1 / 46

Chapter 09 Monoprotic Acid-Base Equilibria

Chapter 09 Monoprotic Acid-Base Equilibria. Contents in Chapter09. 1. pH/pOH Calculations 1) Strong Acids and Strong Bases 2) Weak Acids and Bases 2. Fraction of Dissociation 3. Acid-Base Buffers 1) General information 2) Calculation the pH of a Buffer

lucius
Download Presentation

Chapter 09 Monoprotic 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 09 Monoprotic Acid-Base Equilibria QCA7e Chapter09

  2. Contents in Chapter09 1. pH/pOH Calculations 1) Strong Acids and Strong Bases 2) Weak Acids and Bases 2. Fraction of Dissociation 3. Acid-Base Buffers 1) General information 2) Calculation the pH of a Buffer 3) Henderson-Hasselbalch equation 4) Buffer Capacity 5) Buffer Preparation 6) Buffer pH Calculation by Systematic Approach 7) Spreadsheet Estimating pH of Buffer without Approximation QCA7e Chapter09

  3. 1. pH Calculations • Strong Acids and Strong Bases • i) General information: • Strong acids: • e.g., HCl(aq) + H2O → H3O+(aq) + Cl–(aq) • Strong bases: e.g., NaOH(aq) → Na+(aq) + OH–(aq) • At 25 oC: Kw = 1 x 10–14 = [H3O+][OH–] pKw = 14.0 = pH + pOH • pH = –logAH3O+ = –logγH3O+ [H3O+] QCA7e Chapter09

  4. pH of Strong Acids and Strong Bases • KOH and HBr for examples: • High conc. (> 10–6 M): • The pH is controlled by added strong acids and/or bases. • b) Low conc. (< 10–8 M): • pH ≈ 7.00, pH controlled by autoprotolysis of water • c) Intermediate conc. (10–8~ 10–6 M): pH obtained by systematic approach calculations. (Slide 29/30, Chapter08) QCA7e Chapter09

  5. iii) 水溶液中的 H3O+和 OH–來自何方？) • Example: (assume activity = concentration) • 計算 1.0×10–3 M 之 HBr(aq)的 pH值=? • 續，來自HBr之[H3O+] = ? • 續，來自H2O之[OH–] = ? • d) 續，來自H2O之[H3O+] = ? Solution: assume 來自H2O 之 [H3O+] << 來自 HBr 之 [H3O+] a) pH = –log[H3O+] = –log(1.0×10–3) = 3 b) 來自 HBr 之 [H3O+] = 1.0×10–3 M c) 來自 H2O 之 [OH–] = Kw/(1.0×10–3) = 1.0×10–11 M d)來自 H2O 之 [H3O+] = 來自 H2O 之 [OH–] = 1.0×10–11 M QCA7e Chapter09

  6. 2) Weak Acids and Bases • pH Calculations ofWeak Acids • HA + H2O  H3O+ + A– • Derivation an equation for calculating the pH of HA with analytical concentration F • (without approximation) Charge balance: [H3O+] = [A–] + [OH–] Mass balance: F = [HA] + [A–] Therefore, [A–] = [H3O+] – [OH–] [HA] = F – [A–] = F – [H3O+] + [OH–] ----------- QCA7e Chapter09

  7. Without approximation QCA7e Chapter09

  8. ***** b) Approximation method 1 for calculating the pH of HA with analytical concentration F (with approximation) Charge balance: [H3O+] = [A–] + [OH–] Mass balance: F = [HA] + [A–] HA + H2O  H3O+ + A– 通常： 來自HA 之[H3O+]>>來自H2O之[H3O+] [A–]>>[OH–], Therefore, the simplified charge balance: [H3O+] ≈ [A–] QCA7e Chapter09

  9. Solving quadratic equation: For ax2 + bx +c = 0 HA + H2O  H3O+ + A– F–x x x 一元二次方程式解 或用連續代入法解..... [H3O+]2 + Ka[H3O+] – KaF = 0 QCA7e Chapter09

  10. c) Approximation method 2 for calculating the pH of HA with analytical concentration F (with more approximation) HA + H2O  H3O+ + A– F–x x x Assume F >> x QCA7e Chapter09

