Manipal MET2011ChemistryIonic Equilibrium
Solubility product of BaCl ₂ is 4 10⁻⁹ . Its solubility in mol / L would be
Options
- A1 10⁻³
- B1 10⁻⁹
- C4 10⁻²⁷
- D1 10⁻²⁷
Correct answer
A. 1 10⁻³
Step-by-step solution
gathered BaCl ₂ 2+ Ba s + 2 2 ~s Cl ⁻ K_ sp =4 s^3, s^3= 4 10⁻⁹ 4 =10⁻³ s=10⁻³ M gathered