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<div class="moz-cite-prefix">On 04/16/2015 10:14 AM, wrote:<br>
</div>
<blockquote
cite="mid:C40B2F181831EF44A88CD73525827803130F177C33@MERCERMAIL.MercerU.local"
type="cite">
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<p>Dr. Pounds,<br>
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<p><br>
</p>
<p>On problem 16.68, why do you take the square root of (k<sub>b</sub>x
M)?<br>
</p>
<p><br>
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<br>
The law of mass action here is..<br>
<br>
<img style="vertical-align: middle"
src="cid:part1.06030503.06050608@mercer.edu"
title="\documentclass{article}
\usepackage[utf8x]{inputenc}
\pagestyle{empty}
\begin{document}
\mathrm{K_b = \frac{[NH_4^+][OH^-]}{[NH_3]}}
\end{document}
" alt="\documentclass{article}
\usepackage[utf8x]{inputenc}
\pagestyle{empty}
\begin{document}
\mathrm{K_b = \frac{[NH_4^+][OH^-]}{[NH_3]}}
\end{document}
"><br>
<br>
or<br>
<br>
<img style="vertical-align: middle"
src="cid:part2.02070903.09040806@mercer.edu" alt="$k_b =
\frac{(x)(x)}{(.1-x)}$"><br>
<br>
if x is small compared to .1...<br>
<br>
<img style="vertical-align: middle"
src="cid:part3.06060005.04080804@mercer.edu" alt="$k_b =
\frac{x^2}{.1}$"><br>
<br>
solving for <img style="vertical-align: middle"
src="cid:part4.02000809.07060105@mercer.edu" alt="$x$"><br>
<br>
<img style="vertical-align: middle"
src="cid:part5.08080309.08090708@mercer.edu" alt="$x =
\sqrt{(k_b)(.1)}$"><br>
<br>
<br>
Got it?<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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