NaOH / KOH Catalysts and the Chemistry behind them.

[Ross.Dunmall]

Mix Methoxide
Add the titration figure to your base. Most Sodium Hydroxide users use a base of 5.0g/l. In this example, 1.4 + 5.0 = 6.4g/l. This figure is the amount of Sodium Hydroxide needed per litre of oil you are going to process (Potassium Hydroxide users use a base of 7.0g/l, 1.4 + 7.0 = 8.4g/l)
Measure out the Sodium or Potassium Hydroxide
Measure out the Methanol (20% of the oil volume)
Mix the Methanol and Sodium or Potassium Hydroxide to make Methoxide

I have read numerous different articles / posts from small batch producers that seem to give very different accounts of the amount of catalyst that is actually required for the reaction. In addition there are again more different figures relating the strength of NaOH to the Strength of KOH as a base.

Sadly with all the information posted out there I am yet to find a post that gives any form of citation or validation for the figures they state.

Does anyone have something that delves deeper into the actual chemistry, backed up quantifiable reasoning?

As an illustration:

I've read many places that the base catalyst amount needed (NaoH) as 3.5g/l but I've also read upto 5g/l.

For KOH conversion I've read factors of 140% and 155.7% KOH w/w. using the 140% as an example would meaning using 1.4x the amount of KOH as you would normally use for NaOH.

This means that if I use the 140% Then the base KOH amount would be 4.9g/l (for 3.5g/l NaOH) or 7g/l (for 5g/l NaOH)

Now if I use the 155.7% then I get 5.45g/l KOH and 7.71g/l respectively.

This leads to many different titration formulae and no real consensus.

Can anyone shed any light on this?

[Tony]

Let me prepend this post by stating that I'm not an organic chemist

The amounts vary depending on whether the catalyst used is bought in Sodium/Potassium Methylate or has been mixed by the brewer (IE adding catalyst to Methanol).

In the former case no water is present, but water is generated by the latter to the detriment of the reaction (some of the catalyst is consumed by the water and oil to make soaps, consequently requiring more catalyst).

The figures used by home brewers come mostly from empirical evidence.  We use the 3/27 test to judge completeness of reaction, which gives us some indication of how much conversion has taken place and how much oil remains to be reacted (although the 3/27 is qualitative rather than quantitative it does provide a good guideline).

There are a few professional organic chemists that also homebrew - I'll ask them if they have comments regarding this thread.

[Ross.Dunmall]

Thanks tony.

I obviously do have a good understanding of Organic Chemistry, problem was I found nothing relating to the chemistry and more just statements of belief rather than fact. Seems somewhere along the lines people have simply recycled information from other sources and the calculations have been lost in translation.

I've already found out one answer and can show that the conversion rate 140% (1.4 times) is correct.

In organic chemistry we react moles of different substances together to make a product:

Example: 2H₂ + 10₂ = 2H₂O

In this example we are reacting 2 mole of Hydrogen with one mole of Oxygen to make 2 moles of water.

NaOH and KOH have different RMM (Relative Molecular Masses).  So one mole of Sodium Hydroxide actually weighs less than one mole of Potossium Hydroxide.

RMM NaOH = 40g/mole
RMM KOH = 56g/mole

Now for the conversion RMM KOH / RMM NaOH  = Conversion needed to change from NaOH to KOH.

56 / 40  = 1.4 = Conversion needed to change from NaOH to KOH.

I'll carry on working on the base amounts and report back when I find something.

[Tony]

Thank you - if we can include that information in the wiki that would be very beneficial.  I'm much happier when practise is based on solid theory.

[Ross.Dunmall]

Sure thing,

Much like you, I'm far happier too. Although I take it to the next level and am unhappy until it's there

[countrypaul]

The reason that the figures for the equivalence of KOH to NaOH varies is that the purity of KOH is usually significantly less than that of NaOH. NaOH is typically >99% pure, but KOH is often only 90% pure. In this case you would need an additional amount 1.111 (1/.9)times as much KOH, which give a figure of 1.56 time the amount of NaOH. This is not organic chemistry at this point its inorganic chemistry (pur simply theres no carbon involved in this part).

