Determination of one or more substances based on qualitative reactions. §5

Solving qualitative problems of identifying substances found in bottles without labels involves carrying out a number of operations, the results of which can be used to determine which substance is in a particular bottle.

The first stage of the solution is a thought experiment, which is a plan of action and its expected results. To record a thought experiment, a special table-matrix is ​​used, in which the formulas of the substances being determined are indicated horizontally and vertically. In places where the formulas of interacting substances intersect, the expected results of observations are recorded: - gas evolution, - precipitation, changes in color, odor or the absence of visible changes are indicated. If, according to the conditions of the problem, it is possible to use additional reagents, then it is better to write down the results of their use before compiling the table - the number of substances to be determined in the table can thus be reduced.
The solution to the problem will therefore consist of the following steps:
- preliminary discussion of individual reactions and external characteristics of substances;
- recording formulas and expected results of pairwise reactions in a table,
- conducting an experiment in accordance with the table (in the case of an experimental task);
- analysis of reaction results and correlating them with specific substances;
- formulation of the answer to the problem.

It must be emphasized that a thought experiment and reality do not always completely coincide, since real reactions take place at certain concentrations, temperatures, and lighting (for example, under electric light, AgCl and AgBr are identical). A thought experiment often leaves out many small details. For example, Br 2 /aq is perfectly decolorized with solutions of Na 2 CO 3, Na 2 SiO 3, CH 3 COONa; the formation of Ag 3 PO 4 precipitate does not occur in a strongly acidic environment, since the acid itself does not give this reaction; glycerol forms a complex with Cu (OH) 2, but does not form with (CuOH) 2 SO 4, if there is no excess alkali, etc. The real situation does not always agree with the theoretical prediction, and in this chapter there are “ideal” matrix tables and "realities" will sometimes be different. And in order to understand what is really happening, look for every opportunity to work with your hands experimentally in a lesson or elective (remember the safety requirements).

Example 1. The numbered bottles contain solutions of the following substances: silver nitrate, hydrochloric acid, silver sulfate, lead nitrate, ammonia and sodium hydroxide. Without using other reagents, determine which bottle contains the solution of which substance.

Solution. To solve the problem, we will compose a matrix table in which we will enter in the appropriate squares below the diagonal that intersects it the observation data of the results of merging substances from one test tube with another.

Observation of the results of sequentially pouring the contents of some numbered test tubes into all others:

1 + 2 - a white precipitate forms; ;
1 + 3 - no visible changes are observed;

Substances	1. AgNO 3,	2. HCl	3. Pb(NO 3) 2,	4.NH4OH	5.NaOH
1. AgNO3	X	AgCl white	-	the precipitate that falls dissolves	Ag 2 O brown
2. HCl	white	X	PbCl 2 white,	-	_
3. Pb(NO 3) 2	-	white PbCl 2	X	Pb(OH) 2 turbidity)	Pb(OH) 2 white
4.NH4OH	-	-	(turbidity)		-
S.NaOH	brown	-	white	-	X

1 + 4 - depending on the order in which the solutions are drained, a precipitate may form;
1 + 5 - a brown precipitate forms;
2+3 - a white precipitate forms;
2+4 - no visible changes are observed;
2+5 - no visible changes are observed;
3+4 - turbidity is observed;
3+5 - a white precipitate forms;
4+5 - no visible changes are observed.

Let us further write down the equations of the ongoing reactions in cases where changes are observed in the reaction system (emission of gas, sediment, color change) and enter the formula of the observed substance and the corresponding square of the matrix table above the diagonal that intersects it:

I. 1+2:	AgNO 3 + HCl		AgCl + HNO 3 ;
II. 1+5:	2AgNO3 + 2NaOH		Ag 2 O + 2NaNO 3 + H 2 O;
			brown (2AgOH Ag 2 O + H 2 O)
III. 2+3:	2HCl + Pb(NO 3) 2		PbCl 2 + 2HNO 3;
			white
IV. 3+4:	Pb(NO 3) 2 + 2NH 4 OH		Pb(OH) 2 + 2NH 4 NO 3 ;
			cloudiness
V.3+5:	Pb(NO 3) 2 + 2NaOH		Pb(OH) 2 + 2NaNO 3
			white

(when lead nitrate is added to excess alkali, the precipitate can immediately dissolve).
Thus, based on five experiments, we distinguish the substances in the numbered test tubes.

