can someone tell me the basis for this formula
FFA% = 0.546 x t for titration with 1g KOH per litre
When there is a 1:1 ratio of FFA to KOH?
RCOOH (FFA) + KOH ---> RCOO-K+ + H2O
and can someone explain the basis for saying that the acid value is equivalent to (1/2)FFA percentage by weight?
I think the formula you are talking about came from Kenito Oko's method, right? I think HE would be about the only guy who could answer, as I don't understand what all he does in that method either!
Only thing I can think of is b/c titration determines acidity, and FFA's are not the only acids in the sample (triglycerides break down with water into FFA's and glycerin), glycerin has acidity to it. But still, assuming FFA is half the total value seems somewhat inaccurate to me...there's got to be a more accurate method.
I'd suggest it's to allow calculation of v/v% or w/w% FFA from the number of moles of FFA (given by the titration). I'm not sure if it's valid or not, as it would depend on the specific FFA composition.
Not sure where the acid value comment has come from.
this is a very old post, but i'd like to comment that i have found the aocs method and it would seem that acid value has nothing to do with it.
acid value will tell you g KOH/g oil needed to neutralize the acids.
ffa you need to use the average molecular weight of all free fatty acids in the profile, which i found to be somewhere around 270-290 g/mole.
since the molar ratio of KOH/ is 1:1, you just take the MW of FFA (280) and substitute it where you would put the MW of KOH in the acid value. If you do this, it would seem you get more not less than the acid value, as much as 5 times more.
so who came up with the idea that FFA is half the acid value?
The titration solution is not a molecular weight based solution. The titration result, therefore, does not report the weight of the FFA, only the % of FFA based on the arbitrary nature of the titration solution concentration.
emily e: Here is my attempt at answering your question mathematically.
The "standard" titration solution is formulated with 1 gm of NaOH (KOH) mixed in 1 liter of water. This solution is then used to titrate 1 ml of oil to determine how much extra catalyst is needed to compensate for the FFA in the oil. Assuming we titrate one ml of oil and find that it takes exactly 1 ml of our titration solution to neutralize the FFA, then we determine our catalyst requirement as 5 gm/l base plus 1 gm/liter to neutralize the FFA.
This is a simple and straightforward method used by home brewers. It works well. The beauty of the procedure is that we don't really have to determine the amount of FFA in the oil. We only need determine the "relative" amount of FFA in the oil.
But, the above data can be used to calculate the actual % FFA in the oil.
One gram of NaOH (KOH) in 1 liter of water is equal to 0.000025 (0.0000178) moles of NaOH (KOH) in every ml of the water.
Using your number of 280 gm/mole as a reasonable estimate for the weight of FFA, then 1 ml of NaOH will neutralize 0.007 (0.0049) gm of FFA in 1 ml of the oil sample.
Assuming 1 ml of oil weighs 0.9 gm, then the percent of FFA in the oil neutralized with the NaOH is 0.78% (0.007/0.9 x 100 = 0.78%).
Similarly the percent of FFA in the oil neutralized with KOH is 0.55% (0.00499/0.9 x 100 = 0.55%).
The above calculations are based on an average weight for NaOH of 40 gm/mole and for KOH of 56.1 gm/mole.
The conversion between grams and moles for KOH and NaOH are not the same! But your definitely on the right track.
You have posted a very general statement.
Other than the different molecular weights why isn't the conversion from grams to moles the same for NaOH and KOH? Can you be more specific?
This is an interesting thread. I suggest that you start at the top of the thread.
You will note that a certain person was apparently detecting sock puppets way back then (Naughty Kenito Oko)
read this post
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2006 Jeep Liberty CRD - the wife's
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everything run B100 when its warm enough
i think i finally figured it out!
producer - i see that is one way to go about it, too. here's how i do it.
%FFA = g FFA/g oil
= (Vtitrant mL soln/1000)*(N moles KOH/L soln)*(1 mole FFA/1 mole KOH)*(280 g FFA/mole FFA)/(weight of oil sample, g)
generally this is the formula i use, but i figured out doing it the homebrewer way gives you the same answer but in a different way. im using a standardized molarity titrant rather than weighing out KOH by hand and mixing it with a volume of water.
1 mole FFA/1 mole KOH is there because in the neutralization it takes one mole of KOH to neutralize one mole of FFA and become a soap/ffa salt KOOC-CH2CH3 and water.
acid value, on the other hand, depending on what base you are using as a titrant goes like this:
AV = g base/g oil to neutralize acids
AV =(Vtitrant mL soln/1000)*(N moles KOH/L soln)*(56 g KOH/mole KOH)/(weight of oil sample, g)
AV =(Vtitrant mL soln/1000)*(N moles NaOH/L soln)*(40 g KOH/mole KOH)/(weight of oil sample, g)
to get moles of base/L solution, you get the g base/L solution you measured and divide it by the MW (56 for KOH, 40 for NaOH). so the above relation could essentially be calculated by
AV =(Vtitrant mL soln/1000)*(1 g KOH/L soln)/(weight of oil sample, g)
AV =(Vtitrant mL soln/1000)*(1 g NaOH/L soln)/(weight of oil sample, g)
Ok, so if i take this value above and convert it to FFA
=(Vtitrant mL soln/1000)*(1 g KOH/L soln)*(280 g FFA/mole FFA)(1 mole FFA/mole KOH)/(56.1 g KOH/mole KOH)*(g of oil)
since it's a percentage you would also mulitply by 100, and divide by 1000.
reduced it gives me
V*C*28/56*Ws for KOH
V*C*28/40*Ws for NaOH
which is about half for KOH and seven tenths for NaOH. so i guess it is about half, if you go about it using concentration rather than molarity.
I follow what you said and agree with your results. Good work.
You provided results for both Na and K catalysts, and presented info' for both % acid and acid value.
I am not a chemist so it always helps me to understand things when people take the time to write out their full results so the derivations can be followed.
Again, good work.
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