The solubility of lead (ii) iodide is 0.064 g/100 ml at 20oC. What is the solubility product for lead (ii) iodide?
Please show work, thanks!
Answer : 0.064 g/100 mL means 0.064 x 10 g / 100 x 10^-3 x 10 mL = 0.64 g/1 Litre
Now next step is to find moles using molar mass of PbI2 = 461.02 g/mol:
number of moles = weight of PbI2 / Molar mass of PbI2 = 0.64 grams/L of PbI2 / 461.02 g/mol
= 1.388 e-3 moles/L PbI2
1.388 e-3 moles/L PbI2 releases an equal number of moles of Pb+2
= 1.388 e-3 moles/L of Pb+2
1.388 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-1
Ksp = [Pb]* [I]^2
Substitute values of [Pb] and [I] from above
Ksp = [1.388 e-3 ] [2.78 e-3]^2
K = 1.07 e -8 ~ 1.1 X 10^-8
The solubility of lead (ii) iodide is 0.064 g/100 ml at 20oC. What is the solubility...
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