Hydrolysis k2s. Hydrolysis of potassium sulfide Hydrolysis of the salt is expressed by the equation

Class: 11

Goal: To create conditions for awareness and comprehension of new information, to provide an opportunity to apply the acquired theoretical knowledge in practice.

Lesson type: combined:

Methods: reproductive, partially search (heuristic), problem-based, laboratory work, explanatory and illustrative.

The end result of training.

Need to know:

1. The concept of hydrolysis.
2. 4 cases of hydrolysis.
3. Rules of hydrolysis.

You must be able to:

1. Draw up hydrolysis schemes.
2. Predict the nature of the medium and the effect of the indicator on a given salt solution based on the composition of the salt.

During the classes

Ι. Organizing time.

Didactic task: creating a psychological climate

- Hello! Take a mood sheet and mark your mood at the beginning of the lesson. Annex 1

Smile! OK, thank you.

II. Preparing to learn new material.

The epigraph of our lesson will be the words Kozma Prutkova

Always stay alert.

III. Updating students' knowledge.

But first, let's remember: the classification of electrolytes, writing the dissociation equations of electrolytes. (At the board, three people complete the task using cards.)

Frontal class survey on the following questions:

1. What substances are called electrolytes?
2. What do we call the degree of electrolytic dissociation?
3. What substances are called acids from the point of view of TED?
4. What substances are called bases from the point of view of TED?
5. What substances are called salts from the point of view of TED?
6. What substances are called ampholytes?
7. What reactions are called neutralization reactions?

We check the answers at the board. (Announce grades.)

Okay, now remember what indicators are? What indicators do you know?

How do they change color in solutions of acids and alkalis? Let's check the answers with the table.

Discussion of experience. (Hang the laboratory experiment table on the board.Annex 3 (II))

Does sodium carbonate solution work on indicators?

Use colored paper to show how the color of the indicators changes. (One student from the 1st row at the board.)

Does aluminum sulfate solution work on indicators?

(One student from the 2nd row at the board completes the previous task for aluminum sulfate solution).

Does sodium chloride solution work on indicators?

(Using colored paper, show in a table on the board the change in color of the indicator).

Fill out the same table in the worksheet for everyone. Appendix 3 (II)

Now compare the two tables on the board and draw a conclusion about the nature of the environment of the proposed salts.

ΙV. Learning new material.

Why can there be very different environments in salt solutions?

The topic of our lesson today will help answer this question. What do you think will be discussed? ( Students determine the topic of the lesson).

Let's try to decipher the word "HYDRO - LIZ". Comes from two Greek words “hydor” - water, “lysis” - decomposition, decay. (Formulate your own definitions)

HYDROLYSIS OF SALT is a reaction of ion exchange interaction of salts with water, leading to their decomposition.

In this lesson, what will we learn? ( Together with the students, we formulate the main goal of the lesson).

What is hydrolysis? Let’s get acquainted with four cases of hydrolysis and the rules of hydrolysis. Let's learn how to draw up hydrolysis schemes, predict the nature of the medium from the composition of the salt and the effect of the indicator on a given salt solution.

The salt dissociates into ions, and the resulting ions interact with water ions.

Let's turn to the salt, Na 2 CO 3, as a result of the interaction of which base and which acid, a salt was formed? (NaOH + H 2 CO 3).

Let us recall the classification of electrolytes

NaOH is a strong electrolyte, and H 2 CO 3 is a weak one. What is the nature of the medium of this salt? What conclusion can be drawn?

As a result of the interaction, what base and what acid formed a salt - AI 2 (SO 4) 3? (AI(OH) 3 + H 2 SO 4). Where is the weak and where is the strong electrolyte? What conclusion do we draw?

As a result of the interaction of which base and which acid, a salt was formed - NaCI? (NaOH + HCI). Determine the strength of these electrolytes.

What pattern did you notice? Record your findings on the worksheets.

An example of which case of hydrolysis was not given in a laboratory experiment? ( When a salt is formed by a weak base and a weak acid.) What is the nature of the environment in this case?

Record your findings on the worksheets. Appendix 3 (III). Say them again.

According to the direction of hydrolysis reactions, they can be divided into reversible and irreversible

According to the algorithm, they must learn to draw up diagrams of hydrolysis equations. ( Appendix 4).

Let's look at the example of salt, K 2 S - teacher at the blackboard.

As a result of the interaction, what base and what acid is this salt formed? Let's make a note:

1. K 2 S→KOH strong
↓
H 2 S weak

What is the nature of the medium of this salt?

2. Write the salt dissociation equation: K 2 S↔2K + + S 2-

3. We emphasize the weak electrolyte ion.

4. We write down the ion of a weak electrolyte from a new line, add HOH to it, put a sign ↔ write the ion OH - , because alkaline environment.

5. We put a “+” sign and write down an ion consisting of a salt ion S 2– and an ion remaining from a water molecule – NS -.

We write the final hydrolysis equation:

K 2 S + H 2 O ↔ KOH + KHS

What was formed as a result of hydrolysis? So why is the nature of the environment of this salt alkaline?

Record hydrolysis of ZnCl 2, (all independently in notebooks, one student at the blackboard).

Let's look at the textbook example Al 2 S 3.( p.150)

When is the hydrolysis scheme not written down? (For salts with a neutral environment.)

And so we analyzed four cases of hydrolysis.

We got acquainted with the rules of hydrolysis: this is a reversible process,

a special case of an ion exchange reaction, hydrolysis Always leaks by cation or anion weak electrolyte.

We learned to draw up hydrolysis schemes, predict the nature of the medium from the composition of the salt and the effect of the indicator on a given salt solution.

Using the algorithm, independently draw up salt hydrolysis schemes. ( Appendix 3 (IV)

After completion, we check the neighbor’s task and evaluate the work.

Physical education minute

V. Consolidation of the studied material

On the worksheet you have questions to consolidate, we will answer them. ( Appendix 3 (V)).

