1 / 45

Chapter 17 Additional Aspects of Acid–Base Equilibria

Chapter 17 Additional Aspects of Acid–Base Equilibria. Contents in Chapter 17. 17-1 Common-Ion Effect in Acid–Base Equilibria 17-2 Buffer Solutions 17-3 Acid–Base Indicators 17-4 Neutralization Reactions and Titration Curves 17-5 Solutions of Salts of Polyprotic Acids

bianca-cole
Download Presentation

Chapter 17 Additional Aspects of Acid–Base Equilibria

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Chapter 17Additional Aspects ofAcid–Base Equilibria

  2. Contents in Chapter 17 17-1 Common-Ion Effect in Acid–Base Equilibria 17-2 Buffer Solutions 17-3 Acid–Base Indicators 17-4 Neutralization Reactions and Titration Curves 17-5 Solutions of Salts of Polyprotic Acids 17-6 Acid–Base Equilibrium Calculations: A Summary

  3. 17-1 Common-Ion Effect in Acid-Base Equilibria • Common Ion: The ion which is formed in two different ionization processes. • e.g.1, H3O+ is the common ion, formed in ionization processes, for CH3COOH and HCl. • e.g.2, CH3COO– is the common ion, formed in ionization processes, forCH3COOH and CH3COONa.

  4. 17-2 Buffer Solutions • Buffer solution: A solution that changes pH only slightly when small amounts of a strong acid or a strong base are added. • Components of buffer solution: • a weak acid and its conjugate base, e.g., CH3COOH + CH3COONa. • a weak base and its conjugate acid, e.g., NH3 + NH4Cl

  5. 17-2 (Continuous) • Common ion effect for acid-base buffer (CH3COOH/CH3COO– for example):

  6. 17-2 (Continuous) • An Equation for Buffer Solutions HA(aq) + H2O  H3O+(aq) + A–(aq) Therefore, (1) Henderson-Hasselbalch equation: (2) • Either equation (1) or (2) available to estimate pH of the buffer • In general: [HA]  CHA, [A–]  CNaA • Effective buffer: • Buffer range: pH = pKa 1 (i.e., [HA]/[A–]: 0.1~ 10) • Molarity of each buffer component > Ka at least 100 fold

  7. 17-2 (Continuous) • Buffer capacity andBuffer range • Buffer capacity: The amount of strong acid and/or strong base that a buffer solution can neutralize while maintaining an essentially constant pH. • Buffer range: The range of pH values over which a buffer solution can maintain a fairly constant pH. (pH = pKa 1)

  8. 17-2 (Continuous) • Preparing Buffer Solutions • Making a buffer solution with a desired pH

  9. Six methods for preparing buffer solutions

  10. New pH of a buffer after strong acid or base is added

  11. 17-3 Acid–Base Indicators • Acid-base indicators: A weak organic acid or a weak organic base whose undissociated formdiffers in color from its conjugate base or its conjugate acid. HIn + H2O  In– + H3O+ Color in Color in acidic solution basic solution • Generally, the indicator range is expressed as pH = pKa 1:

  12. mol mmol mol/1000 M = = = L mL L/1000 17-4 Neutralization Reactions and Titration Curves • General description • Equivalence point: The point in a titration in which the reactants are in stoichiometric proportions. They consume each other, and neither reactant is in excess. • Endpoint: The point in a titration at which indicator changes color. • Titration curve: The graph of pH versus volume of titrant. • To choose a proper indicator, the end point should as closely as possible to the equivalence point of the titration. • The millimole:

  13. (Continuous) • Choosing a proper acid-base indicator-1

  14. (Continuous) • Choosing a proper acid-base indicator-2

  15. Titrating strong acids with strong bases • Example : Titrating HCl with NaOH (1) At beginning: [H3O+] ≡ CHCl (2) Before the equivalence point: calculating net excess moles of HCl, then [H3O+] (3) At the equivalence point: [H3O+] = 1.0 x 10−7, pH = 7.00 (4) After the equivalence point: calculating net excess moles of NaOH, then [OH–], then, [H3O+]

  16. (a) pH = –log[H3O+] = –log(0.100) = 1.00

  17. (b) Initial amount of H3O+: 25.00 mL x 0.100 mmol H3O+/mL = 2.50 mmol H3O+ Amount of OH– added: 24.00 mL x 0.100 mmol OH–/mL = 2.40 mmol OH– H3O+ + OH–→ 2 H2O Initial, mmol: 2.50 2.40 Changes, mmol: –2.40 –2.40 Equilibrium, mmol: 0.10 – Total volume is 25.00 mL + 24.00 mL = 49.00 mL

