VmlAL2(SO4)3溶液中含AL3+ag,现取V/4ml该溶液稀释至4Vml,反应后溶液中SO2+的物质的量浓度是原溶液n(Al3+)=m/27 mol所以C(Al3+)=(m/27)/(V*0.001)=1000m/27V所以C(SO42-)=1.5*C(Al3+)=500m/9V取出1/4 V ml
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