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JEE Main 2016 · Chemistry · Ionic Equilibrium · hard

Solid AgNO3 is added slowly (no volume change) to a solution that is 0.010 M in Cl^- and 0.010 M in CrO4^2-. Given Ksp(AgCl)=1.8×10^-10 and…

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Solid AgNO3\mathrm{AgNO_3} is added slowly (no volume change) to a solution that is 0.010 M0.010\ \mathrm{M} in Cl−\mathrm{Cl^-} and 0.010 M0.010\ \mathrm{M} in CrO42−\mathrm{CrO_4^{2-}}. Given Ksp(AgCl)=1.8×10−10K_{sp}(\mathrm{AgCl})=1.8\times10^{-10} and Ksp(Ag2CrO4)=1.0×10−12K_{sp}(\mathrm{Ag_2CrO_4})=1.0\times10^{-12}, the concentration of Cl−\mathrm{Cl^-} remaining in solution when Ag2CrO4\mathrm{Ag_2CrO_4} just begins to precipitate is
  1. A. 1.0×10−5 M1.0\times10^{-5}\ \mathrm{M}
  2. B. 1.8×10−8 M1.8\times10^{-8}\ \mathrm{M}
  3. C. 1.8×10−3 M1.8\times10^{-3}\ \mathrm{M}
  4. D. 1.8×10−5 M1.8\times10^{-5}\ \mathrm{M}
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Answer: (D) 1.8×10−5 M1.8\times10^{-5}\ \mathrm{M}

AgCl needs [Ag+]=1.8×10−10/0.010=1.8×10−8[\mathrm{Ag^+}]=1.8\times10^{-10}/0.010=1.8\times10^{-8} M; Ag2CrO4\mathrm{Ag_2CrO_4} needs [Ag+]=10−12/0.010=1.0×10−5[\mathrm{Ag^+}]=\sqrt{10^{-12}/0.010}=1.0\times10^{-5} M. The much smaller requirement means AgCl precipitates first. Chromate begins to precipitate only when [Ag+][\mathrm{Ag^+}] reaches 1.0×10−51.0\times10^{-5} M, at which point [Cl−]=Ksp(AgCl)/[Ag+]=1.8×10−10/10−5=1.8×10−5[\mathrm{Cl^-}]=K_{sp}(\mathrm{AgCl})/[\mathrm{Ag^+}]=1.8\times10^{-10}/10^{-5}=1.8\times10^{-5} M, i.e. only 0.18%0.18\% of the original chloride is left. The value 1.0×10−51.0\times10^{-5} is the silver ion concentration, and 1.8×10−81.8\times10^{-8} is the initial [Ag+][\mathrm{Ag^+}] for AgCl.
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Original question written in the JEE Main 2016 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake