Various inorganic substances undergo hydrolysis. Hydrolysis
Hydrolysis is the exchange reaction of a salt with water ( solvolysis with water ). In this case, the original substance is destroyed by water, with the formation of new substances.
Since hydrolysis is an ion exchange reaction, its driving force is the formation of a weak electrolyte (precipitation or (and) gas evolution). It is important to remember that the hydrolysis reaction is a reversible reaction (in most cases), but there is also an irreversible hydrolysis (it proceeds to the end, there will be no starting substance in the solution). Hydrolysis is an endothermic process (with an increase in temperature, both the rate of hydrolysis and the yield of hydrolysis products increase).
As can be seen from the definition that hydrolysis is an exchange reaction, it can be assumed that an OH group goes to the metal (+ a possible acid residue if a basic salt is formed (during the hydrolysis of a salt formed by a strong acid and a weak polyacid base)), and to the acid residue there is a hydrogen proton H + (+ a possible metal ion and a hydrogen ion, with the formation of an acid salt, if a salt formed by a weak polybasic acid is hydrolyzed)).
There are 4 types of hydrolysis:
1. Salt formed by a strong base and a strong acid. Since it has already been mentioned above, hydrolysis is an ion exchange reaction, and it proceeds only in the case of the formation of a weak electrolyte. As described above, an OH group goes to the metal, and a hydrogen proton H + goes to the acid residue, but neither a strong base nor a strong acid are weak electrolytes, therefore hydrolysis does not occur in this case:
NaCl+HOH≠NaOH+HCl
Medium reaction is close to neutral: pH≈7
2. Salt is formed by a weak base and a strong acid. As stated above: an OH group goes to the metal, and a hydrogen proton H + goes to the acidic residue. For example:
NH4Cl+HOH↔NH4OH+HCl
NH 4 + +Cl - +HOH↔NH 4 OH+H + +Cl -
NH 4 + +HOH↔NH 4 OH+H +
As can be seen from the example, hydrolysis proceeds along the cation, the reaction of the medium is acidic pH < 7.При написании уравнений гидролиза для солей, образованных сильной кислотой и слабым многокислотным основанием, то в правой части следует писать основную соль, так как гидролиз идёт только по первой ступени:
FeCl 2 + HOH ↔ FeOHCl + HCl
Fe 2+ +2Cl - +HOH↔FeO + +H + +2Cl -
Fe 2+ + HOH ↔ FeOH + + H +
3. The salt is formed by a weak acid and a strong base. As mentioned above: an OH group goes to the metal, and a hydrogen proton H + goes to the acid residue. For example:
CH 3 COONa+HOH↔NaOH+CH 3 COOH
СH 3 COO - +Na + +HOH↔Na + +CH 3 COOH+OH -
СH 3 COO - +HOH↔+CH 3 COOH+OH -
Hydrolysis proceeds along the anion, the reaction of the medium is alkaline, pH > 7. When writing the equations for the hydrolysis of a salt formed by a weak polybasic acid and a strong base, the formation of an acid salt should be written on the right side, hydrolysis proceeds in 1 step. For example:
Na 2 CO 3 + HOH ↔ NaOH + NaHCO 3
2Na + +CO 3 2- +HOH↔HCO 3 - +2Na + +OH -
CO 3 2- +HOH↔HCO 3 - +OH -
4. Salt is formed by a weak base and a weak acid. This is the only case when hydrolysis goes to the end, is irreversible (until the initial salt is completely consumed). For example:
СH 3 COONH 4 +HOH↔NH 4 OH+CH 3 COOH
This is the only case when hydrolysis goes to the end. Hydrolysis occurs both in the anion and in the cation; it is difficult to predict the reaction of the medium, but it is close to neutral: pH ≈ 7.
There is also a hydrolysis constant, consider it using the example of an acetate ion, denoting it Ac- . As can be seen from the examples above, acetic (ethanoic) acid is a weak acid, and, therefore, its salts are hydrolyzed according to the scheme:
Ac-+HOH↔HAc+OH-
Let's find the equilibrium constant for this system:
Knowing ionic product of water, we can express the concentration through it [ OH] - ,
Substituting this expression into the equation for the hydrolysis constant, we get:
Substituting the water ionization constant into the equation, we get:
But the constant dissociation of the acid (on the example of hydrochloric acid) is equal to:
Where is a hydrated hydrogen proton: . Similarly for acetic acid, as in the example. Substituting the value for the acid dissociation constant into the hydrolysis constant equation, we get:
As follows from the example, if the salt is formed by a weak base, then the denominator will be the dissociation constant of the base, calculated on the same basis as the dissociation constant of the acid. If the salt is formed by a weak base and a weak acid, then the denominator will be the product of the dissociation constants of the acid and the base.
degree of hydrolysis.
There is also another value that characterizes hydrolysis - the degree of hydrolysis -α. Which is equal to the ratio of the amount (concentration) of salt undergoing hydrolysis to the total amount (concentration) of dissolved saltThe degree of hydrolysis depends on the concentration of salt, the temperature of the solution. It increases with dilution of the salt solution and with an increase in the temperature of the solution. Recall that the more dilute the solution, the lower the molar concentration of the original salt; and the degree of hydrolysis increases with increasing temperature, since hydrolysis is an endothermic process, as mentioned above.
The degree of salt hydrolysis is the higher, the weaker the acid or base that forms it. As follows from the equation for the degree of hydrolysis and types of hydrolysis: with irreversible hydrolysisα≈1.
The degree of hydrolysis and the hydrolysis constant are interconnected through the Ostwald equation (Wilhelm Friedrich Ostwald-sdilution akon Ostwald, bred in 1888year).The dilution law shows that the degree of electrolyte dissociation depends on its concentration and dissociation constant. Let us take the initial concentration of the substance asC 0 , and the dissociated part of the substance - forγ, recall the scheme of dissociation of a substance in solution:
AB↔A + +B -
Then Ostwald's law can be expressed as follows:
Recall that the equation contains concentrations at the moment of equilibrium. But if the substance is slightly dissociated, then (1-γ) → 1, which brings the Ostwald equation into the form: K d \u003d γ 2 C 0.
