In electrolyte solutions, dissociation proceeds irreversibly. Electrolytic dissociation

The abbreviated ionic equation H + + OH - \u003d H 2 O corresponds to the interaction of nitric acid with:

1) sodium oxide

2) copper hydroxide

3) sodium hydroxide

Answer: 3

Explanation:

Nitric acid is a strong acid, therefore, almost all of its molecules dissociate into H + cations and NO 3 - anions. Strong water-soluble bases dissociate into hydroxide ions OH −, i.e. alkali. Of all the answers presented in the task, sodium hydroxide is suitable, which decomposes into Na + and OH - in an aqueous solution.

The complete ionic equation for the reaction of NaOH and HNO 3: Na + + OH − + H + + NO 3 − = Na + + NO 3 − + H 2 O. Reducing the same ions on the left and right in the equation, we obtain the reduced ionic equation presented in the task . This reaction proceeds due to the formation of a low-dissociating substance - water.

Sodium oxide does not dissociate in water, but reacts with it to form alkali:

Na 2 O + H 2 O \u003d 2 NaOH.

Copper hydroxide is an insoluble base and therefore does not dissociate in water.

Full ionic equation Cu(OH) 2 + 2H + + 2NO 3 − = Cu 2+ + 2NO 3 − + 2H 2 O

Abbreviated ionic equation: Cu(OH) 2 + 2H + = Cu 2+ + 2H 2 O

The water-soluble KNO 3 salt does not give hydroxide ions upon dissociation. Being a strong electrolyte, it decomposes into K + cations and NO 3 anions -

A precipitate forms when sulfuric acid is added to a solution containing ions:

1) NH 4 + and NO 3 -

2) K + and SiO 3 2−

Answer: 2

Explanation:

Sulfuric acid is a strong electrolyte and dissociates in water into ions: H + and SO 4 2-. When H + cations interact with SiO 3 2− anions, water-insoluble silicic acid H 2 SiO 3 is formed.

The acid residue of sulfuric acid SO 4 2- does not form precipitates with the proposed cations, as can be checked from the table of the solubility of acids, bases and salts in water.

The H + cation, except with SiO 3 2− , also does not form precipitates with the proposed anions.

The abbreviated ionic equation Cu 2+ + 2OH - = Cu(OH) 2 corresponds to the interaction between:

1) CuSO 4 (p-p) and Fe (OH) 3

2) CuS and Ba (OH) 2 (p-p)

3) CuCl 2 (p-p) and NaOH (p-p)

Answer: 3

Explanation:

In the first case, the reaction between copper sulfate CuSO 4 and iron (III) hydroxide Fe (OH) 3 does not proceed, since iron hydroxide is an insoluble base and does not dissociate in an aqueous solution.

In the second case, the reaction also does not proceed due to the insolubility of copper sulfide CuS.

In the third variant, the exchange reaction between copper chloride (II) and NaOH proceeds due to precipitation of Cu(OH) 2 .

The reaction equation in molecular form is as follows:

CuCl 2 + 2NaOH \u003d Cu (OH) 2 ↓ + 2NaCl.

The equation for this reaction in full ionic form is:

Cu 2+ + 2Cl − + 2Na + + 2OH − = Cu(OH) 2 ↓ + 2Na + + 2Cl − .

Reducing the same ions Na + and Cl - in the left and right parts of the full ionic equation, we obtain the reduced ionic equation:

Cu 2+ + 2OH - \u003d Cu (OH) 2 ↓

Copper oxide CuO (II), being an oxide of a transition metal (group IA), does not interact with water, since it does not form a soluble base.

The interaction of solutions of copper(II) chloride and potassium hydroxide corresponds to the reduced ionic equation:

1) Cl - + K + = KCl

2) CuCl 2 + 2OH - \u003d Cu (OH) 2 + 2Cl -

3) Cu 2+ + 2KOH = Cu(OH) 2 + 2K +

Answer: 4

Explanation:

The exchange reaction between solutions of copper (II) chloride and potassium hydroxide in molecular form is written as follows:

CuCl 2 + 2KOH = Cu(OH) 2 ↓ + 2KCl

The reaction takes place due to the precipitation of a blue precipitate of Cu(OH) 2 .

CuCl 2 and KOH are soluble compounds, therefore, in solution they decompose into ions.

We write the reaction in full ionic form:

Cu 2+ + 2Cl − + 2K + + 2OH − = Cu(OH) 2 ↓ + 2Cl − + 2K +

We reduce identical ions 2Cl − and 2K +

left and right of the full ionic equation and we get the reduced ionic equation:

Cu 2+ + 2OH - \u003d Cu (OH) 2 ↓

KCl, CuCl 2 and KOH are soluble substances and in an aqueous solution dissociate into cations and anions almost completely. In other proposed answers, these compounds appear in an undissociated form, so options 1, 2 and 3 are not correct.

