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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Alexander on August 04, 2010, 05:01:21 PM

Title: Henderson-Hasselbalch equation
Post by: Alexander on August 04, 2010, 05:01:21 PM

As this famous equation states:

pH=pK+lg([base]/[acid])

Now let us take sample. Phosphate buffer with pH=7. You can find anywhere that for this case you need the ratio of K₂HPO₄/KH2PO4=61.5/38.5. and this values are correct. I have done myself. (base => K₂HPO₄)

pK in this case, as you can also find almost everywhere is 7.21

therefore:
lg([base]/[acid])=-0.21
and
[base]/[acid]=10^-0.21=0.616595...

Please note that this value is reverse of above mentioned ratio. The proper (above given) ratio you can get if pK=6.79.
well, at some places I also found that pK is 6.86 but those sources are less reliable and anyway, this value does not lead to proper solution.

Any Ideas? ???

Title: Re: Henderson-Hasselbalch equation
Post by: Borek on August 04, 2010, 05:32:44 PM

http://www.chembuddy.com/?left=pH-calculation&right=ionic-strength-activity-coefficients

Title: Re: Henderson-Hasselbalch equation
Post by: RandoFlyer on August 04, 2010, 05:35:19 PM

Isn't the formula pH=pK +log ([conj/ base]/[acid])  ?