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The solubility product of AgCl " is " 4....

The solubility product of `AgCl " is " 4.0xx10^(-10)` at 298 K. The solubility of AgCl in `0.04 M CaCl_(2)` will be

A

`2.0xx10^(-5)M`

B

`1.0xx10^(-4)M`

C

`5.0xx10^(-9)M`

D

`2.2xx10^(-4)M`

Text Solution

Verified by Experts

The correct Answer is:
C
Solubility of AgCl
`[Ag^(+)]=(K_(sp))/([Cl^(-)])=(4xx10^(-10))/(0.08)=5.0xx10^(-9)M`
[0.08 M is concentration of `[Cl^(-)]` from `CaCl_(2)]`
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