The anhydrous crystals will be absorbing moisture from the air anyway, because it is not a stable form at ambient conditions. If you need to use it as a drying agent, you can heat it in vacuum to remove the crystallized water.
No this is merely an effluent disposal application. But the anhydrous form leads to less landfill costs.
The other very confusing part in this solubility literature is how solubilities are reported. It seems we all agree that the crystals will be hydrated even above 32C so long as they are crystallized in contact with mother liquor.
Now see the solubility curve in the Fig. below. This seems a standard curve reported without controversy in many sources. Reading the curve, solubility at approx. 70°C is roughly 45 gm of Na2SO4 in 100 gm water.
Let's say I take 100 gm pure H2O at 70°C & slowly start adding anhydrous salt to it. Technically I should be able to add 45 gm & then I will see crystals. Say I added 50 gm anhydr. Na2SO4 (MW=142) we are 5 gm above the solubility limit so I expect 5 gm crystals & 105 mother liquor which ought to be a saturated soln. of Na2SO4 in H2O (31% w/w) .
On a dry basis. But if these are indeed Na2SO4.10H2O (MW=322) crystals, I'd be seeing 11.3 gms of decahydrate crystals. So would I be left with mother liquor that is 48% solution? i.e. technically above the saturation solubility at that temperature? Or does it mean I'd be getting less than 5 gm of dry crystals out?
Perhaps this seems pedantic, but in reporting stoichiometry an accurate way of communication seems essential. I'm only wondering if this field has a standard convention of how these things are reported & I'm not getting it?
Besides a difference of only a few percent might mean tons of extra solids at scale. Finally I could always do the whole experiments myself but Na2SO4 seems such a commonly studied crystallization that it seems a waste to ignore the prior work & re-invent the wheel.