Page 350 - Modern Analytical Chemistry

1400-CH09 9/9/99 2:13 PM Page 333

Chapter 9 Titrimetric Methods of Analysis 333

Calculating the Titration Curve As an example, let’s calculate the titration curve
for the titration of 50.0 mL of 0.100 M Fe 2+ with 0.100 M Ce 4+ in a matrix of 1 M
HClO 4 . The reaction in this case is
4+
2+
3+
3+
Fe (aq)+Ce (aq) t Ce (aq)+Fe (aq) 9.16
The equilibrium constant for this reaction is quite large (it is approximately
15
6 ´10 ), so we may assume that the analyte and titrant react completely.
The first task is to calculate the volume of Ce 4+ needed to reach the equivalence
point. From the stoichiometry of the reaction we know
Moles Fe 2+ = moles Ce 4+
or

M Fe V Fe = M Ce V Ce
Solving for the volume of Ce 4+
(.100 M)(50.0 mL)
0
Fe
MV Fe
50
V Ce = = = .0 mL
M Ce (0.100 M)
gives the equivalence point volume as 50.0 mL.
Before the equivalence point the concentration of unreacted Fe 2+ and the con-
centration of Fe 3+ produced by reaction 9.16 are easy to calculate. For this reason
we find the potential using the Nernst equation for the analyte’s half-reaction
[ Fe 2 + ]
E = EFe / Fe 2 + -0 05916 ln 9.17
.
° 3
+
3
+
[ Fe ]
2+
3+
The concentrations of Fe and Fe after adding 5.0 mL of titrant are
moles unreacted Fe 2 + MV Ce Ce
Fe Fe -M V
[Fe 2 + ] = =
total volume V Fe +V Ce
(. 0 100 M)(50.0 mL ) - ( . 0 100 M)(5.0 mL) 2
= = . 818 ´ 10 - M
. mL
50.0 mL + 50
4+
moles Ce added MV
Ce Ce
[Fe 3+ ] = =
total volume V Fe +V Ce
(. 0 100 M)(5.0 mL) 3
= = . 909 ´ 10 - M
50.0 ml + 5.0 mL
Substituting these concentrations into equation 9.17 along with the formal potential
2+
3+
for the Fe /Fe half-reaction from Appendix 3D, we find that the potential is
æ 818. ´10 -2 ö
.
E =+0 767 V -0 05916 log =+ 0 711. V
.
ç -3 ÷
è 909. ´10 ø
At the equivalence point, the moles of Fe 2+ initially present and the moles of
Ce 4+ added are equal. Because the equilibrium constant for reaction 9.16 is large,
the concentrations of Fe 2+ and Ce 4+ are exceedingly small and difficult to calculate
without resorting to a complex equilibrium problem. Consequently, we cannot cal-
culate the potential at the equivalence point, E eq , using just the Nernst equation for
the analyte’s half-reaction or the titrant’s half-reaction. We can, however, calculate