# why are Ca(OH)₂ and Al(OH)₃ less soluble than Na(OH) in water

Even though it has more $\ce{OH}$ groups and still it is very weak in terms of dissolving?

please clarify to me in terms of polar and non polar solubility, how to determine whether a compound is soluble or not in water.

$\ce{OH}$ is small anion and bonds effectively with small cations. Thus, in general, the bigger (and less specific charge) of the cation, the higher is solubility of the corresponding base in polar solvents, meaning that $\ce{Ca(OH)_2}$ and $\ce{Al(OH)_3}$ having much smaller and more charged cations are almost covalent in nature and thus, do not dissociate.
Please, note: solubility for molecular compounds, like sugars and some non-dissociating compounds that looks like salts (like $\ce{Cu(CH_3COO)_2*H_2O}$) is an entirely different matter and determined by relative strength of intermolecular interactions (read: the molecule have to be able to form hydrogen bonds)