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Chapter 10 Polyprotic Acid-Base Equilibria

Chapter 10 Polyprotic Acid-Base Equilibria. Contents in Chapter10. 1. Species/Concentrations Calculations of Polyprotic Acids 1) Species/Conc. Calculations of Diprotic Acids 2) pH Calculations of Triprotic Acids 3) pH Calculations of Amino Acids Polyprotic Buffers

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Chapter 10 Polyprotic Acid-Base Equilibria

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  1. Chapter 10 Polyprotic Acid-Base Equilibria QCA7e Chapter10

  2. Contents in Chapter10 • 1. Species/Concentrations Calculations of Polyprotic Acids • 1) Species/Conc. Calculations of Diprotic Acids • 2) pH Calculations of Triprotic Acids • 3) pH Calculations of Amino Acids • Polyprotic Buffers • Conditional Principal Species • 1) Ladder Diagrams • 2) Fractional Composition Diagram • 4. Isoelectric Point QCA7e Chapter10

  3. Species/Concentrations Calculations of Polyprotic Acids • 1) Species/Conc. Calculations of Diprotic Acids • Hydrolysis constants revisit (for H2A, NaHA, Na2A) • H2A + H2O  H3O+ + HA–Ka1 • A2– + H2O  OH– + HA–Kb1 • HA– + H2O  H3O+ + A2– Ka2 • HA– + H2O  OH– + H2A Kb2 QCA7e Chapter10

  4. 1) Species/Conc. Calculations of Diprotic Acids Example 1: 計算0.1 F之H2A的pH值，以及[H3O+]、[OH–]、[H2A]、[HA–]、[A2–]=? (Ka1=1x10–3, Ka2=1x10–7) H2A + H2O  H3O+ + HA- Ka1 = 1x10-3 x x 0.1-x 設[H3O+] = [HA–] [H3O+] = [HA–] = 0.0095 M, pH=2.02 [OH–]=10–(14–2.02)=1.05x10–12 M HA– + H2O  H3O+ + A2– ∵已設[H3O+] = [HA–]∴[A2–] = 1.0x10–7 M [H2A] = 0.1 – 0.0095 - 1.0x10–7 = 0.0905 M QCA7e Chapter10

  5. Example 2: 計算0.1 F之Na2A的pH值，以及[H3O+]、[OH–]、[H2A]、[HA–]、[A2–]=? (Ka1=1x10–3, Ka2=1x10–7) A2- + H2O  OH– + HA– Kb1= 1x10–7 x x 0.1-x 設[OH–] = [HA–] [OH–] = [HA–] = 1.0x10–4 M [H3O+]= 1.0x10–10M, pH=10 HA– + H2O  OH– + H2A ∵已設[OH–] = [HA–]∴[H2A]=1.0x10–11 M [A2–] = 0.1 – 1.0x10–4 – 1.0x10–11 = 0.1 M QCA7e Chapter10

  6. Example 3: 計算0.1 F之 NaHA 的pH值，以及[H3O+]、[OH–]、[H2A]、[HA–]、[A2–]=? (Ka1=1x10–3, Ka2=1x10–7) (NaHA 為 amphiprotic sbustance, 需導公式了) HA− + H2O  H3O+ + A2− Ka2 HA− + H2O  OH− + H2A Kb2 Mass balance: [Na+]=0.1 (1) [H2A] + [HA–] + [A2–] = 0.1 (2) Charge balance: [Na+] + [H3O+] = [OH–] + [HA–] +2[A2–] (3) (2)+(3) [H2A] + [Na+] + [H3O+] = 0.1 + [OH–] + [A2–] 又由 (1) 知 [Na+] = 0.1 ∴ [H2A] + [H3O+] = [OH–] + [A2–] (4) QCA7e Chapter10

  7. ∵ and 代入(4)得 (5) 只剩下[H3O+]和[HA-]兩個未知數了 (5) × [H3O+]得 (6) 整理一下 (7) (8) QCA7e Chapter10

