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Triprotic acid?

  1. Dec 30, 2006 #1
    Why isn't ammonia and hyronium considered triprotic acids? Both nitrogen and oxygen are highly electronegative so other negative charges would like to snactch their hydrogens. Is it because it would take too much energy due to its tetrahedral geometry which allows their charges to be distributed so they don't become very polar molecules?

    I understand that In reality, the former is a weak base and the latter is a weak acid.
  2. jcsd
  3. Dec 31, 2006 #2


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    What is hyronium? Are you refering to the hydronium ion, H3O+?
    That is not a complete argument. Youu need to specifically determine what the partial positive charge on the hydrogen is, due to this electronegativity, and that's not all. In the case of NH3, there is a partial positive charge on the H atoms, though it isn't very big, being shared by three of them. But more important is the lone pair on the N atom, which would be happy to attract a proton. It is this desire to accept protons that makes NH3 a Bronsted base.

    It doesn't have very much to do with the actual geometry, though it is important that the partial positive charge is being shared by the three H atoms.
  4. Dec 31, 2006 #3


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    With H3PO4 for instance, the resulting negative charges of the conjugate base is delocalized within the phosphate.......this is not the case with Hydronium.
  5. Jan 2, 2007 #4
    This a molecule with a delocalised electron tend to give up its Hydrogen more easily? The base of hydronium is H2O which doesn't have a delocalised electron.
  6. Jan 2, 2007 #5
    Delocalization increases the stability (lowers the potential energy) of the conjugate base,
    thus decreasing the pKA of your acid (i.e., your acid is stronger).

    Examples of acids with delocalized electrons in their conjugate bases include:
    -Nitric acid (HNO3)
    -Sulphuric acid (H2SO4)
    -Carbonic acid (H2CO3)
    -Perchloric acid (HClO3)

    Delocalized electrons in conjugate bases, however, do not make acids "strong";
    Take carboxylic acids (e.g., acetic acid), for example...
    Well, GCT was referring to acids whose conjugate bases are charged molecules.
    Given two such acids, the acid on whose conjugate base the charge is also delocalized will generally be stronger than the other acid.
    Last edited: Jan 3, 2007
  7. Jan 3, 2007 #6

    How does the acid know (before it dontes H+) that after it donates the H+, it (the conjugate base) will be more stable (i.e have lower potential energy).
  8. Jan 3, 2007 #7
    (Certainly seems like chemistry nowadays is taught anthropomorphically, doesn't it? :biggrin:)

    *pivoxa15, two slightly more general questions:
    How do any chemical reagents "know" how to react?
    How do they "know" what will happen afterwards?

    A more complete answer may involve some quantum physics, and a workable definition of what it means for a simple compound to "know" something (perhaps have some analogous "molecular cognition")...

    The explanation you may be looking for (concerning reactions proceeding to increase thermodynamic stability) may simply be an application of spontaneity-->a property of chemical reactions that release Gibbs free energy (usually as heat) and move toward a lower, more thermodynamically stable, energy state. Such reactions will proceed without any extra energy input from the surroundings into the system.

    At constant temperature and pressure, the net change in the free energy ([itex]\Delta G[/itex]) of your system can be calculated as [tex]\Delta G = \Delta H _{net} - T \Delta S _{net}[/tex],
    where [itex]\Delta H _{net}[/itex] and [itex]\Delta S _{net}[/itex] are the net changes in enthalpy and entropy, respectively, of your system (and T is the absolute temperature of your system (units Kelvin, obviously)).

    For example, the dissociation of a general acid HA (i.e., HA→H+(aq) + A-(aq))
    in aqueous solution can be considered as the sum of reactions
    {\text{HA}} \to {\text{H}}^ + + {\text{A}}^ - \hfill \\
    {\text{H}}^ + + {\text{H}}_{\text{2}} {\text{O}} \to {\text{H}}^ + \left( {aq} \right) \hfill \\
    {\text{A}}^ - + {\text{H}}_2 {\text{O}} \to {\text{A}}^ - \left( {aq} \right) \hfill \\ \end{gathered} [/tex]

    and so
    \Delta H_{net} = \Delta H_{{\text{HA}} \to {\text{H}}^ + + {\text{A}}^ - } + \Delta H_{{\text{H}}^ + + {\text{H}}_{\text{2}} {\text{O}} \to {\text{H}}^ + \left( {aq} \right)} + \Delta H_{{\text{A}}^ - + {\text{H}}_2 {\text{O}} \to {\text{A}}^ - \left( {aq} \right)} \hfill \\
    \Delta S_{net} = \Delta S_{{\text{HA}} \to {\text{H}}^ + + {\text{A}}^ - } + \Delta S_{{\text{H}}^ + + {\text{H}}_{\text{2}} {\text{O}} \to {\text{H}}^ + \left( {aq} \right)} + \Delta S_{{\text{A}}^ - + {\text{H}}_2 {\text{O}} \to {\text{A}}^ - \left( {aq} \right)} \hfill \\
    \end{gathered} [/tex]

    After measuring your solution's temperature, you can determine via
    [tex]\Delta G = \Delta H_{net} - T\Delta S_{net} [/tex]
    the change in free energy of your system caused by dissociation.

    A spontaneous reaction corresponds to a release of free energy, which implies [itex]\Delta G < 0[/itex].
    If this is the case in your solution, your acid will dissociate.

