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Understanding Titrations and Buffers in Chemistry: Key Concepts and Practice Problems**

Join us for the Chemistry Department Seminar today at 4 PM, featuring valuable lectures leading up to Exam 2. We will delve into Chapter 18, focusing on polyprotic acid titrations with plenty of examples to clarify the concepts. Key topics include how to prepare buffers using weak acids and conjugate bases, the relationship between pH and pKa, and solving titration problems involving weak acids and bases. Ensure you're well-prepared for Exam 2, which covers Chapters 16 to 18. Don’t miss this essential learning opportunity!

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Understanding Titrations and Buffers in Chemistry: Key Concepts and Practice Problems**

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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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