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Home Q & A Board Science-Related Homework Help Chemistry (Moderator: Laser_3)
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11 years ago
Derive the Henderson-Hasselbalch equation for a solution of a weak acid, HA, answer its salt NaA...

Please help anyone? Frowning Face
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#1
11 years ago
For the equilibrium
HA <> H+ + A-
Ka = [H+][A-]/ [HA]
Ka = [H+] x [A-]/ [HA]
- log Ka = - log [H+] + log [A-]/ [HA]
pKa = pH + log [A-]/[HA]
wrote...
#2
11 years ago
http://en.wikipedia.org/wiki/Henderson-Hasselbalch_equation

start with the acid-base equilibrium:

HA + H2O <=> A- + H3O+

for this equilibrium you have that

Ka = [H3O+][A-] / [HA]

where Ka is the acid dissociation constant

take the 'p' of both sides: p(x) = -log10(x)

pKa = pH - log10([A-]/[HA])

rearrange in terms of pH, and you have the required equation:

pH = pKa + log10([A-]/[HA])
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