Ok, so NaOH can act as a nucleophile but the equilibrium lies so much to the left that the NaOH ends up acting as a base to form the corresponding enolate!?
I would argue that it is not the position of the equilibrium for intermediate formation, but the productivity of the intermediate that is important. Remember, the position of equilibrium for the deprotonation of a ketone (pKa ~ 20) with NaOH (pKa of water ~16) lies far to the left as well.
You cannot normally isolate hydrated ketones, and you cannot normally isolate enolates by treating ketones with NaOH (you can if you use stronger bases though).
Point is, the enolate intermediate reacts to form more stable products, whereas a hydrated ketone is usually a dead end intermediate.
I gave the example of a Cannizzaro for an important reaction in which addition of hydroxide to an aldehyde is productive. Also, it occurs to me now that the reverse-Claisen condensation (which is a ketone [specifically a
b-ketoester] hydrolysis reaction) is an example in which this type of addition of hydroxide to a ketone does lead to a thermodynamic minimum.
However, reacting a ketone with water will yield the diol we want quite easily because the equilibrium lies far to the right?
No, it still lies to the left.