[CHM 111] lab report

[Andrew J. Pounds]

On 09/23/2018 03:45 PM,  wrote:
> Hello Dr. Pounds,
> 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?
> Thank you very much!

We covered this briefly in class... but there's nothing like a
"refresher"....

The dilution equation is $M_1V_1=M_2V_2$, but if you remember, each side
of the equation is the number of moles of substance.  In other words,
molarity times volume = moles.

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...

(moles of KHP) = (Molarity of NaOH)(Volume of NaOH)

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.

((mass of NaOH)/(204.22 g/mol)) / (volume of NaOH in liters)  = Molarity
of NaOH


  

-- 
Andrew J. Pounds, Ph.D.  (pounds_aj at mercer.edu)
Professor of Chemistry and Computer Science
Mercer University,  Macon, GA 31207   (478) 301-5627
http://faculty.mercer.edu/pounds_aj

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