Example 2. Eight numbered test tubes (from 1 to 8) without inscriptions contain dry substances: silver nitrate (1), aluminum chloride (2), sodium sulfide (3), barium chloride (4), potassium nitrate (5), phosphate potassium (6), as well as solutions of sulfuric (7) and hydrochloric (8) acids. How, without any additional reagents other than water, can you distinguish between these substances?

Solution. First of all, let's dissolve the solids in water and mark the test tubes where they ended up. Let's create a matrix table (as in the previous example), in which we will enter data from observations of the results of merging substances from one test tube with another below and above the diagonal that intersects it. On the right side of the table we will introduce an additional column “general result of observation”, which we will fill in after completing all experiments and summing up the results of observations horizontally from left to right (see, for example, p. 178).

1+2:	3AgNO3 + A1C1,		3AgCl white	+ Al(NO 3) 3 ;
1 + 3:	2AgNO3 + Na2S		Ag 2 S black	+ 2NaNO 3 ;
1 + 4:	2AgNO3 + BaCl2		2AgCl white	+ Ba(NO 3) 2 ;
1 + 6:	3AgN0 3 + K 3 PO 4		Ag 3 PO 4 yellow	+ 3KNO 3 ;
1 + 7:	2AgNO3 + H2SO4		Ag,SO 4 white	+ 2HNO S;
1 + 8:	AgNO3 + HCl		AgCl white	+ HNO3;
2 + 3:	2AlCl 3 + 3Na 2 S + 6H 2 O		2Al(OH)3,	+ 3H 2 S + 6NaCl;
		(Na 2 S + H 2 O NaOH + NaHS, hydrolysis);
2 + 6:	AlCl 3 + K 3 PO 4		A1PO 4 white	+ 3KCl;
3 + 7:	Na 2 S + H 2 SO 4		Na2SO4	+H2S
3 + 8:	Na 2 S + 2HCl		-2NaCl	+H2S;
4 + 6:	3BaCl2 + 2K3PO4		Ba 3 (PO 4) 2 white	+ 6KC1;
4 + 7	BaCl 2 + H 2 SO 4		BaSO 4 white	+ 2HC1.

Visible changes do not occur only with potassium nitrate.

Based on the number of times a precipitate forms and gas is released, all reagents are uniquely identified. In addition, BaCl 2 and K 3 PO 4 are distinguished by the color of the precipitate with AgNO 3: AgCl is white, and Ag 3 PO 4 is yellow. In this problem, the solution may be simpler - any of the acid solutions allows you to immediately isolate sodium sulfide, which determines silver nitrate and aluminum chloride. Among the remaining three solid substances, barium chloride and potassium phosphate are determined by silver nitrate; hydrochloric and sulfuric acids are distinguished by barium chloride.

Example 3. Four unlabeled test tubes contain benzene, chlorhexane, hexane and hexene. Using the minimum quantities and number of reagents, propose a method for determining each of the specified substances.

Solution. The substances being determined do not react with each other; there is no point in compiling a table of pairwise reactions.
There are several methods for determining these substances, one of them is given below.
Only hexene immediately discolors bromine water:

C 6 H 12 + Br 2 = C 6 H 12 Br 2.

Chlorhexane can be distinguished from hexane by passing their combustion products through a solution of silver nitrate (in the case of chlorhexane, a white precipitate of silver chloride precipitates, insoluble in nitric acid, unlike silver carbonate):

2C 6 H 14 + 19O 2 = 12CO 2 + 14H 2 O;
C 6 H 13 Cl + 9O 2 = 6 CO 2 + 6 H 2 O + HC1;
HCl + AgNO 3 = AgCl + HNO 3.

