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26 August, 14:29

Phosphoric acid is a triprotic acid with Ka1 = 6.9 x 10-3, Ka2 = 6.2 x 10-8, and Ka3 = 4.8 x 10-13. Calculate the pH of a solution created by dissolving 23.4 g of KH2PO4 (s) and 38.4 g of Na2HPO4 (s) in 2 L of water.

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  1. 26 August, 14:50
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    The pH of the solution created is 7.21

    Explanation:

    Ka1 = 6.9 X 10⁻³

    Ka2 = 6.2 X 10⁻⁸

    Ka3 = 4.8 X 10⁻¹³

    pKa = - log (Ka)

    Thus,

    pKa1 = - log (6.9 X 10⁻³) = 2.16

    pKa2 = - log (6.2 X 10⁻⁸) = 7.21

    pKa3 = - log (4.8 X 10⁻¹³) = 12.3

    KH₂PO₄ + Na₂HPO₄ = KH₃PO₄ + Na₂PO₄

    Ka2 = [ KH₃PO₄] [Na₂PO₄] / [KH₂PO₄] [Na₂HPO₄]

    log (Ka2) = log [[ KH₃PO₄] [Na₂PO₄] / [KH₂PO₄] [Na₂HPO₄]]

    pH = pKa2 + log [KH₂PO₄] / [Na₂HPO₄] = 7.21

    Therefore, the pH of the solution created is 7.21
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