Solubility is the ability of substances to dissolve in water. Some substances dissolve very well in water, some even in unlimited quantities. Others - only in small quantities, and still others - hardly dissolve at all. Therefore, substances are divided into soluble, slightly soluble and practically insoluble.

Soluble substances include those substances that are dissolved in 100 g of water in an amount of more than 1 g (NaCl, sugar, HCl, KNO 3). Slightly soluble substances dissolve in an amount of 0.01 g to 1 g in 100 g of water (Ca (OH) 2, CaSO 4). Practically insoluble substances cannot dissolve in 100 g of water in an amount greater than 0.01 g (metals, CaCO 3 , BaSO 4).

During the course of chemical reactions in aqueous solutions, insoluble substances can form, which precipitate or are in suspension, making the solution cloudy.

There is a table of solubility in water of acids, bases and salts, which reflects whether the compound is soluble. All salts of potassium and sodium, as well as all nitrates (salts of nitric acid) are highly soluble in water. From sulfates (salts of sulfuric acid), calcium sulfate is sparingly soluble, barium and lead sulfates are insoluble. Lead chloride is sparingly soluble, while silver chloride is insoluble.

If there is a dash in the cells of the solubility table, this means that the compound reacts with water, resulting in the formation of other substances, i.e. the compound does not exist in water (for example, aluminum carbonate).

All solids, even those highly soluble in water, dissolve only in certain quantities. The solubility of substances is expressed as a number that indicates the largest mass of a substance that can be dissolved in 100 g of water under certain conditions (usually temperature). So at 20 ° C, 36 g of table salt (sodium chloride NaCl), more than 200 g of sugar are dissolved in water.

On the other hand, there are no insoluble substances at all. Any practically insoluble substance, even in very small quantities, but dissolves in water. For example, chalk dissolves in 100 g of water at room temperature in an amount of 0.007 g.

Most substances dissolve better in water with increasing temperature. However, NaCl is almost equally soluble at any temperature, while Ca(OH)2 (lime) is more soluble at lower temperatures. Based on the dependence of the solubility of substances on temperature, solubility curves are built.

If a certain amount of a substance can still be dissolved in a solution at a given temperature, then such a solution is called unsaturated. If the limit of solubility is reached, and no more substance can be dissolved, then they say that the solution is saturated.

When a saturated solution is cooled, the solubility of the substance decreases, and, consequently, it begins to precipitate. Often the substance is released in the form of crystals. For different salts, crystals have their own shape. So the crystals of table salt are cubic in shape, in potassium nitrate they look like needles.

Class: 8

Presentation for the lesson

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Attention! The slide preview is for informational purposes only and may not represent the full extent of the presentation. If you are interested in this work, please download the full version.

Textbook: Rudzitis G.E., Feldman F.G. Chemistry: a textbook for grade 8 educational institutions / G.E. Rudzitis, F.G. Feldman. – 12th ed. - M .: Education, JSC "Moscow textbooks", 2009. - 176 p.

Target: to form students' ideas about the solubility of substances, solutions, concentrations of solutions.

Tasks:

• contribute to the systematization of the conceptual apparatus: solvent, solute, solution, solubility of substances in water, concentration of solutions
• « 5 » - substantiate, prove; " 4 » - characterize, apply; " 3 " - tell;
• contribute to the improvement of special subject skills: solve and compose tasks on the topic "Solutions"
• contribute to the formation of general educational skills:
• a) educational and intellectual (analyze facts, establish causal relationships; put forward a hypothesis; compare, classify, draw conclusions);
• b) educational and informational (work with text, convert a text task into a sign task);
• c) educational and organizational (understand the meaning of the task, allocate time to complete tasks, plan work on organizing work, exercise self-control);
• contribute to the formation of critical thinking of students (critically evaluate their own knowledge on the topic and compare them with scientific ones);

Conduct form: lesson with the use of ICT, the inclusion of paired, individual forms of organization of educational and cognitive activity of students.

Duration of the training session: 90 minutes.

The use of pedagogical technologies: heuristic learning method, collaborative learning

DURING THE CLASSES

I. Organizational moment - 3 minutes: mobilizing beginning (greeting, checking readiness for the lesson, organizing the attention of students), information about the purpose and course of the lesson, motivation

II. Frontal conversation (12 min)

– How often do we encounter solutions in life? What solutions do we know? (Sea, rivers, oceans; household solutions: salt solution, sugar solution, washing powder solution, etc.; medical solutions, etc.)
– What is the basis of most of the solutions known to us? (Water)
- Let's think about how the solution is formed? ( Attachment 1 , slide 2)

Where did the dissolution take place? (In the case of table salt and potassium oxide)
Where did the chemical reaction take place? (In the case of potassium oxide, a new substance was formed)
– What is the similarity between the formation of a mixture (suspension and emulsion) with the formation of a solution?
What is the difference between a dissolution process and a chemical reaction? (No new substances are formed)

III. Learning new material. Explanation of the teacher with elements of frontal conversation and problem solving. 30 minutes.

1. Let's try to formulate what is a solution? (slide 3)

Definition: solutions are homogeneous systems consisting of solvent molecules and solute particles, between which physical and chemical interactions occur.

2. Solubility b (slide 4) - the ability of a substance to form homogeneous systems with other substances (solvents) - solutions

• From the nature of the solute
• From temperature

3. Dependence on the nature of solutes (slide 5). All substances are divided into:

• well soluble,
• sparingly soluble,
• practically insoluble.

*Working with the solubility table

4. The dependence of the solubility of substances on temperature (slide 6)

*Work with the graph of the solubility of substances.
* In the Gulf of Kara-Bogaz-Gol (Turkmenistan), at a water temperature of +50C, a white precipitate of Na2SO4 salt precipitates at the bottom, and above this temperature the precipitate disappears. How do you think this can be explained?

5. Thus, solutions are (slide 7):

6. Solubility factor is the mass of a substance (g) that can be dissolved in one liter of solvent (l)

For example, the solubility of NANO3 is 80.5 g/l at 100C. This means that at a given temperature, 80.5 g of sodium nitrate can dissolve in one liter of water.

IV. Let's solve the problem (slide 8)

In 400 ml. water at 200C can dissolve 48 g of potassium sulfate. What is the solubility of potassium sulfate at a given temperature?

*** Interesting fact. Because the potassium sulfate recognized as a safe dietary supplement, it is approved for use in the countries of the European Union and on the territory of the Russian Federation. Most often, potassium sulfate finds its use as an additive as a salt substitute. In addition, it acts as an acidity regulator in beverages.

Solve the problem (slide 9).

Students solve the problem in pairs.

