Tài liệu Chapter 10: Polyprotic Acid-Base Equilibria pptx

Chapter 11:
Acid-Base Titrations
Chapter 10:
Polyprotic Acid-Base Equilibria
1. Treating Complex Acid-Base Systems

Complex systems are defined as solutions made up of:




1) Two acids or bases of different strengths
HCl + CH
3
COOH
NaOH + CH
3
COO
-



2) An acid or base that has two or more acidic
protons or basic functional groups

H
3
PO
4

Ca(OH)

2


3) An amphiprotic substance that is capable of acting
as both acid and base

HCO
3
-
+ H
2
O <=> CO
3
2-
+ H
3
O
+

HCO
3
-
+ H
2
O <=> H
2
CO
3
+ OH
-

N
H
3
+
CH
2
COO
-
+ H
2
O <=> NH
2
CH
2
COO
-
+ H
3
O
+


N
H
3
+
CH
2
COO

-
+ H
2
O <=> NH
3
+
CH
2
COOH + OH
-


2. Equations for more than one equilibrium are
required to describe the characteristics of any of
these systems.

3. Applications

1) We encounter these systems in most biological
and environmental matrices

2) We need to predict which species will be
present (and in what amounts) is important in
defining biological acid/base buffering.

3) We can use the titration curve to examine how
the species will change as we slowly add titran
t
(in other words, change the pH).

4. Examples
A. Mixtures of Strong and Weak Acids or Strong and Weak Bases
B. Polyfunctional Acids and Bases:
C. Buffer solutions Involving Polyprotic Acids:
D. Calculation of the pH of Solutions of Amphiprotic Salts (NaHA):
E. Titration Curves of Polyfunctional Acids:
A. Mixtures of Strong and Weak Acids or Strong and Weak Bases
4. Examples
A. Mixtures of Strong and Weak Acids or Strong and Weak Bases
Example 11-1A: Derive a titration curve for a titration of a 50.00 mL
solution containing 0.1000 M strong acid, HCl, and 0.1000 M weak aci
d
HA (K
a
= 1 X 10
-4
) with 0.1000 NaOH. (similar to Example 11-1)


The equilibria:

HCl + H
2
O => H
3
O
+
+ Cl
-
[1]

HA + H
2
O <=> H
3
O
+
+ A
-
[2]
2 H
2
O <=> H
3
O
+
+
OH
-
[3]

[H
3
O
+
] = C
HCl
+ [A
-
] + [OH
-

]

Since the HCl, which is completely dissociated, will repress the dissociatio
n
of HA and H
2
O.

Assume that [OH
-
] and [A
-
]<<C
HCl
so that [H
3
O+] = C
HCl
.

1. V
NaOH
=0: The pH before the addition of titrant is determined by the
concentration of HCl alone: [H
3
O+] = C
HCl

2. 0 <V
NaOH

<50.00 mL: After titrant has been added, the titration
curve will be identical to that for the titration of the HCl alone an
d
p
H will be determined b
y
the remainin
g
HCl in the solution.

total
added3
3
V
]OH[ of Moles]OH[ of Moles
]OH[
−+
+
−
=

3. V
NaOH
=50.00 mL: When the HCl has been neutralized (the
first equivalence-point), the presence of the weak acid must be
considered. At this point, the equilibrium described in eq. [2]:


]AH[
]HA][OH[

K
2
3
a1
−+
=

100
K
C
KC]OH[
a
HA
aHA3
>=
+


4. 50.00 mL<V
NaOH
<100.00mL HA is now reactin
g
with the titran
t
and the titration curve will be identical to that of the titration of a
weak acid and

HA
NaA
a

-
a
-
a
C
C
logpK
]HA[
][A
logpK
]A[
][HA
logpKpH +=+=−=

5. V
NaOH
=100.00 mL At the point where both HCl and HA are
neutralized (the second equivalence-point), the titration solution now
contains A
-
which reacts with water
A
-
+ H
2
O <=> HA + OH
-


and the pH at this equivalence point is determined by concentration o

f

[A
-
] and K
b
(1.0 x 10
-10
)


100
K
C
KC]OH[
b
b
bb
>=
−

6. V
NaOH
>100.00 mL After all of the acid has been neutralized,
further addition of titrant results in a mixture of a weak and stron
g
b
ase.
p
H of the solution is determined b

y
the concentration of the stron
g

b
ase.
total
3added
V
]OH[ of Moles]OH[ of Moles
]OH[
+−
−
−
=

In summary, the titration curve includes a titration curve of
HCl (a strong acid) with NaOH and a titration curve of HA
(weak acid) with NaOH.

You still can apply the theories and equations of neutralization
titration in Chapter 9 into this complex system.
≈ C
HCL
= 0.1200 M
≈ C
HCL
= 0.1200 M
C
HCL

V
HCL
=C
KOH
V
KOH
V
1st Eq
.
KOH
= 25.00x0.1200/0.1000= 30.00mL
C
HA
V
HA
=C
KOH
V
KOH
V
2nd Eq
.= V
1st Eq
.
KOH
+25.00x0.0800/0.1000= 50.00m
L
and (c) 29.00 mL.
total
added3

3
V
]OH[ of Moles]OH[ of Moles
]OH[
−+
+
−
=
Before the 1
st
equivalence point: 5.00 mL <30.00 mL
(c) Upon addition of 29.00 mL of 0.1000 M KOH
total
added3
3
V
]OH[ of Moles]OH[ of Moles
]OH[
−+
+
−
=
Before the 1
st
equivalence point: 29.00 mL <30.00 mL
A. Mixtures of Strong and Weak Acids or Strong and Weak Bases
Any Questions?
B. Polyfunctional Acids:
These acids may donate two or more H

+
, and the bases may accept
two or more H
+
.

An example of a polyfunctional acid is H
3
PO
4
, with the following
equilibria:
H
3
PO
4
+ H
2
O <=> H
3
O
+
+ H
2
PO
4
-


3

43
423
a1
10x11.7
]POH[
]POH][OH[
K
−
−
+
==

H
2
PO
4
-
+ H
2
O <=> H
3
O
+
+ HPO
4
2-


8
42

2
43
a2
10x34.6
]POH[
]HPO][OH[
K
−
−
−
+
==

HPO
4
-2
+ H
2
O <=> H
3
O
+
+ PO
4
3-



13
2

4
3
43
a3
10x5.4
]HPO[
]PO][OH[
K
−
−
−
+
==

H
3
PO
4
+ H
2
O <=> H
3
O
+
+ H
2
PO
4
-
[Eq. 1]

3
43
423
a1
10x11.7
]POH[
]POH][OH[
K
−
−
+
==

H
2
PO
4
-
+ H
2
O <=> H
3
O
+
+ HPO
4
2-
[Eq. 2]

8
42
2
43
a2
10x34.6
]POH[
]HPO][OH[
K
−
−
−
+
==

HPO
4
2-
+ H
2
O <=> H
3
O
+
+ PO
4
3-
[Eq. 3]

13
2
4
3
43
a3
10x5.4
]HPO[
]PO][OH[
K
−
−
−
+
==

[Eq. 1] +[Eq. 2] + [Eq. 3] :
H
3
PO
4
+ H
2
O <=> 3H
3
O
+
+ PO
4
3-

22
43
3
4
3
3
a3a2a1
10x0.2
]POH[
]PO[]OH[
KKK
−
−
+
==

Equilibria Constants for Overall Reactions
K
a
H
3
PO
4
+ H
2
O <=> H
3
O

+
+ H
2
PO
4
-
K
a1


H
2
PO
4
-
+ H
2
O <=> H
3
O
+
+ HPO
4
2-
K
a2


HPO
4

2-
+ H
2
O <=> H
3
O
+
+ PO
4
3-
K
a3


K
b

PO
4
3-
+ H
2
O <=> HPO
4
2-
+ OH
-
K
b1

HPO
4
2-
+ H
2
O <=> H
2
PO
4
-
+ OH
-
K
b2

H
2
PO
4
-
+ H
2
O <=> H
3
PO
4
+ OH
-
K
b3

Definition of K
a
and K
b
for Polyfunctional Acids
K
a1
K
b3
= K
a2
K
b2
= K
a3
K
b1
= [H
3
O
+
][OH
-
] = K
w
Any Questions?
C. Buffer solutions Involving Polyprotic Acids:
(A) Three buffer s
y
stems can be described when usin

g

a
weak diprotic acid, H
3
A, and its salts, H
2
A
-
,
HA
2-
and A
3-
.


H
3
A/ H
2
A
-



H
2
A
-

/HA
2-



HA
2-
/A
3-


When K
a1,
K
a2,
K
a3
are well separated

3
a2
a1
10
K
K
>

3
a3
a2

10
K
K
>

AH
AH
a1
3
-
2
C
C
logpKpH +=
-
2
-2
AH
HA
a2
C
C
logpKpH +=
-2
-3
HA
A
a3
C
C

logpKpH +=
(B) Two buffer systems can be described when using a weak
diprotic acid, H
2
A, and its salts, HA
-
and A
2-
.



H
2
A/HA
-




HA
-
/A
2-



When K
a1
is w ell-separated from K

a2
:


3
a2
a1
10
K
K
>

W e can treat these systems as separated one for calculation o
f

pH and species present.
AH
HA
a1
2
-
C
C
logpKpH +=
-
-2
HA
A
a2
C

C
logpKpH +=
AH
AH
a1
3
-
2
C
C
logpKpH +=
Example 11-2:
D. Calculation of the pH of Solutions of Amphiprotic Salts (NaHA)


The pH is determined by the following equilibria:

HA
-
+ H
2
O <=> H
2
A + OH
-

a1
w
b2
K

K
K =


HA
-
+ H
2
O <=> A
2-
+ H
3
O
+

K
a2


At the first equivalence point, the solution will be either acidic or basic
b
ased upon the relative magnitude of the equilibrium constants (K
b2
an
d

K
a2
)