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The solubility product of AgCl is `4.0xx10^(-10)` at 298 K. The solubility of AgCl in 0.04 m `CaCl_(2)` will beA. `2.0xx10^(-5) m`B. `1.0xx10^(-4)m`C. `5.0xx10^(-9)m`D. `2.2xx10^(-4)`m

Answer» Correct Answer - 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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