`EMF` of the cell
`|Ag|AgNO_(3)(0.1M)||K Br(1 N),AgBr(s)|Ag is -0.6V` at `298K`
`AgNO_(3) ` is `80%` and `KBr` is `60%` dissociated.
Calculate `a.` Solubility and
`b. K_(sp)` of `AgBr` at `298 K`.

Explanation

The above cell is a concentration cell.
`:. E^(c-)._(cell)=0.0V`
The half cell reactions are
Anode reaction `: Ag(anode) hArrAg^(o+)(0.1M)+e^(-)` and Cathode reaction `:`
`Ag^(o+)(cathode) +e^(-)hArrAgs(cathode)`
Cell reaction `ulbar( :Ag^(o+)(cathode)hArrAg^(o+)(0.1 M) _(anode))`
Since `AgNO_(3)` is `80%` ionized.
`:. [Ag^(o+)]_(a)=0.1xx(80)/(100)=0.08M`
`:.E_(cell)=E^(c-)._(cell)-(0.06)/(1)log.([Ag^(o+)]_(a))/([Ag^(o+)]_(c))(` Take `0.059~~0.06)`
`-0.6 V=0-0.06log.(0.08)/([Ag^(o+)]_(c))`
`log[Ag^(o+)]_(c)=log0.08-10=(log_(2)^(3)-log 100)-10`
`=0.9-2-10=-11.1`
`:. [Ag^(o+)]_(c)=Antil o g(-11.1)=Antil og(bar(12).9)=7.9xx10^(-12)`
`~~8xx10^(-12)`
Since `K Br` is `60%` dissociated,
`:.[Br^(c-)]=1xx(60)/(100)=0.6M`
`:. K_(sp)=[Ag^(o+)][Br^(c-)]`
`=8 xx 10^(-12)xx0.6xx10^(-12)M^(2)`
Solubility `(S) = sqrt(K_(sp))=sqrt(4.8)xx10^(-6)M^(2)`
`=2.19xx10^(-6)M`