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Chapter 17 Additional Aspects of Aqueous Equilibria

Chapter 17 Additional Aspects of Aqueous Equilibria. Various aqueous solutions encountered in nature typically contain many solutes. Many equilibria can take place simultaneously in these solutions. The understanding of complex solutions is a further application of acid-base equilibria.

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Chapter 17 Additional Aspects of Aqueous Equilibria

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  1. Chapter 17Additional Aspects of Aqueous Equilibria

  2. Various aqueous solutions encountered in nature typically contain many solutes. • Many equilibria can take place simultaneously in these solutions. • The understanding of complex solutions is a further application of acid-base equilibria.

  3. HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2−(aq) 17.1 The Common-Ion Effect • Def: A shift of an equilibrium induced by an ion common to the equilibrium • Consider a solution of acetic acid: • If acetate ion is added to the solution, Le Châtelier says the equilibrium will shift to the left. • The dissociation of the weak acid HC2H3O2 decreases when the strong electrolyte NaC2H3O2, which has an ion in common with it, is added.

  4. 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. • Equilibrium concentrations may be calculated like those for weak acids and weak bases in Ch 16.

  5. Example 1: What is the pH of a solution made by adding 0.30 mol of acetic acid (HC2H3O2) and 0.30 mol of sodium acetate (NaC2H3O2) to enough water to make 1.0 L of solution?

  6. Ka = [H3O+] [F−] [HF] = 6.8  10-4 Example 2: Calculate the fluoride ion concentration and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. Ka for HF is 6.8  10−4.

  7. H3O+(aq) + F−(aq) HF(aq) + H2O(l) Because HCl, a strong acid, is also present, the initial [H3O+] is not 0, but rather 0.10 M.

  8. (0.10) (x) (0.20) 6.8  10−4 = (0.20) (6.8  10−4) (0.10) x = x = 1.4  10−3 • Therefore, [F−] = x = 1.4  10−3 M [H3O+] = 0.10 + x = 0.10 + 1.4  10−3 = 0.10 M • So, pH = −log (0.10) pH = 1.00

  9. 17.2 Buffered Solutions • Solutions of a weak conjugate acid-base pair. • They are particularly resistant to pH changes, even when strong acid or base is added.

  10. Composition and Action of Buffered Solutions • A buffer resists changes in pH because it contains both an acidic species to neutralize OH- ions and a basic species to neutralize H+ ions. • The acidic and basic species must not consume themselves through neutralization. • These requirements are met by using a weak conjugate acid-base pair:

  11. 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.

  12. If acid is added, the F− reacts to form HF and water.

  13. Ka = [H3O+] [A−] [HA] H3O+ + A− HA + H2O Calculating the pH of a Buffer • Because conjugate acid-base pairs share a common ion, we can use the same procedures to calculate the pH of a buffer that were used for th common-ion effect. • Consider the equilibrium constant expression for the dissociation of a generic acid, HA:

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

  15. [base] [acid] pKa = pH − log [base] [acid] pH = pKa + log • So • Rearranging, this becomes • This is the Henderson–Hasselbalchequation.

  16. (0.10) (0.12) [base] [acid] pH = pKa + log pH = −log (1.4  10−4) + log Henderson–Hasselbalch Equation What is the pH of a buffer that is 0.12 M in lactic acid, HC3H5O3, and 0.10 M in sodium lactate? Ka for lactic acid is 1.4  10−4. pH = 3.85 + (−0.08) pH = 3.77

  17. Buffer Capacity and pH Range • Buffer capacity is the amount of acid or base the buffer can neutralize before the pH begins to change to an appreciable degree. • The greater the amounts of the conjugate acid-base pair, the more resistant the ratio of their concentrations, and hence the pH, is to change.

  18. The pH range is the range of pH values over which a buffer system works effectively. • It is best to choose an acid with a pKa close to the desired pH.

  19. Addition of Strong Acids or Bases to Buffers It is safe to assume that all of the strong acid or base is consumed in the reaction.

  20. 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.

  21. Calculating pH Change in a Buffer • 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. • Calculate the pH of this solution after 0.020 mol of NaOH is added.

  22. Before the reaction, since mol HC2H3O2 = mol C2H3O2− pH = pKa = −log (1.8  10−5) = 4.74

  23. 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)

  24. (0.320) (0. 280) pH = 4.74 + log Now use the Henderson–Hasselbalch equation to calculate the new pH: pH = 4.74 + 0.06 pH = 4.80 Note: Read “Blood as a Buffered Solution” on page 731

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