SO 4
Purpose: to obtain a complex copper sulfate–tetroamino salt from copper sulfate CuSO 4 ∙5H 2 O and a concentrated solution of ammonia NH 4 OH.
Safety precautions:
1. Glass chemical containers require careful handling; before starting work, you should check them for cracks.
2.Before starting work, you should check the serviceability of electrical appliances.
3. Heat only in heat-resistant containers.
4. Use chemicals carefully and sparingly. reagents. Do not taste them, do not smell them.
5.Work should be carried out in dressing gowns.
6. Ammonia is poisonous and its vapors irritate the mucous membrane.
Reagents and equipment:
Concentrated ammonia solution - NH 4 OH
Ethyl alcohol – C 2 H 5 OH
Copper sulfate - CuSO 4 ∙ 5H 2 O
Distilled water
Graduated cylinders
Petri dishes
Vacuum pump (water jet vacuum pump)
Glass funnels
Theoretical background:
Complex compounds are substances containing a complexing agent with which a certain number of ions or molecules called addends or legends are associated. The complexing agent with addends constitutes the inner sphere of the complex compound. In the outer sphere of complex compounds there is an ion bound to the complex ion.
Complex compounds are obtained by the interaction of substances of simpler composition. In aqueous solutions they dissociate to form a positively or negatively charged complex ion and the corresponding anion or cation.
SO 4 = 2+ + SO 4 2-
2+ = Cu 2+ + 4NH 3 –
Complex 2+ colors the solution cornflower blue, but Cu2+ and 4NH3 taken separately do not give such a color. Complex compounds are of great importance in applied chemistry.
SO4 - dark purple crystals, soluble in water, but not soluble in alcohol. When heated to 1200C, it loses water and part of the ammonia, and at 2600C, it loses all ammonia. When stored in air, the salt decomposes.
Synthesis equation:
CuSO4 ∙ 5H2O +4NH4OH = SO4 ∙ H2O +8H2O
CuSO4 ∙ 5H2O + 4NH4OH= SO4 ∙ H2O +8H2O
Mm CuSO4∙5H2O = 250 g/mol
mm SO4 ∙ H2O = 246 g/mol
6g CuSO4∙5H2O - Xg
250 g CuSO4∙5H2O - 246 SO4∙H2O
Х=246∙6/250= 5.9 g SO4 ∙ H2O
Progress:
Dissolve 6 g of copper sulfate in 10 ml of distilled water in a heat-resistant glass. Heat the solution. Stir vigorously until completely dissolved, then add concentrated ammonia solution in small portions until a purple complex salt solution appears.
Then transfer the solution to a Petri dish or porcelain dish and precipitate crystals of the complex salt with ethyl alcohol, which is poured in with a burette for 30-40 minutes, the volume of ethyl alcohol is 5-8 ml.
Filter the resulting complex salt crystals on a Buchner funnel and leave to dry until the next day. Then weigh the crystals and calculate the % yield.
5.9g SO4 ∙ H2O - 100%
m of sample – X
X = m sample ∙100% / 5.9 g
Control questions:
1.What type of chemical bonds are in complex salts?
2.What is the mechanism of formation of a complex ion?
3.How to determine the charge of a complexing agent and a complex ion?
4.How does a complex salt dissociate?
5. Make up formulas for complex compounds dicyano - sodium argentate.
Laboratory work No. 6
Preparation of orthoboric acid
Target: obtain orthoboric acid from borax and hydrochloric acid.
Safety precautions:
1. Glass chemical containers require careful handling and should be checked for cracks before use.
2. Before starting work, you should check the serviceability of electrical appliances.
3. Heat only in heat-resistant containers.
4. Use chemicals carefully and sparingly. Do not taste them, do not smell them.
5. Work should be carried out in dressing gowns.
Equipment and reagents:
Sodium tetraborate (decahydrate) – Na 2 B 4 O 7 *10H 2 O
Hydrochloric acid (conc.) – HCl
Distilled water
Electric stove, vacuum pump (water jet vacuum pump), beakers, filter paper, porcelain cups, glass rods, glass funnels.
