The solubility product of aluminium hydroxide at 298 K is 4.2 10⁻¹⁴ . Calculate its solubility in mol/L at same temperature.

Step-by-step solution

The dissociation of aluminium hydroxide is given by: Al(OH)₃ Al³⁺ + 3OH^- Let the solubility of Al(OH)₃ be s mol/L. [Al³⁺] = s [OH^-] = 3s The solubility product K_ sp is given by: K_ sp = [Al³⁺][OH^-]^3 K_ sp = (s)(3s)^3 = 27s^4 Given K_ sp = 4.2 10⁻¹⁴ : 27s^4 = 4.2 10⁻¹⁴ s^4 = 4.2 10⁻¹⁴ 27 = 15.55 10⁻¹⁶ s = (15.55)^ 1/4 10⁻⁴ 1.987 10⁻⁴ mol/L Answer: 1.987 10⁻⁴