CN115925086A - A method for degrading organic matter in wastewater - Google Patents
A method for degrading organic matter in wastewater Download PDFInfo
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Abstract
本发明提供了一种降解废水中有机物的方法,属于废水处理技术领域,该方案通过在废水中形成二价锰离子/金属配合物体系,利用二价锰离子/金属配合物体系激活四价硫离子形成硫酸根自由基,通过硫酸根自由基对有机物进行降解。该方法可有效解决现有技术中采用二价锰激活四价硫离子时存在的激活效率低的问题。The invention provides a method for degrading organic matter in wastewater, which belongs to the technical field of wastewater treatment. The scheme forms a divalent manganese ion/metal complex system in wastewater, and utilizes the divalent manganese ion/metal complex system to activate tetravalent sulfur The ions form sulfate radicals, which degrade organic matter through sulfate radicals. The method can effectively solve the problem of low activation efficiency when divalent manganese is used to activate tetravalent sulfur ions in the prior art.
Description
技术领域technical field
本发明属于有机物降解技术领域,具体涉及一种降解废水中有机物的方法。The invention belongs to the technical field of organic matter degradation, and in particular relates to a method for degrading organic matter in waste water.
背景技术Background technique
基于SO4·-自由基的高级氧化工艺(SR-AOPs)具有比传统Fenton氧化工艺中的羟基自由基具有更广的应用pH范围、更高的氧化还原电位(SO4·-自由基为2.5~3.1V,OH·自由基为1.8~2.7V)以及更长的半衰期(SO4·-自由基为30~40μs,OH·自由基为1μs)从而得到了广泛的研究和开发。在SR-AOPs中,过硫酸盐(过二硫酸盐(PDS)和过一硫酸氢盐(PMS))经常被用作产生SO4·-自由基的前驱体。然而,过硫酸盐的应用仍然面临着成本高、酸度大和残留生物毒性的缺陷。亚硫酸盐(S(IV))是一种常见的源于二氧化硫的工业副产物,由于其环境友好性、价格低廉且同样可以被活化生成SO4·-自由基,因而如何高效地活化S(IV)产生SO4·-自由基越来越成为研究的重点。The advanced oxidation process (SR-AOPs) based on SO4 - radicals has a wider application pH range and higher redox potential than the hydroxyl radicals in the traditional Fenton oxidation process (SO4 - radicals are 2.5 to 3.1 V, OH · free radical is 1.8 ~ 2.7V) and longer half-life (SO4 · - free radical is 30 ~ 40μs, OH · free radical is 1μs), which has been extensively researched and developed. In SR-AOPs, persulfates (peroxydisulfate (PDS) and permonomonosulfate (PMS)) are often used as precursors to generate SO4 ·- radicals. However, the application of persulfate still faces the defects of high cost, high acidity and residual biological toxicity. Sulfite (S(IV)) is a common industrial by-product derived from sulfur dioxide. Due to its environmental friendliness, low price and the ability to be activated to generate SO4 ·-free radicals, how to efficiently activate S(IV) ) to produce SO4 ·-free radicals has increasingly become the focus of research.
常见的活化S(IV)的方法主要包括热活化、光活化、电活化和过渡金属活化。锰(Mn)是一种常见的过渡金属,具有毒性低,自然储量丰富等特点。Mn常被用于SR-AOPs,被认为是一种优秀的金属催化剂。目前研究多致力于将高锰酸盐、单一或复合锰氧化物与S(IV)结合。然而,因为Mn(II)活化S(IV)时,存在Mn(II)和Mn(III)之间的转化,而处在中碱性溶液中的Mn(III)非常不稳定,会发生歧化反应生成Mn(II)和MnO2(平衡常数log K=7~9),导致水溶液中Mn(II)活化S(IV)的研究与应用受到限制。Common methods for activating S(IV) mainly include thermal activation, photoactivation, electrical activation, and transition metal activation. Manganese (Mn) is a common transition metal with low toxicity and abundant natural reserves. Mn is often used in SR-AOPs and is considered as an excellent metal catalyst. At present, most studies are devoted to combining permanganate, single or composite manganese oxides with S(IV). However, because Mn(II) activates S(IV), there is a conversion between Mn(II) and Mn(III), and Mn(III) in neutral alkaline solution is very unstable, and disproportionation will occur The formation of Mn(II) and MnO 2 (equilibrium constant log K=7~9) has limited the research and application of Mn(II) activation of S(IV) in aqueous solution.
发明内容Contents of the invention
针对现有技术中存在的上述问题,本发明提供一种降解废水中有机物的方法,该方法可有效解决现有技术中采用二价锰激活四价硫离子时存在的激活效率低的问题。Aiming at the above-mentioned problems in the prior art, the present invention provides a method for degrading organic matter in wastewater, which can effectively solve the problem of low activation efficiency in the prior art when divalent manganese is used to activate tetravalent sulfide ions.
