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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`. |
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Answer» 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)` |
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