Question

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`.

Answer 1

Correct Answer - `2.287xx10^(-12)M^(3)`
Cell circuit is given as below `:`
`Ag|Ag^(o+)(` saturated `)Ag_(2)CrO_(4)||Ag^(o+)(0.1M)|Ag`
`E_(cell)=0.164` at `298K`
For the given cell `(` concentration cell `)`
`E_(cell)=(0.0591)/(n)log .(c_(2))/(c_(1))`
`c_(1)=[Ag^(o+)](` in `Ag_(2)CrO_(4)(` left side `)`
`c_(2)=(Ag^(o+))(` right side `)`
`n=1(` number of exchange electrodn `)`
So, `0.164=(0.0591)/(1)log.(0.1)/(c_(1))`
or `log.(0.1)/(c_(1))=(0.164)/(0.05910=2.774`
On solving
`c_(1)=[Ag^(o+)]` in `Ag_(2)CrO_(4)=1.66xx10^(-4)M`
We get
`[CrO_(4)^(2-)]([Ag^(o+)])/(2)`
So, `[CrO_(4)^(2-)]=(1.66xx10^(-4))/(2)M=0.83xx10^(-4)M`
`:gtK_(sp)` of `Ag_(2)CrO_(4)=[Ag^(o+)]^(2)[CrO_(4)^(2-)]`
`=(1.66xx10^(-4))(0.83xx10^(-4))`
`=2.287xx10^(-12)M^(3)`