Kd in the unit of mol g-1

[Yoshio Takahashi]

About the Kd approach to treat sorbing solutes, it is written that we must convert the unit of Kd from cm3/g to mol/g. Could you please explain how we can do that? 

In the textbook of the online academy (https://academy.gwb.com/sorbing_solutes.php), for example, the Kd for Pb2+ was given as 0.16 cm3/g, which was converted to 0.0003 mol/g.  I would like to know details of the conversion, when we assume that (i) dissolved Pb2+ is 10 micromolar, (ii) the activity coefficient of Pb2+ in the goundwater is 0.66, and (iii) the free Pb2+ proportion is 80% of total dissolved Pb, as written in the textbook.

[Jia Wang]

Hello Yoshio,

Thank you for your question. To calculate Kd' I would use the original sorption isotherm, S = Kd*C, where the sorbed concentration (mole/g dry sediment) is equal to the product of the Kd(cm3/g) and concentration of the free ion (mole/cm3). I would first convert the concentration of dissolved Pb2+ to mole/cm3 which is approximately 10^-8 mole/cm3. Then using the Kd and concentration of dissolved Pb2+, calculate the concentration of Pb sorbed:

S = 0.16 [cm3/g] * 10^-8 [mole/cm3] = 1.6 * 10 ^-9 mole of Pb2+ sorbed per gram of sediment. 

The Kd' term accounts for the activity instead of the concentration. To do so you, you will need to multiply the activity coefficient by the concentration of free ions in solution. Using the activity coefficient (0.66), fraction of free ions in the fluid (0.8), and the concentration of dissolved Pb2+ (~10^-5 molal), you can calculate Kd' like so:

1.6*10^-9 = Kd' * (0.66*0.8*10^-5)

where the Kd' is calculated to be approximately 0.0003 mol/g. 

Hope this helps,

Jia Wang