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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 $Pb^{2+}$ at a concentration of 0.10 M. Aqueous HCl is added to this solution until the $Cl^-$ concentration is 0.10 M. What will the concentrations of $Ag^+$ and $Pb^{2+}$ be at equilibrium? ($K_{sp}$ for AgCl = $1.8\times10^{-10}$, $K_{sp}$ for $PbCl_2$ = $1.7\times10^{-5}$)
A
$[Ag^+] = 1.8\times10^{-9}$ M; $[Pb^{2+}] = 1.7\times10^{-3}$ M
B
$[Ag^+] = 1.8\times10^{-11}$ M; $[Pb^{2+}] = 1.7\times10^{-4}$ M
C
$[Ag^+] = 1.8\times10^{-7}$ M; $[Pb^{2+}] = 1.7\times10^{-6}$ M
D
$[Ag^+] = 1.8\times10^{-11}$ M; $[Pb^{2+}] = 8.5\times10^{-5}$ M
Detailed Solution
At equilibrium the solution is saturated with both chlorides and $[Cl^-] = 0.10$ M.
For AgCl: $K_{sp} = [Ag^+][Cl^-]$
$[Ag^+] = \frac{K_{sp}}{[Cl^-]} = \frac{1.8\times10^{-10}}{10^{-1}} = 1.8\times10^{-9}$ M
For $PbCl_2$: $K_{sp} = [Pb^{2+}][Cl^-]^2$
$[Pb^{2+}] = \frac{K_{sp}}{[Cl^-]^2} = \frac{1.7\times10^{-5}}{(10^{-1})^2} = \frac{1.7\times10^{-5}}{10^{-2}}$
$[Pb^{2+}] = 1.7\times10^{-3}$ M
So $[Ag^+] = 1.8\times10^{-9}$ M and $[Pb^{2+}] = 1.7\times10^{-3}$ M.
For AgCl: $K_{sp} = [Ag^+][Cl^-]$
$[Ag^+] = \frac{K_{sp}}{[Cl^-]} = \frac{1.8\times10^{-10}}{10^{-1}} = 1.8\times10^{-9}$ M
For $PbCl_2$: $K_{sp} = [Pb^{2+}][Cl^-]^2$
$[Pb^{2+}] = \frac{K_{sp}}{[Cl^-]^2} = \frac{1.7\times10^{-5}}{(10^{-1})^2} = \frac{1.7\times10^{-5}}{10^{-2}}$
$[Pb^{2+}] = 1.7\times10^{-3}$ M
So $[Ag^+] = 1.8\times10^{-9}$ M and $[Pb^{2+}] = 1.7\times10^{-3}$ M.