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for the following equilibrium, if the concentration of b− is 9.3×10−7 m, what is the solubility product for ab3? ab3(s)↽−−⇀a3 (aq) 3b−(aq)

User Arbme
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2 Answers

1 vote

Final answer:

The solubility product (Ksp) for AB3 is calculated by taking the stoichiometry of the dissolution reaction into account. It involves finding the molarity of A3+ (1/3 of B− concentration) and then using these concentrations to calculate Ksp.

Step-by-step explanation:

To find the solubility product (Ksp) for AB3 when the concentration of B⁻ is 9.3×10⁻⁷ M, we use the stoichiometry of the dissolution reaction AB3(s) ⇔ A³⁺(aq) + 3B⁻(aq). The solubility (s) of AB3 represents the molarity of A³⁺ in solution, and in this equilibrium, it is 1/3 of the concentration of B⁻. Therefore, s is 3.1×10⁻⁷ M (9.3×10⁻⁷ M / 3). Since each formula unit of AB3 produces three ions of B⁻, we raise the concentration of B⁻ to the third power and multiply by the concentration of A³⁺ to get the Ksp. Thus, Ksp for AB3 is (s)(s×3)³ = (3.1×10⁻⁷)(9.3×10⁻⁷)³.

User Arnelle Balane
by
8.7k points
6 votes

Answer:

2.5E-25

Step-by-step explanation:

Ksp = [A3+][B-]^3 = (1/3[B-])[B-]^3

Ksp = 1/3(9.3E-7)(9.3E-7)^3 = 2.5E-25

User Blelump
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7.7k points