The tiger has cooked 20 o C 2 solutions: 5 liters of copper (II) chloride solution - (blue solution) and 3 liters of iron (III) chloride solution - (yellow solution). To prepare solutions, he took 2.8 kg. FeCl 3 and 3.2 kg. CuCl 2 . Which of the solutions did he turn out to be saturated, and which - not?
At 20 o C the solubility of CuCl 2 is 730 g/l, the solubility of FeCl 3 is 920 g/l

Solution:

The solubility of CuCl 2 is 730 g / l, therefore, to prepare 5 liters of a saturated solution, he needs 730 x 5 \u003d 3650, he took 3.2 kg \u003d 3200 g. This means that the solution unsaturated.
The solubility of FeCl 3 is 920 g / l, therefore, to prepare 3 liters of a saturated solution, he needs 920 x 3 \u003d 2760, he took 2.8 kg \u003d 2800 g. Hence, the solution saturated.

These concepts are relative, for example
25% HCl solution is concentrated, and
25% H 2 SO 4 solution - diluted

8. Expression of the concentration of solutions (slide 11)

One way to express the concentration of solutions is the mass fraction (w)

9. Solve problems (slide 12):.

Task 1. Calculate the mass fraction of the solution in%, which will be obtained if 50 g of the substance is dissolved in 450 g of water.

Z hell 2. Calculate the mass of water and mass of salt that must be taken to prepare 300 g of a solution with a mass fraction of 15%.

10. Solve the problems (slides 13, 14, 15).

Tasks are solved in pairs - 30 min.

Task 1. In order to process flowers, Winnie the Pooh needs to prepare 2 kg. 2% sodium nitrate solution. Help him calculate the mass of water and mass of salt that he needs to take?

Task 2. The heroes of this cartoon need to treat musical instruments with some 20% secret solution. They have 700 g of this solution at a concentration of 45%. How much water do they need to add to get what they need?

Task 3. Complete Aunt Owl's task. Calculate the mass fraction of the solution that will be obtained if 120 g of salt is dissolved in 1.4 kg. water.

Task 4. The healer mixed two solutions: 150 g of a 25% solution and 400 g of a 42% solution. Help him calculate the mass fraction of the resulting solution.

Task 5. Masha took 700 g of water for the broth, added 1.5 teaspoons of salt (15 g), tried it - the solution seemed too salty to her, and she added 500 g of water. A solution with what mass fraction of salt did Mashenka end up with?

Task 6. The mice helped Cinderella prepare the magic solution. They took two solutions: 200 g of a 10% solution of a secret substance and 250 g of a 25% solution of the same substance. Then they added 30 g of the substance to the resulting solution. How much water should be added to Cinderella so that the mass fraction of the solution is 15%?

V. Checking solved problems on the board– 14 min. ( Appendix 2 )

VI. Homework(slide 16) - 1 min.

1. Solve problems 1,2,3,4 page 81
2. Make up your problem on the topic "Solutions". Write it down on a 12 cm x 7 cm card made of white paper.

In the next lesson, we will make a lottery of your tasks. You will solve each other's problems and give each other marks.

A solution is a homogeneous system consisting of two or more substances, the content of which can be changed within certain limits without disturbing the homogeneity.

Aquatic solutions are made up of water(solvent) and solute. The state of substances in an aqueous solution, if necessary, is indicated by a subscript (p), for example, KNO 3 in solution - KNO 3 (p) .

Solutions that contain a small amount of solute are often referred to as diluted while solutions with high solute content concentrated. A solution in which further dissolution of a substance is possible is called unsaturated and a solution in which a substance ceases to dissolve under given conditions is saturated. The last solution is always in contact (in heterogeneous equilibrium) with the undissolved substance (one or more crystals).

Under special conditions, such as gentle (without stirring) cooling of a hot unsaturated solution solid substances can form supersaturated solution. When a crystal of a substance is introduced, such a solution is separated into a saturated solution and a precipitate of the substance.

In accordance with chemical theory of solutions D. I. Mendeleev, the dissolution of a substance in water is accompanied, firstly, destruction chemical bonds between molecules (intermolecular bonds in covalent substances) or between ions (in ionic substances), and, thus, particles of a substance mix with water (in which part of the hydrogen bonds between molecules is also destroyed). Chemical bonds are broken due to the thermal energy of the movement of water molecules, and in this case cost energy in the form of heat.

Secondly, once in the water, the particles (molecules or ions) of the substance are subjected to hydration. As a result, hydrates- compounds of indefinite composition between particles of a substance and water molecules (the internal composition of the particles of a substance itself does not change when dissolved). This process is accompanied highlighting energy in the form of heat due to the formation of new chemical bonds in hydrates.

In general, a solution cools down(if the cost of heat exceeds its release), or heats up (otherwise); sometimes - if the cost of heat and its release are equal - the temperature of the solution remains unchanged.

Many hydrates are so stable that they do not break down even when the solution is completely evaporated. So, solid crystal hydrates of salts CuSO 4 5H 2 O, Na 2 CO 3 10H 2 O, KAl (SO 4) 2 12H 2 O, etc. are known.

The content of a substance in a saturated solution at T= const quantifies solubility this substance. Solubility is usually expressed as the mass of solute per 100 g of water, for example 65.2 g KBr/100 g H 2 O at 20 °C. Therefore, if 70 g of solid potassium bromide is introduced into 100 g of water at 20 °C, then 65.2 g of salt will go into solution (which will be saturated), and 4.8 g of solid KBr (excess) will remain at the bottom of the beaker.

It should be remembered that the solute content in rich solution equals, in unsaturated solution less and in supersaturated solution more its solubility at a given temperature. So, a solution prepared at 20 ° C from 100 g of water and sodium sulfate Na 2 SO 4 (solubility 19.2 g / 100 g H 2 O), with a content

15.7 g of salt - unsaturated;

19.2 g salt - saturated;

2O.3 g of salt is supersaturated.

The solubility of solids (Table 14) usually increases with increasing temperature (KBr, NaCl), and only for some substances (CaSO 4 , Li 2 CO 3) is the opposite observed.

The solubility of gases decreases with increasing temperature, and increases with increasing pressure; for example, at a pressure of 1 atm, the solubility of ammonia is 52.6 (20 ° C) and 15.4 g / 100 g H 2 O (80 ° C), and at 20 ° C and 9 atm it is 93.5 g / 100 g H 2 O.

In accordance with the solubility values, substances are distinguished:

– well soluble, the mass of which in a saturated solution is commensurate with the mass of water (for example, KBr - at 20 ° C the solubility is 65.2 g / 100 g H 2 O; 4.6 M solution), they form saturated solutions with a molarity of more than 0.1 M;

– sparingly soluble, the mass of which in a saturated solution is much less than the mass of water (for example, CaSO 4 - at 20 ° C the solubility is 0.206 g / 100 g H 2 O; 0.015 M solution), they form saturated solutions with a molarity of 0.1–0.001 M;

– practically insoluble the mass of which in a saturated solution is negligible compared to the mass of the solvent (for example, AgCl - at 20 ° C, the solubility is 0.00019 g per 100 g of H 2 O; 0.0000134 M solution), they form saturated solutions with a molarity of less than 0.001 M.

