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Ksp of Pb3(AsO4)2

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The following cell was found to have a potential of 0.416 V. Calculate the k for Pb3(ASO4)2.

Pb3(s)| Pb3(AsO)2(saturated) AsO4^3-(1.00x10^-3 M) || S.H.E.

Pb^2+ +2e- <---> Pb(s) E^o = -0.126 V

H3AsO4 + 2H+ + 2e- <---> H3AsO3 + H2O E^o = 0.559 V

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Solution Summary

This solution shows the calculation of solubility product constant from electrochemical data.

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The oxidation half-reaction of the cell is the oxidation of Pb to Pb2+, and the reduction half reaction is the SHE half reaction.
OHR: Pb(s) <=> Pb2+(aq) + 2e- E° = 0.126 V
RHR: 2H+ + 2e- <=> H2O E° = 0.000 V
Overall cell ...

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