I titrated 1 gram of oil with 8 ml of NaOH (tickell method).

Acid value (mgr./gr.) is described that Milli Equivalent Gram NaOH to neutralize 1 gram FFA.

And Acid Value = 1.99 x FFA %

Now, 8 ml is equal to 0.008 gram NaOH. So, In order to neutralize 1 gram oil, it is required 0.008 gram NaOH. Acid value of this is 8.
So FFA% is 4 %.
But, conversion of tickell titration to FFA is given by titration x 0.766 in our forum.
8 X 0.766 = 6,28 FFA % and Acid Value is 12,56

which one is true? or Does our math lye to me?

first 8ml of NaOH should not be equal to 0.008g...

8ml = 0.008 is for water.. pure water.

second I would think the acid value would use a standard titration method. say 0.1N instead on 0.1% these aren't the same thing.

-dkenny

0.1 Normality is equal to 4 gram NaOH for 1 liter pure water solution. Thus 1 gram must be equal to O.025 N.

In Neutral's previous post;

The formula used in industry for calculation of free fatty acid percentage is:

FFA% = (v - b) x N x 28.2 / w

where v is the volume in ml of titration solution
b is the volume in ml of the blank
N is the normality of the titration solution
w is the weight of the sample of oil in grams

Let's put parameters in equation for 8 Ml of titration
FFA % = (8 X 0.025 X 28.2)/ 1 gram

FFA % = 5.64 there is different result too.

FFA % = ml of titration x 0.766 = 6.128

which one is true?

at what temps are the volumes measured?

you measured 1 gram of oil or 1 ml of oil?

normal homebrew(tickell) uses 1 ml of oil not 1 gram.

you might note that 5.64/6.128 = 0.92 this about the specific gravity of vegetable oil.


-dkenny

I measured 1 gram oil at room temperature.

how were you derive the density using acid value and FFA %. Mathematically, it does not seem to be logical. Ampirically, Is there relation between density and acid value, FFA% I measured 100 ml of oil and weighed it 93.57 gram of oil.
thanks.

just a whim I decided to try.
but the tickell method uses 1 ml not 1gram...
so your initial titration you used too much oil cause the titration to measure high.

1ml of oil/ 1 gram would be 0.92 or so.. the exact number depends on the oil and %FFA.

I hope helps

-dkenny

Dear Dkenny;

Let's assume we produced a new oil and let it's FFA % be 1,65. This new oil mainly consist of monoglyceride and diglyceride.

Regardless of titration, Based on just FFA %, Do you think what we should use NaOH catalsyt?

This new oil is neither waste vegi-oil or fresh frier oil. In our forum, we add 5 gram NaOH onto titration value of ml.