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First, i apologize for the obvious mistakes you will find in the
calculus. they were transcripts errors, so everything was basically accurate
( was very tired when i posted ).
Thanks to George Slusarczuk for the answer !
I post part of it because it is interesting for everyone who likes the
water chemistry things.
«For simplicity sakes water hardness is expressed as milligrams/Liter of
_Ca_CO3, even when we are talking about _Mg_CO3! Thus the results should
differ by the atomic weight ratio 40.1/24.3 = 1.65. Your ratio is
13.54/8.54 = 1.59. The error is 3.6% -- not bad! (As a matter of fact,
it's pretty good!)»
«Just to be sure,as i understand it, when i want the TRUE level of Mg++, i
have to divide my result by 1.65, is that it ? ( so 13,54/1,65 = 8,2 ppm of
«Yes, if you have _only_ magnesium compounds present. In a Ca/Mg mixture
the results are not 100% -- that is why, for purposes of water hardness,
it is assumed that all alkaline earth metals are calcium. It simplifies
the analysis (and the result) tremendously.»
Thanks again !