http://tinyurl.com/7rgj3ke 关於Karl Fischer reagent的计算，题目如上，想请问版大们计算方式对不对，谢谢 想法： 一.由题目第5行：when pure "dry" methanol was titrated, 25.00 ml of methanol reacted with 3.18 ml of the same Karl Fischer reagent. 知道纯无水甲醇25ml消耗3.18ml的Karl Fischer reagent →1ml无水甲醇消耗(3.18/25)ml的Karl Fischer reagent 二.由题目第4行：A 25.00 ml aliquot of Karl Fischer reagentreacted with 34.61 ml of methanol to which was added 4.163 mg of H2O per ml. 想说它的意思是：34.61ml的甲醇(含水量是4.163mg/每ml甲醇)需要消耗25ml的 Karl Fischer reagent →甲醇总含水量是(34.61*4.163)mg 三.34.61ml的无水甲醇需要消耗34.61*(3.18/25)ml=4.40ml的Karl Fischer reagent 所以第二点中25ml的Karl Fischer reagent只有25-4.4=20.6ml是被(34.61*4.163)mg H2O消耗掉→1ml Karl Fischer reagent 可以消耗6.99mg的H2O 四.由题目第6行：A suspension of 1.000 g of a hydrated crystalline salt in 25.00 ml of methanol consumed a total of 38.12 ml of Karl Fischer reagent. 38.12ml的Karl Fischer reagent有25.00*(3.18/25)=3.18ml是被甲醇消耗 所以实际被水消耗的Karl Fischer reagent只有38.12-3.18=34.94ml 样品中H2O含量为34.94*6.99=244.38mg 占样品比例：﹝(244.38*10^-3)/1﹞*100%=24.44% --

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