Which abbreviated ionic equation corresponds to the interaction of sodium silicate with nitric acid?

1) K + + NO 3 - = KNO 3

2) H + + NO 3 - = HNO 3

3) 2H + + SiO 3 2- = H 2 SiO 3

Answer: 3

Explanation:

The reaction of interaction of sodium silicate with nitric acid (exchange reaction) in molecular form is written as follows:

Na 2 SiO 3 + 2HNO 3 \u003d H 2 SiO 3 ↓ + 2NaNO 3

Since sodium silicate is a soluble salt and nitric acid is strong, both substances in solution dissociate into ions. We write the reaction in full ionic form:

2Na + + SiO 3 2− + 2H + + 2NO 3 − = H 2 SiO 3 ↓ + 2Na + + 2NO 3 −

SiO 3 2- + 2H + = H 2 SiO 3 ↓

The rest of the proposed options do not reflect the sign of the reaction - precipitation. In addition, in the presented answer options, the soluble salts of KNO 3 and K 2 SiO 3 and the strong acid HNO 3 are presented in an undissociated form, which, of course, is not true, since these substances are strong electrolytes.

The abbreviated ionic equation Ba 2+ + SO 4 2− =BaSO 4 corresponds to the interaction

1) Ba(NO 3) 2 and Na 2 SO 4

2) Ba (OH) 2 and CuSO 4

3) BaO and H 2 SO 4

Answer: 1

Explanation:

The reaction of interaction of barium nitrate with sodium sulfate (exchange reaction) in molecular form is written as follows:

Ba(NO 3) 2 + Na 2 SO 4 = BaSO 4 ↓ + 2NaNO 3

Since barium nitrate and sodium sulfate are soluble salts, both substances in solution dissociate into ions. We write the reaction in full ionic form:

Ba 2+ + 2NO 3 − + 2Na + + SO 4 2− = BaSO 4 ↓ + 2Na + + 2NO 3 −

Having reduced the Na + and NO 3 − ions in the left and right parts of the equation, we obtain the reduced ionic equation:

Ba 2+ + SO 4 2− = BaSO 4 ↓

The reaction of interaction of barium hydroxide with copper sulfate (exchange reaction) in molecular form is written as follows:

Ba(OH) 2 + CuSO 4 = BaSO 4 ↓ + Cu(OH) 2 ↓

Two precipitates are formed. Since barium hydroxide and copper sulfate are soluble substances, both dissociate into ions in solution. We write the reaction in full ionic form:

Ba 2+ + 2OH − + Cu 2+ + SO 4 2− = BaSO 4 ↓ + Cu(OH) 2 ↓

The reaction of interaction of barium oxide with sulfuric acid (exchange reaction) in molecular form is written as follows:

BaO + H 2 SO 4 \u003d BaSO 4 ↓ + H 2 O

Since BaO is an oxide, it does not dissociate in water (BaO interacts with water to form alkali), we write the BaO formula in undissociated form. Sulfuric acid is strong, therefore, in solution it dissociates into H + cations and SO 4 2− anions. The reaction proceeds due to the precipitation of barium sulfate and the formation of a low-dissociating substance. We write the reaction in full ionic form:

BaO + 2H + + SO 4 2− = BaSO 4 ↓ + 2H 2 O

Here, too, there are no identical ions in the left and right parts of the equation and it is impossible to reduce anything, then the reduced ionic equation looks the same as the complete one.
The reaction of interaction of barium carbonate with sulfuric acid (exchange reaction) in molecular form is written as follows:

BaCO 3 + H 2 SO 4 = BaSO 4 ↓ + CO 2 + H 2 O

The reaction proceeds due to the formation of a precipitate, gas evolution and the formation of a low-dissociating compound - water. Since BaCO 3 is an insoluble salt, therefore, it does not decompose into ions in solution, we write the formula BaCO 3 in molecular form. Sulfuric acid is strong, therefore, in solution it dissociates into H + cations and SO 4 2− anions. We write the reaction in full ionic form:

BaCO 3 + 2H + + SO 4 2− = BaSO 4 ↓ + CO 2 + H 2 O

The full ionic equation coincides with the reduced one, since there are no identical ions on the left and right sides of the equation.

The reduced ionic equation Ba 2+ + CO 3 2− = BaCO 3 corresponds to the interaction

1) barium sulfate and potassium carbonate

2) barium hydroxide and carbon dioxide

3) barium chloride and sodium carbonate

4) barium nitrate and carbon dioxide

Answer: 3

Explanation:

The reaction between barium sulfate BaSO 4 and potassium carbonate K 2 CO 3 does not proceed, since barium sulfate is an insoluble salt. A necessary condition for the exchange of two salts is the solubility of both salts.

