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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!

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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)

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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:

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

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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!

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

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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

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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

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of H added Buffer region weak base

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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

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Titration of a weak base…

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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

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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

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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

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Titration of a weak base…

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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

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Titration of a weak base…

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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

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Titration of a weak base…

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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

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Titration of a weak base…

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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

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Titration of a weak base…

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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…

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Worksheet #8 practice… Finish the rest at home…

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