Benzene differs from hexane in freezing in ice water (C 6 H has 6 mp. = +5.5 ° C, and C 6 H has 14 mp. = -95.3 ° C).

1. Equal volumes are poured into two identical beakers: one of water, the other of a dilute solution of sulfuric acid. How can you distinguish between these liquids without having any chemical reagents at hand (you cannot taste the solutions)?

2. Four test tubes contain powders of copper(II) oxide, iron(III) oxide, silver, and iron. How to recognize these substances using only one chemical reagent? Recognition by appearance excluded.

3. Four numbered test tubes contain dry copper(II) oxide, carbon black, sodium chloride, and barium chloride. How, using a minimum amount of reagents, can you determine which test tube contains which substance? Justify your answer and confirm it with the equations of the corresponding chemical reactions.

4. Six unlabeled test tubes contain anhydrous compounds: phosphorus(V) oxide, sodium chloride, copper sulfate, aluminum chloride, aluminum sulfide, ammonium chloride. How can you determine the contents of each test tube if all you have is a set of empty test tubes, water, and a burner? Propose an analysis plan.

5 . Four unmarked test tubes contain aqueous solutions of sodium hydroxide, hydrochloric acid, potash and aluminum sulfate. Suggest a way to determine the contents of each test tube without using additional reagents.

6 . The numbered test tubes contain solutions of sodium hydroxide, sulfuric acid, sodium sulfate and phenolphthalein. How to distinguish between these solutions without using additional reagents?

7. Unlabeled jars contain the following individual substances: powders of iron, zinc, calcium carbonate, potassium carbonate, sodium sulfate, sodium chloride, sodium nitrate, as well as solutions of sodium hydroxide and barium hydroxide. There are no other chemical reagents at your disposal, including water. Make a plan to determine the contents of each jar.

8 . Four numbered jars without labels contain solid phosphorus (V) oxide (1), calcium oxide (2), lead nitrate (3), calcium chloride (4). Determine which jar contains each from of the indicated compounds, if it is known that substances (1) and (2) react violently with water, and substances (3) and (4) dissolve in water, and the resulting solutions (1) and (3) can react with all other solutions with formation of precipitation.

9 . Five test tubes without labels contain solutions of hydroxide, sulfide, chloride, sodium iodide and ammonia. How to determine these substances using one additional reagent? Give equations for chemical reactions.

10. How to recognize solutions of sodium chloride, ammonium chloride, barium hydroxide, sodium hydroxide contained in vessels without labels, using only these solutions?

11. . Eight numbered test tubes contain aqueous solutions of hydrochloric acid, sodium hydroxide, sodium sulfate, sodium carbonate, ammonium chloride, lead nitrate, barium chloride, and silver nitrate. Using indicator paper and carrying out any reactions between solutions in test tubes, determine what substance is contained in each of them.

12. Two test tubes contain solutions of sodium hydroxide and aluminum sulfate. How to distinguish them, if possible, without the use of additional substances, having only one empty test tube or even without it?

13. Five numbered test tubes contain solutions of potassium permanganate, sodium sulfide, bromine water, toluene and benzene. How can you distinguish between them using only the named reagents? Use their characteristic features to detect each of the five substances (indicate them); give a plan for the analysis. Write diagrams of the necessary reactions.

14. Six unnamed bottles contain glycerin, an aqueous solution of glucose, butyraldehyde (butanal), 1-hexene, an aqueous solution of sodium acetate and 1,2-dichloroethane. With only anhydrous sodium hydroxide and copper sulfate as additional chemicals, determine what is in each bottle.

1. To determine water and sulfuric acid, you can use the difference in physical properties: boiling and freezing points, density, electrical conductivity, refractive index, etc. The strongest difference will be in electrical conductivity.

