Ionic EquilibriumMedium
Question
pH of a saturated solution of Ba(OH)2 is 12. The value of solubility product (Ksp) of Ba(OH)2 is
Options
A.5.0 × 10-6
B.3.3 × 10-7
C.5.0 × 10-7
D.4.0 × 10-6
Solution
pH = 12
pOH = 12 ⇒ [OH-] = 10-2
Ba(OH)2 ⇋ Ba2+ + 2OH- ⇒ 2x = 10-2
x 2x ⇒ x = 5 × 10-3
Ksp = x(2x)2 = 5 × 10-3 × (10-2)2 = 5 × 10-7
pOH = 12 ⇒ [OH-] = 10-2
Ba(OH)2 ⇋ Ba2+ + 2OH- ⇒ 2x = 10-2
x 2x ⇒ x = 5 × 10-3
Ksp = x(2x)2 = 5 × 10-3 × (10-2)2 = 5 × 10-7
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