While deriving the Henderson–Hasselbalch equation why don't we consider the hydrolysis reaction? For example if we consider an acidic buffer of acetic acid and sodium acetate, why do we not consider the hydrolysis of the acetate ion furnished by sodium acetate?

  • $\begingroup$ Unclear what you mean here. Isn't the whole point that you are accounting for the interaction of the acid and conjugate base with water via the equilibrium? $\endgroup$
    – Zhe
    Mar 6 '19 at 18:50

Just because it's negligible.

Well, I tried to make some calculations. I was too lazy to make them using all this internal stackexchange math system, so I hope you'll understand my handwriting.

I also need to say what conditions I used: activity=equilibrium concentration; I didn't care much about hydroxonium, because it doesn't change anything; I used 0.1M solutions of both acetate and acid just because I wanted to.

The conclusion is: really, if you have 0.1M solutions, a change in 10-10M will change absolutely nothing, so why suffer and not neglect this?

P. S. All calculations made in Excel 2013 :)

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  • 1
    $\begingroup$ Do you mind using MathJax? Images are not easily searchable? $\endgroup$ Sep 29 '17 at 10:46
  • $\begingroup$ @PrittBalagopal I probably will, but not in the middle of working day :) $\endgroup$ Sep 29 '17 at 12:25

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