Acidity constant of hydrochloric acid. Acidity and basicity in water

Self-ionization of water

Water, even after repeated distillation, retains the ability to conduct electricity. This ability of water is due to its self-ionization.

$2H_2O ↔ H_3O^+ + OH^-$

The thermodynamic equilibrium constant has the form:

Picture 1.

where $a_X^(rel)=\frac(a_X^(equal))(a_X^0)$ is the relative activity of the particle $X$ in the equilibrium system;

$aX^(equal)$ - absolute activity of the particle $X$ in the equilibrium system;

$(a_x)^0$ - absolute activity $X$ in the thermodynamic state of the system.

The relative activity of water at equilibrium is practically equal to unity, since the degree of reaction is very small (if theoretically non-ionized water is taken as the standard state.

The activity coefficients of $OH^-$ and $H_3O^+$ ions will be close to unity in pure water. The equilibrium of the reaction is strongly shifted to the left. The relative activities of $OH^-$ and $H_3O^+$ are practically equal to their molar concentrations. Where

$(K_a)^0 \sim K_(auto) = $

where $ and $ are molar concentrations;

$K_(auto)$ - water autopropolis constant equal to $1.00\cdot 10^(-14) \ mol^2/l^2$ at $25^\circ \ C.$

In pure water, the concentrations of $ and $ will be equal, so

$==\sqrt(10^(-14))=10^(-7)$ for $25^\circ \ C.$

For ease of calculation, the concentration is indicated as a negative logarithm, denoted as $pH$:

$pH=-lg $

$pH$ values ​​for pure water are $7$, in acidic solutions $pH $7$.

Acid dissociation and acidity constant

For the acid $AH$, the dissociation can be expressed by the equation:

$AH + H_2O ↔ A^- + H_3O^+$

In a state of equilibrium, the relative density of water changes insignificantly when passing from one acid to another, and with infinite dilution it approaches zero. Therefore, the thermodynamic acidity constant $K_a^0$ ($AH$) is used.

The ratio of activity coefficients is the same for all acids and is equal to one if the processes proceed in dilute solutions.

Then, in a dilute aqueous solution, the acidity constant $Ka (AH$) is used as a measure of acid strength, which can be determined by the formula:

$Ka (AH)=\frac()()$

The formula displays the molar concentration of particles at a fixed temperature $(25^\circ \ C)$ in the equilibrium state.

The higher the acidity constant, the higher the degree of dissociation, the stronger the acid. For calculations and characteristics of acidity, the negative logarithm of the acidity constant $pKa$ is used.

$pKa (AH)= -lgKa (AH)$

The larger the value of the acidity constant, the weaker the acid.

The value of the acidity constant is equal to the $pH$ value of the solution at which the acid will be ionized by half:

$pKa (AH) = pH - lg \frac()()$

The value characterizing the acidity of water molecules in an aqueous solution is:

$Ka=\frac()()=\frac(Ka_(auto))()=\frac(10^(-14))(55.5)$

Thus, at a temperature of $25^\circ C$, $pKa (H_2O) = 15.7$. This value characterizes the acidity of water molecules in solution.

For the hydroxonium ion $pKa (H_3O^+) = pK_(auto) - pKa = 14-15.7 = -1.7.$

The $pKa$ values ​​are tabular data. However, for acids with $pKa 0$ the table data will be inaccurate.

It is possible to determine the acidity constants in water by directly measuring the concentrations of $A^-$ and $AH$ only when acid dissociation occurs at least to some extent, even barely noticeable.

If the acid is very weak, which practically does not dissociate, then the concentration of $A^-$ cannot be accurately measured. If, on the contrary, the acid is so strong that it dissociates almost completely, then it is impossible to measure the concentration of $AH$. In this case, indirect methods for determining acidity will be used.