为实现上述目的,本发明解决其技术问题所采用的技术方案是:In order to achieve the above object, the technical solution adopted by the present invention to solve the technical problems is:
一种降解废水中有机物的方法,其特征在于,通过在废水中形成二价锰/金属配合物体系,利用二价锰/金属配合物体系激活四价硫形成硫酸根自由基,通过硫酸根自由基对有机物进行降解。A method for degrading organic matter in wastewater, characterized in that, by forming a divalent manganese/metal complex system in wastewater, utilizing the divalent manganese/metal complex system to activate tetravalent sulfur to form sulfate radicals, base to degrade organic matter.
进一步地,具体操作过程如下:将含有机物的水体、二价锰盐和金属配合剂混合,得混合溶液,调节混合溶液pH值为4-10,然后向其中添加亚硫酸盐,搅拌反应即可。Further, the specific operation process is as follows: mix the water body containing organic matter, divalent manganese salt and metal complexing agent to obtain a mixed solution, adjust the pH value of the mixed solution to 4-10, then add sulfite to it, and stir to react .
进一步地,混合溶液中二价锰盐与金属配合剂的摩尔比为1:1-10。Further, the molar ratio of the divalent manganese salt to the metal complexing agent in the mixed solution is 1:1-10.
进一步地,混合溶液中二价锰盐的浓度为10-100μM。Further, the concentration of the divalent manganese salt in the mixed solution is 10-100 μM.
进一步地,四价硫离子在混合溶液中的浓度为100-800μM。Further, the concentration of tetravalent sulfide ions in the mixed solution is 100-800 μM.
进一步地,调节混合溶液pH值为6-7.5。Further, adjust the pH value of the mixed solution to 6-7.5.
进一步地,于25-35℃条件下反应即可。Further, the reaction can be carried out under the condition of 25-35°C.
进一步地,二价锰盐包括一水合硫酸锰、氯化锰和硝酸锰中的至少一种。Further, the divalent manganese salt includes at least one of manganese sulfate monohydrate, manganese chloride and manganese nitrate.
进一步地,金属配合剂为次氮基三乙酸钠盐。Further, the metal complexing agent is nitrilotriacetic acid sodium salt.
进一步地,亚硫酸盐包括无水亚硫酸钠、亚硫酸钾、亚硫酸钙、亚硫酸氢钠和亚硫酸氢钾中的至少一种。Further, the sulfite includes at least one of anhydrous sodium sulfite, potassium sulfite, calcium sulfite, sodium bisulfite and potassium bisulfite.
本发明所产生的有益效果为:The beneficial effects produced by the present invention are:
1、本申请的方法中向废水中添加二价锰(Mn(II))和金属配合物(NTA),在NTA的存在下,使得Mn(II)形成稳定的配位形态,不易发生水解歧化反应,配位形态的Mn(II)激发活化四价硫(S(IV)),使其形成硫酸根自由基(SO4 ·-),利用SO4 ·-的强氧化性将水体中的有机物进行降解;由于有金属配合物的存在,可大大提高Mn(II)在水体中的稳定状态,进而提高对S(IV)的激发作用,提高有机物的降解效果。1. In the method of the present application, divalent manganese (Mn(II)) and metal complexes (NTA) are added to the wastewater. In the presence of NTA, Mn(II) forms a stable coordination form, which is not prone to hydrolysis and disproportionation Reaction, the coordination form of Mn(II) excites and activates tetravalent sulfur (S(IV)) to form sulfate radicals (SO 4 ·- ), and utilizes the strong oxidation of SO 4 ·- to decompose organic matter in water Degradation; due to the existence of metal complexes, the stable state of Mn(II) in water can be greatly improved, and then the excitation of S(IV) can be improved, and the degradation effect of organic matter can be improved.
2、本申请的方法中以亚硫酸盐为原料实现对水体中有机污染物进行降解,亚硫酸盐性质稳定、便于运输和储存,价格低廉,来源广泛,无生物毒性作用,大大提高使用的安全性。2. In the method of this application, sulfite is used as a raw material to degrade organic pollutants in water bodies. The sulfite is stable in nature, convenient for transportation and storage, low in price, wide in source, and has no biological toxicity, which greatly improves the safety of use sex.
3、本申请中以二价锰为原料,二价锰广泛存在于地下水和天然水中,在水体中分布均匀,可减少原料的使用;且二价锰来源广泛,价格低廉,可降低处理成本;二价锰能够快速、充分的活化亚硫酸盐,弥补了二价锰离子高效活化亚硫酸盐的理论空缺。3. In this application, divalent manganese is used as raw material. Divalent manganese widely exists in groundwater and natural water, and is evenly distributed in water bodies, which can reduce the use of raw materials; and divalent manganese has a wide range of sources and low prices, which can reduce processing costs; Divalent manganese can quickly and fully activate sulfite, which makes up for the theoretical gap in the efficient activation of sulfite by divalent manganese ions.
