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

`Ksp=2.2x10^(-32)`

Step-by-step 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)(molOH^-)/(L)*(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.