Guys, please note that this topic appears in the Unified State Exam assignment in all three parts. Let's look at a selection of tasks and determine how difficult the questions in these tasks are? ( Appendix 5).

What is the importance of hydrolysis of organic substances in industry?

Obtaining hydrolytic alcohol and obtaining soap. ( Student message)

Guys, remember what goals we had?

Have we achieved them?

What conclusion of the lesson will we draw?

LESSON CONCLUSIONS.

1. If a salt is formed by a strong base and a strong acid, then hydrolysis does not occur in the salt solution, because no ion binding occurs. The indicators do not change their color.

2. If a salt is formed by a strong base and a weak acid, then hydrolysis occurs along the anion. The environment is alkaline.

3. If a salt is formed by neutralizing a weak metal base with a strong acid, then hydrolysis occurs along the cation. The environment is acidic.

4. If a salt is formed by a weak base and a weak acid, then hydrolysis can occur at both the cation and the anion. The indicators do not change their color. The environment depends on the degree of dissociation of the resulting cation and anion.

V. Reflection.

Mark your mood at the end of the lesson on the mood scale. (Annex 1)

Has your mood changed? How do you evaluate the knowledge gained, on the back you will find an anonymous, monosyllable answer to 6 questions.

1. Are you satisfied with how the lesson went?
2. Were you interested?
3. Were you active in class?
4. Were you able to demonstrate your existing knowledge and acquire new ones?
5. Have you learned a lot of new things?
6. What did you like best?

VΙ. Homework.

• § 18, p. 154 No. 3, 8, 11, individual task cards.
• Study on your own how food hydrolysis occurs in the human body ( p.154).
• Find assignments on the topic “Hydrolysis” in the Unified State Examination materials 2009-2012 and complete them in your notebook.

Hydrolysis is a chemical interaction of dissolved salt ions with water, leading to the formation of weakly dissociating products (molecules of weak acids or bases, acid anions or basic salt cations) and accompanied by a changepH of the environment.
1. Na3 P.O.4 it is a salt of a strong base (alkali) NaOH and a medium acid (phosphoric) H3PO4. Salt hydrolysis occurs according to the anionic type, because The Na+ cation, binding to the hydroxyl anion OH¯, forms a strong electrolyte NaOH, which dissociates into ions.
Phosphoric tribasic acid forms three types of salts:
NaH2PO4 – primary Na phosphate, highly soluble
Na2HPO4 – secondary Na phosphate, practically insoluble
Na3PO4 is a tertiary Na phosphate, practically insoluble.
From this it is clear that during the hydrolysis of Na3PO4, i.e. the reaction proceeding until the formation of a weakly dissociating (poorly soluble) salt will form secondary sodium phosphate Na2HPO4.
1st stage
Ionic-molecular equation
PO4¯³ + H2O ↔ HPO4¯² + OH¯
Molecular equation:
Na3PO4 + H2O ↔ Na2HPO4 + NaOH
2nd stage
Ionic-molecular equation
Na2HPO4 + H2O↔ H2PO4¯² +OH¯
Molecular equation
Na2HPO4 + H2O↔ NaH2PO4 + NaOH
3rd stage
Ionic-molecular equation
H2PO4¯+ H2O = H3PO4 + OH¯
Molecular equation
NaH2PO4 + H2O = H3PO4 + NaOH
Usually the reaction proceeds in the first stage, then hydroxyl ions OH¯ accumulate and prevent the reaction from proceeding to completion.
Since an acidic salt and a strong base (alkali) are formed, the reaction of the solution will be alkaline, i.e. pH>7.
2.SaltK2 S, potassium sulfide is a salt of a strong base and a weak hydrofluoric acid H2S. Hydrolysis of the salt will occur in two stages, because Hydrogen sulfide acid is dibasic, of the anionic type. Salt K2S, when dissolved in water, dissociates into the K+ cation and the sulfide anion S¯². The K+ cation cannot bind the hydroxyl anion, because in this case, a strong electrolyte KOH is formed, which immediately dissociates into ions, and the sulfide anion S¯² of a weak acid binds to the hydroxyl group into a weakly dissociating compound.
1st stage

S¯² + H2O = HS¯ + OH¯
Molecular equation
K2S + H2O = KHS + KOH
2nd stage
Ionic-molecular equation
HS¯ + H2O = H2S + OH¯
Molecular equation
KHS + H2O = H2S + KOH
Hydrolysis proceeds in the first stage with the formation of a highly alkaline reaction, pH>7.

3. CuSO4, copper sulfate– salt of a strong acid and a weak polyacid base. Cu(OH)2. Hydrolysis of the salt will proceed with the formation of cations of the main salt CuOH+.
1st stage
Ionic-molecular equation
Cu+² + H2O↔ CuOH+ + H+
Molecular equation
CuSO4+ H2O ↔ (CuOH)2SO4 + H2SO4
The reaction will not proceed at stage 2 due to the resulting excess hydrogen ions of strong sulfuric acid. The medium is acidic, pH<7.

Hydrolysis is the interaction of a salt with water, as a result of which the hydrogen ions of water combine with the anions of the acidic residue of the salt, and the hydroxyl ions combine with the metal cation of the salt. This produces an acid (or acid salt) and a base (basic salt). When drawing up hydrolysis equations, it is necessary to determine which salt ions can bind water ions (H + or OH -) into a weakly dissociating compound. These can be either weak acid ions or weak base ions.

Strong bases include alkalis (bases of alkali and alkaline earth metals): LiOH, NaOH, KOH, CsOH, FrOH, Ca(OH) 2, Ba(OH) 2, Sr(OH) 2, Ra(OH) 2. The remaining bases are weak electrolytes (NH 4 OH, Fe(OH) 3, Cu(OH) 2, Pb(OH) 2, Zn(OH) 2, etc.).

