The H^{+} and OH^{–} log concentration lines are
the same ones that we saw in Fig. 1. The other two lines show how the concentrations
of CH_{3}COOH and of the the acetate ion vary with the pH of the
solution.

How do we construct the plots for [HAc] and [Ac^{–}]? If
you look carefully at Fig 2, you will observe that each line is horizontal
at the top, and then bends to become diagonal. There are thus three parameters
that define these two lines: the location of their top, horizontal parts,
their crossing points with the other lines, and the slopes of their diagonal
parts.

The horizonal sections of these lines are placed at 3 on the ordinate
scale, corresponding to the nominal acid concentration of 10^{–3}
M. This value corresponds to

C_{a} = [HAc] + [Ac^{–}]

which you will recognize as the mass balance condition saying that "acetate"
is conserved; C_{a} is the nominal "acid concentration"
of the solution, and is to be distinguished from the concentration of the
actual acidic species HAc.

At low pH values (strongly acidic solution) the acetate species is completely
protonated, so [HAc] = 10^{–3} M and [Ac^{–}]=0.
Similarly, at high pH, –log [Ac^{–}]=3$ and [HAc]=0. If
the solution had some other nominal concentration, such as 0.1 M or
10^{–5}, we would simply move the pair of lines up or down.

The diagonal parts of the lines have slopes of equal magnitude but opposite
sign. It can easily be shown that these slopes d(–log [HAc]}/d{pH}
etc.are +-1, corresponding to the slopes of the [OH^{–}] and
[H^{+}] lines. Using the latter as a guide, the diagonal portions
of lines 3 and 4 Can easily be drawn.

The crossing point of the plots for the acid and base forms corresponds
to the condition [HAc]=[Ac^{–}]. You already know that this
condition holds when the pH is the same as the pK_{a} of the acid,
so the the pH coordinate of the crossing point must be 4.75 for acetic acid.
The vertical location of the crossing point is found as follows: When [HAc]
= [Ac], the concentration of each species must be C_{a}/2 or in
this case 0.0005 M. The logarithm of 1/2 is 0.3, so a 50% reduction in the
concentration of a species will shift its location down on the log concentration
scale by 0.3 unit. The crossing point therefore falls at a log-C value of
(–3) – 0.3 = –3.3. Knowing the value of log_{10}0.5
is one of the few new "facts" that must be learned in order to
construct these graphs.