Figure 6.6
Potentiometric titration curves. (a) Normal curve. (b) First derivative curve.
(b) Second derivative curve.eliminated by titrating standards. Figures 6.6(b) and (c) show the first and second derivative curves of a
normal symmetrical titration curve. These may offer some advantage in locating the equivalence point.
However, precise location of the inflection point is dependent on the collection of data in its immediate
vicinity where the cell potential is likely to be unstable due to the low concentrations of electroactive
species. Derivative curves should therefore be used with caution and are probably not worth the extra
effort involved in their computation although some automated titration systems can produce a
derivative curve directly.
Potentiometric titrations yield more precise results than direct measurements, better than 0.2% if
required. This is because the data collected effectively averages Ecell over a large number of readings.
The principal types of potentiometric titration are summarized in the following sections.
Acid–base Titrations
The magnitude of the potential change or 'break' at the equivalence point depends on the strengths of
the acid and base involved. Curves for the titration of acids with pKa values ranging from 4 to 10 using
a strong base as titrant are shown in Figure 6.7. The size of the inflection diminishes as the acids
become weaker (larger pKa) until for a pKa > 10 it is difficult to detect the equivalence point at all. A
similar effect occurs for the titration of weak bases with a strong acid. The potentiometric method can
be very useful for titrating mixtures of acids or a polyprotic acid where a visual indicator would show
only one colour change. It can also be used to titrate the salt of a weak acid or base with a strong acid or
base, e.g. in the standardization of a hydrobromic acid solution with sodium acetate for an epoxide
determination. The titration data can be used to calculate the pKa and equivalent weight of an unknown
partly dissociated acid, HA. Thus, HA = H+ + A– and at a point half-way through the titration,