1. Compare by structure and properties:
a) Ca0 and Ca2+
b) Cu2+ (hydr) and Cu2+ (nonhydr);
c) H0₂ and H+.
2. Using the solubility table, give examples of five substances that form sulfate - SO₄2- ions - in solutions. Write down the equations for the electrolytic dissociation of these substances. 
3. What information does the following equation contain:
Al(NO)= Al3++3NO₃-?
Give the names of the substance and ions.
Al(NO)= Al3++3NO₃-
This equation suggests that the substance aluminum nitrate is a strong electrolyte and in solution dissociates into ions: aluminum cation and nitrate ion.
4. Write down the dissociation equations: iron (III) sulfate, potassium carbonate, ammonium phosphate, copper (II) nitrate, barium hydroxide, of hydrochloric acid, potassium hydroxide, iron (II) chloride. Give the names of the ions. 
5. Which of the following substances will dissociate: iron (II) hydroxide, potassium hydroxide, silicic acid, Nitric acid, sulfur (IV) oxide, silicon (IV) oxide, sodium sulfide, iron (II) sulfide, sulfuric acid? Why? Write down possible dissociation equations. 
6. In writing the equations for stepwise dissociation of sulfuric acid, the equal sign is used for the first step, and the reversibility sign for the second. Why?
H₂SO₄= H++HSO₄-
HSO₄-=H++SO₄2-
The dissociation of sulfuric acid occurs completely in the first stage, and partially in the second stage.
Electrolytic dissociation of electrolytes in aqueous solutions. Weak and strong electrolytes.
1. Dissociation in three stages is possible in solution
1) aluminum chloride
2) aluminum nitrate
3) potassium orthophosphate
4) phosphoric acid
2. Ions I - are formed during dissociation
1) KIO 3 2) KI 3) C 2 H 5 I 4) NaIO 4
3. A substance, upon dissociation of which Na +, H + cations are formed, as well as SO 4 2- anions, is
1) acid 2) alkali 3) average salt 4) acid salt
4. Electricity conducts
1) alcohol solution of iodine
2) paraffin melt
3) melt sodium acetate
4) aqueous glucose solution
5. The weakest electrolyte is
I) HF 2) HCI 3) HBg 4) HI
6. As anions, only OH ions - dissociations are formed
1) CH 3 OH 2) ZnOHBr 3) NaOH 4) CH 3 COOH
7. Each substance in the series is an electrolyte:
1) C 2 H 6, Ca(OH) 2, H 2 S, ZnSO 4
2) BaCl 2, CH 3 OCH 3, NaNO 3, H 2 SO 4
3) KOH, H 3 PO 4, MgF 2, CH 3 COONa
4) PbCO 3, AIBr 3, C 12 H 22 O 11, H 2 SO 3
8. The light bulb will light up when the electrodes are lowered into the aqueous solution
1) formaldehyde
2) sodium acetate
3) glucose
4) methyl alcohol
9. Which of the statements about the dissociation of bases in aqueous solutions are correct?
A. Bases in water dissociate into metal cations (or a similar cation NH 4 +) and hydroxide anions OH -.
B. No other anions except OH - form bases.
1) only A is correct
2) only B is correct
3) both statements are true
4) both statements are incorrect
10. They are not electrolytes.
1) soluble salts 2) alkalis 3) soluble acids 4) oxides
11. The lamp of the device for testing electrical conductivity burns brightest in solution
I) acetic acid 2) ethyl alcohol 3) sugar 4) sodium chloride
12. 2 moles of ions are formed upon complete dissociation of 1 mole
1) K 3 PO 4 2) Na 2 S 3) K 2 CO 3 4) NaCl
13. Electrolytic dissociation of 1 mol of aluminum nitrate A1(NO 3) 3 leads to the formation
1) 1 mol A1 and 3 mol NO 3 -
2) 1 mol A1 3+ and 1 mol NO 3 -
3) 1 mol Al 3+ and 3 mol NO -
4) 3 mol AI 3+, 3 mol N 5+ and 9 mol O 2-
14. From the above statements:
A. The degree of dissociation shows what part of the total
molecules dissociated.
