Molar solubility refers to the maximum amount of solute that can dissolve in a solvent at a given temperature. In the case of AL(OH)3 (aluminum hydroxide), the molar solubility can be increased in the presence of both NaOH (sodium hydroxide) and HCl (hydrochloric acid).

- Effect of NaOH:
When NaOH is added to a solution of AL(OH)3, it reacts with the aluminum hydroxide to form a soluble complex ion. The reaction can be represented as follows:
2Al(OH)3 + 6NaOH → 2Na3Al(OH)6
The resulting compound, Na3Al(OH)6, is more soluble in water compared to pure AL(OH)3. This increase in solubility is due to the formation of a complex ion that can dissociate more easily into its constituent ions in the solution.

- Effect of HCl:
Similarly, when HCl is added to a solution of AL(OH)3, it reacts with the aluminum hydroxide to form a soluble aluminum chloride complex. The reaction can be represented as follows:
Al(OH)3 + 3HCl → AlCl3 + 3H2O
The resulting compound, AlCl3, is also more soluble in water compared to pure AL(OH)3. The formation of the aluminum chloride complex increases the solubility of AL(OH)3 by breaking down the solid compound into its constituent ions.

- Effect of Both NaOH and HCl:
When both NaOH and HCl are present in a solution of AL(OH)3, the solubility of the compound is further increased. This is because the complex ions formed by the reaction of AL(OH)3 with NaOH and HCl can coexist in the solution. The presence of both complex ions increases the overall solubility of AL(OH)3.

In conclusion, the molar solubility of AL(OH)3 is increased in the presence of both NaOH and HCl. The formation of soluble complex ions through the reaction with these compounds enhances the dissociation of AL(OH)3 into its constituent ions in the solution, resulting in a higher solubility.