Outline:Outline: 3/9/073/9/07 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!

Ways to prepare a buffer:

Weak acid and limited OH(or weak base & limited H3O)

Weak acid and conjugate base (or weak base & conjugate acid)

pH = pKa + log [conj. base]

[acid]

HA + OH A + H2O 1.0 0.5 0.0 0.0 (init) 0.5 0.0 0.5 0.5 (init)

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

Titration….

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

Conjugate

Base

Stong

Acid

CAPA-14 problem #1:

weak acid

of OH added

Buffer region

Equivalence point

Mid-point

At the midpoint of a titration...Exactly half of the weak acid is used up and turned into conjugate base

pH = pKa + log ([conj base]/[acid])

pH= 4.75 + log (0. 5/0. 5) or pH = 4.75 + 0.0 = pKa

1.0 M HA becomes

0.50 M HA and 0.50 M A

pH = pKa at midpoint!

weak acid

of OH added

Buffer region

Equivalence point

Mid-point

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 = pKa + log ([conj base]/[acid])

solve pH for a solution of A

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 = pKa + log ([conj base]/[acid])

Or if given grams, can calculate g/mol…

Like C

APA

#14&

15

of H added

Buffer region

weak base

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) = 1.8 105

0.0 mL added: Weak base problem:

NH3 + H2O NH4+ + OH

-xx x

pH =10.87

Titration of a weak base…

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) = 1.8 105

10.0 mL added: acid base reaction:

NH3 + H+ NH4+

0.030×0.0300.025×0.010 0.00090 mol 0.00025 mol

What is K for acid base rxns?LARGE

acid base reaction: NH3 + H+ NH4

+

0.00090 .000250.00065 mol 0.0 mol 0.00025 mol Just a buffer….NH3 and NH4

+

pOH = pKb + log (acid/base)

Volume = 10mL + 30 mL = 0.040 L

10mL:

Just a buffer….NH3 and NH4+

pOH = pKb + log (acid/base)

pH = 14 - 4.32 = 9.67

pOH = 4.74 + log (.00625/0.01625)

= 4.74 – 0.41 = 4.32

Titration of a weak base…

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) = 1.8 105

20.0 mL added:

pH = 9.16

Titration of a weak base…

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) = 1.8 105

35.0 mL added:

pH = 7.7

Titration of a weak base…

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) = 1.8 105

36.0 mL added:

Equivalence: pH = 5.56

Titration of a weak base…

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) = 1.8 105

37.0 mL added:

Strong Acid calc: pH = 3.4

Titration of a weak base…

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?

Exam 2 after 2 more days…

Worksheet #8 practice…

Finish the rest at home…