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Cubic equation for weak acid hydronium

  1. Apr 8, 2007 #1

    symbolipoint

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    If one would make no assumtions to try to simplify, except for just assuming
    OH- were negligible in a solution of weak diprotic acid, H2A, we can create four
    equations to represent concentrations for :
    H2A == H + HA-
    HA- == H + A-

    Each of those has a K value for equilibrium constant expression. They would be

    K1 = [H][HA]/[H2A]
    K2 = [H][A]/[HA]

    Letting F equal the formality of the H2A then allows the mass balance and charge
    balance equations:

    F = [H2A] + [HA] + [A]
    [H] = [HA] + 2[A] + [OH] but we might usually justly assume hydroxide OH is negligible.

    An analytical textbook stated that the system of those four equations can be solved for each of the variables. I then tried doing this to obtain an equation having H as the only variable, and
    I obtain this equation:

    H^3 + K1H^2 + (K1K2 - FK1)H - 2(FK1K2) = 0
    At this point I am not using brackets, but H stands for molarity of hydronium; and the
    "2" shown in front of the last term is a factor in order to distinguish it from the subscripts on the K values (since I am unable to more conventionally type-set my work here).

    The point is that the last equation is a cubic equation, not easy to solve. Best to be done, I guess
    would be use either graphing calculator, or create a computer program to help find the roots for the variable, H.
    Is that cubic equation practical or useful? Certainly, once a good solution is found for H, the other unknowns
    will be easy. But does anyone actually work with a cubic equation in a weak-acid equilibrium situation?
     
  2. jcsd
  3. Apr 8, 2007 #2

    Borek

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  4. Apr 8, 2007 #3

    symbolipoint

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    Thanks, Borek.
    I am just trying to understand the mathematics and the practicality more thoroughly. The textbook said that the equilibria constant equations and the mass and charge balance equations can be formed, and then the variables can be solved analytically, and then the discussion of the assumptions and associated simplifications were very confusing.

    Then, I wanted to see what I would obtain if I try to solve the system analytically for at least the hydronium ion concentration.
     
  5. Apr 9, 2007 #4

    Borek

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    You always start with the full set of equations (mass balances, charge balance, Kw and dissociation constants). When solved analytically this leads to quadratic equation (in [H+]) for strong acid (full dissociation), plus one degree for every dissociation step (thus cubic equation for acetic acid and 5th degree polynomial for phosphoric acid).

    While I know people able to routinely solve cubic equations, most people knows how to deal with quadratics at best. Thus you have to look for simplifications - or use software.
     
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