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Major Spinka DVM

Feb 20, 2025

What is the solubility product for lead(II) iodide?

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

Feb 20, 2025

0.064 g/100 mL has 0.64 g/1 Litre find moles, using molar mass:0.64 grams/L of PbI2 @ 461.02 g/mol = 1.388 e-3 moles/L PbI21.388 e-3 moles/L PbI2 releases an equal number of moles of Pb+2 = 1.388 e-3 moles/L of Pb+21.388 e-3 moles/L PbI2 releases twices as many moles of I-1 = 2.78 e-3 moles/L of Iidide PbI2 <=> Pb+2 & 2 I-1Ksp = [Pb] [I]^2Ksp = [1.388 e-3 ] [2.78 e-3]^2K = 1.07 e -8your answer,rounded to 2 sig figs is1.1 X 10^-8... Show More
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