The solubility product of the sparingly soluble salt A B₂ is 2.56 10⁻¹⁰ at 298 K. Calculate its solubility in mol~ dm ⁻³ at the same temperature ?

Step-by-step solution

The salt AB₂ dissociates as AB₂(s) A²⁺(aq) + 2B⁻(aq) , with K_ sp = [A²⁺][B⁻]^2 . Letting s represent the solubility in mol~dm⁻³ , the ion concentrations are [A²⁺] = s and [B⁻] = 2s . Substituting into the K_ sp expression gives K_ sp = (s)(2s)^2 = 4s^3 . Given K_ sp = 2.56 10⁻¹⁰ , solving 4s^3 = 2.56 10⁻¹⁰ yields s^3 = 6.4 10⁻¹¹ , or s^3 = 64 10⁻¹² . Taking the cube root, s = [3] 64 10⁻¹² = 4 10⁻⁴ ~mol~dm⁻³ , which corresponds to option C .