In qualitative analysis, the metals of group I can be separated from other ions by precipitating them as…

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 $\mathrm{PbCl}_2=1.7 \times 10^{-5}$ )
  1. $\left[\mathrm{Ag}^{+}\right]=1.8 \times 10^{-7} \mathrm{M} ;\left[\mathrm{Pb}^{2+}\right]=1.7 \times 10^{-6} \mathrm{M}$
  2. $\left[\mathrm{Ag}^{+}\right]=1.8 \times 10^{-11} \mathrm{M} ;\left[\mathrm{Pb}^{2+}\right]=8.5 \times 10^{-5} \mathrm{M}$
  3. $\left[\mathrm{Ag}^{+}\right]=1.8 \times 10^{-9} \mathrm{M} ;\left[\mathrm{Pb}^{2+}\right]=1.7 \times 10^{-3} \mathrm{M}$
  4. $\left[\mathrm{Ag}^{+}\right]=1.8 \times 10^{-11} \mathrm{M} ;\left[\mathrm{Pb}^{2+}\right]=8.5 \times 10^{-4} \mathrm{M}$

Solution

$K_{\mathrm{sp}}$ for $\mathrm{AgCl}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]$ $\begin{aligned} \therefore \quad\left[\mathrm{Ag}^{+}\right] & =\frac{1.8 \times 10^{-10}}{10^{-1}} \\ & =1.8 \times 10^{-9} \mathrm{M} . \\ K_{\mathrm{sp}} \text { for } \mathrm{PbCl}_2 & =\left[\mathrm{Pb}^{2+}\right]\left[\mathrm{Cl}^{-}\right]^2 \\ \therefore \quad\left[\mathrm{Pb}^{2+}\right] & =\frac{1.7 \times 10^{-5}}{10^{-1} \times 10^{-1}} \\ & =1.7 \times 10^{-3} \mathrm{M} \end{aligned}$

Asked in: NEET 2011 (Mains)

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