The EMF of cell
`Ag | AgCl, 0.05 M KCl || 0.05 M AgNO_(3) |Ag` is 0.788 Volt.
Find the solubility product of AgCl`.

Explanation

For the cell reaction: `Ag^(+)(0.05) = Ag^(+)(c)`,
where c is the Ag concentration in LHS half cell,
`E_("cell") = (0.0591)/(1) log (0.05)/(c0 , K_(sp) (AgCl) = c xx 0.05`
`1.16 xx 10^(-16)`