Xque - 14-11-2005 at 08:34
I know this is probably extremely simple stuff for most of you, but I was a lazy ass in my younger days and I can't find anything in my
textbooks.
BaSO4 (s) -> Ba2+ (HCl) + SO42- (HCl)
I know the sol. product for BaSO4, but how does the calculation of solubility for a solution in acid differ from that of water? :S
Magpie - 14-11-2005 at 19:30
IIRC the solubility product is only a function of [Ba++] and [SO4--], that is the molarity of these two ions. But then we have the little issue of
"activity." I am not a good enough inorganic chemist to address that. But I'm sure there are forum members who are.
[Edited on 15-11-2005 by Magpie]
Magpie - 15-11-2005 at 09:06
I had a couple more speculations on this interesting question.
On reading a little on the subject of "activity" I learned that it is not a factor for very dilute solutions where the ions don't bond
with any other ions/molecules. Since any solution of BaSO4 is bound to be very dilute then activity would not be a factor I wouldn't think.
The equilibrium of H+ with SO4-- to form HSO4- would have to be satisfied. Since an acidic solution is rich in H+ this would affect the molarity of
SO4--.