Referring to this, when "hydrolysis of a salt" is its reaction with water:

How is it that both the cation and anion of the salt are reacting in an equal amount with water?

To state this specifically in reference to the link provided, how is it that there is a unique $h$ (degree of hydrolysis)?

  • $\begingroup$ I feel a bit icky about the site you linked. And why did you delete your previous question to ask it again in another account? Doesn't all of the $CH_3COONH_4$ get hydrolysed? remember that all ammonium salts are soluble in water. $\endgroup$
    – Jerry
    Apr 12, 2013 at 17:33
  • $\begingroup$ I don't see much value in that particular method of viewing this phenomena. $\endgroup$
    – Lighthart
    Apr 12, 2013 at 19:17
  • $\begingroup$ @Lighthart: I dunno... the textbook way(seen it multiple times) of explaining hydrolysis of weak acid-weak base salts assumes that the degree of hydrolysis is the same for both ions. It was a question that occurred to me and many of my friends when I was taught this. IIRC the answer has to do with the dissociation of water limiting the degrees of hydrolysis. So it seems like a legit question to me :) If you feel like this should be closed, just let me know with a more detailed explanation. $\endgroup$ Apr 12, 2013 at 19:40
  • $\begingroup$ The question is legit, I just never found much educational value in this analysis. I always preferred to look at this in the context of solvolysis and free energy. Tim Toady. $\endgroup$
    – Lighthart
    Apr 12, 2013 at 19:43

1 Answer 1


I think @ManishEarth has it right. The actual phenomenon involves not only the hydrolysis of the salt, but the autoionization of water. To keep things simple the autoionization of water in this case is often omitted in elementary textbooks.

In reality the hydrolysis of the anion and the cation are not normally identical and the resulting solution is slightly acidic or basic, depending on the nature of the ions.

Nevertheless, for each such reaction there is a definite amount of hydrolysis (which you call $h$ in the question). This constant will vary somewhat with temperature.


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