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Enhanced Bromate Formation during Chlorination of Bromide- Containing Waters in the Presence of CuO: Catalytic Disproportionate of Hypobromous Acid

机译:CuO存在下的含溴水氯化过程中增强的溴酸盐形成:次溴酸的催化歧化

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摘要

Bromate (BrO_3~- ) in drinking water is traditionally seen as an ozonation byproduct from the oxidation of bromide (Br~), and its formation during chlorination is usually not significant This study shows enhanced bromate formation during chlorination of bromide-containing waters in the presence of cupric oxide (CuO). CuO was effective to catalyze hypochlorous acid (HOC1) or hypobromous acid (HOBr) decay (e.g., at least 104 times enhancement for HOBr at pH 8.6 by 0.2 g L~(-1) CuO). Significant halate concentrations were formed from a CuO-catalyzed hypohalite disproportjonation pathway. For example, the chlorate concentration was 2.7 ± 0.2 μM (225.5 ± 16.7 n% L~(-1)) after 90 min for HOC1 (C_o = 37 μM, 2.6 mg L~(-1) Cl_2) in the presence of 0.2 g L~(-1) CuO at pH 7.6, and the bromate concentration was 6.6 ± 0.5 μM (844.8 ± 64 μg L~(-1)) after 180 min for HOBr (C_o = 35 μM) in the presence of 0.2 g L~(-1) CuO at pH 8.6. The maximum halate formation was at pHs 7.6 and 8.6 for HOC1 or HOBr, respectively, which are dose to their corresponding pK_a values. In a HOCl-Br~-CuO system, BrO_3~- formation increases with increasing CuO doses and initial HOC1 and Br~- concentrations. A molar conversion (Br~ -to BrO_3~-) of up to (90 ±1)% could be achieved in the HOC1-Br~-CuO system because of recycling of Br" to HOBr by HOC1, whereas the maximum BrO_3~- yield in HOBr-CuO is only 26%. Bromate formation is initiated by the formation of a complex between CuO and HOBr/ OBr~-, which then reacts with HOBr to generate bromite. Bromite is further oxidized to BrO_3~- by a second CuO-catalyzed process. These novel findings may have implications for bromate formation during chlorination of bromide-containing drinking waters in copper pipes.
机译:传统上将饮用水中的溴酸盐(BrO_3〜-)视为溴化物(Br〜)氧化的臭氧化副产物,其在氯化过程中的形成通常不显着。该研究表明,在含溴化物的水中氯化过程中,溴酸盐的形成增强。氧化铜(CuO)的存在。 CuO可以有效催化次氯酸(HOCl)或次溴酸(HOBr)的衰变(例如,0.2 g L〜(-1)CuO对pH 8.6的HOBr具有至少104倍的增强作用)。从CuO催化的次卤酸盐歧化途径形成大量的卤化物。例如,HOC1(C_o = 37μM,2.6 mg L〜(-1)Cl_2)在0.2存在下90分钟后氯酸盐浓度为2.7±0.2μM(225.5±16.7 n%L〜(-1)) pH 7.6时的g L〜(-1)CuO,在0.2 g存在下180分钟后,HOBr(C_o = 35μM)的溴酸盐浓度为6.6±0.5μM(844.8±64μgL〜(-1)) pH 8.6的L〜(-1)CuO。对于HOCl或HOBr,最大的卤化物形成分别在pH 7.6和8.6,其剂量达到其相应的pK_a值。在HOCl-Br〜-CuO系统中,BrO_3-的形成随CuO剂量的增加以及初始HOC1和Br〜-浓度的增加而增加。在HOC1-Br〜-CuO系统中,由于HOCl将Br“转化为HOBr,而摩尔转化率(Br〜-至BrO_3〜-)可以达到(90±1)%。 HOBr-CuO的收率只有26%。溴酸盐的形成是通过在CuO和HOBr / OBr〜-之间形成络合物而引发的,然后与HOBr反应生成溴酸盐,溴酸盐被第二种CuO进一步氧化为BrO_3〜-。催化过程:这些新颖的发现可能对铜管中含溴化物的饮用水进行氯化过程中溴酸盐的形成有影响。

著录项

  • 来源
    《Environmental Science & Technology》 |2012年第20期|p.11054-11061|共8页
  • 作者单位

    Water Desalination and Reuse Center, King Abdullah University of Science and Technology (KAUST), Thuwal 23955-6900, Saudi Arabia;

    Eawag, Swiss Federal Institute of Aquatic Science and Technology, Ueberlandstrasse 133, CH-8600 Duebendorf, Switzerland,School of Architecture, Civil and Environmental Engineering (ENAC), Ecole Polytechnique Federale de Lausanne (EPFL), CH-1015 Lausanne, Switzerland;

    Water Desalination and Reuse Center, King Abdullah University of Science and Technology (KAUST), Thuwal 23955-6900, Saudi Arabia;

  • 收录信息 美国《科学引文索引》(SCI);美国《工程索引》(EI);美国《生物学医学文摘》(MEDLINE);美国《化学文摘》(CA);
  • 原文格式 PDF
  • 正文语种 eng
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  • 入库时间 2022-08-17 14:03:06

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