For the measurement of the solubility product of AgCl the following cell is constructed :
`Ag|AgCl||KCl (0.1 M)||AgNO_(3) (0.1 M)|Ag`
The emf of the cell is 0.45 volt. In the cell, KCl is dissociated to the extent of 83% and `AgNO_(3)` is dissociated to the extent of 86%. Calculate the solubility product of AgCl at 298 K.

Answer

`1.735xx10^(-10)`

Explanation

`E_(cell)=E_(cell)^(@)-0.0591 "log "([Ag^(+)]_("Anode"))/([Ag^(+)]_("Cathode"))`