Final answer:
i) The solubility of Bi₂S₃ is 2.08 x 10⁻³² M and the solubility of Hgs is 2 x 10⁻²⁷ M. ii) Hgs is the most soluble compound.
Step-by-step explanation:
i) To calculate the solubility of Bi₂S₃ and Hgs, we need to first determine the equilibrium concentration of each ion in the compound. The solubility product expression for Bi₂S₃ is Ksp = [Bi³⁺][S²⁻]² = 1 x 10⁻⁹⁷. Since the stoichiometry of the compound is 2:3, the equilibrium concentration of Bi³⁺ is 2x and the equilibrium concentration of S²⁻ is 3x. Substituting these values into the expression, we get 4(3x)³ = 1 x 10⁻⁹⁷. Solving for x gives x = 2.08 x 10⁻³² M, which is the solubility of Bi₂S₃.
The solubility product expression for Hgs is Ksp = [Hg₂²⁺][S²⁻] = 4 x 10⁻⁵³. Since the stoichiometry of the compound is 1:1, the equilibrium concentration of Hg₂²⁺ is x and the equilibrium concentration of S²⁻ is x. Substituting these values into the expression, we get x² = 4 x 10⁻⁵³. Solving for x gives x = 2 x 10⁻²⁷ M, which is the solubility of Hgs.
ii) To determine the most soluble compound, we compare the solubility products. The compound with a higher Ksp value is more soluble. In this case, since the solubility product of Hgs (Ksp = 4 x 10⁻⁵³) is larger than that of Bi₂S₃ (Ksp = 1 x 10⁻⁹⁷), Hgs is the most soluble compound.