ACIDS and BASES:
Arrhenius Theory: In 1884, Svante Arrhenius defined acids and bases as follows...
 An Arrhenius acid contains H and produces H+ (or H3O+) in water (e.g., HCl).
 An Arrhenius base contains the OH group and produces OH- in water (e.g., NaOH).
 Neutralization is the combination of H+ (or H3O+) and OH- ions to form HOH.
The Arrhenius theory explains reactions of protonic acids with metal hydroxides (hydroxy
bases), e.g., HCl (aq) + NaOH  HOH + NaCl
Bronsted-Lowry Theory: In 1923, J. N. Bronsted and T. M. Lowry defined acids and bases as
 A Bronsted acid is a proton (H+) donor
 A Bronsted base is a proton acceptor.
 Neutralization was defined as the transfer of a proton from a proton donor (acid) to a proton
acceptor (base).
The Bronsted Theory is broader than the Arrhenius and includes non aqueous reactions,
e.g., HCl (g) + :NH3 (g)  NH4Cl (s)
e.g., H-CC-H + Na+NH2-  H-CC: - Na+ + :NH3
Conjugate acid / base pairs are defined as species that differ by a proton. Consider ...
Na+ OH-
+
CH3COOH
base

acid
HOH
conj. acid
+
Na+ CH3COOconj. base
Strength of Acids and Bases:
 A strong acid (HCl) has a weak conjugate base (Cl-), i.e., if the acid releases a proton readily, its
conjugate base will not attract a proton strongly.
 A strong base (NaOH) has a weak conjugate acid (H2O), i.e., if the base accepts a proton
readily, its conjugate base will not readily give up the proton.
 Acid strength is quantified by Ka and pKa ....
HA + HOH  H3O+ + A-
and the degree of dissociation is given by...
[H 3 O + ] [A - ]
but in dilute soln., [HOH] is a constant (55.5 M) and this is combined
K eq =
[ HA] [HOH]
with the equilibrium constant (Keq) giving another constant, Ka, which is a measure of the
[ H 3O  ] [A - ]
strength of an acid... Ka = Keq  [HOH] = Keq  55.5 =
[ HA ]
and note that:
pKa = -log10 Ka
1
 Similarly for bases: B + HOH  BH+ + OH- and the degree of dissociation is ...
[BH + ] [OH - ]
but in dilute soln., [HOH] is a constant (55.5 M) and this is combined
Keq =
[ B] [HOH]
with the equilibrium constant (Keq) giving another constant, Kb, which is a measure of the
[ BH  ] [OH - ]
strength of a base... Kb = Keq  [HOH] = Keq  55.5 =
[ B]
and
pKb = -log10 Kb
 The relationships that apply to conjugate acid base pairs are ...
Ka  Kb = 10-14
or
pKa + pKb = 14
For example, CH3COOH has pKa = 4.7. Its conjugate base, CH3COO- has pKb = (14 – 4.7) = 9.3
 It is important that students of organic chemistry be familiar with pK values of acids and bases.
Not only do they provide a quantitative measure of the strength of an acid or base but, as we
will soon see, these values can also combined to predict the extent of reaction of acid/base
reactions
 In the following tables, the approximate strengths versus pK value is given.
pKa
<1
1-5
5 - 15
> 15
acid strength
strong
moderate
weak
very weak
example
H2SO4
H3PO4
H2CO3
HOH
pKb
<1
1-5
5 - 15
> 15
base strength
strong
moderate
weak
very weak
example
NaOH
Na3PO4
Na2CO3
HOH
Consider Acids Stronger Than H3O+: H3O+ has a pKa = -1.74. Stronger acids such as HCl,
with pKa's lower than -1.74, protonate HOH completely producing H3O+.
HCl
+
HOH

H3O+
Cl-
+
pka = -7
pkb = 15.74
pKa = -1.74
pKb = 21
Thus H3O+ is the strongest acid which can exist in water. Any stronger acid will be 'levelled'
(reduced) in strength to pKa = -1.74 by the solvent water. Water is said to be a levelling solvent.
Consider Bases Stronger Than OH-: OH- has a pKb = -1.74. Stronger bases such as NaNH2,
with pKb's lower than -1.74, are protonated completely by HOH producing OH-.
NaNH2
+
HOH

