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25 July, 02:42

How many grams of solid potassium hypochlorite should be added to 2.00 L of a 0.244 M hypochlorous acid solution to prepare a buffer with a pH of 6.733? ka = 3.5*10-8

grams potassium hypochlorite = g.

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  1. 25 July, 02:58
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    8.2763 g

    Explanation:

    Considering the Henderson - Hasselbalch equation for the calculation of the pH of the buffer solution as:

    pH=pKa+log[base]/[acid]

    Where Ka is the dissociation constant of the acid.

    Given that the acid dissociation constant = 3.5*10⁻⁸

    pKa = - log (Ka) = - log (3.5*10⁻⁸) = 7.46

    Given concentration of acid = [acid] = 0.244 M

    pH = 6.733

    So,

    6.733 = 7.46+log[base]/0.244

    [Base] = 0.0457 M

    Given that Volume = 2 L

    So, Moles = Molarity * Volume

    Moles = 0.0457 * 2 = 0.0914 moles

    Molar mass of potassium hypochlorite = 90.55 g/mol

    Mass = Moles * Molar mass = (0.0914 * 90.55) g = 8.2763 g
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