附图说明Description of drawings
图1为不同反应体系中有机物的降解效果统计图;Fig. 1 is the statistical diagram of the degradation effect of organic matter in different reaction systems;
图2为Mn(III)-NTA和Mn(II)/NTA/S(IV)体系在t=1min时的紫外-可见光光谱图;Fig. 2 is the UV-Vis spectrogram of Mn(III)-NTA and Mn(II)/NTA/S(IV) system at t=1min;
图3为RhB在Mn(III)-NTA和Mn(III)-NTA/S(IV)体系中的降解统计图;Fig. 3 is the degradation statistics diagram of RhB in Mn(III)-NTA and Mn(III)-NTA/S(IV) systems;
图4中a为EtOH和TBA对Mn(II)/NTA/S(IV)体系中RhB降解的统计图;b为EtOH和TBA对Mn(III)-NTA/RhB体系中RhB降解的统计图;Among Fig. 4, a is the statistical diagram of EtOH and TBA to RhB degradation in the Mn(II)/NTA/S(IV) system; b is the statistical diagram of EtOH and TBA to the RhB degradation in the Mn(III)-NTA/RhB system;
图5中a为改变Mn(II)浓度对RhB降解的统计图;b为改变[Mn(II)]与[NTA]的比例后对RhB降解的统计图;Among Fig. 5, a is a statistical diagram of changing the concentration of Mn(II) on RhB degradation; b is a statistical diagram of RhB degradation after changing the ratio of [Mn(II)] and [NTA];
图6为S(IV)投加量对Mn(II)/NTA/S(IV)体系降解RhB的统计图;Fig. 6 is the statistical diagram of S(IV) dosage on the degradation of RhB by Mn(II)/NTA/S(IV) system;
图7中a为pH对Mn(II)/NTA/S(IV)体系降解RhB的统计图;b为Mn(II)-NTA和NTA物种在溶液中的百分含量随pH变化图;Among Fig. 7, a is the statistical diagram of pH on Mn(II)/NTA/S(IV) system degrading RhB; b is the percentage content of Mn(II)-NTA and NTA species in the solution with pH change diagram;
图8为无机阴离子对Mn(II)/NTA/S(IV)体系降解RhB的统计图。Fig. 8 is a statistical diagram of the degradation of RhB by inorganic anions on the Mn(II)/NTA/S(IV) system.
具体实施方式Detailed ways
下面结合附图对本发明的具体实施方式做详细的说明。The specific embodiments of the present invention will be described in detail below in conjunction with the accompanying drawings.
实施例1Example 1
一种降解废水中有机物的方法,具体操作过程如下:将含有机物的水体、MnSO4·H2O和NTA混合,得混合溶液,混合溶液中MnSO4·H2O和NTA的摩尔比为1:3,且混合溶液中MnSO4·H2O的浓度为50μM,使用硼酸-硼酸钠缓冲液调节混合溶液pH值为7.5,然后向其中添加无水亚硫酸钠(S(IV)),使其在混合溶液中的浓度为500μM,于30℃条件下以2000r/min搅拌反应即可。A method for degrading organic matter in wastewater, the specific operation process is as follows: the water body containing organic matter, MnSO 4 ·H 2 O and NTA are mixed to obtain a mixed solution, and the molar ratio of MnSO 4 ·H 2 O and NTA in the mixed solution is 1 :3, and the concentration of MnSO 4 ·H 2 O in the mixed solution is 50 μM, use boric acid-sodium borate buffer to adjust the pH value of the mixed solution to 7.5, and then add anhydrous sodium sulfite (S(IV)) to it to make it in The concentration in the mixed solution is 500 μM, and it is sufficient to stir and react at 2000 r/min at 30° C.
实施例2Example 2
一种降解废水中有机物的方法,具体操作过程如下:将含有机物的水体、氯化锰和NTA混合,得混合溶液,混合溶液中氯化锰和NTA的摩尔比为1:5,且混合溶液中氯化锰的浓度为30μM,使用硼酸-硼酸钠调节混合溶液pH值为6,然后向其中添加无水亚硫酸钠(S(IV)),使其在混合溶液中的浓度为300μM,于25℃条件下以2000r/min搅拌反应即可。A method for degrading organic matter in wastewater, the specific operation process is as follows: the water body containing organic matter, manganese chloride and NTA are mixed to obtain a mixed solution, the molar ratio of manganese chloride and NTA in the mixed solution is 1:5, and the mixed solution The concentration of manganese chloride in the medium is 30 μM, and the pH value of the mixed solution is adjusted to 6 by using boric acid-sodium borate, and then anhydrous sodium sulfite (S(IV)) is added therein so that the concentration in the mixed solution is 300 μM. Under the conditions, the reaction can be stirred at 2000r/min.
实施例3Example 3
一种降解废水中有机物的方法,具体操作过程如下:将含有机物的水体、硝酸锰和NTA混合,得混合溶液,混合溶液中硝酸锰和NTA的摩尔比为1:1,且混合溶液中硝酸锰的浓度为20μM,使用硼酸-硼酸钠缓冲液调节混合溶液pH值为7,然后向其中添加无水亚硫酸钠(S(IV)),使其在混合溶液中的浓度为200μM,于35℃条件下以2000r/min搅拌反应即可。A method for degrading organic matter in wastewater, the specific operation process is as follows: the water body containing organic matter, manganese nitrate and NTA are mixed to obtain a mixed solution, the molar ratio of manganese nitrate and NTA in the mixed solution is 1:1, and nitric acid in the mixed solution The concentration of manganese is 20 μM, and the pH value of the mixed solution is adjusted to 7 using boric acid-sodium borate buffer solution, and then anhydrous sodium sulfite (S(IV)) is added thereto so that the concentration in the mixed solution is 200 μM. The reaction can be stirred at 2000r/min.