The reaction between barium hydroxide Ba(OH) 2 and carbon dioxide CO 2 (acid oxide) occurs due to the formation of an insoluble salt BaCO 3 . This is the reaction of alkali with acid oxide to form salt and water. Let's write the reaction in molecular form:

Ba(OH) 2 + CO 2 = BaCO 3 ↓ + H 2 O

Since barium hydroxide is a soluble base, in solution it dissociates into Ba 2+ cations and OH − hydroxide ions. Carbon monoxide does not dissociate in water, therefore, in ionic equations, its formula should be written in molecular form. Barium carbonate is an insoluble salt, therefore, in the ionic reaction equation, it is also written in molecular form. Thus, the reaction of interaction of barium hydroxide and carbon dioxide in full ionic form is as follows:

Ba 2+ + 2OH − + CO 2 = BaCO 3 ↓ + H 2 O

Since there are no identical ions on the left and right sides of the equation and it is impossible to reduce anything, the reduced ionic equation looks the same as the complete one.

The reaction of interaction of barium chloride with sodium carbonate (exchange reaction) in molecular form is written as follows:

BaCl 2 + Na 2 CO 3 \u003d BaCO 3 ↓ + 2NaCl

Since barium chloride and sodium carbonate are soluble salts, both substances in solution dissociate into ions. We write the reaction in full ionic form:

Ba 2+ + 2Cl − + 2Na + + CO 3 2- = BaCO 3 ↓ + 2Na + + 2Cl −

Reducing the Na + and Cl − ions in the left and right parts of the equation, we obtain the reduced ionic equation:

Ba 2+ + CO 3 2- \u003d BaCO 3 ↓

The reaction between barium nitrate Ba (NO 3) 2 and carbon dioxide CO 2 (acidic oxide) in an aqueous solution does not proceed. Carbon dioxide CO 2 in an aqueous solution forms a weak unstable carbonic acid H 2 CO 3 , which is not able to displace strong HNO 3 from a Ba(NO 3) 2 salt solution.

Formulas for calculation.

1. Calculate the normal concentration of an acid solution (op. No. 1) or an alkali solution (op. No. 2) from the formula of the law of equivalents for solutions:

2. Calculate the mass of acid (op. No. 1) or alkali (op. No. 2) contained in 10 ml of the corresponding solution from the normal concentration formula:

3. Calculate the mass of water (solvent) in 10 ml of solution, assuming the density of the solution is 1:

4. Using the obtained data, calculate the given concentrations using the appropriate formulas.

Lab #5

Objective: to study the conditions for the occurrence of ion exchange reactions and the rules for writing ion exchange reactions in molecular and ion-molecular forms.

Theoretical part.

electrolytic dissociation called partial or complete decomposition of electrolyte molecules into ions under the action of polar solvent molecules. Dissociation proceeds as a result of a complex physicochemical interaction of electrolyte molecules with polar solvent molecules. The interaction of ions with polar solvent molecules is called solvation (for aqueous solutions - hydration) of ions. Solvated ions are formed in electrolyte solutions.

Electrolytes conduct electric current, since there are charged particles in solutions: cations and anions.

Quantitatively, the dissociation process is characterized the degree of electrolytic dissociation α. The degree of dissociation is the ratio of the number of molecules that have decayed into ions n to the total number of molecules N of the solute:

The degree of dissociation is expressed as a percentage or fractions of a unit.

Electrolytes are divided into three groups: a) strong (α> 30%), b) medium (3<α<30%), в) слабые (α<3%).

The educational literature contains tables of the degrees of dissociation of acids, bases and salts. The degree of dissociation depends on the nature of the solute and solvents, temperature, concentration and presence in the solution of the same ions. For weak electrolytes, the degree of dissociation significantly depends on the concentration: the lower the concentration of the solution, the greater the degree of electrolytic dissociation.

It is much more convenient to characterize the ability of electrolytes to dissociate to a solution dissociation constant K , which does not depend on the concentration of the solution. The dissociation constant K is the equilibrium constant of the reversible process of dissociation of a weak electrolyte - acid or base. The dissociation constant of acids is also called the acidity constant, and bases - the basicity constant. The values of the dissociation constants of weak electrolytes are given in the tables for standard conditions.

