pH of saturated solution of Ba(OH)2 is 12.The value of solubility product of Ksp of Ba(OH)2 is

Given pH of Ba(OH)2=12pOH=14-pH=14-12=2We know that, pOH=-logOH-2=-logOH-OH-=10-2Ba(OH)2 dissolve in water as Ba(OH)2(s)⇌Ba2++2OH-S mol L-1  S  2SOH-=2s=10-2Ksp=Ba2+OH-2=10-22×10-22=0.5×10-6=5×10-7