实施例4Example 4
一种降解废水中有机物的方法,具体操作过程如下:将含有机物的水体、MnSO4·H2O和NTA混合,得混合溶液,混合溶液中MnSO4·H2O和NTA的摩尔比为1:10,且混合溶液中MnSO4·H2O的浓度为100μM,使用硼酸-硼酸钠或乙酸-乙酸钠缓冲液调节混合溶液pH值为6.5,然后向其中添加无水亚硫酸钠(S(IV)),使其在混合溶液中的浓度为800μM,于30℃条件下以2000r/min搅拌反应即可。A method for degrading organic matter in wastewater, the specific operation process is as follows: the water body containing organic matter, MnSO 4 ·H 2 O and NTA are mixed to obtain a mixed solution, and the molar ratio of MnSO 4 ·H 2 O and NTA in the mixed solution is 1 :10, and the concentration of MnSO 4 ·H 2 O in the mixed solution is 100 μM, use boric acid-sodium borate or acetic acid-sodium acetate buffer to adjust the pH value of the mixed solution to 6.5, and then add anhydrous sodium sulfite (S(IV) ), so that the concentration in the mixed solution is 800 μM, and the reaction can be stirred at 2000 r/min under the condition of 30 ° C.
试验例Test case
1、配置有机物溶液1. Configure organic solution
罗丹明B(RhB)是一种常见的人工染料,因为其结构和大多有机污染物相似,具有苯环和多种活性位点,并且检测方法便捷,因此将RhB作为模型有机污染物,将其加入水中制成RhB浓度为1mM的RhB溶液,备用。Rhodamine B (RhB) is a common artificial dye, because its structure is similar to most organic pollutants, it has a benzene ring and a variety of active sites, and the detection method is convenient, so RhB is used as a model organic pollutant, and its Add water to prepare a RhB solution with a RhB concentration of 1 mM, and set aside.
2、试验2. Test
2.1、对照实验2.1. Control experiment
取RhB溶液,采用实施例1中的方法对溶液中的RhB进行降解处理,该组作为实验组;Get the RhB solution, adopt the method in
取RhB溶液,向其中添加MnSO4·H2O(Mn(II))对溶液中的RhB进行降解处理,该组作为对照组1;Take the RhB solution, add MnSO 4 ·H 2 O (Mn(II)) to it to degrade the RhB in the solution, and this group is used as the
取RhB溶液,向其中添加金属配合剂(NTA)对溶液中的RhB进行降解处理,该组作为对照组2;Get the RhB solution, add metal complexing agent (NTA) therein to degrade the RhB in the solution, and this group is used as the
取RhB溶液,向其中添加无水亚硫酸钠(S(IV))对溶液中的RhB进行降解处理,该组作为对照组3;Take the RhB solution, add anhydrous sodium sulfite (S(IV)) to it to degrade the RhB in the solution, and this group is used as the
取RhB溶液,向其中添加MnSO4·H2O(Mn(II))和无水亚硫酸钠(S(IV))对溶液中的RhB进行降解处理,该组作为对照组4;Take the RhB solution, add MnSO 4 ·H 2 O (Mn(II)) and anhydrous sodium sulfite (S(IV)) to it to degrade RhB in the solution, and this group is used as the
取RhB溶液,向其中添加MnSO4·H2O(Mn(II))和金属配合剂(NTA)对溶液中的RhB进行降解处理,该组作为对照组5;Take the RhB solution, add MnSO 4 ·H 2 O (Mn(II)) and metal complexing agent (NTA) to it to degrade the RhB in the solution, and this group is used as the
取RhB溶液,向其中添加金属配合剂(NTA)和无水亚硫酸钠(S(IV))对溶液中的RhB进行降解处理,该组作为对照组6;Take the RhB solution, add metal complexing agent (NTA) and anhydrous sodium sulfite (S(IV)) to it to degrade the RhB in the solution, and this group is used as the
投入药剂后分别于0、1、2、3、4、5、7、10、15min时分别取样,立刻加入至含有1mL200mM的Na2S2O3溶液的比色管中。Na2S2O3溶液的作用是猝灭体系中可能存在的活性物种,从而停止RhB进一步被氧化。使用紫外可见分光光度计在554nm处测定RhB的吸光度,同样采用紫外可见分光光度计测量反应过程中不同体系的紫外-可见光吸收光谱图;统计反应后溶液中RhB的含量,具体结果见图1。Samples were taken at 0, 1, 2, 3, 4, 5, 7, 10, and 15 minutes after the drug was injected, and immediately added to a colorimetric tube containing 1 mL of 200 mM Na 2 S 2 O 3 solution. The role of Na 2 S 2 O 3 solution is to quench the active species that may exist in the system, thereby stopping the further oxidation of RhB. Use a UV-visible spectrophotometer to measure the absorbance of RhB at 554nm, and also use a UV-visible spectrophotometer to measure the UV-visible light absorption spectra of different systems in the reaction process; count the content of RhB in the solution after the reaction, and the specific results are shown in Figure 1.