The dissociation constant (basicity) is expressed as the ratio of the product of the equilibrium concentrations of ions in a solution of a given weak electrolyte to the concentration of undissociated molecules:

The dissociation constant is a measure of the relative strength of weak electrolytes: the smaller it is, the weaker the electrolyte. The relationship between the constant and the degree of dissociation of a weak binary electrolyte obeys Ostwald's breeding law:

From the point of view of electrolytic dissociation, acids are called electrolytes that form positively charged hydrogen nones and anions of the acid residue in aqueous solutions. Hydrogen ions are characteristic of acids and determine their properties. Acids that are strong electrolytes: nitric HNO 3, hydrochloric HCl, hydrobromic HBr, hydroiodic HJ, sulfuric H 2 SO 4, manganese HMnO 4 and others.

There are more weak electrolytes than strong ones. Weak electrolytes are acids: sulfurous H 2 SO 3, hydrofluoric HF, coal H 2 CO 3, hydrogen sulfide H 2 S, acetic CH 3 COOH, etc. Polybasic acids dissociate in steps. Examples of acid dissociation:

HCl = H + + Cl-

CH 3 COOH CH 3 COO - + H +

Stage I: H 2 SO 3 H + + HSO 3 -

or H 2 SO 3 2H + + SO 3 2-,

Stage II: HSO 3 - H + + SO 3 2 -

From the point of view of electrolytic dissociation, bases are electrolytes that form negatively charged OH hydroxide ions and metal cations in aqueous solutions. Hydroxide ions determine the general properties of bases. Bases with a cation valency greater than one dissociate in steps. Strong electrolytes are bases in which alkali and alkaline earth metals are cations, with the exception of Be (OH) 2 and Mg (OH) 2.

In general, bases are weak electrolytes, especially those formed by amphoteric metals. Amphoteric hydroxides dissociate as bases in an acidic medium and as acids in an alkaline medium. Examples of dissociation of bases and amphoteric hydroxides:

NaOH \u003d Na + + OH -

1 st. Fe(OH) 2 FeOH + +OH -

II Art. FeOH + Fe 2+ + OH - or Fe (OH) 2 Fe 2+ + 2OH -

Zn 2+ + 2OH - Zn(OH) 2 H 2 ZnО 2 2H + + ZnO 2 2-

Salts are electrolytes that dissociate in water into positive metal ions and negative ions of the acid residue. All salts that are readily soluble in water are strong electrolytes. Examples of dissociation of normal (medium), acidic, basic, complex and double salts:

KVg \u003d K + + Vg -; K 3 \u003d 3K + + 3-;

NaHCO 3 \u003d Na + + HCO 3 -; KAl(SO 4) 2 = K + + Al 3+ + 2SO 4 2-.

AlOHCl 2 \u003d AlOH 2+ + 2C1 -;

The study of various reactions, mainly in non-aqueous media, led to the creation of more general ideas about acids and bases. The most important of the modern theories of acids and bases is the proton theory, according to which an acid is a proton donor, that is, a particle (molecule or ion) that is capable of donating a hydrogen ion - a proton, and a base is a proton acceptor, i.e. a particle (molecule or ion) capable of accepting a proton. For example, in react:

HC1 + NH 3 \u003d NH 4 + + Cl -

the C1 ion is the base conjugated to the HCl acid, and the NH 4 + ion is the acid conjugate to the NH 3 base. Reactions in electrolyte solutions proceed between ions, into which molecules of dissolved substances break up. Reactions are written in three forms: molecular, complete ionic-molecular and reduced ionic-molecular. Strong electrolytes are written in the form of ions, medium and weak electrolytes, precipitation and gases - in the form of molecules. The essence of the reaction is reflected in the abbreviated ion-molecular equation, in which only particles that directly enter into the reaction are indicated and ions and molecules whose concentration does not change significantly are not indicated. Reactions between electrolytes proceed towards the formation of a gas, a precipitate, or a weaker

electrolyte.

An example of a reaction in electrolyte solutions: neutralization of strong nitric acid with a weak base (ammonium hydroxide). Molecular reaction equation:

HNO 3 + NH 4 OH \u003d NH 4 NO 3 + H 2 O.

In this reaction, strong electrolytes are nitric acid and the resulting ammonium nitrate salt, which are written in the form of ions, and weak electrolytes are ammonium hydroxide and water, which are written in the form of molecules. The complete ion-molecular equation has the form:

H + + NO 3 - + NH 4 OH \u003d NH 4 + + NO 3 - + H 2 O.

As you can see, only NO 3 - ions do not undergo changes during the reaction, excluding them, we write down the reduced ion-molecular equation:

H + + NH 4 OH \u003d NH 4 + + H 2 O.

Practical part

Ion - molecular exchange reactions

Carry out reactions between electrolyte solutions according to the task. To do this, pour 7-8 drops of one reagent into a test tube and add 7-8 drops of another reagent. Note the signs of the reaction: precipitation, gas evolution or a change in odor (which indicates the formation of a low-dissociating substance).

