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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…

NEET UG 2011 — Chemistry Physical Chemistry

2011mcqhard

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+\mathrm{Ag}^{+}Ag+and Pb2+\mathrm{Pb}^{2+}Pb2+ at a concentration of 0.10M0.10 \mathrm{M}0.10M. Aqueous HCl\mathrm{HCl}HCl is added to this solution until the Cl−\mathrm{Cl}^{-}Cl−concentration is 0.10M0.10 \mathrm{M}0.10M. What will be the concentration of Ag+\mathrm{Ag}^{+}Ag+and Pb2+\mathrm{Pb}^{2+}Pb2+ be at equilibrium? ( KspK_{\mathrm{sp}}Ksp​ for AgCl=1.8×10−10,Ksp\mathrm{AgCl}=1.8 \times 10^{-10}, K_{\mathrm{sp}}AgCl=1.8×10−10,Ksp​ for PbCl2=1.7×10−5\mathrm{PbCl}_2=1.7 \times 10^{-5}PbCl2​=1.7×10−5 )

Official previous-year question

Held on 30 Apr 2011 · Verified 9 Jul 2026.

Options

  1. A

    [Ag+]=1.8×10−7M;[Pb2+]=1.7×10−6M\left[\mathrm{Ag}^{+}\right]=1.8 \times 10^{-7} \mathrm{M} ;\left[\mathrm{Pb}^{2+}\right]=1.7 \times 10^{-6} \mathrm{M}[Ag+]=1.8×10−7M;[Pb2+]=1.7×10−6M

  2. B

    [Ag+]=1.8×10−11M;[Pb2+]=8.5×10−5M\left[\mathrm{Ag}^{+}\right]=1.8 \times 10^{-11} \mathrm{M} ;\left[\mathrm{Pb}^{2+}\right]=8.5 \times 10^{-5} \mathrm{M}[Ag+]=1.8×10−11M;[Pb2+]=8.5×10−5M

  3. C

    [Ag+]=1.8×10−9M;[Pb2+]=1.7×10−3M\left[\mathrm{Ag}^{+}\right]=1.8 \times 10^{-9} \mathrm{M} ;\left[\mathrm{Pb}^{2+}\right]=1.7 \times 10^{-3} \mathrm{M}[Ag+]=1.8×10−9M;[Pb2+]=1.7×10−3M

  4. D

    [Ag+]=1.8×10−11M;[Pb2+]=8.5×10−4M\left[\mathrm{Ag}^{+}\right]=1.8 \times 10^{-11} \mathrm{M} ;\left[\mathrm{Pb}^{2+}\right]=8.5 \times 10^{-4} \mathrm{M}[Ag+]=1.8×10−11M;[Pb2+]=8.5×10−4M

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