Base ionization constant

To express the dissociation constant of a base in water, we use the equation:

$B + H_2O ↔ BH^+ + OH^-$

The basicity constant is:

$Kb=\frac()([B])$

Recently, basicity constants are practically not used in calculations, since the acidity constant of the conjugate acid can be used to obtain all the necessary information about the base $BH^+.$

$BH^+ + H_2O ↔ B + H_3O^+$

$Ka (BH^+) = \frac([B])()$

The acidity constant of an acid will be a measure of strength:

• $AH$ or $BH^+$ as proton donors;
• $A^-$ or $B$ as proton acceptors;
• a strong acid $AH$ or $BH^+$ corresponds to a weak conjugate base $A^-$ or $B$, and then $pKa$ is small or negative;
• a strong base $A^-$ or $B$ corresponds to a weak acid $AH$ or $BH^+$ and the acidity constant will be positive

It is possible to directly measure the strength of acids or bases only in a narrow range of $pKa (BH^+).$ Outside the interval, basicity will be determined by indirect methods. $pka (BH^+)$ values ​​outside the range $-2$ to $17$ will be inaccurate.

Correlation between structure and strength of acids

The relative strength of acids can be predicted based on the nature of the central atom and the structure of the acid molecule.

The strength of the oxygen-free acid $HX$ and $H_2X$ (where $X$ is halogen) is the higher, the weaker the bond $X-H$, that is, the greater the radius of the $X$ atom.

In the series $HF - HCl - HBr - HI$ and $H_2S - H_2Se - H_2Te$, the strength of acids increases.

For oxygen-containing acids, the stronger the value of m in the compound $E(OH)nOm$, the stronger the acid.

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; 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, donating a proton, turns into a base, which is conjugate to this acid. The concepts of "acid" and "base" are relative concepts, since the same particles - molecules or ions - can exhibit both basic and acidic properties, depending on the partner.

In 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 the usual transition of protons from acid to base.

Particle A formed after the separation of the hydrogen ion

pH value

Water as a weak electrolyte undergoes ionization to a small extent:

H 2 O ↔ H + + OH -.

Ions in aqueous solution undergo hydration (aq.)

Water is characterized by protolytic amphotericity. The self-ionization reaction (autoprotolysis) of water, during which a proton from one water molecule (acid) passes to another water molecule (base), is described by the equation:

H 2 O + H 2 O ↔ H 3 O + + OH -.

The equilibrium constant of water autoprotolysis is equal to:

The law of mass action applies to the ionization constant:

where a is activity.

For brevity, instead of H 3 O + in acid-base equilibrium, we write

Since water is in solution in large excess and undergoes ionization to a small extent, it can be noted that its concentration is constant and equal to 55.6 mol (1000 g: 18 g / mol \u003d 56 mol) per liter of water.

Therefore, the product of K and (H 2 O) and the concentration of water is 1.8 10 -16 mol / l 55.6 mol / l \u003d 10 -14 mol 2 / l 2. Thus, \u003d 10 -14 (at 25 ° C) is a constant value, denoted Kw and called constant of water autoprotolysis. Sometimes they use an outdated name - the ionic product of water.

Solutions in which the concentration of hydrogen ions and hydroxide ions are the same are called neutral solutions = = = 10 -7 mol / l. In acidic solutions > , > 10 -7 mol / l, and in alkaline > , > 10 -7 mol / l.

For simplicity, the pH value is taken as the basis - the decimal logarithm of the concentration of hydrogen ions, taken with the opposite sign: pH \u003d -lg.

Interesting facts:

Violation of the state of isohydria ( pH constancy) observed in cardiovascular diseases, with ischemia, diabetes mellitus (acidosis develops). Acid-base balance is maintained by breathing, urination, sweating. These systems work slowly, and the immediate neutralization of acidic and alkaline metabolic products is carried out by the buffer systems of the body. The state of isohydria is provided by the joint action of a number of physicochemical and physiological mechanisms. The buffering action is provided by combining several protolytic equilibria.

The strength of acids is determined by their ability to donate a proton. The measure of this ability is acidity constant (Ka).

The larger the acidity constant, the stronger the acid. For example, acetic acid is stronger than hydrocyanic acid, since Ka (CH 3 COOH) \u003d 1.74 10 -5, Ka (HCN) \u003d 1 10 -9. For the convenience of calculations and recording, they often use not the constants themselves, but their negative decimal logarithms: pKa = -lgKa. The pKa value is called acid strength. The larger the pKa value, the weaker the acid.

Strong acids almost completely donate their proton to water molecules, so the acid present in the solution is actually the hydronium ion.

In this regard, when calculating the pH of a solution of a strong monobasic acid, the concentration of protons is equated to the concentration of the acid

c(H 3 O +) = c(HB).

In solutions of weak acids, the concentration of hydronium ions is much lower than the concentration of the acid. It is calculated based on

both parts of this equation gives a formula for calculating the pH of solutions of weak acids: pH = 0.5 (pKa - lg c(HB)).

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 the concentration of water ions × is ionic product of water.

It is equal to: × = K equal. × 2 = 1 × 10 - 14

Therefore: K H 2O \u003d × \u003d 10 - 14 or simplified K H 2O \u003d × \u003d 10 - 14

K H 2 O is the ionic product of water, the constant of autoprotolysis of water, or simply the constant of water. K H 2 O depends on temperature. As t°C rises, it increases.

In chemically pure water = = = 1×10 – 7 . This is a neutral environment.

There may be > in the solution - the medium is acidic or< – среда щелочная

= ; =

pH value

To quantify 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 = – lg ⇒ = 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.

pOH \u003d - lg [OH -] ⇒ [OH -] \u003d 10 - pOH

pH + pOH = 14 Þ pH = 14 - pOH and pOH = 14 - pH

Formulas for calculating pH for solutions of acids and bases.

pH = – lg

1. Strong acids: \u003d C (1 / z acids)

Calculate the pH of an HCl solution with С(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: [OH - ] \u003d C (1 / z bases)

Calculate the pH of the NaOH solution under the same conditions.

C(NaOH) = 0.1 mol/l; = = 10 – 13; pH \u003d - lg 10 - 13 \u003d 13

3. Weak acids

Calculate the pH of a solution of acetic acid with a molar concentration of 0.5 mol/l. To CH 3COOH \u003d 1.8 × 10 - 5.

3×10 - 3

pH \u003d - lg 3 × 10 - 3 \u003d 2.5

4. Weak bases

Calculate the pH of an ammonia solution with a molar concentration of 0.2 mol/L.

K NH 3 \u003d 1.76 × 10 - 5

1.88×10 - 3

0.53 × 10 - 11; pH \u003d - lg 0.53 × 10 - 11 \u003d 11.3

5. C (H +) \u003d [H +] \u003d 10 - pH

At pH = 7, [H + ] = 10 - 7

There are various methods for determining pH: using indicators and ionomer devices.

pH value for chemical reactions and biochemical processes of the body.

Many reactions in order to proceed in a certain direction require a strictly defined pH value of the medium.

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 the state of digestion - 1.5-2

The deviation of pH from the norm has a diagnostic (determination of the disease) and prognostic (the course of the disease) value.

Acidosis is a shift in pH to the acid side, the pH decreases, the concentration of hydrogen ions increases.

Alkalosis - pH shift to the alkaline region, pH increases, the concentration of hydrogen ions decreases.

A temporary deviation of blood pH from the norm by tenths leads to serious disturbances in the body. Prolonged fluctuations in blood pH can be fatal. Deviations in blood pH can be 6.8 - 8, changes outside this interval in any direction are incompatible with life.

Combined and isolated protolytic equilibria.

Protolytic processes are reversible reactions. Protolytic equilibria are biased towards the formation of weaker acids and bases. They can be viewed as competition between bases of different strength 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 position of equilibrium 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 that is 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, methylamine will mainly undergo protonation.

And also: the strongest acid suppresses the ionization of weak acids. So, hydrochloric acid, found in gastric juice, inhibits the ionization of acetic acid (comes with food) or acetylsalicylic acid (drug).

______________________________________________________________

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

where all activities are equilibrium. Let's represent this ratio in the form:

Denote, as in the previous case, the product of two constants To and a (H 2 O) through (H 2 O) \u003d 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 To a apply strength indicator (or just an indicator):

RK a =- lg To a .

