Buffers
What is a buffer solution?


Solution that resist pH change with the addition of small quantities of acid or base.
Composition: Weak acid and its salt or Weak base and its salt
examples: 1) CH3COOH (weak acid) and NaCH3COO (salt)
2) NH3 (weak base) and NH4Cl (salt)
How does a buffer solution work?
Example: CH3COOH and NaCH3COO
NaCH3COO + CH3COOH <----------> 2 CH3COO – +


Na + +
H+
When an acid is added to this solution (excess H+) the equilibrium shifts to the left to use up the excess H+
that has been added (allowing pH to remain constant)
When a base is added to this solution, (excess OH-) the OH- gets together with the H+ to make water; as the
H+ is used up the equilibrium shifts to the right (allowing pH to remain constant)
QUESTION:
Describe how the below buffer system (weak base and its salt) will resist changes in pH when: 1) an acid is added
and 2) a base is added.
NH3(aq) + NH4Cl(aq) + H2O(l) <----------> 2NH4+
weak base
salt
(aq)
+ Cl-
(aq)
+ OH-(aq)
Example of a biological Buffer systems, Hemoglobin (HHb) binding to oxygen,:
Blood pH requires to be ~ 7.35 for proper functioning.
HHb + H2O + O2 < === > H3O
+
+ HbO2
–

A reduction of pH of blood (pH < 7.2) causes a shift towards the left, so oxygen does not bind very
well to hemoglobin.

An increase of pH (pH > 7.5) causes a shift towards the right, so oxygen remains attached to
hemoglobin and will not release for cellular action.
To control the pH of blood, there are two other buffer systems:
CO2/ HCO3 – system:
CO2 (g) < ==== > CO2 (aq)
CO2 (aq) + 2 H2O < ==== > H3O+ + HCO3 –
 When CO2 is not removed from the blood (choking), the equilibrium shifts to the right, therefore
the pH of blood increases.
 When CO2 is expelled very fast from the blood (hyperventilating), the equilibrium shifts to the left
and the blood becomes alkaline (increase in pH).
H2PO4 -/HPO4
2-
system:
H2PO4 - + H2O < ==== > H3O

+
+ HPO4
2–
This is buffer system exists mostly at the cellular level, and stabilise the pH during the metabolic
processes of cells.
Note that if the buffer system did not exist, the simple act of drinking a glass of orange juice would
change the pH of blood enough to kill a human being.
4. Calculating the pH of buffers:
Buffer in the acidic range:
Calculate the pH of a buffer solution consisting of 0.10 M NaCH 3COO and 0.15 M CH3COOH. If the Ka of
CH3COOH is 1.8 x 10 –5, calculate the pH of this solution.
Buffer in the basic range:
A buffer solution consists of 0.10 M NH4Cl and 0.10 M NH 3. If the Kb for ammonia is 1.8 x 10
the pH of this solution.
, calculate
–5
3. Henderson-Hasselbach equation: used to determine the pH of a buffer
solution:
For a given a weak acid, HA,
HA < === > H
Ka =
[H+][A-]
----------[HA]
-
take the “ –log “ of both sides;
pH = pKa –
pH = pKa –
For a base,
log
log
[HA]
------[A-]
[A -]
------[HA]
[H+] =
+
+A
–
Ka [HA]
--------[A-]
pOH = pKb –
log
[HA]
------[A -]
Notes on the equations:

To quickly determine the type of weak acid needed to prepare a solution of a desired pH, choose a
weak acid that has:
pH = pKa of the acid.
e.g. if you want a buffer of pH = 9, choose a weak acid that has a Ka~ 10
–9

Since the volume of the buffer solution is constant, the ratio between the concentration of the acid and
the base can actually be simplified, and the ratio of the moles of the base and the acid can be used for
the calculations.

If the concentration of the acid is equal to the concentration of the base, then pH of the buffer is equal
to pKa.
Example:
How would you prepare a buffer solution that has a pH of 4.00?
Choose an acid that has pKa ~ pH. A good choice is CH3COOH.
4. Buffer Capacity:



There are limitations to the abilities of buffer solution to resist pH change.
Higher concentration of the weak acid and it’s conjugate base produces a buffer solution that can have
higher capacity for the buffer to resist pH changes.
pH changes greater than 0.2 units suggests that the buffer has been exhausted.
Example:
A 1.00 L buffer solution is made up of 1.00 M NaCH3COOH, and 1.00 M CH3COOH.
a) Calculate the pH of the solution.
b) Calculate the pH of the solution when 0.10 mol of HCl is added to this solution (assume no volume
change).
Old Exam Questions:
1993 A
CH3NH2 + H2O <=> CH3NH3+ + OHMethylamine, CH3NH2, is a weak base that reacts according to the equation above. The value of the
ionization constant, Kb, is 5.25x10-4. Methylamine forms salts such as methylammonium nitrate,
(CH3NH3+)(NO3-).
(a) Calculate the hydroxide ion concentration, [OH-], of a 0.225-molar aqueous solution of methylamine.
(b) Calculate the pH of a solution made by adding 0.0100 mole of solid methylammonium nitrate to 120.0
milliliters of a 0.225-molar solution of methylamine. Assume no volume change occurs.
(c) How many moles of either NaOH or HCl (state clearly which you choose) should be added to the solution
in (b) to produce a solution that has a pH of 11.00? Assume that no volume change occurs.
(d) A volume of 100. milliliters of distilled water is added to the solution in (c). How is the pH of the solution
affected? Explain.
1996 A
HOCl <=> OCl- + H+
Hypochlorous acid, HOCl, is a weak acid commonly used as a bleaching agent. The acid-dissociation constant,
Ka, for the reaction represented above is 3.2x10-8.
(a) Calculate the [H+] of a 0.14-molar solution of HOCl.
(b) Write the correctly balanced net ionic equation for the reaction that occurs when NaOCl is dissolved in
water and calculate the numerical value of the equilibrium constant for the reaction.
(c) Calculate the pH of a solution made by combining 40.0 milliliters of 0.14-molar HOCl and 10.0 milliliters
of 0.56-molar NaOH.
(d) How many millimoles of solid NaOH must be added to 50.0 milliliters of 0.20-molar HOCl to obtain a
buffer solution that has a pH of 7.49? Assume that the addition of the solid NaOH results in a negligible
change in volume.
(e) Household bleach is made by dissolving chlorine gas in water, as represented below.
Cl2(g) + H2O --> H+ + Cl- + HOCl(aq)
Calculate the pH of such a solution if the concentration of HOCl in the solution is 0.065 molar.