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18 December, 14:38

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]?

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  1. 18 December, 16:42
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    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
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