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Chapter 15

Chapter 15. Applications of Aqueous Equilibria. The Common-Ion Effect. Common-Ion Effect : The shift in the position of an equilibrium on addition of a substance that provides an ion in common with one of the ions already involved in the equilibrium

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Chapter 15

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  1. Chapter 15 Applications of Aqueous Equilibria

  2. The Common-Ion Effect Common-Ion Effect: The shift in the position of an equilibrium on addition of a substance that provides an ion in common with one of the ions already involved in the equilibrium This applies to weak acids, weak bases and solubility of salts when a common ion is added to the equilibrium reaction. You can change the pH, by adding salt. HF (aq) + H2O(aq) H3O+ (aq) + F- (aq) Ka = 7.2 x 10-4 What happened to the equilibirium as NaF salt was added to the solution? NaF (s) → Na+ (aq) + F– (aq) When F- is added (from NaF), then [H+] must decrease (Le Chatelier’s principle). The pH increases

  3. 15.3 Buffer Solutions Buffer Solution: A solution which contains a weak acid and its conjugate base and resists drastic changes in pH. Weak acid + Conjugate base For Example: CH3CO2H + CH3CO21- HF + F1- NH41+ + NH3 H2PO41- + HPO42- Buffer Solutions: contain a common ion and are important in biochemical and physiological processes Organisms (and humans) have built-in buffers to protect them against changes in pH. Human blood is maintained by a combination of CO3-2, PO4-3 and protein buffers. Blood: (pH 7.4)Death = 7.0 <pH > 7.8 = Death

  4. Example • Calculate the pH of a solution that is prepared by dissolved 0.10 mol of acetic acid and 0.10 mol sodium acetate in enough water to make 1.00 L of solution. Ka = 1.8 x 10-5

  5. Example • Calculate the pH of 0.15M HF and 0.25NaF mixture. Is this a buffer solution?

  6. Example • Calculate the pH of 100.0 mL DI water • Calculate the new pH after adding 1.0 mL of 0.10M HCl to the above water solution.

  7. HA(aq) + H2O(l) H3O1+(aq) + A1-(aq) 100% HA(aq) + OH1-(aq) H2O(l) + A1-(aq) Buffer Solutions • Add a small amount of base (-OH) to a buffer solution • Acid component of solution neutralizes the added base Weak acid Conjugate base (M+A-) Addition of OH1- to a buffer:

  8. HA(aq) + H2O(l) H3O1+(aq) + A1-(aq) 100% A1-(aq) + H3O1+(aq) H2O(l) + HA(aq) Buffer Solutions • Add a small amount of acid (H3O+) to a buffer solution • Base component of solution neutralizes the added acid Conjugate base (M+A-) Weak acid H3O+ added Addition of H3O1+ to a buffer: The H3O+ added changed A- to HA, but [HA] and [A-] are large compared to the [H3O+]added

  9. Buffer Solutions • The addition of –OH or H3O+ to a buffer solution will change the pH of the solution, but not as drastically as the addition of –OH or H3O+ to a non-buffered solution

  10. Example • pH of human blood (pH = 7.4) controlled by conjugated acid-base pairs (H2CO3/HCO3-). Write an equation for this buffer mixture then neutralization equation for the following effects • With addition of HCl • With addition of NaOH

  11. Example • 50.0 mL of 0.100 M HCl was added to a .100L buffer consisting of 0.025 moles of sodium acetate and 0.030 moles of acetic acid. What is the pH of the buffer before and after the addition of the acid? Ka of acetic acid is 1.7 x 10-5. Assume the volume is constant

  12. Example • Calculate the pH of 0.100L of a buffer solution that is 0.25M in HF and 0.50 M in NaF. • What is the change in pH on addition of 0.010 moles KOH • Calculate the pH after addition of 0.080 moles HBr • Assume the volume remains constant • Ka = 3.5 x 10-4

  13. Example • Calculate the pH of the buffer that results from mixing 60.0mL of 0.250M HCHO2 and 15.0 mL of 0.500M NaCHO2 Ka = 1.7 x 10-4 • Calculate the pH after addition of 10.0 mL of 0.150 MHBr. Assume volume is additive

  14. Buffer Capacity • A measure of amount of acid or base that the solution can absorb without a significant change in pH. • Depends on how many moles of weak acid and conjugated base are present. • For an equal volume of solution: the more concentrated the solution, the greater buffer capacity • For an equal concentration: the greater the volume, the greater the buffer capacity • The capacity of a buffer depends on the total (formal) concentration of the buffer species ( [acid] + [conjugate base] ), and for a given total concentration, buffer capacity will be greatest when [acid] = [conjugate base] that is, when pH = pKa for that buffer's acidic form.

  15. Example • The following pictures represent solutions that contain a weak acid HA and/ or its sodium salt NaA. (Na+ ions and solvent water molecules have been omitted for clarity • Which of the solutions are buffer solution? • Which solution has the greatest buffer capacity?

  16. Example • What is the maximum amount of acid that can be added to a buffer made by the mixing of 0.35 moles of sodium hydrogen carbonate with 0.50 moles of sodium carbonate? How much base can be added before the pH will begin to show a significant change?

  17. Acid(aq) + H2O(l) H3O1+(aq) + Base(aq) [Base] [Acid] [H3O1+][Base] [Base] [Acid] [Acid] 15.4 The Henderson-Hasselbalch Equation Weak acid Conjugate base [H3O1+] = Ka Ka = pH = pKa + log

  18. Examples • Calculate the pH of a buffer solution that is 0.50 M in benzoic acid (HC7H5O2) and 0.150 M in sodium benzoate (NaC7H5O2). Ka = 6.5 x 10-5

  19. Example • How would you prepare a NaHCO3-Na2CO3 buffer solution that has pH = 10.40 Ka2 = 4.7 x 10-11

  20. Example • You prepare a buffer solution of .323 M NH3 and (NH4)2SO4. What molarity of (NH4)2SO4 is necessary to have a pH of 8.6? (pKb NH3= 4.74)

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