[sodium.dioxid]

Does dissolution go to completion when the solute is not enough to produce saturation. I am thinking that the answer is yes because if we were to get a pure ocean of water and put a bag of salt in it, surely it would dissolve and not crystallize.

Basically, here is what I m trying to get at: Equilibrium does not exist for an unsaturated solution; dissolution goes to completion. Is this correct or not?

[JustinCh3m]

solvation of a solute depends on the solvent.  If the solvent is water and the solute "goes well with it" (i.e. "like dissolves like"), then solvolysis = hydrolysis.

do you have a specific solute/solvent combination in mind when asking  your question?

[sodium.dioxid]

This is a conceptual question more than anything else. Lets say Lead Chloride and water.

[ramboacid]

Saturation is determined by the K_sp of the solute, which is the equilibrium constant for solvation. The reaction that the equilibrium describes is
XY_(s) X⁺_(aq) + Y^-_(aq)
K_sp = [X⁺][Y^-]

If the solute contained a +2 cation and two -1 anions, then consequently the reaction would be
XY_2(s) X²⁺_(aq) + 2Y^-_(aq)
K_sp = [X⁺][Y^-]²
and so on so forth for other combinations.

Using normal equilibrium logic, we can compare the reaction quotient Q to the equilibrium constant K to determine the direction of the reaction. If Q>K, then the reaction shifts towards the reactants. If Q<K, then the reaction shifts towards the products. If K=Q, then the reaction is at equilibrium. With solvation equilibrium, we denote the K and Q as K_sp and Q_sp, just like we denote acid-base equilibrium with an "a" subscript in K_a.

With NaCl, for example, the K_sp is so high that the reaction shifts towards the products so much it is called "soluble." So in short, yes, the solute will dissolve until the reaction reaches equilibrium. If the maximium value for the reaction quotient Q_sp at the current conditions is less than the K_sp, then it will dissolve completely.

[sodium.dioxid]

I just noticed that the equilibrium constant expression brings up a weird implication.
Consider this: A(s) B(aq) + 2C(aq)
K_sp = BC²
Assume the solution is saturated
Here is the weird thing - the solubility product constant expression implies that if we were to remove one C, then we could not re-saturate the solution by adding one B instead. We would have to add an amount of B greater than the amount lost of C. This is weird because common intuition tells us that the removal of an amount of an ion can be replaced by the same amount of another ion (it is a matter of space not rate law). So how can this be?

[Jorriss]

. This is weird because common intuition tells us that the removal of an amount of an ion can be replaced by the same amount of another ion (it is a matter of space not rate law).

You can't do this in general if I understand you.

If you are looking at the reaction A -> B + 2C, you can't remove, say, 1 mole of C and add 1 mole of Q and expect everything to be the same. The identity of the ion matters.

[Borek]

Does dissolution go to completion when the solute is not enough to produce saturation.

Yes.

[fledarmus]

I just noticed that the equilibrium constant expression brings up a weird implication.
Consider this: A(s) B(aq) + 2C(aq)
K_sp = BC²
Assume the solution is saturated
Here is the weird thing - the solubility product constant expression implies that if we were to remove one C, then we could not re-saturate the solution by adding one B instead. We would have to add an amount of B greater than the amount lost of C. This is weird because common intuition tells us that the removal of an amount of an ion can be replaced by the same amount of another ion (it is a matter of space not rate law). So how can this be?

No, it isn't a matter of space, it is a matter of rate law. When you have a sparingly soluble salt, you are balancing the rate of a solid dissolving with the rate of separate ions in solution getting together and precipitating. It is a kinetic process. Dissolution doesn't just stop when you reach equilibrium - it is still occurring as fast as it was, it's just that reprecipitation is occurring as fast as the dissolution.

You can actually add more B and resaturate the solution. Not necessarily one atom for one atom; instead, enough to comply with the rate law. You might be interested in reading up on the common ion effect:

http://en.wikipedia.org/wiki/Common_ion_effect