Answer:
Ksp = 3.47x10⁻⁹
Explanation:
When Lead (II) iodide (PbI₂) is added to water, the equilbrium produced is:
PbI₂(s) → Pb²⁺(aq) + 2 I⁻(aq)
And solubility product constant, ksp is:
Ksp = [Pb²⁺] [I⁻]²
A solubility of 9.54x10⁻⁴ M means the maximum concentration of Pb²⁺ is 9.54x10⁻⁴ M and 9.54x10⁻⁴ M×2 of I⁻. Replacing in ksp formula:
Ksp = [9.54x10⁻⁴] [2×9.54x10⁻⁴]²
Ksp = 3.47x10⁻⁹
