MHT CET202521 Apr 2025Evening ShiftChemistryIonic EquilibriumActual
The solubility of calcium carbonate at 298 K is 6.4 10⁻⁵ ~mol~ dm ⁻³ . Calculate the value of solubility product at the same temperature?
Options
- A5.06 10⁻¹⁰
- B4.096 10⁻⁹
- C3.05 10⁻¹⁰
- D2.8 10⁻⁹
Correct answer
B. 4.096 10⁻⁹
Step-by-step solution
The dissolution equilibrium for calcium carbonate is CaCO3(s) Ca²⁺(aq) + CO3²⁻(aq) , with the solubility product given by K_ sp = [ Ca²⁺ ][ CO3²⁻ ] . For a saturated solution with molar solubility s , the ion concentrations are [ Ca²⁺ ] = s and [ CO3²⁻ ] = s . Substituting into the K_ sp expression yields K_ sp = s^2 . Given s = 6.4 10⁻⁵ mol/dm^3 , we compute K_ sp = (6.4 10⁻⁵)^2 = 4.096 10⁻⁹ . This calculated value corresponds to option B from the provided choices.