  11. Summary • (Calculating the pH of monoprotic weak acid) Without approximation Assume [H3O+] = [A–] [H3O+]2 + Ka[H3O+] – KaF = 0 Approximation Method 1 ***** Assume F >> x Approximation Method 2 QCA7e Chapter09

  12. ii) pOH Calculations ofWeak Bases Calculating the pOH of monobasic weak base B + H2O  BH+ + OH– Without approximation Assume [OH–] = [BH+] [OH–]2 + Kb[OH–] – KbF = 0 Approximation Method 1 ***** Assume F >> x Approximation Method 2 QCA7e Chapter09

  13. iii) Conversion Ka and Kb for conjugate acid-base pair For monoprotic acid: HA + H2O  A– + H3O+ Its conjugate base: A–+ H2O  HA + OH– Ka x Kb = Kw Therefore, Kb = Kw/ Ka QCA7e Chapter09

  14. 2. Fraction of Dissociation 1) Fraction dissociation of monoprotic acid (HA) For HA with formal concentration F: F = [HA] + [A–] Fraction of dissociated HA: 2) Fraction dissociation of monobasic base (B) QCA7e Chapter09

  15. 3. Acid-Base Buffers • General information • Define Acid-Base Buffer: • A solution containing a conjugate weak acid/weak base pair that is resistant to a change in pH when a strong acid or strong base is added when the conjugate acid-base pair still present. QCA7e Chapter09

  16. ii) Common ion effect in acid-base equilibrium • A solution containing a weak electrolyte (e.g., weak acid HA) and a strong electrolyte (e.g., NaA) with one of the same (common) ion in both electrolyte. • The acid component (conjugate acid) of the buffer can neutralize small added amounts of OH–: OH– + HA  H2O + A– • The basic component (conjugate base) can neutralize small added amounts of H3O+: H3O+ + A– H2O + HA QCA7e Chapter09

  17. Example 1 (without common ion effect) Q1: What is the pH of 0.2 M CH3COOH (Ka=1.8x10–5) solution? A1: CH3COOH + H2O  CH3COO– + H3O+ initial 0.2 0 – change –x +x +x final 0.2–x +x +x QCA7e Chapter09

  18. Q2: What is the pH if 0.0050 mol NaOH added to 0.5 L of 0.2 M CH3COOH (Ka=1.8x10–5) solution? A2: Neutralization reaction: CH3COOH + OH– CH3COO– + H2O initial 0.2 M x 0.5 L 0 – = 0.1 mol change –0.005 mol +0.005 mol – final 0.095 mol 0.005 mol – final conc.0.19 M 0.01 M – QCA7e Chapter09

  19. CH3COOH + H2O  CH3COO– + H3O+ initial 0.19 0.01 – change –x +x +x Final 0.19–x 0.01+x +x QCA7e Chapter09

  20. Example 2 (with common ion effect/a buffer solution) Q1: What is the pH of a buffer solution containing 0.2 M CH3COOH (Ka=1.8x10–5) and 0.2 M CH3COONa? A1: CH3COOH + H2O  CH3COO– + H3O+ initial 0.2 0.2 – change ≈0≈0 – final 0.2 0.2 +x QCA7e Chapter09

  21. Q2: What is the pH if 0.0050 mol NaOH added to a 0.5 L buffer solution containing 0.2 M CH3COOH (pKa=4.74) and 0.2 M CH3COONa? A2: Neutralization reaction: CH3COOH + OH– CH3COO– + H2O initial 0.2 M x 0.5 L 0.2 M x 0.5 L– = 0.1 mol = 0.1 mol change –0.005 mol +0.005 mol – final 0.095 mol 0.105 mol – Final Conc.0.19 M 0.21 M – QCA7e Chapter09

  22. CH3COOH + H2O  CH3COO– + H3O+ initial 0.19 0.21 – change ≈0≈0 – final 0.19 0.21 +x QCA7e Chapter09

  23. 2. Henderson-Hasselbalch Equation i) Calculating pH of HA, from known Ka HA(aq) + H2O  H3O+(aq) + A–(aq) Therefore, Henderson-Hasselbalch equation: Effective buffer range: pH = pKa 1 ([A–]/[HA]: 0.1~ 10) QCA7e Chapter09