2. Add hydrochloric acid to the powders in test tubes. Silver will not react. When iron dissolves, gas will be released: Fe + 2HCl = FeCl 2 + H 2
Iron (III) oxide and copper (II) oxide dissolve without releasing gas, forming yellow-brown and blue-green solutions: Fe 2 O 3 + 6HCl = 2FeCl 3 + 3H 2 O; CuO + 2HCl = CuCl 2 + H 2 O.

3. CuO and C are black, NaCl and BaBr 2 are white. The only reagent may be, for example, dilute sulfuric acid H 2 SO 4:

CuO + H 2 SO 4 = CuSO 4 + H 2 O (blue solution); BaCl 2 + H 2 SO 4 = BaSO 4 + 2HCl (white precipitate).
Dilute sulfuric acid does not interact with soot and NaCl.

4 . Place a small amount of each substance in water:

CuSO 4 +5H 2 O = CuSO 4 5H 2 O	(a blue solution and crystals are formed);
Al 2 S 3 + 6H 2 O = 2Al(OH) 3 + 3H 2 S	(a precipitate forms and a gas with an unpleasant odor is released);
AlCl 3 + 6H 2 O = A1C1 3 6H 2 O + Q AlCl 3 + H 2 O AlOHCl 2 + HCl AlOHC1 2 + H 2 0 = Al (OH) 2 Cl + HCl A1(OH) 2 C1 + H 2 O = A1(OH) 2 + HCl	(a violent reaction occurs, precipitates of basic salts and aluminum hydroxide are formed);
P 2 O 5 + H 2 O = 2HPO 3 HPO 3 +H 2 O = H 3 PO 4	(violent reaction with release large quantity heat, a clear solution is formed).

Two substances - sodium chloride and ammonium chloride - dissolve without reacting with water; they can be distinguished by heating the dry salts (ammonium chloride sublimes without residue): NH 4 Cl NH 3 + HCl; or by the color of the flame with solutions of these salts (sodium compounds color the flame yellow).

5. Let's make a table of pairwise interactions of the indicated reagents

Substances	1.NaOH	2 HCl	3. K 2 CO 3	4. Al 2 (SO 4) 3	Overall result observations
1, NaOH		-	-	Al(OH)3	1 sediment
2. NS1	_		CO2	__	1 gas
3. K 2 CO 3	-	CO2		Al(OH)3 CO2	1 sediment and 2 gases
4. Al 2 (S0 4) 3	A1(OH) 3	-	A1(OH) 3 CO2		2 sediments and 1 gas
NaOH + HCl = NaCl + H2O
K 2 CO 3 + 2HC1 = 2KS1 + H 2 O + CO 2

3K 2 CO 3 + Al 2 (SO 4) 3 + 3H 2 O = 2 Al (OH) 3 + 3CO 2 + 3K 2 SO 4 ;

Based on the table presented, all substances can be determined by the number of precipitation and gas evolution.

6. All solutions are mixed in pairs. A pair of solutions that gives a raspberry color is NaOH and phenolphthalein. The raspberry solution is added to the two remaining test tubes. Where the color disappears is sulfuric acid, in the other - sodium sulfate. It remains to distinguish between NaOH and phenolphthalein (test tubes 1 and 2).
A. From test tube 1, add a drop of solution to a large amount of solution 2.
B. From test tube 2, a drop of solution is added to a large amount of solution 1. In both cases, the color is crimson.
Add 2 drops of sulfuric acid solution to solutions A and B. Where the color disappears, a drop of NaOH was contained. (If the color disappears in solution A, then NaOH - in test tube 1).

