The `EMF` of the cell `:`
`Ag|AgCl,0.1 MKCl||0.1 M AgNO_(3)|Ag` is `0.45V. 0.1 M KCl `is `85%` dissociated and `0.1 M AgNO_(3)` is `82%` dissociated. Calculate the solubility product of `AgCl` at `25^(@)C`.

Explanation

It is a concentration cell.
`:. E^(c-)._(cell)=0`
`Ag(Anode ) rarr Ag^(o+)+(c_(1))+e^(-)`
`Ag^(o+)(c_(2))+e^(-) rarrAg(Anode)`
`ulbar(Ag^(o+)(c_(2)) rarr Ag^(o+)(c_(1)))`
`0.1 M AgNO_(3)=0.1M Ag^(o+)`
Since it is `82%` ionized
`:. Ag^(o+)(` cathode `)=0.1xx0.82`
`c_(2)=0.082 M`
`:. E_(cell)=0-(0.0591)/(1)log.([Ag^(o+)(c_(1))])/([Ag^(o+)(c_(2))])`
`0.45=-(0.0591)/(1)log.([Ag^(o+)(c_(1))])/(0.082)`
`=(0.45 xx1)/(0.0591)=log .(0.082)/([Ag^(o+)(c_(1))])=7.627`
Antilog`(7.627xx10^(7))=3.99xx10^(7)`
`:. [Ag^(o+)(c_(1))]=(0.082)/(3.99xx10^(7))=2.055xx10^(-9)M`
`0.1 M KCl=0.1 MCl^(c-)`
Since it is `85%` dissociated
`:. [Cl^(c-)]=0.1xx0.85=0.085 M`
`:. K_(sp)(AgCl)=[Ag^(o+)][Cl^(c-)]`
`=(1.9 xx10^(-9))(0.085)`
`:. K_(sp)=1.615 xx 10^(-1)`