Compiled according to reference data solubility table common acids, bases and salts (Table 15), in which the type of solubility is indicated, substances are noted that are not known to science (not obtained) or completely decomposed by water.

The ability of a substance to dissolve in water or another solvent is called solubility. The quantitative characteristic of solubility is the solubility coefficient, which shows what is the maximum mass of a substance that can be dissolved in 1000 or 100 g of water at a given temperature. The solubility of a substance depends on the nature of the solvent and substance, on temperature and pressure (for gases). The solubility of solids generally increases with increasing temperature. The solubility of gases decreases with increasing temperature, but increases with increasing pressure.

According to their solubility in water, substances are divided into three groups:

• 1. Highly soluble (p.). The solubility of substances is more than 10 g in 1000 g of water. For example, 2000 g of sugar dissolves in 1000 g of water, or 1 liter of water.
• 2. Slightly soluble (m.). The solubility of substances is from 0.01 g to 10 g of a substance in 1000 g of water. For example, 2 g of gypsum (CaSO4 * 2H20) is dissolved in 1000 g of water.
• 3. Practically insoluble (n.). The solubility of substances is less than 0.01 g of a substance in 1000 g of water. For example, 1.5 * 10_3 g of AgCl dissolves in 1000 g of water.

When substances are dissolved, saturated, unsaturated and supersaturated solutions can be formed.

A saturated solution is a solution that contains the maximum amount of solute under given conditions. When a substance is added to such a solution, the substance no longer dissolves.

An unsaturated solution is a solution that contains less solute than a saturated solution under given conditions. When a substance is added to such a solution, the substance still dissolves.

Sometimes it is possible to obtain a solution in which the solute contains more than in a saturated solution at a given temperature. Such a solution is called supersaturated. This solution is obtained by carefully cooling the saturated solution to room temperature. Supersaturated solutions are very unstable. Crystallization of a substance in such a solution can be caused by rubbing the walls of the vessel in which the solution is located with a glass rod. This method is used when performing some qualitative reactions.

The solubility of a substance can also be expressed by the molar concentration of its saturated solution.

The rate of the dissolution process depends on the substances being dissolved, the state of their surfaces, the temperature of the solvent, and the concentration of the final solution.

Do not confuse the concepts of "saturated" and "dilute" solution. For example, a saturated solution of silver chloride (1.5 * 10-3g / l) is yavl. very dilute, and unsaturated solution of sugar (1000 g / l) - concentrated.

Concentration of solutions and methods of its expression

According to modern concepts, the quantitative composition of a solution can be expressed both with the help of dimensionless quantities and quantities with dimensions. Dimensionless quantities are usually called fractions. 3 types of fractions are known: mass (u), volume (c), molar (h)

The mass fraction of a solute is the ratio of the mass of the solute X to the total mass of the solution:

u (X) \u003d t (X) / t

where w(X) is the mass fraction of the dissolved substance X, expressed in fractions of a unit; m(X) -- mass of solute X, g; m is the total mass of the solution, g.

If the mass fraction of dissolved sodium chloride in the solution is 0.03, or 3%, then this means that 100 g of the solution contains 3 g of sodium chloride and 97 g of water.

Volume fraction of a substance in a solution - the ratio of the volume of a solute to the sum of the volumes of all substances involved in the formation of a solution (before mixing)

c(X)= V(X)/?V

The molar fraction of a substance in a solution is the ratio of the amount of the substance to the sum of the amounts of all the substances in the solution.

h(X)=p(X)/ ?p

Of all types of fractions in analytical chemistry, the mass fraction is most often used. The volume fraction is usually used for solutions of gaseous substances and liquids (in pharmacy for solutions of ethyl alcohol) The numerical value is expressed in fractions of a unit and ranges from 0 (pure solvent) to 1 (pure substance. As you know, a hundredth of a unit is called a percentage. Percent - this is not a unit of measurement, but just a synonym for the concept of "one hundredth". For example, if the mass fraction of NaOH in a certain solution is 0.05, then instead of five hundredths, you can use the value 5%. Percentages cannot be mass, volume or molar, and can only be calculated by mass, volume or amount of substance.

The mass fraction can also be expressed as a percentage.

For example, a 10% sodium hydroxide solution contains 10 g of NaOH and 90 g of water in 100 g of a solution.

Cmas(X) = m(X)/tcm 100%.

Volume percentage - the percentage of the volume of a substance contained in the total volume of the mixture. Indicates the number of milliliters of the substance in 100 ml of the volume of the mixture.

Sob% \u003d V / Vcm * 100

The relationship between the volume and mass of the solution (t) is expressed by the formula

where c is the density of the solution, g/ml; V is the volume of the solution, ml.

The dimensional quantities used to describe the quantitative composition of solutions include the concentration of a substance in a solution (mass, molar) and the molality of a solute. If earlier any methods of describing the quantitative composition of a solution were called concentrations of a substance, then today this concept has become narrower.

Concentration is the ratio of the mass or amount of a solute to the volume of a solution. Thus, the mass fraction, according to the modern approach, is no longer a concentration and should not be called a percentage concentration.

Mass concentration is the ratio of the mass of a solute to the volume of a solution. This type of concentration is denoted as g (X), s (X) or not to be confused with the density of the solution, s * (X)

The mass concentration unit is kg/m3 or equivalently, g/l. The mass concentration, which has the dimension g / ml, is called the titer of the solution

Molar concentration - C (X) - is the ratio of the amount of a solute (mol) to the volume of a solution (1 l) Calculated as the ratio of the amount of substance p (X) contained in a solution to the volume of this solution V:

C(X) = n(X)/ Vp= m(X)/M(X)V

where m(X) is the mass of the dissolved substance, g; M(X) is the molar mass of the solute, g/mol. The molar concentration is expressed in mol/dm3 (mol/l). The most commonly used unit is mol/l. If 1 liter of a solution contains 1 mole of a solute, then the solution is called molar (1 M). If 1 liter of a solution contains 0.1 mol or 0.01 mol of a solute, then the solution is respectively called decimolar (0.1 M), centimolar (0.01 M), 0.001 mol-millimolar (0.001M)

The unit of measurement of molar concentration is mol/m3, but in practice a multiple of the unit, mol/l, is usually used. Instead of the designation “mol / l”, you can use “M” (and the word solution no longer needs to be written) For example, 0.1 M NaOH means the same as C (NaOH) \u003d 0.1 mol / l

A mole is a unit of chemical quantity of a substance. A mole is a portion of a substance (i.e., such an amount) that contains as many structural units as there are atoms in 0.012 kg of carbon. 0.012 kg of carbon contains 6.02*1023 carbon atoms. And this portion is 1 mol. The same number of structural units is contained in 1 mole of any substance. that is, a mole is the amount of a substance containing 6.02 * 1023 particles. This value is called the Avogadro constant.