Substances	Fe	Zn	CaCO 3	K 2 CO 3	Na2SO4	NaCl	NaNO3
Ba(OH) 2				sediment	sediment	solution	solution
NaOH		hydrogen evolution possible		solution	solution	solution	solution
There is no precipitate in the case of two salts in Ba(OH) 2 and in the case of four salts in NaOH	dark powders (alkali-soluble - Zn, alkali-insoluble - Fe)		CaCO 3 gives a precipitate with both alkalis	give one precipitate, differ in flame color: K + - violet, Na + - yellow		there is no precipitation; differ in behavior when heated (NaNO 3 melts and then decomposes to release O 2, then NO 2

8 . React violently with water: P 2 O 5 and CaO with the formation of H 3 PO 4 and Ca(OH) 2, respectively:

P 2 O 5 + 3H 2 O = 2H 3 PO 4, CaO + H 2 O = Ca(OH) 2.
Substances (3) and (4) - Pb(NO 3) 2 and CaCl 2 - dissolve in water. Solutions can react with each other as follows:

Substances	1. N 3 RO 4	2. Ca(OH) 2,	3. Pb(NO 3) 2	4.CaCl2
1. N 3 RO 4		CaHPO 4	PbHPO4	CaHPO 4
2. Ca(OH) 2	SaNRO 4		Pb(OH)2	-
3. Pb(NO 3) 2	PbNPO 4	Pb(OH)2		РbСl 2
4. CaC1 2	CaHPO 4		PbCl2

Thus, solution 1 (H 3 PO 4) forms precipitates with all other solutions upon interaction. Solution 3 - Pb(NO 3) 2 also forms precipitates with all other solutions. Substances: I -P 2 O 5, II -CaO, III -Pb(NO 3) 2, IV-CaCl 2.
In general, the occurrence of most precipitation will depend on the order in which the solutions are drained and the excess of one of them (in a large excess of H 3 PO 4, lead and calcium phosphates are soluble).

9. The problem has several solutions, two of which are given below.
A. Add copper sulfate solution to all test tubes:
2NaOH + CuSO 4 = Na 2 SO 4 + Cu(OH) 2 (blue precipitate);
Na 2 S + CuSO 4 = Na 2 SO 4 + CuS (black precipitate);
NaCl + CuSO 4 (no changes in a dilute solution);
4NaI+2CuSO 4 = 2Na 2 SO 4 + 2CuI+I 2 (brown precipitate);
4NH 3 + CuSO 4 = Cu(NH 3) 4 SO 4 (blue solution or blue precipitate, soluble in excess ammonia solution).

b. Add silver nitrate solution to all test tubes:
2NaOH + 2AgNO 3 = 2NaNO 3 + H 2 O + Ag 2 O (brown precipitate);
Na 2 S + 2AgNO 3 = 2NaNO 3 + Ag 2 S (black precipitate);
NaCl + AgNO 3 = NaN0 3 + AgCl (white precipitate);
NaI + AgNO 3 = NaNO 3 + AgI (yellow precipitate);
2NH 3 + 2AgNO 3 + H 2 O = 2NH 4 NO 3 + Ag 2 O (brown precipitate).
Ag 2 O dissolves in excess ammonia solution: Ag 2 0 + 4NH 3 + H 2 O = 2OH.

10 . To recognize these substances, all solutions should be reacted with each other:

Substances	1. NaCl	2.NH4C1	3. Ba(OH),	4. NaOH	General observation result
1. NaCl		___	_	_	no interaction observed
2.NH4Cl	_	X	NH 3	NH 3	in two cases gas is released
3. Ba(OH) 2	-	NH 3	X	-
4. NaOH	-	NH 3	-	X	in one case gas is released

NaOH and Ba(OH) 2 can be distinguished by their different flame colors (Na+ is yellow, and Ba 2+ is green).

11. Determine the acidity of solutions using indicator paper:
1) acidic environment -HCl, NH 4 C1, Pb(NO 3) 2;
2) neutral medium - Na 2 SO 4, BaCl 2, AgNO 3;
3) alkaline environment - Na 2 CO 3, NaOH. Let's make a table.

Problem 9-1.

Two flasks are balanced on the scales, into which 100 ml of the same sulfuric acid solution is poured. 1 g of aluminum was dropped into one of the flasks, which completely dissolved. What mass of magnesium carbonate must be added to the second flask to restore the disturbed equilibrium?
What should be the minimum molar concentration of acid in the solution used for this conclusion to be unambiguous? (10 points).

Problem 9-2.

