Class: 11

Purpose: To create conditions for understanding and comprehending new information, to provide an opportunity to apply the acquired theoretical knowledge in practice.

• Training:

formation of basic concepts (hydrolysis, classification of salts according to the strength of the acid and the base of their constituents, types of hydrolysis) on a differentiated basis; the formation of skills to write down the equations of hydrolysis reactions in molecular, complete and abbreviated form, to predict and explain the change in the solution environment, the formation of acidic and basic salts.

Developing:

the development of creative thinking, experimental skills, the development of the ability to put forward a hypothesis, test it, establish patterns, look for new facts that would confirm the correctness of the hypothesis put forward, the development of the emotional sphere of students, cognitive activity, the ability to observe the world around them, to think about the inner essence of what they see.

Educational:

the formation of skills to apply the studied material in practical situations, to defend one's beliefs, to work in a group.

Lesson type: combined:

Methods: reproductive, partially search (heuristic), problematic, laboratory work, explanatory - illustrative.

The end result of learning.

Need to know:

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

You need to be able to:

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

During the classes

Ι. Organizing time.

Didactic task: creating a psychological climate

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

Smile! Well thank you.

II. Preparing to learn new material.

The epigraph of our lesson will be the words Kozma Prutkov

Always stay alert.

III. Updating students' knowledge.

But first, let's remember: the classification of electrolytes, the recording of the equations of dissociation of electrolytes. (At the board, three people complete the task on the 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 in terms of TED?
4. What substances are called bases in terms of TED?
5. What substances are called salts in terms 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, alkalis? Let's check the answers with the table.

Discussion of experience. (Hang up a table of laboratory experiments on the board.Annex 3(II))

Does sodium carbonate solution work on indicators?

Using colored paper, show how the color of the indicators changes. (One student from the 1st row at the blackboard.)

Does aluminum sulfate solution work on indicators?

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

Does sodium chloride solution work on indicators?

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

Fill in the same table in the worksheets for everyone. Annex 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 today's lesson will help answer this question. What do you think it will be about? ( The students decide the topic of the lesson.

Let's try to decipher the word "HYDRO - LIZ". It comes from two Greek words "hydor" - water, "lysis" - decomposition, decay. (Your own definitions)

HYDROLYSIS OF SALTS 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, the rules of hydrolysis. We will learn how to draw up hydrolysis schemes, predict the nature of the medium and the effect of the indicator on a given salt solution by the composition of the salt.

Salt dissociates into ions, and the resulting ions interact with water ions.

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

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 environment of this salt? What conclusion can be drawn?

As a result of the interaction, which base and which 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 do we conclude?

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

What pattern did you notice? Record findings on 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 findings on worksheets. Annex 3 (III). Speak them again.

According to the direction of the hydrolysis reaction can be divided into reversible and irreversible

According to the algorithm, they must learn to draw up schemes 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? Making a record:

1. K 2 S→KOH strong
↓
H 2 S weak

What is the nature of the environment of this salt?

2. We write down the salt dissociation equation: K 2 S↔2K + + S2-

3. We emphasize the ion of a weak electrolyte.

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

5. We put the “+” sign, write down the ion consisting of the salt ion S 2– and the ion remaining from the water molecule - HS -.

We write the final hydrolysis equation:

K 2 S + H 2 O ↔ KOH + KHS

What is 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).

Consider the textbook example Al 2 S 3 .( p.150)

When is a hydrolysis scheme not recorded? (For salts with a neutral character of the 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 how to draw up hydrolysis schemes, predict the nature of the medium and the effect of the indicator on a given salt solution by the composition of the salt.

Using the algorithm, independently draw up schemes for the hydrolysis of salts. ( Annex 3 (IV)

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

Physical education minute

V. Consolidation of the studied material

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

Guys, please note that this topic is found in the assignment for the exam in all three parts. Let's look at a selection of tasks and determine what complexity the questions in these tasks contain? ( Appendix 5).

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

Obtaining hydrolysis alcohol and obtaining soap. ( Student message)

Guys, remember what goals we had before us?

Have we reached them?

What is the conclusion of the lesson?

LESSON CONCLUSIONS.

1. If the salt is formed by a strong base and a strong acid, then hydrolysis in a salt solution does not occur, because ion binding does not occur. Indicators do not change their color.

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

3. If a salt is formed by neutralizing a weak base of a metal with a strong acid, then hydrolysis proceeds along the cation. Wednesday is acidic.

4. If the salt is formed by a weak base and a weak acid, then hydrolysis can proceed both along the cation and the anion. 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. (Attachment 1)

Has your mood changed? How do you evaluate the knowledge gained, on the back of the anonymous, monosyllabic answer to 6 questions.

1. Are you satisfied with the lesson?
2. Were you interested?
3. Were you active in class?
4. Have you managed to show what you already have and acquire new ones?
5. Did you learn a lot?
6. What did you like more?

VI. Homework.

• § 18, p. 154 No. 3, 8, 11, individual task cards.
• To study independently how food is hydrolyzed in the human body ( p.154).
• Find assignments on the topic “Hydrolysis” in the USE materials for 2009-2012 and complete them in your notebook.

DEFINITION

potassium sulfide- an average salt formed by a strong base - potassium hydroxide (KOH) and a weak acid - hydrogen sulfide (H 2 S). Formula - K 2 S.

Molar mass - 110g / mol. It is colorless cubic crystals.

Hydrolysis of potassium sulfide

Hydrolyzed at the anion. The nature of the medium is alkaline. The hydrolysis equation looks like this:

First stage:

K 2 S ↔ 2K + + S 2- (salt dissociation);

S 2- + HOH ↔ HS - + OH - (anion hydrolysis);

2K + + S 2- + HOH ↔ HS - + 2K + + OH - (equation in ionic form);

K 2 S + H 2 O ↔ KHS + KOH (molecular equation).

Second step:

KHS ↔ K + +HS - (salt dissociation);

HS - + HOH ↔H 2 S + OH - (anion hydrolysis);

K + + 2HS - + HOH ↔ H 2 S + K + + OH - (equation in ionic form);

KHS + H 2 O ↔ H 2 S + KOH (molecular equation).

Examples of problem solving

EXAMPLE 1

Exercise	Potassium sulfide is obtained by heating a mixture of potassium and sulfur at a temperature of 100-200 o C. What mass of the reaction product is formed if 11 g of potassium and 16 g of sulfur interact?
Solution	We write the reaction equation for the interaction of sulfur and potassium: Let's find the number of moles of the starting substances using the data indicated in the condition of the problem. The molar mass of potassium is -39 g / mol, sulfur - 32 g / mol. υ (K) \u003d m (K) / M (K) \u003d 11/39 \u003d 0.28 mol; υ (S) \u003d m (S) / M (S) \u003d 16/32 \u003d 0.5 mol. Potassium deficiency (υ(K)< υ(S)). Согласно уравнению υ (K 2 S) \u003d 2 × υ (K) \u003d 2 × 0.28 \u003d 0.56 mol. Find the mass of potassium sulfide (molar mass - 110 g / mol): m (K 2 S) \u003d υ (K 2 S) × M (K 2 S) \u003d 0.56 × 110 \u003d 61.6 g.
Answer	The mass of potassium sulfide is 61.6 g.

Hydrolysis is the interaction of salt with water, as a result of which the hydrogen ions of water combine with the anions of the acid residue of the salt, and the hydroxyl ions with the metal cation of the salt. This produces an acid (or acid salt) and a base (basic salt). When compiling 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 rest of the 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 acidic 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 holding H + or OH - ions, disturb the balance between water molecules and its ions, causing an acidic or alkaline reaction of the salt solution. Therefore, those salts, which include ions of a weak electrolyte, are subjected to hydrolysis, i.e. 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 COOH 4).

Salts of strong acids and strong bases do not undergo hydrolysis (for example, KNO 3).

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

Examples of problem solving

Example 1 Hydrolysis of potassium sulfide K 2 S.

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

Molecular form of the reaction:

K 2 S+H 2 O=KHS+KOH

Ionic equations:

Full ionic form:

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

Shortened ionic form:

S 2- + H 2 O \u003d HS - + OH -

Because as a result of hydrolysis in a salt solution, an excess of OH - ions is formed, 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 \u003d H 2 S + K + + OH -

Shortened ionic form:

HS - + H 2 O \u003d H 2 S + OH -

Alkaline medium, pH>7.

