Question

What is the Ksp of Al(OH)3 if the concentration of OH– in a saturated solution of Al(OH)3 is 1.6 × 10–8 M?

Answer 1

Answer:

$Ksp=2.2x10^{-32}$

Explanation:

Hello,

In this case, since the dissociation of aluminum hydroxide is:

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

The equilibrium expression is:

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

Thus, given the concentration of hydroxyl ions in the solution, and the 3:1 mole ratio with the aluminum ions, the concentration of those turn out:

$[Al^{3+}]=1.6x10^{-8}\frac{molOH^-}{L}*\frac{1molAl^{3+}}{3molOH^-} =5.3x10^{-9}M$

Therefore, the solubility product, Ksp turns out:

$Ksp=(5.3x10^{-9}M)[1.6x10^{-8}M]^3\\\\Ksp=2.2x10^{-32}$

Regards.