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Find the solution product of `Ag_(2)CrO_(4)` in water at 298 K if the e.m.f. of the cell `Ag//Ag^(+)` (saturated `Ag_(2)CrO_(4)soln.)||Ag^(+)(0.1M)//Ag" is "0.164V" at "298K`. |
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Answer» The cell is `Ag|Ag^(+)(c_(1))||Ag^(+)(0.1M)|Ag` `E_(cell)=(0.0591)/(n)"log"(c_(2))/(c_(1))` (for concentration cell) `0.164=(0.0591)/(1)"log"(0.1)/(c_(1))` or log`(0.1)/(c_(1))=2.7750` or `0.1//c_(1)=5.957xx10^(2)` o `c_(1)=1.679xx10^(-4)` i.e., `[Ag^(+)]` in saturated `Ag_(2)CrO_(4)` sol.=`1.679xx10^(-4)M` `Ag_(2)CrO_(4)hArr2Ag^(+)+CrO_(4)^(2-)` `[Ag^(+)]=1.679xx10^(-4)M,[CrO_(4)^(2-)]=([Ag^(+)])/(2)=(1.679xx10^(-4))/(2)M` `K_(sp)=[Ag^(+)]^(2)[CrO_(4)^(2-)]=(1.679xx10^(-4))^(2)xx((1.679xx10^(-4))/(2))=2.37xx10^(-12)`. |
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