Calculating meq/l from meq/kg CEC (fractured rock)

[Felix]

Hello David and Phreeqc community!

I am trying to calculate a CEC for a paper and think my problem may be of interest to other readers:

A sandstone has 2.5% matrix porosity and is a fractured aquifer. The literature gives 50 meq/kg rock (50 mmol(c)/kg).
Appelo & Parkhurst (2005 Groundwater and Pollution, p. 248) give a calculation method that I adapt, but which seems to provide unrealistic values:

Quartz has 2.65 g/cm3, so 1 kg of dry sediment contains 1000/2.65 = 377mL.
Total sediment volume is 377/ (1-0.025) = 387mL
pore water: 387*0.025 = 9.7mL.
CEC = 50 meq/kg = 50/9.7 meq/mL = 5155 meq/L pore water (note conversion from mL to L).

But 5 moles CEC for the pore fluid seems crazy big. Appelo & Postma have 100 meq/l in their example (at 20% porosity). If the porosity becomes even smaller, 25 moles arise, and so on... (Isn't it the opposite: The less porosity, the less rock-fluid-contact?)
How would someone judge this value? Is there a more elegant way to calculate the CEC for my paper from the fracture volume*?

*I have about 6 fractures per meter with an opening of 0.5mm/fracture. This also gives very low values for the porosity...

I would be very grateful for some comments!

Kind regards
Felix

[dlparkhurst]

I would do the calculation a bit differently:

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0.975 (L rock) * 2.65 (kg/L rock) * 50 meq/kg  = 129 meq

If you divide by the volume of water (0.025), you get essentially your result, 5167 meq/L water. If you are doing a batch calculation with ~1 L of water, this calculation would indicate a very large CEC per liter of water, and the number gets larger as the porosity decreases. The calculation assumes that all of the sites in the volume are accessible even as the volume of water decreases.

I think you are right to consider how many sites are available considering a rather limited surface that is exposed to the water. So, how was the value of 50 meq/kg rock measured? If you can estimate the surface area and that applies to that estimate (meq/m^2), you could make the calculation based on the surface area of the fractures relative to a liter of water.

You probably have a sufficiently difficult problem with the surface area, but I'll just mention that a TRANSPORT calculation can consider additional diffusion/exchange from the mobile zone (fracture) into a secondary porosity (side pores).