    Regarding my previous post,
    _as delocalization would decrease [tex]\Delta H_{{\text{HA}} \to {\text{H}}^ + + {\text{A}}^ - } [/tex]
    Last edited: Jan 3, 2007
  9. Jan 4, 2007 #8
    I think its more me not having properly learned chemistry.

    So its to do with there being more probable the fact that dissociation occurs and it will occur. Its more probable because after occuring, it will be at a lower energy state hence its likely that it will spontaeously occur. There are reaons (at the atomic level) why it will be at a lower energy state. These reasons are why reformation occurs and the spontaeous nature of it.

    Can you explain
    For example, the dissociation of a general acid HA (i.e., HA→H+(aq) + A-(aq))
    in aqueous solution can be considered as the sum of reactions
    {\text{HA}} \to {\text{H}}^ + + {\text{A}}^ - \hfill \\
    {\text{H}}^ + + {\text{H}}_{\text{2}} {\text{O}} \to {\text{H}}^ + \left( {aq} \right) \hfill \\
    {\text{A}}^ - + {\text{H}}_2 {\text{O}} \to {\text{A}}^ - \left( {aq} \right) \hfill \\ \end{gathered} [/tex]

    Wouldn't you get hyrodium ions and hydroxide ions as products with the last two reactions?
  10. Jan 4, 2007 #9


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    Continuing this line of thinking (NH3 being triprotic acid) looks like CH4 is tetraprotic ;)

    Last edited by a moderator: Aug 13, 2013
  11. Jan 4, 2007 #10
    But NH3 is not a triprotic acid and CH4 is not acidic not basic because the C and H are too close in electronegativities.
  12. Jan 4, 2007 #11
    The second reaction listed
    [tex]{\text{H}}^ + + {\text{H}}_{\text{2}} {\text{O}} \to {\text{H}}^ + \left( {aq} \right)[/tex]
    does produce hydronium ions.

    The label (aq) denotes that the species is in aqueous solution (it is hydrated);
    H+(aq) is essentially H3O+. You can usually omit the (aq) label if you write out the formula of the hydrate (H+(aq) is equivalently H3O+)
    Why? From what would you "produce hydroxide" ?

    Remember that distilled water already contains hydronium and hydroxide
    (in equal concentrations of 10-7M :wink:), as
    [tex]{\text{H}}_2 {\text{O}} \rightleftharpoons {\text{H}}^ + \left( {aq} \right) + {\text{OH}}^ - \left( {aq} \right)[/tex]
    This is why distilled water has a pH = -log(10-7) = 7,
    which we define as neutral.

    Adding an acid does not "produce hydroxide";
    The second reaction
    [tex]{\text{H}}^ + + {\text{H}}_{\text{2}} {\text{O}} \to {\text{H}}^ + \left( {aq} \right)[/tex]
    produces hydronium (your "hydrogen cation hydrate")
    and the third reaction
    HA→H+(aq) + A-(aq)<<(sorry no LaTeX)
    produces a conjugate base hydrate.
    Last edited: Jan 5, 2007
  13. Jan 5, 2007 #12

    In your previous post you had for your third reaction
    [tex]{\text{A}}^ {-} + {\text{H}}_{\text{2}} {\text{O}} \to {\text{A}}^ - \left( {aq} \right)[/tex]
    Is your the one in your recent post wrong?

    Would A- have hydronium ions surrounding it? If so should you show it in the formula?
    Last edited: Jan 5, 2007
  14. Jan 5, 2007 #13
    No, that's just a typo from copy->pasting the [tex]\LaTeX[/tex] too much :devil:...
    (but thanks for noticing :wink:)

    The third reaction (correct as originally posted) is
    [tex]{\text{A}}^ {-} + {\text{H}}_{\text{2}} {\text{O}} \to {\text{A}}^ - \left( {aq} \right)[/tex]
    and it produces a conjugate base hydrate.

    Regarding your other question,
    some hydrogen ions dissociated from acids may still be "close" to the conjugate bases simply due to charge difference.

    However, remember that
    strong acids have weak conjugate bases; while [tex]\Delta H_{{\text{HA}} \to {\text{H}}^ + + {\text{A}}^ - } [/tex] may be relatively lower, it may not be favorable (in terms of enthalpy) for many cations (i.e., hydronium) to be so close to one another, especially around such a weak anion (i.e., your conjugate bases).

    Weak acids, on the other hand, have strong conjugate bases. As [tex]\Delta H_{{\text{HA}} \to {\text{H}}^ + + {\text{A}}^ - } [/tex] would be somewhat (relatively speaking) large, only a few hydrogen will dissociate from the acid. Compared to the previous case (stronger acid), you have a stronger anion and fewer cations; hence, your hydroniums may well be (somewhat) closer to your conjugate bases.

    Remember that, without energy input from the environment, spontaneity generally decides how much dissociation will take place. While there may exist some 'optimal' distance between given hydronium and conjugate base ions (weighing the relative enthalpies and entropies associated with hydroniums existing near one another, hydroniums existing near the conjugate anion, hydroniums' distance from other dissociated conjugate base anions, etc...), dissociated hydroniums are generally free to travel anywhere within your solution.
    Last edited: Jan 6, 2007
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