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

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3.3×10-7

5.0×10-7

4.0×10-4

5.0×106

answer is B.

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Detailed Solution

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

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