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Chemistry · Ionic Equilibrium

JEE Main 2024 — 1 February, Shift 2 — Question 75

Solubility of calcium phosphate (molecular mass, M) in water is Wg\mathrm{W}_{\mathrm{g}} per 100 mL at 25∘C25^{\circ} \mathrm{C}. Its solubility product at 25∘C25^{\circ} \mathrm{C} will be approximately.

  1. Option A:

    107( WM)310^{7}\left(\frac{\mathrm{~W}}{\mathrm{M}}\right)^{3}

  2. Option B:

    107( WM)510^{7}\left(\frac{\mathrm{~W}}{\mathrm{M}}\right)^{5}

    Correct
  3. Option C:

    103( WM)510^{3}\left(\frac{\mathrm{~W}}{\mathrm{M}}\right)^{5}

  4. Option D:

    105( WM)510^{5}\left(\frac{\mathrm{~W}}{\mathrm{M}}\right)^{5}

Answer: B

Step-by-step solution

S=W×10MS=\frac{W \times 10}{M}

Ca3(PO4)2( s)⇌3Ca2+\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(\mathrm{~s}) \rightleftharpoons 3 \mathrm{Ca}^{2+} (aq.) +2PO43−+2 \mathrm{PO}_{4}^{3-} (aq.) S=W×1000M×100=W×10MS=\frac{W \times 1000}{M \times 100}=\frac{W \times 10}{M}

Ksp=(3 s)3(2 s)2\mathrm{K}_{\mathrm{sp}}=(3 \mathrm{~s})^{3}(2 \mathrm{~s})^{2} =108 s5=108 \mathrm{~s}^{5}

=108×105×(WM)5=108 \times 10^{5} \times\left(\frac{W}{M}\right)^{5} =1.08×107(WM)5=1.08 \times 10^{7}\left(\frac{W}{M}\right)^{5}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
1 February, Shift 2
Subject
Chemistry
Chapter
Ionic Equilibrium
Topic
Sparingly Soluble Salts, Solubility Product & Precipitation Conditions