Amy Winehouse - 5-5-2012 at 16:05
This is just something I've been wondering,
If I am trying to make a solution with a specific molarity of anhydrous Sodium Carbonate (which is hydroscopic) will I have to account for the
addition of H2O molecules attaching to the salts when trying to determine the volume? In other words, when the H2O molecules mozey on over to the
dihydrate in the salt, will everything be conserved volumetrically? And if this is true, is it true for all hygroscopic salts? (That you don't need to
worry when standardizing)
Thanks everyone
sargent1015 - 5-5-2012 at 20:18
If it is "anhydrous" you just have to weigh out the amount you need and place it in a volumetric flask. Any hydrating will be accounted for, so don't
worry about it. If it is NOT anhydrous, you will need to dry whatever salt first (as long as it will not decompose), most likely in an oven (100-160C
depending on the compound). This is the best way to be analytical, or else you could have a mix of mono, di, poly hydrate species present, making your
calculations inaccurate.
But if you don't really need to be that accurate, don't worry about it.
Pyridinium - 6-5-2012 at 10:37
Sodium carbonate when heated to the anhydrous form does not take on water that fast (hygroscopic as opposed to deliquescent), so it should not pick up
significant water mass while you're weighing it out. Once it gets into the volumetric flask, it's not a concern; it could sit there for days.
sargent1015 was right, I double checked and 160 C is the good number. Two hours at that temperature and it should be down to the 0-hydrate.