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27 July, 07:50

What is the concentration of h + in a 0.150 m solution of the diprotic acid h2x? for h2x, ka1=4.5*10-6 and ka2=1.2*10-11?

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  1. 27 July, 10:18
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    The equation of the first dissociation

    H₂X (aq) + H₂O (l) ⇄ HX⁻ (aq) + H₃O⁺ (aq)

    The acid dissociation is constant Kₐ1 is 4.5 * 10⁻⁶

    To construct ICE table and equilibrium concentration

    H₂X (aq) + H₂O (l) ⇄ HX⁻ (aq) + H₃O⁺ (aq)

    I (M) : 0.150 0 0

    C (M) : ₋x ₊x + x

    E (M) : 0.150₋x x x

    Acidiozation constant

    4.5 * 10⁻⁶ = (x) (x) / (0.150 - x)

    x = 0.000820M

    From equilibrium table

    [H₃O⁺] = [4x⁻] = x = 0.000820M

    [H₂x] = (0.0150 - 0.000820) = 0.0143M

    Equation of second dissociation

    HX⁻ (aq) + H₂O (l) ⇆ X⁻ (aq) ₊ H₃O⁺ (aq)

    acid dissociation Ka₂ is 1.2 * 10⁻¹¹

    The ICE table and equilibrium concentration

    HX⁻ (aq) + H₂O (l) ⇆ X⁻ (aq) ₊ H₃O⁺ (aq)

    I (M) : 0.000820 0 0.00820

    C (M) : - Y + Y + Y

    E (M) : 0.000820-Y Y 0.000820+Y

    Acidiozation constant

    1.2 * 10⁻¹¹ = (y) (0.000820₊y/0.000820-y

    y = 1.20 * 10⁻¹¹ M

    From equilibrium table

    [H₃O⁺] = 0.000820 + 1.20 * 10⁻¹¹

    = 8.20 * 10⁻4 M
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