作者darkrate (無名)
看板Chemistry
標題[分析] Karl Fisher 滴定
時間Mon Apr 2 16:17:55 2012
http://tinyurl.com/7rgj3ke
關於Karl Fischer reagent的計算，題目如上，想請問版大們計算方式對不對，謝謝
想法：
一.由題目第5行：when pure "dry" methanol was titrated, 25.00 ml of methanol
reacted with 3.18 ml of the same Karl Fischer reagent.
知道純無水甲醇25ml消耗3.18ml的Karl Fischer reagent
→1ml無水甲醇消耗(3.18/25)ml的Karl Fischer reagent
二.由題目第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.
想說它的意思是：34.61ml的甲醇(含水量是4.163mg/每ml甲醇)需要消耗25ml的
Karl Fischer reagent
→甲醇總含水量是(34.61*4.163)mg
三.34.61ml的無水甲醇需要消耗34.61*(3.18/25)ml=4.40ml的Karl Fischer reagent
所以第二點中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.由題目第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
所以實際被水消耗的Karl Fischer reagent只有38.12-3.18=34.94ml
樣品中H2O含量為34.94*6.99=244.38mg
佔樣品比例：﹝(244.38*10^-3)/1﹞*100%=24.44%
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※ 編輯: darkrate 來自: 111.252.230.199 (04/02 16:19)