Strong acids include HNO 3, HCl, HBr, HJ, H 2 SO 4, H 2 SeO 4, HClO 3, HCLO 4, HMnO 4, H 2 CrO 4, H 2 Cr 2 O 7. The remaining acids are weak electrolytes (H 2 CO 3, H 2 SO 3, H 2 SiO 3, H 2 S, HCN, CH 3 COOH, HNO 2, H 3 PO 4, etc.). Since strong acids and strong bases completely dissociate into ions in solution, only ions of acid residues of weak acids and metal ions that form weak bases can combine with water ions into weakly dissociating compounds. These weak electrolytes, by binding and retaining H + or OH - ions, upset the balance between water molecules and its ions, causing an acidic or alkaline reaction of the salt solution. Therefore, those salts that contain weak electrolyte ions, i.e., undergo hydrolysis. salts formed:

1) a weak acid and a strong base (for example, K 2 SiO 3);

2) a weak base and a strong acid (for example, CuSO 4);

3) a weak base and a weak acid (for example, CH 3 COONH 4).

Salts of a strong acid and a strong base do not undergo hydrolysis (for example, KNO 3).

Ionic equations for hydrolysis reactions are compiled according to the same rules as ionic equations for ordinary exchange reactions. If the salt is formed by a polyacidic weak acid or a polyacidic weak base, then hydrolysis proceeds stepwise with the formation of acidic and basic salts.

Examples of problem solving

Example 1. Hydrolysis of potassium sulfide K 2 S.

Stage I of hydrolysis: weakly dissociating ions HS - are formed.

Molecular form of the reaction:

K2S+H2O=KHS+KOH

Ionic equations:

Full ionic form:

2K + +S 2- +H 2 O=K + +HS - +K + +OH -

Abbreviated ionic form:

S 2- +H 2 O=HS - +OH -

Because As a result of hydrolysis, an excess of OH - ions is formed in the salt solution, then the reaction of the solution is alkaline pH>7.

Stage II: weakly dissociating H 2 S molecules are formed.

Molecular form of the reaction

KHS+H 2 O=H 2 S+KOH

Ionic equations

Full ionic form:

K + +HS - +H 2 O=H 2 S+K + +OH -

Abbreviated ionic form:

HS - +H 2 O=H 2 S+OH -

The environment is alkaline, pH>7.

Example 2. Hydrolysis of copper sulfate CuSO 4.

Stage I of hydrolysis: weakly dissociating ions (CuOH) + are formed.

Molecular form of the reaction:

2CuSO 4 +2H 2 O= 2 SO 4 +H 2 SO 4

Ionic equations

Full ionic form:

2Cu 2+ +2SO 4 2- +2H 2 O=2(CuOH) + +SO 4 2- +2H + +SO 4 2-

Abbreviated ionic form:

Cu 2+ +H 2 O=(CuOH) + +H +

Because As a result of hydrolysis in a salt solution, an excess of H + ions is formed, then the reaction of the solution is acidic pH<7.

Stage II of hydrolysis: weakly dissociating Cu(OH) 2 molecules are formed.

Molecular form of the reaction

2 SO 4 +2H 2 O=2Cu(OH) 2 +H 2 SO 4

Ionic equations

Full ionic form:

2(CuOH) + +SO 4 2- +2H 2 O= 2Cu(OH) 2 +2H + +SO 4 2-

Abbreviated ionic form:

(CuOH) + +H 2 O=Cu(OH) 2 +H +

Acidic medium, pH<7.

Example 3. Hydrolysis of lead acetate Pb(CH 3 COO) 2.

Stage I of hydrolysis: weakly dissociating ions (PbOH) + and weak acid CH 3 COOH are formed.

Molecular form of the reaction:

Pb(CH 3 COO) 2 +H 2 O=Pb(OH)CH 3 COO+CH 3 COOH

Ionic equations

Full ionic form:

Pb 2+ +2CH 3 COO - +H 2 O=(PbOH) + +CH 3 COO - +CH 3 COOH

Abbreviated ionic form:

Pb 2+ +CH 3 COO - +H 2 O=(PbOH) + +CH 3 COOH

When the solution is boiled, hydrolysis is almost complete, and a precipitate of Pb(OH) 2 is formed

II stage of hydrolysis:

Pb(OH)CH 3 COO+H 2 O=Pb(OH) 2 +CH 3 COOH

Hydrolysis of salts- this is the chemical interaction of salt ions with water ions, leading to the formation of a weak electrolyte.

If we consider a salt as a product of neutralization of a base with an acid, then we can divide the salts into four groups, for each of which hydrolysis will proceed in its own way.

1). Hydrolysis is not possible

A salt formed by a strong base and a strong acid ( KBr, NaCl, NaNO3), will not undergo hydrolysis, since in this case a weak electrolyte is not formed.

pH of such solutions = 7. The reaction of the medium remains neutral.

2). Hydrolysis by cation (only the cation reacts with water)

In a salt formed by a weak base and a strong acid ( FeCl2,NH4Cl, Al 2 (SO 4) 3, MgSO4) the cation undergoes hydrolysis:

FeCl2 + HOH<=>Fe(OH)Cl + HCl
Fe 2+ + 2Cl - + H + + OH -<=>FeOH + + 2Cl - + N +

As a result of hydrolysis, a weak electrolyte, H + ion and other ions are formed.

solution pH< 7 (раствор приобретает кислую реакцию).

3).Hydrolysis by anion (only the anion reacts with water)

A salt formed by a strong base and a weak acid ( KClO, K2SiO3, Na 2 CO 3, CH 3 COONa) undergoes hydrolysis at the anion, resulting in the formation of a weak electrolyte, the hydroxide ion OH - and other ions.

K 2 SiO 3 + HOH<=>KHSiO 3 + KOH
2K + +SiO 3 2- + H + + OH -<=>NSiO 3 - + 2K + + OH -

The pH of such solutions is > 7 (the solution becomes alkaline).

