The solubility of Ca(OH)2 means just number of moles per litre of Ca(OH)2 (equal also to [Ca2+] ) in solution when you mix an excess of solid calcium hydroxide with water and allow it to come to equilibrium, that is allowed all that is ever going to dissolve to do so. (At that temperature).
That refers to calcium hydroxide by itself with water. If you add acid more will dissolve. Because you are removing OH- so displacing an equilibrium. So you could say the solubility of calcium hydroxide changes when you add acid. On the other hand the solubility constant, which is defined in your second equation, is constant, doesn't change according to these circumstances. (It does change with the temperature).
If you knew the solubility constant Ksp you could calculate the solubility from the very broad hint in your first equation. That would be a purely theoretical calculation. (.But now I have reminded you of the difference between solubility and solubility product, you will probably find it better explained in your textbook)
However it looks as if you are trying to calculate something from experimental measurements. I have to guess what this is about, maybe just you have tou determine the total amount of Ca(OH)2 by titration because if I remember it is hygroscopic and not a very good primary standard, so unreliable to do it by weighing? Anyway to calculate anything you would have to either know the Ksp or a measured solubility - it is not very clear to me what the question is.
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