Example 2 Hydrolysis of copper sulfate CuSO 4 .

I stage of hydrolysis: weakly dissociating ions (СuOH) + are formed.

Molecular form of the reaction:

2CuSO 4 + 2H 2 O \u003d 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-

Shortened ionic form:

Cu 2+ + H 2 O \u003d (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.

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

Molecular form of the reaction

2 SO 4 +2H 2 O \u003d 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-

Shortened ionic form:

(CuOH) + + H 2 O \u003d Cu (OH) 2 + H +

Medium acidic, pH<7.

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

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

Molecular form of the reaction:

Pb (CH 3 COO) 2 + H 2 O \u003d Pb (OH)CH 3 COO + CH 3 COOH

Ionic equations

Full ionic form:

Pb 2+ +2CH 3 COO - +H 2 O \u003d (PbOH) + +CH 3 COO - +CH 3 COOH

Shortened ionic form:

Pb 2+ +CH 3 COO - +H 2 O \u003d (PbOH) + +CH 3 COOH

When the solution is boiled, the hydrolysis practically goes to the end, a precipitate of Pb (OH) 2 is formed

II stage of hydrolysis:

Pb (OH) CH 3 COO + H 2 O \u003d Pb (OH) 2 +CH 3 COOH

The chemical interaction of salt ions with water ions, leading to the formation of a weak electrolyte and accompanied by a change in the pH of the solution, is called salt hydrolysis.

Any salt can be thought of as the reaction product of an acid and a base. The type of salt hydrolysis depends on the nature of the base and acid forming the salt. There are 3 types of hydrolysis of salts.

Anion hydrolysis goes if the salt is formed by a cation of a strong base and an anion of a weak acid.

For example, the salt CH 3 COOHa is formed by the strong base NaOH and the weak monobasic acid CH 3 COOH. The ion of a weak electrolyte CH 3 COO - is subjected to hydrolysis.

Ionic-molecular equation of salt hydrolysis:

CH 3 COO - + NON "CH 3 COOH + OH -

H + water ions bind with CH 3 COO anions - into a weak electrolyte CH 3 COOH, OH ions - accumulate in solution, creating an alkaline environment (pH> 7).

Molecular equation of salt hydrolysis:

CH 3 COONa + H 2 O "CH 3 COOH + NaOH

The hydrolysis of salts of polybasic acids proceeds in stages, forming acidic salts as intermediate products.

For example, the salt K 2 S is formed by the strong base KOH and the weak dibasic acid H 2 S. The hydrolysis of this salt proceeds in two stages.

Stage 1: S 2– + HOH « HS – + OH –

K 2 S + H 2 O "KHS + KOH

Stage 2: HS - - + HOH "H 2 S + OH -

KHS + H 2 O « H 2 S + KOH

The reaction of the medium is alkaline (pH> 7), because OH - ions accumulate in the solution. The hydrolysis of the salt is the stronger, the smaller the dissociation constant formed during the hydrolysis of the weak acid (table 3). Thus, aqueous solutions of salts formed by a strong base and a weak acid are characterized by an alkaline reaction of the medium.

Hydrolysis by cation goes if the salt is formed by a cation of a weak base and an anion of a strong acid. For example, the CuSO 4 salt is formed by the weak diacid base Cu(OH) 2 and the strong acid H 2 SO 4 . Hydrolysis proceeds along the Cu 2+ cation and proceeds in two stages with the formation of a basic salt as an intermediate product.

Stage 1: Cu 2+ + HOH « CuOH + + H +

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

Stage 2: CuOH + + HOH « Cu (OH) 2 + H +

(CuOH) 2 SO 4 + 2H 2 O « 2Cu(OH) 2 + H 2 SO 4

Hydrogen ions H + accumulate in solution, creating an acidic environment (pH<7). Чем меньше константа диссоциации образующегося при гидролизе основания, тем сильнее идет гидролиз.

Thus, aqueous solutions of salts formed by a weak base and a strong acid are characterized by an acid reaction of the medium.

