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<div class="moz-cite-prefix">In slide 8 we did not have any
conjugate base in the system at the beginning of the system. This
is an unbuffered system. I was demonstrating hoe the pH would
drop quickly in this case if we added a small amount of acid. We
just dumped hydronium ions into an unbuffered system so they had
to be represented in the initial condtions of the reaction.<br>
<br>
Understand this -- when you add strong base strong acid to a
buffer system the system is going to react almost instantaneosly
to that "stress" (that is why we do the calculations for the new
initial conditions first) and then the system will respond to
these new conditions to attain equilibrium.<br>
<br>
In slide 12 we are working with a buffer. There is weak acid
present and the conjugate base of the weak acid. When we ADD a
strong acid it will go into the system as hydronium ions (on the
right) and the system will respond by making more weak acid (by
consuming the conjugate base of the waak acid). That is why in
that case we subtracted from the conjugate base and added to the
weak acid.<br>
<br>
On slide 35 the <img alt="$x$" style="vertical-align: middle;"
src="cid:part1.4158BFC0.61AF0D4D@mercer.edu"> value is the
amount of conjugate base added to the system in liters. 0.0624
liters is 62.4 ml.<br>
<br>
<img alt="$1.66 = \frac{(.50)(.0624)}{(.50)(.100-0.0624)}
=\frac{0.0312}{0.0188}= 1.66$" style="vertical-align: middle;"
src="cid:part2.7563CB8D.7F537E27@mercer.edu"><br>
<br>
On 04/19/2018 03:41 PM, wrote:<br>
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<p style="margin-top:0;margin-bottom:0">Hello Dr. Pounds I have
some questions about the information on our powerpoint for
chapter 17. On slide 12 of the powerpoint, the strong acid HCl
is added to a buffer solution. In order for us to calculate
the new initial concentration, we subtracted the number of
moles of H^+ by the initial concentrations in the solution
before dividing by the total volume. My question is why do we
do it that way on slide 12 but on slide 8 we do it another
way? When .50M of HCl is added to the solution in slide 8 we
just put the added acid in under H30^+ in the ICE table. </p>
<p style="margin-top:0;margin-bottom:0"><br>
</p>
<p style="margin-top:0;margin-bottom:0">My next question is on
slide 35 how did you find the buffer mix in ml? when I plug in
the values of x into the equation of each concentration I
don't get the same answers you got. </p>
<p style="margin-top:0;margin-bottom:0"><br>
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<pre class="moz-signature" cols="72">--
Andrew J. Pounds, Ph.D. (<a class="moz-txt-link-abbreviated" href="mailto:pounds_aj@mercer.edu">pounds_aj@mercer.edu</a>)
Professor of Chemistry and Computer Science
Mercer University, Macon, GA 31207 (478) 301-5627
<a class="moz-txt-link-freetext" href="http://faculty.mercer.edu/pounds_aj">http://faculty.mercer.edu/pounds_aj</a>
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