JEE Main20241 Feb 2024Evening ShiftChemistryIonic EquilibriumActual
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
- A10 7 W M 3
- B10 7 W M 5
- C10 3 W M 5
- D10 5 W M 5
Correct answer
B. 10 7 W M 5
Step-by-step solution
Given the solubility of Ca 3 PO 4 2 is Wg / 100 mL . The solubility in mol / L is S = W M × 1000 100 mol / L . The equilibrium reaction of calcium phosphate is Ca 3 PO 4 2 ⇌ 3 Ca 2 + + 2 PO 4 3 - At equilibrium 3 S 2 S Now, K sp = Ca 2 + 3 PO 4 3 - 2 ⇒ K sp = 3 S 3 2 S 2 = 108 × S 5 ⇒ K sp = 108 × W M × 10 5 ⇒ K sp = 108 × 10 5 × W M 5 = 1 . 08 × 10 7 W M 5