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<div class="moz-cite-prefix">On 03/31/2015 09:32 PM, wrote:<br>
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<p>Dr. Pounds, I'm a little confused by number 28. If
pOH=-log[OH], and since KOH is a strong base, then
pOH=-log[KOH]. This makes the equation pOH=-log[.36],
resulting in .44. But wouldn't that be acidic on the pH scale?</p>
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<br>
We make our determination about acid or base by using the pH, not
the pOH. Convert this to pH we have to use the equation<br>
<br>
pH + pOH = pKw = 14<br>
<br>
or..<br>
<br>
<br>
pH = 14 - pOH<br>
<br>
in this particular case...<br>
<br>
pH = 14 - 0.44 <br>
<br>
= 13.56<br>
<br>
<br>
That's definitely basic!<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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