通过图1可以看出,在单独的Mn(II),NTA和S(IV)的一元体系以及Mn(II)/S(IV),Mn(II)/NTA和NTA/S(IV)的二元体系中,RhB均没有明显的降解。而在实验组即实施例1中的Mn(II)/NTA/S(IV)体系中,RhB的降解率达到了86.05%。因此,引入配体NTA与Mn(II)形成的Mn(II)-NTA能够有效与S(IV)发生反应,并且产生活性氧化物种降解RhB。It can be seen from Figure 1 that in the single system of Mn(II), NTA and S(IV) and the binary system of Mn(II)/S(IV), Mn(II)/NTA and NTA/S(IV), In the element system, RhB was not significantly degraded. In the experimental group, namely the Mn(II)/NTA/S(IV) system in Example 1, the degradation rate of RhB reached 86.05%. Therefore, Mn(II)-NTA formed by introducing ligand NTA and Mn(II) can effectively react with S(IV) and generate active oxidative species to degrade RhB.
2.2、Mn(III)的生成2.2. Formation of Mn(III)
Mn(II)在活化S(IV)产生SO4 ·-自由基的过程中,自身也会被氧化为Mn(III),Mn(III)自身具有很强的氧化性,Mn(III)遇水容易发生歧化(式(1)),导致溶液中存在的Mn(III)含量甚微,无法实现降解RhB的目的。During the process of activating S(IV) to generate SO 4 ·- free radicals, Mn(II) itself will also be oxidized to Mn(III). Mn(III) itself has strong oxidizing properties, and Mn(III) Prone to disproportionation (formula (1)), resulting in very little Mn(III) content in the solution, unable to achieve the purpose of degrading RhB.
2Mn(III)+2H2O=MnO2+Mn(II)+4H+ (1)2Mn(III)+2H 2 O=MnO 2 +Mn(II)+4H + (1)
通过以下式(2)中的反应过程原位制备Mn(III)-NTA体系,利用紫外可见分光光度计扫描了原位制备的Mn(III)-NTA体系与实施例1中的Mn(II)/NTA/S(IV)体系在t=1min时的紫外-可见光光谱图,结果见图2。The Mn(III)-NTA system was prepared in situ through the reaction process in the following formula (2), and the Mn(III)-NTA system prepared in situ and the Mn(II) in Example 1 were scanned using a UV-visible spectrophotometer The UV-Vis spectrum of the /NTA/S(IV) system at t=1min, the results are shown in Figure 2.
通过图2可以看出,原位制备的Mn(III)-NTA光谱中大约在280mm处出现了一个特征吸收峰,这与配位形态的Mn(III)的吸收峰一致。结果表明在NTA存在条件下,通过式(2)原位制备的Mn(III)-NTA可以稳定存在。同时,Mn(II)/NTA/S(IV)体系的光谱图中,在同样的位置观察到了Mn(III)-NTA的特征吸收峰,因此可以确认Mn(II)/NTA/S(IV)体系中产生了Mn(III)-NTA。It can be seen from Figure 2 that a characteristic absorption peak appears at about 280 mm in the spectrum of the in situ prepared Mn(III)-NTA, which is consistent with the absorption peak of the coordinated form of Mn(III). The results show that the Mn(III)-NTA prepared in situ by formula (2) can exist stably in the presence of NTA. At the same time, in the spectrogram of the Mn(II)/NTA/S(IV) system, the characteristic absorption peak of Mn(III)-NTA was observed at the same position, so it can be confirmed that the Mn(II)/NTA/S(IV) Mn(III)-NTA was produced in the system.
分别将实施例1中的Mn(II)/NTA/S(IV)体系、原位制备的Mn(III)-NTA体系以及Mn(III)-NTA/S(IV)体系直接与RhB进行反应,Mn(III)-NTA体系中的RhB并未明显发生降解,而Mn(III)-NTA/S(IV)体系中的RhB降解,具体结果见图3。The Mn(II)/NTA/S(IV) system in Example 1, the Mn(III)-NTA system prepared in situ and the Mn(III)-NTA/S(IV) system were directly reacted with RhB respectively, RhB in the Mn(III)-NTA system did not obviously degrade, but RhB in the Mn(III)-NTA/S(IV) system degraded. The specific results are shown in Figure 3.
