Asked by slamj
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?
Answers
Answered by
denny
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
[Pb]=.10M*(10ml/(40mL+10mL+50mL)=.01M
[I]=.10M*(40/(40mL+10mL+50mL))=.04M
Ksp=1.6E-05
This is what i got
Answered by
DrBob222
PbSO4 is insoluble. How do you make 1M PbSO4?
Answered by
karl
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
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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