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<div class="moz-cite-prefix">On 04/22/2015 08:01 AM, wrote:<br>
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cite="mid:C40B2F181831EF44A88CD73525827803130F177D5A@MERCERMAIL.MercerU.local"
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<p>Dr. Pounds, <br>
</p>
<p>Can you explain in Problem 17.32, how you get [Na] and [H+]?<br>
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<p><br>
</p>
<p>I understand all of the math. I'm just confused on the final
step. Thanks!<br>
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<br>
Since we added an excess of NaOH to the solution all of the weak
acid is used up. This therefore changes our system to one in which
we will have the conjugate base of a weak acid and a base in the
equilibrium expression. Because of that I have to also switch to an
equilibrium expression in which I use Kb.<br>
<br>
Once I determined my initial concentrations based on dilution
(remember to use m1v1=m2v2 because we mixed the two solutions
together and are now in a 1 L solution) I have the initial
concentrations shown in the online solutions.<br>
<br>
Now, one thing that might be confusing you is that in the law of
mass action numerator I switched the positions of OH- and CH3COOH,
but make no mistake, x is equal to the concentration of the weak
acid.<br>
<br>
<br>
<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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