1.

Calculate the solubility product ofAg_(2)CrO_(4) at 298k, if the EMF of the concentration cell, Ag, Ag^(+)("solid" 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^(+)]_("cathode"))/([Ag^(+)]_("anode"))=(0.164)/(0.059)=2.78 = log. (0.1)/([Ag^(+)]_("anode"))`
`[Ag^(+)]_("anod")=[Ag^(+)]` from saturated
`Ag_(2)CrO_(4) =1.66xx10^(-4)M`.
Solubility PRODUCT of `Ag_(2)CrO_(4)` is given as, `[Ag^(+)]^(2)[CrO_(4)^(2-)]`.
Solubility product of `Ag_(2)CrO_(4)`
`=(1.66xx10^(-4))^(2)(1.66xx10^(-4)//2)`
`=2.28xx10^(-12)"mol"^(3)"lit"^(-3)`


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