Hydrolysis by cation and anion goes if the salt is formed by a cation of a weak base and an anion of a weak acid. For example, the salt CH 3 COONH 4 is formed by the weak base NH 4 OH and the weak acid CH 3 COOH. Hydrolysis proceeds along the NH 4 + cation and the CH 3 COO - anion:

NH 4 + + CH 3 COO - + HOH "NH 4 OH + CH 3 COOH

Aqueous solutions of this type of salts, depending on the degree of dissociation of the resulting weak electrolytes, have a neutral, slightly acidic or slightly alkaline environment.

When mixing solutions of salts, such as CrCl 3 and Na 2 S, each of the salts is hydrolyzed irreversibly to the end with the formation of a weak base and a weak acid.

Hydrolysis of the CrCl 3 salt proceeds along the cation:

Cr 3+ + HOH « CrOH 2+ + H +

The hydrolysis of the Na 2 S salt proceeds along the anion:

S 2– + HOH « HS – + OH –

When mixing solutions of salts CrCl 3 and Na 2 S, the hydrolysis of each of the salts is mutually enhanced, since the H + and OH ions form a weak electrolyte H 2 O and the ionic equilibrium of each salt shifts towards the formation of end products: chromium hydroxide Cr (OH) 3 and hydrosulfide acid H 2 S.

Ionic-molecular equation of joint hydrolysis of salts:

2Cr 3+ + 3S 2– + 6H 2 O = 2Cr(OH) 3 ¯ + 3H 2 S

Molecular equation:

2CrCl 3 + 3Na 2 S + 6H 2 O \u003d 2Cr (OH) 3 + 3H 2 S + 6NaCl

Salts formed by cations of strong bases and anions of strong acids do not undergo hydrolysis, since none of the salt ions form weak electrolytes with H + and OH ions. Aqueous solutions of such salts have a neutral environment.

A prerequisite for the occurrence of reactions between electrolytes is the removal of certain ions from the solution, due to the formation of weakly dissociating substances, or substances released from the solution in the form of a precipitate or gas. To correctly reflect the nature and mechanism of ion exchange reactions, the reaction equations must be written in ion-molecular form. Whereinstrong electrolytes are written in the form of ions, weak and sparingly soluble - in molecular form.

EXAMPLE 5. Neutralization reaction. Reaction involving strong electrolytes.

HNO 3 + NaOH = NaNO 3 + H 2 O

Full ion-molecular equation: H+ + NO 3 - + Na+ + Oh- = Na+ + NO 3 - + H 2 O

Brief ion-molecular equation: H+ + Oh- = H 2 O(expresses the chemical nature of the reaction).

Conclusion: in solutions of strong electrolytes, the reaction proceeds as a result of the binding of ions with the formation of a weak electrolyte(in this case, water).

EXAMPLE 6.Reaction involving weak electrolytes.HCN + NH 4 Oh = NH 4 CN + H 2 O

: HCN + NH 4 Oh = NH 4 + + CN- + H 2 O

The reaction involving weak electrolytes (example 6) includes two stages: the dissociation of weak (or sparingly soluble) electrolytes into ions and the binding of ions to form a weaker electrolyte. Since the processes of decomposition into ions and binding of ions are reversible, the reactions of ion exchange are reversible.

The direction of ion exchange reactions is determined by the change in the Gibbs energy . A reaction can proceed spontaneously only in the direction for which DG< 0 until a state of equilibrium is reached, when DG = 0. A quantitative measure of the extent to which a reaction proceeds from left to right is the equilibrium constant To FROM. For the reaction shown in example 6: To FROM = [ NH 4 +][ CN- ]/[ HCN][ NH 4 Oh].

The equilibrium constant is related to the change in the Gibbs energy by the equation:

DG0 T = - 2,3 RTlgK C (15)

If a To FROM > 1 , DG < 0 A direct reaction proceeds spontaneously if To FROM < 1, DG > 0 the reaction proceeds in the opposite direction.

Equilibrium constant To FROM calculated through the dissociation constants of weakly dissociating electrolytes:

To FROM =K ref. in-in /TO prod. (16)

For the reaction shown in example 6, the equilibrium constant is calculated by the equation:

To FROM = K HCN . K NH 4 Oh / K H 2 O\u003d 4.9.10-9.!, 76.10-5 / 1014 \u003d 8.67.K C\u003e 1 , track. the reaction proceeds in the forward direction.

