid substances.
In fact, equation (1) is valid for any reversible reaction, regardless of which of the listed groups it belongs to. But in a heterophase reaction, the equilibrium concentrations of substances in a more ordered phase are constant values and can be combined in an equilibrium constant (see Chapter 9, § 5).
So, for a heterophase reaction
a A g + b B cr d D g + f F crthe law of mass action will be expressed by the relation
The type of this relationship depends on which substances participating in the reaction are in a solid or liquid state (liquid if the remaining substances are gases).
In the expressions of the law of mass action (1) and (2), the formulas of molecules or ions in square brackets mean the equilibrium concentration of these particles in a gas or solution. In this case, the concentrations should not be high (no more than 0.1 mol/l), since these ratios are valid only for ideal gases and ideal solutions. (At high concentrations, the law of mass action remains valid, but instead of concentration it is necessary to use another physical quantity (the so-called activity), which takes into account the interactions between particles of a gas or solution. Activity is not proportional to concentration).
The law of mass action is applicable not only to reversible chemical reactions; many reversible physical processes are also subject to it, for example, interphase equilibria of individual substances during their transition from one state of aggregation to another. Thus, the reversible process of evaporation - condensation of water can be expressed by the equation
H 2 O f H 2 O g
For this process, we can write the equilibrium constant equation:
The resulting relationship confirms, in particular, the statement known to you from physics that air humidity depends on temperature and pressure.
20.2. Autoprotolysis constant (ion product)
Another application of the law of mass action known to you is the quantitative description of autoprotolysis (Chapter X § 5). Do you know that homophase equilibrium is observed in pure water?
2H 2 O H 3 O + + OH -
for a quantitative description of which we can use the law of mass action, the mathematical expression of which is autoprotolysis constant(ion product) of water
Autoprotolysis is characteristic not only of water, but also of many other liquids whose molecules are interconnected by hydrogen bonds, for example, ammonia, methanol and hydrogen fluoride:
| 2NH 3 NH 4 + + NH 2 - | K(NH 3) = 1.91. 10 –33 (at –50 o C); |
| 2CH 3 OH CH 3 OH 2 + + CH 3 O - | K(CH 3 OH) = 4.90. 10 –18 (at 25 o C); |
| 2HF H 2 F + + F - | K(HF) = 2.00. 10 –12 (at 0 o C). |
For these and many other substances, autoprotolysis constants are known, which are taken into account when choosing a solvent for certain chemical reactions.
The symbol is often used to denote the autoprotolysis constant K S.
The autoprotolysis constant does not depend on the theory within which autoprotolysis is considered. The values of the equilibrium constants, on the contrary, depend on the adopted model. Let us verify this by comparing the description of water autoprotolysis according to the protolytic theory (column on the left) and according to the outdated, but still widely used theory of electrolytic dissociation (column on the right):
According to the theory of electrolytic dissociation, it was assumed that water molecules partially dissociate (break up) into hydrogen ions and hydroxide ions. The theory explained neither the reasons nor the mechanism of this “decay.” The name "autoprotolysis constant" is usually used in the protolytic theory, and the "ion product" is used in the theory of electrolytic dissociation.
20.3. Acidity and basicity constants. pH value
The law of mass action is also used to quantitatively characterize the acid-base properties of various substances. In the protolytic theory, acidity and basicity constants are used for this, and in the theory of electrolytic dissociation - dissociation constants.
You already know how the protolytic theory explains the acid-base properties of chemical substances (Chapter XII § 4). Let's compare this approach with the approach of the theory of electrolytic dissociation using the example of a reversible homophase reaction with water of hydrocyanic acid HCN - a weak acid (on the left - according to the protolytic theory, on the right - according to the theory of electrolytic dissociation):
HCN + H 2 O H 3 O + + CN -
KK(HCN) = K C. == 4.93. 10–10 mol/l |
HCN H + + CN –
Equilibrium constant K C in this case is called dissociation constant(or ionization constant), denoted TO and is equal to the acidity constant in the protolytic theory. K = 4.93. 10–10 mol/l |
The degree of protolysis of a weak acid () in the theory of electrolytic dissociation is called degree of dissociation(unless this theory considers the substance to be an acid).
