10/23/2013
Polyprotic Acid/Base Equilibria
General considerations are the same as monoprotic acids/bases:
H2A = H+ + HA-
K a1 
HA- = H+ + A-
K a2 
[H ][HA  ]
[H2 A ]
[H  ][ A  ]
[HA  ]
Lets look at this one species at a time, we'll use sulfurous acid (H2SO3)
as a model compound:
H2SO3 =
H+
+ HSO3
HSO3- = H+ + SO32-
-
[H  ][HSO 3 ]
K a1 
 1.23 x10 2
[H2 SO 3 ]
K a2 
[H ][SO 23 ]
[HSO 3 ]
 6.6 x10 8
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Polyprotic Acid/Base Equilibria…Chemistry!
K a1 
[H  ][HSO 3 ]
[H2 SO 3 ]
 1.23 x10 2
K a2 
[H  ][SO 23 ]
[HSO 3 ]
 6.6 x10 8
What is the pH of a solution prepared by dissolving 0.10 mol H2SO3 in
1.00 L water?
• What does the chemistry tell you?
• Look at the Ka's, H2SO3 is a much stronger acid than HSO3-, what
does this mean??
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10/23/2013
Polyprotic Acid/Base Equilibria…Chemistry!
What is the pH of a solution prepared by dissolving 0.10 mol Na2SO3 in
1.00 L water?
• SO32- is the dibasic (fully deprotonated) form of this weak acid, so
let's look at Kb's
SO32- + H2O = OH- + HSO3HSO3- + H2O = OH- + H2SO3
K b1 
K b2 
Kw
 1.51x10 7
K a2
Kw
 8.13 x10 13
K a1
• Since SO32- is a much stronger base than HSO3-, we can solve a
monoprotic base problem:
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Amphiprotic Species
What is the pH of a solution prepared by dissolving 0.10 mol NaHSO3
in 1.00 L water?
• HSO3- is the amphiprotic form of this weak acid, the problem is a
little tougher.
[H  ][SO 23 ]
 6.6 x10 8
K a2 
+
2HSO3 = H + SO3
[HSO 3 ]
HSO3- + H2O = OH- + H2SO3
K b2 
Kw
 8.13 x10 13
K a1
• How do we approach this?
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10/23/2013
Amphiprotic Species
We need to take a more systematic approach! One starting point is a
mass balance relationship called the “proton condition”
proton condition: The concentration of H+ is a result of the difference
in concentration of the species that liberate H+ minus the
concentrations of species that consume H+
Lets look at the proton condition for our system:
Consume H+
Liberate H+
HSO3- + H+ = H2SO3
HSO3- = SO32- + H+
H2O = H+ + OH-
So, the proton condition is:
[H+] = [SO32-] + [OH-] – [H2SO3]
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Amphiprotic Species
[H+] = [SO32-] + [OH-] – [H2SO3]
• We can substitute expressions for [SO32-], [OH-] and [H2SO3] from
the appropriate equilibrium constant expressions to arrive at an
expression in terms of [H+] and [HSO3-].

H  HSO 3 
Kw
3 
H    K a 2 HSO


K a1
H  
H  
• With the help of the god’s of algebra we can come to this:
H   


K a 1 K a 2 HSO 3  K a 1 K w

K a 1  HSO 3

• We still can’t solve this completely (we don’t know [HSO3-])….
What DO we know?
REMEMBER THIS IS CHEMISTRY!
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10/23/2013
Chemistry Can Be Helpful!
1. We know that HSO3- is acting as both an acid and a base in this
case
2. Every time an SO32- is formed, a H+ is liberated. Every time a
H2SO3 is formed OH- is produced.
3. The H+ and OH- can react to reform the original HSO34. Therefore [HSO3-] will not vary much from original concentration, F.
So, working under this assumption, the equation becomes:
H   
K a 1K a 2 F  K a 1K w
K a1 F
We know all of these values, if we plug and chug, we discover that
[H+] = 2.69 x 10-5M, with pH = 4.57
Can we make things easier? Maybe!
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Chemistry Can Be (more) Helpful!
H   
K a 1K a 2 F  K a 1K w
K a1 F
Two considerations:
1. If Ka2F >> Kw, the numerator in the fraction becomes Ka1Ka2F
2. If Ka1 << F, the denominator becomes F
The result of these assumptions is a much simpler equation:
H   
OR:
pH 
K a 1K a 2 F
F

K a 1K a 2
1
 pK a 1  pK a 2 
2
Using this simplification, our problem becomes: pH  1.91  7.18  4.55
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10/23/2013
Handling Polyprotic Acids/Bases
(Read text carefully)
As long as Ka’s aren’t too close (~1000x)
1. Treat fully protonated acid as a weak monoprotic acid
Ka 
x2
Fx
2. Treat fully deprotonated acid as weak monobasic base
3. Treat intermediate forms by looking at the Ka’s for the surrounding
equilibria
H   
K ax K ax 1 F  K ax K w
K ax  F
Complete systematic approach will always work, but will require more
effort.
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Fraction of Dissociation
K a1 
H2A = H+ + HA-
[ H  ][ HA  ]
[H 2 A ]
HA- = H+ + A-
K a2 
[ H  ][ A  ]
[ HA  ]
Calculate H2A as a function of pH.
H2A =
How do we get things in terms of K's and [H+]? Look at equilibrium
expressions:
From Ka1;
[HA] =
From Ka2;
[A2-] =
Combine Terms:
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10/23/2013
Fraction of Dissociation
 H2 A 
H 2 A 
K a 1 [ H 2 A ] K a 1K a 2 [ H 2 A ]

H 2 A  
[H  ]
[H  ] 2
With some cancellation and manipulation:
 H2 A 
[H  ] 2
[ H  ] 2  K a 1 [ H  ]  K a 1K a 2
We can do the same treatment for HA-, and A2-.
 HA  
K a1 [H  ]
 2

[ H ]  K a 1 [ H ]  K a 1K a 2
 A 2 
K a 1K a 2
 2
[ H ]  K a 1 [ H  ]  K a 1K a 2
Is there a pattern emerging?
For triprotic acid, denominator becomes:
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