Solution to redox titration challenge
Anal Bioanal Chem
Solution to redox titration challenge
Tadeusz Michałowski 0 1 2
Anna Maria Michałowska-Kaczmarczyk 0 1 2
Juris Meija 0 1 2
0 Measurement Science and Standards, National Research Council Canada , Ottawa, ON , Canada
1 Department of Oncology, The University Hospital in Kraków , Kraków , Poland
2 Faculty of Engineering and Chemical Technology, Technical University of Kraków , Kraków , Poland
Inspection of Fig. 3 allows us to evaluate the course of many processes during the titration. At the beginning of titration (Φ = 0) we see that the concentrations of IO- and HIO are vanishingly small (HIO is formed by hydrolysis IO- + H2O = HIO + OH- and [HIO] > [IO-] because HIO is a very weak acid). This suggests that both IO- and HIO have disproportionated. As a result, the iodine species with oxidation numbers below and above +1 are formed simultaneously. We see from Fig. 3 that initially I- and IO3- are the two predominant species. They are formed from the following half-reactions:
-
The problem at hand is to simulate the process of
titrating V0 mL of a NaIO solution (C0 = 0.01 mol/L) with V
mL of an HCl solution (C = 0.10 mol/L) [1]. The
mathematical setup of the problem is given in Fig. 1 and the
resulting graphs for the titration plots are shown in
Figs. 2 and 3.
From Fig. 2 we can establish that the equivalence
point, corresponding to the inflection point of the
titration curves E = E(Φ) and pH = pH(Φ), occurs at Φ =
0.801. Knowing the four parameters—pH, pI, pCl, and
E—at each titration point allows us to calculate the
concentration of any other species. For example, we
can determine that solid iodine emerges in the
equilibrium solid phase at Φ > 0.465.
This article is the solution to the Analytical Challenge to be found at
http://link.springer.com/article/10.1007/s00216-016-0020-0
IO−−4e– þ 2H2O ¼ IO3– þ 4Hþ
HIO–4e– þ 2H2O ¼ IO3– þ 5Hþ
From here we obtain the schemes of predominating
reactions of the disproportionating species, HIO and
IO–, at the start and in close vicinity of Φ = 0:
3IO– ¼ IO3– þ 2I– ðfrom 1 and 1aÞ
3HIO ¼ IO3– þ 2I– þ 3Hþ ðfrom 2 and 2aÞ
With the increase of Φ, the share of I2 grows and, at Φ =
0.465, the excess of I2 forms the precipitate.
Fig. 1 The R code for simulating
the titration of NaIO with HCl
Fig. 2 Changes of pH and E during the titration of NaIO with HCl
Fig. 3 Changes of chlorine and iodine species concentration during the titration of NaIO with HCl
2IO– þ 2e– þ 4Hþ ¼ I2ðsÞ þ 2H2O
From Eqs. (1), (3a), and (3b) we have:
5IO– þ 4Hþ ¼ IO3– þ 2ðI2; I2ðsÞÞ þ 2H2O
This equation explains the inflection point of the
titration curve at Φ ≈ 4/5, although several other reactions
clearly take place. In this system, chloride ions
(introduced by HCl) could also be considered a priori as a
reducing agent. Such a possibility was assumed a priori
when the balances involving all (known) products of
chloride oxidation and complexation (I2Cl–, ICl, and
ICl2–) were included. This way, full “democracy” was
assumed with no simplifications. However, from the
calc u l a t i o n s w e s e e t h a t H C l a c t s p r i m a r i l y a s a
disproportionating, and not reducing agent.
1. Meija J , Michałowska-Kaczmarczyk AM , Michałowski T . Anal Bioanal Chem . 2017 ; 409 : 11 - 13 .