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Chemistry · Ionic Equilibrium

JEE Main 2025 — 2 April, Morning Shift — Question 11

If equal volumes of AB2A B_{2} and XY (both are salts) aqueous solutions are mixed, which of the following combination will give a precipitate of AY2\mathrm{AY}_{2} at 300 K ? (Given Ksp (\mathrm{K}_{\text {sp }}\left(\right. at 300 K ) for AY2−=5.2×10−7\mathrm{AY}_{2-}=5.2 \times 10^{-7} )

  1. Option A:

    2.0×10−2MAB,2.0×10−2MXY2.0 \times 10^{-2} \mathrm{M} \mathrm{AB}, 2.0 \times 10^{-2} \mathrm{M} \mathrm{XY}

    Correct
  2. Option B:

    2.0×10−4MAB2,0.8×10−3MXY2.0 \times 10^{-4} \mathrm{M} \mathrm{AB} 2,0.8 \times 10^{-3} \mathrm{M} \mathrm{XY}

  3. Option C:

    3.6×10−3MAB2,5.0×10−4MXY3.6 \times 10^{-3} \mathrm{M} \mathrm{AB} 2,5.0 \times 10^{-4} \mathrm{M} \mathrm{XY}

  4. Option D:

    1.5×10−4MAB,1.5×10−3MXY1.5 \times 10^{-4} \mathrm{M} \mathrm{AB}, 1.5 \times 10^{-3} \mathrm{M} \mathrm{XY}

Answer: A

Step-by-step solution

For precipitation

[A2+][Y−]2≥5.2×10−7\left[A^{2+}\right]\left[Y^{-}\right]^{2} \geq 5.2 \times 10^{-7}

In option (1)

[A2+]=10−2M\left[\mathrm{A}^{2+}\right]=10^{-2} \mathrm{M}

[Y−]=10−2M\left[\mathrm{Y}^{-}\right]=10^{-2} \mathrm{M}

[A2+][Y−]2=10−6\left[A^{2+}\right][Y-]^{2}=10^{-6} which is greater than 5.2×10−75.2 \times 10^{-7}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
2 April, Morning Shift
Subject
Chemistry
Chapter
Ionic Equilibrium
Topic
Sparingly Soluble Salts, Solubility Product & Precipitation Conditions

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