Transcript Chapter 10: Polyprotic Acids & Bases

Polyprotic Acid-Base Equilibria
Introduction
1.) Polyprotic systems


Acid or bases that can donate or accept more than one proton
Proteins are a common example of a polyprotic system
-

why activity of proteins are pH dependent
Polymer of amino acids
-
Some amino acids have acidic or basic substituents
Polyprotic Acid-Base Equilibria
Introduction
1.) Polyprotic systems

Amino acids
(basic)
(acidic)
-
Carboxyl group is stronger acid of ammonium group
R is different group for each amino acid
Overall charge is still neutral

Amino acids are zwitterion – molecule with both positive and negative charge
-
At low pH, both ammonium and carboxy group are protonated
At high pH, neither group is protonated
Stabilized by interaction with solvent
Polyprotic Acid-Base Equilibria
Diprotic Acids and Bases
2.) Multiple Equilibriums

Illustration with amino acid leucine (HL)
high pH
low pH
Carboxyl group
Loses H+

ammonium group
Loses H+
Equilibrium reactions
Diprotic acid:
K a1  K1
K a2  K 2
Polyprotic Acid-Base Equilibria
Diprotic Acids and Bases
2.) Multiple Equilibriums

Equilibrium reactions
Diprotic base:
K b1
K b2
Relationship between Ka and Kb
K a1  K b2  Kw
K a2  K b1  Kw
pKa of carboxy
and ammonium
group vary
depending on
substituents
Largest
variations
Polyprotic Acid-Base Equilibria
Diprotic Acids and Bases
3.) General Process to Determine pH

Three components to the process
high pH
low pH
Carboxyl group
Loses H+

Acid Form [H2L+]

Basic Form [L-]

Intermediate Form [HL]
ammonium group
Loses H+
Polyprotic Acid-Base Equilibria
Diprotic Acids and Bases
3.) General Process to Determine pH

Acid Form (H2L+)

Illustration with amino acid leucine
K1=4.70x10-3

K2=1.80x10-10
H2L+ is a weak acid and HL is a very weak acid
K1  K 2
Assume H2L+ behaves as a monoprotic acid
K a  K1
Polyprotic Acid-Base Equilibria
Diprotic Acids and Bases
3.) General Process to Determine pH

0.050 M leucine hydrochloride
K1=4.70x10-3
+ H+
H2L+
HL
H+
0.0500 - x
x
x
Determine [H+] from Ka:
K a  4.7  10
3
[HL][H  ]
x2
2




x

1
.
32
x
10
M

[
HL
]

[
H
]
Fx
[H2 L ]
Determine pH from [H+]:
pH   log[ H  ]   log( 1.23  10 2 M )  1.88
Determine [H2L+]:
[ H2 L ]  F  x  3.68  10 2 M
Polyprotic Acid-Base Equilibria
Diprotic Acids and Bases
3.) General Process to Determine pH

Acid Form (H2L+)
What is the concentration of L- in the solution?
[L-] is very small, but non-zero. Calculate from Ka2
K [HL]
[H  ][L ]
K a2 
 [L ]  a2 
[HL]
[H ]

[L ] 
( 1.80  10 -10 )( 1.32  10 - 2 )
( 1.32  10 - 2 )
 1.80  10 -10 ( K a2 )
Approximation [H+] ≈ [HL], reduces Ka2 equation to [L-]=Ka2
[ L ]  1.80  10 10  1.32  10 2  [ HL ]
Validates assumption
Polyprotic Acid-Base Equilibria
Diprotic Acids and Bases
3.) General Process to Determine pH

For most diprotic acids, K1 >> K2
-

Even if K1 is just 10x larger than K2
-

Assumption that diprotic acid behaves as monoprotic is valid
Ka ≈ Ka1
Error in pH is only 4% or 0.01 pH units
Basic Form (L-)
K b1  Kw / K a2  5.55  10 5
K b2  Kw / K a2  2.13  10 12

L- is a weak base and HL is an extremely weak base
K b1  K b2
Assume L- behaves as a monoprotic base
K b  K b1
Polyprotic Acid-Base Equilibria
Diprotic Acids and Bases
3.) General Process to Determine pH

0.050 M leucine salt (sodium leucinate)
L-
HL
0.0500 - x
x
OHx
Determine [OH-] from Kb:
K b  5.55  10
5

[HL][OH - ]
[L- ]
x2

 x  1.64 x10  3 M  [ HL ]  [ OH  ]
Fx
Determine pH and [H+] from Kw:
[H ] 
Kw
1  10
 
[ OH ]
14
12

6
.
10

10
M  pH  11.21
1.64  10 3
Determine [L-]:
[ L ]  F  x  4.84  10 2 M
Polyprotic Acid-Base Equilibria
Diprotic Acids and Bases
3.) General Process to Determine pH

