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Outline:3/9/07 è Chem. Dept. Seminar 4pm è 3 more lectures until Exam 2… è Chemistry Advising – 4pm Today: è More Chapter 18 Titrations.

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Presentation on theme: "Outline:3/9/07 è Chem. Dept. Seminar 4pm è 3 more lectures until Exam 2… è Chemistry Advising – 4pm Today: è More Chapter 18 Titrations."— Presentation transcript:

1 Outline:3/9/07 è Chem. Dept. Seminar 4pm è 3 more lectures until Exam 2… è Chemistry Advising – 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  (init)  (init)

3 n Which is a buffer? 0.10L of 0.25 M NaCH 3 CO L of 0.25 M HCl or 0.10L of 0.25 M NaCH 3 CO L of 0.25 M HCl Titration…. n Which is a buffer? 0.10L of 0.25 M NaCH 3 CO L of 0.25 M HCl 0.10L of 0.25 M NaCH 3 CO L 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= log (0. 5/0. 5) or pH = = 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 M NH 3 with M HCl. What is the pH after adding 0, 10, 20, 35, 36, 37 mL? K b (NH 3 ) = 1.8  10  mL added: Weak base problem: NH 3 + H 2 O  NH OH   -x  x x pH =10.87

11 Titration of a weak base…

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

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

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

15 Titration of a weak base…

16 Try a titration problem: Titrate 30 mL of M NH 3 with M HCl. What is the pH after adding 0, 10, 20, 35, 36, 37 mL? K b (NH 3 ) = 1.8  10  mL added: pH = 9.16

17 Titration of a weak base…

18 Try a titration problem: Titrate 30 mL of M NH 3 with M HCl. What is the pH after adding 0, 10, 20, 35, 36, 37 mL? K b (NH 3 ) = 1.8  10  mL added: pH = 7.7

19 Titration of a weak base…

20 Try a titration problem: Titrate 30 mL of M NH 3 with M HCl. What is the pH after adding 0, 10, 20, 35, 36, 37 mL? K b (NH 3 ) = 1.8  10  mL added: Equivalence: pH = 5.56

21 Titration of a weak base…

22 Try a titration problem: Titrate 30 mL of M NH 3 with M HCl. What is the pH after adding 0, 10, 20, 35, 36, 37 mL? K b (NH 3 ) = 1.8  10  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…


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