Find the solubility product of a saturated solution of `Ag_(2)CrO_(4)` in water at `298K`, if the `EMF` of the cell `:`
`Ag|Ag^(o+)(satAg_(2)CrO_(4)sol)||Ag(0.1M)|Agis 0.164V` at `298K`.

Explanation

For the cell
`Ag|Ag^(+)(Ag_(2)CrO_(4) "sol. Saturated") ||(Ag^(+)),(0.1M)|Ag`:
`E_(cell) = 0.164 V` at `298 K`
We have `E_(cell) = E_(OP_(Ag//Ag^(+)))^(@)+E_(RP_(Ag^(+)//Ag))^(@)`
`+ (0.059)/(1)log_(10).([Ag^(+)]_(R.H.S.))/([Ag^(+)]_(L.H.S.))`
or ` 0.164 = 0 + (0.059)/(1)log_(10).(0.1)/([Ag^(+)]_(L.H.S.))`
`:. [Ag^(+)]_(L.H.S.) = 1.66 xx 10^(-4)M`
Now `K_(SP)` for `Ag_(2)CrO hArr 2Ag^(+) +CrO_(4)^(2-)`
`K_(SP) = [Ag^(+)]^(2)[CrO_(4)^(2-)]`
Since, `[Ag^(+)]_(L.H.S.) = 1.66 xx 10^(-4)M`
`:. [CrO_(4)^(2-)]_(L.H.S.) = (1.66 xx 10^(-4))/(2) M`
`:. K_(SP) = [1.6 xx 10^(-4)]^(2)[(1.66 xx 10^(-4))/(2)]`
`K_(SP) = 2.287 xx 10^(-12) mol^(3) litre^(-3)`