MHT CET202527 Apr 2025Evening ShiftChemistryIonic EquilibriumActual
Calculate the ionic concentration of sparingly soluble salt BA in mol dm ⁻³ at 300 K when equilibrium is attained if solubility product of salt is 2.7 10⁻¹⁰ at same temperature.
Options
- A1.123 10⁻⁵
- B2.051 10⁻⁵
- C2.263 10⁻⁵
- D1.643 10⁻⁵
Correct answer
A. 1.123 10⁻⁵
Step-by-step solution
The dissociation equilibrium for the sparingly soluble salt BA is: BA(s) B ^+ (aq) + A ^- (aq) Let s represent the molar solubility in mol dm ⁻³ . At equilibrium, concentrations are [ B ^+] = s and [ A ^-] = s . The solubility product is K_ sp = [ B ^+][ A ^-] = s^2 . Given K_ sp = 2.7 10⁻¹⁰ , solving for s gives: s^2 = 2.7 10⁻¹⁰ s = 2.7 10⁻¹⁰ 1.64 10⁻⁵ mol dm ⁻³ This value corresponds to option D .