The degree of hydrolysis is similarly related to its constant:
In the vast majority of cases, this formula is used. But if necessary, you can express the degree of hydrolysis through the following formula:
Special cases of hydrolysis:
1) Hydrolysis of hydrides (compounds of hydrogen with elements (here we will consider only metals of groups 1 and 2 and metam), where hydrogen exhibits an oxidation state of -1):
NaH+HOH→NaOH+H 2
CaH 2 + 2HOH → Ca (OH) 2 + 2H 2
CH 4 +HOH→CO+3H 2
The reaction with methane is one of the industrial methods for producing hydrogen.
2) Hydrolysis of peroxides.Peroxides of alkali and alkaline earth metals are decomposed by water, with the formation of the corresponding hydroxide and hydrogen peroxide (or oxygen):
Na 2 O 2 +2 H 2 O → 2 NaOH + H 2 O 2
Na 2 O 2 + 2H 2 O → 2NaOH + O 2
3) Hydrolysis of nitrides.
Ca 3 N 2 + 6HOH → 3Ca (OH) 2 + 2NH 3
4) Hydrolysis of phosphides.
K 3 P+3HOH→3KOH+PH 3
escaping gas PH 3 -phosphine, very poisonous, affects the nervous system. It is also capable of spontaneous combustion upon contact with oxygen. Have you ever walked through a swamp at night or walked past cemeteries? We saw rare bursts of lights - "wandering lights", appear as phosphine burns.
5) Hydrolysis of carbides. Two reactions that have practical application will be given here, since with their help 1 members of the homologous series of alkanes (reaction 1) and alkynes (reaction 2) are obtained:
Al 4 C 3 +12 HOH →4 Al (OH) 3 +3CH 4 (reaction 1)
CaC 2 +2 HOH →Ca(OH) 2 +2C 2 H 2 (reaction 2, the product is acylene, according to UPA With Ethyne)
6) Hydrolysis of silicides. As a result of this reaction, 1 representative of the homologous series of silanes is formed (there are 8 in total) SiH 4 is a monomeric covalent hydride.
Mg 2 Si + 4HOH → 2Mg (OH) 2 + SiH 4
7) Hydrolysis of phosphorus halides. Phosphorus chlorides 3 and 5, which are acid chlorides of phosphorous and phosphoric acids, respectively, will be considered here:
PCl 3 + 3H 2 O \u003d H 3 PO 3 + 3HCl
PCl 5 + 4H 2 O \u003d H 3 PO 4 + 5HCl
8) Hydrolysis of organic substances. Fats are hydrolyzed, with the formation of glycerol (C 3 H 5 (OH) 3) and carboxylic acid (an example of limiting carboxylic acid) (C n H (2n + 1) COOH)
Esters:
CH 3 COOCH 3 + H 2 O↔CH 3 COOH + CH 3 OH
Alcohol:
C 2 H 5 ONa+H 2 O↔C 2 H 5 OH+NaOH
Living organisms carry out the hydrolysis of various organic substances in the course of reactions catabolism with the participation enzymes. For example, during hydrolysis with the participation of digestive enzymes proteins are broken down into amino acids, fats into glycerol and fatty acids, polysaccharides into monosaccharides (for example, into glucose).
When fats are hydrolyzed in the presence of alkalis, soap; hydrolysis of fats in the presence catalysts applied to obtain glycine and fatty acids.
Tasks
1) The degree of dissociation a of acetic acid in a 0.1 M solution at 18 ° C is 1.4 10 -2. Calculate the acid dissociation constant K d. (Hint - use the Ostwald equation.)
2) What mass of calcium hydride must be dissolved in water in order to reduce the released gas to iron 6.96 g of iron oxide ( II, III)?
3) Write the equation for the reaction Fe 2 (SO 4) 3 + Na 2 CO 3 + H 2 O
4) Calculate the degree, the constant of the hydrolysis of the Na 2 SO 3 salt for the concentration Cm = 0.03 M, taking into account only the 1st stage of hydrolysis. (The dissociation constant of sulfurous acid is taken equal to 6.3∙10 -8)
Solutions:
a) Substitute these problems into the Ostwald dilution law:
b) K d \u003d [C] \u003d (1.4 10 -2) 0.1 / (1 - 0.014) \u003d 1.99 10 -5
Answer. K d \u003d 1.99 10 -5.
c) Fe 3 O 4 + 4H 2 → 4H 2 O + 3Fe
CaH 2 +HOH→Ca(OH) 2 +2H 2
We find the number of moles of iron oxide (II, III), it is equal to the ratio of the mass of a given substance to its molar mass, we get 0.03 (mol). According to the CRS, we find that the moles of calcium hydride are 0.06 (mol). This means the mass of calcium hydride equals 2.52 (grams).
Answer: 2.52(grams).
d) Fe 2 (SO 4) 3 + 3Na 2 CO 3 + 3H 2 O → 3СO2 + 2Fe (OH) 3 ↓ + 3Na 2 SO 4
e) Sodium sulfite undergoes anion hydrolysis, the reaction of the salt solution medium is alkaline (pH > 7):
SO 3 2- + H 2 O<-->OH - + HSO 3 -
The hydrolysis constant (see equation above) is: 10 -14 / 6.3 * 10 -8 \u003d 1.58 * 10 -7
The degree of hydrolysis is calculated by the formula α 2 /(1 - α) = K h /C 0 .
So, α \u003d (K h / C 0) 1/2 \u003d (1.58 * 10 -7 / 0.03) 1/2 \u003d 2.3 * 10 -3
Answer: K h \u003d 1.58 * 10 -7; α \u003d 2.3 * 10 -3
Editor: Kharlamova Galina Nikolaevna
Hydrolysis occupies a special place among the exchange reactions. In general, hydrolysis is the decomposition of substances by water. Water is one of the most active substances. It acts on a wide variety of classes of compounds: salts, carbohydrates, proteins, esters, fats, etc. During the hydrolysis of non-metal compounds, two acids are usually formed, for example:
PCl 3 + 3 H 2 O \u003d H 3 PO 3 + 3 HCl
In this case, the acidity of the solutions changes in comparison with the acidity of the solvent.
