compound weighing correction factors for salts of acids

[analchem]

My question is for compound weighing purposes. If you have a salt (e.g., sodium) of an acid (1:1 stoichiometric ratio) and you want to correct for the salt, do you "put the proton back on" to determine the free acid equivalent? This weight would then be used to prepare standard solutions and ultimately a calibration curve.

For example, you want to measure and report acetic acid (mw = 60.05) but you only have a sodium acetate (mw 82.03) standard, do you use a correction factor of 60.05/82.03 or 59.04/82.03?

We don't have any controversy about salts of bases, because salt correction there only gives the free base which is neutral.

[tom jupille]

You do it all in moles, and then convert with whatever molecular weight is appropriate for your purpose.

[analchem]

You do it all in moles, and then convert with whatever molecular weight is appropriate for your purpose.

Thanks Tom. I should have been more specific that I was preparing a w/v solution, e.g., I need a 1 mg/mL solution of acetic acid, prepared using Na-Acetate. If I am preparing molar solutions, no correction is necessary since we are normalizing by molecular weight:

So 5 g of salt = 5 g/82.03 g/mol = 0.06095 mol
and correcting for free acid, 5 g x 60.05/82.03 = 3.66 g acid; 3.66 g acid/ 60.05 g/mol = 0.06095 mol acid.

However a huge error is incurred if a w/v solution is being prepared. So if you have addressed my original question as to which value in the numerator I should use, could you perhaps be a bit more pedantic? I want to report my data as "acetic acid" and so my standard solution is reported in mg/mL of acetic acid.

[tom jupille]

I can be very pedantic and point out that, since "1 mg/mL" is a one significant figure specification, it really doesn't matter.

However, assuming you mean "1.0 mg/mL" (to two significant figures) of acetic acid, that would be 1.0/60. = 0.017 millimoles/mL. The equivalent solution prepared using sodium acetate would then be 0.017*82 = 1.4 mg/mL of sodium acetate.

[analchem]

Ouch, I got out-pedanticized! Thanks Tom!

[tom jupille]

You're welcome !

And, to paraphrase Tigger from the Winnie-the-Pooh books "Pedantic is what I do best!"