4). Joint hydrolysis (both the cation and the anion react with water)

A salt formed by a weak base and a weak acid ( CH 3 COONH 4, (NH 4) 2 CO 3, Al 2 S 3), hydrolyzes both the cation and the anion. As a result, a slightly dissociating base and acid are formed. The pH of solutions of such salts depends on the relative strength of the acid and base. A measure of the strength of an acid and a base is the dissociation constant of the corresponding reagent.

The reaction of the medium of these solutions can be neutral, slightly acidic or slightly alkaline:

Al 2 S 3 + 6H 2 O =>2Al(OH) 3 ↓+ 3H 2 S

Hydrolysis is a reversible process.

Hydrolysis is irreversible if the reaction results in the formation of an insoluble base and (or) a volatile acid

Algorithm for composing salt hydrolysis equations

Na 2 CO 3 + HOH ↔ NaHCO 3 + NaOH

Practical use.

In practice, the teacher has to deal with hydrolysis, for example, when preparing solutions of hydrolyzed salts (lead acetate, for example). The usual “method”: water is poured into the flask, salt is added, and shaken. A white precipitate remains. Add more water, shake, the sediment does not disappear. We add hot water from the kettle - there seems to be even more sediment... And the reason is that, simultaneously with dissolution, hydrolysis of the salt occurs, and the white precipitate that we see is already the products of hydrolysis - poorly soluble basic salts. All our further actions, dilution, heating, only increase the degree of hydrolysis. How to suppress hydrolysis? Do not heat, do not prepare too dilute solutions, and since hydrolysis at the cation mainly interferes, add acid. Better than the corresponding one, that is, vinegar.

Hydrolysis plays an important role in the process of deferrization of water by aeration. When water is saturated with oxygen, the iron(II) bicarbonate contained in it is oxidized to an iron(III) salt, which is much more susceptible to hydrolysis. As a result, complete hydrolysis occurs and iron is separated in the form of a precipitate of iron(III) hydroxide.

This is also the basis for the use of aluminum salts as coagulants in water purification processes. Aluminum salts added to water in the presence of bicarbonate ions are completely hydrolyzed and bulky aluminum hydroxide coagulates, carrying various impurities with it into the sediment."Increased hydrolysis of salts when heated"

ASSIGNMENT TASKS

№1.Write down the equations for the hydrolysis of salts and determine the medium of aqueous solutions (pH) and the type of hydrolysis:
Na 2 SiO 3, AlCl 3, K 2 S.

No. 2. Draw up equations for the hydrolysis of salts, determine the type of hydrolysis and the solution medium:
Potassium sulfite, sodium chloride, iron (III) bromide

No. 3. Create hydrolysis equations, determine the type of hydrolysis and the medium of an aqueous salt solution for the following substances:
Potassium sulfide - K 2 S, Aluminum bromide - AlBr 3, Lithium chloride - LiCl, Sodium phosphate - Na 3 PO 4, Potassium sulfate - K 2 SO 4, Zinc chloride - ZnCl 2, Sodium sulfite - Na 2 SO 3, Ammonium sulfate - (NH 4) 2 SO 4, Barium bromide - BaBr 2.

Task 201.
Compose ionic-molecular and molecular equations for hydrolysis that occurs when mixing solutions of K 2 S and CrC1 3 . Each of the salts taken is hydrolyzed irreversibly to the end with the formation of the corresponding base and acid.
Solution:
K 2 S - a salt of a strong base and a weak acid is hydrolyzed by the anion, and CrCl 3 - a salt of a weak base and a strong acid is hydrolyzed by the cation:

K 2 S ⇔ 2K + + S 2- ; CrCl3 ⇔ Cr 3+ + 3Cl - ;
a) S 2- + H 2 O ⇔ HS - + OH -;
b) Cr 3+ + H 2 O ⇔ CrOH 2+ + H +.

If solutions of these salts are in the same vessel, then there is a mutual enhancement of the hydrolysis of each of them, because the H+ and OH- ions, bonding with each other, form molecules of the weak electrolyte H 2 O (H + + OH - ⇔ H 2 O). With the formation of additional water, the hydrolytic equilibrium of both salts shifts to the right, and the hydrolysis of each salt proceeds to completion with the formation of a precipitate and gas:

3S 2- + 2Cr 3+ + 6H 2 O ⇔ 2Cr(OH) 3 ↓ + 3H 2 S (ionic molecular form);
3K 2 S + 2CrCl 3 + 6H 2 O ⇔ 2Cr(OH) 3 ↓ + 3H 2 S + 6KCl (molecular form).

Task 202.
The following substances were added to the FeCl 3 solution: a) HCl; b) CON; c) ZnCl 2; d) Na 2 CO 3. In what cases will hydrolysis of iron (III) chloride increase? Why? Write ionic-molecular equations for the hydrolysis of the corresponding salts.
Solution:
a) The FeCl 3 salt hydrolyzes into the cation, and HCl dissociates in an aqueous solution:

FeCl 3 ⇔ Fe 3+ + 3Cl - ;

HCl ⇔ H + + Cl -

If solutions of these substances are in the same vessel, then the hydrolysis of the FeCl 3 salt is inhibited, because an excess of hydrogen ions H + is formed and the hydrolysis equilibrium shifts to the left:
b) The FeCl 3 salt is hydrolyzed into the cation, and KOH dissociates in an aqueous solution to form OH -:

FeCl 3 ⇔ Fe 3+ + 3Cl - ;
Fe 3+ + H 2 O ⇔ FeOH 2+ + H + ;
KOH ⇔ K + + OH -

If solutions of these substances are in the same vessel, then hydrolysis of the FeCl3 salt and dissociation of KOH occurs, because the H+ and OH- ions, bonding with each other, form molecules of the weak electrolyte H 2 O (H + + OH - ⇔ H 2 O). In this case, the hydrolytic equilibrium of the FeCl 3 salt and the dissociation of KOH shift to the right and the hydrolysis of the salt and the dissociation of the base proceed to the end with the formation of a Fe(OH) 3 precipitate. Essentially, when FeCl3 and KOH are mixed, an exchange reaction occurs. Ionic

