In principle, the concentrations of all species in a monoprotic acid-base system can be found by solving the appropriate cubic equation.  Although simplifications are often possible that reduce this to a quadratic or sometimes only a first-order equation, there is still an element of mathematical complexity that tends to get in the way of a real understanding of what is going on in such a solution.


There is an alternative method that, while lacking precision, affords one a far clearer view of the relationships between the various species related to a given acid-base system.  This starts out with the graph shown in \figref{AB10}.  This graph is no more than a definition of pH and pOH; for example, when the pH is 4, &#150;log [H<sub>3</sub>O<sup>+</sup>] = 4$.  (Notice that the ordinate is the negative of the log concentration, so the smaller numbers near the top of the scale refer to larger concentrations.)

You should be able to construct this kind of graph from memory; all
you need for the two straight lines are three points: the two at the
top (0,0) and (14,0), and the other at (7,7) where the two lines
intersect; let us hope that you understand the significance of this
third point!


The above graph of is of no use by itself, but it forms the basis for the construction of other graphs specific to a given acid-base system. For example, suppose that we want to see how the species concentrations vary with pH in a 0.001 solution of acetic acid, K<sub>a</sub> = 10<sup>&#150;4.74</sup>}$.  The graph in the top part of Fig. 2 describes this system. The H<sub>3</sub>O>sup>+</sup> and OH<sup>&#150;</sup> log concentration lines are the same ones that we saw earlier.  The other two lines show how the concentrations of CH<sub>3</sub>COOH ("HAc") and of the the acetate ion vary with the pH of the solution.


Locating the lines on the graph

How can we construct the plots for [HAc] and [Ac$^{-}$]?  It is easier than you might think.  If you look carefully at \figref{AB11}a, 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<sup>&#150;3</sup>.  This value corresponds to

C<sub>a</sub> =  [HAc] + [Ac<sup>&#150;</sup>]

which you will recognize as the mass balance condition saying that "acetate" is conserved; C<sub>a</sub>\ 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<sup>&#150;3</sup> and $\rm [Ac<sup>&#150;</sup>]=0. Similarly, at high pH, $&#150;log \rm [Ac<sup>&#150;</sup>]=3$ and $\rm [HAc]=0$.  If the solution had some other nominal concentration, such as 0.1\molar\ or $10^{-5}$\molar, 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 C<sub>a</sub>n easily be shown that these slopes $d\/{\rm \&#150;log [HAc]}/d{\rm pH}$ etc.\ are $\pm 1$, corresponding to the slopes of the [OH<sup>&#150;</sup>]\ and [H<sub>3</sub>O<sup>+</sup>]\ lines.  Using the latter as a guide, the diagonal portions of lines 3 and 4 C<sub>a</sub>n easily be drawn.

The crossing point of the plots for the acid and base forms corresponds to the condition $\rm [HAc]=[\Ac]$.  You already know that this condition holds when the pH is the same as the \pka\ of the acid, so the the pH coordinate of the crossing point must be 4.75 for acetic acid.  The vertiC<sub>a</sub>l loC<sub>a</sub>tion of the crossing point is found as follows:  When $\rm [HAc]= [\Ac]$, the concentration of each species must be $\half \C<sub>a</sub>$, or in this C<sub>a</sub>se 0.0005\molar .  The logarithm of $\half$ is 0.3, so a 50\% reduction in the concentration of a species (from an initial value of \C<sub>a</sub> ) will shift its loC<sub>a</sub>tion down on the log concentration sC<sub>a</sub>le by 0.3 unit.  The crossing point therefore falls at a log-$C$ value of $(-3) - .3 = -3.3$,

By following this example, you should be able to construct a similar
diagram for any monoprotic acid, weak or strong; the only numbers you
need are the pK of the acid, and its nominal concentration
Ca.  The resulting graph will provide a reasonably
precise indication of the concentrations of all four
related species in the solution over the entire pH range.  The only
major uncertainty occurs within about one pH unit of the pKa , where
the lines undergo changes of slope from 0 (horizontal) to $\pm 1$.

Estimating the pH on $&#150;log C$ vs pH diagrams

Of special interest in acid-base chemistry are the pH values of a
solution of an acid and of its conjugate base in pure water; as you
know, these correspond to the beginning and equivalence points in the
titration of an acid with a strong base.

subsubsection{pH of an acid in pure water

Except for the special cases of extremely dilute solutions or very
weak acids in which the autoprotolysis of water is a major
contributor to the [H<sub>3</sub>O<sup>+</sup>] , the pH of a
solution of an acid in pure water will be determined largely by the
extent of the reaction


HAc \ps H<sub>2</sub> \yields \hm \ps \Ac \] so that at equilibrium,
the approximate relation $[H<sub>3</sub>O<sup>+</sup>] = [\Ac ]$ will
hold.  The equivalence of these two concentrations corresponds to the
point labeled {\bf 1} in \figref{AB11}a; this occurs at a pH of about
3.7, and this is the pH of a 0.001\molar\ solution of acetic acid in
pure water.

