The emf of the cell `Ag|KI(0.05M)||AgNO_(3)(0.05M)|Ag` is `0.788V`. Calculate the solubility product of `AgI`.

Answer

`E_(sp) = 1.16 xx 10^(-16)`

Explanation

`K_(sp) = [Ag^(+)] [0.05]`
`E_(cell) = E_(OP_(Ag)) +E_(RP_(Ag))`
`E_(cell) = (E_(OP_(Ag))^(@) -(0.0591)/(1)log[Ag^(+)]_(L.H.S))`
`+(E_(RP_(Ag))^(@) +(0.0591)/(1)log[Ag^(+)]_(R.H.S))`
`E_(cell) =(0.0591)/(1)log.([Ag^(+)]_(R.H.S))/([Ag^(+)]_(L.H.S))`
`0.788 = 0.0591 log.(0.05)/([Ag^(+)]_(L.H.S))`
`[Ag^(+)]L.H.S = 2.203 xx 10^(-15)`
`K_(sp) = 2.203 xx 10^(-5) xx 0.05 = 1.1 xx 10^(-16)`