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Please ponder on an equilibrium of a very weak base, say $\ce{MOH}$. This base undergoes a very low ionization, even in its pure form (pure form implies that there isn't the presence of a solvent). The concentration of $\ce{OH-}$ ions, that our base generates, can be assumed to be $ \le 10^{-6} $. Now, let us add some water to $\ce{MOH}$ so as to dilute it. Ostwald's dilution law states that if the degree of dissociation ($ \alpha $) is very less as compared to unity, which in this case, is, it can be calculated by the equation $ \alpha = \sqrt{\frac{K_a}{C}} $. This leads to the conclusion that as I make the solution less concentrated (by dilution), $ \alpha $ increases.

But if I think of it in a different way, I get confused. For instance, please consider the simultaneous equilibria: $$ \ce{MOH <=> M+ + OH-} \tag{1}$$ $$ \ce{H_{2}O <=> H+ + OH-} \tag{2}$$ Since the concentration of $ \ce{OH-} $ ions our base generated in its pure form was comparable to the concentration of $ \ce{OH-} $ ions generated by water, they may cause the first equilibrium to be shifted in the backward direction, thus leading to a decrease in $ \alpha $ upon dilution. Is there any point I'm missing which makes me conclude the opposite of what one concludes from Ostwald's dilution law?

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Your argument is valid, though I believe the scenario you describe is not within the range of applicability of Ostwald's law. Allow me to recast your example in a more obvious form.

You have a solution of weak base. You add a stronger base. The increase in hydroxide concentration leads, by Le Chatelier's principle, to a lower extent of dissociation.*

This is beyond the scope of Ostwald's law, which tacitly assumes the lack of what amounts to a common-ion effect. Ostwald's law holds if the solvent which you're using to dilute your solution does not interfere (or interferes only weakly) with your electrolyte's solubility; the added solvent must serve to decrease the concentration of your electrolyte. Evidently, then, adding a stronger base will interfere with the solubility of your weak base, and is not governed by Ostwald's law.


*There is perhaps some subtlety here. As an example, consider the salt $\ce{MX}$, which dissociates weakly into $\ce{M+}$ and $\ce{X-}$. The intuition behind the greater degree of dissociation is that (1) upon dilution, the equilibrium constant for dissociation doesn't change, whereas (2) the concentration of each species decreases. Because dissociation produces more ions than was originally present, this always leads to $Q < K$, and more dissociation will occur.

If we add a salt $\ce{MY}$, then, we are introducing two effects: (1) decreased concentration of $\ce{X-}$ and (2) increased concentration of $\ce{M+}$. Because these effects pull the equilibrium in opposite directions, we can't conclusively determine what exactly will happen. Going back to your example, if we add a slightly stronger base, then we get exactly these two contrasting effects, and the change in the degree of dissociation is indeterminate. In contrast, if we add a much stronger base, then certainly (2) dominates and the degree of dissociation is reduced.

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