Then, in accordance with the observed signs, attribute the reaction to one of 3 types:

1) ion-exchange reactions with the formation of a poorly soluble substance (precipitate);

2) ion-exchange reactions with gas evolution;

3) ion-exchange reactions with the formation of a weak electrolyte.

Write each reaction in 3 forms:

a) molecular

b) complete ionic - molecular,

c) abbreviated ion-molecular.

Make a conclusion about the direction of the ion exchange reactions.

Task list:

1. CH 3 COONa + H 2 SO 4 2. NaNO 2 + H 2 SO 4 3. MgCl 2 + Na 3 PO 4 4. NH 4 Cl + KOH 5. Na 2 CO 3 + HCl 6. Na 2 CO 3 + Ba (NO 3) 2 7. (CH 3 COO) 2 Pb + HCl 8. Hg (NO 3) 2 + NaOH 9. H 2 SO 4 + BaCl 2 10. NaCl + Pb (NO 3) 2 11. NiSO 4 +KOH 12. NaNO 2 +HCl 13. Bi(NO 3) 3 +KOH 14. Na 2 S+CdCl 2 15. Bi(NO 3) 3 +Na 2 S 16. CoSO 4 +KOH 17. CuSO 4 +KOH 18. Na 2 CO 3 + HNO 3 19. K 2 CrO 4 + CuSO 4 20. K 2 CrO 4 + MnSO 4 21. K 2 CrO 4 + NiSO 4 22. K 2 CO 3 + MnSO 4 23. Na 2 SO 3+HCl

24. Hg(NO 3) 2 + Na 2 S 25. NiSO 4 + NH 4 OH 26. NiSO 4 + NH 4 OH 27. AlCl 3 +KOH 28. FeCl 3 + Na 3 PO 4 29. K 2 CrO 4 + Ba(NO 3) 2 30. NaNO 2 + HNO 3 31. MgCl 2 + NaOH 32. CuSO 4 + NH 4 OH 33. CuSO 4 + NH 4 OH 34. AlCl 3 +KOH 35. Pb (NO 3 ) 2 + KI 36. CH 3 COOK+ HCl 37. Al 2 (SO 4) 3 + NaOH 38. Al 2 (SO 4) 3 + NaOH ex 39. CoSO 4 + Na 2 S 40. Pb (NO 3) 2 + Na 3 PO 4 41. Na 3 PO 4 + CuSO 4 42. CH 3 COOK + HNO 3 43. CH 3 COOH + KOH 44. CoSO 4 + NH 4 OH 45. CoSO 4 + NH 4 OH ex 46. Hg (NO 3 ) 2 + KI

47. Hg(NO 3) 2 + KI 48. CdCl 2 + NH 4 OH 49. CdCl 2 + NH 4 OH 50. NaHCO 3 + HNO 3 51. ZnSO 4 + BaCl 2 52. ZnSO 4 +KOH 53. ZnSO 4 + KOH ex 54. (CH 3 COO) 2 Pb + H 2 SO 4 55. NaHCO 3 + H 2 SO 4 56. (NH 4) 2 SO 4 +KOH 57. K 2 CO 3 + H 2 SO 4 58 .(NH 4) 2 SO 4 +NaOH 59. K 2 CO 3 + HCl 60. CrCl 3 +KOH 61. CrCl 3 +KOH ex 62. ZnCl 2 + NaOH 63. ZnCl 2 + NaOH ex 64. MnSO 4 +KOH 65. MnSO 4 + Na 3 PO 4 66. Na 2 SO 3 + H 2 SO 4 67. K 2 CO 3 + CH 3 COOH 68. Na 2 CO 3 + CH 3 COOH 69. NaHCO 3 + CH 3 COOH

Lab #6

Electrolytic dissociation - this is the process of decomposition of electrolyte molecules into ions under the action of polar solvent molecules.

electrolytes- These are substances whose melts or aqueous solutions conduct an electric current. These include solutions of acids, melts and solutions of alkalis and salts. Non-electrolytes are substances that do not conduct electricity. These include many organic substances.

Electrolytes that almost completely dissociate into ions are called strong; electrolytes that partially dissociate into ions are called weak. To quantify the completeness of dissociation, the concept of the degree of dissociation is introduced. Degree of dissociation electrolyte called the ratio of the number of molecules that have decayed into ions, to the total number of molecules in solution.

Usually the degree of dissociation ( α ) are expressed in fractions of a unit or%:

where n is the number of particles subjected to electrolytic dissociation;

n 0 is the total number of particles in the solution.

Strong electrolytes - almost all salts, soluble bases ( NaOH, KOH, Ba(Oh) 2 etc.), inorganic acids ( H 2 SO 4 , HCl, HNO 3 , HBr, HI and etc) .