  24. ii) Calculating pOH of A–, from known Kb A–(aq) + H2O  OH–(aq) + HA(aq) Therefore, (1) Henderson-Hasselbalch equation: (2) Effective buffer range: pOH = pKb 1 ([HA]/[A–]: 0.1~ 10) QCA7e Chapter09

  25. 4) Buffer Capacity • Define buffer capacity: • The ability of how well a solution resists changes in pH when acid or base is added ii) Quantitative equation of buffer capacity Thenumber of molesof astrong acid or a strongbase that causes1.00 L of the bufferto undergo a1.00 pH unitchange: QCA7e Chapter09

  26. Cb (mole of strong base/L) vs. pH for a solution containing 0.100 F HA with pKa = 5.00 Buffer capacity vs. pH for same system QCA7e Chapter09

  27. iii) Buffer capacity dependence: • Concentrations of its components: Thehigher concentrationsof its components (conjugate acid-base pair) the higher buffer capacity. • Ratio of the conjugate acid-base: The higher buffer capacity at the[HA]/[A–]=1, i.e.,pH  pKa. • Buffer range: rule of thumb,pKa 1 QCA7e Chapter09

  28. Buffer Preparation • i) Remarks for the real pH value/practical buffer preparation : • Concentration equilibrium constant (e.g., Ka’ and Kb’) changed when either temperature changed or ionic strength of the solution changed. • pH measured pH meter response to aH3O+ rather than [H3O+]. • Activity is the effective concentration, and the real pH equal to –log(aH3O+) rather than –log[H3O+]. • Example of practical buffer preparation procedure: Adding NaOH solution to CH3COOH solution until desired pH monitored by a pH meter. QCA7e Chapter09

  29. ii) Calculations for preparing a buffer solution Example: How many mL of 1.0 M NaOH should be added to 500 mL of 0.5 M CH3COOH (Ka=1.8x10–5) to give a pH 4.90 buffer? Solution: Neutralization reaction: CH3COOH + OH– CH3COO– + H2O initial 0.5 M x 0.5 L –– = 0.25 mol change –0.001x mol +0.001x mol – final 0.25 – 0.001x mol +0.001x mol – final Conc.(0.25 – 0.001x)/0.5 L 0.001x/0.5 L – QCA7e Chapter09

  30. Ans QCA7e Chapter09

  31. QCA7e Chapter09

  32. QCA7e Chapter09

  33. QCA7e Chapter09

  34. QCA7e Chapter09

  35. 6) Buffer pH Calculation by Systematic Approach For a buffer prepared from HA (formal conc. FHA) and NaA(formal conc. FA–) (2) 帶入 (1) (3) 帶入 (4)，整理 (5) 帶入 (1)，整理 QCA7e Chapter09

  36. Approximation method 1: • FHA≈ [HA]，FA– ≈ [A–]，帶入下式運算 ii) Approximation method 2: Assume [H3O+]>>[OH–] QCA7e Chapter09

  37. iii) Without approximation 帶入下式運算： 以 spreadsheet 輔助為佳。 QCA7e Chapter09

  38. 6) Spreadsheet Estimating pH of Buffer without Approximation Example: A monoprotic weak acid, Ka = 1x10–2, Kw = 1x10–14, FHA = 0.01 M, FA– = 0.01 M. QCA7e Chapter09

  39. QCA7e Chapter09

  40. QCA7e Chapter09

  41. QCA7e Chapter09

  42. QCA7e Chapter09

  43. QCA7e Chapter09

  44. B5 Cell 已定義為 H QCA7e Chapter09

  45. QCA7e Chapter09

  46. Homework: Problem 9-42 (b)/p.179, Due 2009/12/23，請用學號作為檔名upload，檔名勿含中文。 Examples: All Exercise： A-D, F-J Problems: 1-8, 10-16, 19-24, 26-35 End of Chapter09 QCA7e Chapter09

More Related