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Calculate the molar solubility of barium fluoride (BaF2). For barium fluoride, Ksp=2.45×10−5.

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Answer:

BaF₂ when it dissolves, dissociates as follows;

BaF₂ --> Ba²⁺ + 2F⁻Molar solubility is the number of moles that can be dissolved in 1 L of solution.
If molar solubility of BaF₂ is x, then molar solubility of Ba²⁺ is x and solubility of

F⁻ is 2x.ksp = [Ba²⁺][F⁻]²ksp = (x)(2x)²2.45 x 10⁻⁵ = 4x³x³ = 0.6125 x 10⁻⁵x = 0.0183 mol/L is molar solubility of BaF₂ -blahblahmali

Step-by-step explanation:

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