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Homework Statement
Find the concentration of Pb2+ in 0.0010 M KI with PbI2. Include activity coefficients in your solubility-product expression. The Ksp of PbI2 is 9.8*10^-9
Homework Equations
Ksp= [Pb2+]\gammaPb2+[I-]2\gamma2I-
Assume that [I-]=0.0010 M. The ionic strength of 0.0010 M Ki can be calculated, then used to find the activity coefficients from a table.
The Attempt at a Solution
-The ionic strength is 0.0010 M.
-Based on that ionic strength the activity coefficients are 0.868 for lead and 0.965 for iodide
9.8*10^-9=[lead](.868)[iodide]^2(.965)^2
9.8*10^-9=[x](.868)[.001+2x]^2(.965)^2
solve for x=.0011345
That is wrong and I have no idea why. Could anyone please help me.