The more To a(i.e., the smaller 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 ions [Н 3 О + ] and [В - ] will be equal to [Н 3 О + ] = [В - ] = αс a , a equilibrium acid concentration [HB] = 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 reciprocal V- dilution (dilution), expressed in l / mol, V=1/With a , then the formula for To a will look like:

This ratio, as well as the expression

describe dilution law (or dilution law) of Ostwald for a weak binary electrolyte. For 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 BUT 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 RO 4 (P = 3,

t= 1), which decays in total into ions according to the scheme

.

For a binary electrolyte, the relation turns into (4.11). For a1 we have:

Let us find the equilibrium value of the pH of a solution of a monobasic acid HB. 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 multibasic 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 BO 3 + H 2 O \u003d H 3 O + +, To 1 =

H 2 O \u003d H 3 O + +, To 2 =

H 2 O \u003d 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=

is the total 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 \u003d 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\u003d const and [H 2 O] \u003d const.

Let's call the quantity K b , equal, therefore,

K b = , (4.13)

dissociation (ionization) constant of a weak single acid baseorjust a basicity constant this base, and the value

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 - rOH = 14 - rOH.

Let us determine the value of pOH = [OH - ]. Obviously

[OH -] = =

Using indicators pON = [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 \u003d 14 - pOH \u003d 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)

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 the relation (4.15):

pH = 14 - 0.5 (p To b+ p) =

A similar consideration can be carried out for any weak polyacid grounds. True, in this case more cumbersome expressions are obtained.

Weak polyacid bases, like weak polybasic acids, dissociate in steps, and each dissociation step also has its own stepwise dissociation constant of the base - the stepwise basicity constant.

So, 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, accepts it from a water molecule:

In this case, the stepwise basicity constants can be represented as:

With such a record of these equilibria, it is assumed that the proton from the water molecule passes to the hydroxyl group with the formation of the 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- full dissociation constant according to the scheme

or according to another scheme

which ultimately leads to the same result.

Another example is the organic base ethylenediamine undergoing ionization in aqueous solution in two steps. First stage:

Second step:

Work -

total dissociation constant. It matches the balance

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 smaller than the dissociation constant of the previous step.

In table. 4.2 shows the numerical values ​​of the constants of acidity and basicity 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 stage,

To 2 - dissociation constant for the second stage, etc.

