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Langmuir Isotherm Constant Conversion (Example 19)


Hello All,

I am researching novel adsorbents and I am attempting to model sorption of multiple metal species using experimentally-measured values and fitted Langmuir isotherm constants.

In example 19, the fitted langmuir constant (often denoted by K_L) is converted to a PHREEQC-friendly equilibrium constant (log_k), where log_k is equal to the base ten log of the molecular weight of the adsorbate divided by K_L. In this case, log_k = log(112.4e6/30.9) = 6.56.

In my mind, it seems a "better"-performing adsorbent would have a higher K_L, i.e., an adsorbent with a higher K_L should be expected adsorb more milligrams (or micrograms as in example 19) of adsorbate per gram of adsorbent. But in the case of example 19, an adsorbent with a higher K_L would perform "worse". E.g., given a K_L of 50, log(112.4e6/50) = 6.35, which is lower than the 6.56 used in the example, and Cd+2 would therefore have a lower propensity to sorb (see attachments). Am I missing something? Any advice would be greatly appreciated.

Thank you,

I think there are other ways to write the Langmuir equation, but the way it is written in the PHREEQC manual, KLangmuir is inversely related to (LangmuirCd+2).

--- Code: ---(LangmuirCd+2)       Langmuir total sites
-------------------  =  ----------------------
(Cd+2)                     KLangmuir + (Cd+2)

--- End code ---

So, a bigger KLangmuir would produce a smaller (LangmuirCd+2) for the same total sites and (Cd+2). Because the PHREEQC log K is the inverse of KLangmuir, it follows that a larger PHREEQC log K would have the opposite effect; it would produce a larger LangmuirCd+2.

That cleared things up and the results I'm getting are making more sense too. Thank you, dlparkhurst!


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