Solution to redox titration challenge

Analytical and Bioanalytical Chemistry, Jun 2017

Tadeusz Michałowski, Anna Maria Michałowska-Kaczmarczyk, Juris Meija

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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 .


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Tadeusz Michałowski, Anna Maria Michałowska-Kaczmarczyk, Juris Meija. Solution to redox titration challenge, Analytical and Bioanalytical Chemistry, 2017, 4113-4115, DOI: 10.1007/s00216-017-0308-8