In the previous lecture, we were talking about
acid and base chemistry, specifically we were
talking about how you create acids, how you
create bases and more specifically, the correlation
between weak and strong and concentrated and
Now, what we’re going to be doing is carrying
out revision exercise considering some of
the more common ionic chemistry reactions,
so the direct application of what you should
have learnt in Module I.
So. if we take neutralisation reactions as
a starting point, it should come as no surprise
that acids react with bases. And here we have
an example of such. Sodium carbonate, Na2CO3
reacting in this case with hydrochloric acid
to create a salt plus water plus carbon dioxide.
Bear in mind, the driving force for this particular
reaction is going to be the generation of
water and also the carbon dioxide, in this
case, which is a gas and obviously escapes.
If we look at sulphides, it could also react
with acids to form SO2 or sulphur dioxide
gas, sodium sulphide, SO3, not to be confused
with sulphate, would be... which would be
SO4, reacts with hydrochloric acid to, again,
provide us with a salt plus water, in this
case, a gas, sulphur dioxide. The reality
is... is that the sulphur dioxide will actually
partially dissolve in the water that you’ve
generated and form an acidic solution of sulphurous
acid. But, that’s beyond our terms of reference.
The neutralisation reaction for the first,
the reaction of sodium carbonate with hydrochloric
acid, can essentially be broken down into
a number of different actual molecular events.
Note, you should be careful when using the
term ‘molecule’ and ‘ions’ in the same sentence,
since ions in their compounds are formed with
the units and are never considered to be molecules.
However, as you can see, where we have hydrogen
bound to carbon, bound to oxygen, there’s
an element of molecular identity to these.
So, let us start with the first step for this
The neutralisation reaction converts sodium
carbonate in the presence of one equivalent
of hydrochloric acid into sodium chloride,
a salt and sodium hydrogen carbonate, NaHCO3.
The complex ion HCO3 or hydrogen carbonate
is a formal charge of -1 whereas if you recall
from Module I, Na existed only two possible
oxidation states, either 0 or +1. Hence, the
reason why the stoichiometry there is 1 equivalent
of Na+ to 1 equivalent of HCO3-. The NaHCO3
or sodium hydrogen carbonate then splits into
Na+ and hydrogen carbonate in solution and
water, in this particular case.
This is broken down further into hydroxide
and carbon dioxide and it’s the hydroxide
OH- which then goes on to react with the second
equivalent of hydrochloric acid resulting
in the formation of water, which we know is
an outcome of this reaction and another
equivalent of sodium chloride.
Hence, the reason why the equation, as given
below, right at the bottom of the board is
the fully-balanced equivalent of all of those
events I’ve just discussed. Notice how on
one side of the equation we have exactly the
same components as we have on the other side.
And that is absolutely essential when balancing
Note also that remember what we said about
the oxidation state in the previous module
of certain atoms within the periodic table.
In group 1, they form +1 ions; in group 2,
So, in this case, because sodium is in group
one, it can only exist as Na+. That is why
we need two Na+ in order to form a complete
uncharged formula unit with the complex ion,
carbonate or CO3, which on itself carries
a -2 charge. Thus, two times Na+, plus CO3,
2- results in a net charge of zero.
Sulphides could also react with acids to
form hydrogen sulphide gas, a rather smelly
experiment in fairness since hydrogen sulphide
really, really smells bad. Sodium sulphide,
which is shown as the reactant in this equation
reacts with two equivalents of HCl, hydrochloric
acid. And knowing what I’ve just taught, being
aware of what I’ve just discussed about the
nature of sodium, you should be able to deduce
that the oxidation state or the charge on
the sulphur, in this case, must therefore
be 2-. The driving force for this neutralisation
reaction is the removal of ions from solution,
in this case, to form either water or, in
this case, hydrogen sulphide gas and is also
results in the increase in entropy associated
with the production of this gas.