In the protolytic theory, to characterize a base, you can use its basicity constant, or you can get by with the acidity constant of the conjugate acid. In the theory of electrolytic dissociation, only substances dissociating in solution into cation and hydroxide ions were considered bases, therefore, for example, it was assumed that an ammonia solution contains “ammonium hydroxide”, and later – ammonia hydrate
NH 3 + H 2 O NH 4 + + OH -
K O (NH 3) = K C .
= |
NH3. H 2 O NH 4 + + OH –
Equilibrium constant K C and in this case is called the dissociation constant, denoted TO and is equal to the basicity constant. K = 1.74. 10–5 mol/l There is no concept of a conjugate acid in this theory. Ammonium ion is not considered an acid. The acidic environment in solutions of ammonium salts is explained by hydrolysis. |
1.74. 10 –5 mol/l
Even greater difficulties in the theory of electrolytic dissociation are caused by the description of the basic properties of other substances that do not contain hydroxyls, for example, amines (methylamine CH 3 NH 2, aniline C 6 H 5 NH 2, etc.).
To characterize the acidic and basic properties of solutions, another physical quantity is used - pH value(indicated by pH, read “peh”). Within the framework of the theory of electrolytic dissociation, the hydrogen index was determined as follows:
pH = –lg
A more accurate definition, taking into account the absence of hydrogen ions in the solution and the impossibility of taking logarithmic units of measurement:
pH = –lg()
It would be more correct to call this quantity the "oxonium" rather than the hydrogen index, but this name is not used.
Determined similarly to hydrogen hydroxide index(denoted pOH, read "pe oash").
pOH = -lg()
Curly brackets indicating the numerical value of a quantity in expressions for hydrogen and hydroxide indicators are very often not placed, forgetting that it is impossible to logarithm physical quantities.
Since the ionic product of water is a constant value not only in pure water, but also in dilute solutions of acids and bases, the hydrogen and hydroxide indicators are related:
K (H 2 O) \u003d \u003d 10 -14 mol 2 / l 2
lg() = lg() + lg() = -14
pH + pOH = 14
In pure water = = 10 –7 mol/l, therefore, pH = pOH = 7.
In an acid solution (in an acidic solution) there is an excess of oxonium ions, their concentration is greater than 10 –7 mol/l and, therefore, pH< 7.
In a base solution (alkaline solution), on the contrary, there is an excess of hydroxide ions, and, therefore, the concentration of oxonium ions is less than 10 –7 mol/l; in this case pH > 7.
20.4. Hydrolysis constant
Within the framework of the theory of electrolytic dissociation, reversible hydrolysis (hydrolysis of salts) is considered as a separate process, and cases of hydrolysis are distinguished
- salts of a strong base and a weak acid,
- salts of a weak base and a strong acid, as well as
- salts of a weak base and a weak acid.
Let us consider these cases in parallel within the framework of the protolytic theory and within the framework of the theory of electrolytic dissociation.
Salt of a strong base and a weak acid
As a first example, consider the hydrolysis of KNO 2, a salt of a strong base and a weak monobasic acid.
| K + , NO 2 - and H 2 O. NO 2 - is a weak base, and H 2 O is an ampholyte, therefore, a reversible reaction is possible NO 2 - + H 2 O HNO 2 + OH - , the equilibrium of which is described by the basicity constant of the nitrite ion and can be expressed in terms of the acidity constant of nitrous acid: K o (NO 2 -) = |
When this substance dissolves, it irreversibly dissociates into K + and NO 2 - ions: KNO 2 = K + + NO 2 - H 2 O H + + OH - With the simultaneous presence of H + and NO 2 - ions in the solution, a reversible reaction occurs H + + NO 2 - HNO 2 NO 2 - + H 2 O HNO 2 + OH - The equilibrium of the hydrolysis reaction is described by the hydrolysis constant ( K h) and can be expressed through the dissociation constant ( TO e) nitrous acid: K h = Kc .
= |
As you can see, in this case the hydrolysis constant is equal to the basicity constant of the base particle.