In inorganic chemistry, most often one has to deal with the hydrolysis of salts, i.e. with the exchange interaction of salt ions with water molecules, as a result of which the equilibrium of electrolytic dissociation of water shifts.
Salt hydrolysis called the reversible interaction of salt ions with water ions, leading to a change in the equilibrium between hydrogen and hydroxide ions in solution.
Hydrolysis is the result of the polarization interaction of salt ions with their hydration shell in an aqueous solution. The more significant this interaction, the more intense the hydrolysis proceeds. Simplistically, the essence of the hydrolysis process can be represented as follows.
K n + cations bind in solution with water molecules that hydrate them by a donor-acceptor bond; the donor is the oxygen atoms of the water molecule, which have two lone electron pairs, the acceptor is the cations, which have free atomic orbitals. The greater the charge of the cation and the smaller its size, the greater the polarizing effect of K n + on H 2 O.
The An‾ anions bind to water molecules by hydrogen bonding. A strong effect of anions can lead to the complete separation of the proton from the H 2 O molecule - the hydrogen bond becomes covalent. As a result, an acid or an anion of the type HS‾, HCO 3 ‾, etc. is formed.
The interaction of An‾ anions with protons is the more significant, the greater the charge of the anion and the smaller its radius. Thus, the intensity of the interaction of a substance with water is determined by the strength of the polarizing influence of K n+ and An‾ on H 2 O molecules. Thus, the cations of the elements of the side subgroups and the elements immediately following them undergo more intense hydrolysis than other ions of the same charge and radius, since the nuclei of the former are less effectively screened by d-electrons.
Hydrolysis - process is the reverse of the neutralization reaction. If the neutralization reaction is an exothermic and irreversible process, then hydrolysis is an endothermic and reversible process.
Neutralization reaction:
2 KOH + H 2 SO 3 → K 2 SO 3 + 2 H 2 O
strong weak strong weak
2 OH‾ + H 2 SO 3 \u003d SO 3 2- + 2 H 2 O
Hydrolysis reaction:
K 2 SO 3 + H 2 O ↔ KOH + KHSO 3
SO 3 2- + HOH ↔ HSO 3 ‾ + Oh‾
During hydrolysis, the equilibrium of water dissociation is shifted due to the binding of one of its ions (H + or OH -) into a weak salt electrolyte. When H + ions are bound, OH - ions accumulate in the solution, the reaction of the medium will be alkaline, and when OH - ions are bound, H + ions accumulate - the environment will be acidic.
There are four variants of the action of water on salt.
1. If cations and anions have small charges and large sizes, then their polarizing effect on water molecules is small, i.e., the interaction of the salt with H 2 O practically does not occur. This applies to cations whose hydroxides are alkalis (for example, K + and Ca 2+) and to anions of strong acids (for example, Cl‾ and NO 3 ‾). Consequently, salts formed by a strong base and a strong acid do not undergo hydrolysis. In this case, the water dissociation equilibrium
H 2 O ↔ H + + OH‾
in the presence of salt ions is practically not disturbed. Therefore, solutions of such salts are neutral (pH ≈ 7).
2. If salt is formed by a cation of a strong base and an anion of a weak acid(S 2-, CO 3 2-, CN‾, etc.), then anion hydrolysis occurs. An example is the hydrolysis of a CH 3 COOK salt. Salt ions CH 3 COO - and K + interact with ions H + and OH - from water. At the same time, acetate ions (CH 3 COO -) bind with hydrogen ions (H +) into molecules of a weak electrolyte - acetic acid (CH 3 COOH), and OH ions - accumulate in the solution, giving it an alkaline reaction, since K + ions cannot bind OH − ions (KOH is a strong electrolyte), pH > 7 .
Molecular equation of hydrolysis:
CH 3 SOOK + H 2 O ↔ KOH + CH 3 UN
Full ionic hydrolysis equation:
K + + CH 3 COO − + HOH ↔ K + + OH − + CH 3 COOH
reduced ionic hydrolysis equation:
CH 3 SOO − + H HE ↔ OH − + CH 3 UNSD
Hydrolysis of Na 2 salt S proceeds in steps. The salt is formed by a strong base (NaOH) and a weak dibasic acid (H 2 S). In this case, the salt anion S 2− binds H + ions of water, OH − ions accumulate in the solution. The equation in abbreviated ionic and molecular form is:
I. S 2− + H HE ↔ H.S. − + OH −
Na 2 S + H 2 O ↔ NaHS+NaOH
II. HS − + H HE ↔ H 2 S+ OH −
NaHS + H 2 O ↔ NaOH + H2S
The second stage of hydrolysis practically does not pass under normal conditions, since, accumulating, OH ions give the solution a strongly alkaline reaction, which leads to a neutralization reaction, a shift of equilibrium to the left in accordance with the Le Chatelier principle. Therefore, the hydrolysis of salts formed by a strong base and a weak acid is suppressed by the addition of an alkali.
The greater the polarizing effect of anions, the more intense the hydrolysis. In accordance with the law of mass action, this means that hydrolysis proceeds more intensively, the weaker the acid.
3. If salt is formed by a cation of a weak base and an anion of a strong acid, then hydrolysis occurs at the cation. For example, this occurs during the hydrolysis of the NH 4 Cl salt (NH 4 OH is a weak base, HCl is a strong acid). We discard the Cl - ion, since it gives a strong electrolyte with the water cation, then the hydrolysis equation will take the following form:
NH 4 + + H HE ↔ NH 4 Oh+H+ (abbreviated ionic equation)
NH 4 Cl + H 2 O ↔ NH 4 OH + HCl (molecular equation)
It can be seen from the reduced equation that OH - water ions bind into a weak electrolyte, H + ions accumulate in solution and the medium becomes acidic (pH< 7). Добавление кислоты к раствору (введение продукта реакции катионов H +) сдвигает равновесие влево.
