Question

9.54 x 10^-4 mol/L is the solubility of lead(II) iodide in water at 0 C. What is the solubility product constant for lead(II) iodide at the same temperature?

Answer 1

Ksp = 3.47x10⁻⁹

Step-by-step 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