Progress:
Dissolve 5 g of sodium tetraborate decahydrate in 12.5 ml of boiling water, add 6 ml of hydrochloric acid solution and leave to stand for 24 hours.
Na 2 B 4 O 7 *10H 2 O + 2HCl + 5H 2 O = 4H 3 BO 3 + 2NaCl
The resulting precipitate of orthoboric acid is decanted, washed with a small amount of water, filtered under vacuum and dried between sheets of filter paper at 50-60 0 C in an oven.
To obtain purer crystals, orthoboric acid is recrystallized. Calculate theoretical and practical output
Control questions:
1. Structural formula of borax, boric acid.
2. Dissociation of borax, boric acid.
3. Create a formula for sodium tetraborate acid.
Laboratory work No. 7
Preparation of copper(II) oxide
Target: obtain copper (II) oxide CuO from copper sulfate.
Reagents:
Copper (II) sulfate CuSO 4 2- * 5H 2 O.
Potassium and sodium hydroxide.
Ammonia solution (p=0.91 g/cm3)
Distilled water
Equipment: technochemical scales, filters, glasses, cylinders, vacuum pump(water jet vacuum pump) , thermometers, electric stove, Buchner funnel, Bunsen flask.
Theoretical part:
Copper (II) oxide CuO is a black-brown powder, at 1026 0 C it decomposes into Cu 2 O and O 2, almost insoluble in water, soluble in ammonia. Copper(II) oxide CuO occurs naturally as a black, earthy weathering product of copper ores (melaconite). In the lava of Vesuvius it was found crystallized in the form of black triclinic tablets (tenorite).
Artificially, copper oxide is obtained by heating copper in the form of shavings or wire in air, at a red-hot temperature (200-375 0 C) or by calcining carbonate nitrate. The copper oxide obtained in this way is amorphous and has a pronounced ability to adsorb gases. When calcined, at a higher temperature, a two-layer scale is formed on the surface of copper: the surface layer is copper (II) oxide, and the inner layer is red copper (I) oxide Cu 2 O.
Copper oxide is used in the production of glass enamels to impart a green or blue color; in addition, CuO is used in the production of copper-ruby glass. When heated with organic substances, copper oxide oxidizes them, converting carbon and carbon dioxide, and hydrogen into oxide and being reduced into metallic copper. This reaction is used in the elementary analysis of organic substances to determine the content of carbon and hydrogen in them. It is also used in medicine, mainly in the form of ointments.
2. Prepare a saturated solution from the calculated amount of copper sulfate at 40 0 C.
3. Prepare a 6% alkali solution from the calculated amount.
4. Heat the alkali solution to 80-90 0 C and pour the copper sulfate solution into it.
5. The mixture is heated at 90 0 C for 10-15 minutes.
6. The precipitate that forms is allowed to settle and washed with water until the ion is removed. SO 4 2- (sample BaCl 2 + HCl).
Names of salts.
If a metal has a variable valency, then it is indicated after the chemical element by a Roman numeral enclosed in brackets. For example, CuSO 4 is copper (II) sulfate.
Task No. 2.
Conditions for completing the task:
Task No. 2. Draw electronic diagrams of the structure of Na +, Ca 2+, Fe 3+ ions.
Task No. 1. Types of disperse systems. Classification of solutions.
Task No. 2. Indicate the features of the electronic structure of copper atoms (No. 28), chromium (No. 24).
Task No. 1 .
Types of disperse systems
A disperse system is a system where one substance is finely divided into another substance.
The dispersed phase is a crushed substance.
Dispersion medium is a substance in which the dispersed phase is distributed.
According to their state of aggregation, they are distinguished:
– gas systems (air);
– solid systems (metal alloys);
– liquid (dispersion medium - water, benzene, ethyl alcohol).
A solid or liquid homogeneous system consisting of 2 or more components is called a solution.