The hydrolysis of a salt formed by a polyacid base (for example, Zn(NO 3) 2) proceeds stepwise over the cation of a weak base.
I. Zn 2+ + H HE ↔ ZnOH + +H+ (abbreviated ionic equation)
Zn (NO 3) 2 + H 2 O ↔ ZnOHNO 3 + HNO 3 (molecular equation)
OH − ions bind to a weak base ZnOH +, H + ions accumulate.
The second stage of hydrolysis practically does not occur under normal conditions., since as a result of the accumulation of H + ions in the solution, a strongly acidic environment is created and the equilibrium of the hydrolysis reaction in the 2nd stage is shifted to the left:
II. ZnOH + + H HE ↔ Zn(Oh) 2 +H+ (abbreviated ionic equation)
ZnOHNO 3 + H 2 O ↔ Zn(OH) 2 + HNO 3 (molecular equation)
Obviously, the weaker the base, the more complete the hydrolysis.
4. A salt formed by a cation of a weak base and an anion of a weak acid undergoes hydrolysis at the cation and at the anion. An example is the process of hydrolysis of salt CH 3 COOH 4 . We write the equation in ionic form:
NH 4 + + CH 3 COO − + HOH ↔ NH 4 OH + CH 3 COOH
The hydrolysis of such salts proceeds very strongly, since as a result of it both a weak base and a weak acid are formed.
The reaction of the medium in this case depends on the comparative strength of the base and acid, i.e. from their dissociation constants (K D):
if K D (bases) > K D (acids), then pH > 7;
if K D (base)< K Д (кислоты), то pH < 7.
In the case of hydrolysis of CH 3 COONH 4:
K D (NH 4 OH) = 1.8 10 -5; K D (CH 3 COOH) \u003d 1.8 10 -5,
therefore, the reaction of an aqueous solution of this salt will be almost neutral (pH ≈ 7).
If the base and acid that form the salt are not only weak electrolytes, but also poorly soluble or unstable and decompose to form volatile products, then in this case the hydrolysis of the salt proceeds through all stages to the end, i.e. until the formation of a weak, sparingly soluble base and a weak acid. In this case, it is about irreversible or complete hydrolysis.
It is complete hydrolysis that is the reason that aqueous solutions of some salts cannot be prepared, for example, Cr 2 (CO 3) 3, Al 2 S 3, etc. For example:
Al 2 S 3 + 6H 2 O → 2Al(OH) 3 ↓ + 3H 2 S
Therefore, aluminum sulfide cannot exist in the form of aqueous solutions, it can only be obtained by a "dry method", for example, from elements at high temperature:
2Al + 3S - t ° → Al 2 S 3,
and must be stored in sealed containers to prevent moisture from entering.
Such compounds cannot be obtained by an exchange reaction in an aqueous solution. When salts A1 3+, Cr 3+ and Fe 3+ react in solution with sulfides and carbonates, not sulfides and carbonates of these cations precipitate, but their hydroxides:
2AlCl 3 + 3Na 2 S + 6H 2 O → 3H 2 S + 2Al (OH) 3 ↓ + 6NaCl
2CrCl 3 + 3Na 2 CO 3 + 3H 2 O → 2Сr(OH) 3 ↓ + 3СO 2 + 6NaCl
In the examples considered, the hydrolysis of two salts (AlCl 3 and Na 2 S or CrCl 3 and Na 2 CO 3) is mutually enhanced and the reaction goes to the end, since the reaction products are released from the solution in the form of a precipitate and gas.
Hydrolysis of salts in some cases can be very difficult. (Simple hydrolysis reaction equations in common notation are often conditional.) Hydrolysis products can only be established on the basis of an analytical study. For example, hydrolysis products of salts containing multiply charged cations can be polynuclear complexes. So, if the solution of Hg 2+ contains only mononuclear complexes, then in solutions of Fe 3+, in addition to the complexes 2+ and +, a two-nuclear complex 4+ is found; in Be 2+ solutions, mainly polynuclear complexes of the composition [Be 3 (OH) 3 ] 3+ are formed; in solutions of Sn 2+ complex ions 2+ , 2+ , + are formed; in Bi 3+ solutions, along with [ВiOH] 2+, there are complex ions of composition 6+ . Hydrolysis reactions leading to the formation of polynuclear complexes can be represented as follows:
mM k+ + nH 2 O ↔ M m (OH) n (mk - n)+ + nH +,
where m varies from 1 to 9, and n can take values from 1 to 15. Such reactions are possible for cations of more than 30 elements. It has been established that in most cases each charge of the ion corresponds to a certain form of the complex. So, for M 2+ ions, the form of dimers 3+ is characteristic, for M 3+ ions - 4+, and for M 4+ - the form 5+ and more complex, for example 8+.
At high temperatures and high pH values, oxo complexes are also formed:
2MOH ↔ MOM + H 2 O or
For example,
BiCl 3 + H 2 O « Bi (OH) 2 Cl + 2HCl
The Bi(OH) 2 + cation easily loses a water molecule, forming the BiO + bismuthyl cation, which gives a white crystalline precipitate with the chloride ion:
Bi(OH) 2 Cl ®BiOCl↓ + H 2 O.
Structurally, polynuclear complexes can be represented as octahedra connected to each other along a vertex, edge, or face by means of various bridges (O, OH, etc.).
The products of hydrolysis of carbonates of a number of metals have a complex composition. So, when soluble salts Mg 2+, Cu 2+, Zn 2+, Pb 2+ interact with sodium carbonate, not medium carbonates are formed, but less soluble hydroxocarbonates, for example Cu 2 (OH) 2 CO 3, Zn 5 (OH) 6 (CO 3) 2, Pb 3 (OH) 2 (CO 3) 2. An example is the reactions:
5MgSO 4 + 5Na 2 CO 3 + H 2 O → Mg 5 (OH) 2 (CO 3) 4 ↓ + 5Na 2 SO 4 + CO 2
2Cu(NO 3) 2 + 2Na 2 CO 3 + H 2 O → Cu 2 (OH) 2 CO 3 ↓ + 4NaNO 3 + CO 2
Quantitatively, hydrolysis is characterized by the degree of hydrolysis h and the hydrolysis constant K G.