Fe 3+ + 3OH - ⇔ Fe(OH) 3 ↓;

Molecular equation of the process:

FeCl 3 + 3KOH ⇔ Fr(OH) 3 ↓ + 3KCl.

c) The FeCl 3 salt and the ZnCl 2 salt are hydrolyzed by the cation:

Fe 3+ + H 2 O ⇔ FeOH 2+ + H + ;
Zn 2+ + H 2 O ⇔ ZnOH + + H +

If solutions of these salts are in the same vessel, then there is mutual inhibition of the hydrolysis of each of them, because the excess amount of H + ions causes a shift in the hydrolytic equilibrium to the left, towards a decrease in the concentration of hydrogen ions H +.
d) The FeCl 3 salt is hydrolyzed by the cation, and the Na 2 CO 3 salt is hydrolyzed by the anion:

Fe 3+ + H 2 O ⇔ FeOH 2+ + H + ;
CO 3 2- + H 2 O ⇔ HСO 3 - + OH -

If solutions of these salts are in the same vessel, then there is a mutual enhancement of the hydrolysis of each of them, because the H + and OH - ions, bonding with each other, form molecules of the weak electrolyte H 2 O (H + + OH - ⇔ H 2 O). With the formation of an additional amount of water, the hydrolytic equilibrium of both salts shifts to the right, and the hydrolysis of each salt proceeds to completion with the formation of the Fe(OH)3↓ precipitate, a weak electrolyte H 2 CO 3:

2Fe 3+ + 3CO 3 2- + 3H 2 O ⇔ 2Fe(OH) 3 ↓ + 3CO 2 (ionic molecular form);
2FeCl 3 + 3Na 2 CO 3 + 3H 2 O ⇔ 2Fe(OH) 3 ↓ + 3CO 2 + 6NaCl (molecular form).

Task 203.
Which of the salts Al 2 (SO4) 3, K 2 S, Pb(NO 3) 2, KCl undergo hydrolysis? Write ionic-molecular and molecular equations for the hydrolysis of the corresponding salts. What is the pH value (> 7 <) have solutions of these salts?
Solution:

a) Al 2 (SO 4) 3 is a salt of a weak base and a strong acid. In this case, Al 3+ cations bind OH - water ions, forming cations of the main salt AlOH 2+. The formation of Al(OH) 2+ and Al(OH) 3 does not occur because AlOH 2+ ions dissociate much more difficultly than Al(OH) 2+ ions and Al(OH) 3 molecules. Under normal conditions, hydrolysis occurs in the first stage. The salt hydrolyzes into the cation. Ionic-molecular hydrolysis equation:

Al2(SO 4) 3 ⇔ Al 3+ + 3SO 4 2-;

or in molecular form:

Al 2 (SO 4) 3 + 2H 2 O ⇔ 2AlOHSO 4 + H 2 SO 4

An excess of hydrogen ions appears in the solution, which gives the Al2(SO4)3 solution an acidic environment, pH< 7 .

b) K 2 S – salt of strong monoacid base KOH and weak polybasic acid H 2 S. In this case, the S2- anions bind the hydrogen ions H+ of water, forming the anions of the acid salt HS-. The formation of H2S does not occur, since HS- ions dissociate much more difficultly than H2S molecules. Under normal conditions, hydrolysis occurs in the first stage. The salt is hydrolyzed at the anion. Ionic-molecular hydrolysis equation:

K 2 S ⇔ 2К + + S 2- ;
S 2- + H 2 O ⇔ H S- + OH -

or in molecular form:

K 2 S + 2H 2 O ⇔ KNS + KOH

An excess of hydroxide ions appears in the solution, which give the K 2 S solution an alkaline environment, pH > 7.

c) Pb(NO 3) 2 is a salt of a weak base and a strong acid. In this case, Pb 2+ cations bind OH-water ions, forming cations of the main salt PbOH +. The formation of Pb(OH) 2 does not occur because PbOH + ions dissociate much more difficultly than Pb(OH) 2 molecules. Under normal conditions, hydrolysis occurs in the first stage. The salt hydrolyzes into the cation. Ionic-molecular hydrolysis equation:

Pb 2+ + H 2 O ⇔ PbOH + + H +

or in molecular form:

< 7.

d) KCl - a salt of a strong base and a strong acid does not undergo hydrolysis, since the K + and Cl - ions are not bound by water ions H + and OH -. The K + , Cl - , H + and OH - ions will remain in solution. Since equal amounts of H + and OH - ions are present in a salt solution, the solution has a neutral environment, pH = 0.

Task 204.
When mixing solutions of FeCl 3 and Na 2 CO 3, each of the salts taken is hydrolyzed irreversibly to the end with the formation of the corresponding base and acid. Express this joint hydrolysis in terms of ionic and molecular equations.
Solution:
FeCl 3 is a salt of a weak base and a strong acid. In this case, Fe 3+ cations bind OH - water ions, forming cations of the main salt FeOH 2+. The formation of Fe(OH)2+ and Fe(OH)3 does not occur because FeOH 2+ ions dissociate much more difficultly than Fe(OH) 2+ ions and Fe(OH) 3 molecules. Under normal conditions, hydrolysis occurs in the first stage. The salt hydrolyzes into the cation. Ionic-molecular hydrolysis equation:

FeC l3 ⇔ Fe 3+ + 3Cl -
Fe 3+ + H 2 O ⇔ FeOH 2+ + H +

Na 2 CO 3 is a salt of a strong base and a weak acid. In this case, the CO 3 2- anions bind the hydrogen ions H + of water, forming the acid salt anions HCO 3 - . The formation of H 2 CO 3 does not occur, since HCO 3 ions dissociate much more difficultly than H 2 CO 3 molecules. Under normal conditions, hydrolysis occurs in the first stage. The salt is hydrolyzed at the anion. Ionic-molecular hydrolysis equation:

2Fe 3+ + 3CO 3 2- + 3H 2 O  2Fe(OH) 3 ⇔ + 3CO 2 (ionic molecular form);
2FeCl 3 + 3Na 2 CO 3 +3H 2 O ⇔ 2Fe(OH) 3  + + 3CO 2 + 6NaCl.