Incedently Ross, your equation for the formation of water is slightly wrong, you should have 2moles of water being formed not one!

The figures for the base amount of catalyst to use come from differnt sources. Originally most recommended 3.5g/L, but once the methanol test was adopted it was found that this figure often failed to result in a full conversion, so the figure was increased to 5g/L which usually resulted in a complete transesterification as far as the simple test could determine. If the oil is high titrating then it is usual to use a 2 stage process due to the formation of significant amounts of water which promote soap formation. The second stage is performed when the glycerol (and most water/soap) ahs been removed. More recently many are performing a 2 stage process with titrating, instead relying on the amount of dropout in the methanol test to indicate how much catalyst to use in the second stage. It appears most who use this method find they end up using less catalyst ten they would have done using the normal titration method.

HTH

Paul

[Tony]

Nick has also added:

I think Ross may be on the money with the Molar Masses You need the same number of moles of OH ions to deprotonate the same amount of methanol to form the same number of moles of methoxide ions. But a mole of KOH weighs more than a mole of NaOH so more KOH is needed (by weight). That and the fact that KOH is typically less pure than NaOH, are probably the two important factors.

[Ross.Dunmall]

The reason that the figures for the equivalence of KOH to NaOH varies is that the purity of KOH is usually significantly less than that of NaOH. NaOH is typically >99% pure, but KOH is often only 90% pure. In this case you would need an additional amount 1.111 (1/.9)times as much KOH, which give a figure of 1.56 time the amount of NaOH. This is not organic chemistry at this point its inorganic chemistry (pur simply theres no carbon involved in this part).

Yeah I thought it might have something to do with the purity. Personally I always prefer to leave different terms separate in a formula, makes understanding it a lot easier. Formulas I've seen quoting 1.56 are indeed using 90% KOH Purity.

Problem is if you then titrate using 90% Purity KOH Solution by making your own and accurately adding 1g to 1l of distilled water, you're then also multiplying the amount by the scaling factor of 1.4 and then the purity factor of 1.11111 and since you're titrating at the same purity as needed you in fact already ahve the correct amount of KOH in g/l that needs to be added with the 1.1111 Purity factor meaning you are using too much KOH. So either you need to re-jig the formula to account for this or titrate with 0.1M KOH solution.

Incedently Ross, your equation for the formation of water is slightly wrong, you should have 2moles of water being formed not one!

*Cough* I don't know what you mean, I put it as 2 moles Water. *Cough* Edited to update correct change. Normally I would have triple checked it, but as it was a loose example I kind of skipped that part, thanks for the correction!

The figures for the base amount of catalyst to use come from differnt sources. Originally most recommended 3.5g/L, but once the methanol test was adopted it was found that this figure often failed to result in a full conversion, so the figure was increased to 5g/L which usually resulted in a complete transesterification as far as the simple test could determine. If the oil is high titrating then it is usual to use a 2 stage process due to the formation of significant amounts of water which promote soap formation. The second stage is performed when the glycerol (and most water/soap) ahs been removed. More recently many are performing a 2 stage process with titrating, instead relying on the amount of dropout in the methanol test to indicate how much catalyst to use in the second stage. It appears most who use this method find they end up using less catalyst ten they would have done using the normal titration method.

See my big problem with this is "Recommend 3.5g/l". I'm sure there is scientific testing and proof behind the number, it's just I can't find it anywhere on the internets. Maybe I need to do some small scale lab tests and report findings, or just hope I get lucky and actually find some data where someone else has already done that. Come to think of it I did find something on journey to Forever the other week relating to that, maybe I should go and dig some more and request author permission for it to be used here. (with credit being given ofc).

[Tony]

Many of us aren't convinced on the 3.5g/l base with NaOH.  Certainly I have to use higher (more like 5g/l) to get a clear pass on the 3/27 test.

