Hi, I have a doubt about this preparation:
5 mL of a mixture of MeOH/DMSO/H2O (40:40:20, v/v/v) containing 0.1% HCl.
the 0.1% HCL is taking into account that all the volume (5 mL) is acidified with 0.1% HCl ,? or just that one of the three solvents is acidified?
thanks
The volume fractions given MeOH/DMSO/H2O (40:40:20, v/v/v) add to 100 vol%, what suggests that it was possibly not indeed considered to further add 0.1% HCl. In my opinion, that unclearly written formulation ratio seemingly was intended to mean: MeOH/DMSO/HCl (0.1 wt% aq.) (40:40:20, v/v/v). This also seems to be the simplest and most logical way to prepare the intended solvent mixture. Also note that 0.1 [wt]% HCl refers to concentration of a solution, while the volume fractions are formulation ratios for a three-component mixture.
About the HCl concentration ― we should generally take not fully specified % concentrations as wt% ― where wt is often omitted or can be substituted by either wt., wt/wt, wt./wt., or m/m; meaning: (mass of pure solute / mass of solution)·100%; even if this is not clearly stated. Moreover the solution should be understood as aq., unless specified otherwise.
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