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 dissolve 4mL of WVO in 40mL of
isopropanol, and titrate with the 0.1N NaOH solution, then the volume of
NaOH required should equal the number of grams required per liter of
WVO directly.
You're confused about the definition of molarity (or normality). Rather
than get hopelessly entangled in Avogadro's number, let's just agree that
if you dissolve 1 ml of oil in any random amount of alcohol, and titrate
it against 0.1% NaOH solution (NOT 0.1N !!), the number of ml of solution
required to neutralize the oil will be the number of grams of NaOH needed
per liter of oil to neutralize the free fatty acids.
(i.e. -- 0.1% means one thousandth, and 1ml of oil is one thousandth of
one liter of oil.)
You lost me with the factor of 4 part -- just stick with the previous
paragraph :-)
Actually, Ken, I don't think I am. Let me try to break down my logic
into smaller steps:
a) you can skip steps b-d if you remember that normality and molarity
are equal for monoprotic and monobasic reagents. I didn't at first.
b) Normality is defined as number of gram equivalents per liter of water.
c) Gram equivalents is equal to the molecular weight divided by the
number of replaceable hydrogen or hydroxide ions.
d) Sodium hydroxide only has one OH- ion and has a molecular weight of
40. Thus, 1N NaOH = 40g of NaOH per liter water.
e) The JtF titration page says to dissolve 1g of lye in 1 liter of water
to form a 0.1% lye solution. (Note: this really should say a 0.1% (w/v)
solution. 0.1% lye is unclear - it could mean (w/w), (w/v) or (v/v).)
f) Anyway, by definition: 1g NaOH / L = 0.1% (w/v) = 0.025M NaOH =
0.025N NaOH
g) In contrast, my titrator came loaded with 0.1 Normal NaOH.
h) Thus, my titrator is loaded with a 0.4% (w/v) solution.
i) If I titrate the FFAs with a base 4 times as strong, I'll only need a
1/4th as much. Thus, I need multiple the volume of base needed by 4 to
determine the correct amount of lye needed per liter of oil.
j) However, under the JtF Better Titration section, it suggests that 1mL
of WVO is too hard to measure accurately and you might want to try to
measure a larger volume and then divide to obtain the correct amount of
lye required per liter.
k) Indeed, coincidentally, the example given is one of dissolving 4mL
WVO in 40mL isopropanol. However, because you then need to neutralize 4
times as much FFA with your base, you need to divide by 4 to get the
proper amount of lye required per liter.
So where does that leave us:
If you combined i and k, you see the two adjustment factors will cancel
out, right?
In other words, by using 4mL of WVO in 40mL with 0.4% (w/v) base, you're
getting the exact same result as 1mL of WVO in 40mL with 0.1% (w/v)
base, namely a direct relationship between number of mL of base required
and number of grams of lye required per liter. Furthermore, you get a
more accurate measure, *without* needing to do any calculations on the
bench.
That, and I don't have to dump the 0.1N NaOH already in my titrator. :)
jh
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?
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