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Question for Example 14

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**Jeonghwan Hwang**:

Hello, I had a question for Example 11.

At the estimation of mineral amounts, the example explained as below;

'Parkhurst and others (1996) provide data from which it is possible to estimate the moles of calcite, dolomite, and cation-exchange sites in the aquifer per liter of water. The weight percent ranges from 0 to 2 percent for calcite and 0 to 7 percent for dolomite, with dolomite much more abundant. Porosity is stated to be 0.22. Cation-exchange capacity for the clay ranges from 20 to 50 meq/100 g (milliequivalent per 100 grams), with an average clay content of 30 percent. For these example calculations, calcite was assumed to be present at 0.1 weight percent and dolomite, at 3 weight percent; by assuming a rock density of 2.7 kg/L, these percentages correspond to 0.1 mol/L for calcite and 1.6 mol/L for dolomite. The number of cation-exchange sites was estimated to be 1.0 eq/L.'

Here, for example at calcite (CaCO3, 100g/mol) and 1L volume (i.e., 0.78L (1-0.22[porosity]) for geologic material)

The possible calculation is 2.7 kg/L * 0.78L * 0.001 * 1000 g/kg / 100 g/mol = 0.02106 mol/L

However, the input data is estimated to 0.1 mol/L.

Can anyone help me to explain what I missed?

Thank you

Sincerely,

Jeonghwan Hwang

**dlparkhurst**:

Your answer is moles per liter of porous medium. You forgot to divide by 0.22 L of water per L porous medium to get mol/L water.

**Jeonghwan Hwang**:

Thank you for reply.

I have one more question.

Example 14 assumed saturation condition and the water content is 22% of total volume.

If I want to design the vadose zone (imaginary) that the water content is 10% and air is 12%, can I change the amounts of minerals as you advised?

'divided by 0.22L' to 'divided by 0.10L'

It should be assumed that mineral surfaces can be sufficiently covered with water with 10% water contents.

Thank you

Sincerely,

Jeonghwan Hwang

**dlparkhurst**:

If your saturation is constant, then yes you could divide by 0.1. At some point, it might be easier to scale everything to 1 L of representative volume and put the corresponding amounts of water and reactants in each cell.

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