Final answer:
The molar solubility of cerium (III) iodate, Ce(IO₃)₃, is calculated using its Ksp value of 1.40 × 10⁻¹¹, resulting in a solubility of approximately 6.43 × 10⁻⁴ mol/L.
Step-by-step explanation:
To calculate the molar solubility of cerium (III) iodate, Ce(IO₃)₃, with a solubility product constant (Ksp) of 1.40 × 10⁻¹¹, we first write out the equilibrium dissolution reaction:
Ce(IO₃)₃ (s) → Ce³⁺ (aq) + 3 IO₃⁻ (aq)
Next, we set up the expression for Ksp:
Ksp = [Ce³⁺][IO₃⁻]³
If we let s represent the molar solubility of Ce(IO₃)₃ in moles per liter, then at equilibrium, the concentration [Ce³⁺] = s and [IO₃⁻] = 3s. Plugging these values into the Ksp expression gives:
Ksp = s(3s)³
Ksp = 27s⁴
Solving for s:
s = √[1.40 × 10⁻¹¹ / 27]
s ≈ 6.43 × 10⁻⁴ M
Thus, the molar solubility of Ce(IO₃)₃ is approximately 6.43 × 10⁻⁴ mol/L.