B. An electrolyte is a substance that dissociates into ions in melts and solutions
1) only A is correct
2) only B is correct
3) A and B are correct
4) both statements are incorrect
15. 4 moles of ions are formed upon complete dissociation of 1 mole
1) NaCI 2) H 2 S 3) KNO 3 4) K 3 PO 4
16. From the above statements:
A. During dissociation, the electrolyte breaks down into ions.
B. The degree of dissociation decreases when the concentrated solution is diluted.
I) only A is correct
2) only B is correct
3) A and B are correct
4) both statements are incorrect
17. Does not form cations other than H + in an aqueous solution
I) benzene 2) hydrogen chloride 3) potassium hydroxide 4) ethane
18. Not an electrolyte
1) benzene 2) hydrogen chloride 3) potassium hydroxide 4) sodium sulfate
19. Does not form anions other than OH - in an aqueous solution,
1) phenol 2) phosphoric acid 3) potassium hydroxide 4) ethanol
20. In which series are all the indicated substances non-electrolytes?
1) ethanol, potassium chloride, barium sulfate
2) ribose, potassium hydroxide, sodium acetate
3) sucrose, glycerin, methanol
4) sodium sulfate, glucose, acetic acid
21. Large quantity ions are formed during electrolytic dissociation 1 mol
1) potassium chloride
2) aluminum sulfate
3) iron (III) nitrate
4) sodium carbonate
22. Strong electrolytes are
1) HCOOH and Cu(OH) 2
2) Ca 3 (PO 4) 2 and NH 3 H 2 O
3) K 2 CO 3, and CH 3 COOH
4) KNSO 3 and H 2 SO 4
23. Among these acids, the strongest is
1) silicon
2) hydrogen sulfide
3) vinegar
4) hydrochloric
24. Acid is a weak electrolyte
2) sulfurous
3) nitrogen
4) hydrochloric
25. Which particles have the lowest concentration in a solution of H 3 PO 4
1) H + 2) PO 4 3- 3) H 2 PO 4 - 4) HPO 4 2-
26. As cations, only nonons H+ form upon dissociation
I) NaOH 2) Na 3 PO 4 3) H 2 SO 4 4) NaHSO 4
27. Not an electrolyte
1) molten sodium hydroxide
2) nitric acid
3) sodium hydroxide solution
4) ethyl alcohol
28. A weak electrolyte is
2) sulfuric acid (solution)
3) sodium chloride (solution)
4) sodium hydroxide (solution)
29. A weak electrolyte is
1) sodium hydroxide
2) acetic acid
3) nitric acid
4) barium chloride
30. Largest quantity chloride ions are formed in solution upon dissociation of 1 mol
1) copper(II) chloride
2) calcium chloride
3) iron(III) chloride
4) lithium chloride
Answers: 1-4, 2-2, 3-3, 4-3, 5-1, 6-3, 7-3, 8-2, 9-3, 10-4, 11-4, 12-4, 13-1, 14-3, 15-4, 16-1, 17-1, 18-1, 19-3, 20-3, 21-2, 22-4, 23-4, 24-2, 25- 2, 26-3, 27-4, 28-1, 29-3, 30-3.
DEFINITION
Aluminum nitrate– a medium salt formed by a weak base – aluminum hydroxide (Al(OH) 3) and a strong acid – nitric acid (HNO 3). Formula – Al(NO 3) 3.
They are colorless crystals that absorb moisture well and smoke in air. Molar mass – 213 g/mol.
Rice. 1. Aluminum nitrate. Appearance.
Hydrolyzes at the cation. The nature of the environment is acidic. Theoretically, the second and third stages are possible. The hydrolysis equation is as follows:
First stage:
Al(NO 3) 3 ↔ Al 3+ +3NO 3 - (salt dissociation);
Al 3+ + HOH ↔ AlOH 2+ + H + (hydrolysis by cation);
Al 3+ +3NO 3 - + HOH ↔ AlOH 2+ +3NO 3 - +H + (ionic equation);
Al(NO 3) 3 + H 2 O ↔Al(OH)(NO 3) 2 + HNO 3 (molecular equation).
Second stage:
Al(OH)(NO 3) 2 ↔ AlOH 2+ + 2NO 3 - (salt dissociation);
AlOH 2+ + HOH ↔ Al(OH) 2 + + H + (hydrolysis by cation);
AlOH 2+ + 2NO 3 - + HOH ↔Al(OH) 2 + + 2NO 3 - + H + (ionic equation);
Al(OH)(NO 3) 2 + H 2 O ↔ Al(OH) 2 NO 3 + HNO 3 (molecular equation).
Third stage:
Al(OH) 2 NO 3 ↔ Al(OH) 2 + + NO 3 - (salt dissociation);
Al(OH) 2 + + HOH ↔ Al(OH) 3 ↓ + H + (hydrolysis by cation);
Al(OH) 2 + + NO 3 - + HOH ↔ Al(OH) 3 ↓ + NO 3 - + H + (ionic equation);
Al(OH) 2 NO 3 + H 2 O ↔ Al(OH) 3 ↓ + HNO 3 (molecular equation).
EXAMPLE 1
| Exercise | Aluminum nitrate weighing 5.9 g and containing 10% non-volatile impurities was calcined. As a result of this reaction, aluminum oxide was formed and gases were released - oxygen and nitrogen oxide (IV). Determine how much oxygen was released. |
| Solution | Let us write the equation for the calcination reaction of aluminum nitrate: 4Al(NO 3) 3 = 2Al 2 O 3 + 12NO 2 + 3O 2. Let's find the mass fraction of pure (without impurities) aluminum nitrate: ω(Al(NO 3) 3) = 100% - ω impurity = 100-10 = 90% = 0.9. Let's find the mass of aluminum nitrate that does not contain impurities: m(Al(NO 3) 3) = m impurity (Al(NO 3) 3) × ω(Al(NO 3) 3) = 5.9 × 0.9 = 5.31 g. Let us determine the number of moles of aluminum nitrate that does not contain impurities ( molar mass– 213 g/mol): υ (Al(NO 3) 3) = m (Al(NO 3) 3)/M(Al(NO 3) 3) = 5.31/213 = 0.02 mol. According to the equation: 4υ(Al(NO 3) 3) = 3υ(O 2); υ(O 2) = 4/3 × υ (Al(NO 3) 3) = 4/3 × 0.02 = 0.03 mol. Then, the volume of oxygen released will be equal to: V (O 2) = V m × υ (O 2) = 22.4 × 0.03 = 0.672 l. |
| Answer |
The volume of oxygen released is 0.672 liters. |
EXAMPLE 2
Potassium sulfite salt (K 2 SO 3) hydrolyzes at the SO 3 2- anion, since it is formed by a strong base and a weak acid. Hydrolysis equation number 4.
The aluminum nitrate salt (Al(NO 3) 3) hydrolyzes at the Al 3+ cation, since it is formed by a strong acid and a weak base. Hydrolysis equation number 1.
The salt sodium chloride (NaCl) does not undergo hydrolysis because it is formed by a strong base and a strong acid (3).