  8. (9) (10) ***** (11) HA– + H2O  H3O+ + A2– Ka2 HA– + H2O  OH– + H2A Kb2 設[HA–]>>[A2–]，[HA–]>>[H2A]，則[HA–]=FHA– 依此假設代入(11)得 (12) QCA7e Chapter10

  9. (11)式引申： 一般而言，Ka1Ka2[HA–]>>Ka1Kw，[HA–]>>Ka1，所以： (13) 亦即： Remark: Successive approximations can resulted more accurate result, please read p.186/Box 10-1/Harris QCA 7th Ed. QCA7e Chapter10

  10. 2) pH Calculations of Triprotic Acids • Hydrolysis constants revisit (for H3A, NaH2A, Na2HA, Na3A) • H3A + H2O  H3O+ + H2A- Ka1 = Kw/Kb3 • A3– + H2O  OH– + HA2–Kb1 = Kw/Ka3 • H2A– + H2O H3O+ + HA2–Ka2 = Kw/Kb2 • H2A– + H2OOH– + H3A Kb3 = Kw/Ka1 • HA2– + H2OH3O+ + A3– Ka3= Kw/Kb1 • HA2– + H2O  OH– + H2A– Kb2 = Kw/Ka2 QCA7e Chapter10

  11. Example 1: Calculate pH prepared from H3A H3A + H2O  H3O+ + H2A–Ka1 FH3A-x x x pH = ..... QCA7e Chapter10

  12. Example 2: Calculate pH prepared from Na3A A3– + H2O  OH– + HA2–Kb1 FA3–– x x x pOH = ....., pH=..... QCA7e Chapter10

  13. Example 3: Calculate pH prepared from NaH2A H2A– + H2O  H3O+ + HA2–Ka2 H2A– + H2OOH– + H3A Kb3 =Kw/Ka1 QCA7e Chapter10

  14. Example 4: Calculate pH prepared from Na2HA HA2– + H2O  H3O+ + A3– Ka3 HA2– + H2O  OH– + H2A– Kb2 =Kw/Ka2 QCA7e Chapter10

  15. 3) pH Calculations of Amino acids • Terms and Definition • Zwitterion: A dipolar compound with acidic and basic groups in the same molecule (a compound with positive and negative charge in the same molecule). For example, amino acid generally exists in the form of +H3NCHRCOO–, behaves an internal neutralization form. QCA7e Chapter10

  16. Nonionized Amino acid Ionized Amino acid H2O OH– H3O+ H2O QCA7e Chapter10

  17. ii) Acid dissociation constants (Ka) of amino acids Type A2– Type H2A Type HA– H2L+ + H2O  H3O+ + HLKa1 L– + H2O  OH– + HLKb1 HL + H2O  H3O+ + L–Ka2 HL + H2O  OH– + H2L+ Kb2 QCA7e Chapter10

  18. QCA7e Chapter10

  19. QCA7e Chapter10

  20. iii) pH Calculations of Amino acids (Diprotic type) H2L+ + H2O  H3O+ + HLKa1 L– + H2O  OH– + HLKb1 HL + H2O  H3O+ + L–Ka2 HL + H2O  OH– + H2L+ Kb2 Leucine for example: QCA7e Chapter10

  21. Example: Followings are the acid dissociation constants of leucine: • +H3NCHCOOH + H2O  H3O+ + +H3NCHCOO–pKa1= 2.328 • +H3NCHCOO– + H2O H3O+ + H2NCHCOO–pKa2= 9.744 • What is the pH of a solution prepared from 0.05 M H2NCHCOOH·HCl, that is (+H3NCHCOOH)(Cl–)? • What is the pH of a solution prepared from 0.05 M H2NCHCOONa, that is (Na+)(H2NCHCOO–)? • What is the pH of a solution prepared from 0.05 M (H2NCHCOOH)? • Solution: • H2L+, i.e., H2A type • L–, i.e., A2– type • HL, i.e., HA– type Ans: (A) pH = 1.88, (B) pH = 11.22, (C) pH = 6.06 QCA7e Chapter10

