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28 December, 10:25

Determine the pH of a 0.28 M solution of pyridinium nitrate (C5H5NHNO3) at 25°C. [Pyridinium nitrate dissociates in water to give pyridinium ions (C5H5NH+), the conjugate acid of pyridine (Kb = 1.7 * 10-9), and nitrate ions (NO3-).]

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  1. 28 December, 12:54
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    The pH is 2.89

    Explanation:

    Step 1: Dara given

    Molarity of C5H5NHNO3 = 0.28 M

    Kb = 1.7*10^-9

    Step 2: Calculate Ka

    Ka = Kw / Kb

    Ka = (10^-14) / (1.7*10^-9)

    Ka = 5.88 * 10^-6

    C5H5NH + in water C5H5N & H+

    Ka = [C5H5N] [H30+] / [C5H5NH+]

    Step 3: The molarity

    The initial molarity of C5H5NH + (aq) = 0.28M

    The initial molarity of C5H5N and H3O + = 0M

    The mole ratio is 1:1 so there will be consumed X of C5H5NH + and there will be produced x of C5H5N and H3O+

    The molarity at equilibrium for C5H5NH + is (0.28 - X) M

    The molarity for C5H5N and H3O + = X M

    Step 4: Calculate the pH

    Ka = [C5H5N] [H30+] / [C5H5NH+]

    5.88*10^-6 = X*X / (0.28-X)

    5.88*10^-6 * (0.28-X) = X²

    X = [H3O+] = [H+] = 0.00128 M

    pH = - log[H+] = - log (0.00128)

    pH = 2.89
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