NH3
+
NaOH
pKb = -21
pka = 15.74
pKa = 35
pKb = -1.74
Thus OH- is the strongest base which can exist in water. Any stronger base will be 'levelled'
(reduced) in strength to pKb = -1.74 by the solvent water.
2
ACIDITY & BASICITY OF WATER
Water is both weakly acidic and weakly basic, i.e., ‘amphoteric’ or ‘amphiprotic’.
 Water acts as a base by accepting protons from acids...
H-O-H
+
H-Cl
 H3O+
+
Cl-
 Water acts as an acid by donating protons to bases ...
H-O-H + :NH3  NH4+ + OH Water undergoes self-ionization, or ‘autopyrolysis’ to form equal concentrations of H3O+ and
OH- ions ...
H-O-H + H-O-H  H3O+ + OHThe extent of self-ionization of water is very small. At room temperature (25C) the
concentrations of H3O+ and OH_ are 1.0  10-7 moles/L each, i.e., [H3O+]= [OH_] = 10-7 M
 The equilibrium constant, Keq, for this reaction is given by ...
Keq =
[H 3O+ ] [OH - ] [107 ] [10-7 ]
1014


[HOH] [HOH] [55.5] [55.5] [101.74 ] [101.74 ]
where 10-14 is ‘KW’, the ion product
constant for water
 As for all other acids and bases, Ka and Kb for water is calculated as follows ...
Ka = Kb = Keq  [101.74] =
1014
 1015.74
1.74
10
and thus pKa = pKb = 15.74 for water.
THREE CLASSES OF SOLVENTS
1. Amphiprotic solvents: Other ionizable solvents are also amphiprotic, e.g., methanol, ethanol,
acetic acid, and ammonia. The autopyrolysis constants for some are listed.
Solvent
- log Ks or (pKw)
water
14
acetic acid
14.5
ethylenediamine
15.3
methanol
16.7
ethanol
19.1
2. Non-polar Aprotic solvents: These have no protons and are not polar. They are nonionizable
and inert, i.e., neither acidic nor basic, e.g., benzene, CCl4.
3. Polar Aprotic (Basic but not Acidic): These are also nonionizable solvents but can accept a
proton because of the presence of atoms such as O or N which have lone pairs of electrons, e.g.,
ether, dioxane, ketones, and pyridine. The are no known examples of solvents that are acidic
but not basic.
3
Predicting the extent of a reaction using the equation:
(pKeq = pKa + pKb - 14)
Consider an acid in HOH:
Consider a base in HOH:
HA + HOH  H3O+ + A-
B + HOH  BH+ + OH-
[H 3O + ] [A - ]
Ka = Keq  [55.5] =
[HA]
[BH + ] [OH - ]
Kb = Keq  [55.5] =
[B]
Now consider an acid (HA) reacting with a base (B):
HA + B  BH+ + A-
Keq =
[BH + ] [A - ]
[HA] [B]
By manipulation of these expressions we derive an expression for predicting the extent of reaction ...
 [H 3 O + ] [A - ]  [BH + ] [OH - ]
 [A - ] [BH + ]
 
Ka  Kb = 
 
 [H 3 O + ] [OH - ]
[HA]
[ B]

 [ HA] [B] 

 
or
Ka  Kb =
( Keq )

( Kw )
or
Ka  Kb =
( Keq )

( 10-14 )
Taking the log of both sides and then multiplying the equation by (-1) yields ...
pKa + pKb =
pKeq
+
14
Finally rearranging this we have a simple expression for calculating the extent of an acid-base reaction
...
pKeq = pKa + pKb - 14
The equilibrium constant (Keq) can then be determined (if desired) by taking the negative antilog of
pKeq calculated above, however it is easier to convert from pKeq to extent of reaction graphically (see
the appendix of these notes).
One can readily get a sense for extent of reaction from basic
thermodynamics …
Keq = 1 (pKeq = 0) means that the extent of a reaction = 50%
A reaction is said to go to completion when Keq  103 or pKeq  -3
These calculations assume that reagents are 100% pure (not dilute) and yet they work quite well as
estimates even in dilute aqueous solutions. For example, Pasto, et. al. state (p. 274) that "compounds
containing acidic functional groups with pKa's of less than ca. 12 will dissolve in dilute aqueous (5%)
NaOH".
HA + NaOH  Na+ A- + HOH
We can understand this by calculating pKeq, i.e., [12 + (-1.74) - 14] = -3.74 = pKeq or Keq  5500
(quite favorable even in dilute aqueous solution)
On p.275, the same author states that "only acids with pKa's < 6 will dissolve in dilute (5%) NaHCO3"
and we could also have predicted this because pKeq = [ 6 + 7.6 - 14] = -0.4 or Keq = 2.5 Using this
equation, determine if 5% NaHCO3 will dissolve a) acetic acid, b) phenol.
Note: in order to quantitatively titrate an acid base pair in aqueous media, pKeq must be  -8.
4
More on Predicting Acid / Base Reactions: In addition to calculating pKeq and Keq we can also
predict whether an acid/base reaction will occur by making use of a few simple principles. Let's
begin with a familiar example which we know reacts essentially to completion ...
HCl
+
strong
acid
-7
NaOH 
HOH
strong
base
-1.74
weak
acid
15.74
+
NaCl
weak
base
21
 (pKa or pKb)
Note the pK values. In general, an acid and base will react only if the reaction products are a
weaker acid and a weaker base than the reagent acid and reagent base. Stated otherly...
 A stronger acid will donate H+ to a base whose conjugate acid is weaker (higher pKa) or
 An acid with a lower pKa will donate H+ to a base whose conjugate acid has a higher pKa.
Assign pK values to all species in the following reactions and predict which of the following
reactions will proceed as shown ...
CH3CH2O-
+
CH3COOH

CH3CH2OH
H2N-
+
CH3CH2OH

NH3
+
CH3CH2O-
H-CC-H
+
NaNH2

NH3
+
H-CC:-Na+
C6H5OH
+
OH-

C6H5O-
+
HOH
C5H5N
+
H3O+

C5H5NH+
+
HOH
+
CH3COO-
5
Strengths of Acids and Bases (pKa + pKb) = 14 for conjugate acid-base pairs
acid
weakest
acids
Cannot
protonate
water
Determined
in water.
Protonate
water
increasingly.
Protonate
water
completely
strongest
acids
pKa
conjugate base
pKb
CH4
~ 55
~ -41
C2H4
44
:CH3:C2H3C6H5-
-30
C6H6
NH3
43
35
:NH2-
-29
-21
H2
35
:H-
-21
25
:C2H-
-11
(CH3)2CO
(CH3)3COH
20
18
CH3COCH2(CH3)3CO-
-6
-4
CH3CHO
C2H5OH
17
16
CH2- CHO
C2H5O-
-3
-2
H2O
15.74
OH-
-1.74
HPO4-2
CH3NH3+
12.3
PO4-3
1.7
10.6
CH3NH2
3.4
C6H5OH
9.9
C6H5O-
4.1
HCN
NH4+
9.3
9.2
CNNH3
4.7
4.8
CH3SH
H2PO4-
8
7.2
CH3SHPO2-2
6
6.8
H2S
7.1
HS-
6.9
6.4
HCO3-
7.6
C6H5SH
C5H5NH+
6
5.3
C6H5SC5H5N
8
8.7
HN3
CH3COOH
4.7
4.7
N3CH3COO-
9.3
9.3
C6H5NH3+
4.6
C6H5NH2
9.4
C6H5COOH
4.0
COO-
10.0
HF
H3PO4
CF3COOH
3.2
2.1
0.2
CF3COO-
10.8
11.9
13.8
C2H5COH+NH2
HNO3
-1
-1.4
C2H5CONH2
NO3-
15
15.4
H3O+
-1.74
HOH
15.74
CH3CH2OH2+
-3.6
CH3CH2OH
17.6
H2SO4
-5
HSO4-
19
HCl
HBr
HI
C2H5C=NH+
HClO4
HSbF6
-7
-8
-9
-10
-10
-12
ClBrIC2H5CN
ClO4SbF6-
21
22
23
24
24
26
C2H2
H2CO3
C6H5
FH2PO4-
strongest
bases
Completely
protonated
by water
Determined
in water.
Protonated
by water
increasingly.
Not
protonated
by water
weakest
bases
6
Determine the extent of the following reactions
(pKeq)
Base
(pKb)
Acid
(pKa)
Conjugate
Base (pKb)
Congugate
Acid (pKa)
NaOH
CH3COOH

CH3COO-
HOH
NaHCO3
HCl (aq)

NaCl (HOH)
H2CO3
CH3NH2
H3O+

HOH
CH3NH3+
C5H5N
H3O+

HOH
C5H5NH+
C6H5NH2
H3O+

HOH
C6H5NH3+
NaOH
C2H5OH

C2H5O-
HOH
C2H5OH
H3O+

HOH
C2H5OH2+
NO2--NH2
H3O+

HOH
NO2--NH3+
NaHCO3
HA
(6.0)

A-
H2CO3
NaOH
HA
(12.0)

A-
HOH
NaHCO3
-OH

-O-
H2CO3
I-
CH4

CH3-
HI
B
7
HA
7

A-
BH+
B
8
H3O+

HOH
BH+
NaOH
HA
8

A-
HOH
NaOH
-OH

-O- Na+
HOH
(pKeq)
7
Typical pKa values of functional groups
Typical pKb values of functional groups
pKa
group
example
pKb
group
example
55
alkane
CH4
24
nitrile
CH3CN:
45
alkene
CH2=CH2
23
acid chloride
ethanoyl
chloride
35
ammonia
NH3
22
aldehyde
ethanal
25
alkyne
CH CH
21.2
ketone
acetone
20
ketone
acetone
20.5
ester
ethyl acetate
17
amide
methanamide
20
carboxylic acid
acetic acid
17
aldehyde
ethanal
17.6
ether
C2H5OC2H5
15 - 18
alcohol
ethanol
16 - 18
alcohols
CH3CH2OH
15.74
water
H2O
15.74
water
H2O
10
phenol
-OH
15
aliphatic amides
ethanamide
9.3
hydrocyanic acid
HCN
9.3
acetate
CH3COO-
8
aliphatic thiol
CH3SH
9
aromatic amine
-NH2
7
hyrogen sulfide
H2S
7.6
bicarbonate
NaHCO3
6
aromatic thiol
-SH
6.0
aliphatic sulfide
CH3S-
5
carboxylic acid
CH3COOH
4.7
cyanide
CN-
-0.6
sulfonic acid
-SO3H
4.6
ammonia
NH3
-1.74
hydronium ion
H3O+
4
phenoxide
-O-
-2
protonated alcohol
CH3CH2OH2+
3
aliphatic amine
CH3NH2
-7
mineral acid
HCl
-1.74
hydroxide
OH-
-2
alkoxide
CH3CH2O-
borohydride
NaBH4
aluminum hydride
LiAlH4
Grignard
CH3- +MgBr
-21
hydride
Li+ :H-
-21
sodium amide
Na+ :NH2-
-41
1 carbanion
Li+ :CH3-
8
LEWIS ACIDS AND BASES:
A Lewis acid (electrophile, E+) is a substance which accepts an electron pair.
A Lewis base (nucleophile, Nu: -) is a substance which donates an electron pair.
As a result of this electron donation from a base to an acid, a covalent bond is formed.
Lewis Acids:
+
 Lewis acids must have vacant, low energy orbitals, or a polar bond to H, so H can be lost.
 Lewis acids include, but are much broader than Bronsted-Lowry and Arrhenius acids. For
+2
example, metal cations, such as Mg , are Lewis acids because they can accept a pair of
electrons when they form a bond to a base.
 In the same way, compounds of Group 3A elements, such as BF3 and AlCl3, are Lewis acids
because they have unfilled valence orbitals and can accept electron pairs from Lewis bases.
 Similarly, many transition metals salts, such as TiCl4, FeCl3, ZnCl2, and SnCl4 are Lewis acids.
By means of Lewis structures, show the following acid-base reactions ...
1. HCl + H2O
..
.. O
H
+
H
H
Lewis base
..
Cl
.. :
H
.. O +
H
H
+
.. _
: Cl :
..
Lewis acid
Chemical reactions invovle the transfer of electrons from electron donors to electron acceptors
Arrows show the electron transfer (from nucleophile to electrophile) (from Lewis base to Lewis acid).
Arrows do not show movement of molecules, atoms or ions!!!!!!!
The large arrow shows that one of the lone pairs of electrons on oxygen (nucleophile) is used to form a
covalent bond with the H atom in HCl (the electrophile). Notice that the hydronium ion product has one less
pair of electrons than HOH and notice that the oxygen atom has a + charge.
The small arrow shows that the hydrogen to chlorine bond breaks as the shared (bonding) pair of electrons
moves to the chlorine atom which thus becomes a chloride anion. Notice that the chloride anion has one more
lone pair of electrons than the chlorine atom in HCl.
2. BF3 + (CH3)2O (dimethyl ether)
3. AlCl3 + (CH3)3N (trimethyl amine)
9
Some Lewis Acids: (proton donors or electron pair acceptors):
 strong acids:
H2O, HCl, HBr, HNO3, H2SO4
 weak acids
CH3COOH, CH3CH2OH, C6H5OH (phenol)
 metal cations
Li+, Mg+2, Br+
 compounds with vacant orbitals
AlCl3, BF3, TiCl4, FeCl3, ZnCl2
Lewis Bases:
Lewis bases have nonbonding electron pairs which can be donated to Lewis acids. Most nitrogencontaining and oxygen-containing organic compounds have 1 or 2 lone pairs of electrons,
respectively.
Examples of Bases: (proton acceptors or electron pair donors)
CH3CH2
CH3
..
O
..
CH3CH2
H
..
O
..
alcohol
ether
: O:
: O:
C
Cl
CH3
acid chloride
C
:O:
CH2CH3
amine
N
: O:
..
O
..
CH3
H
CH3
CH3
H
C
: O:
..
O
..
CH3
S
CH3
CH3
CH3
C
..
NH2
amide
..
CH3
C
ketone
ester
carboxylic acid
CH2CH3
C
aldehyde
..
CH3CH2
CH3
: O:
H
sulfide
..
..
O
H
water
 Note that some compounds can act as both Lewis acids or Lewis bases, depending upon the
reaction conditions.
 Alcohols and carboxylic acids act as acids by donating a proton but also act as bases when their
oxygen atoms donate an electron pair, thereby accepting a proton.
Draw Lewis structures showing the following acid-base reactions.....
1. CH3OH + HBr
2. (CH3)2O + H2SO4
 Note that some Lewis bases, such as carboxylic acids, esters, and amides, have more than one
atom with a lone pair of electrons and can therefore react at more than one site. For example,
acetic acid can be protonated on the doubly-bonded or singly-bonded O atoms ...
10
Some Factors Affecting Acidity:
HI > HBr > HCl > HF
       
decreasing acidity
 Decreasing halogen size decreases acid strength, or conversely, increasing halogen size
increases acid strength. In the same vertical group of the periodic table, the larger atoms
(higher molecular weight), can disperse '-' charge over a larger region and thus add stability to
the conjugate base. As the stability of the conjugate base increases (i.e., weaker conjugate
base), the greater is the strength of the acid.
Consider EN also: In the same row of the periodic table, the EN of the atom bonded to H increases
from left to right across the table. Since more EN atoms can carry a negative charge more readily
than a less EN atom, the acidity increases as shown....
(CH3)3C ---- H
   
(CH3)2N ---- H
CH3O ---- H
       
increasing acidity

F ---- H

EN
stability
C
- CH
3
<
<
N
- NH
2
<
<
O
OH -
<
<
F
F-
acidity
H ----CH3
<
H ----NH2
<
H ----OH
<
H ----F
basicity
- CH
3
>
- NH
2
>
OH -
>
F-
 For organic acids, i.e., carboxylic acids, the proximity of an electronegative atom to an acidic H
affects the acidity of the compound ...
(CH2Cl)CH2CH2COOH
   

>

CH3(CHCl)CH2COOH
     
increasing acidity

> CH3CH2(CHCl)COOH
  
 As previously mentioned in the section on resonance, delocalization of electrons in a conjugate
base increases the stability of the anion, therefore increasing the acidity of its conjugate acid.....
CH3SOOOH
 
>
 
CH3COOH
>
     
decreasing acidity
CH3CH2OH
   
11