hasn't the titration already accounted for the 85.9% purity??
mathematical elegance -- desired result achieved with minimal complication
Hlndr1125 and Ant,
The correct amount is 8.64 by the usual method. The base amount is what needs adjusting, the titration is self correcting. With such a good oil you would never notice the difference.
More discussion of calculation of amount of caustic is found under gas chromatograph.
I went over just a bit looks like. Oh well, no harm was done as the little bit I made turned out straw coloured this time. My other batches had an orange colour to them. This was just a trial run using the single stage method, as well as KOH & it seemed to work out okay.
Using your formula, how would I go about calculating the FFA's of this oil??? They don't seem to be too high as the oil I got looks fairly good.
Just look at the first post on this thread.
The only part of the formula I understand is the one dealing with the Normality. Nothing much else is getting through my thick skull.
I copied the following from the first post.
> "To calculate the FFA from a homebrew titration use the following:
FFA% = 0.766 x t for titration with 1g NaOH per litre where t is the volume of solution in ml for 1ml oil
FFA% = 0.546 x t for titration with 1g KOH per litre"
Your homebrew titration was 0.5ml. It was done with KOH so use the second equation. Your FFA% = 0.546 x 0.5 = 0.273%This message has been edited. Last edited by: neutral,
Sorry it seems as though I'm the "prize chump" at the county fair. The calculation is what caused all that trouble. I was looking at the wrong thing. Wow, .273% doesn't seem to be much at all in the way of FFA's. BTW, I did the titration twice & came up with the .5 mls both times.
There is a difference between the titration methods in understanding what you are doing with them. The industrial titration method is a molar titration, and is used when doing synthesis and analysis based on molar weights. However, the homebrew titration is a mass titration, letting us know how much of what we need to add to that to get a pH of whatever.
I think this will help:
industrial titration: molar titration = using 0.1 N solution for molar analyses of chemical reactions (what petrochem companies need)
homebrew titration: mass titration = 1 g/liter solution for mass requirement analysis (what we need)
Of course, I am no chemist so I could be off my rocker, but that's how I understand it.
OK, so I have oil that titrates to 36 with KOH and Turmeric.
So I get: FFA=19.66 and AV=39.11
In the top post of this thread
Kenito Oko calls for (given a 1000 gram sample of such oil)
My oil has 196.6 grams FFA per 1000 gram sample.
I need 442 mL Methanol and 10 grams acid per 1000 grams oil?
Is this correct? Seems like a hell of a lot of methanol.
Neutral and others,
What is the "blank". Also, I will be working with some super high FFA palm oil. Should I use that 270 number in your first post saying the molecular weight of palm oil is 270? THen do I divide that by ten to get 27 instead of 28.2?
If I understand Neutral's post. Equation should be this (I am using NaOH):
FFA% = 0.766 x t for titration with 1g NaOH per litre where t is the volume of solution in ml for 1g oil
.766 x t (however many mL I use to get my tumeric to indicate).
How will that number be adjusted for my palm oil?
Yes, I am confused...
The blank is a titration done in the normal way except that the oil is left out. It is done to make allowance for any acidity which may be present in the alcohol.
An alternative is to add a drop or two of titrating solution to the alcohol beofore adding the oil, just sufficient to get the colour change, then add the oil, which will reverse the colour change, and continue as normal.
I have been asked on another thread how to titrate in such a way as to obtain the FFA%. The post that starts this thread sets out how it is done in industry.
Here is an alternative method:
To calculate the FFA% from a homebrew titration multiply the number of ml by 0.5463 where you have 1g KOH dissolved in 1 litre of water and titrate against 1ml of oil.
So if you dissolved 1 / 0.5463 = 1.83 g KOH in 1 litre of water and titrated against 1ml of oil you would get the FFA% direct.
Note that this assumes the KOH is 100% pure. Allowance must be made if it is not.
I was just rereading this thread..
my how far has acid esterification has come..
this thread has some references to the basis for my method but nothing easy..lots conversion and room for errors..
please don't see this thread as a plug for my method.. at the time of this thread I had only been making BD for maybe 6 months. a newbie!!
this thread planted the seed that let to my method..just the fact that acid esterification works..
its been a fun and interesting learning new stuff since I started in 04..only 4 yrs ago..
'84 bluebird school bus, DD8.2L turbo( 4/2011, the bus tranny has died.. 8.23.11 bus driven to scrap yard )
2006 Jeep Liberty CRD - the wife's
99 dodge 2500 5.9l 24v..-mine
everything run B100 when its warm enough
D.Kenny -SIR I'm so grateful I read your deliberations on the acid method & although my oil is good between .5 and 3.5 KOH I still find 'your' acid method allows faster & better clearing of my bio. I now filter with two 5+1 micron filters instead of using water which seemed to take ages- after bubbling and heating to 90F for 4 or even 5 days. The fuel is gin clear before I filter it....
Primarily I think you were right when you said there may have been a problem for my oil re water although I have never actually seen any molecules of water-I now get rid of water in any case before processing.. Sincere thanks Mick
hi, im not a chemist, im just a student. i not realy understand about i experiment i done which is titration of ffa.Here i asked to use the mix of diethyl ether and 2-propanol.Then weigh some sample which the oil and then pour to the mixture i say just now.put drop indicator phenolpthalein,then start the titration using NaOH.
can somebody explain me hoe the process of titration done?
what is the function of mixture diethyl ethe and 2-propanol in the process?
can u let me know what is the chemical reaction occurs inside the process?
can u write the chemical reaction?
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