The oldest white pigment that has come down to us in works of easel painting formally consists of two related compounds of the same divalent metal; the molar ratio of compounds in the pigment is 1:2. Both compounds dissolve in nitric acid, while one of them does not release gas.

When 15.5 g of pigment is dissolved in nitric acid, 896 ml of gas (n.s.) with a hydrogen density of 22 is released. If the resulting solution is treated with an excess of sodium sulfate solution, then 18.18 g of sediment can be obtained. Set the pigment composition. (10 points)

Problem 9-3.

Chlorophyll is an important pigment that provides the green color of plant leaves and the process of photosynthesis. When 89.2 mg of chlorophyll is burned in excess oxygen, the following four substances are formed: 242 mg of the gas that carbonates drinks, 64.8 mg of the liquid that forms the basis of these drinks, 5.6 mg of gas, which is most abundant in the earth's atmosphere, and 4.00 mg of a white powder, which is a metal oxide , in which the number of protons in the nucleus of the atom is 6 times greater than the number of electrons in the outer layer of the electron shell.

Questions:
a) What substances were formed when chlorophyll was burned?
b) What chemical elements are included in its molecule? Find their masses
c) Calculate the formula of chlorophyll, taking into account that its molecule contains one metal atom.
d) Write the equation for the combustion reaction of chlorophyll.
e) Does chlorophyll contain chlorine? What do they have in common?

(10 points)

Problem 9-4.

You have been given a mixture of the following dry salts: ammonium sulfate, copper sulfate, zinc sulfate and barium sulfate. In addition, you have at your disposal water, diluted solutions of caustic potassium and sulfuric acid, and the necessary laboratory equipment.

Write a description of the work on separating the mixture and obtaining the original salts in their pure form. Write down the equations for the reactions you will carry out in this case.

Make a list of the minimum equipment required.

(10 points)

Problem 9-5.

Four test tubes contain transparent solutions of four substances with a concentration of 0.1 mol/L. It is known that hydrogen, zinc, barium and sodium cations and chloride, sulfate and carbonate anions can be found in these solutions. It is also known that the chloride ion is present in only one solution.

1) What substances can be in each test tube? Is the proposed option the only one? Explain your choice.

2) Describe the sequence of actions that allow you to determine what substance is in each test tube without resorting to the help of other reagents.

3) Write the equations of the reactions you propose in molecular and ionic form and indicate the signs of their occurrence.

(10 points)

Total 50 points.

Hydrogen halides and hydrohalic acids. All hydrogen halides (their general formula can be written as NG) are colorless gases with a pungent odor and are toxic. They dissolve very well in water and smoke in humid air, as they attract water vapor in the air, forming a foggy cloud.

Figure 93 illustrates an experiment that clearly shows the good solubility of hydrogen chloride in water (under normal conditions, about 500 of its volumes are dissolved in one volume of water).

Rice. 93.
Dissolution of hydrogen chloride in water:
a - at the beginning of the experiment; b - some time after it occurs

Solutions of hydrogen halides in water are acids, these are HF - hydrofluoric, or hydrofluoric, acid, HCl - hydrochloric, or hydrochloric acid, HBr - hydrobromic acid, HI - hydroiodic acid. Their ability to electrolytically dissociate with the formation of hydrogen cations increases from HF to HI.

The strongest of the hydrohalic acids is hydroiodic acid, and the weakest is hydrofluoric acid. Great chemical strength H-F connections(therefore, hydrofluoric acid dissociates weakly in water) is due to the small size of the F atom and, accordingly, the small distance between the nuclei of the hydrogen and fluorine atoms. As the atomic radius increases from F to I, so does distance N-G, the strength of the molecules decreases and, accordingly, the ability for electrolytic dissociation increases.

The most technically important are hydrogen chloride and hydrochloric acid. In industry, hydrogen chloride is produced by synthesis from hydrogen and chlorine:

H 2 + Cl 2 = 2HCl.

In laboratory conditions, to produce hydrogen chloride, a reaction carried out by heating is used (Fig. 94):

Rice. 94.
Obtaining hydrogen chloride

The irreversible occurrence of this reaction is facilitated by the volatility of HCl.

Hydrochloric acid is a colorless liquid that fumes in air and is slightly heavier than water. This is a typical acid that reacts with metals, metal oxides and hydroxides and salts (give equations for the corresponding reactions and characterize them in the light of the theory of electrolytic dissociation and oxidation and reduction processes where this occurs).

Hydrochloric acid is widely used in industry (Fig. 95).

Rice. 95.
Application of hydrochloric acid:
1 - cleaning the surface of metals; 2 - soldering; 3 - obtaining salts; 4 - production of plastics and other synthetic materials; 5 - receiving medicines; 6 - paint production

Salts of hydrohalic acids. Hydrohalic acids form salts: fluorides, chlorides, bromides and iodides. Chlorides, bromides and iodides of many metals are highly soluble in water.

To determine chloride, bromide and iodide ions in solution and distinguish them, use the reaction with silver nitrate AgNO 3 (Fig. 96). As a result of the reaction of chlorides (and hydrochloric acid itself) with this reagent, a white cheesy precipitate of silver chloride AgCl precipitates; the abbreviated ionic equation of this reaction is written as follows:

Ag + + Cl - = AgCl↓.

Rice. 96.
Qualitative reactions to halide ions (Cl -, Br -, I -)

In reactions with hydrobromic acid and its salts and with hydroiodic acid and its salts, precipitates are also formed, but only yellow in color, which differ in shades:

Laboratory experiment No. 26
Qualitative reaction to halide ions

But for the recognition of hydrofluoric acid and its salts (fluorides), silver nitrate is unsuitable as a reagent, since the resulting silver fluoride AgF is soluble in water. To prove the presence of F - fluoride ions in the solution, you can use the reaction with calcium ions Ca 2+, since calcium fluoride CaF 2 precipitates (Fig. 97).

Rice. 97.
Qualitative reaction to fluoride ion F -

Hydrofluoric acid got its name because of its unique property: when interacting with silicon (IV) oxide, which is part of the glass, it seems to melt it:

SiO 2 + 4HF = SiF 4 + 2H 2 O.

This reaction is used to make inscriptions and designs on glass. A thin layer of paraffin is applied to the glass, a design is scratched into it, and then the product is immersed in a solution of hydrofluoric acid. Thus, for example, the Lithuanian artist M. Ciurlionis created about 30 works of art (Fig. 98).

Rice. 98.
Reproductions of paintings by M. K. Ciurlionis (1875-1911) from the “Winter” cycle. 1907

Halogens in nature. Halogens exist in nature only in a bound state. Among them, the most common are chlorine (0.19% by weight of the earth’s crust) and fluorine (0.03%).

The most important natural chlorine compound is halite NaCl (Fig. 99), which you became familiar with in detail last year. Halite is mined by mining deposits of rock salt - solid sodium chloride.

Rice. 99.
Rock salt

In addition to halite, natural potassium chloride KCl is found. These are the minerals sylvin (Fig. 100) and sylvinite (a mixture of KCl and NaCl, the composition of which is reflected by the formula KCl NaCl).

Rice. 100.
Silvin

The natural fluorine mineral is fluorite, or fluorspar CaF 2 (Fig. 101).

Rice. 101.
Fluorspar

Bromine and iodine are trace elements and do not form their own minerals. These elements are concentrated in the waters of the oceans and seas, in the waters of drilling wells, as well as in algae (Fig. 102).

Rice. 102.
Kelp seaweed is rich in iodine

New words and concepts

1. Hydrogen halides.
2. Hydrohalic acids: hydrofluoric, or hydrofluoric, hydrochloric, or hydrochloric, hydrobromic, hydroiodic.
3. Halides: fluorides, chlorides, bromides, iodides. Qualitative reactions to halide ions.
4. Natural halogen compounds: halite, sylvin, sylvinite, fluorite.

Tasks for independent work