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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 $\mathrm{Ag}^{+}$and $\mathrm{Pb}^{2+}$ at a concentration of $0.10 \mathrm{M}$. Aqueous $\mathrm{HCl}$ is added to this solution until the $\mathrm{Cl}^{-}$concentration is $0.10 \mathrm{M}$. What will be the concentration of $\mathrm{Ag}^{+}$and $\mathrm{Pb}^{2+}$ be at equilibrium? ( $K_{\mathrm{sp}}$ for $\mathrm{AgCl}=1.8 \times 10^{-10}, K_{\mathrm{sp}}$ for $\mathrm{PbCl}_2=1.7 \times 10^{-5}$ )
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The correct answer is:
$\left[\mathrm{Ag}^{+}\right]=1.8 \times 10^{-9} \mathrm{M} ;\left[\mathrm{Pb}^{2+}\right]=1.7 \times 10^{-3} \mathrm{M}$
$K_{\mathrm{sp}}$ for $\mathrm{AgCl}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]$
$$
\begin{aligned}
\therefore \quad\left[\mathrm{Ag}^{+}\right] & =\frac{1.8 \times 10^{-10}}{10^{-1}} \\
& =1.8 \times 10^{-9} \mathrm{M} . \\
K_{\mathrm{sp}} \text { for } \mathrm{PbCl}_2 & =\left[\mathrm{Pb}^{2+}\right]\left[\mathrm{Cl}^{-}\right]^2 \\
\therefore \quad\left[\mathrm{Pb}^{2+}\right] & =\frac{1.7 \times 10^{-5}}{10^{-1} \times 10^{-1}} \\
& =1.7 \times 10^{-3} \mathrm{M}
\end{aligned}
$$
$$
\begin{aligned}
\therefore \quad\left[\mathrm{Ag}^{+}\right] & =\frac{1.8 \times 10^{-10}}{10^{-1}} \\
& =1.8 \times 10^{-9} \mathrm{M} . \\
K_{\mathrm{sp}} \text { for } \mathrm{PbCl}_2 & =\left[\mathrm{Pb}^{2+}\right]\left[\mathrm{Cl}^{-}\right]^2 \\
\therefore \quad\left[\mathrm{Pb}^{2+}\right] & =\frac{1.7 \times 10^{-5}}{10^{-1} \times 10^{-1}} \\
& =1.7 \times 10^{-3} \mathrm{M}
\end{aligned}
$$
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