Basic Form (L-)
What is the concentration of H2L+ in the solution?
[H2L+] is very small, but non-zero. Calculate from Kb2
[H2 L ][OH  ] [H2 L ]x
K b2 

 [H2 L ]
[HL]
x
[ H2 L ]  2.13  10 12  1.64  10 3  [ HL ]
Validates assumption [OH-] ≈ [HL],
Fully basic form of a diprotic acid can be treated as a monobasic, Kb=Kb1
Polyprotic Acid-Base Equilibria
Diprotic Acids and Bases
3.) General Process to Determine pH

Intermediate Form (HL)
-
More complicated HL is both an acid and base
K a  K a2  1.80  10 10
K b  K b2  2.13  10 12

Amphiprotic – can both donate and accept a proton

Since Ka > Kb, expect solution to be acidic
-

Can not ignore base equilibrium
Need to use Systematic Treatment of Equilibrium
Polyprotic Acid-Base Equilibria
Diprotic Acids and Bases
3.) General Process to Determine pH

Intermediate Form (HL)
Step 1: Pertinent reactions:
K1
K b2
K2
K b1
Step 2: Charge Balance:

[H ]  [H2 L ]  [L- ]  [OH - ]
Step 3: Mass Balance:
F  [ HL]  [ H 2 L ]  [ L- ]
Step 4: Equilibrium constant expression (one for each reaction):
[HL][H  ]
[H2 L ][OH - ] K
[L- ][H  ]
K1 
b2
K2 
K b1 

[HL]
[HL]
[H2 L ]

[HL][OH - ]
[L ]
Polyprotic Acid-Base Equilibria
Diprotic Acids and Bases
3.) General Process to Determine pH

Intermediate Form (HL)
Step 6: Solve:
Substitute Acid Equilibrium Equations into charge balance:

-

-
[H2 L ]  [L ]  [H ]  [OH ]  0
All Terms are
related to [H+]
[HL][H  ] [L- ]  [HL]K 2 [OH - ]  Kw
[H2 L ] 
[H  ]
[H  ]
K1

Kw
[HL][H  ] [HL]K 2



[
H
]

0


K1
[H ]
[H ]
Multiply by [H+]
[HL][H  ]2
 [HL]K 2  [H  ]2  Kw  0
K1
Polyprotic Acid-Base Equilibria
Diprotic Acids and Bases
3.) General Process to Determine pH

Intermediate Form (HL)
Step 6: Solve:
[HL][H  ]2
 [HL]K 2  [H  ]2  Kw  0
K1
Factor out [H+]2:
 2  [HL]

[H ] 
 1   K 2 [HL]  Kw
 K1

Rearrange:
[H  ] 2 
K 2 [HL]  Kw
[HL]
1
K1
Polyprotic Acid-Base Equilibria
Diprotic Acids and Bases
3.) General Process to Determine pH

Intermediate Form (HL)
Step 6: Solve:
[H  ] 2 
K 2 [HL]  Kw
[HL]
1
K1
Multiply by K1 and take square-root:
[H  ] 
K 2 K1 [HL]  K1Kw
K1  [HL]
Assume [HL]=F, minimal dissociation:
(K1 & K2 are small)
[H  ] 
K 2 K1F  K1Kw
K1  F
Polyprotic Acid-Base Equilibria
Diprotic Acids and Bases
3.) General Process to Determine pH
Intermediate Form (HL)

Step 6: Solve:
[H  ] 
K 2 K1F  K1Kw
K1  F
Calculate a pH:

[H ] 
( 4.70 x10  3 )( 1.80 x10 10 )( 0.0500 )  ( 4.70 x10  3 )( 1.0 x10 14 )
4.70 x10  3  0.0500
 8.80 x10 7 M  pH  6.06
Polyprotic Acid-Base Equilibria
Diprotic Acids and Bases
3.) General Process to Determine pH

Intermediate Form (HL)
Step 7: Validate Assumptions
Assume [HL]=F=0.0500M, minimal dissociation (K1 & K2 are small).
Calculate [L-] & [H2L+] from K1 & K2:
[HL][H  ] ( 0.0500 )( 8.80 x10 7 )
6
[H2 L ] 


9
.
36
x
10
K1
4.70 x10  3

-
[L ] 
[HL]K 2
[H  ]

( 0.0500 )( 1.80 x10 10 )
8.80 x10 7
[HL]=0.0500M >> 9.36x10-6 [H2L+] & 1.02x10-5 [L-]
 1.02 x10  5
Assumption Valid
Polyprotic Acid-Base Equilibria
Diprotic Acids and Bases
3.) General Process to Determine pH