The degree of hydrolysis shows what part of the salt contained in the solution (C M) has undergone hydrolysis (C Mhyd) and is calculated as the ratio:
h = S M guide / S M (100%).
Obviously, for a reversible hydrolysis process h < 1 (<100%), а для необратимого гидролиза h= 1 (100%). In addition to the nature of the salt, the degree of hydrolysis depends on the concentration of the salt and the temperature of the solution.
In solutions with a moderate concentration of a solute, the degree of hydrolysis at room temperature is usually small. For salts formed by a strong base and a strong acid, it is practically zero; for salts formed by a weak base and a strong acid or a strong base and a weak acid, it is ≈1%. So, for a 0.01 M solution of NH 4 Cl h= 0.01%; for 0.1 n. CH 3 COONH 4 solution h ≈ 0,5%.
Hydrolysis is a reversible process, so the law of mass action applies to it.
The hydrolysis constant is the equilibrium constant of the hydrolysis process, and in its physical meaning determines the degree of hydrolysis irreversibility. The more KG, the more irreversible the hydrolysis. K G has its own expression for each case of hydrolysis.
Let us derive an expression for the hydrolysis constant of a salt of a weak acid and a strong base using NaCN as an example:
NaCN + H 2 O ↔ NaOH + HCN;
Na + + CN - + H 2 O ↔ Na + + OH - + HCN;
CN - + H 2 O ↔ HCN + OH -
K equals = / .
It has the highest value, which practically does not change during the reaction, so it can be conditionally considered constant. Then multiplying the numerator and denominator by the concentration of protons and introducing a constant concentration of water into a constant, we get:
K equals \u003d K W / K D (sour) \u003d K G
since / \u003d 1 / K D (sour)
Since K W is a constant and equal to 10 -14, it is obvious that the lower the K D of a weak acid, the anion of which is part of the salt, the greater the K G.
Similarly, for a salt hydrolyzed by a cation (for example, NH 4 Cl), we get:
NH 4 + + H 2 O ↔ NH 4 OH + H + (abbreviated hydrolysis equation)
K equals = /
K G \u003d K equal \u003d K W / K D (basic)
In this expression, the numerator and denominator of the fraction are multiplied by . Obviously, the lower the K D of a weak base, the cation of which is part of the salt, the greater the K G.
If the salt is formed by a weak base and a weak acid (for example, NH 4 CN), then the reduced hydrolysis equation is:
NH 4 + + CN - + H 2 O ↔ NH 4 OH + HCN
K equals = / ,
In this expression for K, the numerator and denominator of the fraction are multiplied by ·, so the expression for K G takes the form:
K G \u003d K W / (K D (acids) K D (basic)).
As follows from the above expressions, the hydrolysis constant is inversely proportional to the dissociation constant of a weak electrolyte involved in the formation of the salt (if two weak electrolytes are involved in the formation of the salt, then K G is inversely proportional to the product of their dissociation constants).
Consider the hydrolysis of a multiply charged ion. Take Na 2 CO 3.
I. CO 3 2- + H 2 O "HCO 3 - + OH -
K G (I) = / × ( / ) = K W / K D (II) ,
that is, the second dissociation constant enters into the expression for the hydrolysis constant for the first stage, and for the second stage of hydrolysis
HCO 3 - + H 2 O "H 2 CO 3 + OH -
K G (II) = / × ( / ) = K W / K D (I)
K D (I) = 4 × 10 -7 K D (II) = 2.5 × 10 -8
K G (II) = 5.6 × 10 -11 K G (I) = 1.8 × 10 -4
Thus, K G(I) >> K G(II) , the constant, and hence the degree of the first stage of hydrolysis, is much greater than the subsequent ones.
Degree of hydrolysis is the value of a similar degree of dissociation. The relationship between the degree and the hydrolysis constant is similar to that for the degree and dissociation constant.
If, in the general case, the initial concentration of the anion of a weak acid is denoted by C o (mol / l), then C o h(mol/l) is the concentration of that part of the anion A - that has undergone hydrolysis and formed C o h(mol/l) weak acid HA and C o h(mol/l) hydroxide groups.
A - + H 2 O ↔ HA + OH -,
C o -C o h C o h C o h
then K G \u003d / \u003d C o h· About h/ (C o -C o h) = C o h 2 / (1-h).
At h << 1 K Г = С о h 2 h\u003d √ K D / C o.
Very similar to Ostwald's dilution law.
C o h, we get:
K G \u003d C o h· About h/ C o \u003d 2 / C o, from where
\u003d √ K G · C o.
Similarly, it can be shown that upon hydrolysis at the cation
\u003d √ K G · C o.
Thus, the ability of salts to undergo hydrolysis depends on two factors:
the properties of the ions that form the salt;
external factors.
How to shift the equilibrium of hydrolysis?
1) Addition of ions of the same name. Since a dynamic equilibrium is established during reversible hydrolysis, in accordance with the law of mass action, the equilibrium can be shifted in one direction or another by introducing an acid or base into the solution. The introduction of an acid (H + cations) suppresses the hydrolysis of the cation, the addition of an alkali (OH – anions) suppresses the hydrolysis of the anion. This is often used to enhance or suppress the hydrolysis process.
2) From the formula for h it's clear that dilution promotes hydrolysis. The increase in the degree of hydrolysis of sodium carbonate
Na 2 CO 3 + HOH ↔ NaHCO 3 + NaOH
when diluting the solution illustrates fig. twenty.
