What is the solubility in pure water of ba (io3) 2 in moles per liter at 25 ˚c? [ksp (25 ˚c) = 6.0  10-10]?

Ba (IO₃) ₂ (s) partially dissociates in water as Ba²⁺ (aq) and IO₃⁻ (aq).

Ba (IO₃) ₂ (s) ⇄ Ba²⁺ (aq) + 2IO₃⁻ (aq)

Initial

Change - X + X 2X

Equilibrium X 2X

Ksp = [Ba²⁺ (aq) ] x [IO₃⁻ (aq) ]²

6.0 x 10⁻¹⁰ = X * (2X) ²

6.0 x 10⁻¹⁰ = 4X³

X = 5.313 x 10⁻⁴ mol/L

Hence, the solubility of the Ba (IO₃) ₂ (s) is 5.313 x 10⁻⁴ mol/L