Weak electrolytes- insoluble bases and NH 4 Oh, inorganic acids ( H 2 CO 3, , H 2 S, HNO 2, H 3 PO 4 etc.), organic acids and water H 2 O.

Strong electrolytes dissociate into ions almost completely (i.e., the dissociation process is irreversible) and in one step:

HCl=H + +Cl – H 2 SO 4 = 2H + + SO 4 2–

Weak electrolytes dissociate partially (i.e., the dissociation process is reversible) and stepwise . For example, for polybasic acids, one hydrogen ion is detached at each stage:

1. H 2 SO 3 ⇄ H + + HSO 3 - 2. HSO 3 - ⇄ H + + SO 3 2-

Thus, the number of stages of polybasic acids is determined by the basicity of the acid (the number of hydrogen ions), and the number of stages of polyacid bases will be determined by the acidity of the base (or the number of hydroxyl groups): NH 4 Oh ⇄ NH 4 + + Oh – . The process of electrolytic dissociation ends with the establishment of a state of chemical equilibrium in the system, which is characterized by an equilibrium constant:

The equilibrium constant of the process of electrolytic dissociation is called the dissociation constant - To D. The dissociation constant depends on the nature of the electrolyte, the nature of the solvent, temperature, but does not depend on the concentration of the electrolyte.

Between To D and α There is a quantitative relationship

(13)

Relation (13) is called the Ostwald dilution law: the degree of dissociation of a weak electrolyte increases with dilution of the solution.

For weak electrolytes, when α  1, To D = α 2 WITH.

Water is a weak electrolyte, therefore it dissociates reversibly:

H 2 O ⇄ H + + Oh – ∆ H\u003d + 56.5 kJ / mol

Water dissociation constant:

The degree of dissociation of water is very small (it is a very weak electrolyte). Since water is present in large excess, its concentration can be considered a constant value and is
, then

To D [ H 2 O] = [ H + ]∙[ Oh - ] = 55,6∙1,8∙10 -16 = 10 -14

[ H + ]∙[ Oh - ] = 10 -14 = K W is the ionic product of water

Since the concentrations of hydrogen cations and hydroxide ions are equal in water, then: [ H + ] = [ Oh - ] =
.

The dissolution of other substances (acids, bases, salts) in water changes the concentration of ions H + or IS HE – , and their product always remains constant and equal to 10 -14 at T \u003d 25 0 C. The concentration of ions H + can serve as a measure of the acidity or alkalinity of a solution. Usually, a pH indicator is used for this purpose: pH = - lg[ H + ]. Thus, pH value is the decimal logarithm of the concentration of hydrogen ions, taken with the opposite sign.

Depending on the concentration of hydrogen ions, three media are distinguished.

AT neutral environment [ H + ] = [ Oh - ]= 10 -7 mol/l, pH= –lg 10 -7 = 7 . This medium is typical for both pure water and neutral solutions. AT sour solutions [ H + ] > 10 -7 mol/l, pH< 7 . In acidic environments pH varies within 0 < рН < 7 . AT alkaline environments [ H + ] < [ОН – ] and [ H + ] < 10 -7 mol/l, hence, pH > 7. Limits of pH change: 7 < рН < 14 .

Ion exchange reactions (RIO)- these are reactions between ions occurring in aqueous solutions of electrolytes. A distinctive feature of exchange reactions is that the elements that make up the reactants do not change their oxidation state. Ion exchange reactions are irreversible reactions and proceed given that: 1) the formation of a poorly soluble substance, 2) the release of a gaseous substance, 3) the formation of a weak electrolyte.

When RIO occurs, oppositely charged ions are bound and removed from the reaction sphere. The essence of ion exchange reactions is expressed using ionic equations, which, unlike molecular ones, show the true participants in the reaction. When compiling ionic equations, one should be guided by the fact that low-dissociating, slightly soluble (precipitating) and gaseous substances are written in molecular form. Strong soluble electrolytes are written as ions. Therefore, when writing ionic equations, it is necessary to use the table of solubility of salts and bases in water.

Hydrolysis- this is the process of interaction of salt ions with water molecules, leading to the formation of low-dissociating compounds; is a special case of ion exchange reactions. Hydrolysis undergoes salts formed:

weak acid and strong base ( NaCH 3 COO, Na 2 CO 3 , Na 2 S, );

weak base and strong acid NH 4 Cl, FeCl 3 , AlCl 3 ,);

weak base and weak acid NH 4 CN, NH 4 CH 3 COO).

Salts formed by a strong acid and a strong base do not undergo hydrolysis: Na 2 SO 4 , BaCl 2 , NaCl, NaJ etc.

