this one: http://www.kimble-kontes.com/html/pg-620F.html and it came
already loaded with 0.1 N NaOH. I didn't really feel like dumping it out
and making some fresh 0.1% w/v NaOH so I pulled out a pencil and some
scrap paper.
A quick calculation reveals that 0.1 Normal NaOH is a 0.4% w/v
on 4/8/05 7:46 PM, John Hayes at [EMAIL PROTECTED] wrote:
A quick calculation reveals that 0.1 Normal NaOH is a 0.4% w/v solution.
Thus, whatever my titration amount is in mL, I need to multiple with
value by 4 to get the number of grams of NaOH per liter WVO.
But then I was thinking, if
on 4/8/05 7:46 PM, John Hayes at [EMAIL PROTECTED] wrote:
A quick calculation reveals that 0.1 Normal NaOH is a 0.4% w/v solution.
Thus, whatever my titration amount is in mL, I need to multiple with
value by 4 to get the number of grams of NaOH per liter WVO.
But then I was thinking, if I
days old or had exposure to the atmosphere, it will have
absorbed CO2. this lowers the amount of OH, and will cause
the titration to be off. (you will overestimate the amount
of NaOH needed for the methoxide preparation.
other than that it your assumptions are correct.
John Hayes wrote:
on 4/9/05 5:39 AM, John Hayes at [EMAIL PROTECTED] wrote:
Ken Provost wrote:
You're confused about
Actually, Ken, I don't think I am.
Given your explanation, I don't think you're confused either!
Sorry for jumping to conclusions :-) The fact that you already
have a prepared
snip
ps. maybe I'm being a pedant but Keith, do you think you could
change 0.1% lye to 0.1% (w/v) lye on the JtF titration page?
Not pedantic, you're quite right, I was actually aware of it but
hadn't done anything about it, especially as it's now common usage
(not my doing). But it's
