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Why is acetic acid weak?

  1. Apr 13, 2013 #1
    So according to the rules of solubility all ionic compounds containing acetate are soluble in water. However acetic acid is a weak acid. Why is this if acetate is dissolved in water?
     
  2. jcsd
  3. Apr 14, 2013 #2

    Borek

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    Please elaborate, I am not sure where the problem is. Solubility and dissociation are two separate concepts (and separate processes). Just because something is well soluble doesn't mean it dissociates, just because something dissociates doesn't mean it is well soluble.
     
  4. Apr 14, 2013 #3

    epenguin

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    Probably all the acids you know are dissolved in water.
     
  5. Apr 14, 2013 #4
    You are confusing "polar" with "ionic". If a molecule has an electric dipole, then it can dissolve in a polar liquid. If a molecule can dissociate into ions, then it can dissolve in a polar liquid. However, not all molecules with an electric dipole can dissociate into ions. Acetic acid happens to be able to do both.

    A molecule doesn't have to dissociate to be dissolved. Glocose dissolves in water without dissociating because the molecule has an electric dipole. However, glucose doesn't dissociate into ions. NaCl can dissociate into ions. However, molecules of NaCl are unstable in water. Dissociation and solubility are two separate concepts.

    The CH3COOH molecule is highly polar even if it doesn't associate. The H2COOH molecule has an electric dipole. Therefore, a solid made of H2COOH molecules will dissolve in polar liquids.


    Only some of the CH3COOH molecules will dissociate in water. When the molecule dissociates, in makes two H+ ions and one COO- ion. The number of H+ ions formed by dissociation determine whether a molecule is "acidic".
     
    Last edited: Apr 14, 2013
  6. Apr 14, 2013 #5
    I think you should double check yourself here. I'm not sure what molecule you're trying to denote but Acetic Acid (Ethanoic Acid) is CH3COOH and Formic Acid (Methanoic Acid) is HCOOH. Both are monoprotic acids.

    Also acidity does not refer to the stoichiometric constants in the dissociation chemical equation but to the Keq/pKeq (or Ka/pKA because Chemists have to be difficult) of the dissociation chemical equation. The former being how I understood your post.

    IE:

    HX + H2O -> X- + H3O+ Keq = 10

    H2X + 2H2O -> X2- + 2H3O+ This would happen in 2 steps but lets just say that the first step Keq_1 = 1/10 and the second lets say is the same Keq_2=1/10.

    The strong acid is the first not the second even though the second yields two protons per dissociation while the first yields only 1. The degree of dissociation of the first is much higher than the second and therefore the first is the strong(er) acid.
     
  7. Apr 14, 2013 #6
    Yes, you are right. Thanks for pointing it out.

    My general argument is still correct. The CH3COOH has an electric dipole even before it dissociates. Therefore, it does not have to release hydrogen ions when it is dissolved in water. A few of the molecules do as determined by the pKa.


    I am doomed !-)
     
  8. Apr 14, 2013 #7
    Yes you're general argument sounds okay and I actually like how you brought up glucose and NaCl examples, very clever.

    In hindsight, my language was not very rigorous or specific really but I hope that people won't get confused and understand what I was trying to get at.

    You are not doomed, I say stupid stuff on here all the time and get corrected by the pro's. Always learning, that why I love PF so much.
     
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