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<div class="moz-text-html" lang="x-unicode"> <big><font size="2"><big>I've
been asked the following before by other students:<br>
<br>
**********************<br>
Dr. Zellmer,<br>
<br>
I was reading the book and saw that for the Henderson
Hasselbach<br>
equation, the concentrations of the acid and base equal to
the<br>
concentrations from the equilibrium but not the original
concentrations.<br>
is that true?<br>
<br>
**********************<br>
<br>
In reality there are two answers to this (which I
addressed in<br>
lecture):<br>
<br>
Technically they are the equil. conc. </big></font></big><big><font
size="2"><big><big><font size="2"><big>When asked to calc.
the ratio<br>
of base/acid from the pH and pKa you are actually
getting the ratio<br>
of the equil. conc. of base and acid.<br>
</big></font></big><br>
However, in the normal practice of determining the pH of a
buffer<br>
solution we use the original starting conc. because we
usually ignore<br>
the "x", just as we would when using an ICE table. If we
didn't do<br>
this there would be no point in using the eqn. and we
would just use<br>
an ICE table every time (which you can do). We do it this
way, using<br>
the original conc. of the acid and base, when we calc. the
pH of a soln.<br>
from conc. and the pKa. Just remember, you can only use
this eqn.<br>
in this way if the conc. of the base and acid (the conj.
acid-base pair)<br>
are relatively large and the Ka is relatively small.
That's because only<br>
then can you "ignore the x" and this eqn is valid using it
for determining<br>
the pH.<br>
<br>
Dr. Zellmer</big></font></big> </div>
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