Question

The Ksp of Al(OH)3 is 1.0 x 10-33. What is the solubility of Al(OH)3 in a solution that has pH = 12? Give your answer using scientific notation and to 2 significant figures (i.e., one decimal place).

Answer 1

Answer : The solubility of
`Al(OH)_3` is
`1.0* 10^(-27)M`

Explanation :

First we have to calculate the pOH.

`pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-12\\\\pOH=2`

Now we have to calculate the concentration of
`OH^-`.

`pOH=-\log [OH^-]`

`2=-\log [OH^-]`

`[OH^-]=0.01`

The solubility equilibrium reaction will be:

`Al(OH)_3\rightleftharpoons Al^(3+)+3OH^(-)`

The expression for solubility constant for this reaction will be,

`K_(sp)=[Al^(3+)][OH^(-)]^3`

Now put all the given values in this expression, we get:

`1.0* 10^(-33)=[Al^(3+)]* (0.01)^3`

`[Al^(3+)]=1.0* 10^(-27)M`

As, the solubility of
`Al(OH)_3` =
`[Al^(3+)]` =
`1.0* 10^(-27)M`

Thus, the solubility of
`Al(OH)_3` is
`1.0* 10^(-27)M`