Salt hydrolysis increases ion concentrations H + or IS HE – . This leads to a shift in the ionic equilibrium of water and, depending on the nature of the salt, imparts an acidic or alkaline environment to the solution (see examples of problem solving).

Electrolytic dissociation - this is the process of decomposition of electrolyte molecules into ions under the action of polar solvent molecules.

Usually the degree of dissociation ( α ) are expressed in fractions of a unit or%:

where n is the number of particles subjected to electrolytic dissociation;

n 0 is the total number of particles in the solution.

Strong electrolytes - almost all salts, soluble bases ( NaOH, KOH, Ba(Oh) 2 etc.), inorganic acids ( H 2 SO 4 , HCl, HNO 3 , HBr, HI and etc) .

Weak electrolytes- insoluble bases and NH 4 Oh, inorganic acids ( H 2 CO 3, , H 2 S, HNO 2, H 3 PO 4 etc.), organic acids and water H 2 O.

Strong electrolytes dissociate into ions almost completely (i.e., the dissociation process is irreversible) and in one step:

HCl=H + +Cl – H 2 SO 4 = 2H + + SO 4 2–

Weak electrolytes dissociate partially (i.e., the dissociation process is reversible) and stepwise . For example, for polybasic acids, one hydrogen ion is detached at each stage:

1. H 2 SO 3 ⇄ H + + HSO 3 - 2. HSO 3 - ⇄ H + + SO 3 2-

Between To D and α There is a quantitative relationship

(13)

Relation (13) is called the Ostwald dilution law: the degree of dissociation of a weak electrolyte increases with dilution of the solution.

For weak electrolytes, when α  1, To D = α 2 WITH.

Water is a weak electrolyte, therefore it dissociates reversibly:

H 2 O ⇄ H + + Oh – ∆ H\u003d + 56.5 kJ / mol

Water dissociation constant:

The degree of dissociation of water is very small (it is a very weak electrolyte). Since water is present in large excess, its concentration can be considered a constant value and is
, then

To D [ H 2 O] = [ H + ]∙[ Oh - ] = 55,6∙1,8∙10 -16 = 10 -14

[ H + ]∙[ Oh - ] = 10 -14 = K W is the ionic product of water

Since the concentrations of hydrogen cations and hydroxide ions are equal in water, then: [ H + ] = [ Oh - ] =
.

Depending on the concentration of hydrogen ions, three media are distinguished.

weak acid and strong base ( NaCH 3 COO, Na 2 CO 3 , Na 2 S, );

weak base and strong acid NH 4 Cl, FeCl 3 , AlCl 3 ,);

weak base and weak acid NH 4 CN, NH 4 CH 3 COO).

Salts formed by a strong acid and a strong base do not undergo hydrolysis: Na 2 SO 4 , BaCl 2 , NaCl, NaJ etc.

Lesson: Electrolytic dissociation. Hydrogen index. Ion exchange reactions
Goals: to systematize students' knowledge about electrolytic dissociation. Show the scientific feat of the founders of the theory. Show the dependence of the properties of substances on their structure. To bring the knowledge gained by students on the topic into a single system.
Tasks: Improve the skills and abilities of compiling dissociation equations, ionic equations, hydrolysis equations. To form the ability to predict the environment of solutions of various salts. To systematize students' knowledge about the hydrolysis of organic substances. Develop the ability to observe, analyze and draw conclusions.
Equipment and reagents : multimedia projector, computer.

During the classes

Organizing time

Updating of basic knowledge:

Students answer according to the plan:
What is the electrical conductivity of solutions?
- Electrolytic dissociation of salts, bases and acids.
- Mechanism of electrolytic dissociation of substances with ionic bond.

Introduction to the study of a new topic:Why do solutions of acids, salts and alkalis conduct electricity?

Why will the boiling point of an electrolyte solution always be higher than the boiling point of a non-electrolyte solution of the same concentration?

Learning new material:

1. The concept of electrolytic dissociation

In 1887 a Swedish physicistchemist Svante Arrhenius, investigating the electrical conductivity of aqueous solutions, he suggested that in such solutions substances decompose into charged particles - ions that can move to the electrodes - a negatively charged cathode and a positively charged anode.

This is the reason for the electric current in solutions. This process is calledelectrolytic dissociation (literal translation - splitting, decomposition under the influence of electricity). This name also suggests that dissociation occurs under the action of an electric current. Further studies showed that this is not so: ions are only charge carriers in solution and exist in it regardless of whether current passes through the solution or not. With the active participation of Svante Arrhenius, the theory of electrolytic dissociation was formulated, which is often named after this scientist. The main idea of this theory is that electrolytes under the action of a solvent spontaneously decompose into ions. And it is these ions that are charge carriers and are responsible for the electrical conductivity of the solution.