The chemical amount of any substances contains the same number of structural units. But for each substance, its structural unit has its own mass. Therefore, the masses of the same chemical quantities of different substances will also be different.

Molar mass is the mass of a portion of a substance with a chemical quantity of 1 mole. It is equal to the ratio of the mass m of a substance to the corresponding amount of substance n

In the International System of Units, molar mass is expressed in kg/mol, but g/mol is more commonly used in chemistry.

It should be noted. That the molar mass numerically coincides with the masses of atoms and molecules (in amu) and with the relative atomic and molecular masses.

Unlike solids and liquids, all gaseous substances with a chemical quantity of 1 mol occupy the same volume (under the same conditions) This value is called the molar volume and is denoted

Because Since the volume of gas depends on temperature and pressure, then when performing calculations, the volumes of gases are taken under normal conditions (0? C and a pressure of 101.325 kPa). the ratio of the volume of any portion of gas to the chemical amount of gas is a constant value equal to 22.4 dm3/mol, i.e. Molar volume of any gas under normal conditions = 22.4 dm3/mol

Relationship between molar mass, molar volume and density (mass of a liter)

c= M/ Vm, g/dm3

The concept of molar concentration can refer to both the molecule or formula unit of a solute, and its equivalent. From a fundamental point of view, it does not matter what we are talking about: the concentration of sulfuric acid molecules - C (H2SO4) or "halves of sulfuric acid molecules" - C (1/2 H2SO4). The molar concentration of the equivalent of a substance used to be called the normal concentration. In addition, the molar concentration was often called molarity, although such a term is not recommended (it can be confused with molality)

The molality of a solute is the ratio of the amount of a substance in solution to the mass of the solvent. Designate molality as m(X), b(X), Cm(X):

Cm(X)= n(X)/mS

The unit of molality is mol/kg. Molality, according to modern terminology, is not a concentration. It is used in cases where the solution is under non-isothermal conditions. A change in temperature affects the volume of the solution and thereby leads to a change in concentration - while the molality remains constant.

For the quantitative characterization of standard solutions, the molar concentration (of a substance or equivalent of a substance) is usually used

Normality of solutions. Gram equivalent.

The concentration of solutions in titrimetric analysis is often expressed in terms of titer, i.e. indicate how many grams of a solute are contained in 1 ml of a solution. It is even more convenient to express it in terms of normality.

Normality is a number that indicates how many gram equivalents of a solute are contained in 1 liter of a solution.

The gram-equivalent (g-equiv) of a substance is the number of grams of it, chemically equivalent (equivalent) to one gram-atom of hydrogen in this reaction.

Cn \u003d peq / V; Cn = z n/V,

Where peq is the number of equivalents of the solute, peq = z n, V is the volume of the solution in liters, n is the number of moles of the solute, z is the effective valency of the solute

To find the gram equivalent, you need to write the reaction equation and calculate how many grams of a given substance corresponds to 1 gram of a hydrogen atom in it.

For example:

HCl + KOH KCl +H2O

One gram equivalent of an acid is equal to one gram molecule - a mole (36.46 g) of HCl, since it is this amount of acid that corresponds to one gram of hydrogen atom interacting with alkali hydroxyl ions during the reaction.

Accordingly, a gram-molecule of H2SO4 in the reactions:

H2SO4 + 2NaOH Na2SO4 + 2H2O

Corresponds to two grams of hydrogen atoms. Therefore, the gram equivalent of H2SO4 is? gram molecules (49.04 g).

Unlike a gram-molecule, a gram-atom, this number is not constant, but depends on the reaction in which the given substance is involved.

Since one gram-atom of OH- reacts with one gram-atom of H + and, therefore, is equivalent to the latter, the gram-equivalents of the bases are found similarly, but with the only difference that in this case they have to be divided by the number of gram-molecules participating in the reaction OH- ions.

Along with the gram equivalent in analytical chemistry, the concept of milligram equivalent is often used. A milligram equivalent (mg equivalent) is equal to a thousandth of a gram equivalent (E:1000) and is the equivalent weight of a substance expressed in milligrams. For example, 1 g-eq HCl is 36.46 g, and 1 meq HCl is 36.46 mg.

From the concept of an equivalent as a chemically equivalent quantity, it follows that the gram equivalents are precisely those weight quantities with which they react with each other.

It is obvious that 1 mg-eq of these substances, which is 0.001 g-eq, is in 1 ml of one-normal solutions of these substances. Therefore, the normality of a solution shows how many gram equivalents of a substance are contained in 1 liter or how many milligram equivalents of it are contained in 1 ml of a solution. The normality of solutions is denoted by the letter n. If 1 liter of solution contains 1 g-eq. substances, then such a solution is called 1 normal (1 n), 2 g-eq - two-normal (2 n), 0.5 g-eq - semi-normal, 0.1 g-eq - decinormal (0.1n), 0.01 g-eq - centinormal, 0.001 g-equiv - millinormal (0.001n). Of course, the normality of the solution, in addition, shows the number of milligram equivalents of the solute in 1 ml of the solution. For example, 1n solution contains 1 mEq, and 0.5 n - 0.5 mEq of a solute per 1 ml. The preparation of normal solutions requires the ability to calculate the gram equivalents of an acid, base or salt.

Gram-equivalent is the number of grams of a substance that is chemically equivalent (i.e. equivalent) to one gram-atom or gram-ion of hydrogen in a given reaction.

Np: HCl + NaOH= NaCl+H2O

It can be seen that one HCl gram molecule participates in the reaction with one H+ gram ion interacting with the OH- ion. Obviously, in this case, the gram equivalent of HCl is equal to its gram molecule and is 36.46 g. However, the gram equivalent of acids, bases and salts depends on the course of the reactions in which they participate. To calculate them, in each case, an equation is written and it is determined how many grams of the substance correspond to 1 gram-atom of hydrogen in this reaction. H-P, molecules of phosphoric acid H3PO4, participating in the reaction

H3PO4 + NaOH=NaH2PO4+ H2O

Gives only one H + ion and its gram equivalent is equal to a gram molecule (98.0 g). In the reaction

H3PO4 + 2NaOH = Na2HPO4+ 2H2O

each molecule corresponds to two grams of hydrogen ions. Therefore, gram-equiv. Is she equal? gram molecules, i.e. 98:2=49g

Finally, the H3PO4 molecule can also participate in the reaction with three hydrogen ions:

H3PO4 + 3NaOH=Na3PO4+ 3H2O

it is clear that in this reaction the H3PO4 gram molecule is equivalent to three H+ gram ions and the gram equivalent of the acid is 1/3 of the gram molecule, i.e. 98:3=32.67g

Gram-equiv-you bases also depend on the nature of the reaction. When calculating the gram equivalent of a base, one usually divides its gram molecule by the number of OH- ions participating in the reaction, because one OH- gram ion is equivalent to one H+ gram ion, Therefore, based on the equations

The order of conversion from one type of concentration to another. Calculations using molar concentration

In most cases, when calculating using the molar concentration, one proceeds from the proportions relating the molar concentration and the molar mass

Where C (X) is the concentration of the solution in mol / l; M is the molar mass, g / mol; m(X)/ is the mass of the solute in grams, p(X) is the amount of the solute in moles, Vp is the volume of the solution in liters. Example, calculate the molar concentration of 2 liters of 80 g of NaOH.

