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Question #44474

Difficulty: 8/10 | 🟢 Free

The Question

Acetylene (\ceHC≡CH\ce{HC \equiv CH}) exhibits terminal acidity due to its sp-hybridized carbon atoms. It reacts with Tollens' reagent to form a highly insoluble silver acetylide (\ceAg2C2\ce{Ag2C2}) precipitate. To utilize \ceAg2C2\ce{Ag2C2} as a controlled source of the \ceC22−\ce{C2^{2-}} nucleophile, excess solid \ceAg2C2(s)\ce{Ag2C2(s)} is added to 500 mL500 \text{ mL} of pure water, and the system is allowed to reach equilibrium. Subsequently, 500 mL500 \text{ mL} of a 0.04 M0.04 \text{ M} \ceAgNO3(aq)\ce{AgNO3(aq)} solution is added to this suspension at 298 K298 \text{ K}. Given that the solubility product (KspK_{sp}) of \ceAg2C2\ce{Ag2C2} is 4.0×10−244.0 \times 10^{-24} at 298 K298 \text{ K}, calculate the new equilibrium concentration of the acetylide ion (\ceC22−\ce{C2^{2-}}) in the final mixture. (Assume the volume is strictly additive and no hydrolysis of \ceC22−\ce{C2^{2-}} occurs).

Options

  • A.

    1.0×10−20 M1.0 \times 10^{-20} \text{ M}

  • B.

    2.5×10−21 M2.5 \times 10^{-21} \text{ M}

  • C.

    2.0×10−22 M2.0 \times 10^{-22} \text{ M}

  • D.

    1.0×10−8 M1.0 \times 10^{-8} \text{ M}

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Exception_Exposure

Learn to identify edge cases and boundary exceptions that JEE examiners hide in math problems.

Concepts Behind This Question

Alkynes Terminal Acidity|Metal Activity Series Logic|Common Ion Effect on Solubility

Why This Question Is Tricky

Common Mistakes

The question tests the student's ability to integrate the concept of terminal alkyne acidity with the common ion effect in solubility equilibria.

The primary trap is the KspK_{sp} expression for silver acetylide (\ceAg2C2\ce{Ag2C2}), which requires squaring the silver ion concentration (Ksp=[\ceAg+]2[\ceC22−]K_{sp} = [\ce{Ag+}]^2[\ce{C2^{2-}}]).

Students frequently forget to square the common ion concentration or fail to account for the dilution of the \ceAgNO3\ce{AgNO3} solution upon mixing (halving the concentration from 0.04 M0.04 \text{ M} to 0.02 M0.02 \text{ M}).

Distractor B catches those who forget the dilution step, Distractor C catches those who forget the square term, and Distractor D catches those who calculate the pure water solubility, ignoring the common ion effect entirely.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterHydrocarbons - Class 11 Part 2
ConceptAlkynes Terminal Acidity
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree