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

Explanation

Given cell : `Ag|Ag^(+)("sat "Ag_(2)CrO_(4)" sol")||Ag (0.1M)|Ag`
and `E_("cell")=0.164V` at 298K.
`{:("At anode: "Ag(s)" "rarr""Ag^(+)("sat "Ag_(2)CrO_(4))+e),("At cathode: "Ag^(+)(aq)+e" "rarr" "Ag(s)),(bar("Cell reation: "Ag^(+)(aq)rarr Ag^(+)("sat "Ag_(2)CrO_(4)))):}`
Let molar concentration of `Ag^(+)` ions in sat `Ag_(2)CrO_(4)` be `C_(1)M`
`therefore" "E_("cell")=E_("cell")^(@)-(0.059)/(1)log.(C_(1))/([Ag^(+)])`
`E_("cell")^(@)=0` as both the electrodes are the same
`therefore" "0.164=(0.059)/(1)log.(0.1)/(C_(1))" or, "log.(0.1)/(C_(1))=(0.164)/(0.059)=2.774`
`"or, "C_(1)=[Ag^(+)]Ag_(2)CrO_(4)=1.66xx10^(-4)(M)`
`"In, "Ag_(2)CrO_(4)(s) rArr 2Ag^(+)(aq)+CrO_(4)^(2-)(aq)`
`[CrO_(4)^(2-)]=([Ag^(+)])/(2)=(1.66xx10^(-4))/(2)(M)=0.83xx10^(-4)(M)`
`therefore" "K_(sp)" of "Ag_(2)CrO_(4)=[Ag^(+)]^(2)[CrO_(4)^(2-)]`
`=(1.66xx10^(-4))^(2)xx0.83xx10^(-4)=2.287xx10^(-12)`