  18. (c) Initial amount of H3O+: 25.00 mL x 0.100 mmol H3O+/mL = 2.50 mmol H3O+ Amount of OH– added: 25.00 mL x 0.100 mmol OH–/mL = 2.50 mmol OH– [H3O+][OH−] = [H3O+]2 = 1.00 x 10−14 [H3O+] = 1.00 x 10−7 pH = 7.00

  19. (d) Initial amount of H3O+: 25.00 mL x 0.100 mmol H3O+/mL = 2.50 mmol H3O+ Amount of OH– added: 26.00 mL x 01500 mmol OH–/mL = 2.60 mmol OH– H3O+ + OH–→ 2 H2O Initial, mmol: 2.50 2.60 Changes, mmol: –2.50 –2.50 Equilibrium, mmol: – 0.1 Total volume is 25.00 mL + 26.00 mL = 51.00 mL

  20. Titration curve for 25.00 mL of 0.100 M HCl with 0.100 M NaOH

  21. Titration of a Weak Acid with a Strong Base • Example : Titrating CH3COOH with NaOH • At the begining: • HA + H2O  A– + H3O+ • CHA–x x x x = [H3O+] • Before the equivalence point: • Calculating net excess moles of HA and the produced A–, then, [HA], [A–], then, or

  22. (3) At the equivalence point:*** Calculating the expected moles of A–, then, [A–], then, A– + H2O  HA + OH– [A–]–x x x x = [OH–], then [H3O+], then pH • After the equivalence point: • Calculating the net excess moles of OH–, then, [OH–], then, [H3O+], then pH

  23. Solution: (a) Addition of 0 mL of 0.100 M NaOH:

  24. (b) Addition of 10.00 mL of 0.100 M NaOH:

  25. (c) Addition of 12.50 mL of 0.100 M NaOH:

  26. (d) Addition of 25.00 mL of 0.100 M NaOH: CH3COOH + OH–  H2O + CH3COO– Initial, mmol: 2.50 2.50 – Changes, mmol: –2.50 –2.50 +2.50 Equilib, mmol: – – 2.50 [CH3COO−] = 2.50 mmol/(25 mL + 25 mL)= 0.05 M CH3COO– + H2O  CH3COOH + OH– Initial, M: 0.05 – – Changes, M: –x +x +x Equilib, M: (0.05−x) x x

  27. (e) Addition of 26.00 mL of 0.100 M NaOH: CH3COOH + OH–  H2O + CH3COO– Initial, mmol: 2.50 2.60 – Changes, mmol: –2.50 –2.50 +2.50 Equilib, mmol: – 0.10 2.50

  28. Titration curve for 25.00 mL of 0.100 M CH3COOH with 0.100 M NaOH • The equivalence-point pH is NOT 7.00. It is > 7.00 for weak acid titrated with strong base • Phenolphthalein is selected as the indicator for weak acid titrated with strong base • The best buffering occurred at pH ≈ pKa

  29. pH values at Equivalence point

  30. Titration of a Weak Polyprotic Acid Example : Titrating H3PO4 with NaOH H3PO4 + H2O  H2PO4– + H3O+ H2PO4– + H2O  HPO42– + H3O+ HPO42– + H2O  PO43– + H3O+ PO43– + H2O  HPO42– + OH– HPO42– + H2O  H2PO4–+ OH– H2PO4– + H2O  HP3O4 + OH– Ka1 x Kb3 = Kw, Ka2 x Kb2 = Kw , Ka3 x Kb1 = Kw

  31. (Continuous) Titration of a weak polyprotic acid - 10.0 mL of 0.100 M H3PO4 with 0.100 M NaOH First equivalence point, Ka2 > Kb3, [H3O+] > [OH], acidic solution Second equivalence point, Kb2 > Ka3, [OH] > [H3O+], basic solution

  32. 17-5 Solutions of Salts of Polyprotic Acids (for examples, Na3PO4, Na2HPO4, and NaH2PO4) • For Na3PO4

  33. (Continuous) • For Na2HPO4 and NaH2PO4

  34. 17-6 Acid-Base Equilibrium Calculations: A Summary • Determine which species are potentially present in solution, and how large their concentrations are likely to be. • Identify possible reactions between components and determine their stoichiometry. • Identify which equilibrium equations apply to the particular situation and which are most significant.

  35. End of Chapter 17

More Related