Page 34 - Modern Analytical Chemistry

1400-CH02 9/8/99 3:48 PM Page 17

Chapter 2 Basic Tools of Analytical Chemistry 17

The number of equivalents, n, is based on a reaction unit, which is that part of
equivalent
a chemical species involved in a reaction. In a precipitation reaction, for example, The moles of a species that can donate
the reaction unit is the charge of the cation or anion involved in the reaction; thus one reaction unit.
for the reaction
2+
–
Pb (aq) +2I (aq) t PbI 2 (s)
–
n =2 for Pb 2+ and n = 1 for I . In an acid–base reaction, the reaction unit is the
+
number of H ions donated by an acid or accepted by a base. For the reaction be-
tween sulfuric acid and ammonia
+
2–
H 2 SO 4 (aq) +2NH 3 (aq) t 2NH 4 (aq) +SO 4 (aq)
we find that n =2 for H 2 SO 4 and n =1 for NH 3 . For a complexation reaction, the
reaction unit is the number of electron pairs that can be accepted by the metal or
+
donated by the ligand. In the reaction between Ag and NH 3
+
+
Ag (aq) +2NH 3 (aq) t Ag(NH 3 ) 2 (aq)
+
the value of n for Ag is 2 and that for NH 3 is 1. Finally, in an oxidation–reduction
reaction the reaction unit is the number of electrons released by the reducing agent
or accepted by the oxidizing agent; thus, for the reaction
4+
3+
2+
2+
2Fe (aq) +Sn (aq) t Sn (aq) +2Fe (aq)
2+
n = 1 for Fe 3+ and n = 2 for Sn . Clearly, determining the number of equivalents
for a chemical species requires an understanding of how it reacts.
Normality is the number of equivalent weights (EW) per unit volume and, equivalent weight
like formality, is independent of speciation. An equivalent weight is defined as the The mass of a compound containing one
ratio of a chemical species’ formula weight (FW) to the number of its equivalents equivalent (EW).
FW formula weight
EW =
n The mass of a compound containing one
mole (FW).
Consequently, the following simple relationship exists between normality and
molarity.
N =n ´M
Example 2.1 illustrates the relationship among chemical reactivity, equivalent
weight, and normality.

EXAMPLE 2.1

Calculate the equivalent weight and normality for a solution of 6.0 M H 3 PO 4
given the following reactions:
–
3–
(a) H 3 PO 4 (aq) +3OH (aq) t PO 4 (aq) +3H 2 O(l)
2–
+
(b) H 3 PO 4 (aq) +2NH 3 (aq) t HPO 4 (aq) +2NH 4 (aq)
–
–
(c) H 3 PO 4 (aq) +F (aq) t H 2 PO 4 (aq) +HF(aq)
SOLUTION
+
For phosphoric acid, the number of equivalents is the number of H ions
donated to the base. For the reactions in (a), (b), and (c) the number of
equivalents are 3, 2, and 1, respectively. Thus, the calculated equivalent weights
and normalities are