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<div class="moz-cite-prefix">On 07/21/13 23:36, Mary Caroline Logan
wrote:<br>
</div>
<blockquote
cite="mid:C40B2F181831EF44A88CD735258278030261C41E4A@MERCERMAIL.MercerU.local"
type="cite">
<pre wrap="">I have multiple questions about your solutions for these problems. But in general, I'm having trouble figuring out when to use the reaction producing OH- versus H3O+ (17.32), figuring out where/how you are coming up with various mole to mole ratios (17.30), and various Ksp values the book said was previously given and it wasn't written in your solutions (17.50).
Caroline Logan
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</blockquote>
<font face="serif">In 17.32 you have to figure out what is in
solution (i.e. -- what is left over after the strong base and the
weak acid react) before you can set up the equilibration. Since
the acid was completely consumed, and I had excess OH in solution,
I had to write the equilibration reaction as shown.<br>
<br>
In 17.30 they are simply doing a titration. The only caveat is
the the Ba(OH)2 is going to give you two OH-.<br>
<br>
In 17.50 when I divide the mass by the molar mass to come up with
the moles that are in solution, I finding the value of x that then
has to be plugged into the Ksp expression. In other words you are
using the solubility data to find Ksp (Ksp was not given in this
case). Look back at my solutions -- if you plug in the value I
determined for the solubility in moles/L as "x" in the Ksp
expression you should get the Ksp value I listed in the solution.<br>
<br>
</font><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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