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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 1+ ( aq ) + OH 1- ( aq ). H 1+ ( aq ) + OH 1- ( aq ). H 2 O( l ). 2H 2 O( l ). Neutralization Reactions. Strong Acid-Strong Base. HCl( 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. H3O1+(aq) + OH1-(aq) H1+(aq) + OH1-(aq) H2O(l) 2H2O(l) Neutralization Reactions Strong Acid-Strong Base HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq) Assuming complete dissociation: or (net ionic equation) After neutralization: pH = 7

  4. H3O1+(aq) + OH1-(aq) 2H2O(l) 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) + OH1-(aq) H2O(l) + CH3CO21-(aq) Neutralization Reactions Weak Acid-Strong Base CH3CO2H(aq) + NaOH(aq) H2O(l) + NaCH3CO2(aq) Assuming complete dissociation: (net ionic equation) After neutralization: pH > 7

  6. CH3CO2H(aq) + OH1-(aq) H2O(l) + CH3CO21-(aq) 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. H1+(aq) + NH3(aq) NH41+(aq) Neutralization Reactions Strong Acid-Weak Base HCl(aq) + NH3(aq) NH4Cl(aq) Assuming complete dissociation: or (net ionic equation) H3O1+(aq) + NH3 (aq) H2O(l) + NH41+(aq) After neutralization: pH < 7

  8. 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 H3O1+(aq) + NH3 (aq) H2O(l) + NH41+(aq)

  9. CH3CO2H(aq) + NH3(aq) NH41+(aq) + CH3CO21-(aq) Neutralization Reactions Weak Acid-Weak Base CH3CO2H(aq) + NH3(aq) NH4CH3CO2(aq) (net ionic equation) After neutralization: pH = ?

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

  11. Examples • Write balanced net ionic equations for the neutralization of equal molar amounts of the following acids and bases. Indicate whether the pH after the neutralization is greater than, equal to or less than 7 • HNO2 and KOH Ka HNO2 = 4.5 x 10-4

  12. 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 CH3CO2H(aq) + H2O(l) H3O+(aq) + CH3CO2–(aq)

  13. The Common-Ion Effect The pH of 0.10 M acetic acid is 2.89. 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.

  14. The Common-Ion Effect

  15. The Common-Ion Effect Le Châtelier’s Principle CH3CO2H(aq) + H2O(l) H3O+(aq) + CH3CO2–(aq) The addition of acetate ion to a solution of acetic acid suppresses the dissociation of the acid. The equilibrium shifts to the left.

  16. The Common-Ion Effect

  17. Example • In 0.15 M NH3, the pH is 11.21 and the percent dissociation is 1.1%. Calculate the concentrations of NH3, pH and percent dissociation of ammonia in a solution that is 0.15M and 0.45 MNH4Cl

  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 + CH3CO21- HF + F1- NH41+ + NH3 H2PO42- + HPO42- For Example:

  19. 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

  20. HA(aq) + H2O(l) H3O1+(aq) + A1-(aq) 100% 100% A1-(aq) + H3O1+(aq) HA(aq) + OH1-(aq) H2O(l) + A1-(aq) H2O(l) + HA(aq) Buffer Solutions Weak acid Conjugate base (M+A-) Addition of OH1- to a buffer: Addition of H3O1+ to a buffer:

  21. Buffer Solutions

  22. Example • pH of human blood (pH = 7.4) controlled by conjugated acid-base pairs (H2CO3/HCO3-). Write the neutralization equation fro the following effects • With addition of HCl • With addition of NaOH

  23. 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 0.002 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 * Assume the volume remains constant

  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 • Calculate the pH after addition of 10.0 mL of 0.150 MHBr. Assume volume is additive

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

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

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

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