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<div class="moz-cite-prefix">On 09/23/2018 03:45 PM, wrote:<br>
</div>
<blockquote type="cite"
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<div>Hello Dr. Pounds,</div>
<div style="direction: ltr;">I have looked over my notes about
how to find the molarity of the NaOH for each trial in our
experiment, but I’m still a little confused because I’m not
sure how we can find the total volume of the solution to plug
into the dilution equation. We have the volume of the NaOH
used in each trial, but isn’t that only the volume of the
solvent, not the volume of the solution?</div>
<div style="direction: ltr;">Thank you very much!</div>
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</blockquote>
<br>
We covered this briefly in class... but there's nothing like a
"refresher"....<br>
<br>
The dilution equation is <img alt="$M_1V_1=M_2V_2$"
style="vertical-align: middle;"
src="cid:part1.794626BA.9B7CD007@mercer.edu">, but if you
remember, each side of the equation is the number of moles of
substance. In other words, molarity times volume = moles.<br>
<br>
KHP and NaOH react on a one to one basis, so the moles of KHP is
equal to the moles of NaOH at the equivalence point so...<br>
<br>
(moles of KHP) = (Molarity of NaOH)(Volume of NaOH)<br>
<br>
you determine the moles of KHP based on the mass of KHP and the
molar mass of KHP (204.22 g/mol). The molarity of NaOH is then
found by rearranging the equation above to isolate the molarity.<br>
<br>
((mass of NaOH)/(204.22 g/mol)) / (volume of NaOH in liters) =
Molarity of NaOH<br>
<br>
<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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