Despite the fact that reversible hydrolysis occurs only in solution, when water is removed it is completely “suppressed”, and, therefore, the products of this reaction cannot be obtained, within the framework of the theory of electrolytic dissociation the molecular equation of hydrolysis is written:
KNO 2 + H 2 O KOH + HNO 2
As another example, consider the hydrolysis of Na 2 CO 3 - a salt of a strong base and a weak dibasic acid. The line of reasoning here is completely similar. Within the framework of both theories, the ionic equation is obtained:
CO 3 2- + H 2 O HCO 3 - + OH -
Within the framework of the protolytic theory, it is called the equation of protolysis of the carbonate ion, and within the framework of the theory of electrolytic dissociation, it is called the ionic equation of hydrolysis of sodium carbonate.
Na 2 CO 3 + H 2 O NaHCO 3 + NaOH
Within the framework of TED, the basicity constant of the carbonate ion is called the hydrolysis constant and is expressed through the “second-stage dissociation constant of carbonic acid,” that is, through the acidity constant of the bicarbonate ion.
It should be noted that under these conditions, HCO 3 -, being a very weak base, practically does not react with water, since possible protolysis is suppressed by the presence of very strong base particles in the solution - hydroxide ions.
Salt of a weak base and a strong acid
Let's consider the hydrolysis of NH 4 Cl. Within the framework of TED, it is a salt of a weak one-acid base and a strong acid.
| The solution of this substance contains particles: NH 4 + , Cl - and H 2 O. NH 4 + is a weak acid, and H 2 O is an ampholyte, therefore, a reversible reaction is possible NH 4 + + H 2 O NH 3 + H 3 O + , the equilibrium of which is described by the acidity constant of the ammonium ion and can be expressed in terms of the basicity constant of ammonia: KK(NH4+) = |
When this substance dissolves, it irreversibly dissociates into NH 4 + and Cl - ions: NH 4 Cl = NH 4 + + Cl - Water is a weak electrolyte and dissociates reversibly: H 2 O H + + OH - NH 4 + + OH - NH 3 . H2O Adding the equations of these two reversible reactions and adding similar terms, we obtain the ionic equation of hydrolysis NH 4 + + H 2 O NH 3 . H2O+H+ The equilibrium of the hydrolysis reaction is described by the hydrolysis constant and can be expressed in terms of the dissociation constant of ammonia hydrate: K h = |
In this case, the hydrolysis constant is equal to the acidity constant of the ammonium ion. The dissociation constant of ammonia hydrate is equal to the basicity constant of ammonia.
Molecular equation of hydrolysis (in the framework of TED): NH 4 Cl + H 2 O NH 3 . H2O + HCl
Another example of a salt hydrolysis reaction of this type is the hydrolysis of ZnCl 2 .
| The solution of this substance contains particles: Zn 2+ aq, Cl - and H 2 O. Zinc ions are 2+ aquacations and are weak cationic acids, and H 2 O is an ampholyte, therefore a reversible reaction is possible 2= + H 2 O + + H 3 O + , the equilibrium of which is described by the acidity constant of the zinc aquacation and can be expressed through the basicity constant of the triaquahydroxozinc ion: K K ( 2+ ) = = | When this substance dissolves, it irreversibly dissociates into Zn 2+ and Cl - ions: ZnCl 2 = Zn 2+ + 2Cl - Water is a weak electrolyte and dissociates reversibly: H 2 O H + + OH - With the simultaneous presence of OH - and Zn 2+ ions in the solution, a reversible reaction occurs Zn 2+ + OH - ZnOH + Adding the equations of these two reversible reactions and adding similar terms, we obtain the ionic equation of hydrolysis Zn 2+ + H 2 O ZnOH + + H + The equilibrium of the hydrolysis reaction is described by the hydrolysis constant and can be expressed through the “second-stage dissociation constant of zinc hydroxide”: K h = |
The hydrolysis constant of this salt is equal to the acidity constant of the zinc aquacation, and the dissociation constant of zinc hydroxide in the second step is the basicity constant of the + ion.
The .+ ion is a weaker acid than the 2+ ion, so it practically does not react with water, since this reaction is suppressed due to the presence of oxonium ions in the solution. Within the framework of the TED, this statement sounds like this: “the hydrolysis of zinc chloride in the second stage practically does not occur.”
Molecular equation of hydrolysis (within TED):
ZnCl 2 + H 2 O Zn(OH)Cl + HCl.
