Lecture 11
Solving buffer
s
and neutralization problems
Deriving the Buffer Equation
You have had it hammered into your head that there are three simplified equations that can be used to calculate proton or
hydroxide concentration in acidbase equilibria:
!
the strong acid and base equation
!
the weak acid and base equation
!
the buffer equation
You have seen the strong and weak cases, now it is time to derive the buffer equation.
But first, what is a buffer? Very simply, a buffer is the mixture of
a weak acid and conjugate base
or
a weak base and conjugate acid
Recall that the word
conjugate
means that the pair differs by a single proton. For example:
acid
base
Buffer?
Example 1
Acetic acid
HC
2
H
3
O
2
Sodium acetate
NaC
2
H
3
O
2
Yes, weak acid and base differ by one proton
Example 2
Ammonium ion
NH
4
+
ammonia
NH
3
Yes, weak acid and base differ by one proton
Example 3
Hydrochloric acid
HCl
Chloride
Cl

No, strong acids and their cases can’t be buffers
Example 4
Carbonic acid
H
2
CO
3
Carbonate ion
CO
3
=
No, the acid and base differ by two protons
H+=Ka(Ca/Cb)
OH=Kb(Cb/Ca)
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Derivation of the buffer equation using the RICE expression
We start with the general case of a acid equilibrium
HA
!
H
+
+ A

with
K
a
= [H
+
] [A

]/[HA]
and then construct a RICE expression with the following steps:
1.
Start with C
a
for the amount weak acid HA
and
C
b
for the amount of the conjugate weak base
2.
We assume 0M is the amount of strong acid,
H
+
, to start
3.
Since Q< K the reaction shifts right so we add signs to the change array accordingly
4.
Assume that x is the concentration of protons that forms
5.
Doing the math, at equilibrium the concentrations are C
a
x, x and C
b
+x respectively
Reaction: HA
!
H+
+
A
____ ___
Substituting these amounts into the equilibrium expression
we get:
K
a
= [x] [C
b
+ x]/[ C
a
x]
But now we make an important approximation.
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 Spring '07
 Holcombe
 Chemistry, Proton, Neutralization, pH, Weak acid, kA

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