JEE Main 2024ChemistryIonic EquilibriumHardMCQ

JEE Main 2024 — Ionic Equilibrium Question with Solution

JEE Main 2024 (01 Feb Shift 2)

Question

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

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Show full solutionCorrect option: B
Correct answer
B107WM5

Step-by-step explanation

Given the solubility of Ca3PO42 is Wg/100 mL. The solubility in mol/L is S=WM×1000100mol/L.

The equilibrium reaction of calcium phosphate is 

Ca3PO42⇌3Ca2++2PO43-At equilibrium3S2S

Now, Ksp=Ca2+3PO43-2

⇒Ksp=3S32S2=108×S5

⇒Ksp=108×WM×105⇒Ksp=108×105×WM5

=1.08×107WM5

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About this question

This is a previous-year question from JEE Main 2024, covering the Ionic Equilibrium chapter of Chemistry. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.