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2 October, 13:11

A diprotic acid, H 2 A, has acid dissociation constants of K a1 = 2.27 * 10 - 4 and K a2 = 4.00 * 10 - 11. Calculate the pH and molar concentrations of H 2 A, HA -, and A 2 - at equilibrium for each of the solutions. A 0.103 M solution of H 2 A.

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  1. 2 October, 16:20
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    What is meant by diprotic acid?

    A diprotic acid is the type of acid that can undergo one or two dissociations depending on the pH.

    A diprotic acid is represented by H2A.

    Dissociation of a diprotic acid does not happen all at once.

    It has two steps of dissociation and each dissociation step has its own Ka value. The step 1 and step 2 Ka values are designated Ka1 and Ka2 respectively.

    Back to the question,

    At equilibrium,

    Step 1

    pKa1 = - log (Ka1) = - log (2.27 * 10^-4) = 3.64

    Step 2

    pKa2 = - log (Ka2) = - log (4.00 x 10^-11) = 10.40

    The pH of H2A is given by

    -log (molar concentration)

    pH = - log (0.103) = - (-0.987) = 0.987. Approximately equals to 1
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