Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
NEET 2011
Exam
NEET
Year
2011
Subject
Chemistry
Chapter
Ionic Equilibrium

NEET 2011 · Chemistry · Ionic Equilibrium

In qualitative analysis, the metals of group I can be separated...

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

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}
  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}
  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}
  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}

Correct answer

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}

Step-by-step solution

KspK_{\mathrm{sp}} for AgCl=[Ag+][Cl−]\mathrm{AgCl}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]∴[Ag+]=1.8×10−1010−1=1.8×10−9M.Ksp for PbCl2=[Pb2+][Cl−]2∴[Pb2+]=1.7×10−510−1×10−1=1.7×10−3M\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}
Practice Ionic Equilibrium on Quantrex Academy →

More from Ionic Equilibrium

What amount of NaOH should be dissolved in a 500 mL solution, so that the p^H of solution… MHT CET Medical 2026Which of the following buffer solutions is used for precipitation of cations of the III A… MHT CET Medical 2026The degree of dissociation of weak acid is directly proportional to MHT CET Medical 2026Which from the following is Lewis base? MHT CET Medical 2026The solubility of Fe(OH)_3 is x mole per liter. Find its K_ sp . MHT CET Medical 2026The Basic buffer solution is made with 0.1\ M\ NH_4OH and 0.2\ M\ NH_4Cl . If K_b for the… MHT CET Medical 2026The hydrogen ion concentration of the ocean is about 2 × 10^ -9 \ M . What is the pOH of… MHT CET Medical 2026Calculate the dissociation constant of a weak monobasic acid if it is 0.04% dissociated… MHT CET Medical 2026 Full Ionic Equilibrium list All NEET PYQs