The solute is uniformly distributed in the form of molecules, atoms or ions in another - solvent.
Depending on the size of dissolved particles, the following are distinguished:
1. Coarse dispersed systems:
– suspensions - solid dispersed phase (clay solution);
– emulsions - liquid dispersed phase (milk).
2. Colloidal solutions (sols) - consist of very small particles (10 -5 - 10 -7 cm), evenly distributed in any medium:
– in water (hydrosols),
– in organic liquid (organosols),
– in air or other gas (aerosols).
Sols occupy an intermediate position between true solutions and coarse systems.
3. True solutions - solutions in which particles cannot be detected optically.
Diameter of dispersed particles in I.r. less than 10 -7 cm.
Liquid solutions consist of a solute, a solvent, and the products of their interaction.
Task No. 2. Indicate the features of the electronic structure of copper atoms (No. 28), chromium (No. 24).
Energy diagrams of valence sublevels of chromium and copper atoms.
The chromium atom has 4 s-there are not two sublevels, as one would expect, but only one electron. But at 3 d-sublevel has five electrons, but this sublevel is filled after 4 s-sublevel. Each of five 3 d-clouds in this case are formed by one electron. The total electron cloud of these five electrons has a spherical shape, or, as they say, spherically symmetrical. According to the nature of the distribution of electron density in different directions, it is similar to 1 s-EO. The energy of the sublevel whose electrons form such a cloud turns out to be less than in the case of a less symmetrical cloud. In this case, the orbital energy is 3 d-sublevel is equal to energy 4 s-orbitals. When symmetry is broken, for example, when a sixth electron appears, the energy of the orbitals is 3 d-the sublevel again becomes greater than energy 4 s-orbitals. Therefore, the manganese atom again has a second electron at 4 s-AO. The general cloud of any sublevel, filled with electrons either half or completely, has spherical symmetry. The decrease in energy in these cases is of a general nature and does not depend on whether any sublevel is half or completely filled with electrons. And if so, then we must look for the next violation in the atom in whose electron shell the ninth one “arrives” last d-electron. Indeed, the copper atom has 3 d-sublevel has 10 electrons, and 4 s-there is only one sublevel. A decrease in the energy of the orbitals of a fully or half-filled sublevel is the cause of a number of important chemical phenomena.
Task No. 1. Methods of expressing the concentration of solutions.
Conditions for completing the task:
Task No. 1 . Answer the question posed.
Methods of expressing the concentration of solutions
1. Percentage concentration - the number of g of a substance present in 100 g of solution.
5% solution C 6 H 12 O 6
100g solution – 5 g C 6 H 12 O 6, i.e.
5g C 6 H 12 O 6 +95g H 2 O
The percentage concentration is related to mass units.
2. Molar concentration - the number of moles present in 1 liter of solution:
5m HCl NaCl=23+35.5=58.5
3. Normal or equivalent concentration - the number of g equivalents contained in 1 liter of solution
Acid equivalent = ;
E(HCl) = , E(H 2 SO 4)= ,
Base equivalent = 
;
E(NaOH) = , E(Al(OH) 3)= ,
Salt equivalent = 
;
E(NaCl) = , E(Na 2 CO 3) = ,
E(Al 2 (SO 4) 3) = ;
Oxide equivalent = 
2n Al 2 (SO 4) 3, equivalent to Al 2 (SO 4) 3 =
For example, in 1 liter of solution 2
Task No. 2. Give examples of the following types of chemical reactions: decomposition reactions; exchange reactions
Task No. 2. Decomposition reactions:
AgNO 3 +NaCl=AgCl +NaNO 3
CaCO 3 =CaO+CO 2
Task for the examinee No. 23
Task No. 1. Theory of electrolytic dissociation.
Task No. 2. Compose molecular, full ionic and abbreviated ionic equations for the reactions of the following salts: a) chromium(III) chloride and silver nitrate; b) barium chloride and manganese sulfate; c) iron (III) nitrate and potassium hydroxide.
