Buffer solutions, a common buffer solution,
which is relatively easy to explain and easy
to understand because it consists only of
one deprotonation step, would be ethanoic
acid and sodium ethanoic solution, otherwise
known as acetic acid, in old terminology - sodium
The buffer is prepared, in this particular
case, as 1.0 M, or moles per litre, acetic
acid or ethanoic acid and sodium acetate or
sodium ethanoate. The equilibrium that I showed
you before in structural form is shown here
where we have the ethanoic acid plus an equivalent
of water giving us the hydronium ion and CH3CO2Na.
On one side, we have the [H+]; on the other
side, we have the sodium acetate.
If we rearrange this equation, so instead
of using our model examples of H+ and A-
we now have a real-world structure to consider
– the Ka, as we said before, for acetic
acid is 1.8 × 10 to the -5.
If we rearrange this equation to determine
the concentration of H+ in our solution, we
get the following equation on the board which
is shown right at the bottom, where Ka multiplied
by the concentration of acetic acid divided
by the concentration of the sodium acetate
results in a calculation for the concentration
of H+ in the form of the hydronium ion.
So, let’s go back to our equation. Let’s
assume that the equilibrium concentrations
do not differ much from the initial concentrations.
So, in this particular case, if we were to
have a 1 M equivalent of acetic acid and sodium
acetate, this would give us a concentration
of [H+] of 1.8 × 10 to the -5.
What that means in pH is, if we do the negative
log to the base 10 of this value, we get a
value of 4.74. This is shown on the board
there. Good. So far so good.
We have a solution. We know that the pH is
4.74. Now, let us add a strong base to neutralise
2% of the acid. So, we’ll add a strong base
so we know it’s going to be completely dissociated
into, let’s say, sodium hydroxide NA+ and
OH-. This amount of base changed the pure
water, as we saw earlier on the scale, from
pH 7.0 to 12.3. The OH- ions will react with
the ethanoic acid reducing its concentration.
However, the acid molecules that react will
become ethanoate ions. In buffers, the neutralisation
of one buffer component converts it to the
So, let us apply that. We’ve neutralised
2%. This results now in a concentration of
acetic acid of 0.98 M and a concentration
of sodium acetate of 1.02 M. Now, let’s
use our knowledge of those concentrations
to then determine what the pH is going to
be now we’ve made that change.
As you can see, we now have a situation where
a very, very small change in concentration
of our sodium acetate and our acetic acid
results in a minute change, in fact is barely
observable, change in the concentration of
[H+] which is available 1.7 × 10 to the -5. This
equates to a pH of 4.77.
So, whereas before in the case of the water,
where we saw a substantial change in pH with
a minute amount of a strong acid or strong
base, here where we add a small amount of
very strong base, we see a fundamentally unchanged
buffer solution. The pH has hardly been affected.
That is an example of a buffer scenario. However,
let’s repeat with the same concentration
of acid. So, know what we did beforehand.
We’re actually adding some more acid to
the scenario, so we’re converting more of
the sodium ethanoate into the ethanoic acid.
So, we’re moving the equilibrium back towards
the reactants. This is represented in the
equation here - 1.02/0.98.
This results in a marginally increased concentration
of [H+] of 1.9 × 10 to the -5.
Now, if we do the negative log to the base
10 of that concentration of [H+], we see that
we end up with a pH value of 4.72.
Again, not very different to the original
pH that we started off with when we used 1
mole equivalent of either the sodium acetate
and the acetic acid.
The equilibrium in the buffer between acid
and conjugate base has prevented the pH from
changing much and this is effectively the
Right. Okay. So, let’s bring this into play.
How does it work and what are the assumptions
that we make when we do these reactions when
we set these buffers up?
We can adjust the equation to actually work
out what the concentration of [H+], therefore,
the pH would be, by virtue of assuming a number
One of the principal assumptions, and this
is why you need either a weak acid or a weak
base, is that the amount of sodium acetate
that you add pretty much stays as sodium acetate.
And the amount of acetic acid you add pretty
much stays as acetic acid. That’s pretty
much your assumption, your principal assumption.
So, you can control precisely how much [H+]
is in that solution.
If we take the negative log to the base 10
of each of these particular terms, as you
can see on the board, we can determine pH
which, of course, is the negative log to the
base 10 of the [H+] component.
And, by doing the log at the base 10 of the
others, we can see that pH = pKa. So, this
is the log of the acidity constant or dissociation
constant for the acid plus the log of the
concentration of sodium acetate over acetic
In its general form, the Henderson–Hasselbach
equation is given thus.
So, where we have a concentration of a conjugate
base, in this case our acetate, over our concentration
of weak acid, in this case our acetic acid,
we add the log of that relationship or ratio
to our log of pKa to give us the pH, a log
to the negative base 10 of hydrogen or [H+].
This equation is very useful for buffer calculations.
So, if you’re ever unsure and you’re using,
let’s say a monoprotic acid, a weak acid,
maybe it was, for example, dichloro- or monochloroacetic
acid, as long as you know the Ka and as long
as you know approximately what’s being added
in terms of the conjugate base, in this case
our sodium acetate, and the amount of free
acid you’re adding, you can calculate approximately
what the buffer pH will be.
The biggest assumption is that the equilibrium
concentrations are the same as the concentrations
of the components which we mixed to make the
buffer. So, in other words, they remain principally
In the previous example of a 1.00 M solution
of ethanoic acid and ethanoic... ethanoate...
sodium ethanoate are mixed to make the buffer.
The assumption, in this case, is that the
concentrations will be the same in dynamic
equilibrium. To ensure that the effects of
this assumption are minimised, the following
should be observed.
When the concentration of conjugate base is
divided by the concentration of the weak acid,
the resulting value should be between 0.10
Concentrations of the components should exceed
Ka by at least 100 times.