Following cell has EMF 0.7995 V .
Pt`| H_(2)` (1) atm) |`HNO_(3) (1M)|| AgNO_(3) (1 M)| Ag` If we add enough KCl to Ag cell so that the final `Cl^(-)` is 1 M . Now the measured emf of the cell is `0.222 V `. The `K_(sp)` of AgCl would be -

Options

A. `1 xx 10^(-9.8)`
B. `1 xx 10^(-19.6)`
C. `2 xx 10^(-10)`
D. `2.64 xx 10^(-14)`

Answer

A

Explanation

`2 Ag^(+) + H_(2) to 2 H^(+) + 2 Ag`
`E = E^(@) - (0.0591)/(2) "log" ([H^+]^(2))/(P_(H_(2)) xx [Ag^(+)]^(2))`
`0.222 = 0.7995 - (0.0591)/(2) "log" (1)/([Ag^(+)]^(2))`
`[Ag^(+)] = 10^(-9.8)`
`K_(sp) = [Ag^(+)] [Cl^(-)] = (10^(-9.8)) xx (1) = 10^(-9.8)`