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Outline: 3/9/07

Outline: 3/9/07. Chem. Dept. Seminar today @ 4pm 3 more lectures until Exam 2 … Chemistry Advising – Monday @ 4pm. Today: More Chapter 18. Titrations Polyprotic acid titrations Lots of examples!. [conj. base]. [acid]. Ways to prepare a buffer:.

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Outline: 3/9/07

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  1. Outline: 3/9/07 • Chem. Dept. Seminar today @ 4pm • 3 more lectures until Exam 2… • Chemistry Advising – Monday @ 4pm Today: • More Chapter 18 Titrations Polyprotic acid titrations Lots of examples!

  2. [conj. base] [acid] Ways to prepare a buffer: • Weak acid and conjugate base (or weak base & conjugate acid) pH = pKa + log • Weak acid and limited OH- (or weak base & limited H3O+) HA + OH- A- + H2O 1.0 0.5  0.0 0.0 (init) 0.5 0.0  0.5 0.5 (init)

  3. Conjugate Base Stong Acid CAPA-14 problem #1: • Which is a buffer? • 0.10L of 0.25 M NaCH3CO2 • +0.05L of 0.25 M HCl or • 0.10L of 0.25 M NaCH3CO2 • +0.15L of 0.25 M HCl • Which is a buffer? • 0.10L of 0.25 M NaCH3CO2 • +0.05L of 0.25 M HCl • 0.10L of 0.25 M NaCH3CO2 • +0.15L of 0.25 M HCl Titration….

  4. weak acid Buffer region Equivalence point Mid-point of OH- added

  5. pH = pKa + log ([conj base]/[acid]) pH= 4.75 + log (0. 5/0. 5) or pH = 4.75 + 0.0 = pKa At the midpoint of a titration... Exactly half of the weak acid is used up and turned into conjugate base 1.0 M HA becomes 0.50 M HA and 0.50 M A- pH = pKa at midpoint!

  6. weak acid Buffer region Equivalence point Mid-point of OH- added

  7. pH = pKa + log ([conj base]/[acid]) At the equivalence point... All of the weak acid is used up and turned into conjugate base 1.0 M HA becomes 1.0 M A- mols of acid = mols of base at equivalence point! solve pH for a solution of A-

  8. pH = pKa + log ([conj base]/[acid]) At the equivalence point... All of the weak acid is used up and turned into conjugate base 1.0 M HA becomes 1.0 M A- Like CAPA #14& 15 mols of acid = mols of base at equivalence point! Or if given grams, can calculate g/mol…

  9. weak base Buffer region of H+ added

  10. Try a titration problem: Titrate 30 mL of 0.030 M NH3 with 0.025 M HCl. What is the pH after adding 0, 10, 20, 35, 36, 37 mL? Kb (NH3) = 1.8  10-5 0.0 mL added: Weak base problem: NH3 + H2O  NH4+ + OH- 0.030-xx x pH =10.87

  11. Titration of a weak base…

  12. acid base reaction: NH3 + H+ NH4+ 0.030×0.0300.025×0.010 Try a titration problem: Titrate 30 mL of 0.030 M NH3 with 0.025 M HCl. What is the pH after adding 0, 10, 20, 35, 36, 37 mL? Kb (NH3) = 1.8  10-5 10.0 mL added: 0.00090 mol 0.00025 mol

  13. 0.00065 mol 0.0 mol 0.00025 mol Just a buffer….NH3 and NH4+ pOH = pKb + log (acid/base) What is K for acid base rxns? LARGE acid base reaction: NH3 + H+ NH4+ 0.00090 .00025 Volume = 10mL + 30 mL = 0.040 L

  14. Just a buffer….NH3 and NH4+ pOH = pKb + log (acid/base) 10mL: pOH = 4.74 + log (.00625/0.01625) = 4.74 – 0.41 = 4.32 pH = 14 - 4.32 = 9.67

  15. Titration of a weak base…

  16. Try a titration problem: Titrate 30 mL of 0.030 M NH3 with 0.025 M HCl. What is the pH after adding 0, 10, 20, 35, 36, 37 mL? Kb (NH3) = 1.8  10-5 20.0 mL added: pH = 9.16

  17. Titration of a weak base…

  18. Try a titration problem: Titrate 30 mL of 0.030 M NH3 with 0.025 M HCl. What is the pH after adding 0, 10, 20, 35, 36, 37 mL? Kb (NH3) = 1.8  10-5 35.0 mL added: pH = 7.7

  19. Titration of a weak base…

  20. Try a titration problem: Titrate 30 mL of 0.030 M NH3 with 0.025 M HCl. What is the pH after adding 0, 10, 20, 35, 36, 37 mL? Kb (NH3) = 1.8  10-5 36.0 mL added: Equivalence: pH = 5.56

  21. Titration of a weak base…

  22. Try a titration problem: Titrate 30 mL of 0.030 M NH3 with 0.025 M HCl. What is the pH after adding 0, 10, 20, 35, 36, 37 mL? Kb (NH3) = 1.8  10-5 37.0 mL added: Strong Acid calc: pH = 3.4

  23. Titration of a weak base…

  24. Exam 2 after 2 more days… Exam 2 in 21 days… • Covers chapter 16 (Equilibrium) • Covers chapter 17 (Acids & Bases) • Covers chapter 18 (Buffers, Ksp) Have you studied for it already?

  25. Worksheet #8 practice… Finish the rest at home…

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