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DIA [1.3K]
3 years ago
15

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]?

Chemistry
1 answer:
diamong [38]3 years ago
6 0
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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