Help converting Langmuir constants

[A.Schneidt]

Hello all!

I'm a master student, planning to model sorption using langmuir isotherms, and as such trying my best to wrap my head around example 19 from the manual. From what I can tell from the example, the K(L) is converted to a phreeqc accepted K(L, phreeqc) by dividing the atomic weight microgram/mol by the K(L) microgram/liter.

However, the vast majority of K(L) values found in the literature (at least as far as I could tell) are given in L/mol. Because of this unit, converting these values almost ubiquitously results in very high K(L, phreeqc) values which don't seem realistic (nor anywhere near the values for Cd from the example.). I feel as though I'm missing something or misunderstanding something about example 19.

Thank you so much for any insight.
Alexander

[dlparkhurst]

I think Langmuir Ks are usually L/g, at least that is what example 19 assumes. So part of the conversion to a PHREEQC log K is to convert from grams to moles. If you already have a K as L/mol, then you do not need to use the gram formula weight in the calculation.

There is another example in the post https://phreeqcusers.org/index.php/topic,6.msg9.html#msg9, which cites Zhu and Anderson (2002). Perhaps that reference could shed some light.

[A.Schneidt]

Thank you for your reply! I'm still having some difficulty understanding the example (though the book they reference is definitely super helpful). In your response you say example 19 assumes K_L is L/g, which confuses me because equation 40 from what I can tell presents the K_L 30.9 from example 19 as in microgram/L water.

What confuses me further is this sentence:
"The Langmuir isotherm (equation 35) and its PHREEQC formulation (equation 38) are equivalent for sorption of Cd on 1 g soil when (Langmuir_total_sites) / 112.4?10 6 = molesLangmuir sites, PHREEQC, and (KLangmuir / 112.4?10 6 )-1 = KLangmuir, PHREEQC, where 112.4?10 6 μg/mol (microgram per mole) is the atomic weight of Cd."

which (I think) implies K(L, Phreeqc) = (K_L / atomic weight)-1 but the later formulation from equation 43 seems to contradict that. Do I understand your earlier answer correctly that by removing gram formula weight you would calculate K if K_L was in L/mol by using:

1 / K_L = K(L, PHREEQC)

[dlparkhurst]

II believe the -1 is actually an exponent.

(Klangmuir / 112.4e6)^-1 = KLangmuir,PHREEQC