通过图3可以看出,Mn(III)-NTA/S(IV)体系中,RhB被快速降解,甚至降解效率比Mn(II)/NTA/S(IV)体系更高,达到了95.72%。可知,Mn(III)-NTA并不能直接氧化降解RhB,而加入S(IV)与Mn(III)-NTA反应后可能生成其他的活性氧化物种从而降解RhB。推测可能是与NTA配位状态的Mn(III)较为稳定,能够比Mn(II)更能快速活化S(IV)从而导致了RhB的降解。因此,稳定的Mn(III)-NTA有促进Mn物种的循环和有效激活S(IV)的重要作用。It can be seen from Figure 3 that in the Mn(III)-NTA/S(IV) system, RhB was rapidly degraded, and even the degradation efficiency was higher than that of the Mn(II)/NTA/S(IV) system, reaching 95.72%. It can be seen that Mn(III)-NTA cannot directly oxidize and degrade RhB, but after adding S(IV) to react with Mn(III)-NTA, other active oxidative species may be generated to degrade RhB. It is speculated that Mn(III) in the coordination state with NTA is more stable and can activate S(IV) faster than Mn(II), thus leading to the degradation of RhB. Therefore, the stable Mn(III)-NTA plays an important role in promoting the circulation of Mn species and effectively activating S(IV).
3、活性氧化物种的探究3. Exploration of active oxide species
推测Mn(II)/NTA/S(IV)反应体系中可能产生硫酸根自由基(SO4 ·-)和羟基自由基(OH·),自由基捕获剂与SO4 ·-自由基和OH·自由基反应具有不同的二阶速率常数,因此,利用叔丁醇(TBA)作为OH·自由基的特定捕获剂(k(TBA,SO4·-)=8.5×105M-1s-1,k(TBA,OH·)=6.0×108M-1s-1)。而无水乙醇(EtOH)可以同时淬灭SO4 ·-自由基和OH·自由基(k(EtOH,SO4·-)=(1.6~7.7)×107M-1s-1,k(EtOH,OH·)=(1.2~2.8)×109M-1s-1),具体结果见图4。It is speculated that sulfate radicals (SO 4 ·- ) and hydroxyl radicals (OH · ) may be generated in the Mn(II)/NTA/S(IV) reaction system, and the free radical scavengers interact with SO 4 ·- radicals and OH· Free radical reactions have different second-order rate constants. Therefore, using tert-butanol (TBA) as a specific scavenger for OH radicals (k (TBA,SO4·-) = 8.5×10 5 M -1 s -1 , k( TBA,OH· )=6.0×10 8 M −1 s −1 ). However, absolute ethanol (EtOH) can simultaneously quench SO 4 ·- radicals and OH · radicals (k (EtOH,SO4·-) = (1.6~7.7)×10 7 M -1 s -1 , k (EtOH ,OH·) = (1.2~2.8)×10 9 M -1 s -1 ), see Figure 4 for the specific results.
如图4a和图4b所示,当添加20mM TBA到Mn(II)/NTA/S(IV)和Mn(III)-NTA/S(IV)体系中时,RhB的降解率仅分别降低了6.13%和5.18%,这说明OH·自由基并不是起主要作用的活性氧化物种。然而,添加20mM EtOH作为自由基捕获剂到两种体系中时,RhB的降解几乎完全被抑制,因此可以确认SO4 ·-自由基是两种体系中的决定性氧化物种。As shown in Figure 4a and Figure 4b, when 20mM TBA was added to Mn(II)/NTA/S(IV) and Mn(III)-NTA/S(IV) systems, the degradation rate of RhB was only reduced by 6.13 % and 5.18%, which shows that OH · free radicals are not the main active oxidation species. However, when 20mM EtOH was added as a radical scavenger to both systems, the degradation of RhB was almost completely inhibited, thus confirming that SO 4 · -radical is the decisive oxidative species in both systems.
4、Mn(II)和NTA浓度的影响4. Effect of Mn(II) and NTA concentration
固定[Mn(II)]:[NTA]=1:3,逐步提高Mn(II)的浓度时,RhB的降解率的变化如图5a所示,RhB的降解率的变化逐渐提升,但是,当Mn(II)浓度增加到100μM时,降解效率急剧下降至53.27%。推测可能是因为Mn(II)浓度的增加会提高后续Mn-NTA配合物浓度,利于S(IV)的活化作用从而降解RhB。然而当Mn(II)浓度持续增加至100μM时,引入的大量配体会与RhB竞争消耗SO4 ·-自由基,导致RhB的降解受到阻碍。Fixed [Mn(II)]:[NTA]=1:3, when increasing the concentration of Mn(II) step by step, the change of the degradation rate of RhB is shown in Figure 5a, the change of the degradation rate of RhB is gradually improved, but, when When the concentration of Mn(II) increased to 100μM, the degradation efficiency dropped sharply to 53.27%. It is speculated that the increase of Mn(II) concentration will increase the concentration of subsequent Mn-NTA complexes, which is conducive to the activation of S(IV) and the degradation of RhB. However, when the concentration of Mn(II) continued to increase to 100 μM, a large number of ligands introduced would compete with RhB to consume SO 4 ·- free radicals, resulting in hindering the degradation of RhB.