Electric current is the directed movement of free charged particles. You already know that solutions and melts of salts and alkalis are electrically conductive, since they do not consist of neutral molecules, but of charged particles - ions. When melted or dissolved, ions becomefree carriers of electric charge.

The process of disintegration of a substance into free ions during its dissolution or melting is called electrolytic dissociation.

2. The essence of the process of electrolytic dissociation of salts

The essence of electrolytic dissociation is that ions become free under the influence of a water molecule. Fig.1. The process of decomposition of the electrolyte into ions is displayed using a chemical equation. Let us write the dissociation equation for sodium chloride and calcium bromide. The dissociation of one mole of sodium chloride produces one mole of sodium cations and one mole of chloride anions.NaCl ⇄ Na+ + Cl-

The dissociation of one mole of calcium bromide produces one mole of sodium cations and two moles of bromide anions.

CaBr2 ⇄ Ca2+ + 2Br-

Please note: since the formula of an electrically neutral particle is written on the left side of the equation, the total charge of the ions must be equal to zero.

Conclusion : during the dissociation of salts, metal cations and anions of the acid residue are formed.

3. The essence of the process of electrolytic dissociation of alkalis

Consider the process of electrolytic dissociation of alkalis. Let us write the dissociation equation in a solution of potassium hydroxide and barium hydroxide.

The dissociation of one mole of potassium hydroxide produces one mole of potassium cations and one mole of hydroxide anions.KOH ⇄ K+ + OH-

The dissociation of one mole of barium hydroxide produces one mole of barium cations and two moles of hydroxide anions.Ba(OH)2 ⇄ Ba2+ + 2OH-

Conclusion: during the electrolytic dissociation of alkalis, metal cations and hydroxide anions are formed.

Water-insoluble bases practically do not undergo electrolytic dissociation, since they are practically insoluble in water, and when heated, they decompose, so that they cannot be melted.

4. The essence of the process of electrolytic dissociation of acids

Consider the process of electrolytic dissociation of acids. Acid molecules are formed by a polar covalent bond, which means that acids do not consist of ions, but of molecules.

The question arises - how then does the acid dissociate, i.e. how do free charged particles form in acids? It turns out that ions are formed in acid solutions precisely during dissolution.

Consider the process of electrolytic dissociation of hydrogen chloride in water, but for this we will write down the structure of the molecules of hydrogen chloride and water. Both molecules are formed by a covalent polar bond. The electron density in the hydrogen chloride molecule is shifted to the chlorine atom, and in the water molecule - to the oxygen atom. The water molecule is able to tear off the hydrogen cation from the hydrogen chloride molecule, thus forming the hydronium cation H3O+.

Then the equation for the dissociation of hydrogen chloride looks like this:HCl ⇄ H+ + Cl-

5. Stepwise dissociation of acids

Stepwise dissociation of sulfuric acid

Consider the process of electrolytic dissociation of sulfuric acid. Sulfuric acid dissociates stepwise, in two stages.

I–I stage of dissociation

In the first stage, one hydrogen cation is detached and a hydrosulfate anion is formed.

H2SO4 ⇄ H+ + HSO4-

hydrosulfate anion.

II - I stage of dissociation

At the second stage, further dissociation of hydrosulfate anions occurs.HSO4- ⇄ H+ + SO42-

This stage is reversible, that is, the resulting sulfate - ions can attach hydrogen cations to themselves and turn into hydrosulfate - anions. This is shown by the sign of reversibility.

There are acids that do not completely dissociate even at the first stage - such acids are weak. For example, carbonic acid H2CO3.

The hydrogen index characterizes the concentration of free hydrogen ions in water.

For convenience of display, a special indicator was introduced, called pH, which is the logarithm of the concentration of hydrogen ions, taken with the opposite sign, i.e. pH = -log.

Simply put, the pH value is determined by the quantitative ratio of H ions in water + and he - formed during the dissociation of water. If the water has a reduced content of free hydrogen ions (pH> 7) compared to OH ions - , then the water will have an alkaline reaction, and with an increased content of H ions + (pH<7)- кислую. В идеально чистой дистиллированной воде эти ионы будут уравновешивать друг друга. В таких случаях вода нейтральна и рН=7. При растворении в воде различных химических веществ этот баланс может быть нарушен, что приводит к изменению уровня рН.

Reflection: compose a cinquain

D/W:

Summing up the lesson

In this lesson, you learned that solutions of acids, salts and alkalis are electrically conductive, since when they are dissolved, charged particles are formed - ions. This process is called electrolytic dissociation. During the dissociation of salts, metal cations and anions of acidic residues are formed. During the dissociation of alkalis, metal cations and hydroxide anions are formed. During the dissociation of acids, hydrogen cations and anions of the acid residue are formed.

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