3)pH=44m,y=0.95 IH2Y2=c8m,y=0.02×0.95=0.02 推论:Na2IH2Y2H2O→Na++H2Y 溶液pH=44 (4)H4Y:EDTA溶解度小0.02g/100mH2O 22 c c=7×10+moL Na2H2Y溶解度大11g/100mH2O 22℃c EDTA =03molL 配成0.01~0.02mQo的水溶液pH=44 上一页 页|返国0上一页 下一页 返回10 3)pH=4.4 δH2Y =0.95 [H2Y2-]=c δH2Y =0.02×0.95=0.02 推论： Na2H2Y·2H2O → Na+ +H2Y 溶液pH=4.4 (4)H4Y: EDTA溶解度小0.02g/100mlH2O 22℃ cEDTA =7×10-4mol/L Na2H2Y溶解度大 11.1g/100mlH2O 22℃ cEDTA =0.3mol/L 配成0.01~~0.02mol/L的水溶液 pH=4.4