Dissociation constants of weak acids
Acid	To a	R To a=-lg To a
nitrogenous Aminoacetic benzoic Boric (orthoboric) Tetrabornaya Chapter 20 20.1. Law of acting masses You got acquainted with the law of mass action by studying the equilibrium of reversible chemical reactions (Chapter 9 § 5). Recall that at a constant temperature for a reversible reaction a A+ b B d D+ f F the 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 participating in the reaction are. But not only that: the number and ratio of phases in a given chemical system is important. According to the number of phases, the reactions are divided into homophasic, and heterophase. Among the heterophasic ones, solid phase reactions. Homophasic reaction A chemical reaction in which all participants are in the same phase. Such a 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) have the ability to perform chaotic motion independently of each other, and the reversible reaction proceeds 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 the reactions of ammonia synthesis, the 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 rest of the substances, then the reversible reaction proceeds only at the interface and is called a heterophase reaction. heterophasic reaction- a chemical reaction, the participants of which 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 barium sulfate solution or the reaction of zinc with hydrochloric acid), as well as gaseous and liquid substances. A special case of heterophase reactions are solid-phase reactions, that is, reactions in which all participants are solids. 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 constants 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 r+ f F cr the law of mass action will be expressed by the relation The type of this ratio depends on which substances participating in the reaction are in a solid or liquid state (liquid, if the rest of the 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 large (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, one has to use another physical quantity (the so-called activity), which takes into account interactions between gas particles or solutions. Activity is not proportional to concentration). The law of mass action is applicable not only for reversible chemical reactions, but many reversible physical processes also obey it, for example, the interfacial equilibrium of individual substances during their transition from one state of aggregation to another. So, 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 equation of the equilibrium constant: The resulting ratio confirms, in particular, the assertion known to you from physics that air humidity depends on temperature and pressure. 20.2. Autoprotolysis constant (ionic 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 pure water is in homophase equilibrium? 2H 2 OH 3 O + + OH - for a quantitative description of which you can use the law of mass action, the mathematical expression of which is autoprotolysis constant(ion product) of water Autoprotolysis is characteristic not only for water, but also for many other liquids, the molecules of which are interconnected by hydrogen bonds, for example, for ammonia, methanol and hydrogen fluoride: 2NH 3 NH 4 + + NH 2 - K(NH 3) = 1.91. 10 –33 (at –50 o С); 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 various chemical reactions. The symbol often used to denote the autoprotolysis constant is K S. The autoprotolysis constant does not depend on the theory in which autoprotolysis is considered. The values ​​of the equilibrium constants, on the contrary, depend on the accepted model. We will verify this by comparing the description of water autoprotolysis according to the protolytic theory (left column) and according to the outdated, but still widely used theory of electrolytic dissociation (right column): According to the theory of electrolytic dissociation, it was assumed that water molecules partially dissociate (decompose) into hydrogen ions and hydroxide ions. The theory did not explain either the reasons or the mechanism of this "disintegration". The name "autoprotolysis constant" is usually used in the protolytic theory, and "ionic product" in the theory of electrolytic dissociation. 20.3. Acidity and basicity constants. Hydrogen indicator The law of mass action is also used to quantify 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. How the protolytic theory explains the acid-base properties of chemicals, you already know (ch. 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 - K K(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(if only this theory considers the given substance as an acid). In the protolytic theory, to characterize the 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 ammonia solution contains "ammonium hydroxide", and later - ammonia hydrate NH 3 + H 2 O NH 4 + + OH - K O (NH 3) \u003d K C . = 1.74. 10–5 mol/l 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. The ammonium ion is not considered an acid. The acidic environment in solutions of ammonium salts is explained by hydrolysis. Even more difficult in the theory of electrolytic dissociation is 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(denoted by pH, read "ph"). In the framework of the theory of electrolytic dissociation, the hydrogen index was determined as follows: pH = –lg A more precise definition, taking into account the absence of hydrogen ions in the solution and the impossibility of taking logarithms of units of measurement: pH = –lg() It would be more correct to call this value "oxonium", and not the pH value, but this name is not used. It is defined similarly to hydrogen hydroxide index(denoted by pOH, read "pe oash"). pOH = -lg() Curly brackets denoting the numerical value of a quantity in expressions for the hydrogen and hydroxide indices are very often not put, forgetting that it is impossible to take the logarithm of 