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Solubility product of AgI at 25^@C is 4x...

Solubility product of AgI at `25^@C` is `4xx10^(-18)"mole"^2 //L^2` ​. The solubility of AgI in presence of `2xx10^(-4)` M Kl solution at `25^@C` is approximately equal to : (in mol/L)

A

`8xx10^(-8)`

B

`2xx10^(-16)`

C

`2xx10^(-14)`

D

`0.5xx10^(-12)`

Text Solution

Verified by Experts

The correct Answer is:
C
`K_"sp"=[Ag^+][I^-]`
`4xx10^(-18)=[Ag^+]xx2xx10^(-4)`
`[Ag^+] = 2 xx 10^(-14)` M
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