Buffers

 Use AP Review Drills # 60-62

 iWBAT
 set up and solve buffer calculations.
 Explain what is required for a buffer situation.

 A solution that resists a change in pH.
 Buffers are:
– A solution that contains a weak acidweak base conjugate pair.
– Often prepared by mixing a weak acid, or a weak base, with a salt of that acid or base.

 The solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base.
 We can make a buffer of any pH by varying the concentrations of these solutions.

 Buffers resist changes in pH because they contain both an acidic species to neutralize OH -1 ions and a basic one to neutralize H +1 ions.
 It is important that the acidic and basic species of the buffer do not consume each other through a neutralization reaction.


HA H
+
+ A
-
 K a
= [H
+
] [A
-
]
[HA]
 Buffers most effectively resist a change in pH in either direction when the concentrations of
HA and A are about the same .
When [HA] equals [A
-
] then [H
+
] equals K a
.
 Scientists usually try to select a buffer whose acid has a pK a close to the desired pH .

 Ka = [H + ] [A ]
[HA]
 so [H + ] = K a
[HA]
[A ]
 The [H + ] depends on the ratio [HA]/[A ]
 Take the negative log of both sides


 pH = -log(K a
[HA]/[A ]) pH = -log(K a
)-log([HA]/[A ]) pH = pK a
+ log([A ]/[HA])

Or Use the
Henderson-Hasselbalch Equation Instead
[base] pH = pK a
+ log [acid]
[ 0.10] pH = -log (1.4 x 10 -4 ) + log [0.12] pH = 3.85 + (-0.08) = 3.77

This is called the
Henderson-Hasselbalch equation

 pH = pK a
+ log([A ]/[HA]) pH = pK a
+ log(base/acid)
 Or
 pOH = pK b
+ log([B + ]/[BOH])

 There are 3 classes of Buffer problems.
 Take a look at the Acid-Base Buffer problems packet.

Look at each “Class” of buffer problem.
 What is the difference between each
“Class”?

C
E
What is the pH of a buffer that is 0.12M in lactic acid,
HC
3
H
5
O
3
, and 0.10M in sodium lactate?
For lactic acid, K = 1.4 x 10 -4
I
HC
3
H
5
O
3
0.12 M
↔ H +
0
+ C
3
H
5
O
3
-1
0.10 M
-x
0.12 – x
+x
+x
+x
0.10 + x

[ H +1 ] [C
3
H
5
O
3
-1 ]
Ka = [HC
3
H
5
O
3
]
1.4 x 10 -4 = (x) (0.10 – x)
(0.12 – x)
Because of the small K a and the presence of the common ion, it is expected that x will be small relative to 0.12 or 0.10M.
x = 1.7 x 10 -4 M therefore [H +1 ] = 1.7 x 10 -4 M pH = -log (1.7 x 10 -4 M) = 3.77

Or Use the
Henderson-Hasselbalch Equation Instead
[base] pH = pK a
+ log [acid]
[ 0.10] pH = -log (1.4 x 10 -4 ) + log [0.12] pH = 3.85 + (-0.08) = 3.77

Practice Problem
Calculate the pH of a buffer composed of
0.12M benzoic acid (HC
7
H
5
O
2
) and 0.20M sodium benzoate.
K a
= 6.3 x 10 -5 ans: 4.42
Remember the Aqueous Equilibrium Constants are located in Appendix D of your textbook.

 In your textbook, try

More Practice
 Try AP Solutions problems
Acids and Bases IV
 # 4, 5, 6

Website to Look At
 http://www.chembuddy.com/?left=pHcalculation&right=pH-buffers-hendersonhasselbalch

Wrap Up
 Explain what conditions are required to create a buffer situation.