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2 February, 18:22

Determine the molar solubility of baf2 in pure water. ksp for baf2 = 2.45 x 10-5. 1.83 x 10-2 m 1.23 x 10-5 m 2.90 x 10-2 m 4.95 x 10-3 m 6.13 x 10-6 m

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  1. 2 February, 18:23
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    The Molar solubility of baf2 in pure water is 1.83 x 10 (₋₂) if ksp for baf2 = 2.45 x 10⁻⁵.

    solubility product equilibrium reaction from the balance equation of reaction is:

    k (sp₎ = [Ba⁺²] [F⁻]²

    using mole ratios from one to another, [Ba⁺²] = x and [F⁻]² = 2x

    k (sp₎ = [Ba⁺²] [F⁻]²

    k (sp₎ = [x][2x]²

    ksp = 2.45 x 10⁻⁵ then,

    2.45 x 10⁻⁵ = [x][2x]²

    4x³ = 2.45 x 10⁻⁵

    x = ∛ (2.45 x 10⁻⁵) / 4 = 1.83 x 10⁻²m

    so, x is molar solubility which is 1.83 x 10⁻²m
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