How to Calculate pH of a Buffer Solution
For a better
understanding of how to calculate pH of a buffer solution, you have to know a
buffer solution. What is a buffer solution? It is a solution that resists vicissitudes
in solution pH when small amounts of either acid solution or basic solution are
added to that solution (a buffer solution). Therefore, a buffer solution obeys
Le Chatelier’s Principle. Le Chatelier’s Principle states that when a system at
equilibrium is subjected to any change, the system readjusts itself to resist
the effect (s) of the change. This results in the establishment of a new
equilibrium. The change can either be pressure or temperature or concentration.
A buffer solution can
be an acid or a base. An acidic buffer solution resists the removal of hydrogen
ions. On the other hand, an alkaline buffer resists the removal of hydroxide
ions.
Up to this point, you
can now calculate the pH of a buffer solution. However, you have to identify
whether the buffer solution is an alkaline or an acidic buffer solution.
Let us consider an
example of a buffer solution. A buffer solution consists of 0.10 mol dm-3
of methanoic acid and 0.20 mol dm-3 sodium methanoate. Calculate the
pH of this buffer solution.
In a solution having a
weak acid, equilibrium exists between undissociated acid and its ions. Hence,
methanoic acid is represented in an equilibrium system as:
H3COOH (aq) = H3COO- (aq) + H+
(aq)
Thus, the addition of methanoate
ions will shift the equilibrium to the left. Though, the equilibrium has been
shifted to the left, but the equilibrium will still occur.
The equilibrium
constant of the above equation can be written as:
Ka=
[H3COO-][H+]/[H3COOH]
In calculations
involving weak acid, it is assumed that the concentration of the species in the
numerator of the equilibrium constant equation is the same. However, for equilibrium
constant equation of a buffer solution, these species are not the same. Presence
of methanoate ions further moves equilibrium to the left. But the methanoate
ions originating from methanoic acid are insignificant than methanoate ions
coming from sodium methanoate.
Therefore, this leads
to the assumption that methanoate concentration is equal to the concentration
of sodium methanoate.
Recommended books for
further reading:
In calculation
involving weak acid, it is presumed that acid hardly ionized. Hence, at the
equilibrium, the concentration of acid is equal to the concentration of acid
used. So, this assumption aptly applies in this case (calculations of buffer
solution).
To calculate the pH of
a buffer solution containing 0.10 dm-3 of methanoic
acid and 0.20 mol dm-3 sodium methanoate, you have to know the Ka of
methanoic acid.
Ka of methanoic is 1.8
x 10-4 mol dm-3. Inserting the right values in the
equilibrium constant equation gives the pH of the buffer solution by finding
–log [H+].
Ka=
[0.20 mol dm-3][H+]/0.10 mol dm-3
[H+]= 1.8 x
10-4 mol dm-3 x (0.1/0.20)
= 9 x 10-5
mol dm-3
pH=-log [= 9 x 10-5]
= 4.0 (in two
significant figures)
The calculations of pH
for alkaline buffer solution are the same as calculations involving acidic
buffer solutions.
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further reading:
NB:
The above pH can also be obtained using the Henderson-Hasselbalch equation. The
Henderson-Hasselbalch equation is derived from the equilibrium constant
equation as follows:
Let HB represent an
acid while B- to denote conjugate base. Now, Ka (equilibrium
constant equation) becomes:
Ka= [H+][B-]/[HB]
Calculation of hydrogen
concentration [H+] gives:
[H+]= Ka[HB]/[B-]
By taking negative
logarithms on both sides of the equation results in:
-log [H+]=-logKa
- log[HB]/[B-]
Now, by substituting pH
for log [H+] and pKa for –logKa gives:
pH=pKa – log[HB]/[B-]
Inverting part of the
above equation (– log [HB]/ [B-]), which embroils changing its sign
results in:
pH=pKa + log[B-]/[HB]
( Henderson-Hasselbalch
equation)
Recommended books for
further reading:
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