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

2. Ways to prepare a buffer: Weak acid and limited OH  (or weak base & limited H 3 O  ) n Weak acid and conjugate base (or weak base & conjugate acid) pH = pKa + log [conj. base] [acid] HA + OH   A  + H 2 O 1.0 0.5  0.0 0.0 (init) 0.5 0.0  0.5 0.5 (init)

3. n Which is a buffer? 0.10L of 0.25 M NaCH 3 CO 2 +0.05L of 0.25 M HCl or 0.10L of 0.25 M NaCH 3 CO 2 +0.15L of 0.25 M HCl Titration…. n Which is a buffer? 0.10L of 0.25 M NaCH 3 CO 2 +0.05L of 0.25 M HCl 0.10L of 0.25 M NaCH 3 CO 2 +0.15L of 0.25 M HCl Conjugate Base Stong Acid CAPA-14 problem #1:

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

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

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

7. 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! pH = pK a + log ([conj base]/[acid]) solve pH for a solution of A 

8. 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! pH = pK a + log ([conj base]/[acid]) Or if given grams, can calculate g/mol… Like CAPA #14& 15

9. of H  added Buffer region weak base

10. Try a titration problem: Titrate 30 mL of 0.030 M NH 3 with 0.025 M HCl. What is the pH after adding 0, 10, 20, 35, 36, 37 mL? K b (NH 3 ) = 1.8  10  5 0.0 mL added: Weak base problem: NH 3 + H 2 O  NH 4 + + OH   -x  x x pH =10.87

11. Titration of a weak base…

12. Try a titration problem: Titrate 30 mL of 0.030 M NH 3 with 0.025 M HCl. What is the pH after adding 0, 10, 20, 35, 36, 37 mL? K b (NH 3 ) = 1.8  10  5 10.0 mL added: acid base reaction: NH 3 + H +  NH 4 +  0.030×0.030  0.025×0.010 0.00090 mol 0.00025 mol

13. What is K for acid base rxns? LARGE acid base reaction: NH 3 + H+  NH 4 +  0.00090.00025 0.00065 mol 0.0 mol 0.00025 mol Just a buffer….NH 3 and NH 4 + pOH = pK b + log (acid/base) Volume = 10mL + 30 mL = 0.040 L

14. 10mL: Just a buffer….NH 3 and NH 4 + pOH = pK b + log (acid/base) pH = 14 - 4.32 = 9.67 pOH = 4.74 + log (.00625/0.01625) = 4.74 – 0.41 = 4.32

15. Titration of a weak base…

16. Try a titration problem: Titrate 30 mL of 0.030 M NH 3 with 0.025 M HCl. What is the pH after adding 0, 10, 20, 35, 36, 37 mL? K b (NH 3 ) = 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 NH 3 with 0.025 M HCl. What is the pH after adding 0, 10, 20, 35, 36, 37 mL? K b (NH 3 ) = 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 NH 3 with 0.025 M HCl. What is the pH after adding 0, 10, 20, 35, 36, 37 mL? K b (NH 3 ) = 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 NH 3 with 0.025 M HCl. What is the pH after adding 0, 10, 20, 35, 36, 37 mL? K b (NH 3 ) = 1.8  10  5 37.0 mL added: Strong Acid calc: pH = 3.4

23. Titration of a weak base…

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

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