Task 205.
The following substances were added to the Na 2 CO 3 solution: a) HCl; b)NaOH; c) Cu(NO 3) 2; d) K 2 S. In what cases will the hydrolysis of sodium carbonate increase? Why? Write ionic-molecular equations for the hydrolysis of the corresponding salts.
Solution:

a) The Na 2 CO 3 salt is hydrolyzed at the anion, and HCl dissociates in an aqueous solution:

Na 2 CO 3 ⇔ 2Na + + CO 3 2- ;
CO 3 2- + H 2 O ⇔ HCO 3 - + OH - ;
HCl ⇔ H + + Cl -

If the solutions of these substances are in the same vessel, then there is a mutual enhancement of the hydrolysis of each of them, because the H + and OH - ions, bonding with each other, form molecules of the weak electrolyte H 2 O (H + + OH - ⇔ H 2 O). In this case, the hydrolytic equilibrium of the Na 2 CO 3 salt and the dissociation of HCl shift to the right and the hydrolysis of the salt and the dissociation of the acid proceed to completion with the formation of gaseous carbon dioxide. Ion-molecular equation of the process:

CO 3 2- + 2H + ⇔ CO 2 + H 2 O

Molecular equation of the process:

Na 2 CO 3 + 2HCl ⇔ 2NaCl + CO 2 + H 2 O

b) The Na 2 CO 3 salt is hydrolyzed at the anion, and NaOH dissociates in an aqueous solution:

NaOH ⇔ Na + + OH - .

If solutions of these substances are mixed, an excess of OH - ions is formed, which shifts the equilibrium of Na 2 CO 3 hydrolysis to the left and the hydrolysis of the salt will be inhibited.

c) The Na 2 CO 3 salt is hydrolyzed by the anion, and the Cu(NO 3) 2 salt is hydrolyzed by the cation:

CO 3 2- + H 2 O ⇔ HCO 3 - + OH - ;
Сu 2+ + H 2 O ⇔ CuOH + + H + .

If solutions of these salts are in the same vessel, then there is a mutual enhancement of the hydrolysis of each of them, because the H + and OH - ions, bonding with each other, form molecules of the weak electrolyte H 2 O (H + + OH - ⇔ H 2 O). With the formation of additional water, the hydrolytic equilibrium of both salts shifts to the right, and the hydrolysis of each salt proceeds to completion with the formation of a precipitate and gas:

Cu 2+ + CO 3 2- + H 2 O ⇔ Cu(OH) 2 ↓ + CO 2 (ionic molecular form);
Cu(NO 3) 2 + Na 2 CO 3 + H 2 O ⇔ Cu(OH) 2 ↓ + CO 2 + 2NaNO 3 (molecular form).

d) Na 2 CO 3 and K 2 S are salts of a strong base and a weak acid, therefore both are hydrolyzed at the anion:

CO 3 2- + H 2 O ⇔ HCO 3 - + OH - ;
S 2- + H 2 O ⇔ HS - + OH - .

If solutions of these salts are in the same vessel, then there is mutual inhibition of the hydrolysis of each of them, because an excess of OH - ions, according to Le Chatelier’s principle, shifts the equilibrium of the hydrolysis of both salts to the left, towards a decrease in the concentration of OH - ions, i.e. hydrolysis of both salts will be inhibited.

Task 206.
What pH value (> 7<) имеют растворы солей Na 2 S, АlСl 3 , NiSO 4 ? Составьте ионно-молекулярные и молекулярные уравнения гидролиза этих солей.
Solution:
a) Na 2 S – salt of strong monoacid base NaOH and weak polybasic acid H 2 S. In this case, the S 2- anions bind the hydrogen ions H+ of water, forming the anions of the acid salt HS-. The formation of H 2 S does not occur, since HS - ions dissociate much more difficultly than H 2 S molecules. Under normal conditions, hydrolysis occurs in the first step. The salt is hydrolyzed at the anion. Ionic-molecular hydrolysis equation:

Na 2 S ⇔ 2Na + + S 2- ;
S 2- + H 2 O ⇔ NS - + OH -

or in molecular form:

Na 2 S + 2H 2 O ⇔ NaHS + KOH

An excess of hydroxide ions appears in the solution, which give the Na2S solution an alkaline environment, pH > 7.

b) AlCl 3 is a salt of a weak base and a strong acid. In this case, Al3+ cations bind OH- ions of water, forming cations of the main salt AlOH2+. The formation of Al(OH) 2+ and Al(OH) 3 does not occur because AlOH 2+ ions dissociate much more difficultly than Al(OH) 2+ ions and Al(OH) 3 molecules. Under normal conditions, hydrolysis occurs in the first stage. The salt hydrolyzes into the cation. Ionic-molecular hydrolysis equation:

AlCl 3 ⇔ Al 3+ + 3Cl - ;
Al 3+ + H 2 O ⇔ AlOH 2+ + H +

or in molecular form:

AlCl 3 + H 2 O ⇔ 2AlOHCl 2 + HCl

An excess of hydrogen ions appears in the solution, which gives the Al2(SO4)3 solution an acidic environment, pH< 7.

c) NiSO4 is a salt of a weak polyacid base Ni(OH)2 and a strong dimonobasic acid H2SO4. In this case, Ni2+ cations bind OH- ions of water, forming cations of the main salt NiOH+. The formation of Ni(OH)2 does not occur because NiOH+ ions dissociate much more difficultly than Ni(OH)2 molecules. Under normal conditions, hydrolysis occurs in the first stage. The salt hydrolyzes into the cation. Ionic-molecular hydrolysis equation:

Ni(NO 3) 2 ⇔ Ni 2+ + 2NO 3 - ;
Ni 2+ + H 2 O ⇔ NiOH + + H +

or in molecular form:

2NiSO 4 + 2H 2 O  (NiOH) 2 SO 4 + H 2 SO 4

An excess of hydrogen ions appears in the solution, which gives the NiSO 4 solution an acidic environment, pH< 7.

