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Chapter 15 Applications of Aqueous Equilibria

Chapter 15 Applications of Aqueous Equilibria. Neutralization Reaction. General Formula Acid + Base  Water + Salt. H 3 O + ( aq ) + OH – ( aq ). 2H 2 O( l ). Neutralization Reactions. Strong Acid- Strong Base. HCl( aq ). +. NaOH( aq ). H 2 O( l ) +. NaCl( aq ).

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Chapter 15 Applications of Aqueous Equilibria

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  1. Chapter 15Applications of Aqueous Equilibria

  2. Neutralization Reaction • General Formula Acid + Base  Water + Salt

  3. H3O+(aq) + OH–(aq) 2H2O(l) Neutralization Reactions Strong Acid-Strong Base HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq) Assume complete dissociation: (net ionic equation) After neutralization: pH = 7

  4. Strong acid-Strong base neutralization • When the number moles of acid and base are mixed together [H3O+] = [-OH] = 1.0 x 10-7M • Reaction proceeds far to the right

  5. CH3CO2H(aq) + OH–(aq) H2O(l) + CH3CO2–(aq) Neutralization Reactions Weak Acid - Strong Base CH3CO2H(aq) + NaCH3CO2(aq) NaOH(aq) H2O(l) + Assume complete dissociation: (net ionic equation) After neutralization: pH > 7

  6. Weak acid-strong base neutralization • Neutralization of any weak acid by a strong base goes 100% to completion • -OH has a great infinity for protons

  7. Neutralization Reactions Weak Acid - Strong Base CH3CO2H(aq) + NaOH(aq) H2O(l) + NaCH3CO2(aq)

  8. Neutralization Reactions Strong Acid - Weak Base HCl(aq) + NH3(aq) NH4Cl(aq) Assume complete dissociation: H3O+(aq) + NH3(aq) H2O(l) + NH4+(aq) (net ionic equation) After neutralization: pH < 7

  9. Strong acid-weak base neutralization • Neutralization of any weak base by a strong acid goes 100% to completion • H3O+ has a great infinity for protons

  10. Neutralization Reactions Strong Acid - Weak Base HCl(aq) + NH3(aq) NH4Cl(aq)

  11. Neutralization Reactions Weak Acid-Weak Base CH3CO2H(aq) + NH3(aq) NH4CH3CO2(aq) Assume complete dissociation: CH3CO2H(aq) + NH3(aq) NH4+(aq) + CH3CO2–(aq) (net ionic equation) After neutralization: pH = ?

  12. Weak acid-weak base neutralization • Less tendency to proceed to completion than neutralization involving strong acids and strong bases

  13. CH3CO2H(aq) + H2O(l) H3O1+(aq) + CH3CO21-(aq) 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. • Example of Le Chatelier’s principle • The addition of acetate ion ,CH3CO21-, to a solution of acetic acid suppresses the dissociation of the acid. The equilibrium shifts to the left.

  14. The Common-Ion Effect

  15. CH3CO2H(aq) + H2O(l) H3O1+(aq) + CH3CO21-(aq) The Common-Ion Effect Le Châtelier’s Principle E.g Adding HCl and NaOH to a solution of acetic acid would shift the equilibrium to which direction?

  16. Example • The pH of a 0.10M of acetic is 2.87 at 25oC. Determine the pH at the same temperature of a solution by adding 0.050 mole of sodium acetate to 1.0 L of 0.10 M acetic acid. (Assume that the addition of sodium acetate does not change the volume of the solution) Ka = 1.8 x 10-5

  17. Example • Dtermine the pH at 25oC of a solution prepared by dissolving 0.35 mol of ammonium chloride in 1.0 L of 0.25 M aqueous ammonia solution. No volume changed. Kb = 1.8 x 10-5

  18. 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 CH3CO2H + CH3CO2– HF + F– NH4+ + NH3 H2PO4– + HPO42– For example: © 2012 Pearson Education, Inc.

  19. CH3CO2H(aq) + H2O(l) H3O+(aq) + CH3CO2–(aq) Buffer Solutions Weak acid Conjugate base (NaCH3CO2) Addition of OH1– to a buffer: 100% CH3CO2H(aq) + OH–(aq) H2O(l) + CH3CO2–(aq) Addition of H3O1+ to a buffer: 100% CH3CO2 –(aq) + H3O+(aq) H2O(l) + CH3CO2H(aq)

  20. Buffer Solutions • Add a small amount of base (-OH) to a buffer solution • Acid component of solution neutralizes the added base • Add a small amount of acid (H3O+) to a buffer solution • Base component of solution neutralizes the added acid • 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

  21. Buffer Solutions

  22. Buffer capacity • A measure of the amount of acid or base that a buffer solution can absorb without a significant change in pH • Depends on how much weak acid and conjugated base is present • For equal volume of solution, the more concentration the solution, the greater the buffer capacity • For solution with the same concentration, increasing the volume increases the buffer capacity

  23. Example • pH of human blood (pH = 7.4) controlled by conjugated acid-base pairs (H2CO3/HCO3-). Write a neutralization for each condition • With addition of H3O+ • With addition of -OH

  24. 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 = 1.8 x 10-4

  25. Example • Calculate the pH of 0.100L of a buffer solution that is 0.25M in HF and 0.50 M in NaF, Ka = 6.3 x 10-4 • What is the change in pH on addition of 0.002 mol HCl • What is the change in pH on addition of 0.010 moles KOH • Calculate the pH after addition of 0.080 moles HBr

  26. Example • calculate the pH of a 50.0 ml buffer solution that is 0.50 M in NH3 and 0.20 M NH4Cl. For ammonia, pKb = 4.75 • Calculate the pH after addition of 150.0 mg HCl

  27. CH3CO2H(aq) + H2O(l) Acid(aq) + H2O(l) H3O1+(aq) + Base(aq) H3O1+(aq) + CH3CO21-(aq) [H3O1+][Base] [Acid] [Acid] [Base] The Henderson-Hasselbalch Equation Weak acid Conjugate base Ka = [H3O1+] = Ka

  28. [Base] [Acid] [Acid] [Base] [Base] [Acid] The Henderson-Hasselbalch Equation [H3O1+] = Ka -log([H3O1+]) = -log(Ka) - log pH = pKa + log

  29. Example • 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 • How would you prepare a NaHCO3-Na2CO3 buffer solution that has pH = 10.40 Ka2 = 5.6 x 10-11

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