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In qualitative analysis, the metals of g...

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)`)

A

`[Ag^(+)] = 1.8 xx 10^(-9) M , [Pb^(2+)] = 1.7 xx 10^(-3)M`

B

`[Ag^(+)] = 1.8 xx 10^(-11) M , [Pb^(2+)] = 1.7 xx 10^(-4) M `

C

`[Ag^(+)] = 1.8 xx 10^(7) M , [Pb^(2+)] = 1.7 xx 10^(-6) M `

D

`[Ag^(+)] = 1.8 xx 10^(-11) M , [Pb^(2+)] = 8.5 xx 10^(-5) M `

Text Solution

Verified by Experts

The correct Answer is:
A
`K_(sp) = [Ag^(+)] [0.1] `
`[Ag^(+)] = 1.8 xx10^(-9) M `
`K_(sp) = [ Pb^(+2)][Cl^(-)]^(2)`
`1.7 xx 10^(-5) = [Pb^(+2)] [0.1]^(2)`
`[Pb^(+2)] = 1.7 xx 10^(-3)M `
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