When it comes to water in equilibrium mixtures and Le Châtelier’s principle, things can be confusing, so let’s try to take some of the mystery out of this.
When reading EK6B1a in the AP Chemistry CED, one finds this;
The final few words are what I care about here.
Where water is important
I think that it’s pretty clear that what we are most likely talking about here, are problems in a Ksp context. For example, this saturated equilibrium system,
PbCl2(s) ⇌ Pb2+(aq) + 2Cl–(aq)
on addition of water (that causes the instantaneous dilution, therefore a drop in concentration of the free ions, thus decreasing Q), will shift the equilibrium forward in order to bring Q back to the fixed K value (and vice-versa in the case of water evaporating). An example of such a question would be 2004, 1g. This is a situation where the addition (or removal) of water should be considered in an AP, equilibrium context, but paradoxically, its addition (or removal) actually makes no difference to the concentrations of the ions.
Most students are clear about the idea of leaving pure water (and other pure solids and liquids) OUT of K expressions, on the assumption that their concentrations are effectively fixed.
In the case of Ka’s for all of the weak acid questions that have ever appeared on the AP exam, water will have been (correctly) omitted from the K expression. For example for ethanoic acid in water,
CH3COOH(aq) + H2O(l) ⇌ CH3COO–(aq) + H3O+(aq)
the Ka expression is,
The reason we can leave water out of the equation is that with it being a very dilute solution, with a relatively huge number of water molecules present, any reaction of the acid molecules with water will never significantly change the amount of water present, thus it can be considered ‘fixed’.
Where water is important but probably won’t come up in an AP context
There are circumstances, where the a weak acid is highly concentrated, when water is included in a Ka expression, i.e., when the system is NOT sufficiently dilute. In these circumstances the amount of water is not overwhelming, so any reaction of the acid molecules with the relatively few water molecules could result in an significant change, so it is not considered ‘fixed’. However, this is NOT likely to ever be relevant on the AP exam, so I would not worry about it at all.
Another tricky question arises in the case of the common equilibrium system of pink and blue cobalt(II) complex ions, thus;
[Co(H2O)6]2+(aq) + 4Cl–(aq) ⇌ [CoCl4]2-(aq) + 6H2O(l)
In many given examples of this system, the addition of H2O(l) is shown to create a pink color (as indeed it does in Expt. 17 of the first edition of the Vonderbrink lab manual) and then is accompanied by an explanation of the addition of a product and in terms of Le Châtelier’s principle. One might assume that since water is shown to be important (since it causes a shift), that we are in scenario #1 described above, i.e., that this system has a relatively high concentration, and the number of water molecules is not fixed, but when one learns that the shift also takes place in relatively dilute solutions, one has to think again. In order to do that, let’s consider the dilution factor.
The addition of water to the system will dilute all of the aqueous species, thus reducing their concentration. Since there is a greater change of aqueous species on the left compared to the right (5 moles of (aq) versus 1 mole of (aq)), the reaction will shift to the left hand side. This is analogous with a gaseous equilibrium shifting to the side where there are mole moles of gas when the volume of the vessel is increased.
The final twist in the cobalt system would be the addition of some large amount of non-aqueous solvent such as ethanol. When this happens, water (and all of the other species) become solutes, and the non-aqueous, organic compound is the solvent.
[Co(H2O)6]2+(ethanol) + 4Cl–(ethanol) ⇌ [CoCl4]2-(ethanol) + 6H2O(ethanol)
As in the final paragraph in scenario #2 above, on the addition of the organic solvent, all species (including water) have their concentrations diluted, and the equilibrium shifts to the side where there are a greater number of moles of the solute – this time to the right hand side and the equilibrium mixture turns blue. Why? Well, since water is now considered a solute, there are more moles of solute on the right hand side (7), then there are on the left hand side (5).
In truth, I think that you can probably walk into the AP exam feeling good about always leaving water OUT of Ka expressions for weak acids, knowing that they might ask about addition (or removal) of water in a Ksp situation, and that all of the other situations described above are not going to show up…until they do, I guess!