Buffer Solutions
A buffer is a solution that resists changes in pH
upon addition of small quantities of acids or
bases. In other words, a buffer is a solution that
keeps its pH almost constant. A buffer is a
solution containing a weak acid and its
conjugate base or a weak base and its
conjugate acid. Below is an explanation of how
buffers work:
Let us look at a buffer formed from a weak acid
(like acetic acid) and its conjugate base (e.g.
sodium acetate); we have the following
equilibria:
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HOAc H+ + OAcOAc- + H2O HOAc + OHIn the first equilibrium, acetate is produced from the
dissociation of acetic acid. However, there is a lot of
acetate added to solution from the sodium acetate.
Therefore, the first equilibrium does not occur to any
significant degree. We can fairly assume that HOAc will
practically not dissociate. The same applies for the
second equilibrium where acetic acid is produced. Since
a great excess of acetic acid is present in solution from
the first equilibrium, one can say that acetate will not
practically associate in water.
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Therefore, neither acetic acid nor acetate equilibria
will proceed to any significant extent. The
equilibrium constant for dissociation of acetic
acid can be written as:
Ka = [OAc-][H+]/[HOAc]
Where the [OAc-] = COAc- and [HOAc] = CHOAc
When an acid is added to the buffer above, H+
from the acid will combine with OAc- to form
HOAc which is reflected by an increase in HOAc
and a decrease in OAc-.
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In case the amount of H+ added is not very
large, the ratio [OAc-]/[HOAc] will only
change slightly. A similar argument can be
presented when a small amount of base is
added to the buffer solution. However, we
will treat the buffer problem as we
described for the common ion effect
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Buffer Capacity
The degree of buffer resistance to changes in pH
is referred to as buffer capacity. The buffer
capacity is defined as the concentration of acid
or base ( in moles ) needed to cause a pH
change equal to dpH where:
β = dCbase/dpH = - dCacid/dpH
The minus sign is because addition of an acid
causes a decrease in pH. A more practical
relation to use for calculation of buffer capacity
is:
β = 2.303 Cacid Cconjugate base / (Cacid + Cconjugate base)
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Example
Calculate the buffer capacity of a solution containing 0.10
M acetic acid and 0.10 M acetate. Find the pH change
when you add NaOH so that the solution becomes 0.005
M in NaOH
Solution
β = 2.303 Cacid Cconjugate base / (Cacid + Cconjugate base)
β = 2.303 * 0.10 * 0.10/( 0.10 + 0.10) = 0.115 M
β = dCbase/dpH
0.115 = 0.005/dpH
dpH = 0.005/0.115 = 0.043
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Hasselbalch-Henderson Equation
In the acetic acid/acetate buffer described above,
we have:
Ka = [OAc-][H+]/[HOAc]
Pka = pH – log [OAc-]/[HOAc]
pH = pka + log [OAc-]/[HOAc]
The above equation is referred to as HasselbalchHenderson equation.
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this equation can be useful to describe buffer limits
where the maximum pH limit for a buffer is when
the salt to acid ratio is 10 and the minimum pH
limit of the buffer is when the acid to salt ratio is
10. Inserting these values, one in a time, in the
Hasselbalch equation gives
pH = pka + 1
Therefore, the buffer acts well within two pH
units and the midpoint pH value is equal to
pka. One should first look at the pka or pkb to
choose the correct buffer system which can
be used within a specific pH range.
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Example
Calculate the pH of the buffer containing 0.50 M formic acid
(HA, ka = 1.8x10-4) and 0.25 M sodium formate (NaA).
Solution
First, write the acid equilibrium equation
HA H+ + A-
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Ka = x (0.25 + x)/(0.50 – x)
Assume that 0.25 >> x
1.8x10-4 = 0.25 x /0.50
x = 3.6x10-4
Relative error = (3.6x10-4/0.25) x 100 = 0.14%
The assumption is therefore valid and we have
[H+] = 3.6x10-4 M
pH = 3.44
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Example
Calculate the pH of the buffer solution prepared by
mixing 10 mL 0.10 M HOAc (ka = 1.75x10-5) with
20 mL of 0.20 M sodium acetate.
Solution
Let us first calculate the concentrations after
mixing (final concentrations of the acid and its
conjugate base)
mmol HOAc = 0.10 x 10 = 1.0 mmol
[HOAc] = 1.0/30
mmol OAc- = 0.20 x 20 = 4.0 mmol
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[OAc-] = 4.0/30
The equilibrium equation is
HOAc = H+ + OAc-
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Ka = x (4.0/30 + x)/ (1.0/30 – x)
Assume 1.0/30 >> x
1.75x10-5 = x (4.0/30)/1.0/30
x = 1.75x10-5 /4.0
x = 4.38x10-6
Relative error = {4.38x10-6/(1.0/30)} x 100 =
0.013%
The assumption is valid therefore:
[H+] = 4.38x10-6 M
pH = 5.36
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Example
Calculate the pH of the solution resulting from
adding 25 mL of 0.10 M NaOH to 30 mL of 0.20
M acetic acid.
Solution
Let us find what happens when we mix the two
solutions. Definitely the hydroxide will react with
the acid to form acetate which also results in a
decrease in the acid concentration.
mmol OH- = 0.10 x 25 = 2.5 mmol
mmol HOAc = 0.20 x 30 = 6.0 mmol
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Now the mmol base added will react in a 1:1 mole
ratio with the acid . Therefore we have
mmol OAc- formed = 2.5 mmol
[OAc-] = 2.5/55 M
mmol HOAc left = 6.0 – 2.5 = 3.5 mmol
[HOAc] = 3.5/55 M
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Ka = x (2.5/55 + x)/ (3.5/55 – x)
Assume 2.5/55 >> x
1.75x10-5 = x (2.5/55)/3.5/55
x = 1.75x10-5 x 2.5/3.5
x = 2.45x10-5
Relative error = {2.45x10-5/(2.5/55)} x 100 =
0.054%
The assumption is valid
pH = 4.61
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Example
A buffer solution is 0.20 M in HOAc and in NaOAc.
Find the change in pH after addition of 0.10
mmol of HCl to 10 mL of the buffer without
change in volume.
Solution
We should find the initial pH (before addition of
HCl) and then calculate the pH after addition.
HOAc H+ + OAc-
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Initial pH
Ka = x (0.20 + x)/ (0.20 – x)
Assume 0.20 >> x
1.75x10-5 = 0.20 x/ 0.20
x = 1.75x10-5
Relative error = (1.75x10-5/0.20) x 100 = 8.8x10-3%
[H+] = 1.75x10-5 M
pH = 4.76
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After addition of HCl, the acetate concentration will
decrease while the acetic acid concentration will
increase.
mmol HOAc = 0.20 x 10 + 0.10 = 2.1 mmol
[HOAc] = 2.1/10 = 0.21 M
mmol OAc- left = 0.20 x 10 – 0.10 = 1.9 mmol
[OAc-] = 1.9/10 = 0.19 M
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Ka = x (0.19 + x)/ (0.21 – x)
Assume 0.19 >> x
1.75x10-5 = 0.19 x/ 0.21
x = 1.93x10-5
Relative error = (1.93x10-5/0.19) x 100 = 0.01%
[H+] = 1.93x10-5 M
pH = 4.71
∆pH = 4.71 – 4.76 = - 0.05
Try the same problem replacing NaOH for the HCl and
get the change in pH using the same procedure. You
should realize that in this case the acetic acid will
decrease while acetate will increase.
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