控制Mn(II)的浓度为50μM,改变[Mn(II)]:[NTA]=1:1~1:10时,RhB的降解率的变化如图5b所示,当两者比例从1:1降低到1:3时,RhB降解率从64.23%提升至86.04%。而[Mn(II)]:[NTA]为1:5以及1:10时,RhB的降解率分别为86.3%和85.66%,可见RhB降解受到的影响可以忽略不计。当[Mn(II)]:[NTA]=1:1时,可能是由于在NTA不足的情况下,生成的Mn(II)-NTA和Mn(III)-NTA浓度低,S(IV)的活化效率不高,从而无法有效降解RhB。当两者比例达到1:3~1:10时,在适当过量的NTA条件下,生成的Mn物种浓度能够保持充足且稳定,利于后续S(IV)的激活。Control the concentration of Mn(II) to 50 μM, and change [Mn(II)]:[NTA]=1:1~1:10, the change of the degradation rate of RhB is shown in Figure 5b, when the ratio of the two changes from 1: When 1 was reduced to 1:3, the RhB degradation rate increased from 64.23% to 86.04%. When [Mn(II)]:[NTA] was 1:5 and 1:10, the degradation rates of RhB were 86.3% and 85.66%, respectively, showing that the degradation of RhB was negligible. When [Mn(II)]:[NTA]=1:1, it may be due to the low concentration of Mn(II)-NTA and Mn(III)-NTA generated in the case of insufficient NTA, and the concentration of S(IV) The activation efficiency is not high, so RhB cannot be effectively degraded. When the ratio of the two reaches 1:3-1:10, under the condition of an appropriate excess of NTA, the concentration of the generated Mn species can remain sufficient and stable, which is conducive to the subsequent activation of S(IV).
5、S(IV)投加量的影响5. The effect of S(IV) dosage
如图6所示,S(IV)投加浓度分别为100、200、300以及500μM时,RhB降解率分别为13.79%,56.36%,83.88%以及86.24%。但是,当S(IV)浓度达到800μM时,RhB的降解率反而急剧下降至16.89%。S(IV)与SO4 ·-自由基和Mn(III)的形成密切相关,充足的S(IV)可以被激活产生大量的SO4 ·-自由基参与RhB降解。然而,过量的S(IV)可能会与RhB竞争SO4 ·-自由基(式(3)),从而引起自由基的内耗,降低SO4 ·-自由基的利用率,导致RhB的降解率下降。此外,过量的S(IV)的加入会导致溶解氧的大量消耗,从而降低SO5 ·-自由基和SO4 ·-自由基的产量(式(4)和式(5)),并限制后续活性氧化物种的生成。As shown in Figure 6, when the S(IV) concentration was 100, 200, 300 and 500 μM, the RhB degradation rates were 13.79%, 56.36%, 83.88% and 86.24%, respectively. However, when the concentration of S(IV) reached 800 μM, the degradation rate of RhB dropped sharply to 16.89%. S(IV) is closely related to the formation of SO 4 ·- free radicals and Mn(III), sufficient S(IV) can be activated to produce a large number of SO 4 ·- free radicals to participate in the degradation of RhB. However, excess S(IV) may compete with RhB for SO 4 ·- radicals (Equation (3)), causing internal friction of free radicals, reducing the utilization rate of SO 4 ·- radicals, and resulting in a decrease in the degradation rate of RhB . In addition, the addition of excessive S(IV) will lead to a large consumption of dissolved oxygen, thereby reducing the production of SO 5 ·- radicals and SO 4 ·- radicals (Equation (4) and Equation (5)), and limiting the subsequent Generation of reactive oxidative species.
SO4 ·-+SO3 2-→SO3 ·-+SO4 2-(3)SO 4 ·- +SO 3 2- → SO 3 ·- +SO 4 2- (3)
SO3 ·-+O2-----→SO5 ·-(4)SO 3 ·- +O 2 -----→SO 5 ·- (4)
SO5 ·-+SO3 2-→SO4 ·-+SO4 2-(5)SO 5 ·- +SO 3 2- → SO 4 ·- +SO 4 2- (5)
6、pH的影响6. The influence of pH
pH对Mn(II)/NTA/S(IV)体系降解RhB的影响结果见图7,通过图7a可知,当溶液的pH=4时,RhB的降解率仅有11.79%。当溶液的pH=6时,RhB的降解率显著提高至99.31%。当溶液的pH=7.5时,RhB降解率为86.05%,但在pH=9条件下,RhB的降解率又下降至49.33%,pH对溶液中物质的电离平衡有很大的影响。图7b为专业化学平衡软件Medusa绘制的Mn(II)-NTA和NTA在溶液中组分的百分数随pH的变化图。NTA的分子结构为(CH2COOH)3N,在较低的pH值下,水溶液中大量的NTA以三元酸的形式存在,因此pH=4时,仅有少量的Mn(II)-NTA形成,后续的S(IV)活化反应难以进行,导致RhB的降解效率低;随着pH的增加,Mn(II)-NTA的含量逐渐升高,S(IV)能够被有效激活生成SO4 ·-自由基降解RhB。但是当pH=9时,Mn(II)/NTA/S(IV)体系中产生的Mn(III)的水解歧化反应加剧,不利于Mn(III)-NTA的稳定生成和后续的反应。The effect of pH on the degradation of RhB by the Mn(II)/NTA/S(IV) system is shown in Figure 7. From Figure 7a, it can be seen that when the pH of the solution is 4, the degradation rate of RhB is only 11.79%. When the pH of the solution was 6, the degradation rate of RhB was significantly increased to 99.31%. When the pH of the solution is 7.5, the degradation rate of RhB is 86.05%, but at pH = 9, the degradation rate of RhB drops to 49.33%. The pH has a great influence on the ionization balance of the substances in the solution. Fig. 7b is a graph showing the percentage of Mn(II)-NTA and NTA in solution as a function of pH, drawn by the professional chemical balance software Medusa. The molecular structure of NTA is (CH 2 COOH) 3 N. At a lower pH value, a large amount of NTA exists in the form of tribasic acid in aqueous solution, so when pH=4, there is only a small amount of Mn(II)-NTA Formation, the subsequent S(IV) activation reaction is difficult to carry out, resulting in low degradation efficiency of RhB; with the increase of pH, the content of Mn(II)-NTA gradually increases, and S(IV) can be effectively activated to generate SO 4 · - Free radicals degrade RhB. However, when the pH=9, the hydrolysis and disproportionation reaction of Mn(III) produced in the Mn(II)/NTA/S(IV) system is intensified, which is not conducive to the stable formation of Mn(III)-NTA and subsequent reactions.
