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Chapter 17 Additional Aspects of Aqueous Equilibria

Chapter 17 Additional Aspects of Aqueous Equilibria. 17.1 The Common Ion Effect. CH 3 COOH ( aq ) + H 2 O ( l ). H 3 O + ( aq ) + CH 3 COO − ( aq). The Common-Ion Effect. Consider a solution of acetic acid:

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Chapter 17 Additional Aspects of Aqueous Equilibria

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  1. Chapter 17Additional Aspects of Aqueous Equilibria

  2. 17.1 The Common Ion Effect

  3. CH3COOH(aq) + H2O(l) H3O+(aq)+ CH3COO−(aq) The Common-Ion Effect • Consider a solution of acetic acid: • If acetate ion is added to the solution, Le Châtelier says the equilibrium will shift to the left.

  4. The Common-Ion Effect “The extent of ionization of a weak electrolyte is decreased by adding to the solution a strong electrolyte that has an ion in common with the weak electrolyte.”

  5. Sample Exercise 17.1 • What is the pH of a solution made by adding 0.30 mol of acetic acid and 0.30 mol of sodium acetate to enough water to make 1.0 L of solution?

  6. Practice 17.1 Calculate the pH of a solution containing 0.085 M nitrous acid (HNO2; Ka = 4.5 × 10-4) and 0.10 M potassium nitrite (KNO2).

  7. Sample Exercise 17.2 • Calculate the fluoride ion concentration and pH of a solution that is 0.20 M in HF and 0.10 M in HCl.

  8. Ka = [H+] [HCOO−] [HCOOH] HCOOH(aq) H+(aq) + HCOO−(aq) = 1.8  10-4 Practice 17.2 Calculate the formate ion concentration and pH of a solution that is 0.050M in formic acid (HCOOH; Ka = 1.8 x 10-4) and 0.10 M HNO3.

  9. HCOOH(aq) H+(aq) + HCOO−(aq) Example Because HNO3, a strong acid, is also present, the initial [H+] is not 0, but rather 0.10 M.

  10. Common Ion Effect • Also applies to weak bases & strong electrolytes • NH3 + H2O ↔ NH4+ + OH- • The addition of ammonium chloride, NH4Cl, would cause the reaction to shift left, decreasing [OH-]

  11. p.723 GIST • A mixture of 0.10 mol of NH4Cl and 0.12 mol of NH3 is added to enough water to make 1.0 L of solution. • A) What are the initial concentrations of the major species in the solution? • B) Which of the ions in this solution is a spectator ion in any acid-base chemistry occurring in the solution? • C) What equilibrium reaction determines [OH-] and therefore the pH of the solution?

  12. 17.2 Buffered Solutions

  13. Buffers • Buffers are solutions of a weak conjugate acid-base pair. • They are particularly resistant to pH changes, even when small amounts of strong acid or base are added. • Contains an acid to neutralize OH- and a base to neutralize H+

  14. Buffers If a small amount of hydroxide is added to an equimolar solution of HF in NaF, for example, the HF reacts with the OH− to make F− and water.

  15. Buffers Similarly, if acid is added, the F− reacts with it to form HF and water.

  16. 17.2 GIST • Which of the following conjugate acid-base pairs will not function as a buffer and explain: • C2H5COOH and C2H5COO- • HCO3- and CO32- • HNO3 and NO3- • What happens when NaOH is added to a buffer composed of CH3COOH and CH3COO-? What happens when HCl is added to this buffer?

  17. Ka = [H3O+] [A−] [HA] HA + H2O H3O+ + A− Buffer Calculations Consider the equilibrium constant expression for the dissociation of a generic acid, HA:

  18. [A−] [HA] [H3O+] Ka = base [A−] [HA] −log [H3O+] + −log −log Ka = pKa acid pH Buffer Calculations Rearranging slightly, this becomes Taking the negative log of both side, we get

  19. [base] [acid] pKa = pH − log [base] [acid] pH = pKa + log Buffer Calculations • So • Rearranging, this becomes • This is the Henderson–Hasselbalchequation. • Since the weak acid dissociates very little, • use the initial concentration.

  20. Sample Exercise 17.3 • What is the pH of a buffer that is 0.12 M in lactic acid (HC3H5O3) and 0.10 M in sodium lactate (NaC3H5O3). For lactic acid, Ka = 1.4 x 10-4.

  21. Practice Exercise 17.3 Calculate the pH of a buffer composed of 0.12 M benzoic acid and 0.20 M sodium benzoate.

  22. [base] [acid] (0.20) (0.12) pH = pKa + log pH = −log (6.3  10−5) + log Henderson–Hasselbalch Equation pH = 4.20 + 0.22 pH = 4.42

  23. Sample Exercise 17.4 Preparing a Buffer How many moles of NH4Cl must be added to 2.0 L of 0.10 M NH3 to form a buffer whose pH is 9.00? (Assume that the addition of NH4Cl does not change the volume of the solution.)

