Transcript Titrating Polyfunctional Acids and Bases

Titrating Polyfunctional
Acids and Bases
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1. Treating Complex Acid-Base Systems
• Complex systems are defined as solutions
made up of:
(1) An acid or base that has two or more
acidic protons or basic functional groups
H3PO4
Ca(OH)2
(2) Two acids or bases of different strengths
HCl + CH3COOH
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)3) An amphiprotic substance that is capable
of acting as both acid and base
HCO3- + H2O  CO32- + H3O+
HCO3- + H2O  H2CO3 + OHNH3+CH2COO- + H2O  NH2CH2COO- + H3O+
NH3+CH2COO- + H2O  NH3+CH2COOH + OH-
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•

H 3 PO 4    H 2 PO 4  
  HPO
K a1
Kb3
Ka1×Kb3=Kw
2
Ka 2
Kb2
Ka2×Kb2=Kw
4

  PO 4
Ka 3
Kb1
Ka3×Kb1=Kw
Ka1=1×10-2 >Ka2=1×10-7> Ka3=1×10-12
Ktotal=Ka1×Ka2× Ka3=1×10-21
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3
pH of H3PO4
1. Calculate the pH of 0.100M H3PO4 solution.
K a1
K a2
 1000  K a 2 .is .negligible


K a 1  1  10
2

[ H ].[ H 2 PO 4 ]
0 . 100  H

H+ is not negligible
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pH of HApH of HA- solution
Ka2
HA- A2- + H+

[H ] 
Kb2
HA- H2A + OHKa1
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C HA   K a 2  K w
1
C HA 
K a1
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pH of HACalculate the pH of 0.100M NaHCO3 solution.
Ka1=1×10-6 >Ka2=1×10-10

[H ] 
C HA   K a 2  K w
1
C HA 
K a1
Ka2×CHA- =1×10-10 ×1.00>>Kw…….Kw is negligible
C HA
K a1


1 . 00
1  10
1 .... isnegligib le ...... pH 
6
 1  10
6
pK a 1  pK a 2
8
2
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pH of HACalculate the pH of 0.0100M NaH2PO4 solution.
C HA   K a 2  K w

[H ] 
1
Ka1=1×10-2 >Ka2=1×10-7> K
C HA 
K a1
a3=1×10
-12
Ka2×CHA- =1×10-7 ×0.01>>Kw…….Kw is negligible
C HA

K a1

0 . 01
1  10
2
1

[H ] 
1 .... is .not .negligible
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C HA   K a 2
1
C HA 
K a1
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pH of HACalculate the pH of 1.00×10-3M Na2HPO4 solution.
Ka2=1×10-7> Ka3=1×10-12
Ka1=1×10-2 >

[H ] 
C HA   K a 2  K w
1
C HA 
K a1
Ka2×CHA- =1×10-10 ×0.001=1×10-13 Kw isnot negligible
C HA

K a1

0 . 001
1  10
7
 1  10
4
1 .... is .negligible

[H ] 
C HA   K a 2  K w
C HA 
K a1
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Mixture of weak and strong acids
Sulfuric acid is unusual in that one of its protons behaves as a strong acid in water and the other as a weak acid
(Ka2 = 1.02 X 10-2). Let us consider how the hydronium ion concentration of sulfuric acid solutions is computed
using a 0.0400M solution as an example.
H2SO4 →H+ +HSO4- SO42- + H+
We will first assume that the dissociation of HSO4 is negligible because of the large excess of H30+ resulting
from the complete dissociation of H2SO4. Therefore,
[H+]
≈ [HSO4 ] ≈ 0.0400 M
Ka 
0 . 04  [ SO 4
2
]
0 . 04
This result shows that [SO4- ] is not small relative to [HSO4 ], and a more rigorous solution is required.
From stoichiometric considerations, it is necessary that
[H+] = 0.0400 + [SO42-]
[SO4] = [H+] - 0.0400
CH2SO4, = 0.0400 = [HS04- ] + [SO42-]
[HSO4-] = 0.0800 - [H3O+]
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Mixture of weak and strong acids
Sulfuric acid is unusual in that one of its protons behaves as a strong acid in water and the other as a weak acid
(Ka2 = 1.02 X 10-2). Let us consider how the hydronium ion concentration of sulfuric acid solutions is computed
using a 0.0400M solution as an example.
H2SO4 →H+ +HSO4- SO42- + H+
We will first assume that the dissociation of HSO4 is negligible because of the large excess of H30+ resulting
from the complete dissociation of H2SO4. Therefore,
[H+] = 0.0400 + [SO42-]
[HSO4-] = 0.0400 - [SO42-]

Ka 
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[ H ]  [ SO 4
[ HSO

4
]
2
]

