So let’s have a look at some example calculations.
What is the pH of human blood when [H+] = 4.0
× 10 to the -8 mol dm-3? Okay? This is a molar concentration.
So, let’s see how we would work this out.
First and foremost, we have our concentration
of [H+]. So, relatively, all of it’s been
done for us. All we need to now do is carry
out the minus log to the base 10 operation
on that concentration. And this gives us,
as I said before, a result for... of 7.40.
That is the pH.
Bear in mind that, because we’re talking
about a logarithmic operation, there are no
units, okay? So, if you hear the word pH,
it is merely a number.
Let’s have a look at the second question.
Let’s see what would happen if we used a
strong acid, hydrochloric acid, and we dissolved
it up to make a solution of 0.020 M which
is, as I said before, 0.02 mol dm-3.
The concentration is given for us here as
0.020. So, again, it’s relatively easy to
work out because, as I said before, strong
acids completely dissociated and, therefore,
the concentration that you had correlates
directly in this case to one equivalent of
H+ for every one equivalent of HCl added.
So, where we have a concentration of 0,
we can do the negative log to the base 10
of that and give us a value of 1, generating
a solution which, of course, is a lot more
acidic than 7, which we had in the earlier
Now, let’s look at it in a bit more depth.
What happens if we have a concentration of
hydrochloric acid and the pH is found to be
3.55? How would we work out the concentration
of HCl that we actually added?
What we then do is we take the inverse log
of -3, reversing the equation and giving
us a value of 2.82 × 10 to the -4 M or the other
unit for molar is mol dm-3 or moles per litre.
The inverse log is represented as the antilog
or, sometimes, as 10x on your calculator.
Right. Okay. Let’s have a look at another
The pH of a 0.015 M aqueous solution of HNO2,
nitrous acid, was measured as 2.63. What is
the value of Ka and what is the value of pKa?
Okay. So, this requires a little bit more
application because we have the concentration
of HNO2, that is to say the non-dissociated,
at the point of entry, HNO2, nitrous acid.
We have the pH which, as you should recall,
correlates to the concentration of H+ in solution.
And now, we’re being asked to work out what
is the ratio between the two. What is the
ratio between starting material, the non-dissociated
HNO2, and the H+ and NO2-?
So, let’s first off start off by drawing
this equilibrium. Remember we said before,
an arrow going in one direction and in the
other suggests that we have a system which
is in equilibrium. That is to say at steady
state, the concentrations of both sides are
at a equivalent... remain the same as they
did... as they would do a little later on.
So, HNO2 goes to H+ and NO2-. And so, what
we need to do is put what we know about those
particular components into an equation.
So, the concentrations are [H+] = [NO2-].
We make that assumption because we’re assuming,
in this case, that the majority of H+ comes
from NO2... from the dissociation of HNO2
and also, because with every dissociation
of one equivalent of HNO2, we must therefore
get one equivalent of [H+] and one equivalent
of [NO2-]. By doing this, we’re able to
say that the concentration of [HNO2] actually,
within our equilibrium, is given by our starting
concentration of [HNO2] minus the concentration
If the pH that we’re measuring of the solution
results in a concentration, as we can see
here, of 0.00234 M, it enables us to substitute
those values into the equation to determine
If we look at that a bit more closely, what
we’re saying here is that the concentration
of [H+] equals the concentration of [NO2-].
Therefore, the product of this [HNO2-] and
[H+] must be equal to 0.002334 multiplied
The remaining amount of non-dissociated HNO2
is given by our starting concentration minus
the amount which has been converted, which
we know from the concentration of [H+]. Therefore,
the Ka for this particular equation for this
relatively weak acid is given as 4.33 × 10 to the -4.
And the pKa, as a consequence of the negative
log to the base 10, is given therefore as
Now, I want to draw your attention to the
fact at the moment, we’ve only been dealing
with monoprotic acids. So, that’s hydrochloric
acid, nitric acid and also, as you saw, nitrous
But some acids, and this is very important
when you’re looking at, for example, phosphate-buffered
saline solutions within medical practice and
so forth, is that you can lose more than one
proton. And therefore, it’s possible to
have more than one Ka value for each successive
deprotonation reaction. Successive removals
become more difficult as the molecule, therefore,
bears a greater charge and pKa values, as
a consequence, get smaller. So, let’s have
a look at phosphoric acid here.
Phosphoric acid, that’s H3PO4, when reacting
with water yields sodium H3O+
and a phosphate anion, hydrogen phosphate
anion. And the Ka, in this first case, is
actually reasonably high, 7.1 × 10 to the -3. However,
successive deprotonations, until we eventually
result in the loss of all hydrogens and only
the phosphate remains, is a bit more difficult
to achieve, giving us a Ka value or Ka3 for
the third deprotonation of 4.3 × 10 to the -13.
Polyprotic acids are often used in buffer
solutions. And certainly, when you’re looking
at trying to mimic a plasma or the natural
pH and natural salt concentrations for a given
biological system, you would often use phosphates
in your buffer.