Hydrolysis of organic compounds. 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 indicated 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:

CH 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 contain 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 (hydrogen compounds 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 ethyn)

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

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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 During 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 SOLEY 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 piece 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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Peculiarities

Hydrolysis of organic compounds

Reversible and irreversible hydrolysis

H I D R O L I S S O L E Y

Reaction equations

10.

11. SCHEME OF SODIUM CARBONATE HYDROLYSIS

12. SCHEME OF HYDROLYSIS OF COPPER(II) CHLORIDE

13. ALUMINUM SULFIDE HYDROLYSIS SCHEME

14.

one). Hydrolysis is an endothermic reaction, so an increase in temperature enhances hydrolysis.

2). An increase in the concentration of hydrogen ions weakens the hydrolysis, in the case of hydrolysis by the cation. Similarly, increasing the concentration of hydroxide ions weakens the hydrolysis, in the case of anion hydrolysis.

3). When diluted with water, the equilibrium shifts in the direction of the reaction, i.e. to the right, the degree of hydrolysis increases.

four). Additives of foreign substances can affect the equilibrium position when these substances react with one of the participants in the reaction. So, when copper sulfate is added to a solution

2CuSO4 + 2H2O<=>(CuOH)2SO4 + H2SO4

sodium hydroxide solution, the hydroxide ions contained in it will interact with hydrogen ions. As a result, their concentration will decrease, and, according to Le Chatelier's principle, the equilibrium in the system will shift to the right, the degree of hydrolysis will increase. And if a solution of sodium sulfide is added to the same solution, then the equilibrium will not shift to the right, as one might expect (mutual enhancement of hydrolysis), but, on the contrary, to the left, due to the binding of copper ions into practically insoluble copper sulfide.

5). salt concentration. Consideration of this factor leads to a paradoxical conclusion: the equilibrium in the system shifts to the right, in accordance with Le Chatelier's principle, but the degree of hydrolysis decreases.

Example,

Al(NO 3 ) 3

The salt is hydrolyzed at the cation. It is possible to enhance the hydrolysis of this salt if:

1. heat or dilute the solution with water;
2. add a solution of alkali (NaOH);
3. add a solution of salt hydrolyzed by the anion Na 2 CO 3 ;

1. lead dissolution in the cold;
2. prepare the most concentrated solution of Al(NO 3 ) 3 as possible;
3. add an acid to the solution, such as HCl

Hydrolysis of salts of polyacid bases and polybasic acids proceeds stepwise

For example, the hydrolysis of iron (II) chloride includes two steps:

1st step

FeCl 2 + HOH<=>Fe(OH)Cl + HCl
Fe2+ + 2Cl - + H + + OH -<=>Fe(OH) + + 2Cl - + H +

2nd stage

Fe(OH)Cl + HOH<=>Fe(OH) 2 + HCl
Fe(OH) + + Cl - + H + + OH -<=>Fe( OH) 2 + H + + Cl -

The hydrolysis of sodium carbonate includes two steps:

1st step

Na 2 CO 3 + HOH<=>NaHCO 3 + NaOH
CO 3 2- + 2Na + + H + + OH - => HCO 3 - + OH - + 2Na +

2nd stage

NaHCO 3 + H 2 O<=>NaOH + H 2 CO 3
HCO 3 - + Na + + H + + OH -<=>H 2 CO 3 + OH - + Na +

Hydrolysis is a reversible process. An increase in the concentration of hydrogen ions and hydroxide ions prevents the reaction from proceeding to completion. In parallel with hydrolysis, a neutralization reaction takes place when the resulting weak base (Fe (OH) 2) interacts with a strong acid, and the resulting weak acid (H 2 CO 3) reacts with an alkali.

Hydrolysis proceeds irreversibly if an insoluble base and (or) a volatile acid is formed as a result of the reaction:

Al 2 S 3 + 6H 2 O \u003d\u003e 2Al (OH) 3 ↓ + 3H 2 S

Salts completely decomposed by water - Al2S3 , cannot be obtained by the exchange reaction in aqueous solutions, since instead of the exchange, the reaction of joint hydrolysis proceeds:

2AlCl 3 +3Na 2 S≠Al 2 S 3 +6NaCl

2AlCl 3 +3Na 2 S+6H 2 O=2Al(OH) 3 ↓+6NaCl+3H 2 S(mutual enhancement of hydrolysis)

Therefore, they are obtained in anhydrous media by sintering or other methods, for example:

2Al+3S = t°C\u003d Al 2 S 3

Examples of hydrolysis reactions

(NH 4) 2 CO 3 ammonium carbonate salt, weak acid and weak base. Soluble. Hydrolyzes both cation and anion at the same time. The number of steps is 2.

