1. The problem statement, all variables and given/known data 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 2. Relevant equations Ksp= [Pb2+][itex]\gamma[/itex]Pb2+[I-]2[itex]\gamma[/itex]2I- 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. 3. 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.