Salt of a weak base and a weak acid
With the exception of ammonium salts, such salts are generally insoluble in water. Therefore, let us consider this type of reaction using ammonium cyanide NH 4 CN as an example.
| The solution of this substance contains particles: NH 4 + , CN - and H 2 O. NH 4 + is a weak acid, CN - is a weak base, and H 2 O is an ampholyte, therefore, the following reversible reactions are possible: NH 4 + + H 2 O NH 3 + H 3 O + , (1) CN - + H 2 O HCN + OH - , (2) NH 4 + + CN - NH 3 + HCN. (3) The latter reaction is preferable because, unlike the first two, it produces both a weak acid and a weak base. It is this reaction that predominantly occurs when ammonium cyanide is dissolved in water, but it is impossible to detect this by a change in the acidity of the solution. The slight alkalization of the solution is due to the fact that the second reaction is still somewhat more preferable than the first, since the acidity constant of hydrocyanic acid (HCN) is much less than the basicity constant of ammonia. The equilibrium in this system is characterized by the acidity constant of hydrocyanic acid, the basicity constant of ammonia and the equilibrium constant of the third reaction:
Let us express the equilibrium concentration of hydrocyanic acid from the first equation, and the equilibrium concentration of ammonia from the second equation and substitute these values into the third equation. As a result, we get
|
When this substance is dissolved, it irreversibly dissociates into NH 4 + and CN - ions: NH 4 CN = NH 4 + + CN - Water is a weak electrolyte and dissociates reversibly: H 2 O H + + OH - With the simultaneous presence of OH - and NH 4 + ions in the solution, a reversible reaction occurs NH 4 + + OH - NH 3 . H2O And with the simultaneous presence of H + and CN - ions, another reversible reaction occurs Adding the equations of these three reversible reactions and adding similar terms, we obtain the ionic equation of hydrolysis NH 4 + + CN - + H 2 O NH 3. H2O+HCN The form of the hydrolysis constant in this case is as follows: K h = And it can be expressed through the dissociation constant of ammonia hydrate and the dissociation constant of hydrocyanic acid: K h = |
Molecular equation of hydrolysis (within the framework of TED):
NH 4 CN + H 2 O NH 3 . H2O+HCN
20.5. Solvation constant (solubility product)
The process of chemical dissolution of a solid in water (and not only in water) can be expressed by an equation. For example, in the case of dissolving sodium chloride:
NaCl cr + ( n+m)H 2 O = + + -
This equation clearly shows that the most important reason for the dissolution of sodium chloride is the hydration of Na + and Cl - ions.
In a saturated solution, a heterophase equilibrium is established:
NaCl cr + ( n+m)H 2 O + + - ,
which obeys the law of mass action. But, since the solubility of sodium chloride is quite significant, the expression for the equilibrium constant in this case can only be written using the activities of ions, which are not always known.
In the case of equilibrium in a solution of a slightly soluble (or practically insoluble substance), the expression for the equilibrium constant in a saturated solution can be written using equilibrium concentrations. For example, for equilibrium in a saturated solution of silver chloride
AgCl cr + ( n+m)H 2 O + + -
Since the equilibrium concentration of water in a dilute solution is almost constant, we can write
K G (AgCl) = K C . n+m = .
The same simplified
K G (AgCl) = or K G (AgCl) =
The resulting value ( K D) is named hydration constants(in the case of any, and not just aqueous solutions - solvation constants).
Within the framework of the theory of electrolytic dissociation, the equilibrium in an AgCl solution is written as follows:
AgCl cr Ag + + Cl –
The corresponding constant is called solubility product and is denoted by the letters PR.
PR(AgCl) =
Depending on the ratio of cations and anions in the formula unit, the expression for the solvation constant (solubility product) can be different, for example:
The values of hydration constants (solubility products) of some poorly soluble substances are given in Appendix 15.
Knowing the solubility product, it is easy to calculate the concentration of a substance in a saturated solution. Examples:
1. BaSO 4cr Ba 2+ + SO 4 2-
PR(BaSO 4) = 1.8. 10–10 mol 2 /l 2.
c(BaSO 4) = = = = 
= 1.34. 10 –5 mol/l.
