1. pH Ka Kb Kw Ch. 14/15 A.P. Chemistry ACIDS AND BASES x = Titration
Covalent (acidic)
Ionic (basic)
Nature of Them
As Equilibrium Rxtns. pH
Salts
Oxides
neutral
acidic
Practical Definitions
Conceptual Definitions
Conjugate Pairs
Dissociation Constants
Defined: -log [H3O+]
Calculations
basic
ACIDS
Arrhenius Ka x Kb = Kw
Strong Species
Weak Species
Polyprotic Acids
Sour
Turns litmus red
Reacts with metals to produce H2
A: Liberates H+ ions in sol’n
B: Liberates OH- ions in sol’n
[H+]
[OH-]
[the autoionization of water]
Easiest… concentration on bottle dictates [H3O+]
Harder… Use Ka or Kb values
B-L
Tells about: Strength
A: proton donor
B: proton acceptor
Sol’ns which resist changes in pH with small additions of acids or bases
BASES
Volumetric analysis to determine unknown [ ]’s…
% Dissociation
Changes with dilution
Bitter
Turns litmus blue
slippery
Bond Strength and Electron Withdrawing Capacity explains this
Titration
SA/SB
Henderson-Hasselbalch Eq.
Calculations
How they work
Indicators… used to find end point of titration
pH curves
WA/SB
Buffers
Range pKa ± 1
WB/SA

2. Adding a __________________ salt of the conj. acid
pH =
Ka= Kw/Kb= 2.28x10-11
pOH = 1.00
pH= 13.00
Adding 2.0 mL of a 0.50 M strong base
CH3NH3NO3
To make a 0.10 M sol’n of ______
NaOH
50.0 mL of 2.00 M __________
CH3NH3+ + OH- ↔ CH3NH2 + H2O
Adding a Strong Base
Adding a __________________
salt of the conj. acid
[CH3NH3+] = 100. mmol / mL
= 1.00 M
99. mmol / 102 mL
= 0.97 M
[CH3NH2] = 100. mmol / mL
= 1.00 M
101 mmol / 102 mL
= M
pH = pKa + log [CH3NH2]/ [CH3NH3+]
pH = log (0.990/0.97)
p.H =
pH =
pH = pOH = 7.000
A Beaker of Water
Adding a ___________
Strong Acid
Titration
H2O + H2O ↔ H3O+ + OH-
CH3NH2 + H3O+ ↔ CH3NH3+ + H2O
pH = 5.355
Kw = 1.00 x 10-14
HCl
HCl
33.33 mL of 1.50 M __________
66.66 mL of 1.50 M __________
To make a buffer
Half-Neutralized! (midpoint)
Equivalence Point !!!!
50.0 mmol of H3O+ added
CH3NH3+ + H2O ↔ CH3NH2 + H3O+
pOH = 1.529
[CH3NH2] = 50.0 mmol / 83.3 mL
x x
pH =
= M
= [CH3NH3+] too!
Ka = x2
Adding a Weak Base
Kb= 4.38 x 10-4
pH =
100. mmol
CH3NH2
[OH-] =
116.7 mL
50.0 mL of 2.00 M _______
CH3NH2 + H2O ↔ CH3NH3+ + OH-
Kb = [OH-] [CH3NH3+] / [CH3NH2]
[H3O+] = 4.42 x 10-6
= x2 / x

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