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24 October, 01:02

328. mL of 0.00345 M NaI (aq) is combined with 703. mL of 0.00802 M Pb (NO3) 2 (aq). Determine if a precipitate will form given that the Ksp of Pbl2 is 1.40x10-8.

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  1. 24 October, 01:15
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    The solution will not form a precipitate.

    Explanation:

    The Ksp of PbI₂ is:

    PbI₂ (s) ⇄ 2I⁻ (aq) + Pb²⁺ (aq)

    Ksp = 1.40x10⁻⁸ = [I⁻]²[Pb²⁺] Concentrations in equilibrium

    When 328mL of 0.00345M NaI (aq) is combined with 703mL of 0.00802M Pb (NO₃) ₂. Molar concentration of I⁻ and Pb²⁺ are:

    [I⁻] = 0.00345M * (328mL / (328mL+703mL) = 1.098x10⁻³M

    [Pb²⁺] = 0.00802M * (703mL / (328mL+703mL) = 5.469x10⁻³M

    Q = [I⁻]²[Pb²⁺] Concentrations not necessary in equilibrium

    If Q = Ksp, the solution is saturated, Q > Ksp, the solution will form a precipitate, if Q < Ksp, the solution is not saturated.

    Replacing:

    Q = [1.098x10⁻³M]²[5.469x10⁻³M] = 6.59x10⁻⁹

    As Q < Ksp, the solution is not saturated and will not form a precipitate.
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