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Drill: Calculate the pH of 0.10 M H 2 Z in 0.50 M KHZ. K a1 = 2.0 x 10 -5 K a2 = 5.0 x 10 -9

Drill: Calculate the pH of 0.10 M H 2 Z in 0.50 M KHZ. K a1 = 2.0 x 10 -5 K a2 = 5.0 x 10 -9. Hydrolysis Reactions. Hydrolysis. Any reaction in which water is decomposed with all or part of its decomposition portions combining with the products. Hydrolysis.

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Drill: Calculate the pH of 0.10 M H 2 Z in 0.50 M KHZ. K a1 = 2.0 x 10 -5 K a2 = 5.0 x 10 -9

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  1. Drill: Calculate the pH of 0.10 M H2Z in 0.50 M KHZ.Ka1 = 2.0 x 10-5 Ka2 = 5.0 x 10-9

  2. Hydrolysis Reactions

  3. Hydrolysis Any reaction in which water is decomposed with all or part of its decomposition portions combining with the products

  4. Hydrolysis Water is added to decompose something

  5. Hydrolysis MX + HOH HX + MOH

  6. Salts

  7. Salts • Ionic compounds that dissolve ~ 100 % in water

  8. Salts of Acids • Salts of acids are negative (A-)

  9. Salts of Bases • Salts of bases are Positive (M+ or B+)

  10. Salt Solutions • When salts dissolve, their ions can recombine with water

  11. Salt Solutions • The salts of weak acids can recombine with water producing basic solutions

  12. Salt Solutions • The salts of weak bases can recombine with water producing acidic solutions

  13. Salt Solutions • A- + H2O HA + OH- • B+ + H2O H+ + BOH

  14. Drill: Calculate the salt/acid ratio of a solution of benzoic acid & sodium benzoate at a pH of 5.00.KaHBz = 6.4 x 10-5

  15. Drill:Calculate the pH of a solution of 0.10 M NH3 in 0.20 NH4Cl.Kb NH3 = 1.8 x 10-5

  16. Salt or Hydrolysis Problems

  17. Salt Problem • Calculate the pH of a 0.20 M solution of NaBz • Ka = 6.4 x 10-5

  18. Salt Problem • Calculate the pH of a 0.10 M solution of R-NH3Cl • Kb = 2.5 x 10-5

  19. Drill: Calculate the pH of a 0.18 M solution of KC2H3O2 • Ka = 1.8 x 10-5

  20. AP CHM HW • Read: Chapter 14 • Problems: 43 • Page: 422

  21. CHM II HW • Read: Chapter 18 • Problems: 83 • Page: 790

  22. Drill: Calculate the pH of a 0.16 M solution of KC7H5O2 • Ka = 6.4 x 10-5

  23. Short Test Friday

  24. Salt Applications • Salts of strong acids & weak bases make acidic solutions

  25. Salt Applications • Salts of strong bases & weak acids make basic solutions

  26. Salt Applications • Salts of strong acids & strong bases make neutral solutions

  27. Predict Relative pH • NaAc MnCl2 • KNO3 NH4Br • KHSO4 NH4Ac

  28. Predict Relative pH • KAc NaCl • KClO2 NH4Cl • K2SO4 NaI

  29. Anhydrides • Compounds without water; that when added to water, form other compounds

  30. Acid Anhydrides • Non-metal oxides that form acids when added to water

  31. Basic Anhydrides • Metal oxides that form bases when added to water

  32. Predict Relative pH • Na2O SO2 • NO2 CO2 • CaO Al2O3

  33. Calculate the pH of a solution of 0.30 M KQ. • KaHQ = 3.0 x 10-5

  34. AP CHM HW • Read: Chapter 14 • Problems: 35 • Page: 422

  35. CHM II HW • Read: Chapter 18 • Problems: 67 & 75 • Page: 789-790

  36. Drill: Calculate the pH of a 0.72 M NH4NO3 solution.KbNH3 = 1.8 x 10-5

  37. A/B eq, Buffer & Salt Hydrolysis Problems

  38. AP Test • Thursday

  39. 11.2 g of KOH was added to 2.0 L of 0.075 M H2CO3. Calculate the molarity of all ions present in the solution.Ka1 = 4.4 x 10-7Ka2 = 4.8 x 10-11

  40. You need to make a buffer solution with its greatest buffering capacity at pH ~ 5.4. In general terms, describe what acid or base you would chose, & how you would make the buffer.

  41. Calculate the pH of 0.10 M HF. Ka HF = 6.5 x 10-4

  42. Calculate [H3PO4], [H2PO4-1], [HPO4-2], [PO4-3], [K+], [H+], & pH of 1.0 M KH2PO4 in 0.50 M K2HPO4. Ka1 = 7.5 x 10-3Ka2 = 6.2 x 10-8Ka3 = 4.2 x 10-13

  43. Calculate the pH of 0.10 M HF in 0.20 M NaF. Ka HF = 6.5 x 10-4

  44. Calculate the pH of 5.0 M KCN. KaHCN= 5.0 x 10-10

  45. Calculate pH of: 0.20 M MOHin 0.50 M MCl Kb = 5.0 x 10-5

  46. Calculate pH of: 0.20 M MCl Kb = 5.0 x 10-5

  47. Drill: Calculate the pH of0.20 M KQ. KaHQ = 8.0 x 10-5

  48. 1.5 L of 0.25 M Ba(OH)2 was added to 1.0 L of 0.60 M H2SO3. Calculate [H2SO3], [HSO3-], [SO3-2], [H+], [OH-], & pH of the solution.Ka1 = 1.7 x 10-2Ka2 = 6.0 x 10-8

  49. Review of Acid/Base descriptions and Acid/Base, Buffer, & Salt Equilibria

  50. Arhenius, Bronsted-Lowry, & Lewis Acids & Bases

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