摘要
氯化钠能使正丙醇水溶液分为醇/水两相,在此分相过程中,罗丹明B(RhB)和SbCl-6生成的三元缔合物[SbCl-6][RhB+]可被完全萃取.研究了罗丹明B和盐酸的浓度对萃取的影响.结果表明,当溶液中罗丹明B和盐酸的浓度分别为0.2 g/L和0.4~1.5 mol/L时,Sb5+的萃取率达到99.1%以上,可以使Sb5+与Ti4+、Rh3+、Sn4+和Ag+分离.
The 1-propanol aqueous solution could be well divided into 1-propanol /water two phases in the presence of sodium chloride.The ternary complex [SbCl 6 ][RhB + ]formed with Rhodamine B and SbCl 6 could be completely extracted during this process.The influences of concentrations of Rhodamine B and hydrochloric acid on extraction were studied.The results showed that Sb 5 + could be quantiatively separated from Ti 4 +,Rh 3 +,Sn 4 + and Ag + by extraction,and values of extraction percentage of Sb 5 + reached over 99.1%.when the concentrations of Rhodamine B and hydrochloric acid in solution are 0.2 g / L and 0.4 ~ 1.5 mol / L,respectively.
出处
《信阳师范学院学报(自然科学版)》
CAS
北大核心
2013年第3期392-394,共3页
Journal of Xinyang Normal University(Natural Science Edition)
基金
河南省教育厅自然科学研究计划项目(2009150023)
关键词
正丙醇
罗丹明B
萃取分离
锑
1-propanol
Rhodamine B
extraction separation
antimony