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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 \times 10^{-10},$$ K_{sp}$ for $PbCl_2 = 1.7 \times 10^{-5}$)
$[Ag^+] = 1.8 \times 10^{-7}\, M,$$ [Pb^{2+}] = 1.7 \times 10^{-6}\, M$
$[Ag^+] = 1.8 \times 10^{-11} \,M, $$[Pb^{2+}] = 8.5 \times 10^{-5} \,M$
$[Ag^+] = 1.8 \times 10^{-9}\, M, $$[Pb^{2+}] = 1.7 \times 10^{-3} \,M$
$[Ag^+] = 1.8 \times 10^{-11} \,M, $$[Pb^{2+}] = 1.7 \times 10^{-4}\, M$
Solution
$K_{s p}[\mathrm{AgCl}]=\left[\mathrm{Ag}^{+}\right][\mathrm{Cl}]$
$\left[\mathrm{Ag}^{+}\right]=\frac{1.8 \times 10^{-10}}{10^{-1}}=1.8 \times 10^{-9}\, \mathrm{M}$
$K_{s p}\left[\mathrm{PbCl}_{2}\right]=\left[\mathrm{Pb}^{2+}\right][\mathrm{Cl}]^{2}$
$\left[\mathrm{Pb}^{2+}\right]=\frac{1.7 \times 10^{-5}}{10^{-1} \times 10^{-1}}=1.7 \times 10^{-3} \,\mathrm{M}$
