Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Big-Daddy on April 27, 2014, 05:43:42 PM
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This question has been nagging for me a while. I have only ever seen Ksp applied to dissolution of ionic solids. What about organic substances? What is our equilibrium measure for their dissolution?
I mean, if we wanted to write a Ksp equation it could just be of the form, e.g. ethanol (s) ::equil:: ethanol (aq) i.e. Ksp≥[ethanol (aq)], with equality Ksp=[ethanol (aq)] when the solution is saturated with respect to ethanol.
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Edit: I suppose maybe for ethanol it would be ethanol (l) ::equil:: ethanol (aq)? Depends on whether Ksp is defined as from standard state to dissolved in solution, or from solid crystal to dissolved in solution (seems hard to tell, because the standard state of everything for which we seem to use Ksp, is a solid crystal).
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Since in this case there is no dissociation that needs to be worried about, it's just simpler to deal with more standard formulations for the thermodynamics (enthalpy, entropy) of mixing. I suppose you could still use a Ksp formalism if you wanted to, but I wouldn't see much point to it.
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Since in this case there is no dissociation that needs to be worried about, it's just simpler to deal with more standard formulations for the thermodynamics (enthalpy, entropy) of mixing. I suppose you could still use a Ksp formalism if you wanted to, but I wouldn't see much point to it.
Ah... but then I'm thinking the maths would be a bit odd when it comes to concentrations/activities? How would we try to define the equilibrium constant for the mixing process at all (there must be an equilibrium constant I assume - otherwise you could get infinite amounts of an organic liquid into water)? Something like K = ethanol (l) ::equil:: ethanol (aq) now seems a bit odd given that, in mixing, we essentially are "dissolving" some water in ethanol too, no?