InterviewSolution
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InterviewSolution
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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^(+)]=1.8xx10^(-11)M,` `[Pb^(2+)]=1.7xx10^(-4)M`B. `[Ag^(+)]=1.8xx10^(-7)M,` `[Pb^(2+)]=1.7xx10^(-6)M`C. `[Ag^(+)]=1.8xx10^(-11)M,` `[Pb^(2+)]=8.5xx10^(-5)M`D. `[Ag^(+)]=1.8xx10^(-9)M,` `[Pb^(2+)]=1.7xx10^(-3)M` |
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Answer» Correct Answer - D `[Ag^+][Cl^-]=1.8xx10^-10` `[Ag^+]=(1.8xx10^-10)/(0.1)=1.8xx10^-9M` `[Pb^(+2)][Cl^-]^2=1.7xx10^-5` `[Pb^(+2)]=(1.7xx10^-5)/(0.1xx0.1)=1.7xx10^-3M` |
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