In qualitative analysis, the metals of group I can be separated from other ions by precipitating them as chloride salts. A solution initially contains `Ag^(+) " and " Pb^(+)` at a concentration of 0.10M. Aqueous HCl is added to this solution until be `Cl^(-)` concentration is 0.10M. What will be concentration of `Ag^(+) " and " Pb^(2+)` be at equilibrium ?
(`K_(sp) " for AgCl " = 1.8xx10^(-10)`
`K_(sp) " for " PbCl_(2) = 1.7xx10^(-5)`)

Options

A. `[Ag^(+)]=1.8xx10^(-7) M , [Pb^(2+)]=1.7xx10^(-6)M`
B. `[Ag^(+)]=1.8xx10^(-11) M , [Pb^(2+)]=8.5xx10^(-5)M`
C. `[Ag^(+)]=1.8xx10^(-9) M , [Pb^(2+)]=1.7xx10^(-3)M`
D. `[Ag^(+)]=1.8xx10^(-11) M , [Pb^(2+)]=8.5xx10^(-4)M`

Answer

C

Explanation

`K_(sp) ` for `AgCl = [Ag^(+)][Cl^(-)]`
`:. 1.8xx10^(-10)=[Ag^(+)][0.1]`
or`[Ag^(+)]=1.8xx10^(-9)M`
`K_(sp)` for `PbCl_(2) = [Pb^(2+)][Cl^(-)]^(2)`
`:.1.7xx10^(-5)=[Pb^(2+)][0.1]^(2)`
or `[Pb^(2+)]=1.7xx10^(-3)M`.