Task No. 1 . Answer the question posed.
Electrolytes have different dissociation abilities.
The degree of dissociation (a) is the ratio of the number of molecules disintegrated into ions (n) to the total number of dissolved electrolyte molecules (n 0):
The degree of dissociation is expressed either as a decimal fraction or, more often, as a percentage:
If a = 1, or 100%, the electrolyte completely dissociates into ions.
If a = 0.5, or 50%, then out of every 100 molecules of a given electrolyte, 50 are in a state of dissociation.
Depending on a there are:
Strong electrolytes, their a in 0.1 n. solution above 30%.
They dissociate almost completely.
Relate:
– almost all salts;
– many mineral acids: H 2 SO 4, HNO 3, HCl, HClO 4, HBr, HJ, HMnO 4, etc.
– bases of alkali metals and some alkaline earth metals: Ba(OH) 2 and Ca(OH) 2.
Average electrolytes, their a from 3 to 30%. These include acids H 3 PO 4, H 2 SO 3, HF, etc.
Weak electrolytes in aqueous solutions they are only partially dissociated, their content is less than 3%.
Relate:
– some mineral acids: H 2 CO 3, H 2 S, H 2 SiO 3, HCN;
– almost all organic acids;
– many metal bases (except alkali and alkaline earth metal bases), as well as ammonium hydroxide;
– some salts: HgCl 2, Hg(CN) 2.
Factors influencinga
Nature of the solvent:
The greater the dielectric constant of the solvent, the greater the degree of dissociation of the electrolyte in it.
Solution concentration:
The degree of electrolyte dissociation increases as the solution is diluted.
As the concentration of the solution increases, the degree of dissociation decreases (frequent collisions of ions).
Nature of the electrolyte:
Electrolyte dissociation depends on the degree of dissociation.
Temperature:
For strong electrolytes, a decreases with increasing temperature, because the number of collisions between ions increases.
For weak electrolytes, as the temperature increases, a first increases, and after 60 0 C it begins to decrease.
Electrolytic dissociation constant
In solutions of weak electrolytes, upon dissociation, a dynamic equilibrium is established between molecules and ions:
CH 3 COOH + H 2 O « CH 3 COO - + H 3 O +
. [H 3 O + ] / =K diss
Task No. 2.Compose molecular, full ionic and abbreviated ionic equations for the reactions of the listed salts.
a) CrCl 3 + 3AgNO 3 → Cr(NO 3) 3 + 3AgCl↓
Cr 3+ + 3Cl - + 3Ag + + 3NO 3 → Cr 3+ + 3NO 3 + 3AgCl↓
Cl - + Ag + → AgCl↓
b) BaCl 2 + MnSO 4 → BaSO 4 ↓ + MnCl 2
Ba 2+ + 2Cl - + Mn 2+ + SO 4 2- → BaSO 4 ↓ + Mn 2+ + 2Cl -
Ba 2+ + SO 4 2- → BaSO 4 ↓
c) Fe(NO 3) 3 + 3KOH → Fe(OH) 3 ↓ + 3KNO 3
Fe 3+ + 3NO 3 - + 3K + + 3OH - → Fe(OH) 3 ↓ + 3K + + 3NO 3 -
Fe 3+ + 3OH - → Fe(OH) 3 ↓
Task No. 1. Hydrolysis of salts.
Task No. 1 . Answer the question posed.
Salt hydrolysis is the exchange reaction of salt with water, resulting in the formation of weak electrolytes.
Water, being a weak electrolyte, dissociates into H + and OH - ions:
H 2 O<->OH - + H +
When some salts are dissolved in water, the ions of the dissolved salt interact with the H + and OH - ions of the water.
There is a shift in the equilibrium of water dissociation:
one of the water ions (or both) binds with solute ions to form slightly dissociated, or sparingly soluble, product.
Every salt can be thought of as being formed by a base and an acid.
