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4 September, 01:16

How many grams of solid ammonium bromide should be added to 0.500 L of a 0.135 M ammonia solution to prepare a buffer with a pH of 10.190? ka = 1.8Ã10-5

grams ammonium bromide = g.

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  1. 4 September, 02:23
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    0.7590 g

    Explanation:

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

    pOH = pK + log[acid] / [base]

    Where K is the dissociation constant of the base.

    Given that the base dissociation constant = 1.8*10⁻⁵

    pK = - log (K) = - log (1.8*10⁻⁵) = 4.75

    Given concentration of base = [base] = 0.135 M

    pH = 10.190

    pOH = 14 - pH = 14 - 10.190 = 3.81

    So,

    3.81 = 4.75 + log[acid]/0.135

    [Acid] = 0.0155 M

    Given that Volume = 0.5 L

    So, Moles = Molarity * Volume

    Moles = 0.0155 * 0.5 = 0.00775 moles

    Molar mass of ammonium bromide = 97.94 g/mol

    Mass = Moles * Molar mass = (0.00775 * 97.94) g = 0.7590 g
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