Task 207.
Make up ion-molecular and molecular equations for the hydrolysis of salts Pb(NO 3) 2, Na 2 CO 3, Fe 2 (SO 4) 3. What pH value (> 7<) имеют растворы этих солей?
Solution:
a) Pb(NO 3) 2 is a salt of a weak base and a strong acid. In this case, Pb 2+ cations bind OH - water ions, forming cations of the main salt PbOH +. The formation of Pb(OH) 2 does not occur because PbOH + ions dissociate much more difficultly than Pb(OH) 2 molecules. Under normal conditions, hydrolysis occurs in the first stage. The salt hydrolyzes into the cation. Ionic-molecular hydrolysis equation:

Pb(NO 3) 2 ⇔ Pb 2+ + 2NO 3 - ;
Pb 2+ + H 2 O ⇔ PbOH + + H +

or in molecular form:

Pb(NO 3) 2 + H 2 O ⇔ PbOHNO 3 + HNO 3

An excess of hydrogen ions appears in the solution, which give the Pb(NO 3) 2 solution an acidic environment, pH< 7.

b) Na 2 CO 3 is a salt of a strong base and a weak acid. In this case, the CO 3 2- anions bind the hydrogen ions H + of water, forming the acid salt anions HCO 3 - . The formation of H 2 CO 3 does not occur, since HCO 3 ions dissociate much more difficultly than H 2 CO 3 molecules. Under normal conditions, hydrolysis occurs in the first stage. The salt is hydrolyzed at the anion. Ionic-molecular hydrolysis equation:

Na 2 CO 3 ⇔ 2Na + + CO 3 2- ;
CO 3 2- + H 2 O ⇔ HCO 3 - + OH -

or in molecular form:

Na 2 CO 3 + H 2 O ⇔ CO 2 + 2NaOH

An excess of hydroxide ions appears in the solution, which give the Na2CO3 solution an alkaline environment, pH > 7.

c) Fe 2 (SO 4) 3 is a salt of a weak base and a strong acid. In this case, Fe 3+ cations bind OH - water ions, forming cations of the main salt FeOH 2+. The formation of Fe(OH) 2+ and Fe(OH) 3 does not occur because FeOH 2+ ions dissociate much more difficultly than Fe(OH) 2+ ions and Fe(OH) 3 molecules. Under normal conditions, hydrolysis occurs in the first stage. The salt hydrolyzes into the cation. Ionic-molecular hydrolysis equation:

Fe 2 (SO 4) 3 ⇔ 2Fe 3+ + 3SO 4 2 -
Fe 3+ + H 2 O ⇔ FeOH 2+ + H +

Molecular form of the process:

Fe 2 (SO 4) 3 + 2H 2 O ⇔ 2FeOHSO 4 + H 2 SO 4.

An excess of hydrogen ions appears in the solution, which gives the Fe2(SO4)3 solution an acidic environment, pH< 7.

Task 208.
Make up ion-molecular and molecular equations for the hydrolysis of salts HCOOOK, ZnSO 4, Al(NO 3) 3. What pH value (> 7<) имеют растворы этих солей?
Solution:
a) NSOOC – salt of strong monoacid base KOH and weak monobasic acid UNDC. In this case, the HCOO - anions bind the hydrogen ions H + of water, forming a weak electrolyte HCOOH. Ionic-molecular hydrolysis equation:

NSOOK ⇔ K + + NSOOK - ;
НСОО - + H 2 O ⇔ НСООН + ОH -

or in molecular form:

HCOOC + H 2 O  HCOOH + KOH

An excess of hydroxide ions appears in the solution, which give the HCOOO solution an alkaline environment, pH > 7.

b) ZnSO 4 is a salt of a weak polyacid base Zn(OH)2 and a strong polyacid. In this case, Zn 2+ cations bind OH - water ions, forming cations of the main salt ZnOH +. The formation of Zn(OH) 2 does not occur because CoOH + ions dissociate much more difficultly than Zn(OH) 2 molecules. Under normal conditions, hydrolysis occurs in the first stage. The salt hydrolyzes into the cation. Ionic-molecular hydrolysis equation:

ZnSO 4  Zn 2+ + SO 4 2- ;
Zn 2+ + H 2 O  ZnOH + + H +

or in molecular form:

2ZnSO4 + 2H2O  (ZnOH)2SO4 + H2SO4

An excess of hydrogen ions appears in the solution, which give the ZnSO 4 solution an acidic environment, pH< 7.

c) Al(NO 3) 3 - salt of weak polyacid base Al(OH) 3 and strong monobasic acid HNO3. In this case, Al 3+ cations bind OH - water ions, forming cations of the main salt AlOH2+. The formation of Al(OH) 2+ and Al(OH) 3 does not occur because AlOH 2+ ions dissociate much more difficultly than Al(OH) 2+ ions and Al(OH) 3 molecules. Under normal conditions, hydrolysis occurs in the first stage. The salt hydrolyzes into the cation. Ionic-molecular hydrolysis equation:

Al(NO3) 3 ⇔ Cr 3+ + 3NO 3 -
Al 3+ + H 2 O ⇔ AlOH 2+ + H +

Al(NO 3) 3 + H 2 O ⇔ AlOH(NO 3) 2 + HNO 3

< 7.

