p.01
Acid-Base Eqm (3):
Explain
Explain the
the Strength
Strength of
of
Organic
Organic Acids & Bases
Review Questions
Q.1:
pKb of CH3NH2 (an organic weak base) at 298K is 4.75.
Find the pH of 0.05M CH3NH2 at 298K.
Kb =
x2
0.05 – x
x2
= 10-4.75
= 1.77810-5
x = 9.4310-4
0.05
 pOH = -log(9.43 10-4) = 3.03
pH = 14 – pOH = 11.0
C. Y. Yeung (CHW, 2009)
Q.2:
p.02
25.0 cm3 of a monobasic organic weak base B(aq) requires
20.25 cm3 of 0.50M HCl(aq) for complete neutralization.
Given that the pH of the B(aq) at 298K is 11.25, what is Kb of
B at 298K?
(20.25/1000)0.50
= 0.405 M
conc. of B(aq) =
(25.0/1000)
B + H 2O
HB+ + OH-
at start 0.405
0
0
at eqm 0.405 – x
x
x
As pH = 11.25, i.e. pOH = 14 – 11.25 = 2.75

x = [OH-] = 10-2.75 = 1.77810-3
Kb =
(1.77810-3)2
= 7.8410-6 M
0.405 – 1.77810-3
p.03
Stability of Conjugate
Base …
e.g. CCl3COO- is the conjugate base of CCl3COOH
i.e. CCl3COO- is stable, thus it is less likely to react
with H3O+ to regenerate CCl3COOH.
CCl3COO- is stabilized by
“negative inductive effect”.
CCl3COO- + H3O+
CCl3COOH + H2O
Cl
electronegative
atom !!
Cl
C
Cl
O
C
-
O
The -ve charge is
dispersed /delocalized
over the anion.

The -ve charge is less
concentrated on the O atom.
p.04
Conjugate Base may be stabilized
by “resonance effect”
OH
O
+
+
H2 O
H3 O +
phenoxide
phenol
O-
-
O
O
-
O
-
(resonance structures)
 The -ve charge is delocalized over the aromatic ring.
(resonance effect)
i.e. phenoxide ion is stabilized by resonance effect.
pKa of Phenol = 9.95
pKa of Ethanoic acid = 4.76
How to explain …?
p.05
stronger acid!
Although CH3COO- is destabilized by positive inductive effect,
it is stabilized by resonance effect!
-
O
O
CH3 C
O
CH3 C
O
both are
electronegative!
i.e. –ve charge is
shared equally.
In the case of phenoxide, delocalization of –ve charge by
resonance is less effective, because the O atom is more
electronegative than the carbon atoms in aromatic ring.
The –ve charge is more delocalized in CH3COO-, and
thus it is more stable.  CH3COOH is the stronger acid.
p.06
Strength of Base …
(depends on the availability of lone pair e-.)
availability of lone pair e- is
increased by +I effect
push e-
CH3 N
H
 stronger base (than NH3)!
availability of lone pair e- is
decreased by resonance.
H
Methylamine
(pKb = 3.38)
N
phenylamine / aniline
(pKb = 9.40)
+
N
H
N
H
-
H
H
 weaker base (than NH3)!
H
Ref: pKb of NH3 = 4.80
-
H
-
+
N
H
H
+
N
H
(resonance structures)
H
Advanced Understanding on the Strength of Organic Acid (1)
--- 4-nitrophenol (p-nitrophenol)
How can 4-nitrophenoxide ion be stabilized by resonance effect?
O
O
O
O
N+ O
O
N+ O
O
O
O
-
O
O
-
Therefore, it is stronger than
phenol.
N+
O
+
N
O
O
The -ve charge on the p-nitro
phenoxide ion is more dispersed
than that on the phenoxide ion.
N+
N+ O
O
O
pKa of phenol = 9.95
pKa of p-nitrophenol = 7.14
a.1
Advanced Understanding on the Strength of Organic Acid (2)
--- 3-nitrophenol (m-nitrophenol)
a.2
Is the pKa of m-nitrophenol smaller than p-nitrophenol?
O
O
The -ve charge cannot be
dispersed over the nitro
group.
+
N O
O
+
N O
O
O
O
(But it is still stronger than
phenol due to negative
inductive effect.)
-
+
N O
O
Therefore, it is weaker than
p-nitrophenol, and its pKa is
larger.
+
N O
O
pKa of phenol = 9.95
pKa of p-nitrophenol = 7.15
pKa of m-nitrophenol = 8.35
Advanced Understanding on the Strength of Organic Acid (3)
--- p-hydroxybenzaldehyde
a.3
How can the conjugate base be stabilized by resonance effect?
O
O
O
O
-
C
C
H
O
H
O
C
O
O
-
O
C
H
O
C
O
H
The -ve charge is dispersed
over the anion, p-hydroxy
benzaldehyde is acidic.
Therefore, it is stronger
than phenol.
O
C
H
-
H
pKa of phenol = 9.95
pKa of p-hydroxybenzaldhyde = 7.66
a.4
Look at their pKa values again ……
OH
OH
+
N O
O
N+ O
O
pKa (at 298K)
9.95
7.15
OH
OH
8.35
C
H
O
7.66
nitro group (– NO2) at the p-position
stabilizes the conjugate base by
delocalizing the – ve charge through:
 negative inductive effect
 resonance effect
a.5
Compare the pKb values of m-nitroaniline, p-nitroaniline
and aniline.
NH2
NH2
NH2
+
N O
O
N+ O
O
pKb (at 298K)
9.40
12.9
11.6
The availability of lone pair e- on
the –NH2 group is reduced by :
 negative inductive effect
 resonance effect
p.07
In Conclusion ….
Stronger Acid: conjugate base stabilized by - I effect /
resonance effect.
(delocalization of –ve charge)
Weaker Acid:
conjugate base destabilized by + I effect.
(localization of –ve charge)
Stronger Base: availability of lone pair e- increased by + I
effect. (localization of –ve charge)
Weaker Base: availability of lone pair e- decreased by - I
effect / resonance.
(delocalization of –ve charge)
p.08
More practice on the calculation of Ka, Kb and pH
Q.1:
At body temperature, human blood pH is 7.40. Lactic acid,
CH3CH(OH)COOH, is produced in muscle tissues during
physical exertion. Whether the lactic acid produced exists
mainly in form of “undissociated molecule” or
“dissociated lactate ions”?
(Given: Ka of lactic acid at body temperature = 8.5010-4 mol dm-3)
Ka =
[H3O+][CH3CH(OH)COO-]
[CH3CH(OH)COOH]
pKa = pH – log
2.14 104 =
[CH3CH(OH)COO-]
[CH3CH(OH)COOH]
[CH3CH(OH)COO-]
With a given Ka, the
ratio of [salt] and [acid]
could be found from
the pH value.
[CH3CH(OH)COOH]
 Lactic acid formed is mainly in form of dissociated lactic ions.
Q.2:
Calculate the pH value of the resultant solution for
10cm3 of 0.20M CH3COOH are mixed with 10cm3 of
0.40M CH3COONa. (Ka = 1.7510-5 mol dm-3)
After mixing,
1.7510-5
new [CH3COOH] = 0.10M
new [CH3COONa] = 0.20M
x(0.20+x)
=
0.10 – x
x = 8.75 10-6
 pH = 5.06
An acid-salt mixture (acid and salt have comparable conc.)
 ACIDIC BUFFER!
p.09
Q.3:
Calculate the pH value of the resultant solution for
10cm3 of 1.00M NH3 are mixed with 10cm3 of 1.00M
NH4Cl. (Kb = 1.7810-5 mol dm-3)
After mixing,
1.7810-5
new [NH3] = 0.50M
new [NH4+] = 0.50M
x(0.50+x)
=
0.50 – x
x = 1.78 10-5
 pOH = 4.75
 pH = 9.25
A base-salt mixture (base and salt have comparable conc.)
 BASIC BUFFER!
p.10
p.11
Assignment
Lab report [due date: 6/4(Mon)]
Book 3A p.141 Q.12, 13, 14 [due date: 20/4 (Mon)]
Pre-Lab: Expt. 13 Analysis of Two Commercial
Brands of Bleaching Solution
Next ….
Acidic Buffers and Basic Buffers
(Book 2 p. 153 – 163)