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

Let S be the solubility of Al(OH)3
Al(OH)311-sAl3+0s(aq)+3OH-03s(aq)
Concentration of species at t=0
Concentration of various species at equilibrium
Ksp=[Al3+][OH-]3=(S)(3S)3=27S4S4=Ksp27=2.7×10-1127=1×10-12S=1×10-3 mol L-1
(i) Solubility of Al(OH)3
Molar mass of Al(OH)3 is 78 g. Therefore, 
Solubility of Al(OH)3 in g L-1=1×10-3×78 g L-1=78×10-3 g L-1=7.8×10-2 g L-1
(ii) pH of the solution 
S=1×10-3 mol L-1[OH-]=3S=3×1×10-3=3×10-3pOH=3-log 3