Task 209.
What pH value (> 7<) имеют растворы солей Na 3 PO 4 , K 2 S, CuSO 4 ? Составьте ионно-молекулярные и молекулярные уравнения гидролиза этих солей.
Solution:
a) Sodium orthophosphate Na 3 PO 4 is a salt of a weak polybasic acid H 3 PO 4 and a strong one-acid base. In this case, the anions PO 4 3- bind the hydrogen ions H + of water, forming the anions of the acid salt HPO 4 2- . The formation of H 2 PO 4 - and H 3 PO 4 does not occur, since HPO 4 2 - ions dissociate much more difficultly than H 2 PO 4 - ions and H 3 PO 4 molecules. Under normal conditions, hydrolysis occurs in the first stage. The salt is hydrolyzed at the anion. Ionic-molecular hydrolysis equation:

Na 3 PO 4 ⇔ 3Na + + PO 4 3- ;
PO 4 3- + H 2 O ⇔ HPO 4 2- + OH -

or in molecular form:

Na 3 PO 4 + H 2 O ⇔ Na 2 HPO 4 + NaOH

An excess of hydroxide ions appears in the solution, which give the Na 3 PO 4 solution an alkaline environment, pH > 7.

b) K2S is a salt of a strong monoacid base KOH and a weak polyacid acid H 2 S. In this case, S 2- anions bind hydrogen ions H + of water, forming acid salt anions HS -. The formation of H 2 S does not occur, since HS - ions dissociate much more difficultly than H 2 S molecules. Under normal conditions, hydrolysis occurs in the first step. The salt is hydrolyzed at the anion. Ionic-molecular hydrolysis equation:

K 2 S ⇔ 2К + + S 2- ;
S 2- + H 2 O ⇔ NS - + OH -

or in molecular form:

K2S + 2H 2 O ⇔  KNS + KOH

An excess of hydroxide ions appears in the solution, which give the K2S solution an alkaline environment, pH > 7.

c) CuSO 4 is a salt of a weak base and a strong acid. In this case, Cu 2+ cations bind OH - water ions, forming cations of the main salt CuOH +. The formation of Cu(OH) 2 does not occur because CuOH + ions dissociate much more difficultly than Cu(OH) 2 molecules. Under normal conditions, hydrolysis occurs in the first stage. The salt hydrolyzes into the cation. Ionic-molecular hydrolysis equation:

CuSO 4 ⇔ Cu 2+ + SO 4 2- ;
Cu 2+ + H 2 O ⇔ CuOH + + H +

or in molecular form:

2CuSO 4 + 2H 2 O ⇔ (CuOH) 2 SO 4 + H 2 SO 4

An excess of hydrogen ions appears in the solution, which gives the CuSO 4 solution an acidic environment, pH< 7.

Task 210.
Make up ion-molecular and molecular equations for the hydrolysis of salts CuCl 2, Cs 2 CO 3, Cr(NO 3) 3. What pH value (> 7<) имеют растворы этих солей?
Solution:
a) CuCl 2 is a salt of a weak polyacid base Cu(OH) 2 and a strong monobasic acid HCl. In this case, Cu 2+ cations bind OH - water ions, forming cations of the main salt CuOH +. The formation of Cu(OH) 2 does not occur because CuOH + ions dissociate much more difficultly than Cu(OH) 2 molecules. Under normal conditions, hydrolysis occurs in the first stage. The salt hydrolyzes into the cation. Ionic-molecular hydrolysis equation:

CuCl 2 ⇔ Cu 2+ + 2Cl - ;
Cu 2+ + H 2 O ⇔ CuOH + + H +

or in molecular form:

CuCl 2 + H 2 O ⇔ CuOHCl + HCl

An excess of hydrogen ions H+ appears in the solution, which give the CuCl 2 solution an acidic environment, pH< 7.

b) Cs 2 CO 3 - a salt of a strong monoacid base CsOH and a weak dibasic acid H 2 CO 3. In this case, the CO 3 2- anions bind the hydrogen ions H + of water, forming the acid salt anions HCO 3 - . The formation of H 2 CO 3 does not occur, since HCO 3 ions dissociate much more difficultly than H 2 CO 3 molecules. Under normal conditions, hydrolysis occurs in the first stage. The salt is hydrolyzed at the anion. Ionic-molecular hydrolysis equation:

Cs 2 CO 3 ⇔ 2Cs + + CO 3 2- ;
CO 3 2- + H 2 O ⇔ HCO 3 - + OH -

or in molecular form:

Cs2CO 3 + H 2 O ⇔ CO 2 + 2CsOH

An excess of hydroxide ions appears in the solution, which give the Cs2CO3 solution an alkaline environment, pH > 7.

c) Cr(NO 3) 3 - a salt of a weak polyacid base Cr(OH) 3 and a strong monobasic acid HNO 3. In this case, Cr 3+ cations bind OH - water ions, forming cations of the main salt CrOH 2+. The formation of Cr(OH) 2 + and Cr(OH) 3 does not occur because CrOH 2+ ions dissociate much more difficultly than Cr(OH) 2 + ions and Cr(OH) 3 molecules. Under normal conditions, hydrolysis occurs in the first stage. The salt hydrolyzes into the cation. Ionic-molecular hydrolysis equation:

Cr(NO 3) 3 ⇔ Cr 3+ + 3NO 3 -
Cr 3+ + H 2 O ⇔ CrOH 2+ + H +

Molecular equation of the reaction:

Cr(NO 3) 3 + H 2 O ⇔ CrOH(NO 3) 2 + HNO 3

An excess of hydrogen ions appears in the solution, which gives the Cr(NO 3) 3 solution an acidic environment, pH< 7.