7、无机阴离子的影响7. Influence of inorganic anions
图8所示为常见的无机阴离子对Mn(II)/NTA/S(IV)体系降解RhB的影响。即使在1~10mM Cl-的存在下,RhB的降解也几乎没有受到影响,反而略有增强。这可能是由于Cl-捕获了SO4 ·-自由基,从而产生了Cl·、Cl2 ·-和HOCl等活性氯物种可能参与了RhB的氧化过程。Figure 8 shows the effect of common inorganic anions on the degradation of RhB by the Mn(II)/NTA/S(IV) system. Even in the presence of 1 ~ 10mM Cl - , the degradation of RhB was hardly affected, but slightly enhanced. This may be because Cl - captures SO 4 ·- free radicals, resulting in active chlorine species such as Cl · , Cl 2 ·- and HOCl, which may participate in the oxidation process of RhB.
HCO3 -通常被认为是S(IV)活化体系中的干扰离子,因为SO4 ·-可以与HCO3 -反应生成活性较低的CO3 ·-(式(6))。而高浓度HCO3 -的存在对RhB的降解不仅没有干扰作用,反而略微促进了RhB的降解。这可能是因为SO4 ·-与RhB的反应速率常数比SO4 ·-与HCO3 -的反应速率常数高2个数量级(k(RhB,SO4·-)=3.02×108M-1s-1,k(HCO3-,SO4·-)=2.8×106M-1s-1),因此HCO3 -对RhB降解的影响不显著。HCO3 -存在条件下使得RhB的降解效果略有提高,可能是因为Mn(II)/NTA/S(IV)体系中存在的HSO5 -与HCO3 -发生反应产生了HCO4 -从而氧化了部分有机物(式(7))。SO4 ·-+HCO3 -→H++SO4 2-+CO3 ·- (6)HCO 3 - is generally considered to be an interfering ion in the S(IV) activation system, because SO 4 ·- can react with HCO 3 - to generate CO 3 ·- (formula (6)). However, the presence of high concentration of HCO 3 - not only did not interfere with the degradation of RhB, but slightly promoted the degradation of RhB. This may be because the reaction rate constant of SO 4 ·- with RhB is two orders of magnitude higher than that of SO 4 ·- with HCO 3 - (k (RhB,SO4·-) = 3.02×10 8 M -1 s - 1 , k (HCO3-,SO4·-) = 2.8×10 6 M -1 s -1 ), so the effect of HCO 3 - on RhB degradation is not significant. In the presence of HCO 3 -, the degradation effect of RhB is slightly improved, probably because the HSO 5 - present in the Mn(II)/NTA/S(IV) system reacts with HCO 3 - to produce HCO 4 - , thereby oxidizing Part of the organic matter (formula (7)). SO 4 ·- +HCO 3 - →H + +SO 4 2- +CO 3 ·- (6)
HSO5 -+HCO3 -→SO4 2-+HCO4 -+H+ (7)HSO 5 - +HCO 3 - →SO 4 2- +HCO 4 - +H + (7)
体系中加入NO3 -会轻微抑制RhB的降解效果和反应速率,可能是因为NO3 -会与SO4 ·-反应生成氧化还原电位较低的NO3 ·(式(8),从而抑制了部分RhB的降解。Adding NO 3 - to the system will slightly inhibit the degradation effect and reaction rate of RhB, probably because NO 3 - will react with SO 4 ·- to generate NO 3 · with a lower redox potential (Formula (8), thus inhibiting some Degradation of RhB.
SO4 ·-+NO3 -→SO4 2-+NO3 · (8)。SO 4 ·- +NO 3 − → SO 4 2- +NO 3 · (8).
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