  22. iv) pH Calculations of Amino acids (Triprotic type) QCA7e Chapter10

  23. Polyprotic Buffers 1) Diprotic Buffers, H2A type for example QCA7e Chapter10

  24. Example 1: Find the pH of a solution prepared by dissolving 1.00 potassium hydrogen phthalate (KHP fw 204.221) and 1.20 g of disodium phthalate (Na2P fw 210.094) in 50.0 mL of water. (phthalic acid: pKa1 = 2.950, pKa2 = 5.408) Solution: QCA7e Chapter10

  25. Example 2: How many mL of 0.8000 M KOH should be added to 3.38 g of oxalic acid (H2Ox, MW 90.035) to give a pH of 4.40 when diluted to 500 mL? (oxalic acid: pKa1 = 1.27, pKa2 = 4.266) Solution: pH of solution 4.40: Using [Ox2–]/[HOx–] pair pH (4.40) > pKa2 (4.266): [Ox2–] > [HOx–] Step 1: mL NaOH for converting H2Ox to HOx– QCA7e Chapter10

  26. Step 2: HOx– + OH– Ox2– + H2O Initial 0.03754 mol x – Change –x –x x Final 0.3754 –x 0 x Step 3: 46.92 mL + 27.05 mL = 73.97 mL Ans: 73.97 mL KOH added QCA7e Chapter10

  27. 2) Triprotic Buffers QCA7e Chapter10

  28. 3. Conditional Principal Species 1) Ladder diagrams i) At pH = pKa, [A–]=[HA] QCA7e Chapter10

  29. ii) QCA7e Chapter10

  30. iii) QCA7e Chapter10

  31. 2) Fractional composition diagram i) For monoprotic acid: F = CT: α0 = αHA, α1=αA– ..... QCA7e Chapter10

  32. ii) For diprotic acid: Fractional Compositional H2A Excel Operation QCA7e Chapter10

  33. iii) For triprotic acid: QCA7e Chapter10

  34. Isoelectric Point (Isoelectric pH, called pI) Definition: The pH at which average charge of the polyprotic acid is 0. QCA7e Chapter10

  35. Examples: All Exercise： A-H Problems: 1-4, 6, 9, 11-12, 15, 17-18, 22-27, 36, 38 End of Chapter 10 QCA7e Chapter10

  36. App 1: Systematic pH Calculations of HA Systematic approach for pH calculation of monoprotic acid (HA): Charge balance: [H3O+] = [A–] + [OH–] (1) Mass balance: CHA = [HA] + [A–] (2) Equilibrium constant: (3) Rewrite (1) [A–] = [H3O+] – [OH–] (4) Substituting (4) into (2) and the rewrite [HA] = CHA – [H3O+] + [OH–] (5) Substituting (4) and (5) into (3) (6) QCA7e Chapter10

  37. Rearrange (6) (7) Solving quadratic equation (7) (8) (9) (10) QCA7e Chapter10

  38. App 2: Systematic pH Calculations of H2A Systematic approach for pH calculation of diprotic acid (H2A) Charge balance: [H3O+] = [HA–]+2[A2–] + [OH–] (1) Mass balance: CH2A = [H2A]+[HA–] + [A2–] (2) Equilibrium constant: (3) (4) Rewrite (3) (5) Substituting (4) into (5), the rewrite (6) QCA7e Chapter10

  39. Substituting (5) and (6) into (2) (7) Rearrange (7) (8) Substituting (8) into (5) (9) (10) QCA7e Chapter10

  40. Substituting (9) and (10) into (1), the rearrange (11) Substituting (9) and (10) into (1), the rearrange (12) Solving quadratic equation (12) (13) QCA7e Chapter10

  41. (14) (15) It is commercial available now: Chemical Calculator: http://www.chembuddy.com/ QCA7e Chapter10

  42. App 3: pH Calculations Summary 請自行練習！！ QCA7e Chapter10

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