At a certain temperature, the solubility of strontium arsenate, sr3(aso4)2, is 0.0490 g/l. what is the ksp of this salt at this temperature?

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

Ithink 2 is the correct