Chapter 14/15 Structured Overview• Chapter 14/15 Structured Overview follows on the next slide. You will find that the Chapter 14 material is already filled in. The blanks that will be filled in through the animation are from the content from chapter 15 because I wanted you to focus on that. All content is fair game for the test however.

• As you go through the presentation fill in the blanks on your own structured overview.

• After filling in the structured overview work through the problem on the slide that follows it (the back of your hard copy) by clicking through the presentation.

• After working that problem, info about the on-line test is given on the last slide.

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Ch. 14/15Ch. 14/15

A.P. ChemistryA.P. ChemistryACIDS AND BASESACIDS AND BASES

Nature of ThemNature of Them As Equilibrium Rxtns.As Equilibrium Rxtns.

Practical Practical DefinitionsDefinitions

Conceptual Conceptual DefinitionsDefinitions

Conjugate Conjugate PairsPairs

Dissociation Dissociation ConstantsConstants

ACIDSACIDS

BASESBASES

•SourSour

•Turns litmus redTurns litmus red

•Reacts with Reacts with metals to metals to produce Hproduce H22

•BitterBitter

•Turns litmus blueTurns litmus blue

•slipperyslippery

ArrheniusArrhenius

B-LB-L

LewisLewis

A: Liberates HA: Liberates H++ ions in sol’nions in sol’n

B: Liberates OHB: Liberates OH-- ions in sol’nions in sol’n

A: proton donorA: proton donor

B: proton B: proton acceptoracceptor

A: electron pair A: electron pair acceptoracceptor

B: electron pair B: electron pair donordonor

[H[H++]] [OH[OH--]] [the autoionization [the autoionization of water]of water]

KKaa KKbb KKww

Tells about:Tells about: Strength Strength

pHpH

Defined: Defined: -log [H-log [H33OO++]]

CalculationsCalculations

Strong Strong SpeciesSpecies

Weak Weak SpeciesSpecies

Polyprotic Polyprotic AcidsAcids

% Dissociation% Dissociation Changes Changes with dilutionwith dilution

Easiest… Easiest… concentration concentration on bottle on bottle dictates [Hdictates [H33OO++]]

Harder… Harder… Use KUse Kaa or K or Kbb

valuesvalues

Volumetric Volumetric analysis to analysis to determine determine unknown [ ]’s…unknown [ ]’s…

Indicators… Indicators… used to used to

find end point of titrationfind end point of titration

Range Range pKa ± 1 pKa ± 1

pH curvespH curves

SA/SBSA/SB

WA/SBWA/SB

WB/SAWB/SA

BuffersBuffers

Sol’ns w

hich resist changes in pH

Sol’ns w

hich resist changes in pH

with

small additions of a

cids or bases

with

small additions of a

cids or bases

How they work

How they work

Calculations

CalculationsHenderson-Hasselbalch Eq.Henderson-Hasselbalch Eq.

SaltsSalts OxidesOxides

neutralneutralacidicacidic

basicbasic

Covalent (acidic)Covalent (acidic)

Ionic (basic)Ionic (basic)

Electron Withdrawing Capacity Electron Withdrawing Capacity explains thisexplains this

TitrationTitration

A Beaker of WaterA Beaker of Water

Adding a Strong BaseAdding a Strong Base

Adding a Weak BaseAdding a Weak Base

To make a 0.10 M sol’n of ______

50.0 mL of 2.00 M _______

Adding a ___________Adding a ___________

Adding a __________________Adding a __________________

50.0 mL of 2.0 M __________

33.33 mL of 1.5M __________ 66.66 mL of 1.5M __________

Adding Adding 2.0 mL of 2.0 mL of a 0.10 M a 0.10 M strong strong basebase

To m

ake

a bu

ffer

H2O + H2O ↔ H3O+ + OH-

Kw = 1.0 x 10-14

pH = 7 pOH = 7

NaOH

pOH = 1 pH= 13

CH3NH2

CH3NH2 + H2O ↔ CH3NH3+ + OH-

Kb= 4.38 x 10-4

[OH-] = 0.0296Kb = [OH-] [CH3NH3

+] / [CH3NH2]

= x2 / 2.00 - x

pOH = 1.53

pH = 12.47

salt of the conj. acid

CH3NH3NO3

[CH3NH3+] = 100. mmol / 100.0 mL

[CH3NH2] = 100. mmol / 100.0 mL

= 1.00 M

= 1.00 MpH = pKa + log [CH3NH2]/ [CH3NH3

+]

pH = 10.6 + 0

pH = 10.6 Ka= Kw/Kb= 2.28x10-11

CH3NH3+ + OH- ↔ CH3NH2 + H2O

99.8 mmol / 102.0 mL = 0.978 M

100.2 mmol / 102.0 mL= 0.982 M

pH = 10.6 + log (0.982/0.978) pH = 10.6

Titration

Strong Acid

CH3NH2 + H3O+ ↔ CH3NH3+ + H2O

HCl

50.0 mmol of H3O+ added[CH3NH2] = 50.0 mmol / 83.33 mL

= 0.600 M = [CH3NH3+] too!

pH = 10.6

CH3NH3+ + H2O ↔ CH3NH2 + H3O+

HCl

x x

Ka = x2 100.0 mmol 116.7 mL

[H3O+] = 4.42 x 10-6

Half-Neutralized !!!! Equivalence Point !!!!

pH = 5.35

Chapter 14/15 On-Line Test

• Click on the “On-Line Tests” link on the class web page.

• Read the instructions and click on the Login button.

• User Name: last name (lower case) Password: student ID #

• Read directions and find the link for the answer sheet.

• Solve the problems and put the answers on the answer sheet.

• Email answer sheet to me as instructed. (no attachments)

• Available beginning 3:30 PM on 4-1-11

• Email prior to 8:00 am on 4/5/11

GOOD LUCK !!!!!!