2. Ca(OH) 2cr Ca 2+ + 2OH -
PR = 2 = 6.3. 10 –6 mol 3 /l 3 .
2 PR = (2) 2 = 4 3
c = = 
= 
= 1.16. 10 –2 mol/l.
If, during a chemical reaction, ions appear in the solution that are part of a poorly soluble substance, then, knowing the product of the solubility of this substance, it is easy to determine whether it will precipitate.
Examples:
1. Will a copper hydroxide precipitate form when 100 ml of a 0.01 M solution of calcium hydroxide is added to an equal volume of 0.001 M solution of copper sulfate?
Cu 2+ + 2OH - Cu(OH) 2
A copper hydroxide precipitate is formed if the product of the concentrations of Cu 2+ and OH - ions is greater than the product of the solubility of this poorly soluble hydroxide. After merging solutions of equal volume, the total volume of the solution will become twice as large as the volume of each of the original solutions, therefore the concentration of each of the reacting substances (before the start of the reaction) will decrease by half. Concentration of copper ions in the resulting solution
c(Cu 2+) = (0.001 mol/l) : 2 = 0.0005 mol/l.
Hydroxide ion concentration –
c(OH -) = (2 . 0.01 mol/l) : 2 = 0.01 mol/l.
Copper hydroxide solubility product
PR = 2 = 5.6. 10–20 mol 3 /l 3.
c(Cu 2+) . ( c(OH -)) 2 = 0.0005 mol/l. (0.01 mol/l) 2 = 5. 10 –8 mol 3 /l 3.
The product of concentrations is greater than the product of solubility, therefore, a precipitate will form.
2. Will a silver sulfate precipitate form when equal volumes of 0.02 M sodium sulfate solution and 0.04 M silver nitrate solution are combined?
2Ag + + SO 4 2- Ag 2 SO 4
Concentration of silver ions in the resulting solution
c(Ag +) = (0.04 mol/l) : 2 = 0.02 mol/l.
Concentration of sulfate ions in the resulting solution
c(SO 4 2-) = (0.02 mol/l) : 2 = 0.01 mol/l.
Silver sulfate solubility product
PR(Ag 2 SO 4) = 2. = 1.2. 10 –5 mol 3 /l 3.
Product of ion concentrations in solution
{c(Ag +)) 2. c(SO 4 2-) = (0.02 mol/l) 2. 0.01 mol/l = 4. 10 –6 mol 3 /l 3.
The product of concentrations is less than the product of solubility, therefore, no precipitate is formed.
20.6. Degree of conversion (degree of protolysis, degree of dissociation, degree of hydrolysis)
The efficiency of a reaction is usually assessed by calculating the yield of the reaction product (section 5.11). At the same time, the efficiency of the reaction can also be assessed by determining what part of the most important (usually the most expensive) substance was converted into the target reaction product, for example, what part of SO 2 was converted into SO 3 during the production of sulfuric acid, that is, find degree of conversion original substance.
Cl 2 + 2KOH = KCl + KClO + H 2 O
chlorine (reagent) is converted equally into potassium chloride and potassium hypochlorite. In this reaction, even with a 100% yield of KClO, the degree of conversion of chlorine into it is 50%.
The quantity you know - the degree of protolysis (section 12.4) - is a special case of the degree of conversion:
Within the framework of TED, similar quantities are called degree of dissociation acids or bases (also designated as the degree of protolysis). The degree of dissociation is related to the dissociation constant according to Ostwald's dilution law.
Within the framework of the same theory, the hydrolysis equilibrium is characterized by degree of hydrolysis (h), and the following expressions are used that relate it to the initial concentration of the substance ( With) and dissociation constants of weak acids (K HA) and weak bases formed during hydrolysis ( K MOH):
The first expression is valid for the hydrolysis of a salt of a weak acid, the second - salts of a weak base, and the third - salts of a weak acid and a weak base. All these expressions can only be used for dilute solutions with a degree of hydrolysis of no more than 0.05 (5%).
Law of mass action, homophase reactions, heterophase reactions, solid-phase reactions, Autoprotolysis constant (ionic product), dissociation (ionization) constant, degree of dissociation (ionization), hydrogen index, hydroxide index, hydrolysis constant, solvation constant (solubility product), degree of conversion .
