BUFFERS AND TITRATIONS
Dr. Harris
Ch 21
Suggested HW: Ch 21: 1, 13, 29
Buffers
• Solutions that contain both a weak acid and its conjugate base are called
buffers.
• Buffers have the unique ability to resist sharp changes in pH
• A buffer is able to do this because it contains both acid and base
components
• However, the acidic and basic components of the buffer themselves do
not neutralize each other, a special property fulfilled by mixing conjugate
pairs
• If an external acid or base is added to the system, it is instantly
neutralized by one of the conjugate species
Creating A Buffer
• For example, an HF/F- buffer can be created by mixing 0.1M HF with 0.1 M
NaF. Since NaF will fully dissociate, we ignore the neutral Na+. In solution,
you’d have:
𝑯𝑭 𝒂𝒒 + 𝑭− (𝒂𝒒)
𝑭− 𝒂𝒒 + 𝑯𝑭 𝒂𝒒
𝑵𝑶 𝑹𝑿𝑵
• Although HF is an acid, and F- is a base, the two CAN NOT neutralize each
other because the reaction has no preferred direction.
• Thus, by mixing an acid with its own conjugate base, you create a buffered
solution in which the components do not affect each other.
How Buffers Work
• The acidic component of the buffer neutralizes any base that is added to the
system. The basic component of the buffer neutralizes any added acid. The
resulting changes in pH are very small.
pH slightly
increases
Initial Buffer solution
Base
Added
(NaOH)
HF
pH slightly
decreases
F-
𝐻𝐹 𝑎𝑞 + 𝑶𝑯− 𝑎𝑞 →
𝐹 − 𝑎𝑞 + 𝐻2 𝑂(𝐿)
Acid
Added
(HCl)
HF
F-
HF
F-
𝐹 − 𝑎𝑞 + 𝑯𝟑 𝑶+ 𝑎𝑞 →
𝐻𝐹 𝑎𝑞 + 𝐻2 𝑂 (𝐿)
C.I.R.L.: Blood as a Buffer
• Blood is the best example of a buffer in everyday life.
• The pH of the human body is 7.4. If the pH of the body becomes to
acidic or basic, certain proteins and enzymes are chemically altered,
rendering them inactive
• If the body’s pH drops below 6.5 or above 8.0, death may occur.
• Human blood acts as the body’s buffer to prevent pH changes
Blood as a Buffer
• O2 is carried throughout the body by binding
with the hemoglobin protein found in red
blood cells.
• Hemoglobin, Hb, binds both H+ and O2. These
binding processes are represented below
𝐻𝑏𝐻 + + 𝑂2
𝐻𝑏𝑂2 + 𝐻+
ΔH > 0
• As expected by LeChatlier’s principle, when
blood reaches oxygen-poor tissue, the reaction
shifts left, releasing O2 in that region.
O
O
Blood As A Buffer (ex. exercising)
Need energy fast.
Must burn more
carbs.
Exercise
Breathe faster.
Bring in more O2
Combustion of carbs
yields CO2 and H+ in
the muscles
Heart rate increases to
transport extra O2
Not enough O2? Anaerobic
breakdown of glucose yields
lactic acid
𝐶6 𝐻12 𝑂6 → 𝐶𝐻3
𝐶𝑂𝐶𝑂𝑂−
+ 𝐴𝑇𝑃 + 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠
anaerobic condition
2𝐶𝐻3 𝐶𝐻𝑂𝐻𝐶𝑂𝑂𝐻 (𝐿𝑎𝑐𝑡𝑖𝑐 𝑎𝑐𝑖𝑑)
in O2
𝐶𝑂2 + 𝐴𝑇𝑃 + 𝑁𝑎𝐷𝐻
Concentration Gradients Shift Equilibria
Muscle
H+
CO2
Blood
O2
Blood as a Buffer
• The primary buffer systems in human blood are a carbonic acid/bicarbonate
buffer system and the HbH+/HbO2 system
𝐻 + 𝑎𝑞 + 𝑯𝑪𝑶−
𝟑 𝑎𝑔
𝑯𝒃𝑯+ + 𝑂2
𝑯𝟐 𝑪𝑶𝟑 𝑎𝑞
𝐻2 𝑂 𝐿 + 𝐶𝑂2 (𝑔)
𝑯𝒃𝑂2 + 𝐻+
• A sudden decrease in pH triggers the brain to increase the breathing rate,
releasing CO2 faster and shifting the equilibrium right to consume H+.
