The solubility product of Al(OH) 3 is 2.7 × 10-11.

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

Al(OH)3 ⇌ Al3+ + 3OH-

We know that,

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

=(s) × (3s) 3 =27s4

S4 = Ksp/27

= 2.7x 10-11/27

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.