P. 306
5.2: The Chemistry of Buffers
P. 306
5.2: The Chemistry of Buffers
P. 306
5.2: The Chemistry of Buffers
Adding NaCH3COO ill increase [Na+] and [CH3COO-]…..will causing to shift to
opposite side….the CH3COOH(aq) side.
The Common Ion Effect.
If the equilibrium shifts away from the ions….the H3O+ goes down, therefore pH goes
down and percent ionization goes down.
The two main species are CH3COOH and CH3COO-.
P. 306
The Definition of Buffers
If the pH of your blood deviated by more than 0.1 pH your
body processes would not function properly….you could die.
Buffer:
P. 306
The Components of a Buffer
A buffer is able to compensate and counteract the affects of both
adding acid or adding base.
Example: H3PO4(aq) +
H2PO4 (aq)
By doing this the pH of the solution remains relatively
constant….NOT always 7.00 but CONSTANT TO WHAT IT WAS
ORIGINALLY.
P. 306
How it works to Keep constant pH
H3PO4(aq) + H2PO4-(aq)
If you take in something basic…like drinking coffee, the system cancels out the OH-(aq) by
doing the following:
H3PO4(aq) + OH-(aq)  H2PO4-(ag) + H2O(l)
If you take in something acidic…like an orange, the system cancels out the H3O+(aq) by doing
the following:
H2PO4 (aq) +
+
H3O (aq) 
H3PO4(ag) + H2O(l)
P. 306
The Components of a Buffer
CH3COO (aq) +
CH3COOH(aq)
Because it original pH is around 4.6
So that the buffer can counteract BOTH acid and base additions equally well.
slightly
Decrease
slightly
Increase
Increase slightly
P. 306
The Components of a Buffer
A buffer is able to compensate and counteract the affects of both
adding acid or adding base.
Depending on the natural pH of the solution being buffered you can
have two situations:
1. Acidic Buffers. (buffer in the acid region (pH = 0-6.999999)
2. Basic Buffers. (Buffer in the basic region (pH = 7.000001 - 14
Adding H3O+(aq) to Buffer.
CH3COO (aq) +
Acidic Buffer
CH3COOH(aq)
pH = 4.74
pH = 4.66
CH3COO-(aq) + H3O+(aq)  CH3COOH(ag) + H2O(l)
Adding OH-(aq) to Buffer.
CH3COO (aq) +
Acidic Buffer
CH3COOH(aq)
pH = 4.74
pH = 4.83
CH3COOH(aq) + OH-(aq)  CH3COO-(ag) + H2O(l)
Adding H3O+(aq) or OH-(aq) to Buffer.
Summary
Convert CH3COOH to CH3COO-
Acidic Buffer
Convert CH3COO- to CH3COOH
pH = 4.83
pH = 4.74
pH = 4.74
pH = 4.66
Pg 312: Practice Problem 5.2.1
#’s
1, 2, 3
Adding H3O+(aq) to Buffer.
NH3 (aq) +
Basic Buffer
+
NH4 (aq)
pH = 9.25
pH = 9.17
NH3(aq) + H3O+(aq)  NH4+(aq) + H2O(l)
Adding OH-(aq) to Buffer.
NH3 (aq) +
Basic Buffer
+
NH4 (aq)
pH = 9.25
pH = 9.34
NH4+(aq) + OH-(aq)  NH3(aq) + H2O(l)
Adding H3O+(aq) or OH-(aq) to Buffer.
Summary
Convert NH4+ to NH3
Basic Buffer
Convert NH3 to NH4+
pH = 9.34
pH = 9.25
pH = 9.25
pH = 9.17
If H2N4 is the weak base then its conjugate acid would be N2H5+.
So add any compound with N2H5+……like N2H5Cl or N2H5F.
Pg 316: Practice Problem 5.2.2
#’s
1, 2
Henderson Hasselbach
Useful Buffer Equation
 An American and a Danish Scientist 
Lawrence Joseph
Henderson
1878 -1942
Together they derived a formula that
related the final pH of a solution to the
pKa and HA:A- ratio of the components of
a buffer solution.
Karl Albert
Hasselbach
1874 -1962
The Capacity of a Buffer
The ability of a buffer depends on the ratio between the acidic form
and it conjugate base being 1:1 or a close to it as possible.
The amount of H3O+ or OH- a buffer can neutralize is called:
The higher the concentration of these conjugate pairs the more
H3O+ or OH- they can neutralize before the get “over powered”.
Making a Buffer
The ability of a buffer to neutralize H3O+ or OH- depends on its ratio
between the [HA] and [A-].
= 10
= 0.1
In order for a buffer to function and keep the pH constant..ish. It
needs to keep the [HA] :[A-] ratio between 10 and 0.1.
More About Buffer Choice
When choosing a buffer to use you need to remember that the
buffer itself has a natural pH and a good buffer will keep the pH as
close to that as possible.
Therefore the best choice for a buffer is to use
one that has a pKa closest to the desired pH.
pKa = -log(Ka)
The weak acid with a pKa closest to 5.00 is CH3COOH(aq) with a
Ka of 1.8 x 10-5.
pKa = -log(Ka)
pKa = -log(1.8 x 10-5)
pKa = 4.744727495
To Find out the concentrations of CH3COOH(aq) and CH3COO-(aq)
we use Henderson-Hasselbach equation.
The opposite of a log
is an exponent of
base 10.
Ratio of
1.8 : 1 is
needed.
Pg 319: Practice Problem 5.2.3
#’s
1, 2, 3
Blood pH Buffering
Example
Equilibrium system of how your blood carries oxygen to
your body cells.
7.2
Optimal pH
7.35
Acidosis
(NOT GOOD)
Might Die
7.5
Alkalosis
(NOT GOOD)
Might Die
HPO42-(aq) + H3O+(aq)  H2PO4-(ag) + H2O(l)
H2PO42-(aq) + OH-(aq)  HPO42-(ag) + H2O(l)
Hyperventilating causes the CO2 to drop which would cause the pH to rise
and you would/could suffer alkalosis.
Breathing into a paper bag means you reinhale the CO2 you exhale which
causes the CO2 to rise and thus bring back down the pH to normal.
P. 323
Pg 323 – 325
#’s
1, 4, 5, 6, 7, 8, 9, 11, 12, 13