Answer : The solubility of [tex]Al(OH)_3[/tex] is [tex]1.0\times 10^{-27}M[/tex]
Explanation :
First we have to calculate the pOH.
[tex]pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-12\\\\pOH=2[/tex]
Now we have to calculate the concentration of [tex]OH^-[/tex].
[tex]pOH=-\log [OH^-][/tex]
[tex]2=-\log [OH^-][/tex]
[tex][OH^-]=0.01[/tex]
The solubility equilibrium reaction will be:
[tex]Al(OH)_3\rightleftharpoons Al^{3+}+3OH^{-}[/tex]
The expression for solubility constant for this reaction will be,
[tex]K_{sp}=[Al^{3+}][OH^{-}]^3[/tex]
Now put all the given values in this expression, we get:
[tex]1.0\times 10^{-33}=[Al^{3+}]\times (0.01)^3[/tex]
[tex][Al^{3+}]=1.0\times 10^{-27}M[/tex]
As, the solubility of [tex]Al(OH)_3[/tex] = [tex][Al^{3+}][/tex] = [tex]1.0\times 10^{-27}M[/tex]
Thus, the solubility of [tex]Al(OH)_3[/tex] is [tex]1.0\times 10^{-27}M[/tex]
