Find the solubility of `AgCl` in `0.1 M CaCl_(2)`. `E^(c-)._(Ag^(o+)|Ag)=0.799V` and that of `AgCl|Ag=0.222V`.

Explanation

`AgrarrAg^(o+)+e^(-)`
`AgCl+e^(-)rarr Ag+Cl^(c-)`
`ulbar(AgCl rarr Ag^(o+)=Cl)`
`E^(c-)=0.222-0.799=-0.577`
Ccell is in equilibrium `,,,, ( :. E_(cell)=0)`
`E^(c-)=0.059 log K_(sp)(n=1)`
`-0.577=0.059log K_(sp)`
`:. K_(sp)=1.66xx10^(-10)`
Let `S M` is a solubility of `AgCl` in `0.1 M CaCl_(2)`
`underset(S)(AgCl)rarrunderset(S)(Ag^(o+))+underset(S)(Cl^(c-))`
`{:(CaCl_(2)rarr,Ca^(2+),+2Cl^(c-)),(0.1,0,0),(0,0.1,0.2):}`
Total `Cl^(c-)=(S+0.2)M`
`K_(sp)=[Ag^(o+)][Cl^(c-)]=S(S+0.2)`
`:. 1.66xx10^(-10)=(S^(2)+0.2S)[S^(2)lt1]`
`:. S=(1.66xx10^(-10))/(2)=8.3xx10^(-10)M`