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    <div class="moz-cite-prefix">On 07/17/14 02:41,&nbsp; wrote:<br>
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        <p style="">Hello Dr. Pounds,<br>
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        <p style="">I have two&nbsp;questions about a step in finding the
          pH&nbsp;in question 3 of Quiz 10.&nbsp;</p>
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        <p style="">Firstly, do we add the liters from question 2 (1.5
          L) to that of question 3 (.4 L), making the total&nbsp;1.90 L (1.0
          L + .5 L +&nbsp;.4 L)? Or should this be left as 1.4 L (1.0 L +.4L)<br>
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    Yes.<br>
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        <p style="">Secondly, for #3, do we find the initial
          concentration of HC<span style="font-size: 8pt; ">7</span>H<span
            style="font-size: 8pt; ">5</span>O<span style="font-size:
            8pt; ">2</span> and C<span style="font-size: 8pt; ">7</span>H<span
            style="font-size: 8pt; ">5</span>O<span style="font-size:
            8pt; ">2</span> by following the method in slide 12 of your
          notes (subtracting the moles of HC<span style="font-size: 8pt;
            ">7</span>H<span style="font-size: 8pt; ">5</span>O<span
            style="font-size: 8pt; ">2&nbsp;<span style="font-size: 12pt; ">from
              the moles of NaOH and dividing by the total L) or like in
              slide 15, in which an extra step is added.</span></span><br>
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    Yes, but in this particular case the amount of weak acid is
    decreasing and the amount of conjugate base is increasing.<br>
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        <p style=""><span style="font-size: 8pt; "><span
              style="font-size: 12pt; ">Thank you for your help!</span></span></p>
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        <p style=""><span style="font-size: 8pt; "><span
              style="font-size: 12pt; ">Christine Okaro</span></span></p>
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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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