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<div class="moz-cite-prefix">On 04/09/2013 10:54 PM, wrote:<br>
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cite="mid:C40B2F181831EF44A88CD7352582780301E53DB34F@MERCERMAIL.MercerU.local"
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<div style="direction: ltr;font-family: Tahoma;color:
#000000;font-size: 10pt;">Will you explain to me how you came to
set up the equation for question 2 in chapter 17? "Calculate the
concentration of C6H5COO- in a solution that is 0.015M C6H5CooH
and 0.015M HCl." <br>
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<br>
For this problem I knew I had the conjugate base of a weak acid (C<sub>6</sub>H<sub>5</sub>COO<sup>-</sup>)
and a strong acid (HCl) in solution. This necessitated an equation
that would have these two components (where the HCl, being a strong
acid, is replaced by the hydronium ion) plus the weak acid (C<sub>6</sub>H<sub>5</sub>COOH).
The weak acid reaction equation, for which we also have a K<sub>a</sub>,
works nicely.<br>
<br>
For the equilibrium expression it is simply<br>
<br>
Ka = [products]/[reactants] = [C<sub>6</sub>H<sub>5</sub>COOH][H<sub>3</sub>O<sup>+</sup>]/[C<sub>6</sub>H<sub>5</sub>COOH]
= (x)(0.051+x)/(0.015-x)<br>
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<br>
<br>
<br>
<pre class="moz-signature" cols="72">--
Andrew J. Pounds, Ph.D. (<a class="moz-txt-link-abbreviated" href="mailto:pounds@theochem.mercer.edu">pounds@theochem.mercer.edu</a>)
Associate Professor of Chemistry and Computer Science
Mercer University, Macon, GA 31207 (478) 301-5627
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