electrolytic dissociation. A22. Electrolytic dissociation Equations of dissociation of matter aluminum nitrate
Electrolytic dissociation of electrolytes in aqueous solutions. Weak and strong electrolytes.
1. Dissociation in three steps 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, during the dissociation of which cations Na +, H +, as well as anions SO 4 2- are formed, is
1) acid 2) alkali 3) medium salt 4) acid salt
4. Electrical current conducts
1) alcohol solution of iodine
2) wax melt
3) sodium acetate melt
4) aqueous glucose solution
5. The weakest electrolyte is
I) HF 2) HCI 3) HBr 4) HI
6. As anions, only OH ions - dissociations are formed
1) CH 3 OH 2) ZnOHBr 3) NaOH 4) CH 3 COOH
7. An electrolyte is each substance in the series:
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 NH 4 + cation) and hydroxide anions OH - .
B. No other anions, except for OH -, do not form bases.
1) only A is true
2) only B is true
3) both statements are true
4) both statements are wrong
10. Electrolytes are not
1) soluble salts 2) alkalis 3) soluble acids 4) oxides
11. The lamp of the electrical conductivity tester burns most brightly in solution
I) acetic acid 2) ethyl alcohol 3) sugar 4) sodium chloride
12. 2 mol of ions are formed with complete dissociation of 1 mol
1) K 3 RO 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 number
molecules have dissociated.
B. An electrolyte is a substance in melts and solutions that dissociates into ions
1) only A is true
2) only B is true
3) A and B are correct
4) both statements are wrong
15. 4 mol of ions are formed with complete dissociation of 1 mol
1) NaCI 2) H 2 S 3) KNO 3 4) K 3 PO 4
16. From the above statements:
A. During dissociation, the electrolyte decomposes into ions.
B. The degree of dissociation decreases when a concentrated solution is diluted.
I) only A is true
2) only B is true
3) A and B are correct
4) both statements are wrong
17. Does not form other cations in an aqueous solution, except for H +
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 other anions in an aqueous solution, except for OH -,
1) phenol 2) phosphoric acid 3) potassium hydroxide 4) ethanol
20. In what 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. More ions are formed during electrolytic dissociation of 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) acetic
4) hydrochloric
24. Weak electrolyte is acid
2) sulfurous
3) nitrogen
4) hydrochloric
25. The concentration of which particles in a solution of H 3 PO 4 is the smallest
1) H + 2) PO 4 3- 3) H 2 PO 4 - 4) HPO 4 2-
26. As cations, only H+ nones form upon dissociation
I) NaOH 2) Na 3 PO 4 3) H 2 SO 4 4) NaHSO 4
27. Not an electrolyte
1) sodium hydroxide melt
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. Weak electrolyte is
1) sodium hydroxide
2) acetic acid
3) nitric acid
4) barium chloride
30. The greatest amount of chloride ions is formed in solution during the 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.
1. Compare by structure and properties:
a) Ca0 and Ca2+
b) Сu2+ (hydr) and Cu2+ (non-hydr);
c) H0₂ and H+.
2. Using the solubility table, give examples of five substances that form sulfate in solutions - SO₄2- ions. Write down the equations of electrolytic dissociation of these substances. 
3. What information does the following equation carry:
Al(NO)= Al3++3NO₃-?
Give the names of the substance and ions.
Al(NO)= Al3++3NO₃-
This equation says that the substance aluminum nitrate is a strong electrolyte and dissociates into ions in solution: an aluminum cation and a nitrate ion.
4. Write down the dissociation equations: iron (III) sulfate, potassium carbonate, ammonium phosphate, copper (II) nitrate, barium hydroxide, 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 oxide (IV), silicon oxide (IV), sodium sulfide, iron (II) sulfide, sulfuric acid? Why? Write down possible dissociation equations. 
6. In writing the equations of the stepwise dissociation of sulfuric acid, the equal sign is used for the first stage, and the reversibility sign for the second. Why?
H₂SO₄= H++HSO₄-
HSO₄-=H++SO₄2-
The dissociation of sulfuric acid in the first stage proceeds completely, and in the second stage partially.
DEFINITION
aluminum nitrate- an average salt formed by a weak base - aluminum hydroxide (Al (OH) 3) and a strong acid - nitric (HNO 3). Formula - Al (NO 3) 3.
It is colorless crystals that absorb moisture well and smoke in the air. Molar mass - 213 g / mol.
Rice. 1. Aluminum nitrate. Appearance.
Hydrolysis of aluminum nitrate
Hydrolyzed at the cation. The nature of the medium is acidic. Theoretically, the second and third steps are possible. The hydrolysis equation has the following form:
First stage:
Al (NO 3) 3 ↔ Al 3+ + 3NO 3 - (salt dissociation);
Al 3+ + HOH ↔ AlOH 2+ + H + (cation hydrolysis);
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 step:
Al (OH) (NO 3) 2 ↔ AlOH 2+ + 2NO 3 - (salt dissociation);
AlOH 2+ + HOH ↔ Al(OH) 2 + + H + (cation hydrolysis);
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 step:
Al (OH) 2 NO 3 ↔ Al (OH) 2 + + NO 3 - (salt dissociation);
Al(OH) 2 + + HOH ↔ Al(OH) 3 ↓ + H + (cation hydrolysis);
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).
Examples of problem solving
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 nitric oxide (IV). Determine how much oxygen is released. |
| Solution | We write the reaction equation for the calcination of aluminum nitrate: 4Al(NO 3) 3 \u003d 2Al 2 O 3 + 12NO 2 + 3O 2. Find the mass fraction of pure (without impurities) aluminum nitrate: ω (Al (NO 3) 3) \u003d 100% - ω impurity \u003d 100-10 \u003d 90% \u003d 0.9. Find the mass of aluminum nitrate that does not contain impurities: m (Al (NO 3) 3) \u003d m impurity (Al (NO 3) 3) × ω (Al (NO 3) 3) \u003d 5.9 × 0.9 \u003d 5.31 g. Let's determine the number of moles of aluminum nitrate that does not contain impurities (molar mass - 213 g / mol): υ (Al (NO 3) 3) \u003d m (Al (NO 3) 3) / M (Al (NO 3) 3) \u003d 5.31 / 213 \u003d 0.02 mol. According to the equation: 4υ (Al (NO 3) 3) = 3υ (O 2); υ (O 2) \u003d 4/3 × υ (Al (NO 3) 3) \u003d 4/3 × 0.02 \u003d 0.03 mol. Then, the volume of released oxygen will be equal to: V (O 2) \u003d V m × υ (O 2) \u003d 22.4 × 0.03 \u003d 0.672 l. |
| Answer |
The volume of released oxygen is 0.672 liters. |
EXAMPLE 2
The potassium sulfite salt (K 2 SO 3) is hydrolyzed by 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) is hydrolyzed by the Al 3+ cation, since it is formed by a strong acid and a weak base. Hydrolysis equation number 1.
Salt sodium chloride (NaCl) does not undergo hydrolysis, since it is formed by a strong base and a strong acid (3).
