PbSO4 dissolution equilibrium
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]
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