Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: brandon1323 on June 08, 2004, 01:05:41 PM
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Ahh... it's been too long since high school chemistry :)
I am trying to verify a procedure that involves calculating a HNO3 percent by volume after performing a titration using a hach digital titrator. The current eq. on the procedure is
(reading from titrator at eq.) * 128.5 *.7
--------------------------------------------- = percent HNO3 by volume
1000
I can't figure out how whoever wrote it arrived at this conclusion (or why they didn't simplify it to *.08995). My question is, is this correct and why, or how would you reccomend calculating the concentration percent volume by volume of the unknown nitric solution? The details are:
an unknown nitric acid solution is diluted to 10% then 5mL of 10% sol. put in flask with 95 mL of DI water and titrated.
titrated with 8N NaOH and bromothymol blue as an indicator
the reading from the digital titrator / 800 = mL of titrant dispensed
Also, would large amounts of Ni and Fe in the solution affect the titration results at all, if so, at what level?
Thanks in advance.
-Brandon
"Education is a progressive discovery of our own ignorance."
- Will Durant
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Finally pulled my head out and answered my question... Just in case any other newbies need to know in the future, here it is (I think)
Tirant(L) * Normality of Titrant * molweight(HNO3)
concentration = ---------------------------------------------------
density of aq HNO3 * L of solution tested
-The mystery 128.5 comes from
1ml/800digits * 64.25g mol weight * 1/.5mL solution tested * 8N normality * 100%
(which still begs the ? why use the wrong mol weight, isn't it 61?)
-The factor of .7 comes from multiplying by the inverse of the density (1ml/1.4g HNO3)
-The division by 1000 is the L/mL factor
If any of this is wrong, please correct it in a reply and let me know.
thanx-bd