[brandon1323]

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

[brandon1323]

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