\subsubsection{pH of a solution of the conjugate base}

Now consider a 0.001~molar solution of sodium acetate in pure water.
This, you will reC<sub>a</sub>ll, corresponds to the composition of a
solution of acetic acid that has been titrated to its equivalence
point with sodium hydroxide.  The acetate ion, being the conjugate
base of a weak acid, will undergo hydrolysis according to \[ \rm \Ac
\ps H<sub>2</sub> \yields HAc \ps \oh \] As long as we C<sub>a</sub>n
neglect the contribution of \oh\ from the autoprotolysis of the
solvent, we can expect the relation $\rm [HAc]=[OH<sup>&#150;</sup>]$
to be valid in such a solution.  The equivalence of these two
concentrations corresponds to the intersection point 3 in AB11}a; a
0.001M solution of sodium or potassium acetate should have a pH of
about 8.


Titration curves

If you have carried out similar caculations mathematically, you will
appreciate the simplicity and utility of this graphical approach. 
Even without graph paper and a ruler, you should be able to sketch
out a graph of this kind on the back of an envelope, and without
recourse to any C<sub>a</sub>lculations at all make estimates of the
pH of a solution of any pure acid and its conjugate base to an
accuracy of about one pH unit.

Besides being a useful tool for rough estimations of pH, the log-C
plot provides considerable insight into the factors that determine
the shape of the titration curve. As shown in the lower part of
\figref{AB11}, you C<sub>a</sub>n use the log-C vs pH diagram to
construct an approximate titration curve.



This diagram is just the superposition of the separate diagrams for
the ammonia and formic acid systems.  In a solution of ammonium
formate, stoichiometry requires that [NH<sub>3</sub>] + [NH+] =
[HCOOH] + [HCOO<sup>&#150;</sup>] \] This condition corresponds to
the point indiC<sub>a</sub>ted on the graph. Compare the pH of this
crossing point with the result in the Problem Example on page ??.


Polyprotic acids

Whereas pH calculations for solutions of polyprotic acids become
quite complex, the log-$C$ plots for such systems simply contain a
few more lines but are no more difficult to construct that those for
monoprotic acids.  Phosphoric acid H<sub>3</sub>PO<sub>4</sub>$ is a
triprotic system that is widely used as a buffering agent.  Each of
the \pka 's in \figref{AB20} corresponds to a crossing point of the
lines depicting the concentrations of a pair of conjugate species.

The only really new feature is that the slopes of plots change from
$\pm 1$ to $\pm 2$ and then to $\pm 3$ as they cross \pka\ values
successively more removed from the pH range at which the particular
species predominates. When the slopes reach $\pm 3$ the species
concentrations are generally insignifiC<sub>a</sub>nt.

\begin{figure} %\vspace{2.6in}
\centerline{\psfig{figure=AB20.epsf,height=3in}}
\C<sub>a</sub>ption{Log-$C$ diagram for the phosphate system}
\label{figure:AB20} \end{figure}

{\small The pH of solutions of the individual phosphate species
C<sub>a</sub>n be estimated by simplifying the appropriate proton
balance expressions.  For $\rm H<sub>3</sub>PO<sub>4</sub>$ the
relation

[H<sub>3</sub>PO<sub>4</sub>] \ps [H<sub>3</sub>O<sup>+</sup>] =
[H<sub>2</sub>PO<sub>4</sub><sup>2&#150;</sup>] \ps
[OH<sup>&#150;</sup>] \] --> $\rm [H_{2}PO_{4}^{2-}] =
[OH<sup>&#150;</sup>]$ which corresponds to point 1 in \figref{AB20}.
 Similarly, for a solution of  NaH<sub>2</sub>PO<sub>4</sub> we have

\[\rm [H<sub>3</sub>PO<sub>4</sub>] \ps [H<sub>3</sub>O<sup>+</sup>]
= [HPO_{4}^{2-}] \ps 2[PO_{4}^{3-}] \ps [OH<sup>&#150;</sup>] \]

which gives

[H<sub>3</sub>PO<sub>4</sub>]  = [HPO<sub>4</sub><sup>2&@150;</sup>]

after smaller terms are eliminated (2).  Na<sub>3</sub>PO<sub>4</sub>
is treated straightforwardly and corresponds to point (5}.  The case
of  Na<sub>2</sub>HPO<sub>4</sub> is not quite so simple, however;
the proton-balance expression

2[H<sub>3</sub>PO<sub>4</sub>] + [H<sub>3</sub>O<sup>+</sup>] +
[H<sub>2</sub>PO<sub>4</sub><sup>2&#150;</sup>] = [PO_{4}^{3-}] \ps
[OH<sup>&#150;</sup>] \]

cannot be so greatly simplified because the two terms on the right
will be comparable. The approximation
[H<sub>2</sub>PO<sub>4</sub><sup>2&#150;</sup>] =
[PO<sub>4</sub><sup>3&#150;</sup>] + [OH<sup>&#150;</sup>] \]

does not correspond to any crossing point in Fig 4 but falls at a pH
slightly to the left of (3}.