I completely agree with what you say about titration using a solution made from impure catalyst.  The only correction that gives is to the titration amount and not the base amount.  I've always argued that it is far better to mix the titration solution up from a high purity catalyst (most NaOH is 99% after all) because at least you know where you're starting from then.  Better still from a known concentration lab reference solution.

[Ross.Dunmall]

In regards to the 3.5g/l.

I found this information. Which sort of helps but doesn't give any proof again.

They say for virgin oil you need 3.5g/l and then suggest using more upto 5g/l for higher titration results (again not quantified as what value a high titration result is)

http://journeytoforever.org/biodiesel_make2.html#joetitrate

I know I saw an actual table with calculations and tests relating the amount of NaOH needed to the diffferent oils and FFA's somewhere. I'll find it again one of these days.

In the meantime I found some info that relates to KOH. It's a proper Research Paper too.

http://ljs.academicdirect.org/A10/131_136.htm

Might be worth e-mailing these guys and getting permission to use some of the info.

[Tony]

As Paul points out, the 3.5 grams was (by their own admission) an empirical measure and used by JTF before biodiesellers had a mechanism by which to measure the completeness of conversion (Jan Warnqvist's 3/27 test).  Subsequently it has been found to be a bit too low - really a lot of the information on the JTF site is considered outdated and few UK biodiesellers use it as a point of reference.

The paper above looks very interesting though - I'll have to read it properly when time allows and digest what it contains.

[countrypaul]

Ross,

You are entirely correct about the calculation problems that occur if using 90% KOH, the base amount will need to be adjusted, but the titration value would be correct. Beware that the strength of solution used for titrating is normally 0.1% not 0.1M. Several begginers have fallen foul of that difference in the past and wondered why the results they were getting were so far out!

As to where the value of 3.5g/L came from - I'm sure that must be from an engineer trying things rather than a chemist working it out

Paul

[Ross.Dunmall]

I'll do more diging on the Titration. I know our current procedure works, but I'm currently trying to adapt it to make it more accurate for our needs and titrate using our actual catalyst for more accurate results and hopefully reducing catalyst costs.

Edit: Digging now done.
Yep I knew you were right in what you were saying as soon as you'd said it but needed to run the math myself.

As stated earlier NaOH has an RMM of 40g
Therefore to make 1M solution you dissolve 40g NaOH in 1l of distilled water so for 0.1M that would be 4g/l which obviously makes the titration solution 4x as strong as the 1g/l titration solution we are and should be using. It's no real issue for us as we're using 90% Purity KOH Flakes. So I simply need to dissolve 1.00g of the flakes in 1.00 litres of water. However I think Instead I'm going to dissolve in the ration of 1.00g of our KOH in 0.900 litres of water, to make calculations after titration simpler.

Should give me:

Mass of KOH Needed (Kg) = (Titration Value + Base KOH)*(Catalyst Purity Factor) Volume being Reacted

Where;
Base KOH = 3.5*1.4 = 4.9
Caralyst Purity Factor = 1 / Purity of Catalyst = 1 / 0.9 = 1.111111111

We are actually working on base of 3.5 which providing you use low FFA WVO / UCO is fine, we're making decent bio already using this.

As to where the value of 3.5g/L came from - I'm sure that must be from an engineer trying things rather than a chemist working it out

You don't envision some mad chemist, in his lab with dry ice fog running over his floor testing it and yelling "It lives, IT LIVES"?

Must just be me then

[Tony]

Oh, I'm imagining that right now MWAHAHAAHAHHA (insert sound of Jacob's Ladder arcing here)



NB I do actually still have a Jacob's Ladder back from my Tesla coiling days

[countrypaul]

One question I have about the KOH (which I have never used) is what is the other 10%? Does it say on the packet? If it was NaOH then obviously the results could be very different compared to say KHCO3. In the latter case of course the titration of FFA may still react with the HCO3 ion releasing CO2 and water thereby distorting the expected results.

Paul