Excess of AgCl is added to 0.1 M solution of KBr at 298 K. Calculate the equilibrium concentrations of `Br^(-)` and `Cl^(-)` ions.
`E_(Cl^(-)//AgCl//Ag)^(@)=0.222" volt "E_(Br^(-)//AgBr//Ag)^(@)=0.095` volt

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Correct Answer - `[Cl^(-)]=0.09929` ol `L^(-1), [Br^(-)]=0.00071` mol `L^(-1)`
`{:(,AgCl,+,Br^(-),hArr,AgBr,+,Cl^(-)),("At equilibrium,",,,(0.1-x),,,,x):}`
Apply `log K =E_(cell)^(@)/0.0591`
or `"log"x/((0.1-x))=E_(cell)^(@)/0.0591`

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