MHT CET202520 Apr 2025Morning ShiftChemistryIonic EquilibriumActual
Solubility of Ca ₃ ( PO ₄ )₂ is 'S' mol dm ⁻³ . Find solubility product.
Options
- AS^5
- B108 ~S ^5
- C54 S^5
- D12 S^5
Correct answer
B. 108 ~S ^5
Step-by-step solution
The dissociation equilibrium for Ca ₃ ( PO ₄ )₂ is: Ca ₃ ( PO ₄ )₂ (s) 3 Ca ²⁺ (aq) + 2 PO ₄³⁻ (aq) Let S represent the molar solubility. The equilibrium concentrations become [ Ca ²⁺] = 3S and [ PO ₄³⁻] = 2S . The solubility product expression is K_ sp = [ Ca ²⁺]^3 [ PO ₄³⁻]^2 . Substituting the concentrations yields: K_ sp = (3S)^3 (2S)^2 = 27S^3 4S^2 = 108S^5 The solubility product is B .