( 0 . 04  [ SO 4
2
])  [ SO 4
( 0 . 04  [ SO 4
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2
2
]
])
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Curves for the titration of strong acid / weak acid mixture with 0.1000 M
NaOH. Each titration is on 25.00 ml of a solution that is 0.1200 M in HCl
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and 0.0800 M in HA.
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Curves for the titration of 25.00 ml of polyprotic acid with 0.1000M NaOH solution .
A)0.1000 M H3PO4,
Ka1=1×10-2 >Ka2=1×10-7>
B) 0.1000M oxalic acid,
Ka1 =5.6 × 10-2 and Ka2 = 5.4 x 10-5
C) 0.1000 M H2SO4
Ka2 = 1.02 × 10-2
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Ka3=1×10-12
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Titration curves for
polyfunctional acids
Titration of 20.00 ml of 0.1000 M H2A with 0.1000 M
NaOH.
For H2A, Ka1= 1.00 × 10–3 and Ka2 = 1.00 × 10–
7.
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Titration of 25.00 ml of 0.1000M maleic acid with 0.1000M NaOH.
HOOC-C=C-COOH
pKa1=1.89 ,pKa2=6.23
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E-HOOC-C=C-COOH
Fractional composition diagram for
fumaric acid (trans-butenedioic acid).
Z-HOOC-C=C-COOH
Fractional composition diagram for
maleic acid (Cis-butenedioic acid).
pKa1=3.05 ,pKa2=4.49
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amino acids
alanine
The amine group behaves as a base, while the carboxyl group acts as an acid.
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1-Determining the pK values for amino acids
Amino acids contain both an acidic and a basic group.
NH2-CH2-COOH  +NH3-CH2-COO- Zwitterion formation
+NH -CH -COO3
2
+ H2O  NH2-CH2-COO- + H3O+

[NH2 - CH2 - COO ]  [ H3O ]
-
Ka 

-
[ NH3 - CH2 - COO ]
+NH -CH -COO3
2
+ H2O  +NH3-CH2-COOH + OH
Kb 

[ NH 3 - CH 2 - COOH ]  [ OH ]

-
[ NH 3 - CH 2 - COO ]
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Ka×Kb= ??!!!!33
2-Determining the pK values for amino acids
Amino acids contain both an acidic and a basic group.
NH2-CH2-COOH  +NH3-CH2-COO- Zwitterion formation
Kb
+NH -CH -COOH +NH -CH -COO3
2
3
2
Ka1=5×10-3
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Ka
 NH2-CH2-COO-
Ka2=2×10-10
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A
B
Curves for the titration of 20.00ml of 0.1000M alanine with
A) 0.1000 M NaOH
B) 0.1000M HCl.
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Iso electric point:
The pH at which the average charge of the polyprotic acid is zero
• The zwitterion of an amino acid, containing as it does a positive and a
negative charge, has no tendency to migrate in an electric field,
+NH -CH -COO3
2
•
whereas the singly charged anionic and cationic species are attracted to
electrodes of opposite charge.
•
NH2-CH2-COO-
+NH
3-CH2-COOH
• No net migration of the amino acid occurs in an electric field when the pH
of the solvent is such that [anionic] = [cationic], which is pH dependent.
• The pH at which no net migration occurs is called the isoelectric point; this
point is an important physical constant for characterizing amino acids. The
isoelectric point is readily related to the ionization constants for the species.
Thus, for glycine,
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1-Determining iso electric point for amino acids
+NH3-CH2-COO-
Ka 
[NH
Zwitterion formation
-
2



- CH 2 - COO ]  [ H 3 O ]
Kb 

[ NH 3 - CH 2 - COOH ]  [ OH ]

[ NH 3 - CH 2 - COO ]
In iso electric
-
[ NH 3 - CH 2 - COO ]
-
point  [NH

- CH 2 - COO ]  [ NH 3 - CH 2 - COOH ]
-
2

Ka



[ H 3O ]
Kb
[ OH


[ H 3O ] 
]

[ H 3O ] 
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Ka  Kw
K a  [ OH ]

[ H 3O ] 
K a  [ H 3O ]  K w
Kb

[ H 3O ] 
Kb
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Kb
Ka  Kw
Kb
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2-Determining iso electric point for amino acids
Ka2=9.87
pKa1=2.35
+NH -CH -COOH +NH -CH -COO3
2
3
2
pH 
 NH2-CH2-COO-
pK a 1  pK a 2
2
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Method1=method2
OH

 Kb

Ka

 H 2 A    HA  
 A  H

H 3O 
Ka

Kw
Kb

 Kb2
 K b1
a1
a2
H 2 A  K  HA  

A
K

H 3O 
Ka 1  Ka 2 
pKa 1  pKa 2
2
Ka=Ka2,,,,,,,,,,,,Kb=Kb2

H 3O 
Ka
Kw
Kb



Ka 1  Ka 2
Kw
Kb

Kw
K b2
 K a1
Formol titration
For simple amino acids, Ka and Kb are generally so small that their
quantitative determination by neutralization titrations is impossible.
+NH -CH -COO3
2
+NH -CH -COO3
2
+ OH-  Product
+ HCOH  CH2=NCH2COOH
Amino acids that contain more than one carboxyl or amine group can
sometimes be determined.
If the Ka values are different enough (104 or more), stepwise end points can
be obtained just like other polyfunctional acids or bases as long as the Ka
values
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