Calculate the solubility in mol dm ⁻³ of sparingly soluble salt BA at 293 K if its solubility product is 8.56 10⁻⁵ at same temperature.

Step-by-step solution

The dissociation of the sparingly soluble salt BA is given by: BA B^+ + A^- Let the solubility of the salt be s mol dm ⁻³ . The solubility product K_ sp is given by: K_ sp = [B^+][A^-] = s s = s^2 Given K_ sp = 8.56 10⁻⁵ , we have: s^2 = 8.56 10⁻⁵ = 85.6 10⁻⁶ s = 85.6 10⁻⁶ = 9.252 10⁻³ mol dm ⁻³ Answer: 9.252 10⁻³