In water, the solubility of Bi(OH)3 is 2.87 × 10⁻⁷ M. Calculate the solubility product constant (Ksp) for Bi(OH)3.

• A. Ksp = 5.67 × 10⁻²²
• B. Ksp = 8.23 × 10⁻²¹
• C. Ksp = 2.36 × 10⁻²⁰
• D. Ksp = 1.23 × 10⁻¹⁹

Answer: (C)

Verified step by step guidance

1. Start by writing the balanced dissolution equation for Bi(OH)₃ in water: Bi(OH)₃(s) ⇌ Bi³⁺(aq) + 3OH⁻(aq).
2. Identify the stoichiometry of the dissolution process: 1 mole of Bi(OH)₃ produces 1 mole of Bi³⁺ ions and 3 moles of OH⁻ ions.
3. Let the solubility of Bi(OH)₃ be represented by 's', which is given as 2.87 × 10⁻⁷ M. Therefore, [Bi³⁺] = s and [OH⁻] = 3s.
4. Write the expression for the solubility product constant (Ksp): Ksp = [Bi³⁺][OH⁻]³.
5. Substitute the concentrations into the Ksp expression: Ksp = (s)(3s)³ = s × 27s³ = 27s⁴. Use the given solubility value to calculate Ksp.