Acids and bases are strong and weak electrolytes,
According to this criterion, salts can be divided into four types:
salts formed by a strong base cation and a strong acid anion;
2) salts formed by a strong base cation and a weak acid anion;
3) salts formed by a weak base cation and a strong acid anion;
4) salts formed by a cation of a weak base and an anion of a weak acid.
Salts formed by a strong base cation and a strong acid anion do not undergo hydrolysis.
Such salts completely dissociate into metal ions and an acid residue.
For example:
NaCl salt is formed by the strong base NaOH and the strong acid HCl and completely dissociates into ions.
Salts formed by a strong base cation and a weak acid anion
Hydrolysis of this salt consists of the addition of hydrogen ions from a water molecule by ions of the acidic residue and the release of hydroxide ions, which cause an alkaline reaction of the medium,
Na2S<->2Na + + S 2-
NON<->OH - + H +
S 2- + HOH<->HS - + OH -
Na 2 S + HOH = NaOH + NaHS
Salts formed by a weak base cation and a strong acid anion
Hydrolysis of this salt involves the addition of metal ions or ammonium ions to hydroxide ions from a water molecule and the release of hydrogen ions, which cause an acidic reaction in the medium,
ZnCl2<->Zn 2+ + 2Cl -
HON =OH - +H +
Zn 2+ + HOH<->ZnOH + + H +
ZnCl 2 + HOH<->HCl + ZnOHCl
Salts formed by a weak base cation and a weak acid anion
Hydrolysis of this salt involves the addition of hydroxide ions by metal ions or ammonium ions, and hydrogen ions from a water molecule by acidic ions. The environment's reaction will be neutral.
CH 3 COONH 4<->CH 3 COO - + NH 4 +
HOH = H + + OH -
CH3COOH NH 4 OH
CH 3 COO + NH4+ + HOH<->CH 3 COOH + NH 4 OH
Task No. 2. Characterize the position of elements No. 21, 32, 38 in the Periodic Table of D.I. Mendeleev. Write their electronic formulas and atomic structures.
Copper belongs to the group of seven metals that have been known to man since ancient times. Today, not only copper, but also its compounds are widely used in various industries, agriculture, everyday life and medicine.
The most important copper salt is copper sulfate. The formula of this substance is CuSO4. It is a strong electrolyte and consists of small white crystals, highly soluble in water, without taste or odor. The substance is non-flammable and fireproof; when used, the possibility of spontaneous combustion is completely excluded. Copper sulfate, when exposed to even the smallest amount of moisture from the air, acquires a characteristic blue color with bright blue. In this case, copper sulfate is converted into blue pentahydrate CuSO4 · 5H2O, known as copper sulfate.
In industry, copper sulfate can be obtained in several ways. One of them, the most common, is dissolving copper waste in diluted copper sulfate. In the laboratory, copper sulfate is obtained using a neutralization reaction with sulfuric acid. The process formula is as follows: Cu(OH)2 + H2SO4 → CuSO4 + H2O.
The color changing property of copper sulfate is used to detect the presence of moisture in organic liquids. It is used to dehydrate ethanol and other substances in laboratory conditions.
Copper sulfate or copper sulfate is widely used in agriculture. Its use, first of all, consists of using a weak solution to spray plants and treat cereals before sowing in order to destroy harmful fungal spores. Based on copper sulfate, the well-known Bordeaux mixture and milk of lime are produced, sold through retail outlets and intended to treat plants from fungal diseases and destroy grape aphids.
Copper sulfate is often used in construction. Its use in this area is to neutralize leaks and eliminate rust stains. The substance is also used to remove salts from brick, concrete or plastered surfaces. In addition, it is used to treat wood as an antiseptic to avoid rotting processes.
In official medicine, copper sulfate is a medicine. It is prescribed by doctors for external use as eye drops, solutions for rinsing and douching, and also for the treatment of burns caused by phosphorus. As an internal remedy, it is used to irritate the stomach to induce vomiting if necessary.
In addition, mineral paints are made from copper sulfate; it is used in spinning solutions for making
In the food industry, copper sulfate is registered as food additive E519, used as a color fixative and preservative.
