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Using the information in the preceding problem, calculate the solubility product of Agl in water at `25^(@)C` `[E_((Ag^(+),Ag))^(@)=+0.799`volt]

A

`1.97xx10^(-17)`

B

`8.43xx10^(-17)`

C

`1.79xx10^(-17)`

D

`9.17xx10^(-17)`

Text Solution

Verified by Experts

The correct Answer is:
B

`E_(Ag|AgI|I^(-))^(0)=E_(Ag^(+)|Ag)^(0)-(0.0591)/(1)log((1)/(K_(SP))`
`-0.151=0.799-(0.0591)/(1)log((1)/(K_(sp))implies0.0591logK_(SP)=-0.151-0.799`
`logK_(SP)=-16.074impliesK_(SP)=8.43xx10^(-17)`
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