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The solubility product of `Pb_(3)(AsO_(4))_(2)` is `4.1xx10^(-36). E^(c-)` for the reaction `:` `Pb_(3)(AsO_(4))_(2)(s)+6e^(-)hArr3Pb(s)+2AsO_(4)^(2-)` `E_((Pb)2^(+)|Pb)^(Θ)=-0.13V`A. `+0.478V`B. `-0.13V`C. `-0.478V`D. `+0.13V` |
Answer» Correct Answer - c `E^(c-)._(AsO_(4)^(2-)|Pb_(3)(AsO_(4))^(2)|Pb)=E^(c-)._(Pb^(2+)|Pb)+(0.059)/(6)logK_(sp)` `=-0.13+(0.059)/(6)log4.1xx10^(-36)` `=-0.13-0.348=-0.478V` |
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