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22 July, 03:17

What is the pH of a 1.4 M pyridine solution that has Kb = 1.7 * 10-9? The equation for the dissociation of pyridine is C5H5N (aq) + H2O (l) ⇌ C5H5NH + (aq) + OH - (aq). What is the pH of a 1.4 M pyridine solution that has Kb = 1.7 * 10-9? The equation for the dissociation of pyridine is C5H5N (aq) + H2O (l) ⇌ C5H5NH + (aq) + OH - (aq). 4.31 9.69 8.72 10.69

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  1. 22 July, 03:28
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    pH = 9.69

    Explanation:

    When pyridine (C₅H₅N) is added to water, the equilibrium that occurs is:

    C₅H₅N (aq) + H₂O (l) ⇌ C₅H₅NH⁺ (aq) + OH⁻ (aq) Kb = 1.7x10⁻⁹

    Where Kb is defined as:

    Kb = 1.7x10⁻⁹ = [C₅H₅NH⁺] [OH⁻] / [C₅H₅N]

    If you have a solution of 1.4M C₅H₅N, the equilibrium concentration of each specie is:

    [C₅H₅N] = 1.4 - X

    [C₅H₅NH⁺] = X

    [OH⁻] = X

    Where X represents the reaction coordinate

    Replacing in Kb expression:

    1.7x10⁻⁹ = [X] [X] / [1.4 - X]

    2.38x10⁻⁹ - 1.7x10⁻⁹X = X²

    0 = X² + 1.7x10⁻⁹X - 2.38x10⁻⁹

    Solving for X:

    X = - 0.0000488M → False answer, there is no negative concentrations

    X = 0.0000488M → Right answer

    Thus, [OH⁻] = 0.0000488M. As pOH = - log [OH⁻]

    pOH = 4.31

    Knowing pH = 14 - pOH

    pH = 9.69
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