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<div class="moz-cite-prefix">On 4/14/26 13:30, wrote:<br>
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Dr. Pounds, </div>
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I am studying for our exam tomorrow and I have run into a quick
question. When looking at the problems we worked in class during
chapter 17, I am confused about how we calculated some of the
initial concentrations. For example, on slide 13 of the Chapter
17 lecture, how do I know what to add and subtract from the
initial moles of acetate and acidic acids to get their new
initial concentrations? I know that it says that you calculate
the one that is decreasing first, but how do I know which is
decreasing? Also, what makes this problem different so that I
know that these types of calculations are needed? Earlier in
that lecture, on slide 9, we added HCl to a solution but didn't
have to add or subtract anything extra during the
concentrations. I am assuming that the sodium acetate in the
problem on 13 is the reason for that, but why exactly does that
change it?</div>
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Sorry for the long email, I am hoping it is easier this way but
if you would rather us meet in person, I am free for most of the
afternoon. </div>
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Thank you so much for the help!</div>
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Respectfully, </div>
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<p>Great questions! Let's start with slide 13. Look at the
equilibrium chemical equation. Remember, before we start to add
anything that system has reached equilibrium. If we add a strong
acid we are essentialy adding H3O+ to the system so Le Chatelier's
principle says that the system must shift back toward reactants to
regain equilibrium. If I shift towards reactants then the acetate
ion will be decreasing and the acetic acid will be increasing.</p>
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<p>Now this is the interesting part... in slide 9 I did not have a
common ion (like acetate) it is an unbuffered solution. There was
no acetate ion to combine with the hydrogen from the HCl to reform
acetic acid. What we were demonstrating there was how much the pH
would drop just by adding a small amount of HCl.</p>
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<p>Notice though that the ICE table is always written so that the
acetic acid is dissociating for the H3O+ and the acetate ion --
because that is what the Ka value represents.</p>
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<p>Let me know if you need further clarification.</p>
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<div class="moz-signature"><b><em>Andrew J. Pounds, Ph.D.</em></b><br>
<em>Professor of Chemistry and Computer Science</em><br>
<em>Director of the Computational Science Program</em><br>
<em>Mercer University</em><br>
<em>1501 Mercer University Drive, Macon, GA 31207 </em><br>
<em>(478) 301-5627</em><br>
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