C(X) = m(X)/M Vp; M = 40 g/mol; C (X) \u003d 80g / 40g / mol * 2l \u003d 1 mol / l

Calculations using normality

Where Sp is the concentration of the solution in mol / l; M-molar mass, g/mol; m(X)/ is the mass of the solute in grams, p(X) is the amount of the solute in moles, Vp is the volume of the solution in liters.

Concentration of solutions and methods of its expression (Chemical analysis in thermal power engineering, Moscow. MPEI Publishing House, 2008)

The quantitative ratios between the masses of reacting substances are expressed by the law of equivalents. Chemical elements and their compounds enter into chemical reactions with each other in strictly defined mass quantities corresponding to their chemical equivalents.

Let the following reaction take place in the system:

aX+ b Y > Reaction products.

The reaction equation can also be written as

X + b/a Y > Reaction products,

which means that one particle of substance X is equivalent to b/a particles of substance Y.

Attitude

Equivalence factor, a dimensionless value not exceeding 1. Its use as a fractional value is not very convenient. More often, the reciprocal of the equivalence factor is used - the equivalence number (or equivalent number) z;

The value of z is determined by the chemical reaction in which a given substance participates.

There are two definitions of the equivalent:

• 1. An equivalent is a certain real or conditional particle that can attach, release, or in some other way be equivalent to one hydrogen ion in acid-base reactions or one electron in redox reactions.
• 2. Equivalent - a conditional particle of a substance, z times smaller than its corresponding formula unit. Formula units in chemistry are actually existing particles, such as atoms, molecules, ions, radicals, conditional molecules of crystalline substances and polymers.

The unit of the amount of substance equivalents is mole or mmol (previously g-eq or mg-eq). The value required for calculations is the molar mass of the equivalent of the substance Meq (Y), g / mol, equal to the ratio of the mass of the substance mY to the amount of substance equivalents neq (Y):

Meq(Y) = mY / neq(Y)

since neq

Consequently

Meq(Y) =MY / zY

where MY is the molar mass of substance Y, g/mol; nY is the amount of substance Y, mol; zY is the equivalence number.

The concentration of a substance is a physical quantity (dimensional or dimensionless) that determines the quantitative composition of a solution, mixture or melt. Various methods are used to express the concentration of a solution.

Molar concentration of substance B or concentration of the amount of substance - the ratio of the amount of dissolved substance B to the volume of the solution, mol / dm3,

St = nv / Vp = mv / Mv Vp

where nv is the amount of substance, mol; Vp is the volume of the solution, dm3; MB -- molar mass of the substance, g/mol; mB is the mass of the solute, g.

The abbreviated form of the molar concentration unit M = mol/dm3 is convenient to use.

Molar concentration of equivalents of substance B - the ratio of the number of equivalents of substance B to the volume of the solution, mol / dm3? n:

Seq (V) \u003d n equiv (V) / Vp \u003d mv / Mv Vp \u003d mv zv / Mv Vp

where neq is the amount of substance equivalents, mole; Meq -- molar mass of substance equivalents, g/mol; zB is an equivalence number.

The use of the terms "normality" and "normal concentration" and units of measurement g-eq/dm3, mg-eq/dm3 is not recommended, as well as the symbol N, for the abbreviated designation of the molar concentration of substance equivalents.

Mass concentration of substance B - the ratio of the mass of the dissolved substance B to the volume of the solution, g / dm3,

Mass fraction of solute B is the ratio of the mass of solute B to the mass of the solution:

Sv = mv / mr = mv / s Vp

where mr is the mass of the solution, g; c is the density of the solution, g/cm3.

The use of the term "percent concentration" is not recommended.

The molar fraction of a solute B is the ratio of the amount of this substance to the total amount of all substances that make up the solution, including the solvent,

XV= nV / ? ni, ? ni = nВ + n1 + n2 +.....+ ni

The molality of substance B in solution is the amount of solute B contained in 1 kg of solvent, mol / kg,

Cm \u003d nv / ms \u003d mv / Mv ms

where ms is the mass of the solvent, kg.

Titer - The titer of a solution of substance B is the concentration of a standard solution equal to the mass of substance B contained in 1 cm3 of solution, g / cm3,

At present, the use of many terms is not recommended, but in the practice of water treatment and in production, specialists use these terms and units of measurement, therefore, in order to eliminate discrepancies, the usual terms and units of measurement will be used in the future, and new terminology will be indicated in brackets.

According to the law of equivalents, substances react in equivalent quantities:

neq (X) = neq (Y), and neq (X) = Seq (X) Vx and neq (Y) = Seq (Y) Vy

therefore, one can write

Seq (X) Vx = Seq (Y) Vy

where neqv(X) and neqv(Y) -- the amount of substance equivalents, mol; Seq (X) and Seq (Y) - normal concentrations, g-eq / dm3 (molar concentrations of substance equivalents, mol / dm3); VX and VY are volumes of reacting solutions, dm3.

Let us assume that it is necessary to determine the concentration of a solution of a titrated substance X-- Ceq(X). To do this, accurately measure an aliquot of this VX solution. Then, a titration reaction is carried out with a solution of substance Y with a concentration of Ceq (Y) and note how much solution is used for titration of VY - titrant. Further, according to the law of equivalents, we can calculate the unknown concentration of a solution of substance X:

Equilibrium in solutions. True solutions and suspensions. Equilibrium in the "precipitate - saturated solution" system. Chemical equilibrium

Chemical reactions can proceed in such a way that the substances taken are completely converted into reaction products - as they say, the reaction goes to the end. Such reactions are called irreversible. An example of an irreversible reaction is the decomposition of hydrogen peroxide:

2H2O2 = 2H2O + O2 ^

Reversible reactions proceed simultaneously in 2 opposite directions. because the products obtained as a result of the reaction interact with each other to form the starting substances. For example: when iodine vapor interacts with hydrogen at 300 ° C, hydrogen iodide is formed:

However, at 300?C, hydrogen iodide decomposes:

Both reactions can be expressed by one general equation, replacing the equal sign with the reversibility sign:

The reaction between the starting substances is called a direct reaction, and its rate depends on the concentration of the starting substances. A chemical reaction between products is called a reverse reaction, and its rate depends on the concentration of the starting substances. A chemical reaction between products is called a reverse reaction, and its rate depends on the concentration of the substances obtained. At the beginning of a reversible process, the rate of the forward reaction is maximum, and the rate of the reverse is zero. As the process proceeds, the rate of the direct reaction decreases, because the concentration of the taken substances decreases, and the rate of the reverse reaction increases, as the concentration of the obtained substances increases. When the rates of both reactions become equal, a state called chemical equilibrium sets in. In chemical equilibrium, neither the forward nor the reverse reactions stop; they both move at the same speed. Therefore, the chemical equilibrium is a mobile, dynamic equilibrium. The state of chemical equilibrium is influenced by the concentration of reacting substances, temperature, and for gaseous substances - pressure in the system.