Stage 1: (NH 4) 2 CO 3 + H 2 O ↔ NH 4 OH + NH 4 HCO 3

2 step: NH 4 HCO 3 + H 2 O ↔NH 4 OH + H 2 CO 3

The reaction of the solution is slightly alkaline pH > 7, because ammonium hydroxide is a stronger electrolyte than carbonic acid. K d (NH 4 OH)> K d (H 2 CO 3)

CH 3 COONH 4 ammonium acetate salt, weak acid and weak base. Soluble. Hydrolyzes both cation and anion at the same time. The number of steps is 1.

CH 3 COONH 4 + H 2 O ↔NH 4 OH + CH 3 COOH

The reaction of the solution is neutral pH \u003d 7, because K d (CH 3 COO H) \u003d K d (NH 4 OH)

K2HPO4– potassium hydrogen phosphate salt, weak acid and strong base. Soluble. Hydrolyzed at the anion. The number of steps is 2.

1 step: K 2 HPO 4 +H 2 O ↔KH 2 PO 4 +KOH

2 step: KH 2 PO 4 +H 2 O ↔H 3 PO 4 +KOH

solution reaction 1 step slightly alkalinepH=8,9 , since as a result of hydrolysis, OH - ions accumulate in the solution and the hydrolysis process prevails over the process of dissociation of HPO 4 2- ions, giving H + ions (HPO 4 2- ↔H + + PO 4 3-)

solution reaction 2 stages slightly acidicpH=6,4 , since the process of dissociation of dihydroorthophosphate ions prevails over the process of hydrolysis, while hydrogen ions not only neutralize hydroxide ions, but also remain in excess, which causes a weakly acid reaction of the medium.

Task: Determine the medium of sodium bicarbonate and sodium hydrosulfite solutions.

Solution:

1) Consider the processes in a solution of sodium bicarbonate. The dissociation of this salt proceeds in two stages, hydrogen cations are formed in the second stage:

NaHCO 3 \u003d Na + + HCO 3 - (I)

HCO 3 - ↔ H + + CO 3 2- ( II )

The dissociation constant for the second stage is K 2 of carbonic acid, equal to 4.8∙10 -11.

The hydrolysis of sodium bicarbonate is described by the equation:

NaHCO 3 + H 2 O ↔ H 2 CO 3 + NaOH

HCO 3 - + H 2 O ↔H 2 CO 3 + OH -, whose constant is

K g \u003d K w / K 1 (H 2 CO 3) \u003d 1 ∙ 10 -14 / 4.5 ∙ 10 -7 \u003d 2.2 ∙ 10 -8.

The hydrolysis constant is noticeably larger than the dissociation constant, therefore solutionNaHCO 3 has an alkaline environment.

2) Consider the processes in a solution of sodium hydrosulfite. The dissociation of this salt proceeds in two stages, hydrogen cations are formed in the second stage:

NaHSO 3 \u003d Na + + HSO 3 - (I)

HSO 3 - ↔ H + + SO 3 2- (II)

The dissociation constant for the second stage is K 2 of sulfurous acid, equal to 6.2∙10 -8.

The hydrolysis of sodium hydrosulfite is described by the equation:

NaHSO 3 + H 2 O ↔H 2 SO 3 + NaOH

HSO 3 - + H 2 O ↔H 2 SO 3 + OH -, whose constant is

K g \u003d K w / K 1 (H 2 SO 3) \u003d 1 ∙ 10 -14 / 1.7 ∙ 10 -2 \u003d 5.9 ∙ 10 -13.

In this case, the dissociation constant is greater than the hydrolysis constant, so solution

NaHSO 3 has an acidic environment.

Task: Determine the medium of the ammonium cyanide salt solution.

Solution:

NH 4 CN ↔NH 4 + + CN -

NH 4 + + 2H 2 O ↔NH 3. H 2 O + H 3 O +

CN - + H 2 O ↔HCN + OH -

NH 4 CN + H 2 O↔ NH 4 OH + HCN

K d (HCN) =7.2∙10 -10; K d (NH 4 OH) \u003d 1.8 ∙ 10 -5

Answer: Hydrolysis by cation and anion, because K o > K k, slightly alkaline, pH > 7