preprints.org > chemistry and materials science > physical chemistry > doi: 10.20944/preprints202401.1779.v1

Preprint Article Version 1 Preserved in Portico This version is not peer-reviewed

Version 1 : Received: 12 January 2024 / Approved: 24 January 2024 / Online: 24 January 2024 (13:38:41 CET)

Our investigation focused on the degradation of perfluorooctanoic acid (PFOA) and 6:2 fluorotelomer sulfonamide alkyl betaine (6:2 FTAB, Capstone B), using UV photolysis under various pH conditions. Initially, we used PFOA as a reference, finding a 90% decomposition after 360 minutes at the original (unadjusted) pH 5.6, with a decomposition rate constant of (1.08 ± 0.30) × 10-4 sec-1 and a half-life of 107 ± 2 minutes. At pH 4 and 7, degradation averaged 85% and 80%, respectively, while at pH 10, it reduced to 57%. For 6:2 FTAB at its natural pH 6.5, almost complete decomposition occurred. The primary UV transformation product was identified as 6:2 fluorotelomer sulfonic acid (6:2 FTSA), occasionally accompanied by shorter-chain perfluoroalkyl acids (PFAA). Interestingly, the overall decomposition percentages were unaffected by pH for 6:2 FTAB, though pH influenced rate constants and half-lives. In PFOA degradation, direct photolysis and reaction with hydrated electrons were presumed mechanisms, excluding the involvement of hydroxyl radicals. The role of superoxide radicals remains uncertain. For 6:2 FTAB, both direct and indirect photolysis were observed, with potential involvement of hydroxyl, superoxide radicals, and/or other reactive oxygen species (ROS). Clarification is needed regarding the role of eaq- in the degradation of 6:2 FTAB.

PFOA; 6:2 FTAB; decomposition; photolysis; scavenger

Chemistry and Materials Science, Physical Chemistry

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