The solubility of lead(II)iodide is 0.064 g/100 mL at 20^oC. What is the solubility product for lead(II)iodide?

Name: The solubility of lead(II)iodide is 0.064 g/100 mL at 20 o C. What is the solubility product for lead(II)iodide?
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Quizplus · Accepted Answer

The solubility product (Ksp) is calculated using the solubility of PbI2. The molar mass of PbI2 is approximately 461 g/mol. Solubility in mol/L = 0.064 g/100 mL * (1 mol/461 g) * (1000 mL/1 L) = 0.00139 mol/L. Ksp = [Pb^2+][I^-]^2 = (0.00139)(2*0.00139)^2 = 1.1 * 10^-8.

The Solubility of Lead(II)iodide Is 0