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Question #44281

Difficulty: 8/10 | 🟢 Free

The Question

Consider the sparingly soluble chlorides of the isoelectronic cations \ce{Tl^+} and \ce{Pb^{2+}}. At 298 K, the solubility products are Ksp(\ceTlCl)=1.6×10−4K_{sp}(\ce{TlCl}) = 1.6 \times 10^{-4} and Ksp(\cePbCl2)=3.2×10−5K_{sp}(\ce{PbCl_2}) = 3.2 \times 10^{-5}. Let S\ceTlS_{\ce{Tl}} and S\cePbS_{\ce{Pb}} be their respective molar solubilities in pure water. Let S\ceTl′S'_{\ce{Tl}} and S\cePb′S'_{\ce{Pb}} be their molar solubilities in an aqueous solution of 0.40 M0.40\text{ M} \ce{NaCl}. Which of the following options correctly identifies the solubility order in pure water, the solubility order in 0.40 M0.40\text{ M} \ce{NaCl}, and the exact concentration of \ce{Cl^-} (in M) at which their molar solubilities are identical?

Options

  • A.

    S\ceTl<S\cePbS_{\ce{Tl}} < S_{\ce{Pb}} ; S\ceTl′>S\cePb′S'_{\ce{Tl}} > S'_{\ce{Pb}} ; [\ceCl−]=0.20 M[\ce{Cl^-}] = 0.20\text{ M}

  • B.

    S\ceTl>S\cePbS_{\ce{Tl}} > S_{\ce{Pb}} ; S\ceTl′>S\cePb′S'_{\ce{Tl}} > S'_{\ce{Pb}} ; [\ceCl−]=0.20 M[\ce{Cl^-}] = 0.20\text{ M}

  • C.

    S\ceTl>S\cePbS_{\ce{Tl}} > S_{\ce{Pb}} ; S\ceTl′<S\cePb′S'_{\ce{Tl}} < S'_{\ce{Pb}} ; [\ceCl−]=0.50 M[\ce{Cl^-}] = 0.50\text{ M}

  • D.

    S\ceTl<S\cePbS_{\ce{Tl}} < S_{\ce{Pb}} ; S\ceTl′<S\cePb′S'_{\ce{Tl}} < S'_{\ce{Pb}} ; [\ceCl−]=0.10 M[\ce{Cl^-}] = 0.10\text{ M}

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Numerical_Precision

Improve calculation stamina and avoid rounding errors in high-stakes competitive examinations.

Concepts Behind This Question

Isoelectronic Energy Ordering|Solubility Product (Ksp) vs. Ionic Product (Qsp)|Common Ion Effect on Solubility

Why This Question Is Tricky

Common Mistakes

This problem tests the 'Common Ion Effect Reversal' trap.

Students often rely on two flawed heuristics: (1) Comparing KspK_{sp} values directly to determine solubility (ignoring stoichiometry, leading to the false conclusion that S\ceTl>S\cePbS_{\ce{Tl}} > S_{\ce{Pb}} in water), and (2) Assuming the relative order of solubility remains constant regardless of the solvent environment.

Because \ce{PbCl_2} has a squared dependence on [\ceCl−][\ce{Cl^-}] in its solubility product expression, its solubility is suppressed much more aggressively by a common ion than \ce{TlCl} (which has a linear dependence).

This mathematical asymmetry forces a 'solubility crossover' at [\ceCl−]=0.20 M[\ce{Cl^-}] = 0.20\text{ M}, reversing the textbook trend dictated by their intrinsic lattice and hydration energies.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterStructure of Atom - Class 11 Part 1
ConceptIsoelectronic Energy Ordering
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree