BUFFER SOLUTIONS

A buffer solution is one which resists changes in pH when small
quantities of acid or base are added to it.

It is made from a weak acid and its conjugate base or a weak base and its
conjugate acid.
Look what happens if we add OH ¯ ions to CH3COOH. The acid is a proton
donor:
OH ¯(aq) + CH3COOH (aq)  CH3COO ¯(aq) + H2O
And when we add H3O+ to CH3COO ¯ the base acts as a proton acceptor:
H3O+(aq) + CH3COO ¯(aq)  CH3COOH(aq) + H2O
If we had a mixture of acetic acid and acetate ions it wouldn’t matter whether
we added acid or base, the excess OH ¯ and H3O+ would be neutralised.
The absorbing of base and acid is important in biochemical systems (e.g.
blood) and in industry. It is often important to keep the pH at a certain level to
avoid side reactions which could occur. Using the correct buffer will ensure
the pH is held at the right level. Medical and industrial processes are
sometimes pH sensitive e.g. producing solutions for delivering of drugs or
preparing synthetic or processed food.
The normal pH of blood is about 7.40. If the pH falls outside the 7.35 to 7.45
range a patient becomes seriously ill. Digestion produces a range of products,
many of them acidic. The blood pH is maintained, in spite of the addition of H +
ions, through
HCO3¯/ H2CO3 buffers in the red blood cells and the blood plasma.
H+(aq) + HCO3¯(aq)  H2CO3(aq)
OH ¯(aq) + H2CO3(aq)  HCO3¯(aq)
Phosphate buffers (HPO42¯/H2PO4¯ ) are also important in maintaining
constant pH in cells and urine.
Buffer solutions can be made in two ways:
 by dissolving the calculated amount of salt into the appropriate weak acid
or base.
 by adding a calculated amount of strong base/acid to the weak acid/base
(so that the required salt is formed).
LRY 2010
Buffers and Ka
Remember that for a general acid, HA, we can write the dissociation equation:
HA + H2O  H3O+ + A ¯
and the expression for Ka:
Ka = [H3O+][A ¯ ]
[HA]
rearranging to give [H3O+]:
[H3O+]= Ka[HA]
[A ¯]
From this we can either calculate the pH of a given buffer mixture (if we know
Ka, [HA] and [A ¯]) or we can calculate the amount of acid/base or salt needed
to make a buffer of a given concentration.
If the concentrations of the acid (HA) and its conjugate base (A¯) are equal:
[H3O+] = Ka [HA]
[A ¯ ]
Ka = [H3O+]
pKa = pH
It is possible to make a buffer of any pH from any acid, but the buffer is most
effective when [A ¯] = [HA], and thus we choose an acid with pKa closest to
the desired pH.
Example: A buffer solution is made by adding 3.28g of sodium ethanoate
to 1.00 L of 0.0100 mol L-1 ethanoic acid. What is the pH of the resulting
buffer?
Ka(CH3COOH) = 1.74 x 10-5, M(CH3COONa) = 82.0 g mol-1)
n(CH3COO Na) = m
M
= 3.28 g
82.0 g mol-1
= 0.0400 mol
Since the volume is 1.00L, the concentration of the solution will be 0.0400 mol
L-1 sodium ethanoate.
[H3O+] = Ka [HA]
[A ¯]
LRY 2010
Assume
[CH3COO ¯] = 0.0400 mol L-1
[CH3COOH] = 0.0100 mol L-1
[H3O+] = Ka[HA]
[A ¯]
= 1.74 x10-5 x 0.0100
0.0400
= 4.35 x 10-6 mol L-1
pH = 5.36
Buffer Zone:
Buffer solutions are only effective over a pH range of about one pH unit above
and below the pKa value for the weak acid of the buffer. This range of
effective pH control (pKa ± 1) is known as the buffer zone.
LRY 2010