Rice. 20. Dependence of the degree of hydrolysis of Na 2 CO 3 on dilution at 20°С
3) An increase in temperature promotes hydrolysis. The dissociation constant of water increases with increasing temperature to a greater extent than the dissociation constants of hydrolysis products - weak acids and bases, therefore, when heated, the degree of hydrolysis increases. It is easy to come to this conclusion in another way: since the neutralization reaction is exothermic (DH = -56 kJ / mol), hydrolysis, being the opposite process, is endothermic, therefore, in accordance with the Le Chatelier principle, heating causes an increase in hydrolysis. Rice. 21 illustrates the effect of temperature on the hydrolysis of chromium(III) chloride.
CrCl 3 + HOH ↔ CrOHCl 2 + HCl
Rice. 21. Dependence of the degree of hydrolysis of CrCl 3 on temperature
In chemical practice, cation hydrolysis of salts formed by a multiply charged cation and a singly charged anion, for example, AlCl 3 , is very common. In solutions of these salts, a less dissociated compound is formed as a result of the addition of one hydroxide ion to a metal ion. Given that the Al 3+ ion in the solution is hydrated, the first stage of hydrolysis can be expressed by the equation
3+ + HOH ↔ 2+ + H 3 O +
At ordinary temperature, the hydrolysis of salts of multiply charged cations is practically limited to this stage. When heated, hydrolysis occurs in the second stage:
2+ + HOH ↔ + + H 3 O +
Thus, the acidic reaction of an aqueous salt solution is explained by the fact that the hydrated cation loses a proton and the H 2 O aqua group is converted into the OH‾ hydroxo group. In the considered process, more complex complexes can also be formed, for example 3+, as well as complex ions of the 3- and [АlO 2 (OH) 2 ] 3- types. The content of various hydrolysis products depends on the reaction conditions (solution concentration, temperature, presence of other substances). The duration of the process is also important, since equilibrium during the hydrolysis of salts of multiply charged cations is usually reached slowly.
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1 HYDROLYSIS OF ORGANIC AND INORGANIC SUBSTANCES
2 Hydrolysis (from the ancient Greek “ὕδωρ” water and “λύσις” decomposition) is one of the types of chemical reactions where, when substances interact with water, the initial substance decomposes with the formation of new compounds. The mechanism of hydrolysis of compounds of various classes: - salts, carbohydrates, fats, esters, etc. has significant differences
3 Hydrolysis of organic substances Living organisms carry out the hydrolysis of various organic substances in the course of reactions with the participation of ENZYMES. For example, during hydrolysis, with the participation of digestive enzymes, PROTEINS are broken down into AMINO ACIDS, FATS into GLYCEROL and FATTY ACIDS, POLYSACCHARIDES (for example, starch and cellulose) into MONOSACCHARIDES (for example, into GLUCOSE), NUCLEIC ACIDS into free NUCLEOTIDES. When fats are hydrolyzed in the presence of alkalis, soap is obtained; hydrolysis of fats in the presence of catalysts is used to obtain glycerol and fatty acids. Ethanol is obtained by hydrolysis of wood, and peat hydrolysis products are used in the production of fodder yeast, wax, fertilizers, etc.
4 1. Hydrolysis of organic compounds fats are hydrolyzed to obtain glycerol and carboxylic acids (saponification with NaOH):
5 starch and cellulose are hydrolyzed to glucose:
7 TEST 1. Hydrolysis of fats produces 1) alcohols and mineral acids 2) aldehydes and carboxylic acids 3) monohydric alcohols and carboxylic acids 4) glycerol and carboxylic acids ANSWER: 4 2. Hydrolysis undergoes: 1) Acetylene 2) Cellulose 3) Ethanol 4) Methane ANSWER: 2 3. Hydrolysis undergoes: 1) Glucose 2) Glycerin 3) Fat 4) Acetic acid ANSWER: 3
8 4. During the hydrolysis of esters, the following are formed: 1) Alcohols and aldehydes 2) Carboxylic acids and glucose 3) Starch and glucose 4) Alcohols and carboxylic acids ANSWER: 4 5. When hydrolysis of starch is obtained: 1) Sucrose 2) Fructose 3) Maltose 4) Glucose ANSWER: 4
9 2. Reversible and irreversible hydrolysis Almost all the considered reactions of hydrolysis of organic substances are reversible. But there is also irreversible hydrolysis. The general property of irreversible hydrolysis is that one (preferably both) of the hydrolysis products must be removed from the reaction sphere in the form of: - SEDIMENT, - GAS. CaC₂ + 2H₂O = Ca(OH)₂ + C₂H₂ In the hydrolysis of salts: Al₄C₃ + 12 H₂O = 4 Al(OH)₃ + 3CH₄ Al₂S₃ + 6 H₂O CaH₂ + 2 H₂O = 2 Al(OH)₃ + 3 H₂S = 2Ca(OH )₂ + H₂
10 HYDROLYSIS SALES Hydrolysis of salts is a kind of hydrolysis reactions caused by the occurrence of ion exchange reactions in solutions of (aqueous) soluble electrolyte salts. The driving force of the process is the interaction of ions with water, leading to the formation of a weak electrolyte in ionic or molecular form (“ion binding”). Distinguish between reversible and irreversible hydrolysis of salts. 1. Hydrolysis of a salt of a weak acid and a strong base (anion hydrolysis). 2. Hydrolysis of a salt of a strong acid and a weak base (cation hydrolysis). 3. Hydrolysis of the salt of a weak acid and a weak base (irreversible) The salt of a strong acid and a strong base does not undergo hydrolysis
12 1. Hydrolysis of a salt of a weak acid and a strong base (anion hydrolysis): (solution has an alkaline environment, the reaction is reversible, hydrolysis in the second stage proceeds to an insignificant degree) 2. Hydrolysis of a salt of a strong acid and a weak base (cation hydrolysis): (the solution has an acidic environment, the reaction proceeds reversibly, hydrolysis in the second stage proceeds to an insignificant degree)
13 3. Hydrolysis of a salt of a weak acid and a weak base: (the equilibrium is shifted towards the products, the hydrolysis proceeds almost completely, since both reaction products leave the reaction zone in the form of a precipitate or gas). The salt of a strong acid and a strong base does not undergo hydrolysis and the solution is neutral.
