The solubility product of Al(OH) 3 is 2.7 × 10-11. Calculate its solubility in gL-1 and also find out the pH of this solution. (Atomic mass of Al = 27 u).

Given, Ksp =2.7×10-11

The equation of disassociation of Al(OH)3will be-




We know that,


Ksp= [Al3+] [OH-] 3-


=(s) × (3s) 3


=27s4




s4= 10-12


s= (10-12)1/4 =10-3 mol/L


Now, molar mass of Al(OH)3 =78


Solubility= molar mass ×s


= 78 ×10-3


= 7.8 × 10-2 g/L


NOW, we know that-


pH = 14 - pOH


[OH]= 3s = 3 × 10-3


pOH= 3-log3


pH = 14 – 3 + log3


= 11.4771


Hence the pH of the solution will be 11.4771 and solubility in g/L will be 7.8×10-2 g/L.


1