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The solubility product of a sparingly soluble salt B4I2 is 4 × 10⁻¹². The solubility of B4I2 is

1. 4 × 10⁻⁴
2. 4 × 10⁻²
3. 4 × 10⁻³
4. 1 × 10⁻⁴

Correct answer: 4 × 10⁻⁴

Solution

For a salt B4I2, the dissociation is B4I2 ⇌ 4B⁺ + 2I⁻. Let the solubility be 's'. The solubility product (Ksp) is given by Ksp = [B⁺]⁴[I⁻]² = (4s)⁴(2s)² = 1024s⁶. Solving 1024s⁶ = 4 × 10⁻¹² gives s = 4 × 10⁻⁴.

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