[biting]

Question: There is a homogeneous solution with no solute at the bottom (no residue). This mixture can be in which state(s)?

a) Unsaturated
b) At equilibrium
c) Supersaturated

I know (a) and (c) are definitely right. My chemistry teacher from 23 years ago insisted that (b) is incorrect. Given the state of (a), (b), and (c) are technically a continuum, it doesn't make sense that it CANNOT be at equilibrium? Yes, I understand that you cannot be certain it is at equilibrium, but is it actually IMPOSSIBLE that the mixture in question is in fact, at equilibrium? Please help me find closure. Thanks.

[Corribus]

It seems like a normal, unsaturated solution should be at equilibrium. After all, it seems to be stable, not doing anything. But "stable" is not the definition of equilibrium.

Let's consider solubilization of sucrose, S, defined by the equation S(s) S(aq), with solubility (equilibrium) constant K = 1.97 mol/dm^-3.

Since the activity of a solid is by definition 1, then a solution of S is only at equilibrium when the concentration of S(aq) is 1.97 mol/dm^-3. This is also the point where the solution becomes saturated. Due to the nature of the solubility equilibrium, the concentration of S(aq) is also 1.97 mol/dm^-3 if you keep adding S beyond that point, because the excess crashes out - the presence of precipitated S(s) does not influence the reaction quotient Q because the activity of a solid is always assigned a value of 1.

Look at it from the reaction quotient perspective. Let's say you have an unsaturated solution of S(aq) with a concentration of 1 mol/dm^-3. The reaction quotient is therefore 1 mol/dm^-3. We see that Q < K, which means that the reaction is not at equilibrium - equilibrium is only reached when Q = K. The system should move to the right (produce more S(aq)) to reach equilibrium. It cannot do that, of course, there being no S(s) available. But the fact that a system is prevented from reaching equilibrium does not mean that it is at equilibrium. This I guess is the important point here. Equilibrium is a specifically defined state only reached by satisfying specific conditions; the appearance of stability has nothing to do with it.

(You may also consider the definition of equilibrium being that where the forward rate is equal to the backward rate; in an unsaturated solution, the forward rate is faster than the backward rate, ensuring that no precipitate is ever formed. If the two rates were equal - that is, if the solution was at equilibrium - then the backward rate would be fast enough that at least a minuscule amount of S(s) would be able to linger around.)

The third option is another case of a system that is not at equilibrium and is prevented from reaching equilibrium. Most students don't have a problem recognizing that this situation is not at equilibrium, perhaps because they've seen the demonstration where a nudge or scratch can cause it to fall out of its delicate balance. But this is really no different a situation from the non-equilibrium state in situation b), in that the system is quasi-stable and merely prevented from reaching equilibrium by its circumstances.

[Meter]

In simple solution chemistry, the equilibrium considered is usually

AB  (s) A⁺ (sol) + B^- (sol).

(or any stoichiometric combination of A and B as well as charges, depending on the species in question).

So I think your old chemistry teacher meant that for equilibrium to occur, there must be solid present in the beaker.