14 SCHEME OF SODIUM CARBONATE HYDROLYSIS NaOH strong base Na₂CO₃ H₂CO₃ weak acid > [H]+ BASIC MEDIUM ACID SALT, ANION hydrolysis
15 First hydrolysis stage Na₂CO₃ + H₂O NaOH + NaHCO₃ 2Na+ + CO₃ ² + H₂O Na+ + OH + Na+ + HCO₃ CO₃ ² + H₂O OH + HCO₃ Second hydrolysis stage NaHCO₃ + H₂O = NaOH + H₂CO ₃ CO₂ H₂O Na+ = Na₂O + HCO + OH + CO₂ + H₂O HCO₃ + H₂O = OH + CO₂ + H₂O
16 COPPER(II) CHLORIDE HYDROLYSIS SCHEME Cu(OH)₂ weak base CuCl₂ HCl strong acid< [ H ]+ КИСЛАЯ СРЕДА СОЛЬ ОСНОВНАЯ, гидролиз по КАТИОНУ
17 First stage of hydrolysis CuCl₂ + H₂O (CuOH)Cl + HCl Cu+² + 2 Cl + H₂O (CuOH)+ + Cl + H+ + Cl Cu+² + H₂O (CuOH)+ + H+ Second stage of hydrolysis (СuOH)Cl + H₂O Cu(OH)₂ + HCl (Cu OH)+ + Cl + H₂O Cu(OH)₂ + H+ + Cl (CuOH)+ + H₂O Cu(OH)₂ + H+
18 ALUMINUM SULFIDE HYDROLYSIS SCHEME Al₂S₃ Al(OH)₃ H₂S weak base weak acid = [H]+ NEUTRAL REACTION OF THE MEDIUM irreversible hydrolysis
19 Al₂S₃ + 6 H₂O = 2Al(OH)₃ + 3H₂S HYDROLYSIS OF SODIUM CHLORIDE NaCl NaOH HCl strong base strong acid = [H]+ NEUTRAL REACTION OF THE ENVIRONMENT no hydrolysis occurs NaCl + H₂O = NaOH + HCl Na+ + Cl + H₂O = Na+ + OH + H+ + Cl
20 Transformation of the earth's crust Providing a slightly alkaline environment for sea water THE ROLE OF HYDROLYSIS IN HUMAN LIFE Laundry Washing dishes Washing with soap Digestion processes
21 Write the hydrolysis equations: A) K₂S B) FeCl₂ C) (NH₄)₂S D) BaI₂ K₂S: KOH is a strong base H₂S weak acid HS + K+ + OH S² + H₂O HS + OH FeCl₂ : Fe(OH)₂ - weak base HCL - strong acid FeOH)+ + Cl + H+ + Cl Fe +² + H₂O (FeOH)+ + H+
22 (NH₄)₂S: NH₄OH - weak base; H₂S - weak acid HI - strong acid HYDROLYSIS NO
23 Perform on a sheet of paper. Hand in your work to the teacher at the next lesson.
25 7. An aqueous solution of which of the salts has a neutral environment? a) Al(NO₃)₃ b) ZnCl₂ c) BaCl₂ d) Fe(NO₃)₂ 8. In which solution will the color of litmus be blue? a) Fe₂(SO₄)₃ b) K₂S c) CuCl₂ d) (NH₄)₂SO₄
26 9. Hydrolysis is not subject to 1) potassium carbonate 2) ethane 3) zinc chloride 4) fat 10. During the hydrolysis of fiber (starch), the following can be formed: 1) glucose 2) only sucrose 3) only fructose 4) carbon dioxide and water 11. The solution medium as a result of the hydrolysis of sodium carbonate 1) alkaline 2) strongly acidic 3) acidic 4) neutral 12. Hydrolysis undergoes 1) CH 3 COOK 2) KCI 3) CaCO 3 4) Na 2 SO 4
27 13. Hydrolysis is not subjected to 1) ferrous sulfate 2) alcohols 3) ammonium chloride 4) esters
28 PROBLEM Explain why when pouring solutions - FeCl₃ and Na₂CO₃ - precipitates and gas is released? 2FeCl₃ + 3Na₂CO₃ + 3H₂O = 2Fe(OH)₃ + 6NaCl + 3CO₂
29 Fe+³ + H₂O (FeOH)+² + H+ CO₃ ² + H₂O HCO₃ + OH CO₂ + H₂O Fe(OH)₃
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As we can see, the environment of the first solution is neutral (pH=7), the second one is acidic (pH< 7), третьего щелочная (рН >7). How to explain such an interesting fact? 🙂
First, let's remember what pH is and what it depends on.
pH is a hydrogen indicator, a measure of the concentration of hydrogen ions in a solution (according to the first letters of the Latin words potentia hydrogeni - the strength of hydrogen).
pH is calculated as the negative decimal logarithm of the concentration of hydrogen ions, expressed in moles per liter:
In pure water at 25 °C, the concentrations of hydrogen ions and hydroxide ions are the same and amount to 10 -7 mol/l (pH=7).
When the concentrations of both types of ions in a solution are the same, the solution is neutral. When > the solution is acidic, and when > - alkaline.
Due to what, in some aqueous solutions of salts, is there a violation of the equality of the concentrations of hydrogen ions and hydroxide ions?
The fact is that there is a shift in the equilibrium of water dissociation due to the binding of one of its ions (or) with salt ions with the formation of a poorly dissociated, hardly soluble or volatile product. This is the essence of hydrolysis.
- this is the chemical interaction of salt ions with water ions, leading to the formation of a weak electrolyte - an acid (or acid salt), or a base (or basic salt).
The word "hydrolysis" means decomposition by water ("hydro" - water, "lysis" - decomposition).
Depending on which salt ion interacts with water, there are three types of hydrolysis:
- hydrolysis by cation (only cation reacts with water);
- anion hydrolysis (only anion reacts with water);
- joint hydrolysis - hydrolysis by cation and anion (both cation and anion react with water).
Any salt can be considered as a product formed by the interaction of a base and an acid:
Salt hydrolysis - the interaction of its ions with water, leading to the appearance of an acidic or alkaline environment, but not accompanied by the formation of a precipitate or gas.
