Given the solubility of strontium arsenate is 0.0480 g/l . we have to convert it into mol/L by dividing it over molar mass (540.7 g/mol) Molar solubility = 0.0480 / 540.7 = 8.9 x 10⁻⁵ mol/L Dissociation equation: Sr₃(AsO₄)₂(s) → 3 Sr²⁺(aq) + 2 AsO₄³⁻(aq) 3 s 2 s Ksp = [Sr²⁺]³ [AsO₄³⁻]² = (3s)³ (2s)² = 108 s⁵ Ksp = 108 (8.9 x 10⁻⁵) = 5.95 x 10⁻¹⁹