By changing these conditions, it is possible to shift the equilibrium to the right (in this case, the product yield will increase) or to the left. Offset chem. equilibrium obeys the Le Chatelier principle:

Under steady state equilibrium, the product of the concentrations of the reaction products divided by the product of the concentrations of the starting materials (for a given reaction, T=const) is a constant value called the equilibrium constant.

When external conditions change, the chemical equilibrium shifts in the direction of the reaction that weakens this external influence. So, with an increase in the concentration of reacting substances, the equilibrium shifts towards the formation of reaction products. The introduction of additional amounts of any of the reactants into the equilibrium system accelerates the reaction in which it is consumed. An increase in the concentration of the starting substances shifts the equilibrium towards the formation of reaction products. An increase in the concentration of reaction products shifts the equilibrium towards the formation of starting materials.

Reactions occurring in the process of chemical analysis. Types of reactions. Characteristic. Types of chemical reactions

Chemical reactions can be classified into four main types:

decomposition	connections	substitution
Decomposition reaction- is called such a chem. reaction, in a cat. from one complex thing-va it turns out two or more. simple or complex substances: 2H2O > 2H2^ +O2^3	A compound reaction is such a reaction, in the result of which one more complex substance is formed from two or more simple or complex substances:	A substitution reaction is a reaction that occurs between simple and complex substances, with a cat. atoms is simple. things replace the atoms of one of the elements in a complex substance: Fe+CuCl2> Cu+FeCl2 Zn+CuCl2>ZnCl2+Cu	An exchange reaction is a reaction in which two complex substances exchanges its constituent parts, forming two new substances: NaCl+AgNO3=AgCl+NaNO3

According to the release and absorption of energy, chemical reactions are divided into exothermic, going with the release of heat into the environment and endothermic, going with the absorption of heat from the environment.

The science of methods for analyzing the composition of an analyte, (in a broad sense) and methods for a comprehensive chemical study of substances that surround us on Earth is called analytical chemistry. The subject of analytical chemistry is the theory and practice of various methods of analysis. The analysis of a substance is carried out in order to establish its qualitative or quantitative chemical composition.

The task of qualitative analysis is the discovery of elements, sometimes compounds that make up the substance under study. Quantitative analysis makes it possible to determine the quantitative ratio of these components.

In a qualitative analysis, to establish the composition of the analyte, other substances are added to it, causing such chemical transformations, which are accompanied by the formation of new compounds with specific properties:

• - a certain physical state (precipitate, liquid, gas)
• - known solubility in water, acids, alkalis and other solvents
• - characteristic color
• - crystalline or amorphous structure
• - smell

Qualitative analysis in the study of the composition of an unknown substance always precedes quantitative, because. the choice of method for quantifying the constituents of the analyte depends on the data obtained using a qualitative analysis. The results of a qualitative analysis do not make it possible to judge the properties of the materials under study, since the properties are determined not only by what parts the object under study consists of, but also by their quantitative ratio. When starting a quantitative analysis, it is necessary to know exactly the qualitative composition of the substance under study; knowing the qualitative composition of the substance and the approximate content of the components, it is possible to choose the right method for the quantitative determination of the element of interest to us.

In practice, the task facing the analyst is usually greatly simplified due to the fact that the qualitative composition of most of the studied materials is well known.

Methods of quantitative analysis

Methods of quantitative analysis, depending on the nature of the experimental technique used for the final determination of the constituent parts of the analyte, are divided into 3 groups:

• - chemical
• - physical
• - physico-chemical (instrumental)

Physical methods - methods of analysis with which you can determine the composition of the substance under study, without resorting to the use of chemical reactions. Physical methods include:

• - spectral analysis - based on studies of emission spectra (or emission and absorption of the substances under study)
• - luminescent (fluorescent) - analysis based on the observation of luminescence (glow) of the analyzed substances, caused by the action of ultraviolet rays
• - x-ray structural - based on the use of x-rays to study the structure of matter
• - mass spectrometric analysis
• - methods based on measuring the density of the studied compounds

Physico-chemical methods are based on the study of physical phenomena that occur during chemical reactions, accompanied by a change in the color of the solution, color intensity (colorimetry), electrical conductivity (conductometry)

Chemical methods are based on the use of the chemical properties of elements or ions.

Chemical	Physico-chemical
Gravimetric	Titrimetric	colorimetric	Electrochemical
The method of quantitative analysis consists in the exact measurement of the mass of the analyzed component of the sample, isolated in the form of a compound of known composition or in the form of an element. The classical name of the weight method	The method of quantitative analysis is based on measuring the volume (or mass) of a solution of a reagent of known concentration, consumed for the reaction with the analyte. They are divided according to the type of reactions into 4 methods: - acid-base (alkalinity, acidity) - redox (bichromate - the substance is titrated with a solution of potassium dichromate, permanganatometry, iodometry) - complexometric: (titrant Trilon B)	Method of quantitative analysis based on the assessment of the color intensity of the solution (visually or with the help of appropriate instruments). Photometric determination is possible only if the color of the solutions is not too intense, therefore, highly diluted solutions are used for such measurements. In practice, photometric determinations are especially often used when the content of the corresponding element in the object under study is low and when the methods of gravimetric and titrimetric analysis are unsuitable. The rapidity of the determination contributes to the widespread use of the photometric method.	The method of quantitative analysis, it retains the usual principle of titrimetric determinations, but the moment of completion of the corresponding reaction is set by measuring the electrical conductivity of the solution (conductometric method), or by measuring the potential of one or another electrode immersed in the test solution (potentiometric method)

In quantitative analysis, macro-, micro- and semi-micro methods are distinguished.

In macroanalysis, relatively large (about 0.1 g or more) samples of the investigated solid or large volumes of solutions (several tens of milliliters or more) are taken. The main working tool in this method is an analytical balance, which allows weighing with an accuracy of 0.0001-0.0002 g, depending on the design of the balance (i.e., 0.1-0.2 mg).

In micro- and semi-micro methods of quantitative analysis, weighings from 1 to 50 mg and solution volumes from tenths of a milliliter to several milliliters are used. for these methods, more sensitive balances are used, such as microbalances (weighing accuracy up to 0.001 mg), as well as more accurate equipment for measuring the volumes of solutions.

Volumetric analysis, essence and characteristics of the method. The concept of titration, titre. General titration techniques, titer setting methods

Titrimetric (volumetric) analysis Essence of analysis.

