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

1. 3.3 × 10⁻⁷
2. 5.0 × 10⁻⁷
3. 4.0 × 10⁻⁶
4. 5.0 × 10⁻⁶

Correct answer: 3.3 × 10⁻⁷

Solution

The pH of 12 implies a pOH of 2, so [OH⁻] = 10⁻² M. Since Ba(OH)₂ dissociates as Ba²⁺ + 2OH⁻, [Ba²⁺] = [OH⁻]/2 = 0.005 M. Ksp = [Ba²⁺][OH⁻]² = (0.005)(10⁻²)² = 3.3 × 10⁻⁷.

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