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

a
$10^7\left(\frac{W}{M}\right)^3$
b
$10^7\left(\frac{W}{M}\right)^5$
✓
c
$10^3\left(\frac{W}{M}\right)^5$
d
$10^5\left(\frac{W}{M}\right)^5$

Correct Answerb

$10^7\left(\frac{W}{M}\right)^5$

Detailed Solution

Step 1 — Calcium phosphate = $\mathrm{Ca_3(PO_4)_2}$ : $\mathrm{Ca_3(PO_4)_2 \rightleftharpoons 3Ca^{2+} + 2PO_4^{3-}} K_{sp} = [\mathrm{Ca^{2+}}]^3[\mathrm{PO_4^{3-}}]^2 = (3s)^3(2s)^2 = 27s^3 \cdot 4s^2 = 108s^5$

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

Step 3 — Calculate $K_{sp}$ : $K_{sp} = 108 \times \left(\frac{10W}{M}\right)^5 = 108 \times \frac{10^5 W^5}{M^5} \approx 10^7 \times \left(\frac{W}{M}\right)^5$

(since $108 \times 10^5 \approx 10^7$ )

Answer: Option (2) — $10^7\left(\frac{W}{M}\right)^5$

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