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<p class="MsoNormal"><span style="font-size:12.0pt;font-family:"Arial",sans-serif">I've been asked the following before by students about the Henderson<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:12.0pt;font-family:"Arial",sans-serif">Hasselbach eqn. We covered this in lecture yesterday and will be<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:12.0pt;font-family:"Arial",sans-serif">using again in titrations.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:12.0pt;font-family:"Arial",sans-serif">.<br>
<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 lecture):<br>
<br>
Technically, they are the equil. conc. 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>
<br>
[A-]eq<br>
pH = pKa + log (-----------)<br>
[HA]eq<br>
<br>
If you set up the ICE table you would find for the equil. conc. above <br>
</span><span style="font-family:"Arial",sans-serif"><br>
</span><span style="font-size:12.0pt;font-family:"Arial",sans-serif">[HA]eq = [HA]o - x</span><span style="font-family:"Arial",sans-serif"><br>
</span><span style="font-size:12.0pt;font-family:"Arial",sans-serif">[A-]eq = [A-]o + x<br>
<br>
where [HA]o and [A-]o are the initial conc. of the buffer components<br>
and "x" is [H+]eq. The pH above is -log([H+]eq). Substituting for<br>
[HA]eq and [A-]eq the above we get for the HH eqn,<br>
</span><span style="font-family:"Arial",sans-serif"><br>
</span><span style="font-size:12.0pt;font-family:"Arial",sans-serif"> ( [A-]o + x)<br>
pH = pKa + log (----------------)<br>
([HA]o - x)<br>
<br>
[A-]o<br>
pH = pKa + log (----------) when the "x" can be ignored.<br>
[HA]o<br>
<br>
So in the normal practice of determining the pH of a buffer solution<br>
we usually ignore the "x", just as we would when using an ICE table,<br>
& use the HH eqn. w. the original starting conc. If we didn't do this<br>
there would be no point in using the eqn. and we would just use an<br>
ICE table every time (which you can do). We often do it this way, using<br>
the original conc. of the acid and base components, when we calc. the<br>
pH of a soln. from conc. and the pKa. Just remember, you can only use<br>
this eqn. in this way if the conc. of the base and acid (the conj. acid-base<br>
pair) are relatively large, their ratio is between 10 and 0.1 and the Ka is<br>
relatively small. That's because only then can you "ignore the x" and this<br>
eqn is then valid using it for determining the pH. <br>
<br>
Sample exercise 17.4 shows an example when it's not okay to use the<br>
HH eqn. Notice in this example, while the ratio of the base/acid for the<br>
conj. acid-base pair of the buffer is 0.1, the conc. of each component<br>
is small, [A-] = 1.00 x 10^-4 M and [HA] = 1.00 x 10^-3 M, and close to the<br>
magnitude of Ka (1.8 x 10^-5). The Ka needs to be about 1000 times<br>
smaller than the conc. of acid and base in order to ignore the "x" and<br>
safely use the HH eqn.<br>
<br>
Dr. Zellmer</span><span style="font-family:"Arial",sans-serif"> <o:p></o:p></span></p>
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