Question

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

Answer 1

Answer:

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:

$Al(OH)_3(s)\rightleftharpoons Al^{3+}(aq)+3OH^-(aq)$

Next, we write the corresponding equilibrium expression:

$Ksp=[Al^{3+}][OH^-]^3$

Which in terms of $x$, the reaction extent, is:

$Ksp=x*(3x)^3$

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

$8.1x10^{-13}=x*(3x)^3$

Which can be solved for x as follows:

$x=\sqrt[4]{\frac{8.1x10^{-13}}{27} } \\\\x=4.2x10^{-4}M$

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

$4.2x10^{-4}\frac{mol}{L}*\frac{78g}{1mol} =0.032\frac{g}{L}$

Best regards!