At a certain temperature, the solubility of aluminum hydroxide is (8.1x10^-13) M. What is the Ksp at this temperature?

4.2x10⁻⁴ M or 0.032 g/L.

Explanation:

Hello!

In this case, for solubility product problems, we apply the concepts of equilibrium because an insoluble salt is ionized until a certain point limited by the solubility product constant, Ksp. Thus, we first write the ionization reaction of aluminum hydroxide at equilibrium:

Next, we write the corresponding equilibrium expression:

Which in terms of , the reaction extent, is:

Because also represents the molar solubility of aluminum hydroxide at the considered temperature; now, we can write:

Which can be solved for x as follows:

Thus, the solubility is 4.2x10⁻⁴ M or mol/L and in g/L we have:

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