The hydrolysis process proceeds only with the participation soluble salt and consists of two stages:
1)dissociation salt in solution irreversible reaction (degree of dissociation, or 100%);
2) actually , i.e. interaction of salt ions with water reversible reaction (degree of hydrolysis ˂ 1, or 100%)
The equations of the 1st and 2nd stages - the first of them is irreversible, the second is reversible - cannot be added!
Note that salts formed by cations alkalis and anions strong acids do not undergo hydrolysis, they only dissociate when dissolved in water. In solutions of salts KCl, NaNO 3 , NaSO 4 and BaI, the medium neutral.
Anion hydrolysis
In case of interaction anions dissolved salt with water the process is called salt hydrolysis at the anion.
1) KNO 2 = K + + NO 2 - (dissociation)
2) NO 2 - + H 2 O ↔ HNO 2 + OH - (hydrolysis)
The dissociation of the KNO 2 salt proceeds completely, the hydrolysis of the NO 2 anion - to a very small extent (for a 0.1 M solution - by 0.0014%), but this turns out to be enough for the solution to become alkaline(among the hydrolysis products there is an OH ion -), in it p H = 8.14.
Anions undergo hydrolysis only weak acids (in this example, the nitrite ion NO 2 corresponding to the weak nitrous acid HNO 2). The anion of a weak acid attracts the hydrogen cation present in water to itself and forms a molecule of this acid, while the hydroxide ion remains free:
NO 2 - + H 2 O (H +, OH -) ↔ HNO 2 + OH -
Examples:
a) NaClO \u003d Na + + ClO -
ClO - + H 2 O ↔ HClO + OH -
b) LiCN = Li + + CN -
CN - + H 2 O ↔ HCN + OH -
c) Na 2 CO 3 \u003d 2Na + + CO 3 2-
CO 3 2- + H 2 O ↔ HCO 3 - + OH -
d) K 3 PO 4 \u003d 3K + + PO 4 3-
PO 4 3- + H 2 O ↔ HPO 4 2- + OH -
e) BaS = Ba 2+ + S 2-
S 2- + H 2 O ↔ HS - + OH -
Please note that in examples (c-e) you cannot increase the number of water molecules and instead of hydroanions (HCO 3, HPO 4, HS) write the formulas of the corresponding acids (H 2 CO 3, H 3 PO 4, H 2 S). Hydrolysis is a reversible reaction, and it cannot proceed “to the end” (before the formation of an acid).
If such an unstable acid as H 2 CO 3 were formed in a solution of its NaCO 3 salt, then CO 2 would be released from the gas solution (H 2 CO 3 \u003d CO 2 + H 2 O). However, when soda is dissolved in water, a transparent solution is formed without gas evolution, which is evidence of the incompleteness of the hydrolysis of the anion with the appearance in the solution of only carbonic acid hydranions HCO 3 -.
The degree of salt hydrolysis by the anion depends on the degree of dissociation of the hydrolysis product, the acid. The weaker the acid, the higher the degree of hydrolysis. For example, CO 3 2-, PO 4 3- and S 2- ions undergo hydrolysis to a greater extent than the NO 2 ion, since the dissociation of H 2 CO 3 and H 2 S in the 2nd stage, and H 3 PO 4 in The 3rd stage proceeds much less than the dissociation of the HNO 2 acid. Therefore, solutions, for example, Na 2 CO 3, K 3 PO 4 and BaS will highly alkaline(which is easy to verify by the soapiness of soda to the touch) .
An excess of OH ions in a solution is easy to detect with an indicator or measure with special instruments (pH meters).
If in a concentrated solution of a salt that is strongly hydrolyzed by the anion,
for example, Na 2 CO 3, add aluminum, then the latter (due to amphoterism) will react with alkali and hydrogen evolution will be observed. This is additional evidence of hydrolysis, because we did not add NaOH alkali to the soda solution!
Pay special attention to salts of acids of medium strength - orthophosphoric and sulfurous. In the first stage, these acids dissociate quite well, so their acid salts do not undergo hydrolysis, and the environment of the solution of such salts is acidic (due to the presence of a hydrogen cation in the composition of the salt). And the average salts are hydrolyzed by the anion - the medium is alkaline. So, hydrosulfites, hydrophosphates and dihydrophosphates are not hydrolyzed by the anion, the medium is acidic. Sulfites and phosphates are hydrolyzed by the anion, the environment is alkaline.
Hydrolysis by cation
In the case of the interaction of a cation of a dissolved salt with water, the process is called
salt hydrolysis at the cation
1) Ni(NO 3) 2 = Ni 2+ + 2NO 3 - (dissociation)
2) Ni 2+ + H 2 O ↔ NiOH + + H + (hydrolysis)
The dissociation of the Ni (NO 3) 2 salt proceeds completely, the hydrolysis of the Ni 2+ cation - to a very small extent (for a 0.1 M solution - by 0.001%), but this is enough for the medium to become acidic (among the hydrolysis products there is an H + ion ).
Only cations of poorly soluble basic and amphoteric hydroxides and the ammonium cation undergo hydrolysis. NH4+. The metal cation splits off the hydroxide ion from the water molecule and releases the hydrogen cation H + .
The ammonium cation, as a result of hydrolysis, forms a weak base - ammonia hydrate and a hydrogen cation:
NH 4 + + H 2 O ↔ NH 3 H 2 O + H +
Please note that you cannot increase the number of water molecules and instead of hydroxocations (for example, NiOH +) write hydroxide formulas (for example, Ni (OH) 2). If hydroxides were formed, then precipitates would fall out of salt solutions, which is not observed (these salts form transparent solutions).
An excess of hydrogen cations is easy to detect with an indicator or measure with special instruments. Magnesium or zinc is introduced into a concentrated solution of a salt that is highly hydrolyzed by the cation, then the latter react with the acid with the release of hydrogen.
If the salt is insoluble, then there is no hydrolysis, because the ions do not interact with water.