• The body consumes more O2 which shifts the hemoglobin binding equilibrium
left, removing more H+.
• Other acids are neutralized by bicarbonate, HCO3-. Blood pH remains intact.
Calculating the pH of a Buffer Using the HendersonHasselbalch Equation
• The Henderson-Hasselbalch equation is used to determine the pH of a
buffer system.
𝒑𝑯 = 𝒑𝑲𝒂 + 𝒍𝒐𝒈
[𝒃𝒂𝒔𝒆]
[𝒂𝒄𝒊𝒅]
• Example: Calculate the pH of a formic acid buffer (HCOOH) that is .25M
HCOOH and .15 M of the conjugate base (HCOO-), given that the Ka of
formic acid is 1.8 x 10-4
−4
𝑝𝐻 = − log 1.8 x 10
𝑝𝐻 = 3.53
.15
+ log
.25
Calculating the pH of a Buffer Solution After Exposure to an
Acid/Base
• Let’s calculate the pH of a buffer solution following a response to a pH
disturbance.
• It is important to know that a molecule of acid completely neutralizes a
molecule of base, forming the conjugate base and conjugate acids
base
acid
neutralized!!
• So, to determine changes in concentration of either the acidic or basic
component of a buffer, simply subtract the number of moles that are
neutralized from the active component, and add those to the inactive one
Example
• Ex. You have a 100 mL of a buffer that is 0.10 M in CH3COOH and 0.10 M in
NaCH3COO at a pH of 4.74. If 10 mL of 0.10 M HCl is added to the buffer
solution, what is the new pH? What is the change in pH?
𝑪𝑯𝟑 𝑪𝑶𝑶𝑯 𝑎𝑞 + 𝐻2 𝑂 (𝐿)
𝑪𝑯𝟑 𝑪𝑶𝑶− 𝑎𝑞 + 𝐻3 𝑂+ 𝑎𝑞
𝐾𝑎 = 1.8 𝑥 10−5
• Since HCl is an acid, it will only react with the basic component of the
buffer, CH3COO• There are exactly .001 moles of HCl added to the buffer, so exactly
.001 moles of CH3COO- will be consumed, forming exactly .001 moles
of additional CH3COOH
Example, contd.
Before addition of 10 mL of 0.10 M HCl
.01 mol
.01 mol
After
.011 mol
.009 mol
CH3COOH CH3COO-
CH3COOH CH3COO-
100 mL
110 mL
.001 mol HCl
Example, contd.
• The new concentrations are:
𝐶𝐻3 𝐶𝑂𝑂𝐻 =
.011 𝑚𝑜𝑙
=. 𝟏𝟎 𝑴
.110 𝐿
𝐶𝐻3 𝐶𝑂𝑂− =
.009 𝑚𝑜𝑙
=. 𝟎𝟖𝟏𝟖 𝑴
.110 𝐿
𝑝𝐻 = 𝑝𝐾𝑎 + 𝑙𝑜𝑔
= 4.65
[𝑏𝑎𝑠𝑒]𝑜
. 𝟎𝟖𝟏𝟖
= 4.74 + log
[𝑎𝑐𝑖𝑑]𝑜
. 𝟏𝟎
Buffer works well. This is
a very small drop in pH
considering that HCl is a
strong acid.
ΔpH = -.09
* Since both the acid and base are in the same volume of solution, you
may substitute moles for concentrations in the equation. The ratio is the
same.
Group Example
• You have 100 mL of a buffer that is 0.3M NH3 and 0.45 M NH4Cl. The pH of
the buffer is 9.07. Find the pKa. Calculate the change in pH after the
addition of 5 mL of 4.0 M NaOH?
Group Example
• You have 150 mL of a 1M HCl solution. You add 150 mL of a 1.2 M NaOH
solution. What will the pH be? (This is NOT a buffered solution)