Question

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

Answer 1

4.2x10⁻⁴ M or 0.032 g/L.

Step-by-step 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]{(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)(mol)/(L)*(78g)/(1mol) =0.032(g)/(L)`

Best regards!