I.
Base Theory
A.
Concepts
1) Arrhenius Concept: base produces OH- in water
2) Bronsted-Lowery Model: base is a H+ acceptor
3)
Strong Hydroxide Bases
a) Alkali Metal Hydroxides: NaOH, KOH, etc…
i. Completely Dissociated in Water
ii. NaOH
Na+ + OH- K = very large
b)
Alkaline Earth Hydroxides: Ca(OH)2, Mg(OH)2, etc…
i. Not very soluble in water
ii. What does dissolve is completely dissociated
iii. Ca(OH)2
Ca2+ + 2 OHK = very large
c)
Example: pH of 0.05 M NaOH
pOH  - log(0.05)  1.30  pH  14 - 1.30  12.70
4)
H
H N
Non-Hydroxide Bases (Weak Bases)
a) Any atom with a lone pair of electrons can accept a proton = base
H
b) Ammonia in water:
Kb = 1.8 x10-5
OH
H N H +
+ H O H
H
acid
base
c)
CH2CH3
CH3CH2
H
conjugate acid
conjugate base
Other amine molecules are also bases
N
N
CH2CH3
triethylamine
pyridine
[BH  ][OH  ]
Kb 
[B]
5)
The general base equation
B + H2O
BH+ + OH-
6)
Strong Base: equilibrium lies far to the right ([OH-] ≈ [B]0)
7)
Weak Base: equilibrium lies far to the left ([OH-] << [B]0)
a) Calculations are similar to weak acids
II.
b)
Example: pH of 15.0 M NH3
Kb = 1.8 x 10-5
i. We can find [H+] from KW = [H+][OH-] = 1 x 10-14
or pH + pOH = 14
ii. Percent dissociation still means the same thing
iii. The 5% rule for approximations: x/[B] x 100% < 5%
c)
Example: pH of 1.0 M methylamine Kb = 4.38 x 10-4
Polyprotic Acids
A.
Carbonic Acid is a diprotic acid
1) Polyprotic means there are more than one ionizable proton
2) Carbonic acid is the acid that helps maintain body pH H2CO3

H2CO3
H+ + HCO3-
[H  ][HCO 3 ]
K a1 
 4.3 x 10 -7
[H 2 CO3 ]
2
HCO33)
H+ + CO32- K a 2
[H  ][CO 3 ]
-11


5
.
6
x
10
[HCO 3 ]
Ka1 and Ka2 stand for the loss of the first and second protons, respectively
4)
B.
Usually Ka1 >> Ka2
a) As (-) charge builds up, it is harder to remove the next proton
b) H+ from the first ionization forces the second ionization to the left
c) We can usually ignore all but the first ionization in calculations
Phosphoric Acid is a triprotic acid
1) Ionizations
H3PO4
H+ + H2PO4- Ka = 7.5 x 10-3
H2PO4H+ + HPO42- Ka = 6.2 x 10-8
HPO42H+ + PO43Ka = 4.8 x 10-13
2)
Ka1 >> Ka2 >> Ka3
K a1 7.5 x 10-3
5


1
.
2
x
10
K a2 6.2 x 10-8
3)
K a2 6.2 x 10-8
5


1
.
3
x
10
K a3 4.8 x 10-13
Example: pH of 5.0 M H3PO4, and the concentrations of all of the phosphoric acid
derived species
a) Use Ka1 only to find [H+] and the pH
b) Then, [H+] = [H2PO4-]
c) [H3PO4] = [H3PO4]0 - [H+]
d) Find [HPO42-] and [PO43-] from what is already calculated and Ka2, Ka3
C.
Sulfuric Acid is a unique diprotic acid
1) The first ionization of sulfuric acid is a strong acid (Ka1 = large)
H2SO4
H+ + HSO42)
The second ionization of sulfuric acid is a weak acid
HSO4H+ + SO42Ka2 = 1.2 x 10-2
3)
Example: pH of 1.0 M H2SO4
a) Assume complete dissociation of the first proton
b) Use the equilibrium calculation on Ka2
i. [H+]0 does not = 0 because of Ka1
ii. [H+]0 = 1.0 M
c) Approximate 1 + x ≈ 1 in this case (5% rule says this is ok)
d) When [H2SO4] > 1.0 M, you can ignore Ka2
4)
Example: pH of 0.01 M H2SO4
a) The 5% rule tells us we can’t ignore Ka2 in this case
b) We must use the quadratic equation to solve for x
c) When H2SO4 < 1.0 M, we can’t ignore Ka2
III. Acid-Base Properties of Salts
A)
1)
2)
a)
b)
c)
1)
2)
3)
Simple Salts
Salt = ionic compound = one that completely ionizes in water
Some salts have no effect on pH
Cations of strong bases have no effect on pH
i. Na+, K+, etc…
ii. These cations have no affinity for OH- in water
Anions of strong acids have no effect on on pH
i. Cl-, NO3-, etc…
ii. These anions have no affinity for H+ in water
Solutions of these combined ion salts have pH = 7.00
B) Basic Salts
Salts containing the conjugate base of a weak acid produce basic solutions
The conjugate base must be strong if the acid is weak, so it must have a
strong affinity for H+, which will affect the pH of a solution
[HC 2 H 3O 2 ][OH  ]
Sodium Acetate Example
Kb 
 ???
[C 2 H 3O 2 ]
NaC2H3O2
Na+ + C2H3O2C2H3O2- + H2O
HC2H3O2 + OH-
4)
How do we find Kb for the conjugate base of a weak acid?
[H  ][C 2 H 3O 2 - ]  [HC 2 H 3O 2 ][OH  ] 


Ka x Kb  

[H
][OH
]


[C 2 H 3O 2 ]
 [HC 2 H 3O 2 ]  

[H  ][OH  ]  K W  1 x 10-14
C.
5)
For any weak acid and its conjugate base, Ka x Kb = KW
Kb = KW / Ka = 1 x 10-14 / 1.8 x 10-5 = 5.6 x 10-10 for acetate
6)
Example: pH of 0.30 M NaF Ka for HF = 7.2 x 10-4
Base Strength in Water
1) Any base in water must compete with the hydroxide anion for protons
2) Hydrocyanic acid example:
HCN + H2O
H3O+ + CN- Ka = 6.2 x 10-10
weak acid
strong base (compared to H2O)
CN- + H2O
(compared to OH-)weak base
HCN + OHstrong acid
Kb = KW / Ka = 1.6 x 10-5
OH- > CN- > H2O
D.
Acidic Salts
1) Salts having the conjugate acid of a weak base produce acidic solutions
2) Ammonium chloride = NH4Cl
NH4+ + ClNH4+
NH3 + H+
3)
Since ammonia is a weak base, ammonium is “strong” acid and will effect pH
4)
Example: pH of 0.1 M NH4Cl Kb = 1.8 x 10-5
a) Find Ka from KW
b) Make sure the conjugate acid is stronger than water or it will have no effect
5)
Highly charged metal ions can also be acidic
a) AlCl3 + 6 H2O
Al(H2O)63+ + 3 Clb) Al(H2O)63+ + H2O
Al(H2O)5(OH)2+ + H3O+
c) The higher the metal’s charge the more acidic
Example: pH of 0.01 M AlCl3 Ka = 1.4 x 10-5
E.
Salts containing both acidic and basic components
1) NH4C2H3O2
NH4+ + C2H3O22) The calculations for these compounds are complex
3) We can, however, at least decide if the solution is acidic or basic
a) If Ka > Kb, the solution will be acidic
b) If Kb > Ka, the solution will be basic
c) If Ka = Kb, the solution will be neutral
4) Example: Will the following solutions be acidic or basic?
NH4C2H3O2
NH4CN
Al2(SO4)3
IV. Acid-Base Properties and Molecular Structure
A.
Polarity
1) Not all H containing molecules are acidic
a) CHCl3
H+ + CCl3b) Strong, nonpolar bonds don’t dissociate easily
2)
Polar X—H bonds are easily dissociated (acidic)
a) H—Cl is a polar bond
b) H—OH is a polar bond
3)
Bond strength also plays a part in acidity
a) Hydrohalide Polarity: H—F > H—Cl > H—Br > H—I
b) Bond Strength (kJ.mol) 565
427
363
295
c) Acidity
weak
strong strong strong
4)
Oxyacids: the more O on the central atom, the stronger the acid in a series
a) Electronegative Oxygens remove electrons from the center atom
b) This polarizes and weakens the O—H bond even more
c) HClO4 > HClO3 > HClO2 > HClO
d) H2SO4 > H2SO3
5)
B.
H—O—X Molecules
a) The more electronegative X is, the more acidic the molecule is
b) Electronegative X removes electrons from the H—O bond
c) Acidity:
H—O—Cl > H—O—Br > H—O—I > H—O—CH3
d) Electronegativity: 3.0
2.8
2.5
2.3
Acid-Base Properties of Oxides
1) A generic oxide can be represented as X—O
2) If X—O is a strong and covalent bond, the oxide will be acidic in water
a) H—O—X
H+ + -O—X
b) If X is electronegative, like O, it should form a strong covalent O—X bond
3)
If X—O is weak and ionic, the oxide will be basic in water
a) H—O—X
X+ + OHb) NaOH
Na+ + OHc) The X atom in these oxides is usually not electronegative (Na+)
4)
Examples of Acidic Oxides
a) SO2 + H2O
H2SO3
b) CO2 + H2O
H2CO3
H+ + HSO3H+ + HCO3-
5)
V.
Examples of Basic Oxides
a) CaO + H2O
b) K2O + H2O
Ca(OH)2
KOH
Ca2+ + 2 OHK+ + OH-
Lewis Acid-Base Definition
A.
Definitions
1) Lewis Acid = an electron pair acceptor
2) Lewis Base = an electron pair donor
3) Example: H+ + :NH3
NH4+
Lewis acid Lewis Base
Lewis Acid-Base complex
4)
This model includes the other acid-base concepts
5)
This model accounts for many other chemical reactions that the others don’t
a) BF3
+
:NH3
H3N:BF3
Lewis acid
Lewis Base
Lewis Acid-Base complex
Al(H2O)63+ + 3 Cl-
b)
AlCl3 + 6 H2O
c)
Example: Identify Lewis Acid and Lewis Base
i. Ni2+ + 6 NH3
Ni(NH3)62+
ii. H+ + H2O
H3O+
Summary of Acid-Base Problem Solving
1. List major species in solution
2. Look for reactions that go to completion
a. concentration of product
b. major species left
3. Identify acids and bases
4. Solve the equilibrium problem, check the approximation, find pH