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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?

  • chemistry -

    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

  • chemistry -

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

  • chemistry -

    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

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