Hi there, can anyone help.
I have seen some posts online where coulometric Karl fisher is done with HPLC grade methanol and a blank taken and the value subtracted from the value of the sample in methanol. However, most literature is with anhydrous methanol (which must still absorb moisture) and a blank taken. Is the use of hplc grade methanol less accurate for this test. Why- is it to do with moisture adsorption from the atmosphere
Thanks
Rosie
As with most solvents, MeOH absorbs moisture readily from the atmosphere. The difference b/w dry (anhydrous) methanol and HPLC grade is the amount of moisture already in the MeOH. I find the moisture data pasted on solvent bottles pretty much useless, because it will definitely change once you open the bottle, or if the bottle has been on lengthy storage, or if the bottle was subjected to improper storing conditions. That moisture value is most probably obtained under regulated sampling conditions from a bulk source. Many variables.
Unless youre doing your analysis in a moisture or low humidity atmosphere, chances are that the MeOH, regardless of grade, will still absorb some moisture. I dont have solid evidence for this, however it is easy to prove (or disprove) in your lab.
So to answer your question, using an anhydrous form of MeOH, instead of HPLC grade, will facilitate your titration -this discussion as you indicated is about coulometry, so therefore I assume that your sample you intend to analyze has micro or milligram levels of moisture present. However in my opinion, as long as there is a blank subtraction used, and the same calculation of blank subtraction is employed in all calculations, there will be no significant difference in the end result if you use either form.
N