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The solubility product of AgCl is 4.0 xx...

The solubility product of AgCl is `4.0 xx 10^(-10)` at 298 K . The solubility of AgCl in 0.04 M Ca `Cl_(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
If x is the solubility of AgCl in 0.04 m `CaCl_(2)`, then
`[Ag^(+)]= x "mol" L^(-1), [Cl^(-1)]=(0.04xx2)+x ~~0.8 m`
`:. x (0.08 ) 4xx10^(-10) "or" x = 5.0 xx 10^(-9) m`
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