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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

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Chapter 17 Additional Aspects of Acid–Base Equilibria

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  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

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