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 indices are interconnected: 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, consequently, 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, while cases of hydrolysis are distinguished salts of a strong base and a weak acid salts of a weak base and a strong acid, and 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 -, whose equilibrium 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 -) \u003d When this substance is dissolved, 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 in terms of 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, it is completely "suppressed" when water is removed, and, therefore, the products of this reaction cannot be obtained, within the framework of the theory of electrolytic dissociation, the molecular hydrolysis equation is also 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 exactly the same. Within the framework of both theories, an ionic equation is obtained: CO 3 2- + H 2 O HCO 3 - + OH - In the framework of the protolytic theory, it is called the carbonate ion protolysis equation, and in the framework of the electrolytic dissociation theory, it is called the ionic equation of sodium carbonate hydrolysis. Na 2 CO 3 + H 2 O NaHCO 3 + NaOH The basicity constant of the carbonate ion in the framework of TED is called the hydrolysis constant and is expressed through the "dissociation constant of carbonic acid in the second stage", that is, through the acidity constant of the hydrocarbonate 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, hydroxide ions, in the solution. Salt of a weak base and a strong acid Consider the hydrolysis of NH 4 Cl. Within the framework of TED, it is a salt of a weak monoacid base and a strong acid. In the solution of this substance there are 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 +, whose equilibrium is described by the acidity constant of the ammonium ion and can be expressed in terms of the basicity constant of ammonia: K K (NH 4 +) \u003d When this substance is dissolved, it irreversibly dissociates into NH 4 + and Cl - ions: NH 4 Cl \u003d NH 4 + + Cl - Water is a weak electrolyte and reversibly dissociates: H 2 O H + + OH - NH 4 + + OH - NH 3. H2O Adding the equations of these two reversible reactions and bringing like terms, we obtain the ionic hydrolysis equation 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 (within the framework of TED): NH 4 Cl + H 2 O NH 3. H2O + HCl Another example of a hydrolysis reaction of salts of this type is the hydrolysis of ZnCl 2 . In a solution of this substance there are particles: Zn2+ aq, Cl - and H 2 O. Zinc ions are aquacations 2+ 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 + , whose equilibrium is described by the acidity constant of the zinc aquacation and can be expressed in terms of the basicity constant of the triaquahydroxozinc ion: K K ( 2+ ) = = When this substance is dissolved, it irreversibly dissociates into Zn 2+ and Cl - ions: ZnCl 2 \u003d Zn 2+ + 2Cl - Water is a weak electrolyte and reversibly dissociates: 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 bringing like terms, we obtain the ionic hydrolysis equation Zn 2+ + H 2 O ZnOH + + H + The equilibrium of the hydrolysis reaction is described by the hydrolysis constant and can be expressed in terms of the "dissociation constant of zinc hydroxide in the second stage": 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 equal to the basicity constant of the + ion. The .+ ion is a weaker acid than the 2+ ion, therefore 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 TED, this statement sounds like this: "the hydrolysis of zinc chloride in the second stage practically does not go" . Molecular equation of hydrolysis (within the framework of 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 reactions using ammonium cyanide NH 4 CN as an example. In the solution of this substance there are 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, such 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 last reaction is preferable, because in it, unlike the first two, both a weak acid and a weak base are formed. It is this reaction that predominantly proceeds when ammonium cyanide is dissolved in water, but it is impossible to detect this by changing the acidity of the solution. A slight alkalinization 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: We express from the first equation the equilibrium concentration of hydrocyanic acid, and from the second equation - the equilibrium concentration of ammonia and substitute these quantities 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 \u003d NH 4 + + CN - Water is a weak electrolyte and reversibly dissociates: 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 proceeds Adding the equations of these three reversible reactions and bringing like terms, we obtain the ionic hydrolysis equation 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 in terms of 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 explicitly 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 the ions, which are far from always known. In the case of equilibrium in a solution of a poorly 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). In 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) \u003d \u003d 1.8. 10–10 mol 2 /l 2. c(BaSO4) = = = = = 1.34. 10–5 mol/l. 2. Ca(OH) 2cr Ca 2+ + 2OH - PR \u003d 2 \u003d 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 solubility product of this substance, it is easy to determine whether it will precipitate. Examples: 1. Will copper hydroxide precipitate when 100 ml of 0.01 M calcium hydroxide solution is added to an equal volume of 0.001 M copper sulfate solution? Cu 2+ + 2OH - Cu (OH) 2 A precipitate of copper hydroxide is formed if the product of the concentrations of Cu 2+ and OH - ions is greater than the product of the solubility of this sparingly soluble hydroxide. After pouring solutions of equal volume, the total volume of the solution will become twice as large as the volume of each of the initial solutions, therefore, the concentration of each of the reacting substances (before the start of the reaction) will be halved. The concentration in the resulting solution of copper ions c(Cu 2+) \u003d (0.001 mol / l): 2 \u003d 0.0005 mol / l. The concentration of hydroxide ions - c (OH -) \u003d (2. 