Hydrolysis of organic and inorganic compounds. Hydrolysis of potassium sulfide Potassium sulfide undergoes hydrolysis

Option 1

1. Add brief ionic equations for salt hydrolysis reactions:

2. Write the reaction equations for the hydrolysis of sodium ethoxide and bromoethane. What is common in the composition of the hydrolysis products of these substances? How to shift the chemical equilibrium towards the process of bromoethane hydrolysis?

3*. Write an equation for the reaction of interaction with water (hydrolysis) of calcium carbide CaC₂ and name the products of this reaction.

Option 2

1. Salts are given: potassium sulfide, iron (III) chloride, sodium nitrate. Upon hydrolysis of one of them, the solution medium becomes alkaline. Write the molecular and brief ionic equations for the reactions of the first stage of the hydrolysis of this salt. Which of the salts also undergoes hydrolysis? Write the molecular and brief ionic equations for the reactions of the first stage of its hydrolysis. What is the medium of this salt solution?

2. What substances are formed during the complete hydrolysis of proteins? What types of protein hydrolysis do you know? In which case protein hydrolysis proceeds faster?

3*. Write the equation for the reaction of interaction with water (hydrolysis) of phosphorus (V) chloride PCl₅ and name the products of this reaction.

Option 3

1. Add brief ionic equations for salt hydrolysis reactions:

Compose the corresponding molecular equations for hydrolysis reactions. What is the medium of each salt solution?

2. Write the reaction equation for the acid hydrolysis of tristearin fat. What products are formed during the hydrolysis of this fat? What will be the difference in hydrolysis products if the process is carried out in an alkaline medium?

3*. Write the equation for the reaction of interaction with water (hydrolysis) of silicon (IV) chloride SiCl₄ and name the products of this reaction.

Option 4

1. Salts are given: zinc sulfate, sodium carbonate, potassium chloride. Upon hydrolysis of one of them, the solution medium becomes acidic. Write the molecular and brief ionic equations for the reactions of the first stage of the hydrolysis of this salt. Which of the salts also undergoes hydrolysis? Write the molecular and brief ionic equations for the reactions of the first stage of its hydrolysis. What is the medium of this salt solution?

2. Write the reaction equations for the hydrolysis of cellulose and sucrose. What is common in the composition of the hydrolysis products of these substances? In what environment is this process carried out and why?

3*. Write the equation for the reaction of interaction with water (hydrolysis) of sodium hydride NaH and name the products of this reaction.

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 - (hydrolysis by anion);

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 reactions:

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