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7 November, 23:13

What concentration of the lead ion, pb2+, must be exceeded to precipitate pbf2 from a solution that is 1.00*10-2 m in the fluoride ion, f-? ksp for lead (ii) fluoride is 3.3*10-8?

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  1. 8 November, 01:49
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    Answer is: concentration of Pb²⁺ must be exceeded is 3.3·10⁻⁴ M.

    Chemical reaction : Pb²⁺ (aq) + 2F⁻ (aq) → PbF₂ (s).

    Ksp (PbF ₂) = 3.3·10⁻⁸.

    c (F ⁻) = 0.01 M.

    Ksp (PbF ₂) = c (Pb²⁺) · c (F⁻) ².

    c (Pb² ⁺) = Ksp (PbF₂) : c (Cl⁻) ².

    c (Pb² ⁺) = 3.3·10⁻⁸ : (0.01 M) ².

    c (Pb² ⁺) = 0.000000033 M³ : 0.0001 M².

    c (Pb² ⁺) = 0.00033 M = 3.3·10⁻⁴ M.
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