Titrimetric analysis offers a huge advantage over gravimetric analysis in terms of speed. In titrimetric analysis, the volume of a reagent solution consumed for the reaction is measured, the concentration (or titer) of which is always exactly known. A titer is usually understood as the number of grams or milligrams of a solute contained in 1 ml of a solution. Thus, in titrimetric analysis, the quantitative determination of chemicals is most often carried out by accurately measuring the volumes of solutions of two substances that react with each other.

In the analysis, a titrated reagent solution is placed in a measuring vessel called a burette, and it is gradually poured into the test solution until it is established in one way or another that the spent amount of the reagent is equivalent to the amount of the analyte. This operation is called titration.

A titratable substance is a substance whose solution concentration is to be determined. In this case, the volume of the solution of the titratable substance must be known.

A titrant is a solution of a reagent used for titration, the concentration of which is known with high accuracy. It is often referred to as a standard (working) or titrated solution.

The solution can be prepared in several ways:

• - according to the exact weight of the starting substance (only chemically pure stable compounds, the composition of which strictly corresponds to the chemical formula, as well as easily cleaned substances, can be used as starting substances);
• - according to fixanal (according to a strictly defined amount of a substance, usually 0.1 mol or its fraction, placed in a glass ampoule);
• - by approximate weight with subsequent determination of the concentration according to the primary standard (it is necessary to have a primary standard - a chemically pure substance of exactly known composition that meets the relevant requirements);
• - by diluting a previously prepared solution with a known concentration.

Titration is the main method of titrimetric analysis, which consists in the gradual addition of a reagent solution of known concentration from a burette (titrant) to the analyzed solution until the equivalence point is reached. Often fixing the equivalence point. is possible due to the fact that the colored reagent changes its color during the reaction (during oxidizability titration). Or substances are added to the test solution that undergo any change during titration and thereby allow fixing the equivalence point, these substances are called indicators. The main characteristic of indicators is considered to be not the value of the end point of the titration, but the interval of the indicator color transition. The color change of the indicator becomes noticeable to the human eye not at a specific pT value,

Transition interval of acid-base indicators

Indicator	transition, pH	acid form	Main form
Alizarin yellow					purple
thymolphthalein				Colorless
Phenolphthalein				Colorless
Cresol purple					Purple
Phenol red
Bromothymol blue
methyl red
methyl orange
Bromophenol blue

However, even if indicators are available, their use is not always possible. In general, strongly colored or cloudy solutions should not be titrated with indicators, as the color change of the indicator becomes difficult to distinguish.

In such cases, the equivalence point is sometimes fixed by changing some of the physical properties of the solution during titration. Electrotitrimetric methods of analysis are based on this principle. For example, the conductometric method, in which the equivalence point is found by measuring the electrical conductivity of the solution; potentiometric method based on measuring the redox potential of a solution (potentiometric titration method).

In addition, it is necessary that the added titrated reagent solution be used exclusively for the reaction with the analyte, i.e. during the titration, no side reactions should occur that make an accurate calculation of the results of the analysis impossible. In the same way, the absence of substances in the solution that interfere with the course of the reaction or prevent the fixation of the equivalence point is necessary.

Only those chemical interactions between the titrated substance and the titrant that meet the following requirements can be used as a reaction:

• 1) the reaction must be strictly stoichiometric, i.e. the chemical composition of the titratable substance, titrant and reaction products must be strictly defined and unchanged;
• 2) the reaction must proceed quickly, since changes can occur in the solution for a long time (due to competing reactions), the nature and influence of which on the main titration reaction is quite difficult to predict and take into account;
• 3) the reaction must proceed quantitatively (if possible completely), i.e. the equilibrium constant of the titration reaction should be as high as possible;
• 4) there must be a way to determine the end of the reaction. .

In titrimetry, the following titration options are distinguished:

• - direct titration method. The titrant is directly added to the substance to be titrated. This method is used if all the requirements for the titration reaction are met;
• - back titration method. A known excess of titrant is added to the substance to be titrated, the reaction is brought to completion, and then the excess of unreacted titrant is titrated with another titrant, i.e. the titrant used in the first part of the experiment is itself converted into the titratable substance in the second part of the experiment. This method is used if the reaction rate is low, it is not possible to select an indicator, side effects are observed (for example, losses of the analyte due to its volatility), or the reaction is not stoichiometric; - method of indirect titration by substituent. A stoichiometric reaction of the titratable compound with another reagent is carried out, and the new compound resulting from this reaction is titrated with a suitable titrant. The method is used if the reaction is non-stoichiometric or occurs slowly.

DISSOLUTION.

SOLUBILITY OF SUBSTANCES IN WATER.

I DISSOLUTION AND SOLUTIONS.

DISSOLUTION. SOLUTIONS.

Physical theory (van't Hoff,

Ostwald, Arrhenius).

Dissolution is a diffusion process

a solutions are homogeneous mixtures.

chemical theory (Mendeleev,

Kablukov, Kistyakovsky).

Dissolution is a chemical process

solute interactions

with water, - the process of hydration,

a solutions These compounds are hydrates.

Modern theory.

Dissolution- This is a physico-chemical process that occurs between the solvent and the particles of the solute and is accompanied by the diffusion process.

Solutions- these are homogeneous (homogeneous) systems consisting of particles of a solute, a solvent and the products of their interaction - hydrates.

II SIGNS OF CHEMICAL INTERACTION DURING DISSOLUTION.

1. Thermal phenomena.

ü Exothermic - these are phenomena accompanied by the release of heat /dissolution of concentrated sulfuric acid H2SO4 in water/.

ü Endothermic- these are phenomena accompanied by the absorption of heat /dissolution of crystals of ammonium nitrate NH4NO3 in water/.

2. Color change.

CuSO4 + 5H2O → CuSO4∙ 5H2O

white blue crystals

crystals

3. Volume change.

III DEPENDENCE OF SOLID SUBSTANCES ON DISSOLUTION.

1. From the nature of substances:

ü highly soluble in water / more than 10 g of substance per 100 g of water /;

ü slightly soluble in water /less than 1g/;

ü practically insoluble in water /less than 0.01g/.

2. From temperature.

IV TYPES OF SOLUTIONS BY SOLUBILITY.

Ø According to the degree of solubility:

ü unsaturated solution - a solution in which, at a given temperature and pressure, further dissolution of the substance already contained in it is possible.

ü saturated solution - a solution that is in phase equilibrium with the solute.

ü Supersaturated solution - an unstable solution in which the content of a solute is greater than in a saturated solution of the same substance at those values ​​of temperature and pressure.

Ø According to the ratio of the solute to the solvent:

ü concentrated;

ü diluted.

THEORY OF ELECTROLYTIC DISSOCIATION (TED).

I. The theory of electrolytic dissociation (TED) was proposed by a Swedish scientist Svante Arrhenius in 1887

Later, TED developed and improved. The modern theory of aqueous solutions of electrolytes, in addition to the theory of electrolytic dissociation by S. Arrhenius, includes ideas about the hydration of ions (,), the theory of strong electrolytes (, 1923).