- List the factors that shift chemical equilibrium and change the equilibrium constant.
- What factors allow you to shift the chemical equilibrium without changing the equilibrium constant?
- It is necessary to prepare a solution containing 0.5 mol NaCl, 0.16 mol KCl and 0.24 mol K 2 SO 4 in 1 liter. How to do this with only sodium chloride, potassium chloride and sodium sulfate at your disposal?
- Determine the degree of protolysis of acetic, hydrocyanic and nitric acids in decimolar, centimolar and millimolar solutions.
- The degree of protolysis of butyric acid in a 0.2 M solution is 0.866%. Determine the acidity constant of this substance.
- At what concentration of solution will the degree of protolysis of nitrous acid be equal to 0.2?
- How much water must be added to 300 ml of 0.2 M acetic acid solution so that the degree of acid protolysis doubles?
- Determine the degree of protolysis of hypobromous acid if its solution has pH = 6. What is the concentration of acid in this solution?
- The pH value of the solution is 3. For this, what should be the concentration of a) nitric, b) acetic acid?
- How should the concentration of a) oxonium ions, b) hydroxide ions in a solution be changed so that the pH value of the solution increases by one?
- How many oxonium ions are contained in 1 ml of solution at pH = 12?
- How will the pH value of water change if 0.4 g of NaOH is added to 10 liters?
- Calculate the concentrations of oxonium and hydroxide ions, as well as the values of the hydrogen and hydroxide indicators in the following aqueous solutions: a) 0.01 M HCl solution; b) 0.01 M solution of CH 3 COOH; c) 0.001 M NaOH solution; d) 0.001 M NH 3 solution.
- Using the values of solubility products given in the appendix, determine the concentration and mass fraction of dissolved substances in a solution of a) silver chloride, b) calcium sulfate, c) aluminum phosphate.
- Determine the volume of water required to dissolve barium sulfate weighing 1 g at 25 o C.
- What is the mass of silver present in the form of ions in 1 liter of silver bromide solution saturated at 25 o C?
- What volume of silver sulfide solution saturated at 25 o C contains 1 mg of dissolved substance?
- Will a precipitate form if an equal volume of 0.4 M KCl solution is added to a 0.05 M solution of Pb(NO 3) 2?
- Determine whether a precipitate will form after pouring 5 ml of a 0.004 M CdCl 2 solution and 15 ml of a 0.003 M KOH solution.
- The following substances are available at your disposal: NH 3, KHS, Fe, Al(OH) 3, CaO, NaNO 3, CaCO 3, N 2 O 5, LiOH, Na 2 SO 4. 10H 2 O, Mg(OH)Cl, Na, Ca(NO 2) 2. 4H 2 O, ZnO, NaI. 2H 2 O, CO 2, N 2, Ba(OH) 2. 8H 2 O, AgNO 3. For each of these substances, answer the following questions on a separate card:
1) What is the type of structure of this substance under normal conditions (molecular or non-molecular)?
2) In what state of aggregation is this substance at room temperature?
3) What type of crystals does this substance form?
4) Describe the chemical bond in this substance.
5) What class does this substance belong to according to the traditional classification?
6) How does this substance interact with water? If it dissolves or reacts, give the chemical equation. Can we reverse this process? If we do, then under what conditions? What physical quantities can characterize the state of equilibrium in this process? If a substance is soluble, how can its solubility be increased?
7) Is it possible to react this substance with hydrochloric acid? If possible, then under what conditions? Give the reaction equation. Why does this reaction occur? Is it reversible? If reversible, then under what conditions? How will the yield in this reaction increase? What will change if dry hydrogen chloride is used instead of hydrochloric acid? Give the corresponding reaction equation.
8) Is it possible to react this substance with a solution of sodium hydroxide? If possible, then under what conditions? Give the reaction equation. Why does this reaction occur? Is it reversible? If reversible, then under what conditions? How will the yield in this reaction increase? What changes if you use dry NaOH instead of sodium hydroxide solution? Give the corresponding reaction equation.
9) Give all the methods known to you for obtaining this substance.
10) Give all the names of this substance known to you.
When answering these questions, you can use any reference literature.
