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31 August, 11:40

Telluric acid (H2TeH4O6) is a diprotic acid with Ka1 = 2.0x10-8 and Ka2 = 1.0x10-11. A 0.25 M H2TeH4O6 contains enough HCl so that the pH is 3.00. What is the concentration of HTeH4O6

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  1. 31 August, 14:47
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    5x10⁻⁶ = [HTeH₄O₆⁺]

    Explanation:

    The first dissociation equilibrium of the telluric acid in water is:

    H₂TeH₄O₆ + H₂O ⇄ HTeH₄O₆⁺ + H₃O⁺

    Using H-H equation for telluric acid:

    pH = pKa + log₁₀ [HTeH₄O₆⁺] / [H₂TeH₄O₆]

    pKa of telluric acid is - logKa1

    pKa = - log 2.0x10⁻⁸

    pKa = 7.699

    As concentration of [H₂TeH₄O₆] is 0.25M, replacing in H-H equation:

    3.00 = 7.699 + log₁₀ [HTeH₄O₆⁺] / [0.25M]

    -4.699 = log₁₀ [HTeH₄O₆⁺] / [0.25M]

    2x10⁻⁵ = [HTeH₄O₆⁺] / [0.25M]

    5x10⁻⁶ = [HTeH₄O₆⁺]
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