Two 50mL beaker. One contains 50mL of 1.0M PbSO4 and the other contains 10mL of 0.1M Pb(NO3)2 added with 40mL of .10M KI. What is the Ksp of PbI2?

Answer 1

Ksp=[Pb][I]^2
[Pb]=.10M*(10ml/(40mL+10mL+50mL)=.01M
[I]=.10M*(40/(40mL+10mL+50mL))=.04M
Ksp=1.6E-05
This is what i got

Answer 2

PbSO4 is insoluble. How do you make 1M PbSO4?

Answer 3

there should be an E(V) too. let suppose your E(V)=.128.. your calculation should somewhat look like this if I'm not wrong,
Ecell=E^o cell - (RT/nF)lnQ
.125=0-(.0592/2)lnQ
e^(-.128/.0296)=[I]/.1M
[I]=.001324
[Pb]= [I]x(1/2)=.000662

ksp=[Pb][I^2]=1.16E-09