Intermediate Form (HL)
Summary of results:


[L-] ≈ [H2L+]  two equilibriums proceed equally even though Ka>Kb
Nearly all leucine remained as HL
Solution
pH
[H+] (M)
[H2L+] (M)
[HL] (M)
[L-] (M)
Acid form
0.0500 M H2A
1.88
1.32x10-2
3.68x10-2
1.32x10-2
1.80x10-10
Intermediate form
0.0500 M HA-
6.06
8.80x10-7
9.36x10-6
5.00x10-2
1.02x10-5
Basic form
0.0500 M HA2-
11.21
6.08x10-12 2.13x10-12
1.64x10-3
4.84x10-2

Range of pHs and concentrations for three different forms
Polyprotic Acid-Base Equilibria
Diprotic Acids and Bases
3.) General Process to Determine pH

Simplified Calculation for the Intermediate Form (HL)
[H  ] 
K 2 K1F  K1Kw
K1  F
Assume K2F >> Kw:
[H  ] 
K 2 K1F
K1  F
Assume K1<< F:
[H  ] 
K 2 K1F
F
Polyprotic Acid-Base Equilibria
Diprotic Acids and Bases
3.) General Process to Determine pH

Simplified Calculation for the Intermediate Form (HL)
[H  ] 
K 2 K1F
F
Cancel F:
[H  ]  K2 K1
Take the -log:
- log[H  ]  2 (  log K1  log K2 )
1
Independent of concentration: pH  2 ( pK1  pK2 )
1
pH of intermediate form of a
diprotic acid is close to midway
between pK1 and pK2
Polyprotic Acid-Base Equilibria
Diprotic Buffers
1.) Same Approach as Monoprotic Buffer

Write two Henderson-Hasselbalch equations
 [HA ] 

pH  pK1  log 
 [H2 A] 


 [ A2  ] 

pH  pK 2  log 
 [HA ] 



Both Equations are always true

Solution only has one pH

Choice of equation is based on what is known
-
[H2A] and [HA-] known use pK1 equation
[HA-] and [A2-] known use pK2 equation
Polyprotic Acid-Base Equilibria
Diprotic Buffers
1.) Example 1:

How many grams of Na2CO3 (FM 105.99) should be mixed with 5.00 g of
NaHCO3 (FM 84.01) to produce 100 mL of buffer with pH 10.00?
FM = 62.03
FM = 84.01
pK a1  6.351
FM = 105.99
pK a 2  10.329
Polyprotic Acid-Base Equilibria
Diprotic Buffers
1.) Example 2:

How many milliliters of 0.202 M NaOH should be added to 25.0 mL of 0.0233 M
of salicylic acid (2-hydroxybenzoic acid) to adjust the pH to 3.50?
pK a1  2.972
pK a 2  13.7
Polyprotic Acid-Base Equilibria
Polyprotic Acids and Bases
1.) Extend Treatment of Diprotic Acids and Bases to Polyprotic Systems

Equilibria for triprotic system
Acid equilibria:
Base equilibria:

For a polyprotic system, would simply contain n such equilibria
Polyprotic Acid-Base Equilibria
Polyprotic Acids and Bases
1.) Extend Treatment of Diprotic Acids and Bases to Polyprotic Systems

Rules for triprotic system
1.
2.
H3A is treated as a monoprotic acid, Ka = K1
H2A- is treated similarly as an intermediate form of a diprotic acid
[H  ] 
3.
HA2- is also treated similarly as an intermediate form of a diprotic acid
a.
b.
Surrounded by H2A- and A3Use K2 & K3, instead of K1 & K2
[H  ] 
4.
K 2 K1F  K1Kw
K1  F
K 2 K 3 F  K 2 Kw
K2  F
A3- is treated as monobasic, with Kb=Kb1=Kw/Ka3
Polyprotic Acid-Base Equilibria
Polyprotic Acids and Bases
1.) Extend Treatment of Diprotic Acids and Bases to Polyprotic Systems

Rules for triprotic system
Treat as
Monoprotic
Treat asacid:
Intermediate Forms
Treat
Treatas
as
Intermediate
Monoprotic Forms
base:
For more complex system, just have additional intermediate forms
K s that “bracket”
or contain
form,
“End Forms” Use
of Equilibria
that Bracket
Reactions
are KTreated
1 & K3
2 as Monoprotic
2
in-between the atwo monoprotic acid and base forms
at “ends”
Polyprotic Acid-Base Equilibria
Polyprotic Acids and Bases
2.) Example

How many milliters of 1.00 M KOH should be added to 100 mL of solution
containing 10.0 g of histidine hydrochloride (His.HCl FM 191.62) to get a pH
of 9.30?.
Polyprotic Acid-Base Equilibria
Polyprotic Acids and Bases
3.) Which is the Principal Species?