When copper sulfate is sold in retail stores, it is labeled as a highly hazardous substance. If it enters the human digestive system in an amount of 8 to 30 grams, it can be fatal. Therefore, when using copper sulfate in everyday life, you should be very careful. If the substance gets on your skin or eyes, rinse the area thoroughly with cool running water. If it enters the stomach, it is necessary to do a weak rinse, drink a saline laxative and a diuretic.
When working with copper sulfate at home, use rubber gloves and other protective equipment, including a respirator. It is prohibited to use food containers for preparing solutions. After finishing work, be sure to wash your hands and face, and rinse your mouth.
General concepts about the hydrolysis of copper (II) sulfate
DEFINITION
Copper(II) sulfate- medium salt. Absorbs moisture. Anhydrous copper (II) sulfate is colorless, opaque crystals.
If water is present (the trivial name is copper sulfate), then the crystals are blue. Formula CuSO 4.
Rice. 1. Copper (II) sulfate. Appearance.
Hydrolysis of copper(II) sulfate
Copper (II) sulfate is a salt formed by a strong acid - sulfuric (H 2 SO 4) and a weak base - copper (II) hydroxide (Cu (OH) 2). Hydrolyzes at the cation. The nature of the environment is acidic. Theoretically, a second stage is possible.
First stage:
CuSO 4 ↔ Cu 2+ + SO 4 2- ;
Cu 2+ + SO 4 2- + HOH ↔ CuOH + + SO 4 2- + H + ;
CuSO 4 + HOH ↔ 2 SO 4 + H 2 SO 4.
Second stage:
2 SO 4 ↔ 2CuOH + +SO 4 2- ;
CuOH + + SO 4 2 + HOH ↔ Cu(OH) 2 + SO 4 2 + HOH.
2 SO 4 + HOH ↔Cu(OH) 2 + H 2 SO 4.
Examples of problem solving
EXAMPLE 1
| Exercise | Iron filings (3.1 g) were added to a solution of copper (II) sulfate weighing 25 g. Determine what mass of copper was formed during the reaction. |
| Solution | Let's write the reaction equation: CuSO 4 + Fe = FeSO 4 + Cu↓. Let's calculate the amounts of substances that reacted. Molar masses, which are 160 and 56 g/mol, respectively, for copper (II) and iron sulfate: υ(CuSO 4) = m (CuSO 4)/M(CuSO 4) = 25/160 = 0.16 mol. υ(Fe)= m(Fe)/M(Fe) = 3.1/56 = 0.05 mol. Let's compare the obtained values: υ(CuSO 4)>υ(Fe). We carry out calculations based on the substance that is in short supply. This is iron. According to the reaction equation υ(Fe)=υ(Cu)= 0.05 mol. Then the mass of copper will be equal (molar mass - 64 g/mol): m(Cu)= υ(Cu)× M(Cu)= 0.05×64 =3.2 g. |
| Answer | The mass of copper is 3.2 g. |
EXAMPLE 2
| Exercise | What concentration of copper (II) sulfate solution will be if another 10 g of the same substance is added to 180 g of a 30% solution of this salt? |
| Solution | Let's find the mass of the dissolved copper (II) sulfate in a 30% solution: ω=m solute /m solution ×100%. m solute (CuSO 4) = ω/100% × m solution (CuSO 4) = 30/100 × 180 = 54 g. Let's find the total mass of dissolved copper (II) sulfate in the new solution: m solute (CuSO 4) sum = m solute (CuSO 4) + m(CuSO 4) = 54 + 10 = 64 g. Let's calculate the mass of the new solution: m solution (CuSO 4) sum = m solution (CuSO 4) + m(CuSO 4) = 180+10 = 190 g. Let's determine the mass concentration of the new solution: ω=m solute (CuSO 4) sum / m solution (CuSO 4) sum ×100% = 64/190 ×100% =33.68%. |
| Answer | Solution concentration 33.68% |
