pH of a saturated solution of Ba(OH)2 is 12. The value of solubility product (Ksp) of Ba(OH)2 is
- 3.3 10-7
- 5.0 10-7
- 4.0 10-6
- 5.0 10-6
Answer
(2) 5.0 10-7
Solution
pH = 12 pOH = 2 [OH-]=10-2 M
Ba(OH)2 Ba2++2OH-
[Ba2+]= 10-22=5 10-3 M
Ksp= [Ba2+][OH-]2
=(5 10-3)(10-2)2
=5.0 10-7
Answer: (2)