1.

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

Answer»

`2.0xx10^(-5) m`
`1.0xx10^(-4)m`
`5.0xx10^(-9)m`
`2.2xx10^(-4)`m

Solution :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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