1.

Calculate the solubility product of Ag_(2)CrO_(4) at 298K, if the EMF of the concentration cell, Ag, Ag^(+)(Solide Ag_(2)CrO_4)"//"Ag^(+)(0.1M), Ag is 0.164 V.

Answer»

Solution :`E_("cell") = 0.164 = 0.059 "log" [Ag^+]_("cath")//[Ag^+]_("ANOD")`
`"log" ([Ag^+]_("cath"))/([Ag^+]_("anod")) = (0.164)/(0.059) = 2.78 = "log"(0.1)/([Ag^+]_("anod"))`
`[Ag^+]_("anod") = [Ag^+]` from saturated solution of `Ag_(2)CrO_(4) = (1.66 xx 10^(-4))^(2) (1.66 xx 10^(-4)//2) = 2.28 xx 10^(-12) mol^(3) "LIT"^(-3)`.


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