II. SUBSTANCES

electrolytes - substances, solutions

or whose melts conduct

electricity.

/acids, salts, bases/

Non-electrolytes Substances whose solutions or melts do not conduct electricity.

/simple substances/

IONS are charged particles.

ü cations /kat+/ are positively charged particles.

ü anions /an-/– negatively charged particles

III. MAIN PROVISIONS OF TED:

ü The spontaneous process of decomposition of an electrolyte into ions in a solution or in a melt is called electrolytic dissociation .

ü In aqueous solutions, ions are not in free, but in hydrated state, i.e., surrounded by dipoles of water and chemically associated with them. Ions in the hydrated state differ in properties from ions in the gaseous state of matter.

ü For the same solute, the degree of dissociation increases as the solution is diluted.

ü In solutions or melts of electrolytes, ions move randomly, but when an electric current is passed through a solution or melt of an electrolyte, the ions move in a direction: cations - to the cathode, anions - to the anode.

MECHANISM OF ELECTROLYTIC DISSOCIATION

1. ED of ionic substances:

ü Orientation of water dipoles relative to crystal ions.

ü The disintegration of the crystal into ions (proper dissociation).

ü Hydration of ions.

2. ED of substances with a covalent polar type of chemical bond.

ü Destruction of hydrogen bonds between water molecules, formation of water dipoles.

ü Orientation of water dipoles relative to the dipoles of a polar molecule.

ü Strong bond polarization, as a result of which the common electron pair is completely shifted to the atomic particle of a more electronegative element.

ü The disintegration of matter into ions (proper dissociation).

ü Hydration of ions.

DEGREE OF ELECTROLYTIC DISSOCIATION /α/

1. Degree of ED is the ratio of the number of decayed molecules to the total number of particles in the solution.

α = ─ ∙ 100%

Ntotal

2. According to the magnitude of the degree of ED, substances are divided:

ü strong electrolytes /HCl; H2SO4; NaOH; Na2CO3/

ü medium strength electrolytes /H3PO4/

ü weak electrolytes /H2CO3; H2SO3/.

CHEMICAL DICTATION

ON THE TOPIC: "ELETROLYTIC DISSOCIATION"

1. All water-soluble bases are strong electrolytes.

2. Only water-soluble salts undergo hydrolysis.

3. Dissociation is a reversible process.

4. The essence of the neutralization reaction, CH3COOH + KOH → CH3COOH + H2O, reflected in the form of a short ionic equation of a chemical reaction is: H++ OH- → H2O.

5. BaSO4 ; AgCl are water-insoluble salts, so they do not dissociate into ions.

6. Is the dissociation equation for the following salts correct:

ü Na2SO4 → 2Na+ + SO42-

ü KCl → K+ + Cl-

7. The dissociation equation for sulfurous acid has the following form: H2 SO3 → 2 H+ + SO3 2- .

8. The true degree of dissociation of a strong electrolyte is less than 100%.

9. As a result of the neutralization reaction, salt and water are always formed.

10. Only water-soluble bases - alkalis, are electrolytes.

11. The equations of chemical reactions presented below are ion exchange reactions:

ü 2KOH + SiO2 → K2SiO3 + H2O

ü Al2O3 + 2NaOH → 2NaAlO2 + H2O

ü CuO + 2HCl → CuCl2 + H2O

12. Sulfurous acid is a weak acid, so it decomposes into water (H2O) and sulfur dioxide (SO2).

H2SO3 → H2O + SO2.

THE CODE

1. No /excluding NH3∙H2O/

2. No: Al2S3 + 2H2O → 2AlOHS + H2S

3. No. /Dissociation of only weak electrolytes is a reversible process, strong electrolytes dissociate irreversibly/.

4. No: CH3COOH + OH - → CH3COO= + H2O.

5. No. /These salts are insoluble in relation to water, but they are able to dissociate/.

6. No. /These salts are strong electrolytes, so they dissociate irreversibly/.

7. No. /Polybasic acids dissociate stepwise/.

8. No. /The true degree of dissociation is equal to 100%/.

9. No: NH3(g.) + HCl(g.) → NH4Cl, water formation remains questionable.

10. No. /All bases are electrolytes/.

11. No. /These are exchange reactions, but ionic/.

12. No. /Decomposition of sulphurous acid occurs because it is a fragile acid/.

REGULATIONS

COMPILATION OF IONIC EQUATIONS OF CHEMICAL REACTIONS.

1. Simple substances, oxides, as well as insoluble acids, salts and bases do not decompose into ions.

2. Solutions are used for the ion exchange reaction, so even poorly soluble substances are in solutions in the form of ions. /If a poorly soluble substance is the original compound, then it is decomposed into ions when compiling ionic equations of chemical reactions/.

3. If the poorly soluble is formed as a result of the reaction, then when writing the ionic equation it is considered insoluble.

4. The sum of the electric charges on the left side of the equation must be equal to the sum of the electric charges on the right side.

TERMS

ION EXCHANGE REACTIONS

1. The formation of a low-dissociating substance of water - H2O:

ü HCl + NaOH → NaCl + H2O

H+ + Cl - + Na+ + OH- → Na+ + Cl - + H2O

H+ + OH - → H2O

ü Cu(OH)2 + H2SO4 → CuSO4 + 2H2O

Cu(OH)2 + 2H+ + SO42- → Cu2+ + SO42- + 2H2O

Cu(OH)2 + 2H+ → Cu2+ + 2H2O

2. Precipitation:

ü FeCl3 + 3NaOH → Fe(OH)3↓ + 3NaCl

Fe3++ 3Cl - + 3Na+ + 3OH- → Fe(OH)3↓ + 3Na++ 3Cl-

Fe3++ 3OH - → Fe(OH)3↓

ü BaCl2 + H2SO4 → BaSO4↓ + 2HCl

Ba2++ 2Cl - + 2H++ SO42- → BaSO4↓ + 2H++ 2Cl-

Ba2++ SO42- → BaSO4↓

ü AgNO3 + KBr → AgBr↓ + KNO3

Ag+ + NO3- + K++ Br - → AgBr↓ + K++ NO3-

Ag+ + Br - → AgBr↓

3. Gas release:

ü Na2CO3 + 2HCl → 2NaCl + H2O + CO2

2Na++ CO32-+ 2H++ 2Cl- → 2Na++ 2Cl - + H2O + CO2

CO32-+ 2H+ → H2O + CO2

ü FeS + H2SO4 → FeSO4 + H2S

FeS + 2H++ SO42-→ Fe2++ SO42-+ H2S

FeS + 2H+ → Fe2++ H2S

ü K2SO3 + 2HNO3 → 2KNO3 + H2O + SO2

2K++ SO32-+ 2H++ 2NO3- → 2K++ 2NO3- + H2O + SO2