how to calculation the original concentration of Mg
An urine specimen (containing Mg) was diluted to 2L . And 10ml of aliquot was titrated with 27.32ml of 0.00396M EDTA.
Mg+ EDTA---> MgY
Kf=1.85X10^8
Kf=[MgY]/[Mg][EDTA]
Titraion...calculation
2008-07-09 6:39 pm
回答 (1)
2008-07-10 5:41 am
✔ 最佳答案
no. of mole of EDTA = (27.32/1000)x0.00396 = 1.08187X10^-5no. of mole of Mg2+ = 1.08187X10^-5
no. of mole of Mg2+ in the 2L sample = 1.08187X10^-5 X (2000/10)
= 2.13X10^-3
Conc. = 2.13X10^-3 / 2 = 1.082X10^-3 M
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