  24. Practice Exercise Calculate the concentration of sodium benzoate that must be present in a 0.20 M solution of benzoic acid (C6H5COOH) to produce a pH of 4.00.

  25. Buffer Capacity & pH Range • Buffer capacity is the amount of acid or base the buffer can neutralize before the pH noticeably changes • Depends on how much of the acid/base was used to make the buffer • greater concentration = greater buffering capacity

  26. Buffer Capacity & pH Range • The pH range is the range of pH values over which a buffer system works effectively. • Depends on • Ka • relative concentration of acid & base that make up the buffer

  27. Buffer Capacity & pH Range • Buffers work best when concentrations of acid/base are the same – if the concentrations differ by more than 10 fold, the buffer will not work well in both directions • Select a buffer by choosing one whose pKa is the same or near the desired pH. • Optimal pH of a buffer is when pH = pKa • Buffers usually have a usable range between +1/-1 pKa

  28. p.727 GIST • What is the optimal pH buffered by a solution containing acetic acid & sodium acetate? (Ka for acetic acid is 1.8 x 10-5)

  29. When Strong Acids or Bases Are Added to a Buffer… …it is safe to assume that all of the strong acid or base is consumed in the reaction.

  30. Addition of Strong Acid or Base to a Buffer • Determine how the neutralization reaction affects the amounts of the weak acid and its conjugate base in solution. • Use the Henderson–Hasselbalch equation to determine the new pH of the solution.

  31. Sample Exercise 17.5Calculating pH Changes in Buffers a) A buffer is made by adding 0.300 mol HC2H3O2 and 0.300 mol NaC2H3O2 to enough water to make 1.00 L of solution. The pH of the buffer is 4.74. a) Calculate the pH of this solution after 0.020 mol of NaOH is added. b) For comparison, calculate the pH that would result if 0.020 mol of NaOH were added to pure water.

  32. Calculating pH Changes in Buffers Before the reaction, since mol HC2H3O2 = mol C2H3O2− pH = pKa = −log (1.8  10−5) = 4.74

  33. Calculating pH Changes in Buffers The 0.020 mol NaOH will react with 0.020 mol of the acetic acid: HC2H3O2(aq) + OH−(aq) C2H3O2−(aq) + H2O(l)

  34. (0.320) (0.280) pH = 4.74 + log Calculating pH Changes in Buffers Now use the Henderson–Hasselbalch equation to calculate the new pH: pH = 4.74 + 0.06 pH pH = 4.80

  35. Practice Exercise 17.5 • Determine (a) the pH of the original buffer in the previous example after the addition of 0.020 mol HCl and (b) the pH of the solution that would result from 0.020 mol HCl to 1.00 L of pure water.

  36. Practice • A buffer solution contains 0.10 mol of propionic acid, C2H5OOH and 0.13 mol of sodium propionate, C2H5OONa in 1.50 L. • (a) what is the pH of this buffer? • (b) what is the pH of this buffer after the addition of 0.01 mol of NaOH? • (c) what is the pH of the buffer after the addition of 0.01 mol of HI?

  37. 17.3 Acid-Base Titrations

  38. Titration In this technique a known concentration of base (or acid) is slowly added to a solution of acid (or base). p.730 GIST: For this setup, will pH increase or decrease as titrant is added?

  39. Titration A pH meter or indicators are used to determine when the solution has reached the equivalence point, at which the stoichiometric amount of acid equals that of base.

  40. Titration of a Strong Acid with a Strong Base From the start of the titration to near the equivalence point, the pH goes up slowly.

  41. Titration of a Strong Acid with a Strong Base Just before (and after) the equivalence point, the pH increases rapidly.

  42. Titration of a Strong Acid with a Strong Base At the equivalence point, moles acid = moles base, and the solution contains only water and the salt from the cation of the base and the anion of the acid.

  43. Titration of a Strong Acid with a Strong Base As more base is added, the increase in pH again levels off.

  44. Sample Exercise 17.6 Calculating pH for a Strong Acid-Strong Base Titration Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution: (a) 49.0 mL, (b) 51.0 mL.

  45. Practice Exercise Calculate the pH when the following quantities of 0.100 M HNO3 have been added to 25.0 mL of 0.100 M KOH solution: (a) 24.9 mL, (b) 25.1 mL.

  46. p.733 GIST • What is the pH at the equivalence point when 0.10 M HNO3 is added to a solution containing 0.30 g of KOH?

  47. Titration of a Weak Acid with a Strong Base • Unlike in the previous case, the conjugate base of the acid affects the pH when it is formed. • At the equivalence point the pH is >7. • Phenolphthalein is commonly used as an indicator in these titrations.

  48. Titration of a Weak Acid with a Strong Base At each point below the equivalence point, the pH of the solution during titration is determined from the amounts of the acid and its conjugate base present at that particular time.

  49. Titration of a Weak Acid with a Strong Base With weaker acids, the initial pH is higher and pH changes near the equivalence point are more subtle.

  50. Sample Exercise 17.7 Calculating pH for a Weak Acid-Strong Base Titration Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH (Ka = 1.8 ×10-5).

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