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K(sp) of AgCl is...

`K_(sp)` of `AgCl` is

A

`10^(-13)`

B

`10^(-12)`

C

`10^(-11)`

D

`10^(-10)`

Text Solution

Verified by Experts

The correct Answer is:
d
Moreover, `E_(cell)=E^(c-)._(cell)-(0.059)/(1) log .(1)/([Ag^(o+)][Cl^(c-)])`
`(` At equilibrium `E_(cell)=0]`
`0=E^(c-)._(cell)-0.059log .(1)/(K_(sp))`
`:. log .(1)/(K_(sp))=(E^(c-)._(cell)xxn_(cell))/(0.059)=(0.59xx1)/(0.059)=10`
`:. K_(sp)=10^(-10)`
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