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 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.
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!*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.
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 ;D
| Characteristics | Specifications of 90% KOH |
| Total Alkalinity as KOH | 95% Max |
| KOH | 90% Min |
| K2CO3 | 0.5% Max |
| KCl | 0.015% Max |
| Fe | 3ppm Max |
| NaOH | 1% Max |
| Ni | 5ppm Max |
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 ;D
Paul
I remember falling foul of that difference. I had access to 0.1N NaOH at work and used it thinking that I had the right solution for titrating WVO. I used it for almost 12 months before a chance posting on the VOD was picked up by countrypaul. BTW...this was before the 3/27 was common knowledge.
Nige
I remember falling foul of that difference. I had access to 0.1N NaOH at work and used it thinking that I had the right solution for titrating WVO. I used it for almost 12 months before a chance posting on the VOD was picked up by countrypaul. BTW...this was before the 3/27 was common knowledge.
Nige