-1
$\begingroup$

It's written in my chemistry book ,under weak acids, that the molecules of organic acids are partially ionised in water into ions. But I don't understand how molecules are ionised.

I'm an ol student.

$\endgroup$
  • 1
    $\begingroup$ See this link: chemistry.stackexchange.com/a/17343/79678 . I did not downvote the question or answer, by the way. $\endgroup$ – Ed V Feb 5 at 18:06
  • $\begingroup$ "Note that in the above text I have used the word "ionize" to describe molecular acids because by definition when these acids react, they ionize. Or form ions. On the other hand salts of acids and bases such as sodium hydroxide are said to "dissociate," because ionic compounds are made of ions, and it wouldn't make much sense to say that ions ionize or become ions." Oh god that makes so much sense now! $\endgroup$ – Manar Feb 5 at 18:32
2
$\begingroup$

The $\ce{O}$ atom in the $\ce{O-H}$ bond is more electronegative than $\ce{H}$ atom leading to a slight shift of the electron density towards the $\ce{O}$ atom and the development of a small negative charge, $\delta-$ on oxygen atom and $\delta+$ on hydrogen atom.

Now, a water molecule also has an oxygen atom which attracts some electron density towards itself from the two hydrogen atoms on it's sides causing $2\delta-$ charge on itself and a $\delta+$ each on both the hydrogen atoms.

This $2\delta-$ charge having $\ce{O}$ atoms of the water molecules surround and attract the $\ce{H}$ atoms of $\ce{O-H}$ bond of $\ce{CH3COOH}$ molecules due to the $\delta+$ charge on them. The hydrogen atoms of water in turn surround and attract the $\ce{O}$ atom of the $\ce{O-H}$ bond of $\ce{CH3COOH}$ molecules.

This attraction leads to separation of the $\ce{O-H}$ bond releasing $\ce{H+}$ and $\ce{CH3COO-}$ ions with a shell of water molecules surrounding both of these ions and stabilising them.

| improve this answer | |
$\endgroup$
-1
$\begingroup$

In the molecule $\ce{CH3COOH}$ there is a final bond $\ce{-O-H}$ on the right hand-side of the molecule. This bond is made of a doublet of electrons and the bond is not very strong. Apparently the $\ce{H}$ atom, or better the $\ce{H}$ atom without its electron, i.e the ion $\ce{H+}$, is attracted by the next molecule of the solvent $\ce{H2O}$, and may quit the molecule $\ce{CH3COOH}$ to produce an ion $\ce{H3O+}$, and leaving an ion $\ce{CH_3COO-}$. For some reason, only a small fraction of the $\ce{CH3COOH}$ molecules are behaving this way. This phenomena is called ionisation of the acid.

| improve this answer | |
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.