MHT CET202519 Apr 2025Morning ShiftChemistryIonic EquilibriumActual
The solubility product of a sparingly soluble salt AX is 4.9 10⁻¹³ . What is its solubility in mol~ dm ⁻³ ?
Options
- A2.4 10⁻¹³
- B4.9 10⁻⁷
- C7.0 10⁻⁷
- D7.0 10⁻¹³
Correct answer
C. 7.0 10⁻⁷
Step-by-step solution
The solubility product K_ sp = [ A ^+][ X ^-] for the dissociation equilibrium AX(s) A^+(aq) + X^-(aq) determines the molar solubility. With [ A ^+] = [ X ^-] = s , the expression becomes K_ sp = s^2 . Given K_ sp = 4.9 10⁻¹³ , solving s^2 = 4.9 10⁻¹³ yields s = 4.9 10⁻¹³ . This simplifies to s = 7 10⁻⁷ , mol , dm⁻³ . The calculated solubility 7.0 10⁻⁷ , mol , dm⁻³ corresponds to option C .