Dissolution of high solubility sulfates

[Tom]

I am attempting to simulate the dissolution of high solubility hydrous sulfates such as epsomite and melanterite.

To simulate dissolution in the script below, I have added 50 g of epsomite to 100 ml of water.

SOLUTION 1
temp 25
water 0.1 #100 ml

EQUILIBRIUM_PHASES 1 #50 g of sulfate mineral, mr 246.48
Epsomite 0 0.2029

END


The output suggests the final moles is zero (i.e. fully dissolved) and that the final mass of water is 0.1256 kg.

I am wondering where the discrepancy arises from in terms of the predicted final mass (0.1 kg water and 0.05 kg sample fully dissolved, total 0.15 kg) and the final mass (0.1256 kg)?

[dlparkhurst]

The waters of hydration in epsomite add to the total mass of water in the reacted solution.

[Tom]

Of course!

Thanks David.

Here is a confirmation calculation for anyone interested:

--------------

Epsomite:

MgSO4.7H2O mr: 246.48 g/mol

Proportion of H2O = ((7*18.015)/246.48)*100 = 51.16 %

Mass of H2O in 50 g epsomite 0.5116*50 = 25.58 g = 0.0256 kg

Dissolve 50 g epsomite in 0.1 kg water (as per script above), crystal water adds to final solution.

Mass solution: 0.1 kg + 0.0256 kg = 0.1256 kg

---------------

Implication:

Molalities given for speciation in mol/kgw and mass water > 1 kg. Therefore dissolved S(6) does not equate to full proportion of sulfate in sample when entire 50 g "dissolves". Need to correct for total solution mass.