0.01 mol / l): 2 \u003d 0.01 mol / l. Solubility product of copper hydroxide PR \u003d 2 \u003d 5.6. 10–20 mol 3 /l 3. c(Cu 2+) . ( c(OH -)) 2 \u003d 0.0005 mol / l. (0.01 mol / l) 2 \u003d 5. 10–8 mol 3 /l 3 . The concentration product is greater than the solubility product, so a precipitate will form. 2. Will silver sulfate precipitate when pouring equal volumes of 0.02 M sodium sulfate solution and 0.04 M silver nitrate solution? 2Ag + + SO 4 2- Ag 2 SO 4 The concentration in the resulting solution of silver ions c (Ag +) \u003d (0.04 mol / l): 2 \u003d 0.02 mol / l. The concentration in the resulting solution of sulfate ions c(SO 4 2-) \u003d (0.02 mol / l): 2 \u003d 0.01 mol / l. Solubility product of silver sulfate PR (Ag 2 SO 4) \u003d 2. \u003d 1.2. 10–5 mol 3 /l 3 . The product of the concentrations of ions in solution {c(Ag +)) 2. c(SO 4 2-) \u003d (0.02 mol / l) 2. 0.01 mol / l \u003d 4. 10 –6 mol 3 /l 3 . The concentration product is less than the solubility product, so no precipitate is formed. 20.6. Degree of conversion (degree of protolysis, degree of dissociation, degree of hydrolysis) The efficiency of the reaction is usually evaluated by calculating the yield of the reaction product (Section 5.11). However, you can also evaluate the efficiency of the reaction by determining what part of the most important (usually the most expensive) substance turned into the target reaction product, for example, what part of SO 2 turned into SO 3 during the production of sulfuric acid, that is, find degree of conversion original substance. Cl 2 + 2KOH \u003d KCl + KClO + H 2 O chlorine (reagent) is equally converted 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 known to you - the degree of protolysis (paragraph 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 referred to as the degree of protolysis). The degree of dissociation is related to the dissociation constant according to the Ostwald dilution law. Within the framework of the same theory, the equilibrium of hydrolysis is characterized by degree of hydrolysis (h), while using the following expressions relating 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 for a salt of a weak base, and the third for a salt of a weak acid and a weak base. All these expressions can only be used for dilute solutions with a degree of hydrolysis of not more than 0.05 (5%). Law of mass action, homophasic 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 the chemical equilibrium and change the equilibrium constant. What factors make it possible 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, having at your disposal only sodium chloride, potassium chloride and sodium sulfate? Determine the degree of protolysis of acetic, hydrocyanic and nitric acids in decimolar, centomolar 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 the solution will the degree of protolysis of nitrous acid be 0.2? How much water must be added to 300 ml of 0.2 M acetic acid solution to double the degree of acid protolysis? Determine the degree of protolysis of hypochlorous acid if pH = 6 in its solution. What is the concentration of acid in this solution? The pH of the solution is 3. What should be the concentration of a) nitric acid, b) acetic acid for this? How should the concentration of a) oxonium ions, b) hydroxide ions in a solution be changed so that the pH of the solution increases by one? How many oxonium ions are contained in 1 ml of solution at pH = 12? How will the pH of water change if 0.4 g of NaOH is added to 10 liters of it? Calculate the concentrations of oxonium ions and hydroxide ions, as well as the values ​​of hydrogen and hydroxide indices 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 the solubility products given in the appendix, determine the concentration and mass fraction of solutes 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 in the form of ions in 1 liter of silver bromide solution saturated at 25 o C? What volume of a solution of silver sulfide saturated at 25 o C contains 1 mg of a solute? Does a precipitate form if an equal volume of 0.4 M KCl solution is added to a 0.05 M Pb(NO 3) 2 solution? Determine if a precipitate will form after pouring 5 ml of 0.004 M CdCl 2 solution and 15 ml of 0.003 M KOH solution. The following substances are 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, on a separate card, answer the following questions: 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 according to the traditional classification does this substance belong to? 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 carry out the reaction of this substance with hydrochloric acid? If possible, under what conditions? Give the reaction equation. Why does this reaction take place? Is she reversible? If reversible, then under what conditions? How to increase the yield in this reaction? What will change if we use dry hydrogen chloride instead of hydrochloric acid? Give the corresponding reaction equation. 8) Is it possible to carry out the reaction of this substance with a solution of sodium hydroxide? If possible, under what conditions? Give the reaction equation. Why does this reaction take place? Is she reversible? If reversible, then under what conditions? How to increase the yield in this reaction? What will change if dry NaOH is used instead of sodium hydroxide solution? Give the corresponding reaction equation. 9) Give all 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. Recent section articles: