Homework
Fundamentals of
Analytical Chemistry
7-10, 12, 14, 18,
20, 27
Chapter 15
Polyfunctional Acids and Bases
Polyprotic Acids
Acids that can donate
more than 1 proton
per molecule
Strong acid – H2SO4
Several weak acids
‘Well behaved’
behaved’ dissociation
For most cases we can safely assume that
protons are removed sequentially
Polyprotic Acids
Consider the dissociation of the weak
polyfunctional acid H2A
First step: H2A H+ + HA• K1 = [H+][HA-] / [H2A]
• Subscript 1 indicates the first proton ‘removed’
removed’
Second Step: HA- H+ + A2• K2 = [H+][A2-] / [HA-]
• Subscript 2 indicates the second proton ‘removed’
removed’
• All molecules will have their first proton removed
before any will have the second removed, etc…
etc…
For higher order polyprotics,
polyprotics, subscript indicated the
number of the proton removed from the ‘parent’
parent’
molecule
Polyprotic Acids
Possible
combinations
H2A /
A2• Conjugate base of the weak acid HA• Kb = Kw / K2
HA-
• Buffer
• Henderson Hasselbalch using pK1
HA- / A2• Also a buffer (conjugate weak acid/base pair)
• HendersonHenderson-Hasselbalch using pK2!
H2A
• Since the protons dissociate sequentially, this is
like any other weak acid
• Calculate [H+] like any monoprotic acid using K1.
Polyprotic Acids
H2A / A2•
•
•
Will react
H2A + A2- 2HAMay have either of the two buffers OR HA-
1
Acid Salt
Partially neutralized
polyprotic acid
Only ‘new’
new’ situation relative to monoprotic
acids
Can act as either an acid or a base
[H+] a function of Ka (acting as an acid) and Kb
(acting as a base)
Acid Salt
Deriving
[H + ] =
• Let Kn+1 = Ka as an acid, then Kb = Kw / Kn
• For a diprotic acid, Kn = K1, and Kn+1 = K2
K n +1C HA− + K w
C −

1 +  HA
K n 

If CHA- / K1 is much greater than 1, and also
Kn+1CHA- is much greater than Kw, then
[ H + ] ≈ K n K n +1
Acid Salt
Titration of Polyprotic Acids
Final equation
shows that under the
assumed conditions, [H+] (and therefore
pH) is independent of the concentration of
the acid salt
Legend
Point
A
B (blank)
First buffer region
pH from pK1
Point
Region
Only a weak acid
Treat like any monoprotic acid using K1
Region
Legend
C
Acid Salt
Point
E
Conjugate base of ‘acid salt’
salt’
Kb = Kw / K2
Region
D
Second buffer region
Use pK2 to calculate pH
F
Strong base
2
Titration of 25.0 mL of 0.10 M
o-phthalic acid with 0.10 M NaOH
Calculations
Consider the
titration of 25.0 mL of 0.10 M
o-phthalic acid with 0.10 M NaOH
Assume the formula of the acid is H2A
Before any titrant
[H ] = − K +
+
1
K12 + 4KaCHA
2
is added
=
mL of titrant is added
H2A + OH- HA- + H2O
I 2.50 0.50
0 ~~
∆ -0.50 -0.50 +0.50 ~~
F 2.00 ~0
0.50 ~~
−1.12x10−3 + (1.12x10−3 )2 + 4 ∗1.12x10−3 ∗ 0.10
2
pH = -log (0.0100) = 2.00
Titration of 25.0 mL of 0.10 M
o-phthalic acid with 0.10 M NaOH
[H + ] =
After 5.00
− (C A− + K a ) + (C A− + K a ) 2 + 4 K a C HA
Buffer, but with K1> 10-3, we have to use
the quadratic form for [H+]
Titration of 25.0 mL of 0.10 M
o-phthalic acid with 0.10 M NaOH
After 25.0
2
 0.5 
 0.5 
−3
−3 2
−3  2.0 
− (
 + 1.12 x10 ) + (
 + 1.12 x10 ) + 4 ∗1.12 x10 ∗ 

 30 
 30 
 30 
[H ] =
2
[ H + ] = 3.51x10− 3 ; pH = 2.46
+
mL of titrant is added
H2A + OH- HA- + H2O
I 2.50 2.50
0 ~~
∆ -2.50 -2.50 +2.50 ~~
F ~0
~0
2.50 ~~
Calculating with the ' short ' form :
Acid salt
pH = pK a + log(b / a ) = 2.95 + log(0.5 / 2.0)
pH = 2.35 ( significant difference)
Titration of 25.0 mL of 0.10 M
o-phthalic acid with 0.10 M NaOH
[H + ] =
K n+1C HA− + K w
2.5
; C HA− =
= 0.050
50
C −

1 +  HA

Kn 

(3.91x10 −6 ∗ 0.050) + 1.0 x10 −14
1 + 0.050
1.12 x10−3
+
[ H ] = 6.54 x10 −5 ; pH = 4.18
[H + ] =
(
or with the ' short ' form :
[ H + ] = (1.12 x10−3 )(3.91x10 −6 )
[ H + ] = 6.62 x10 −5 ; pH = 4.18
)
Titration of 25.0 mL of 0.10 M
o-phthalic acid with 0.10 M NaOH
After 35.0
mL of titrant is added
H2A + OH- I 2.50 3.50
∆ -2.50 -2.50
F ~0
1.00
THEN
HA- + OH- I 2.50 1.00
∆ -1.00 -1.00
F 1.50
~0
HA- + H2O
0
~~
+2.50 ~~
2.50 ~~
A2- + H2O
0
~~
+1.00 ~~
1.00 ~~
3
Titration of 25.0 mL of 0.10 M
o-phthalic acid with 0.10 M NaOH
Buffer with
After 50.0
HA- / A2-
pH = pKa + log (b/a)
pH = 5.41 + log (1.00 / 1.50)
pH = 5.23
Titration of 25.0 mL of 0.10 M
o-phthalic acid with 0.10 M NaOH
Only species affecting the
Titration of 25.0 mL of 0.10 M
o-phthalic acid with 0.10 M NaOH
pH is A2-
Kb = KW / K2
CB = 2.5 / 75
 1.0 x10 −14   2.5 
∗
[OH − ] = 

−6  
 3.91x10   75 
[OH − ] = 9.23x10−6 ; pOH = 5.03
pH = 14.00 − pOH = 8.97
[OH-] = 2.5/100
pOH = 1.60
pH = 12.40
different than a monoprotic titration
past the equivalence point
After 75.0
For polyprotic you must be past the last
equivalence point.
A2- + H2O
0
~~
+2.50 ~~
2.50 ~~
mL of titrant is added
H2A + OH- I 2.50 7.50
∆ -2.50 -2.50
F ~0
5.00
THEN
HA- + OH- I 2.50 5.00
∆ -2.50 -2.50
F ~0
2.50
HA- + H2O
0
~~
+2.50 ~~
2.50 ~~
A2- + H2O
0
~~
+2.50 ~~
2.50 ~~
Sulfuric Acid pH
Unique
No
HA- + H2O
0
~~
+2.50 ~~
2.50 ~~
Titration of 25.0 mL of 0.10 M
o-phthalic acid with 0.10 M NaOH
Titration of 25.0 mL of 0.10 M ophthalic acid with 0.10 M NaOH
Strong base
mL of titrant is added
H2A + OH- I 2.50 5.00
∆ -2.50 -2.50
F ~0
2.50
THEN
HA- + OH- I 2.50 2.50
∆ -2.50 -2.50
F ~0
~0
situation
First proton is completely dissociated (strong
acid)
Second proton comes from a weak acid
(partially dissociated)
Dissociation of the first proton affects the
amount of dissociation for the second proton
What
is the pH for a 0.010 M solution of
sulfuric acid?
4
Sulfuric Acid pH
Assume 1 mL of solution, then M = mmol acid
H2SO4 HSO4- + H+
I 0.010
0
0
∆ -0.010
+0.010 +0.010
F ~0
0.010 0.010
THEN
HSO4- SO42- + H+
I 0.010
0
0.010
∆ -x
+x
+x
F 0.010x
0.010+x
0.010-x
K2 = [H+][SO42-] / [HSO4-]; K2 = 0.0102
= (0.010+x)(x) / (0.010 - x)
(0.0102)(0.010 - x) = (0.010+x)(x)
1.02x10-4 - 0.0102x = x2 + 0.010x
Rearranging:
x2 + 0.010x + 0.0102x – 1.02x10-4 = 0
Quadratic – solve for x
x = 0.0042, [H+] = 0.010+0.0042 = 0.0142
pH = 1.85
Assume one proton dissociates, pH = 2.00
Assume both protons dissociate, pH = 1.70
5