At `25^(@)C`, after the addition of `110mL` of `0.1NaCI` solution to `100mL` of `0.1N AgNO_(3)` solution, the reduction potentila of a silver electrode placed in it is `0.36V`. Calculate the `K_(sp)`of `AgCI`. (Given: `E^(Θ) Ag//Ag^(o+) =- 0.799V)`.

Explanation

`110mL` of `1.0NaCI + 100 mL of 0.1 N AgNO_(3)`
`{:(110 xx 0.1 = 11 mEq,,,100xx0.1=10mEq,):}`
So mEq of `AgNO_(3)` reacts with `10mEq` of `NaCI` to given `10 mEq` of AgCI.
mEq of NaCI left `= 11 - 10 = 1` mEq
Total volume `= 100 +110 = 210 mL`
`:.` Concentration of `NaCI` and
`:.` Concentration of `[CI^(Θ)] = (1mEq)/(210mL) = (1)/(210)N` or `M`
The electrode is `CI^(Θ), Ag CI|Ag`
`Ag^(o+) +e^(Θ) rarr Ag`
`E^(Θ) Ag^(o+) |Ag = 0.799V`
`{:[(K_(sp)=[Ag^(o+)][CI^(Θ)]),(:. [Ag^(o+)]=(K_(sp))/([CI^(Θ)]):}`
`E cell = 0.799 - 0.059 "log"([CI^(Θ)])/(K_(sp))`
E cell = `0.799 + (0.059)/(1) "log"(K_(sp))/([CI^(Θ)])`
`= 0.36 = 0.799 + 0.059 log K_(sp) - 0.59 log [CI^(Θ)]`
`0.36 = 0.799 + 0.059 log K_(sp) - 0.059 "log"(1)/(210)`
`0.059 log K_(sp) = 0.36 - 0.799 + 0.059 "log"(1)/(210) =- 0.576`
`log K_(sp) =- (0.576)/(0.059) =- 9.76`
`:. K_(sp) = 1.73 xx 10^(-10)`