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5 January, 18:04

Calculate the [H+] in 1.0 M solution of Na2CO3 (for H2CO3, Ka1 = 4.3 * 10-7; Ka2 = 5.6 * 10-11). 7.5 * 10-6 M 1.3 * 10-2 M 7.5 * 10-13 M 6.6 * 10-4 M None of these choices are correct.

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  1. 5 January, 19:09
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    7.5x10⁻¹³M = [H⁺]

    Explanation:

    When a solution of Na₂CO₃ is dissolved in water, the equilibrium produced is:

    Na₂CO₃ (aq) + H₂O (l) ⇄ HCO₃⁺ (aq) + OH⁻ (aq) + 2Na⁺

    Where Kb is defined from equilibrium concentrations of reactants, thus:

    Kb = [HCO₃⁺][OH⁻] / [Na₂CO₃] (1)

    It is possible to obtain Kb value from Ka2 and Kw thus:

    Kb = Kw / Ka2

    Kb = 1x10⁻¹⁴ / 5.6x10⁻¹¹

    Kb = 1.8x10⁻⁴

    Replacing in (1):

    1.8x10⁻⁴ = [HCO₃⁺][OH⁻] / [Na₂CO₃]

    The equilibrium concentrations are:

    [Na₂CO₃] = 1.0M - X

    [HCO₃⁺] = X

    [OH⁻] = X

    Thus:

    1.8x10⁻⁴ = [X][X] / [1-X]

    1.8x10⁻⁴ - 1.8x10⁻⁴X = X²

    X² + 1.8x10⁻⁴X - 1.8x10⁻⁴ = 0

    Solving for X:

    X = - 0.0135 → False answer, there is no negative concentrations

    X = 0.0133

    As [OH⁻] = X;

    [OH⁻] = 0.0133

    From Kw:

    Kw = [OH⁻] [H⁺]

    1x10⁻¹⁴ = 0.0133[H⁺]

    7.5x10⁻¹³M = [H⁺]
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