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21 January, 14:35

Calculate the pH of 0.65 M H2SO4 (sulfuric acid) solution? (Ka2: 1.1x10-2)

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  1. 21 January, 14:42
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    pH = 0.18

    Explanation:

    The sulfuric acid (H₂SO₄) has the following reactions in aqueous medium:

    H₂SO₄ → HSO₄⁻ + H⁺

    HSO₄⁻ ⇄ SO₄²⁻ + H⁺ Ka = 1.1x10⁻²

    Where Ka is defined as Ka = [SO₄²⁻] [H⁺] / [HSO₄⁻] = 1.1x10⁻²

    Based in the first reaction, [H⁺] = 0.65M and [HSO₄⁻] = 0.65M

    In the second reaction, the two species are in equilibrium, thus, concentrations will be:

    [H⁺] = 0.65M + X

    [HSO₄⁻] = 0.65M - X

    [SO₄²⁻] = X

    Replacing in Ka formula:

    1.1x10⁻² = [X] [0.65 + X] / [0.65M - X]

    7.15x10⁻³ - 1.1x10⁻²X = 0.65X + X²

    0 = X² + 0.661X - 7.15x10⁻³

    Solving for X:

    X = - 0.67M → False solution. There is no negative concentrations.

    X = 0.01065M → Right answer.

    Thus [H⁺] = 0.65M + 0.01065M = 0.66065M

    As pH = - log [H⁺];

    pH = - log 0.66065M = 0.18
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