JEE Main · 2024 · Shift-IIhardIEQ-050

Solubility of calcium phosphate (molecular mass M) in water is W g per 100 mL at 25 °C. Its solubility product at 25 °C…

Ionic Equilibrium · Class 11 · JEE Main Previous Year Question

Question

Solubility of calcium phosphate (molecular mass M) in water is W g per 100 mL at 25 °C. Its solubility product at 25 °C will be approximately:

Options
  1. a

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

  2. b

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

    ✓
  3. c

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

  4. d

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

Correct Answerb

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

Detailed Solution

Step 1 — Calcium phosphate = Ca3(PO4)2\mathrm{Ca_3(PO_4)_2}: Ca3(PO4)2⇌3Ca2++2PO43−Ksp=[Ca2+]3[PO43−]2=(3s)3(2s)2=27s3⋅4s2=108s5\mathrm{Ca_3(PO_4)_2 \rightleftharpoons 3Ca^{2+} + 2PO_4^{3-}} K_{sp} = [\mathrm{Ca^{2+}}]^3[\mathrm{PO_4^{3-}}]^2 = (3s)^3(2s)^2 = 27s^3 \cdot 4s^2 = 108s^5

Step 2 — Convert solubility: W g per 100 mL = W g per 0.1 L = 10W g per L s=10WM mol L−1s = \frac{10W}{M}\ \mathrm{mol\ L^{-1}}

Step 3 — Calculate KspK_{sp}: Ksp=108×(10WM)5=108×105W5M5≈107×(WM)5K_{sp} = 108 \times \left(\frac{10W}{M}\right)^5 = 108 \times \frac{10^5 W^5}{M^5} \approx 10^7 \times \left(\frac{W}{M}\right)^5

(since 108×105≈107108 \times 10^5 \approx 10^7)

Answer: Option (2) — 107(WM)510^7\left(\frac{W}{M}\right)^5

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