Sr₃(AsO₄)₂ partially dissociates into its ions as Sr²⁺ and AsO₄³⁻. The balanced equation is
Sr₃(AsO₄)₂(s) ⇄ 3Sr²⁺(aq) + 2AsO₄³⁻(aq)
change -X +3X +2X
Equilibrium 3X 2X
Since the solubility of Sr₃(AsO₄)₂ is 0.0490 g/L,
X = 0.0490 g/L
molar mass of Sr₃(AsO₄)₂ =540.7 g/mol
Hence, X = (0.0490 g/L) / 540.7 g/mol
= 9.06 x 10⁻⁵ mol/L
Ksp = [Sr²⁺(aq)]³ [AsO₄³⁻(aq)]²
Ksp = (3X)³ (2X)²
Ksp = 108X⁵
Ksp = 108 x (9.06 x 10⁻⁵ mol/L)⁵
= 6.59 x 10⁻¹⁹ mol⁵/L⁻⁵