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The `EMF` of the cell `Ag|0.1(N)AgNO_(3)||1(N) KBr, AgBr(s)|Ag` was found to be `-0.64V` at `298 K. 0.1 N Ag NO_(3)` is `81.3%` dissociated and` 1N KBr` is `75.5%` dissociated. Calculate `:` `a.` Solubility `b.` Solubility product of `AgBr ` at `298 K` |
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Answer» `Ag|0.1N AgNO_(3)||1N KBr, AgBr(s)|Ag` The above cell is concentration cell `:. E^(c-)._(cell)=0` `Ag^(o+)._((anode))rarr Ag^(o+)(0.1N)+e^(-)` `Ag^(o+)._((cathode))+e^(-)rarr Ag_((cathode))` `ulbar(Ag^(o+)._((cathode))rarr Ag^(o+)(0.1N))` Since `AgNO_(3)` is `81.3%` ionized `:. [Ag^(o+)]_(a)=0.1xx0.813=0.0813N` `-0.64=0-0.059log .(0.0813)/([Ag^(o+)._((cathode))])` `:. [Ag^(o+)]_((cathode))=3.2xx10^(-12)M` Since `1 N KBr` is `75.5%` dissociated `:. [Br^(c-)]=1xx0.755=0.755 M` `K_(sp)[Ag^(o+)][Br^(c-)]=3.2xx10^(-12)xx0.755` `2.416xx10^(-12)` `S=sqrt(2.416xx10^(-12))` `=1.544xx10^(-6)M` |
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