The general rule following from the expression for K FROM , is that ion exchange reactions proceed in the direction of stronger binding of ions, i.e. in the direction of formation of electrolytes with lower values ​​of dissociation constants.

7. Hydrolysis of salts.

Salt hydrolysis is an ion exchange reaction between salt and water. Hydrolysis is a reverse reaction of neutralization: KatAn + H 2 OÛ KatOH + HAN (17)

salt base acid

Depending on the strength of the acid and base formed, the salt solution becomes alkaline as a result of hydrolysis. (pH> 7) or sour (pH< 7).

There are four types of hydrolysis:

1. Salts of strong acids and strong bases – hydrolysis is not subjected, since when interacting with water, a weak electrolyte is not formed. Therefore, in solutions of such salts pH=7, those. medium neutral .

2. Salts of strong bases and weak acids – hydrolysis proceeds along the anion. For solutions of salts of strong bases and polybasic acids, hydrolysis proceeds practically in the first stage with the formation of acid salts.

EXAMPLE 7. Determine the pH of a centimole solution of potassium sulfide (FROM K 2 S =0.01mol/l).

K 2 S – salt of a weak dibasic acid H 2 S.

Salt hydrolysis is expressed by the equation:

K 2 S + H 2 OÛ KHS + KOH(an acidic salt is formed - KHS).

Ionic-molecular reaction equation:

S 2- + H 2 OÛ HS - + Oh - (18)

The equilibrium constant of the reaction (hydrolysis constant) is equal to: To G =K H 2 O / K HS - = 10 -14 / 1.2 . 10 - 14 \u003d 0.83, i.e. K g<1, track. the balance is shifted to the left. The resulting excess of OH - ions leads to a change in the nature of the environment. Knowing K G, you can calculate the concentration of OH - ions, and then the pH of the solution. K G \u003d. [ HS - ]/[ S 2- ]. Equation (18) shows that = [ HS- ]. Since salts are weakly hydrolyzed (K G< 1), то можно принять, что = 0,01моль/л, тогда = Ö К Г. = Ö 0,83 . 10 -2 = 9 . 10 - 2 . Из уравнения (6) =10-14/[ OH-]=10 -14 /9 . 10 - 2 = 1,1 . 10 - 11 .

From equation (7) pH = -lg1.1. 10 - 11 = 11.

Conclusion.BecausepH> 7, then the environment is alkaline.

3. Salts of weak bases and strong acids – hydrolysis proceeds through the cation.

For salts formed by strong acids and polyacid bases, hydrolysis proceeds predominantly in the first stage with the formation of a basic salt.

EXAMPLE 8. Hydrolysis of salt of manganese chloride (C salt = 0.01 mol/l).

MnCI 2 + H 2 OÛ MnOHCI + HCI(basic MnOHCI salt is formed).

Ionic-molecular equation: Mn 2+ + H 2 OÛ MnOH + + H + (first step of hydrolysis)

Hydrolysis constant: To G = K H 2 O / K MnOH + = 10 -14 /4 . 10 - 4 = 2,5 . 10 - 11 .

An excess of H + ions leads to a change in the nature of the medium. The calculation of the pH of the solution is carried out analogously to example 7.

The hydrolysis constant is: To G =[ H + ] . [ MnOH + /[ Mn 2+ ]. Since this salt is highly soluble in water and completely dissociated into ions, then FROM salt =[ Mn2+ ] = 0.01 mol/l.

That's why [ H + ] = Ö To G . [ Mn 2+ ] =Ö 2.5 . 10 - 11. 10 - 2 \u003d 5. 10 - 7, pH = 6.3.

Conclusion. BecausepH < 7, then the environment is acidic.

4. Salts of weak bases and weak acids- hydrolysis occurs both in the cation and in the anion.

In most cases, these salts hydrolyze completely to form a base and an acid.

EXAMPLE 9. Hydrolysis of the ammonium acetate salt. CH 3 COONH 4 + H 2 OÛ CH 3 COOH + NH 4 Oh

Ionic-molecular equation: CH 3 COO - + NH 4 + + H 2 OÛ CH 3 COOH + NH 4 Oh .

The hydrolysis constant is: To G = K H 2 O /TO to-you . To main .

The nature of the medium is determined by the relative strength of the acid and base.