frogfot - 13-2-2005 at 13:57
Some time ago I've calculated how much PbSO4 can dissolve in dilute H2SO4, but now I'm not sure if result is right.. Anyone have time to
help?..
So, I have 20% H2SO4, which corresponds to 0,01014 M SO4-- and 2,31 M HSO4-. Nothing could be wrong sofar..
When we add solid PbSO4 we will get following equilibria (anion conc shouldn't change due to small Pb conc):
PbSO4 <==> Pb++ + SO4-- (Ksp = 1,6E-8 at 25oC)
Pb(HSO4)2 <==> Pb++ + 2HSO4- (Ksp = 1,4E-16 at 18oC)
(Temp difference is neglected)
So, here comes the problem.. The way I did it was multiplying together the equilibria equations below:
1,6E-8 = [Pb++]*[SO4--]
1,4E-16 = [Pb++]*[HSO4-]2 (**)
To get following:
2,24E-24 = [Pb++]2*[SO4--]*[HSO4-]2
Typing in SO4-- and HSO4- concentrations in "combined" equation above I got Pb conc of 6,43E-12 M.
Then one could think, since conc of HSO4- is much bigger than SO4--, maby equilibrium could be better described by equation ** ?
That is [Pb++] = 1,4E-16/2,31*2,31 = 2,62E-17 M.
So what would be more correct? Thanks for the time.
EDIT: darn superscripts doesn't work..
[Edited on 13-2-2005 by frogfot]