Depends on the pH of the Sample and the pKa values
-

At pKa, 1:1 mixture of HA and AFor monoprotic, A- is predominant when pH > pKa
For monoprotic, HA is predominant when pH < pKa
Similar for polyprotic, but several pKa values
Triprotic acid
Diprotic acid
pH
Major
Species
pH < pK1
H2A
pK1 < pH < pK2
HA-
pH > pK2
A2-
Determine Principal Species by Comparing the pH of the
Solution with the pKa Values
Polyprotic Acid-Base Equilibria
Polyprotic Acids and Bases
4.) Fractional Composition Equations


Fraction of Each Species at a Given pH
Useful for:
-

Acid-base titrations
EDTA titrations
Electrochemical equilibria
Combine Mass Balance and Equilibrium Constant
-
[H  ][ A  ]
Ka 
[HA]
-
F  [HA ]  [ A ]  [ A ]  F  [HA ]
[H  ](F-[HA])
Ka 
[HA]
Rearrange:
[HA] 
[H  ]F
[H  ]  K a
Polyprotic Acid-Base Equilibria
Polyprotic Acids and Bases
4.) Fractional Composition Equations

Combine Mass Balance and Equilibrium Constant
Recall: fraction of molecule in the form HA is:
[HA] 
 HA 
[H  ]F
[H  ]  K a
Divide by F:
[HA]
[H  ]
 
Fraction in the form HA:  HA 
F
[H ]  K a
Fraction in the form A-:
K
[ A ]
 A 
  a
F
[H ]  K a
[HA]
[HA]

F
[HA]  [ A ]
Polyprotic Acid-Base Equilibria
Polyprotic Acids and Bases
4.) Fractional Composition Equations


Diprotic Systems
Follows same process as monoprotic systems
Fraction in the form H2A:
[H2 A]
[H  ]2
 H2 A 
  2
F
[H ]  [H  ]K1  K1K 2
Fraction in the form HA-:
K1 [H  ]
[HA ]
 HA 
  2
F
[H ]  [H  ]K1  K1K 2
Fraction in the form A2-:
 A2 
K1K 2
[ A2  ]

  2
F
[H ]  [H  ]K1  K1K 2
Polyprotic Acid-Base Equilibria
Isoelectric and Isoionic pH
1.) Isoionic point – is the pH obtained when the pure, neutral polyprotic acid
HA is dissolved in water


Neutral zwitterion
Only ions are H2A+, A-, H+ and OHConcentrations are not equal to each other
Isoionic point: [H  ] 
K1K 2 F  K1Kw
K1  F
pH obtained by simply
dissolving alanine
Remember: Net Charge of Solution is Always Zero!
Polyprotic Acid-Base Equilibria
Isoelectric and Isoionic pH
2.) Isoelectric point – is the pH at which the average charge of the polyprotic
acid is 0

pH at which [H2A+] = [A-]
-
Always some A- and H2A+ in equilibrium with HA

Most of molecule is in uncharged HA form

To go from isoionic point (all HA) to isoelectric point, add acid to decrease [A-]
and increase [H2A+] until equal
-
pK1 < pK2  isoionic point is acidic excess [A-]
Remember: Net Charge of Solution is Always Zero!
Polyprotic Acid-Base Equilibria
Isoelectric and Isoionic pH
2.) Isoelectric point – is the pH at which the average charge of the polyprotic
acid is 0

isoelectric point: [A-] = [H2A+]
[HA][H  ]
[H2 A ] 
K1

[ A- ] 
K 2 [HA]
[H  ]
[HA][H  ] K 2 [HA]



[
H
]  K1K 2

K1
[H ]
Isoelectric point: pH  2 ( pK1  pK2 )
1
Polyprotic Acid-Base Equilibria
Isoelectric and Isoionic pH
3.) Example:

Determine isoelectric and isoionic pH for 0.10 M alanine.
Solution:
For isoionic point:

[H ] 
K1K 2 F  K1Kw
( 10  2.34 )( 10  9.87 )( 0.10 )  ( 10  2.34 )( 1  10 14 )

K1  F
10  2.34  0.10
 7.7  10 7 M  pH  6.11
Polyprotic Acid-Base Equilibria
Isoelectric and Isoionic pH
3.) Example:

Determine isoelectric and isoionic pH for 0.10 M alanine.
Solution:
For isoelectric point:
pH  2 ( pK1  pK2 )  2 ( 2.34  9.87 )  6.10
1
1
Isoelectric and isoionic points for polyprotic acid are almost the same